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https://math.stackexchange.com/questions/1834472/approximate-greatest-common-divisor
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crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00250.warc.gz
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# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
| 1,424
| 4,823
|
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| 3.734375
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.89591
|
# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers. For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ? more details:
I try to find the carrier signal length of a HF signal.
|
This signal is an alternation of high levels and low levels. eg. ----__------____------__-- ... I have the duration of each level, but this time is not accurate. My aim is to find as quick as possible the base time of the signal. My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
|
https://math.stackexchange.com/questions/1834472/approximate-greatest-common-divisor
| 1,708,939,357,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00250.warc.gz
| 387,870,564
| 37,416
|
# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
| 1,424
| 4,823
|
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| 3.734375
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.89591
|
397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it.
|
Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution. How to find the approximate basic frequency or GCD of a list of numbers?
|
https://math.stackexchange.com/questions/1834472/approximate-greatest-common-divisor
| 1,708,939,357,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00250.warc.gz
| 387,870,564
| 37,416
|
# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
| 1,424
| 4,823
|
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| 3.734375
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.89591
|
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$.
|
Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial.
|
https://math.stackexchange.com/questions/1834472/approximate-greatest-common-divisor
| 1,708,939,357,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00250.warc.gz
| 387,870,564
| 37,416
|
# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
| 1,424
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Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval.
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Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers? If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
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# approximate greatest common divisor
I try, without success, to create an algorithm that can compute the average greatest common divisor of a series of integers.
For example, I have the following numbers:
399, 710, 105, 891, 402, 102, 397, ...
As you can see, the average gcd is approximately 100, but how to compute it ?
more details:
I try to find the carrier signal length of a HF signal. This signal is an alternation of high levels and low levels.
eg. ----__------____------__-- ...
I have the duration of each level, but this time is not accurate.
My aim is to find as quick as possible the base time of the signal.
My first idea was to compute the gcd of the first times I get, to find the carrier of the signal. But I cannot use the classical gcd because the values are not very accurate.
With a perfect signal I would have gcd(400, 700, 100, 900, 400, 100, 400) = 100
• What is the average gcd? 397 is a prime, so its gcd with any of the other numbers in the series is 1. Jun 21, 2016 at 13:24
• Does computing all gcds then taking the average not work..? Jun 21, 2016 at 13:25
• @MattSamuel, all gcds give very low values, eg. (399, 710) => 1, etc..., then the average will be far of the expected value Jun 21, 2016 at 13:28
• @Soubok: No matter which word you use, it looks like you will have to describe what it is you want, with greater detail and specificity than just choosing a word to use for it. Jun 21, 2016 at 13:46
• Try to reformulate then, it might be an interesting problem. You'll need some kind of norm or measure, like the maximum of $\gcd(a_1+e_1,\dots,a_n+e_n)$ where $e_1+\cdots e_n<N$...
– Lehs
Jun 21, 2016 at 14:25
I made a similar question here, where I propose a partial solution.
How to find the approximate basic frequency or GCD of a list of numbers?
In summary, I came with this
• being $v$ the list $\{v_1, v_2, \ldots, v_n\}$,
• $\operatorname{mean}_{\sin}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\sin(2\pi v_i/x)$
• $\operatorname{mean}_{\cos}(v, x)$ $= \frac{1}{n}\sum_{i=1}^n\cos(2\pi v_i/x)$
• $\operatorname{gcd}_{appeal}(v, x)$ = $1 - \frac{1}{2}\sqrt{\operatorname{mean}_{\sin}(v, x)^2 + (\operatorname{mean}_{\cos}(v, x) - 1)^2}$
And the goal is to find the $x$ which maximizes the $\operatorname{gcd}_{appeal}$. Using the formulas and code described there, using CoCalc/Sage you can experiment with them and, in the case of your example, find that the optimum GCD is ~100.18867794375123:
testSeq = [399, 710, 105, 891, 402, 102, 397]
gcd = calculateGCDAppeal(x, testSeq)
find_local_maximum(gcd,90,110)
plot(gcd,(x, 10, 200), scale = "semilogx")
One approach: take the minimum of all your numbers, here $102$ as the first trial. Divide it into all your other numbers, choosing the quotient that gives the remainder of smallest absolute value. For your example, this would give $-9,2,3,-27,-6,0,-11$ The fact that your remainders are generally negative says your divisor is too large, so try a number a little smaller. Keep going until the remainders get positive and larger. Depending on how the errors accumulate, you might also add up all the numbers and assume the sum is a multiple of the minimum interval. Here your numbers add to $3006$, so you might think this is $30$ periods of $100.2$ Are your periods constrained to integers?
If you have a stubbornly large remainder, you can think that the smallest number is not one interval but two. You might have a number around $150$, so the fundamental period is $50$, not $100$. The challenge will be that if $100$ fits, any factor will fit as well.
This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it.
Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$.
Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
It is guaranteed that one $$d_{a,b}$$ is equal to $$p$$. Thus, to check which one is the correct one, just compute the "centered modular reduction" $$r_i' := x_i \bmod d_{a,b} \in [-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$ for $$i \ge 3$$.
If $$d_{a,b} = p$$, then each $$r_i'$$ equals $$r_i$$, therefore, all the values $$r_i'$$ that you get are small (e.g., smaller than 50). Otherwise, the values $$r_i'$$ will look randomly distributed in $$[-d_{a,b}/2,~ d_{a,b}/2]\cap\mathbb{Z}$$.
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This problem is actually not new. It is usually called the ACD problem (approximate common divisor problem) or the AGCD problem (approximate greatest common divisor) and there exist several algorithms to solve it. Although no algorithm is efficient in general, in your case, since the integers are tiny, you can simply try to eliminate the noise and compute the gcd.
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Namely, you want to recover $$p$$ given many values of the form $$x_i = pq_i + r_i$$ where $$|r_i|$$ is very small, say, smaller than $$50$$. Then, you can take two of those values, say, $$x_1$$ and $$x_2$$, and compute $$d_{a, b} = \gcd(x_1 - a, x_2 - b)$$ for all $$a, b \in [-50, 50]\cap\mathbb{Z}$$. Notice that this step costs only $$50^2$$ gcds computations, which is very cheap for any computer.
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https://math.stackexchange.com/questions/4591726/fracab4-fracbc4-fraccd4-fracde4-fracea4
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# $(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$
How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II)
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
• someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54
• I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48
Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$
• I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33
• @Freshman'sDream You're right, I just corrected this typo. Thanks!
– NN2
Commented Dec 5, 2022 at 9:34
• ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08
NN2 gave a simple and very elegamt proof. I tried another way.
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas?
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(Source: 1984 British Math Olympiad #3 part II)
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem.
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I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS? • someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54
• I'm kinda new around here, but I was also surprised to see it closed.
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Math for Future Value of Growing Annuity
Am I working this out correctly? I need to verify that my code is correct...
$$1000 \cdot \left(\frac{(1 + 0.1 / 12)^{40 * 12} - (1 + 0.06 / 12)^{40 * 12}}{(0.1 / 12) - (0.06 / 12)}\right)$$
Something like this:
53.700663174244 - 10.957453671655 ( = 42.7432095026 )
/
0.0083333333333333 - 0.005 ( = 0.00333333333 )
*
1000
=
12822 962.8636
ps. could someone please help me with the tag selection * blush*
EDIT: Sorry I know this is a mouthful, but if the math don't add up the code can't add up plus I'm actually a designer... not equal to programmer or mathematician. I'm a creative logician :)
Below is part A which must be added (summed) to part B (original question).
A: $$Future Value (FV) of Lumpsum = PV \cdot (1+i/12)^{b*12}$$
B:
$$FV of Growing Annuity = R1 \cdot \left(\frac{(1 + i / 12)^{b * 12} - (1 + g / 12)^{b * 12}}{(i / 12) - (g / 12)}\right)$$
• Current savings for retirement (Rands) = PV
• Rate of return = i/100
• Retirement age (years) – Current age (years) = b
• Current monthly contribution towards retirement (Rands) = R1
• 6/100 (Annual Growth rate of annuities) = g
This is all I have to offer except for the more complicated formula to work out the rest of "Savings for Retirement", but if my example B is correct then the B they gave me is wrong and it's driving me nuts because I'm also having trouble with:
C: $$PV of an Growing Annuity = \left(\frac{R2 \cdot(1 + g / 12)^{b * 12}}{(i / 12) - (g / 12)}\right) \cdot \left(1- \left( \frac{(1 + g / 12)^{b * 12}}{(1 + i / 12)^{n * 12}}\right)\right)$$
• Rate of return = i/100
• Retirement age (years) – Current age (years) = b
• 95 (Assumed age of death) - Retirement age (years) = n
• Monthly income need at retirement (Rands) = R2
• 6/100 (Annual Growth rate of annuities) = g
Which then must be: $$C-(A+B)$$ And finally, let me just give it all...
D: $$FV of Growing Annuity = R3 \cdot \left(\frac{((1 + i / 12)^{b * 12} - (1 + g / 12)^{b * 12} )}{(i / 12) - (g / 12)}\right)$$
• Answer of C-(A + B) = FV of Growing Annuity
• Rate of return = i/100
• Retirement age (years) – Current age (years) = b
• 6/100 (Annual Growth rate of annuities) = g
• You should state what problem you are trying to solve. It appears you are starting with a deposit of 1000 that draws some amount of interest for some time, but what the subtractions mean I can't guess. I think the first term is $10\%$ annual interest compounded monthly for 40 years. Then you should write it mathematically-we don't necessarily know what the arguments for Math.Pow are. – Ross Millikan Apr 24 '14 at 21:05
• To elaborate on what @RossMillikan meant, you gave a series of numbers and asked "Is this correct?" without specifying what those numbers mean and the goal of the calculation. For instance, $1000(1+0.1/12)^{40*12}$ gives your total money with an initial investment of \$1000, a rate of 10%, monthly compounding and 40 years of time. Why are you then subtracting the same calculation but with a 6% rate? Why are you dividing by the difference of these rates? We can't know if what you're doing is correct if we don't know what you're trying to do. – RandomUser Apr 24 '14 at 21:27
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It appears you are starting with a deposit of 1000 that draws some amount of interest for some time, but what the subtractions mean I can't guess. I think the first term is $10\%$ annual interest compounded monthly for 40 years. Then you should write it mathematically-we don't necessarily know what the arguments for Math.Pow are. – Ross Millikan Apr 24 '14 at 21:05
• To elaborate on what @RossMillikan meant, you gave a series of numbers and asked "Is this correct?"
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without specifying what those numbers mean and the goal of the calculation. For instance, $1000(1+0.1/12)^{40*12}$ gives your total money with an initial investment of \$1000, a rate of 10%, monthly compounding and 40 years of time. Why are you then subtracting the same calculation but with a 6% rate? Why are you dividing by the difference of these rates?
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space? I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving.
|
The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long. The data I have are: (example image below)
1.
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https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions? Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$.
|
The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$.
|
https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$.
|
The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$.
|
https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again. • Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
|
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates. The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d).
|
https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance).
|
Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
|
https://math.stackexchange.com/questions/1921033/how-do-i-plot-a-field-of-view-in-2d-space-and-find-if-a-certain-point-lies-wit
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# How do I plot a 'field of view' in 2D space and find if a certain point lies within the space?
I am trying to model a robot arm (in 2D) that's supposed to have a camera at the end that moves with the arm. Given the field of view of the camera, I need to find whether a certain target is visible in this field of view as the endpoint of the arm starts moving. The horizontal of the camera is aligned with the last link of the robot arm, so the 'triangle' moves up and down, forwards and backwards etc. While this intuitively makes sense, I'm having trouble coming up with a formula for this. Also, as it's a camera field of view, the rays are technically infinitely long.
The data I have are: (example image below)
1. Cartesian coordinates of the last two 'points' of the robot arm (x3, y3) and (x4, y4)
2. Theta angle of the field of view
3. Coordinates of the point of interest (this never changes)
Any suggestions?
Maybe something simple like this would work? Assume you have been able to compute your coordinates $P_3 = (x_3, y_3)$ and $P_4 = (x_4, y_4)$ and you know the position of the point you want to see with the camera $Q = (a, b)$. Form the vectors $\overrightarrow{P_3P_4} = (x_4 - x_3, \, y_4 - y_3)$ and $\overrightarrow{P_4Q} = (a - x_4, \, b - y_4)$. The dot product between the two vectors has the following meaning $$\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big) = \|\overrightarrow{P_3P_4}\| \, \|\overrightarrow{P_4Q} \| \cos{\theta} = \|P_4 - P_3\|\, \|Q - P_4\| \cos{\theta}$$ where $\theta$ is the angle between the oriented line $P_4Q$ and the oriented horizontal of the camera, which is defined by the line $P_3P_4$. If $\theta \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$ then your object is visible. In terms of cosine of the angle, this translates into the condition $\cos{\theta} \in \left(\frac{\sqrt{3}}{2}, 1\right)$. If $\cos{\theta}$ is not in the interval $\left(\frac{\sqrt{3}}{2}, 1\right)$, then your point is not visible. Form the above formula \begin{align}\cos{\theta} &= \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } = \frac{\Big(\, \overrightarrow{P_3P_4} \cdot \overrightarrow{P_4Q} \, \Big)}{\|P_4 - P_3\|\, \|Q - P_4\| } \end{align} so if we plug in the coordinates we obtain the expression
\begin{align} \cos{\theta} = f(x_3,y_3,x_4,y_4,a,b) = \frac{ (x_4 - x_3)(a- x_4) + (y_4 - y_3)(b- y_4)}{\sqrt{ (x_4 - x_3)^2 + (y_4 - y_3)^2} \,\,\, \sqrt{ (a- x_4)^2 + (b- y_4)^2}} \end{align}
If $\,\, f(x_3,y_3,x_4,y_4,a,b) \,\in \, \left(\frac{\sqrt{3}}{2}, 1\right) \,\,$ then the point with coordinates $(a,b)$ is visible. Otherwise, it is not.
Make sure I haven't made a mistake in the condition and the proper orientation of the angle. I have changed them three times already :)
Comment: Regarding the bounds of $\cos{\theta}$, I will try to explain this as follows.
The camera horizontal is the directed line $P_3P_4$, directed from $P_3$ to $P_4$. Since the directed line $P_4Q$, oriented from $P_4$ to $Q$, determines of the angle of the point $Q$ with respect to the camera horizontal $P_3P_4$, by definition the visibility occurs when the angle $\theta$ between $P_3P_4$ and $P_4Q$ (measured counterclockwise from $P_4Q$ to $P_3P_4$) changes from $-30^{\circ}$ to $30^{\circ}$. The cosine $\cos{\theta}$ is related to the orthogonal projection of $P_4Q$ onto the directed horizontal $P_3P_4$. Let us rotate $P_4Q$ around $P_4$ and see how $\theta$ changes. At first when $\theta = -30^{\circ}$ then $\cos{(-30^{\circ})} = \sqrt{3}/2$. As the direction $P_4Q$ gets more and more aligned with $P_3P_4$, the angle $\theta$ grows from $-30^{\circ}$ to $0$ and thus the cosine grows from $\sqrt{3}/2$ to $1$. After the alignment, when $\theta=0$ and $\cos{0} = 1$, the cosine $\cos{\theta}$ starts to decrease (while the angle $\theta$ keeps growing from $0$ to $30^{\circ}$) until the direction $P_4Q$ reaches $30^{\circ}$ with $P_3P_4$ and the cosine becomes $\cos{\theta} = \sqrt{3}/2$ again.
• Thanks for the detailed explanation! That approach worked. – HighVoltage Sep 10 '16 at 6:12
Here’s an approach that anticipates doing more with the camera view later.
We’ll be working in two dimensions, but the same technique applies in three. We will assume that the camera view is a perspective projection as illustrated here:
This will necessitate working in homogeneous coordinates.
The first thing to do is to switch to the camera’s coordinate system. The origin of this coordinate system is at the camera’s position and by convention, the camera sights along the negative $y'$ direction (negative $z'$ in 3-d). The world to camera transformation is thus a translation to the camera’s position followed by a rotation. The matrix of this translation is easy to produce. It’s simply $$T=\pmatrix{1&0&-x_4\\0&1&-y_4\\0&0&1}.$$ For the angle $\phi$ that the camera’s line of sight makes with the world $x$-axis, we have $$\cos\phi = {x_4-x_3\over\|P_4-P_3\|}\\\sin\phi = {y_4-y_3\over\|P_4-P_3\|}.$$ To get aligned with the camera’s line of sight, we start by rotating clockwise through this angle, but we also need to rotate clockwise by an additional 90 degrees to get it to point down the camera’s negative $y'$-axis. Putting those two rotations together produces the rotation matrix $$R=\pmatrix{-\sin\phi&\cos\phi&0\\-\cos\phi&-\sin\phi&0\\0&0&1},$$ with $\cos\phi$ and $\sin\phi$ as above. Combining these two matrices, we have $$RT = \pmatrix{-\sin\phi&\cos\phi&x_4\sin\phi-y_4\cos\phi\\-\cos\phi&-\sin\phi&x_4\cos\phi+y_4\sin\phi\\0&0&1},$$ i.e., \begin{align}x'&=-(x-x_4)\sin\phi+(y-y_4)\cos\phi\\y'&=-(x-x_4)\cos\phi-(y-y_4)\sin\phi.\end{align}
The line labeled “i” in the above diagram is the image plane, which is perpendicular to the camera’s line of sight and at a distance $f$ from the camera (the focal distance). Note that, since the camera is looking down the negative $y'$-axis, $f<0$. The perspective projection $M$ maps a point in the $x$-$y$ plane onto the intersection of the image plane with the ray emanating from the camera and passing through the point. If we take $f=-1$, then the bounds of the visible region in the image plane are $\pm\tan\theta$, so if the $x'$-coordinate of the projection of a point is in this range, then it’s visible.
In the camera coordinate system, a projection matrix is very simple: $$P=\pmatrix{1&0&0\\0&1&0\\0&\frac1f&0}.$$ Putting this all together, given a point $Q=(x,y)$, we compute $$M(Q)=PRT\pmatrix{x\\y\\1}$$ and recover the projected $x'$-coordinate by dividing the first component of the resulting vector by the third. We can save ourselves a bit of work, though, by taking advantage of $P$’s simple form. Note that $$\pmatrix{1&0&0\\0&1&0\\0&-1&0}\pmatrix{x'\\y'\\1}=\pmatrix{x'\\y'\\-y'},$$ so we really only need to transform the target point into camera coordinates, after which we can just check that $-\tan\theta\le-x'/y'\le\tan\theta$. We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero.
You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection. You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly.
As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices. The rotation matrix will be a bit more complicated, but the translation will still be straightforward. Taking $f=-1$ again, the projection will result in $(x',y',z',-z')$. Assuming that the field of view is a circular cone, the test for visibility will then be $$x'^2+y'^2\le z'^2\tan^2\theta.$$
Postscript: This is, of course, overkill when the field of view is a right circular cone, whether in two dimensions or three. Checking that $(Q-P_4)\cdot(P_4-P_3)\ge\|Q-P_4\|\,\|P_4-P_3\|\cos\theta$ is much simpler and more efficient. However, the procedure that I’ve outlined here applies generally to any size and shape aperture, which becomes much more interesting when you move to three dimensions.
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{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.75
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CC-MAIN-2019-22
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latest
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en
| 0.861014
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We might have $y'=0$, however, so let’s rewrite this as $|x'|\le|y'|\tan\theta$ to avoid dividing by zero. You might object that the projection also maps points behind the camera onto the image plane, but that’s easily dealt with: check the sign of the camera-relative $y'$-coordinate. If it’s positive, the point is behind the camera, so there’s no need to compute its projection.
|
You can eliminate the $y'=0$ case at the same time. If this seems backwards to you, you can always have the camera point in the positive $y'$ direction instead so that visible points have a positive $y'$-coordinate, but you’ll have to modify $R$ and $P$ accordingly. As I mentioned above, the same approach works in 3-d, except that you’ll be working with $4\times4$ matrices.
|
https://mathematica.stackexchange.com/questions/216293/phase-portrait-for-ode-with-ivp?noredirect=1
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text/html
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crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00732.warc.gz
| 414,570,837
| 42,266
|
# Phase Portrait for ODE with IVP
I'm trying to make a phase portrait for the ODE x'' + 16x = 0, with initial conditions x[0]=-1 & x'[0]=0. I know how to solve the ODE and find the integration constants; the solution comes out to be x(t) = -cos(4t) and x'(t) = 4sin(4t). But I don't know how to make a phase portrait out of it. I've looked at this link Plotting a Phase Portrait but I couldn't replicate mine based off of it.
Phase portrait for any second order autonomous ODE can be found as follows.
Convert the ODE to state space. This results in 2 first order ODE's. Then call StreamPlot with these 2 equations.
Let the state variables be $$x_1=x,x_2=x'(t)$$, then taking derivatives w.r.t time gives $$x'{_1}=x_2,x'{_2}=x''(t)=-16 x_1$$. Now, using StreamPlot gives
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -2, 2}]
To see the line that passes through the initial conditions $$x_1(0)=1,x_2(0)=0.1$$, add the option StreamPoints
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
To verify the above is the correct phase plot, you can do
ClearAll[x, t]
ode = x''[t] + 16 x[t] == 0;
ic = {x[0] == 1, x'[0] == 1/10};
sol = x[t] /. First@(DSolve[{ode, ic}, x[t], t]);
ParametricPlot[Evaluate[{sol, D[sol, t]}], {t, 0, 3}, PlotStyle -> Red]
The advatage of phase plot, is that one does not have to solve the ODE first (so it works for nonlinear hard to solve ODE's).
All what you have to do is convert the ODE to state space and use function like StreamPlot
If you want to automate the part of converting the ODE to state space, you can also use Mathematica for that. Simply use StateSpaceModel and just read of the equations.
eq = x''[t] + 16 x[t] == 0;
ss = StateSpaceModel[{eq}, {{x[t], 0}, {x'[t], 0}}, {}, {x[t]}, t]
The above shows the A matrix in $$x'=Ax$$. So first row reads $$x_1'(t)=x_2$$ and second row reads $$x'_2(t)=-16 x_1$$
The following can be done to automate plotting StreamPlot directly from the state space ss result
A = First@Normal[ss];
vars = {x1, x2}; (*state space variables*)
eqs = A . vars;
StreamPlot[eqs, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
• Can you method plot y''[x]+2 y'[x]+3 y[x]==2 x?
– yode
Mar 27, 2022 at 8:59
• @yode Phase portrait are used for homogeneous ode's. Systems of the form $x'=A x$ and not $x'=A x + u$. Since it shows the behaviour of the system itself, independent of any forcing functions (the stuff on the RHS). This behavior is given by phase portrait diagram. The reason is, it is only the $A$ matrix eigenvalues and eigenvectors that determines this behaviour, and $A$ depends only on the system itself, without any external input being there. Mar 27, 2022 at 14:08
• Can we plot your ss in MMA directly?
– yode
Mar 29, 2022 at 10:59
• @yoda Yes. I've updated the above with what I think you are asking for. Hope this helps. Mar 29, 2022 at 15:23
EquationTrekker works for me, but if you are not interested in looking at a range of solutions, it might be easier to just do it with ParametricPlot
x[t_] := -Cos[4 t]
ParametricPlot[{x[t], x'[t]} // Evaluate, {t, 0, 2 π},
Axes -> False, PlotLabel -> PhaseTrajectory, Frame -> True,
FrameLabel -> {x[t], x'[t]}, GridLines -> Automatic]
• What version is this on, Bill? Someone in the QA that OP links to says EquationTrekker is broken for them on v11.0 Mar 15, 2020 at 6:04
• This plot is from ParametricPlot, not EquationTrekker, but in v12.0 EquationTrekker gives me plots, although I do get PropertyValue errors. Mar 15, 2020 at 7:40
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{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
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CC-MAIN-2023-40
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longest
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en
| 0.844324
|
But I don't know how to make a phase portrait out of it. I've looked at this link Plotting a Phase Portrait but I couldn't replicate mine based off of it. Phase portrait for any second order autonomous ODE can be found as follows. Convert the ODE to state space. This results in 2 first order ODE's. Then call StreamPlot with these 2 equations. Let the state variables be $$x_1=x,x_2=x'(t)$$, then taking derivatives w.r.t time gives $$x'{_1}=x_2,x'{_2}=x''(t)=-16 x_1$$.
|
Now, using StreamPlot gives
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -2, 2}]
To see the line that passes through the initial conditions $$x_1(0)=1,x_2(0)=0.1$$, add the option StreamPoints
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
To verify the above is the correct phase plot, you can do
ClearAll[x, t]
ode = x''[t] + 16 x[t] == 0;
ic = {x[0] == 1, x'[0] == 1/10};
sol = x[t] /.
|
https://mathematica.stackexchange.com/questions/216293/phase-portrait-for-ode-with-ivp?noredirect=1
| 1,696,284,912,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00732.warc.gz
| 414,570,837
| 42,266
|
# Phase Portrait for ODE with IVP
I'm trying to make a phase portrait for the ODE x'' + 16x = 0, with initial conditions x[0]=-1 & x'[0]=0. I know how to solve the ODE and find the integration constants; the solution comes out to be x(t) = -cos(4t) and x'(t) = 4sin(4t). But I don't know how to make a phase portrait out of it. I've looked at this link Plotting a Phase Portrait but I couldn't replicate mine based off of it.
Phase portrait for any second order autonomous ODE can be found as follows.
Convert the ODE to state space. This results in 2 first order ODE's. Then call StreamPlot with these 2 equations.
Let the state variables be $$x_1=x,x_2=x'(t)$$, then taking derivatives w.r.t time gives $$x'{_1}=x_2,x'{_2}=x''(t)=-16 x_1$$. Now, using StreamPlot gives
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -2, 2}]
To see the line that passes through the initial conditions $$x_1(0)=1,x_2(0)=0.1$$, add the option StreamPoints
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
To verify the above is the correct phase plot, you can do
ClearAll[x, t]
ode = x''[t] + 16 x[t] == 0;
ic = {x[0] == 1, x'[0] == 1/10};
sol = x[t] /. First@(DSolve[{ode, ic}, x[t], t]);
ParametricPlot[Evaluate[{sol, D[sol, t]}], {t, 0, 3}, PlotStyle -> Red]
The advatage of phase plot, is that one does not have to solve the ODE first (so it works for nonlinear hard to solve ODE's).
All what you have to do is convert the ODE to state space and use function like StreamPlot
If you want to automate the part of converting the ODE to state space, you can also use Mathematica for that. Simply use StateSpaceModel and just read of the equations.
eq = x''[t] + 16 x[t] == 0;
ss = StateSpaceModel[{eq}, {{x[t], 0}, {x'[t], 0}}, {}, {x[t]}, t]
The above shows the A matrix in $$x'=Ax$$. So first row reads $$x_1'(t)=x_2$$ and second row reads $$x'_2(t)=-16 x_1$$
The following can be done to automate plotting StreamPlot directly from the state space ss result
A = First@Normal[ss];
vars = {x1, x2}; (*state space variables*)
eqs = A . vars;
StreamPlot[eqs, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
• Can you method plot y''[x]+2 y'[x]+3 y[x]==2 x?
– yode
Mar 27, 2022 at 8:59
• @yode Phase portrait are used for homogeneous ode's. Systems of the form $x'=A x$ and not $x'=A x + u$. Since it shows the behaviour of the system itself, independent of any forcing functions (the stuff on the RHS). This behavior is given by phase portrait diagram. The reason is, it is only the $A$ matrix eigenvalues and eigenvectors that determines this behaviour, and $A$ depends only on the system itself, without any external input being there. Mar 27, 2022 at 14:08
• Can we plot your ss in MMA directly?
– yode
Mar 29, 2022 at 10:59
• @yoda Yes. I've updated the above with what I think you are asking for. Hope this helps. Mar 29, 2022 at 15:23
EquationTrekker works for me, but if you are not interested in looking at a range of solutions, it might be easier to just do it with ParametricPlot
x[t_] := -Cos[4 t]
ParametricPlot[{x[t], x'[t]} // Evaluate, {t, 0, 2 π},
Axes -> False, PlotLabel -> PhaseTrajectory, Frame -> True,
FrameLabel -> {x[t], x'[t]}, GridLines -> Automatic]
• What version is this on, Bill? Someone in the QA that OP links to says EquationTrekker is broken for them on v11.0 Mar 15, 2020 at 6:04
• This plot is from ParametricPlot, not EquationTrekker, but in v12.0 EquationTrekker gives me plots, although I do get PropertyValue errors. Mar 15, 2020 at 7:40
| 1,140
| 3,557
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625
| 4
|
CC-MAIN-2023-40
|
longest
|
en
| 0.844324
|
First@(DSolve[{ode, ic}, x[t], t]);
ParametricPlot[Evaluate[{sol, D[sol, t]}], {t, 0, 3}, PlotStyle -> Red]
The advatage of phase plot, is that one does not have to solve the ODE first (so it works for nonlinear hard to solve ODE's). All what you have to do is convert the ODE to state space and use function like StreamPlot
If you want to automate the part of converting the ODE to state space, you can also use Mathematica for that.
|
Simply use StateSpaceModel and just read of the equations. eq = x''[t] + 16 x[t] == 0;
ss = StateSpaceModel[{eq}, {{x[t], 0}, {x'[t], 0}}, {}, {x[t]}, t]
The above shows the A matrix in $$x'=Ax$$. So first row reads $$x_1'(t)=x_2$$ and second row reads $$x'_2(t)=-16 x_1$$
The following can be done to automate plotting StreamPlot directly from the state space ss result
A = First@Normal[ss];
vars = {x1, x2}; (*state space variables*)
eqs = A .
|
https://mathematica.stackexchange.com/questions/216293/phase-portrait-for-ode-with-ivp?noredirect=1
| 1,696,284,912,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00732.warc.gz
| 414,570,837
| 42,266
|
# Phase Portrait for ODE with IVP
I'm trying to make a phase portrait for the ODE x'' + 16x = 0, with initial conditions x[0]=-1 & x'[0]=0. I know how to solve the ODE and find the integration constants; the solution comes out to be x(t) = -cos(4t) and x'(t) = 4sin(4t). But I don't know how to make a phase portrait out of it. I've looked at this link Plotting a Phase Portrait but I couldn't replicate mine based off of it.
Phase portrait for any second order autonomous ODE can be found as follows.
Convert the ODE to state space. This results in 2 first order ODE's. Then call StreamPlot with these 2 equations.
Let the state variables be $$x_1=x,x_2=x'(t)$$, then taking derivatives w.r.t time gives $$x'{_1}=x_2,x'{_2}=x''(t)=-16 x_1$$. Now, using StreamPlot gives
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -2, 2}]
To see the line that passes through the initial conditions $$x_1(0)=1,x_2(0)=0.1$$, add the option StreamPoints
StreamPlot[{x2, -16 x1}, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
To verify the above is the correct phase plot, you can do
ClearAll[x, t]
ode = x''[t] + 16 x[t] == 0;
ic = {x[0] == 1, x'[0] == 1/10};
sol = x[t] /. First@(DSolve[{ode, ic}, x[t], t]);
ParametricPlot[Evaluate[{sol, D[sol, t]}], {t, 0, 3}, PlotStyle -> Red]
The advatage of phase plot, is that one does not have to solve the ODE first (so it works for nonlinear hard to solve ODE's).
All what you have to do is convert the ODE to state space and use function like StreamPlot
If you want to automate the part of converting the ODE to state space, you can also use Mathematica for that. Simply use StateSpaceModel and just read of the equations.
eq = x''[t] + 16 x[t] == 0;
ss = StateSpaceModel[{eq}, {{x[t], 0}, {x'[t], 0}}, {}, {x[t]}, t]
The above shows the A matrix in $$x'=Ax$$. So first row reads $$x_1'(t)=x_2$$ and second row reads $$x'_2(t)=-16 x_1$$
The following can be done to automate plotting StreamPlot directly from the state space ss result
A = First@Normal[ss];
vars = {x1, x2}; (*state space variables*)
eqs = A . vars;
StreamPlot[eqs, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
• Can you method plot y''[x]+2 y'[x]+3 y[x]==2 x?
– yode
Mar 27, 2022 at 8:59
• @yode Phase portrait are used for homogeneous ode's. Systems of the form $x'=A x$ and not $x'=A x + u$. Since it shows the behaviour of the system itself, independent of any forcing functions (the stuff on the RHS). This behavior is given by phase portrait diagram. The reason is, it is only the $A$ matrix eigenvalues and eigenvectors that determines this behaviour, and $A$ depends only on the system itself, without any external input being there. Mar 27, 2022 at 14:08
• Can we plot your ss in MMA directly?
– yode
Mar 29, 2022 at 10:59
• @yoda Yes. I've updated the above with what I think you are asking for. Hope this helps. Mar 29, 2022 at 15:23
EquationTrekker works for me, but if you are not interested in looking at a range of solutions, it might be easier to just do it with ParametricPlot
x[t_] := -Cos[4 t]
ParametricPlot[{x[t], x'[t]} // Evaluate, {t, 0, 2 π},
Axes -> False, PlotLabel -> PhaseTrajectory, Frame -> True,
FrameLabel -> {x[t], x'[t]}, GridLines -> Automatic]
• What version is this on, Bill? Someone in the QA that OP links to says EquationTrekker is broken for them on v11.0 Mar 15, 2020 at 6:04
• This plot is from ParametricPlot, not EquationTrekker, but in v12.0 EquationTrekker gives me plots, although I do get PropertyValue errors. Mar 15, 2020 at 7:40
| 1,140
| 3,557
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
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CC-MAIN-2023-40
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longest
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vars;
StreamPlot[eqs, {x1, -2, 2}, {x2, -5, 5},
StreamPoints -> {{{{1, .1}, Red}, Automatic}}]
• Can you method plot y''[x]+2 y'[x]+3 y[x]==2 x? – yode
Mar 27, 2022 at 8:59
• @yode Phase portrait are used for homogeneous ode's. Systems of the form $x'=A x$ and not $x'=A x + u$. Since it shows the behaviour of the system itself, independent of any forcing functions (the stuff on the RHS).
|
This behavior is given by phase portrait diagram. The reason is, it is only the $A$ matrix eigenvalues and eigenvectors that determines this behaviour, and $A$ depends only on the system itself, without any external input being there. Mar 27, 2022 at 14:08
• Can we plot your ss in MMA directly? – yode
Mar 29, 2022 at 10:59
• @yoda Yes. I've updated the above with what I think you are asking for.
|
http://math.stackexchange.com/questions/tagged/vector-spaces+vector-analysis
| 1,398,348,055,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-15/segments/1398223206147.1/warc/CC-MAIN-20140423032006-00167-ip-10-147-4-33.ec2.internal.warc.gz
| 210,644,211
| 24,883
|
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Let $A\in$ $\mathbb{C}^{p\times q}$ with column $u_1,\ldots,u_q$ and rows $\vec{v_1},\ldots,\vec{v_p}$. show that $$\|A\|\geq\max\{\|u_1\|,\ldots,\|u_q\|,\|\vec{v_1}\|,\ldots,\|\vec{v_q}\|\}$$ and ...
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### Find the necessary and sufficient conditions on $A$ such that $\|T(\vec{x})\|=|\det A|\cdot\|\vec{x}\|$ for all $\vec{x}$.
Consider the mapping $T:\mathbb{R}^n\mapsto\mathbb{R}^n$ defined by $T(\vec{x})=A\vec{x}$ where $A$ is a $n\times n$ matrix. Find the necessary and sufficient conditions on $A$ such that ...
58 views
### Dot products of three or more vectors
Can't we construct a mapping from $V^3(R^1)$ to $R$ such that $a.b.c = a_{x}b_{x}c_{x}+a_{y}b_{y}c_{y}+a_{z}b_{z}c_{z}$ (a,b,c are vectors in $V^3(R^1)$ ) and more generally $a^n$ , $a.b.c.d.e...$ ...
365 views
50 views
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# Tagged Questions
23 views
### Why should we expect the divergence operator to be invariant under transformations? A lot of the time with vector calculus identities, something that seems magical at first ends up having a nice and unique proof. For the divergence operator, one can prove that it's invariant under a ...
3 views
### Gentle introduction to discrete vector field [closed]
I am looking for a gentle introduction to discrete vector field.
|
Thanks in advance. 26 views
### Vectors and Planes
Let there be 2 planes: $x-y+z=2, 2x-y-z=1$ Find the equation of the line of the intersection of the two planes, as well as that of another plane which goes through that line. Attempt to solve: the ...
25 views
63 views
### Extrema of a vector norm under two inner-product constraints. If $\langle\vec{A},\vec{V}\rangle=1\; ,\; \langle\vec{B},\vec{V}\rangle=c$, then: \begin{align} max\left \| \vec{V} \right \|_{1}=?\;\;\;min\left \| \vec{V} \right \|_{1}=?
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http://math.stackexchange.com/questions/tagged/vector-spaces+vector-analysis
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| 210,644,211
| 24,883
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# Tagged Questions
23 views
### Why should we expect the divergence operator to be invariant under transformations?
A lot of the time with vector calculus identities, something that seems magical at first ends up having a nice and unique proof. For the divergence operator, one can prove that it's invariant under a ...
3 views
### Gentle introduction to discrete vector field [closed]
I am looking for a gentle introduction to discrete vector field. Thanks in advance.
26 views
### Vectors and Planes
Let there be 2 planes: $x-y+z=2, 2x-y-z=1$ Find the equation of the line of the intersection of the two planes, as well as that of another plane which goes through that line. Attempt to solve: the ...
25 views
63 views
### Extrema of a vector norm under two inner-product constraints.
If $\langle\vec{A},\vec{V}\rangle=1\; ,\; \langle\vec{B},\vec{V}\rangle=c$, then: \begin{align} max\left \| \vec{V} \right \|_{1}=?\;\;\;min\left \| \vec{V} \right \|_{1}=? \end{align} Consider the ...
129 views
### How to rotate two vectors (2d), where their angle is larger than 180.
The rotation matrix $$\begin{bmatrix} \cos\theta & -\sin \theta\\ \sin\theta & \cos\theta \end{bmatrix}$$ cannot process the case that the angle between two vectors is larger than $180$ ...
53 views
### Is this statement about vectors true?
If vectors $A$ and $B$ are parallel, then, $|A-B| = |A| - |B|$ Is the above statement true?
822 views
### Collinearity of three points of vectors
Show that the three vectors $$A\_ = 2i + j - 3k , B\_ = i - 4k , C\_ = 4i + 3j -k$$ are linearly dependent. Determine a relation between them and hence show that the terminal points are collinear. ...
92 views
135 views
### Vectors transformation
Give a necessary and sufficient condition ("if and only if") for when three vectors $a, b, c, \in \mathbb{R^2}$ can be transformed to unit length vectors by a single affine transformation. This is ...
56 views
### To show the inequality $\|A\|\geq\max\{\|u_1\|,\ldots,\|u_q\|,\|\vec{v_1}\|,\ldots,\|\vec{v_q}\|\}$
Let $A\in$ $\mathbb{C}^{p\times q}$ with column $u_1,\ldots,u_q$ and rows $\vec{v_1},\ldots,\vec{v_p}$. show that $$\|A\|\geq\max\{\|u_1\|,\ldots,\|u_q\|,\|\vec{v_1}\|,\ldots,\|\vec{v_q}\|\}$$ and ...
160 views
### Find the necessary and sufficient conditions on $A$ such that $\|T(\vec{x})\|=|\det A|\cdot\|\vec{x}\|$ for all $\vec{x}$.
Consider the mapping $T:\mathbb{R}^n\mapsto\mathbb{R}^n$ defined by $T(\vec{x})=A\vec{x}$ where $A$ is a $n\times n$ matrix. Find the necessary and sufficient conditions on $A$ such that ...
58 views
### Dot products of three or more vectors
Can't we construct a mapping from $V^3(R^1)$ to $R$ such that $a.b.c = a_{x}b_{x}c_{x}+a_{y}b_{y}c_{y}+a_{z}b_{z}c_{z}$ (a,b,c are vectors in $V^3(R^1)$ ) and more generally $a^n$ , $a.b.c.d.e...$ ...
365 views
50 views
| 900
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{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0}
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CC-MAIN-2014-15
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longest
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822 views
### Collinearity of three points of vectors
Show that the three vectors $$A\_ = 2i + j - 3k , B\_ = i - 4k , C\_ = 4i + 3j -k$$ are linearly dependent. Determine a relation between them and hence show that the terminal points are collinear. ...
92 views
135 views
### Vectors transformation
Give a necessary and sufficient condition ("if and only if") for when three vectors $a, b, c, \in \mathbb{R^2}$ can be transformed to unit length vectors by a single affine transformation.
|
This is ...
56 views
### To show the inequality $\|A\|\geq\max\{\|u_1\|,\ldots,\|u_q\|,\|\vec{v_1}\|,\ldots,\|\vec{v_q}\|\}$
Let $A\in$ $\mathbb{C}^{p\times q}$ with column $u_1,\ldots,u_q$ and rows $\vec{v_1},\ldots,\vec{v_p}$. show that $$\|A\|\geq\max\{\|u_1\|,\ldots,\|u_q\|,\|\vec{v_1}\|,\ldots,\|\vec{v_q}\|\}$$ and ...
160 views
### Find the necessary and sufficient conditions on $A$ such that $\|T(\vec{x})\|=|\det A|\cdot\|\vec{x}\|$ for all $\vec{x}$.
|
https://quant.stackexchange.com/questions/68635/characteristics-of-factor-portfolios/68646
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# characteristics of factor portfolios
In the paper Characteristics of Factor Portfolios (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1601414), when it discusses pure factor portfolios, it says that simple style factor portfolios have zero exposure to all other style, country, and industry factors. Could someone help me understand the math for why the style factor portfolios have zero exposure to all other style, country, and industry factors?
So, for example, if we are interested in the return of a P/E factor and a P/B factor, we would gather the P/E and P/B for all of our stocks into a matrix of loadings $$B$$. $$B$$ would have two columns – one containing P/E and one containing P/B for all assets. We then regress $$R$$ (a vector containing the returns of all assets) on $$B$$. OLS regression gives us $$f= (B’B)^{-1} B’R$$ = the returns of the style factors for this particular period. The rows of $$(B’B)^{-1} B’$$ are considered to be the factor portfolios.
So, let’s go one step further and look at the loadings of the portfolio on the individual styles by multiplying the factor portfolios with the matrix of loadings. This gives $$(B’B)^{-1} B’B = I$$ - an identity matrix. Hence, the loadings of each factor portfolio are 1 against the particular style and 0 against any other style.
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# characteristics of factor portfolios
In the paper Characteristics of Factor Portfolios (https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1601414), when it discusses pure factor portfolios, it says that simple style factor portfolios have zero exposure to all other style, country, and industry factors. Could someone help me understand the math for why the style factor portfolios have zero exposure to all other style, country, and industry factors?
|
So, for example, if we are interested in the return of a P/E factor and a P/B factor, we would gather the P/E and P/B for all of our stocks into a matrix of loadings $$B$$. $$B$$ would have two columns – one containing P/E and one containing P/B for all assets. We then regress $$R$$ (a vector containing the returns of all assets) on $$B$$. OLS regression gives us $$f= (B’B)^{-1} B’R$$ = the returns of the style factors for this particular period.
|
http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=votes
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# What is the last digit of $\pi$?
I want to know: what is the last digit of $\pi$?
Some people say there are no such thing, but they fail to mention why.
-
Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.
(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.
Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).
By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.
In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.
-
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit?
One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.
Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)
-
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
-
## protected by Qiaochu YuanJul 21 '11 at 22:17
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
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Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented.
|
To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$.
|
http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=votes
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# What is the last digit of $\pi$?
I want to know: what is the last digit of $\pi$?
Some people say there are no such thing, but they fail to mention why.
-
Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.
(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.
Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).
By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.
In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.
-
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit?
One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.
Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)
-
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
-
## protected by Qiaochu YuanJul 21 '11 at 22:17
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
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If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert. (This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious.
|
I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail. Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion.
|
http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=votes
| 1,461,905,621,000,000,000
|
text/html
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crawl-data/CC-MAIN-2016-18/segments/1461860110372.12/warc/CC-MAIN-20160428161510-00036-ip-10-239-7-51.ec2.internal.warc.gz
| 126,954,852
| 20,812
|
# What is the last digit of $\pi$?
I want to know: what is the last digit of $\pi$?
Some people say there are no such thing, but they fail to mention why.
-
Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.
(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.
Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).
By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.
In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.
-
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit?
One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.
Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)
-
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
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## protected by Qiaochu YuanJul 21 '11 at 22:17
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
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All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers). By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus.
|
The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$.
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http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=votes
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# What is the last digit of $\pi$?
I want to know: what is the last digit of $\pi$?
Some people say there are no such thing, but they fail to mention why.
-
Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.
(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.
Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).
By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.
In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.
-
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit?
One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.
Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)
-
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
-
## protected by Qiaochu YuanJul 21 '11 at 22:17
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
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More generally, he proved that if $x$ is rational, then $\tan x$ is irrational. In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational. -
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
|
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit? One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$.
|
http://math.stackexchange.com/questions/52974/what-is-the-last-digit-of-pi?answertab=votes
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# What is the last digit of $\pi$?
I want to know: what is the last digit of $\pi$?
Some people say there are no such thing, but they fail to mention why.
-
Sometimes even rational numbers don't have a "last digit", think of 1/3=0.33333... – Vhailor Jul 21 '11 at 22:12
Nice Poem. ${}$ – jspecter Jul 21 '11 at 22:13
If $\pi$ has a last digit, then $0.999\ldots\neq 1$. – Asaf Karagila Jul 21 '11 at 22:13
This question should specify "base 10". The "no last digit" phenomenon depends on how $\pi$ is represented. To take a contrived setting, base-$\pi$ numbers, then $\pi$ is written as $1$. I'm not trying to be pedantic here: representation is a fundamental part of this question. – Fixee Jul 22 '11 at 6:09
@jspecter: A slightly enhanced version: $$\text{I desperately want to know}\\ \text{The last digit of \pi}\\ \text{Some people say there's no such thing}\\ \text{They fail to mention why}$$ – joriki Dec 22 '12 at 3:49
There is no "last" digit of $\pi$. If there was a last digit, then there could only be finitely many digits in front so that $\pi$ would be a rational number. However $\pi$ was shown to be irrational in the 18th century by Lambert.
(This Meta.StackExchange post is a joke based on the impossibility of finding such a last digit)
-
Thanks for the link to the Meta.StackExchange post: hilarious. I've got my laughs in for the day. We all need a good dose of humor now and then! – amWhy Jul 21 '11 at 22:16
Since you may have never seen the topics in my colleagues' answers, I'll try to explain them in some detail.
Suppose for the sake of argument that when $\pi$ is written as a decimal expansion ($3.1415 \dots$) it does have a final digit. This would clearly imply that there is a finite number of terms in the expansion. All real numbers with finite decimal expansions can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers (whole numbers).
By this reasoning we conclude that $\pi = \frac{a}{b}$ for some positive integers $a$ and $b$, i.e., that $\pi$ is rational. This is the starting point for this short proof given by I. Niven in 1946, which is especially easy to follow if you've had a little trigonometry and even less differential calculus. The proof concludes with an absurdity like the existence of an integer between $0$ and $1$, which implies that $a$ and $b$ do not exist and $\pi$ is irrational (and has an infinite decimal expansion). It should be noted that the irrationality of $\pi$ was first established by Lambert in 1761 by studying the continued fraction expansion of the tangent function and using the identity $\tan \frac{\pi}{4} = 1$. More generally, he proved that if $x$ is rational, then $\tan x$ is irrational.
In short, there is no final digit in the decimal expansion of $\pi$ because it is irrational.
-
Ok, try it now. – user02138 Jul 22 '11 at 5:25
Proving that $\pi$ is irrational is more difficult than proving that $e$ or $\sqrt{2}$ or $\log_2 3$ is irrational. See http://en.wikipedia.org/wiki/Proof_that_pi_is_irrational .
Proving that an irrational number has no last digit is easier than that: http://en.wikipedia.org/wiki/Irrational_number
-
Even rational numbers usually have no "last digit": what is the last digit of $$0.1313131\dots = 0.\overline{13} = \frac{13}{99} ?$$
So what sort of numbers have a last digit?
One, numbers with a terminating decimal expansion: numbers like $\displaystyle 2.23627 = \frac{223627}{100000}$. As you can see, all such numbers can be written as a fraction with denominator being a power of $10$.
Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.)
-
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
-
## protected by Qiaochu YuanJul 21 '11 at 22:17
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.
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As you can see, all such numbers can be written as a fraction with denominator being a power of $10$. Two, depending on your definition of "last digit", numbers like $0.4677777\dots = \frac1{100}46.77777$ = $\displaystyle \frac1{100}\left(46 + \frac79\right) = \frac1{100}\frac{421}{9}$. These numbers can be written as $\displaystyle \frac1{10^k} \frac{n}9$ for some integers $k$ and $n$.
|
So a number $x$ has a "last digit" if and only if $(9\cdot 10^k)x$ is an integer for some $k$. Only very special numbers are of this form, and it should be no surprise that $\pi$ is not. (Admittedly, I don't actually see how to prove this without invoking $\pi$'s irrationality, but it's a much weaker property.) -
HINT $\rm\ \pi = 3.1415\ \Rightarrow\ 10^4\: \pi = 31415\ \Rightarrow\ \pi = 31415/10^4\$ is rational, contra Lambert's proof.
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https://physics.stackexchange.com/questions/533659/kinetic-energy-operator-in-momentum-basis
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# Kinetic Energy Operator in Momentum Basis
I'm having a little trouble with something that's 'easy to check' according to the script I'm using. I consider the kinetic energy operator
$$\hat T = \frac{p^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$
and the wave function in momentum space, which is obtained using the Fourier transform
$$\tilde \psi(k) = \mathcal{F}[\psi(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \psi(x) e^{-ikx} dx \ .$$
According to the script it's easy to see that
$$\hat T \tilde \psi(k) = \frac{\hbar^2 k^2}{2m} \tilde \psi(k)\ ,$$
i.e. that $$\tilde \psi(k)$$ is an eigenstate of the kinetic energy operator. I don't manage to replicate that result. In particular, using integration by parts a few times I get
\begin{align} \hat T \tilde \psi(k) &= -\frac{\hbar^2}{2m} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \frac{d^2}{dx^2} ( \psi(x) e^{-ikx} ) dx \\ &= … \\ &= -\frac{\hbar^2}{2m} \tilde \psi(k) (-k^2 + 2k^2 - k^2) \\ &= 0 \end{align}
I see that $$\mathcal{F}[\hat T \psi(x)] = \frac{\hbar^2 k^2}{2m} \mathcal{F}[\psi(x)]$$ but $$\mathcal{F}[\hat T \psi(x)] \neq \hat T \tilde \psi(k)$$, right? So I'm confused ...
Can anybody show me where I go wrong?
• If you are doing derivatives with respect to $x$ for the momentum operator, then you are in the position basis, not the momentum basis. Commented Feb 28, 2020 at 20:08
• The momentum operator takes different form. In position basis, momentum is a derivative. In momentum basis, momentum is just multiplying by p. Commented Feb 28, 2020 at 20:14
• oh I see, we use de Broglie's relation $p = \hbar k$ so $\hat T = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$ and we just multiply by it, that's great, thanks to both of you!
– mpr
Commented Feb 28, 2020 at 20:35
First, the expression $$\hat T\bar\psi (k)=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$ does not mean $$\bar\psi(k)$$ is an eigenvalue of $$\hat T$$ because $$\hbar^2k^2/2m$$ is not a constant ($$k$$ is a variable now).
Second, $$\hat T\bar\psi(k)$$ is not an expression that makes much sense. I think you actually mean $$\langle k|\hat T|\psi\rangle$$. To be more formal, you are starting with $$\bar\psi(k)=\langle k|\psi\rangle=\int_{-\infty}^\infty\langle k|x\rangle\langle x|\psi\rangle\,\text dx=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-ikx}\psi(x)\,\text dx$$
But you cannot "operate" $$\hat T$$ on $$\langle k|\psi\rangle$$, since this is technically just a number for some value of $$k$$. This is why I think you actually mean $$\langle k|\hat T|\psi\rangle$$ because we can compute this in the momentum basis $$\langle k|\hat T|\psi\rangle=\int_{-\infty}^\infty\langle k|\hat T|k'\rangle\langle k'|\psi\rangle \,\text dk'=\int_{-\infty}^\infty\frac{\hbar^2k^2}{2m}\delta(k-k')\bar\psi(k') \,\text dk'=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$
This also reveals why $$\bar\psi(k)$$ is not, in general, an eigenfunction of $$\hat T$$. Note that really what we want is the basis-independent statement $$\hat T|\psi\rangle=c|\psi\rangle$$ to be true for $$|\psi\rangle$$ to be an eigenstate of $$\hat T$$. Now, at first glance this seems to be the case here, but the issue is that we needed to move into the $$k$$-basis in order to get this expression. In other words, $$\frac{\hbar^2k^2}{2m}$$ isn't an eigenvalue, it is just what is multiplied by $$\bar\psi(k)$$ when looking at $$\hat T$$ being applied to $$|\psi\rangle$$ specifically in the $$k$$-basis. In other other words, the reason that we get $$\frac{\hbar^2k^2}{2m}$$ is because of the operator and the chosen basis, not because of $$|\psi\rangle$$.
• Thanks Aaron, that's what I was looking for, I'd think in the script they meant $\langle k \vert \hat{T} \vert \psi \rangle$. Just one question, you say because $k$ is a variable it can't be an eigenvalue. Why not? What's in the definition of eigenvectors/eigenvalues that prohibits that?
– mpr
Commented Mar 5, 2020 at 23:06
• @mpr Yeah I agree I was being pretty unclear on that first part. When I have time I'll edit it to be more clear, and I'll let you know when I do that. Commented Mar 6, 2020 at 0:19
• @mpr I added a part at the end. Commented Mar 6, 2020 at 9:59
• that clarifies it well, thank you very much!
– mpr
Commented Mar 6, 2020 at 10:29
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Commented Feb 28, 2020 at 20:08
• The momentum operator takes different form. In position basis, momentum is a derivative. In momentum basis, momentum is just multiplying by p. Commented Feb 28, 2020 at 20:14
• oh I see, we use de Broglie's relation $p = \hbar k$ so $\hat T = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$ and we just multiply by it, that's great, thanks to both of you!
|
– mpr
Commented Feb 28, 2020 at 20:35
First, the expression $$\hat T\bar\psi (k)=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$ does not mean $$\bar\psi(k)$$ is an eigenvalue of $$\hat T$$ because $$\hbar^2k^2/2m$$ is not a constant ($$k$$ is a variable now). Second, $$\hat T\bar\psi(k)$$ is not an expression that makes much sense. I think you actually mean $$\langle k|\hat T|\psi\rangle$$.
|
https://physics.stackexchange.com/questions/533659/kinetic-energy-operator-in-momentum-basis
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# Kinetic Energy Operator in Momentum Basis
I'm having a little trouble with something that's 'easy to check' according to the script I'm using. I consider the kinetic energy operator
$$\hat T = \frac{p^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$
and the wave function in momentum space, which is obtained using the Fourier transform
$$\tilde \psi(k) = \mathcal{F}[\psi(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \psi(x) e^{-ikx} dx \ .$$
According to the script it's easy to see that
$$\hat T \tilde \psi(k) = \frac{\hbar^2 k^2}{2m} \tilde \psi(k)\ ,$$
i.e. that $$\tilde \psi(k)$$ is an eigenstate of the kinetic energy operator. I don't manage to replicate that result. In particular, using integration by parts a few times I get
\begin{align} \hat T \tilde \psi(k) &= -\frac{\hbar^2}{2m} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \frac{d^2}{dx^2} ( \psi(x) e^{-ikx} ) dx \\ &= … \\ &= -\frac{\hbar^2}{2m} \tilde \psi(k) (-k^2 + 2k^2 - k^2) \\ &= 0 \end{align}
I see that $$\mathcal{F}[\hat T \psi(x)] = \frac{\hbar^2 k^2}{2m} \mathcal{F}[\psi(x)]$$ but $$\mathcal{F}[\hat T \psi(x)] \neq \hat T \tilde \psi(k)$$, right? So I'm confused ...
Can anybody show me where I go wrong?
• If you are doing derivatives with respect to $x$ for the momentum operator, then you are in the position basis, not the momentum basis. Commented Feb 28, 2020 at 20:08
• The momentum operator takes different form. In position basis, momentum is a derivative. In momentum basis, momentum is just multiplying by p. Commented Feb 28, 2020 at 20:14
• oh I see, we use de Broglie's relation $p = \hbar k$ so $\hat T = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$ and we just multiply by it, that's great, thanks to both of you!
– mpr
Commented Feb 28, 2020 at 20:35
First, the expression $$\hat T\bar\psi (k)=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$ does not mean $$\bar\psi(k)$$ is an eigenvalue of $$\hat T$$ because $$\hbar^2k^2/2m$$ is not a constant ($$k$$ is a variable now).
Second, $$\hat T\bar\psi(k)$$ is not an expression that makes much sense. I think you actually mean $$\langle k|\hat T|\psi\rangle$$. To be more formal, you are starting with $$\bar\psi(k)=\langle k|\psi\rangle=\int_{-\infty}^\infty\langle k|x\rangle\langle x|\psi\rangle\,\text dx=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-ikx}\psi(x)\,\text dx$$
But you cannot "operate" $$\hat T$$ on $$\langle k|\psi\rangle$$, since this is technically just a number for some value of $$k$$. This is why I think you actually mean $$\langle k|\hat T|\psi\rangle$$ because we can compute this in the momentum basis $$\langle k|\hat T|\psi\rangle=\int_{-\infty}^\infty\langle k|\hat T|k'\rangle\langle k'|\psi\rangle \,\text dk'=\int_{-\infty}^\infty\frac{\hbar^2k^2}{2m}\delta(k-k')\bar\psi(k') \,\text dk'=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$
This also reveals why $$\bar\psi(k)$$ is not, in general, an eigenfunction of $$\hat T$$. Note that really what we want is the basis-independent statement $$\hat T|\psi\rangle=c|\psi\rangle$$ to be true for $$|\psi\rangle$$ to be an eigenstate of $$\hat T$$. Now, at first glance this seems to be the case here, but the issue is that we needed to move into the $$k$$-basis in order to get this expression. In other words, $$\frac{\hbar^2k^2}{2m}$$ isn't an eigenvalue, it is just what is multiplied by $$\bar\psi(k)$$ when looking at $$\hat T$$ being applied to $$|\psi\rangle$$ specifically in the $$k$$-basis. In other other words, the reason that we get $$\frac{\hbar^2k^2}{2m}$$ is because of the operator and the chosen basis, not because of $$|\psi\rangle$$.
• Thanks Aaron, that's what I was looking for, I'd think in the script they meant $\langle k \vert \hat{T} \vert \psi \rangle$. Just one question, you say because $k$ is a variable it can't be an eigenvalue. Why not? What's in the definition of eigenvectors/eigenvalues that prohibits that?
– mpr
Commented Mar 5, 2020 at 23:06
• @mpr Yeah I agree I was being pretty unclear on that first part. When I have time I'll edit it to be more clear, and I'll let you know when I do that. Commented Mar 6, 2020 at 0:19
• @mpr I added a part at the end. Commented Mar 6, 2020 at 9:59
• that clarifies it well, thank you very much!
– mpr
Commented Mar 6, 2020 at 10:29
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Now, at first glance this seems to be the case here, but the issue is that we needed to move into the $$k$$-basis in order to get this expression. In other words, $$\frac{\hbar^2k^2}{2m}$$ isn't an eigenvalue, it is just what is multiplied by $$\bar\psi(k)$$ when looking at $$\hat T$$ being applied to $$|\psi\rangle$$ specifically in the $$k$$-basis.
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In other other words, the reason that we get $$\frac{\hbar^2k^2}{2m}$$ is because of the operator and the chosen basis, not because of $$|\psi\rangle$$. • Thanks Aaron, that's what I was looking for, I'd think in the script they meant $\langle k \vert \hat{T} \vert \psi \rangle$. Just one question, you say because $k$ is a variable it can't be an eigenvalue.
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https://math.stackexchange.com/questions/851072/theorem-on-giuga-number/851114
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# Theorem on Giuga number
Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$
How to prove the following theorem on Giuga numbers
$n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$
## 1 Answer
The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible. So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero. This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$. The chinese theorem now give: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-1\pmod{\prod_{i=1}^{k}p_i}$$ that is just: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv -1\pmod{n}$$ as claimed. For the $\Leftarrow$ part, we have that the congruence $\!\!\!\pmod{n}$ implies the congruence $\!\!\!\pmod{p_i}$, hence $(1)$ must hold, so we must have: $$\frac{n}{p_i}\equiv 1\pmod{p_i}$$ that is equivalent to $p_i\mid\left(\frac{n}{p_i}-1\right).$
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# Theorem on Giuga number
Giuga number : $n$ is a Giuga number $\iff$ For every prime factor $p$ of $n$ , $p | (\frac{n}{p}-1)$
How to prove the following theorem on Giuga numbers
$n$ is a giuga number $\iff$ $\sum_{i=1}^{n-1} i^{\phi(n)} \equiv -1 \mod {n}$
## 1 Answer
The $\Rightarrow$ part. For first, a giuga number must be squarefree, since, by assuming $p^2\mid n$, we have that $p$ divides two consecutive numbers, $\frac{n}{p}$ and $\frac{n}{p}-1$, that is clearly impossible.
|
So we have: $$n = \prod_{i=1}^{k} p_i$$ that implies: $$\phi(n) = \prod_{i=1}^{k} (p_i-1).$$ By considering the sum $$\sum_{i=0}^{n-1}i^{\phi(n)}$$ $\pmod{p_i}$ we have that all the terms contribute with a $1$, except the multiples of $p_i$ that contribute with a zero. This gives: $$\sum_{i=0}^{n-1}i^{\phi(n)}\equiv n-\frac{n}{p_i}\equiv (n-1)\pmod{p_i}\tag{1}$$ that holds for any $i\in[1,k]$.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude. Can someone explain these situations why exactly are they producing different results in AoA?
|
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight.
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question.
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If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous.
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb?
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. (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? ) Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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For more, see this answer to a related question What produces Thrust along the line of flight in a glider? Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space.
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Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack.
|
https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed! From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting.
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Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about. A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle.
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
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Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
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In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced.
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This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions.
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During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs.
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Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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As you establish an angle of climb the angle of attack will begin to decrease. If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase. As you pitch up the thrust vector points upwards and the lift vector points more backwards.
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So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases. "Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
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If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance. If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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Pitch does control speed and will increase AoA. Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
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But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed. The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little.
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Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight.
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image.
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As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector. The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power.
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb."
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[Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.) Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation. • Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes".
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https://aviation.stackexchange.com/questions/56874/are-we-changing-the-angle-of-attack-by-changing-the-pitch-of-an-aircraft?noredirect=1
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted!
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And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag).
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# Are we changing the angle of attack by changing the pitch of an aircraft?
For example:
if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. And yes, I’ll probably be climbing, although there are exceptions, becuase… if I pitch up, slow down, and maintain the same altitude, then undoubtedly the AoA has increased. Flying slower reduces the amount of lift generated. An increased AoA gets that back so I can maintain the altitude.
Can someone explain these situations why exactly are they producing different results in AoA?
Can someone explain these what is going on in terms of forces? Please explain in reference with these forces ( in the pic )
• Any question including such a poorly-drawn diagram as this, really ought not be answered at all, because it is a question with erroneous content included as part of the premise of the question. Thanks a lot, NASA! Commented Nov 8, 2018 at 15:36
• Lift will not change when angle of attack changes and speed is adjusted accordingly to maintain level flight. What does change is the lift coefficient. Commented Nov 8, 2018 at 18:57
• This sentence is unclear: "For example: if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other. " It sounds more like an answer than a question. If you don't tell us anything about the relationship between the change in pitch attitude and the change in climb angle, then it is impossible to know whether or not the angle-of-attack has changed. Only if the two values are identical, has the angle-of-attack remained constant. Commented May 31, 2021 at 22:40
First of all, note that the diagram linked in the question is erroneous. The equations may be right, but the forces are drawn in the wrong proportion-- lift is erroneously shown as greater than weight, when it should be less than weight. Also thrust appears to be about equal to drag, when it should be distinctly greater than drag. Only when lift is less than weight and thrust is greater than drag can we build a closed vector triangle-- meaning that net force is zero-- from weight, lift, and (thrust minus drag ). For more, see this answer to a related question Does lift equal weight in a climb? . (If you want to see a similar diagram from an outside source, see the one included in this answer to a related question Is excess lift or excess power needed for a climb? )
Just as NASA also messes up the proportions in this diagram for gliding flight-- https://www.grc.nasa.gov/www/k-12/airplane/glidvec.html -- again, lift is shown as being greater than weight, when it should be less than weight, so we can build a closed vector triangle from weight, lift, and drag. For more, see this answer to a related question What produces Thrust along the line of flight in a glider?
Now as to your questions-- to a first approximation we can think of our pitch control inputs-- the position we are placing the control stick or yoke in, in the fore-and-aft direction-- as most directly governing angle-of-attack, not pitch attitude in space. Pitch attitude in space is influenced by the climb angle which is influenced by power setting. Now, there are all kinds of inter-relationships that complicate things-- for example on a high-wing plane with flaps down, adding power may produce a strong downwash over the tail that tends to lead to an increase in angle-of-attack. But to a first approximation we can think of our pitch control inputs as governing angle-of-attack. There is a slight delay between a change in pitch control input and a change in angle-of-attack, due to the aircraft's rotational inertia in the pitch axis.
If I tell you that I've increased my aircraft's pitch attitude by 10 degrees but I don't tell you what I did with the elevator control to make that happen, nor do I tell you whether I've added power or not, then you have no way to guess whether I've kept the angle-of-attack constant and started climbing due to added power, or I've managed power as needed to keep altitude constant while transitioning to a higher angle-of-attack and lower airspeed, or any number of other possibilities-- I might even have put the stick or yoke forward to decrease the angle-of-attack and still added enough power that the aircraft transitioned into a climbing flight path resulting an in increase in pitch attitude. For example when a jet fighter aircraft is climbing vertically on raw thrust alone, the control stick is probably forward of the position where it would be during horizontal flight at the same airspeed. Certainly the angle-of-attack is lower in the vertical climb, than in horizontal flight at the same airspeed!
From the standpoint of what is really going on physically with the aircraft, most flight training curricula vastly over-emphasize the idea that the pilot is directly controlling the aircraft's pitch attitude. What he's really doing is controlling angle-of-attack and power setting. Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control) and is simpler to think about.
A key point is that our flight operations are usually conducted on the "front side of the power curve", where for a given power setting, an increase in angle-of-attack usually results in an increased climb rate and an increased climb angle. Therefore moving yoke or stick aft results in an increased angle-of-attack AND an increased climb angle (or a decreased glide angle) AND an increased pitch attitude. On the "back side of the power curve", like just above stall speed, an increase in angle-of-attack will generally lead to a decreased climb angle or increased sink angle, and the aircraft will end up in a more nose-down pitch attitude, so the idea that we're somehow directly controlling pitch attitude with the control yoke or stick no longer works very well.
Your question indicates a desire to better understand some of the physical relationships at play. Lift is proportional to (lift coefficient * airspeed squared). Lift coefficient is determined by angle-of-attack, with higher angles-of-attack creating higher lift coefficients. As shown in the vector diagrams attached to the two links given at the start of this answer, for shallow to moderate climb or dive angles, lift is NEARLY equal to weight. Actually, lift is a little less than weight unless the flight path is exactly horizontal, but for shallow to moderate climb or dive angles, the difference is small. Since weight is staying constant, we can conclude that for shallow to moderate climb or dive angles-- with no other accelerations going on (airspeed is staying constant or changing only slowly, and the flight path isn't curving up or down, and the wings aren't banked so the flight path isn't curving to describe a turn), lift is also staying nearly constant. This means that for shallow to moderate climb or dive angles, airspeed ends up being a pretty good guide to angle-of-attack-- to keep lift nearly constant, if the airspeed is low, the lift coefficient and angle-of-attack must be high. So the airspeed indicator is in essence an angle-of-attack gauge. At very steep climb angles where lift is quite a bit less than weight, things get more complicated-- if the aircraft is climbing straight up, lift must be zero, so lift coefficient must be zero, and angle-of-attack must be nearly zero (actually it must be slightly negative, unless the airfoil is completely symmetrical), no matter what the airspeed indicator reads.
In actual practice in general aviation, commercial aviation, etc, a shallow to moderately steep climb is NORMALLY carried out a higher angle-of-attack and lift coefficient--and therefore a lower airspeed-- than we'd use for high-speed cruising flight. It's more efficient this way, and it also gives us the most climb performance out of a given, limited amount of thrust available. Why? Because a high lift coefficient also correlates with a high ratio of (lift coefficient to drag coefficient), which means a high ratio of lift to drag. For shallow to moderate climb angles, the higher the L/D ratio we can achieve, the steeper we can climb for a given amount of thrust. This is explored in more detail in the first link given in this answer. To look at climb rate rather than climb angle, we'd have to look at a chart of (power-available minus power-required) at various airspeeds or various angles-of-attack, but we'd come to a similar conclusion-- our best climb performance will be achieved at an angle-of=attack well above what we'll be using in high-speed cruising flight.
The diagram in the original question doesn't touch in any way on the relationship between airspeed, angle-of-attack, lift coefficient, magnitude of lift vector, and L/D ratio, so it doesn't help us to understand why a shallow to moderately steep climb is normally carried out at a higher angle-of-attack than we'd use for high-speed cruising flight.
Your question included the statement "if I pitch the airplane up, but also increase power and am able to maintain the same speed, then no, the AoA hasn’t changed, although it may have varied in the transition between one situation and the other." For shallow to moderate climb angles, your statement is true for all practical purposes, but it is not EXACTLY true. If we want to be very precise about it, we could note that since lift is slightly reduced in the climb, if airspeed stayed constant than angle-of-attack must have been slightly reduced, and if angle-of-attack stayed exactly then airspeed must have been slightly reduced. This same idea came up in these two related answers to related questions, though in these cases the lift vector was reduced because the aircraft was in a descent rather than a climb -- 'Gravitational' power vs. engine power and Descending on a given glide slope (e.g. ILS) at a given airspeed-- is the size of the lift vector different in headwind versus tailwind?
• Not only the essential proportions are wrong in the diagram, but also the typical proportions like L/D are way wrong. Normally, L (and W) would be 5-15x greater than D (or F). Sometimes it is useful to exaggerate magnitudes/angles, but not in a such educational diagram...
– Zeus
Commented Nov 8, 2018 at 23:42
• An important point to stress (which everyone just touches) is that you are mostly talking about static, stabilised conditions. During transition, indeed, the pitch control (almost) directly controls AoA, and then this change affects other things, including the final AoA itself. It is important to note because AoA (and pitch, but unlike speed) are 'fast' variables, and their transition/dynamics is a matter of much consideration as well. Assuming static values is a very pilot-centric mindset.
– Zeus
Commented Nov 8, 2018 at 23:53
• "Yet the former way of looking at things works well enough in actual practice (e.g. flying an ILS glide slope by referring to an attitude indicator rather than an angle-of-attack meter as the primary guide to pitch control)..."-- also, trying to maintain a specific pitch attitude rather than a specific airspeed or angle-of-attack helps to avoid accidental "phugoid" oscillations due to the pilot getting a little out of phase as he "chases" the desired airspeed with his pitch inputs. Yet it remains the case that the position of the yoke/stick/elevator mainly governs a-o-a not pitch attitude. Commented Nov 9, 2018 at 19:51
Initially at the point of pitch increase, Yes. The formula to understand the relationship between pitch and Angle of Attack is:
Pitch Attitude + Incidence = Angle of Climb + Angle of Attack
The angle of attack will increase the most when you first increase the pitch. As you establish an angle of climb the angle of attack will begin to decrease.
If you are able to maintain the same airspeed in the climb by increasing power than the angle of attack will return to the trimmed angle of attack. So if your pitch angle matches your angle of climb then angle of attack will be the same as before the pitch increase.
As you pitch up the thrust vector points upwards and the lift vector points more backwards. So thrust has to carry some of the weight of the plane during the climb if the same Angle of Attack is maintained. Generally a decrease in airspeed (thus an increase of angle of attack) is preferred during climb (if on the front side of the power curve) as less excess thrust will be required to maintain a climb. There will also be less parasitic drag at the slower airspeed.
What is not shown on your diagram is the pitch moment caused by the center of thrust not being aligned with the center of gravity. This causes a perturbation in the trim speed so retrimming may be needed with power increases.
"Are we changing the angle of attack by changing the pitch of an aircraft"
We are taught "pitch controls speed, power controls altitude".
If that was the only input, the nose would rise AND the plane would slow down. The increase in AOA causes an increase in lift AND drag. The altimeter would show an increase. Forces out of balance.
If the elevator deflection was full, the AOA would increase until stall. With no power, stall. With full power either stall or loop! Unless there is sufficient power to loop, there will be a power-on stall. Pitch does control speed and will increase AoA.
Lower degrees of elevator deflection will result in a climb and loss of speed. This is why pitch and throttle go hand in hand. Airspeed must be maintained by increasing throttle to continue climb, or the same inexorable process of lower speed and increasing AOA again occurs. However, with a slight pitch change with no power change, the aircraft may merely slow down a little and climb gently.
But this depends on your airspeed and power setting before you apply elevator. IF YOU ARE LOW AND SLOW DO NOT PULL ELEVATOR OR YOU MAY STALL. ADD POWER TO CLIMB. And keep an eye on your airspeed.
The lift component is reduced when you climb more steeply, that means you need less angle of attack and more engine thrust to fight gravity. Take this to the extreme: F18 going vertically, you need zero angle of attack and zero lift, your engines are providing enough thrust to counter the gravitational pull.
• Not necessarily-- yes the lift vector must be smaller at a steeper climb angle. But if we start in shallow climb at 100 mph, and we slow to 80 mph while leaving the power setting unchanged, and our best climb angle occurs at 60 mph, there's no doubt that we've increased the angle-of-attack, even though for the reasons given in your answer and mine, we know the lift vector has decreased a little. Lift depends on airspeed squared, as well as angle-of-attack. More to the point, for SHALLOW climb angles lift stays ALMOST equal to weight even for substantial variations in airspeed. Commented Nov 8, 2018 at 15:49
• Yes going straight up the angle-of-attack MUST be zero. Yet in actual practice in general aviation, commercial aviation, etc, a climb is NORMALLY carried out a higher angle-of-attack and lift coefficient than we'd use for straight-and-level cruising flight. Commented Nov 8, 2018 at 15:51
• For shallow climb angles and dive angles, the airspeed is a pretty good guide to angle-of-attack-- lift is roughly equal to weight (slightly less unless flight path is horizontal, but only slightly less), so a lower airspeed correlates to a higher angle-of-attack. At steep climb angles things get more complicated. Commented Nov 8, 2018 at 15:57
In your Question, you asked for an explanation the NASA image. As pointed out, the image is incorrect because the Lift vector appears to be greater than the Weight vector.
The reason the image may be incorrect is because it may have been drawn by a person who received training as a private pilot. When I received my private pilot training 30 years ago, I was taught that to climb, you increase lift and increase power. The same thing is apparently being taught today. According to the FAA "When an airplane enters a climb, excess lift must be developed to overcome the weight or gravity. This requirement to develop more lift results in more induced drag, which either results in decreased airspeed and/or an increased power setting to maintain a minimum airspeed in the climb." [Airplane Flying Handbook (FAA-H-8083-3B), p. 3-16] (I removed the reference to the Embry Riddle website, because the FAA quote is enough.)
Not every flight school teaches the subject that way. See, e.g., the webpage for Gleim Aviation.
• Re "Regarding the point made in the first answer"-- be aware that different readers will see different answers displayed at the top of the screen depending on whether they have selected to display by "active", "oldest", or "votes". If you click on the "share" button under an answer or question you will be given a link that you can cut and paste anywhere you want to give a link to that answer or question. Commented May 31, 2021 at 22:43
• Clearly, the Embry-Riddle Aeronautical University should not be taken as an authority on piloting airplanes-- or at least upon the actual physics that are at play while an aircraft is being piloted! And likewise, the FAA. Commented May 31, 2021 at 22:45
• @quietflyer - "likewise the FAA" Yes, but if you are taking an FAA test or an engineering test in which the question appears, you had better be prepared to give the appropriate "right" answer :) FYI, I have had some questions about statements made in that NASA series (specifically relating to drag). But I was told that the person who created them has retired and that there is no one in charge of revising them. Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?" (referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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Commented Jun 1, 2021 at 1:43
• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Commented Jun 1, 2021 at 6:58
• @Bianfable - I should have deleted that first sentence - since my Answer did respond to the original Question which was "Can someone explain what is going on in terms of forces?"
|
(referring to the NASA image). I am trying to provide some useful background as to why the image was drawn that way and I have rewritten my answer to (hopefully) do a better job of explaining this. As far as I know, I am not asking for any clarification from the Asker. Let me know if you still feel that I should delete my Answer. Commented Jun 1, 2021 at 8:11
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https://math.meta.stackexchange.com/questions/31929/is-there-any-stack-exchange-site-that-allows-sharing-review-of-interesting-obse/31930
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# Is there any Stack Exchange site that allows sharing, review of interesting observations\results made\obtained by students in Mathematics?
I'm a 10th grader who is extremely interested in Mathematics and I frequently come across some interesting (at least, to me) results while doing some Math problem and sometimes I want to get an expert-level opinion on that result.
For example, I was recently thinking about how one would go on about defining a function that gives a graph like the one given below :
I successfully defined such a function using a combination of the floor function, ceiling function, fractional part function and the signum function. It was pretty interesting for me.
Another time, I discovered a simple derivation for the quadratic formula and once, a derivation for the compound angle identities in Trigonometry
These are some examples of when I wanted to share these and get some reviews/opinions about the results that I had obtained.
So, basically, is there a website for Mathematics like Code Review for Coding in the Stack Exchange Community?
Thanks!
PS : If you're wondering what the functions is, it's given below : $$f(x) = \text{Sign}\Bigg(\Bigg\{\dfrac{\lceil x \rceil}{2} \Bigg \} - a \Bigg) \text{, where } 0 < a < 0.5$$ $$\text{Here, }\{ x \} \text { is the fractional part function which is defined as } \{ x \} = x - \lfloor x \rfloor$$ $$\text{And Sign}(x) \text{ is the signum function, which gives the sign of the input, and } 0 \text{ in case the input is }0$$
Edit : I recently thought of a much simpler version of the function that I talk about above. It is : $$f(x) = \cos(\lfloor x \rfloor \cdot \pi)$$
• A better and clearer definition would be to not insist that it be given by a single formula and to say that $$f(x)=\begin{cases}1&\mbox{ if }2n<x\leq 2n+1\\-1&\mbox{ otherwise}.\end{cases}$$ – Matt Samuel Jun 16 '20 at 20:47
• An addition though : "Where $n \in \Bbb Z$". Actually, the reason that I insisted on a Mathematical definition of the function was so that it can be graphed using a graphing calculator and embedded in a computer program with a mathematical approach. Thanks for the suggestion! – Rajdeep Sindhu Jun 16 '20 at 20:52
• Is "check my work" question not allowed? We have a tag specified for that. – Arctic Char Jun 16 '20 at 20:55
• I am familiar with the solution-verification tag and in fact, have used it a few times too. As far as I know, this question : math.stackexchange.com/questions/3704308/… was closed till some time ago for the reason : Homework and check my work type questions not allowed. It's re-opened now though. Also, wouldn't a separate site (like Code Review for reviewing programs) be nice? – Rajdeep Sindhu Jun 16 '20 at 21:03
• I looked at the timeline and it doesn't appear that the question was ever closed. The reason is invalid in any case, because both homework and check-my-work questions are allowed. – Matt Samuel Jun 16 '20 at 21:34
• @MattSamuel I'm sorry for the misleading info. Looks like I can't recall the question which was closed for being a 'check my work' type question. Maybe (and most probably), it wasn't even on Mathematics SE. – Rajdeep Sindhu Jun 16 '20 at 21:43
• Then get rid of the first sentence of your post, @RajdeepSindhu ! – amWhy Jun 16 '20 at 23:30
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Actually, the reason that I insisted on a Mathematical definition of the function was so that it can be graphed using a graphing calculator and embedded in a computer program with a mathematical approach. Thanks for the suggestion! – Rajdeep Sindhu Jun 16 '20 at 20:52
• Is "check my work" question not allowed? We have a tag specified for that. – Arctic Char Jun 16 '20 at 20:55
• I am familiar with the solution-verification tag and in fact, have used it a few times too.
|
As far as I know, this question : math.stackexchange.com/questions/3704308/… was closed till some time ago for the reason : Homework and check my work type questions not allowed. It's re-opened now though. Also, wouldn't a separate site (like Code Review for reviewing programs) be nice? – Rajdeep Sindhu Jun 16 '20 at 21:03
• I looked at the timeline and it doesn't appear that the question was ever closed.
|
https://math.stackexchange.com/questions/2098138/how-to-lower-upper-bound-n-using-1x-leq-ex
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# How to lower/upper bound $n!$ using $1+x\leq e^x$?
I need to prove for all positive integer $n$
$$e\left(\frac{n}{e}\right)^n\leq n!\leq en\left(\frac{n}{e}\right)^n,$$ using the hint $1+x\leq e^x$ for all $x\in \mathbb{R}$.
I did this:
The hint says
• for $x=0$, $1\leq 1$;
• for $x=1$, $2\leq e$;
• ...
• for $x=n-1$, $n\leq e^{n-1}$.
So I multiplied these $n$ inequalities to get
$$n!\leq e^{n-1+\ldots+1},$$ or $$n!\leq e^{\frac{n(n-1)}{2}},$$ and I get stuck there.
• I will edit to let $n$ be positive. – Zir Jan 15 '17 at 0:15
• Oh, whoops, was I supposed to use $1+x\le e^x$? – Simply Beautiful Art Jan 15 '17 at 0:39
• Yes, that was the hint I am supposed to use. – Zir Jan 15 '17 at 0:44
• Bleh, well, whatever XD. – Simply Beautiful Art Jan 15 '17 at 1:33
Rearranging gives the equivalent inequalities $$1 \le (n-1)! \left(\frac{e}{n}\right)^{n-1} \le n.$$
When $n=1$ both inequalities are equalities. Assuming the statement holds for $n=k$, then we want to prove $$1 \le k! \left(\frac{e}{k+1}\right)^k \le k+1.$$
The first inequality holds since $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\ge 1} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\ge 1 \text{ by induction}}$$ where the first term is $\ge 1$ because the hint gives $$e^{1/k} \frac{k}{k+1} \ge \left(1+\frac{1}{k}\right) \frac{k}{k+1} = 1.$$
The second inequality is due to $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\le \frac{k+1}{k}} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\le k \text{ by induction}}$$ where the first term is $\le \frac{k+1}{k}$ because $$e \cdot \left(\frac{k}{k+1}\right)^{k+1} = e \cdot \left(1-\frac{1}{k+1}\right)^{k+1} \le e \cdot e^{-1} = 1.$$ [It is well known that $\left(1-\frac{1}{k+1}\right)^{k+1}$ converges to $e^{-1}$ as $k \to \infty$, but one can show by induction that this actually increases monotonically to $e^{-1}$.]
• Thanks. In fact the last one is also due using the hint by setting $x=-\frac{1}{k+1}$ in $1+x\leq e^x$. – Zir Jan 15 '17 at 2:43
Take the log of all sides, so we just need to prove that
$$n\ln n-n+1\le\ln(n!)\le (n+1)\ln n-n+1$$
Then, it is easy to see that
$$\ln(n!)=\sum_{k=1}^n\ln(k)$$
And that is the right Riemann sum to the following integral:
$$\sum_{k=1}^n\ln(k)\ge\int_1^n\ln(x)\ dx=n\ln n-n+1$$
So we have proven the left side. Then notice that we have the following trapezoidal sum:
$$\frac{\ln(n)+2\ln((n-1)!)}2=\frac{\ln(n!)+\ln((n-1)!)}2\\=\sum_{k=1}^n\frac{\ln(k+1)+\ln(k)}2\\\le\int_1^n\ln(x)\ dx=n\ln(n)-n+1\\\frac{\ln(n)+2\ln((n-1)!)}2\le n\ln(n)-n+1\\\implies\ln((n-1)!)\le\left(n-\frac12\right)\ln(n)-n+1\\\implies\ln(n!)=\ln(n)+\ln((n-1)!)\le\left(n+\frac12\right)\ln(n)-n+1$$
Thus, when $n\ge1$,
$$\ln(n!)\le\left(n+\frac12\right)\ln(n)-n+1\le(n+1)\ln(n)-n+1$$
and we have everything we wanted.
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# How to lower/upper bound $n!$ using $1+x\leq e^x$? I need to prove for all positive integer $n$
$$e\left(\frac{n}{e}\right)^n\leq n!\leq en\left(\frac{n}{e}\right)^n,$$ using the hint $1+x\leq e^x$ for all $x\in \mathbb{R}$. I did this:
The hint says
• for $x=0$, $1\leq 1$;
• for $x=1$, $2\leq e$;
• ...
• for $x=n-1$, $n\leq e^{n-1}$. So I multiplied these $n$ inequalities to get
$$n!\leq e^{n-1+\ldots+1},$$ or $$n!\leq e^{\frac{n(n-1)}{2}},$$ and I get stuck there.
|
• I will edit to let $n$ be positive. – Zir Jan 15 '17 at 0:15
• Oh, whoops, was I supposed to use $1+x\le e^x$? – Simply Beautiful Art Jan 15 '17 at 0:39
• Yes, that was the hint I am supposed to use. – Zir Jan 15 '17 at 0:44
• Bleh, well, whatever XD. – Simply Beautiful Art Jan 15 '17 at 1:33
Rearranging gives the equivalent inequalities $$1 \le (n-1)!
|
https://math.stackexchange.com/questions/2098138/how-to-lower-upper-bound-n-using-1x-leq-ex
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# How to lower/upper bound $n!$ using $1+x\leq e^x$?
I need to prove for all positive integer $n$
$$e\left(\frac{n}{e}\right)^n\leq n!\leq en\left(\frac{n}{e}\right)^n,$$ using the hint $1+x\leq e^x$ for all $x\in \mathbb{R}$.
I did this:
The hint says
• for $x=0$, $1\leq 1$;
• for $x=1$, $2\leq e$;
• ...
• for $x=n-1$, $n\leq e^{n-1}$.
So I multiplied these $n$ inequalities to get
$$n!\leq e^{n-1+\ldots+1},$$ or $$n!\leq e^{\frac{n(n-1)}{2}},$$ and I get stuck there.
• I will edit to let $n$ be positive. – Zir Jan 15 '17 at 0:15
• Oh, whoops, was I supposed to use $1+x\le e^x$? – Simply Beautiful Art Jan 15 '17 at 0:39
• Yes, that was the hint I am supposed to use. – Zir Jan 15 '17 at 0:44
• Bleh, well, whatever XD. – Simply Beautiful Art Jan 15 '17 at 1:33
Rearranging gives the equivalent inequalities $$1 \le (n-1)! \left(\frac{e}{n}\right)^{n-1} \le n.$$
When $n=1$ both inequalities are equalities. Assuming the statement holds for $n=k$, then we want to prove $$1 \le k! \left(\frac{e}{k+1}\right)^k \le k+1.$$
The first inequality holds since $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\ge 1} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\ge 1 \text{ by induction}}$$ where the first term is $\ge 1$ because the hint gives $$e^{1/k} \frac{k}{k+1} \ge \left(1+\frac{1}{k}\right) \frac{k}{k+1} = 1.$$
The second inequality is due to $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\le \frac{k+1}{k}} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\le k \text{ by induction}}$$ where the first term is $\le \frac{k+1}{k}$ because $$e \cdot \left(\frac{k}{k+1}\right)^{k+1} = e \cdot \left(1-\frac{1}{k+1}\right)^{k+1} \le e \cdot e^{-1} = 1.$$ [It is well known that $\left(1-\frac{1}{k+1}\right)^{k+1}$ converges to $e^{-1}$ as $k \to \infty$, but one can show by induction that this actually increases monotonically to $e^{-1}$.]
• Thanks. In fact the last one is also due using the hint by setting $x=-\frac{1}{k+1}$ in $1+x\leq e^x$. – Zir Jan 15 '17 at 2:43
Take the log of all sides, so we just need to prove that
$$n\ln n-n+1\le\ln(n!)\le (n+1)\ln n-n+1$$
Then, it is easy to see that
$$\ln(n!)=\sum_{k=1}^n\ln(k)$$
And that is the right Riemann sum to the following integral:
$$\sum_{k=1}^n\ln(k)\ge\int_1^n\ln(x)\ dx=n\ln n-n+1$$
So we have proven the left side. Then notice that we have the following trapezoidal sum:
$$\frac{\ln(n)+2\ln((n-1)!)}2=\frac{\ln(n!)+\ln((n-1)!)}2\\=\sum_{k=1}^n\frac{\ln(k+1)+\ln(k)}2\\\le\int_1^n\ln(x)\ dx=n\ln(n)-n+1\\\frac{\ln(n)+2\ln((n-1)!)}2\le n\ln(n)-n+1\\\implies\ln((n-1)!)\le\left(n-\frac12\right)\ln(n)-n+1\\\implies\ln(n!)=\ln(n)+\ln((n-1)!)\le\left(n+\frac12\right)\ln(n)-n+1$$
Thus, when $n\ge1$,
$$\ln(n!)\le\left(n+\frac12\right)\ln(n)-n+1\le(n+1)\ln(n)-n+1$$
and we have everything we wanted.
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\left(\frac{e}{n}\right)^{n-1} \le n.$$
When $n=1$ both inequalities are equalities. Assuming the statement holds for $n=k$, then we want to prove $$1 \le k! \left(\frac{e}{k+1}\right)^k \le k+1.$$
The first inequality holds since $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\ge 1} \cdot \underbrace{(k-1)!
|
\left(\frac{e}{k}\right)^{k-1}}_{\ge 1 \text{ by induction}}$$ where the first term is $\ge 1$ because the hint gives $$e^{1/k} \frac{k}{k+1} \ge \left(1+\frac{1}{k}\right) \frac{k}{k+1} = 1.$$
The second inequality is due to $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\le \frac{k+1}{k}} \cdot \underbrace{(k-1)!
|
https://math.stackexchange.com/questions/2098138/how-to-lower-upper-bound-n-using-1x-leq-ex
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# How to lower/upper bound $n!$ using $1+x\leq e^x$?
I need to prove for all positive integer $n$
$$e\left(\frac{n}{e}\right)^n\leq n!\leq en\left(\frac{n}{e}\right)^n,$$ using the hint $1+x\leq e^x$ for all $x\in \mathbb{R}$.
I did this:
The hint says
• for $x=0$, $1\leq 1$;
• for $x=1$, $2\leq e$;
• ...
• for $x=n-1$, $n\leq e^{n-1}$.
So I multiplied these $n$ inequalities to get
$$n!\leq e^{n-1+\ldots+1},$$ or $$n!\leq e^{\frac{n(n-1)}{2}},$$ and I get stuck there.
• I will edit to let $n$ be positive. – Zir Jan 15 '17 at 0:15
• Oh, whoops, was I supposed to use $1+x\le e^x$? – Simply Beautiful Art Jan 15 '17 at 0:39
• Yes, that was the hint I am supposed to use. – Zir Jan 15 '17 at 0:44
• Bleh, well, whatever XD. – Simply Beautiful Art Jan 15 '17 at 1:33
Rearranging gives the equivalent inequalities $$1 \le (n-1)! \left(\frac{e}{n}\right)^{n-1} \le n.$$
When $n=1$ both inequalities are equalities. Assuming the statement holds for $n=k$, then we want to prove $$1 \le k! \left(\frac{e}{k+1}\right)^k \le k+1.$$
The first inequality holds since $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\ge 1} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\ge 1 \text{ by induction}}$$ where the first term is $\ge 1$ because the hint gives $$e^{1/k} \frac{k}{k+1} \ge \left(1+\frac{1}{k}\right) \frac{k}{k+1} = 1.$$
The second inequality is due to $$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\le \frac{k+1}{k}} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\le k \text{ by induction}}$$ where the first term is $\le \frac{k+1}{k}$ because $$e \cdot \left(\frac{k}{k+1}\right)^{k+1} = e \cdot \left(1-\frac{1}{k+1}\right)^{k+1} \le e \cdot e^{-1} = 1.$$ [It is well known that $\left(1-\frac{1}{k+1}\right)^{k+1}$ converges to $e^{-1}$ as $k \to \infty$, but one can show by induction that this actually increases monotonically to $e^{-1}$.]
• Thanks. In fact the last one is also due using the hint by setting $x=-\frac{1}{k+1}$ in $1+x\leq e^x$. – Zir Jan 15 '17 at 2:43
Take the log of all sides, so we just need to prove that
$$n\ln n-n+1\le\ln(n!)\le (n+1)\ln n-n+1$$
Then, it is easy to see that
$$\ln(n!)=\sum_{k=1}^n\ln(k)$$
And that is the right Riemann sum to the following integral:
$$\sum_{k=1}^n\ln(k)\ge\int_1^n\ln(x)\ dx=n\ln n-n+1$$
So we have proven the left side. Then notice that we have the following trapezoidal sum:
$$\frac{\ln(n)+2\ln((n-1)!)}2=\frac{\ln(n!)+\ln((n-1)!)}2\\=\sum_{k=1}^n\frac{\ln(k+1)+\ln(k)}2\\\le\int_1^n\ln(x)\ dx=n\ln(n)-n+1\\\frac{\ln(n)+2\ln((n-1)!)}2\le n\ln(n)-n+1\\\implies\ln((n-1)!)\le\left(n-\frac12\right)\ln(n)-n+1\\\implies\ln(n!)=\ln(n)+\ln((n-1)!)\le\left(n+\frac12\right)\ln(n)-n+1$$
Thus, when $n\ge1$,
$$\ln(n!)\le\left(n+\frac12\right)\ln(n)-n+1\le(n+1)\ln(n)-n+1$$
and we have everything we wanted.
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\left(\frac{e}{k}\right)^{k-1}}_{\le k \text{ by induction}}$$ where the first term is $\le \frac{k+1}{k}$ because $$e \cdot \left(\frac{k}{k+1}\right)^{k+1} = e \cdot \left(1-\frac{1}{k+1}\right)^{k+1} \le e \cdot e^{-1} = 1.$$ [It is well known that $\left(1-\frac{1}{k+1}\right)^{k+1}$ converges to $e^{-1}$ as $k \to \infty$, but one can show by induction that this actually increases monotonically to $e^{-1}$.]
|
• Thanks. In fact the last one is also due using the hint by setting $x=-\frac{1}{k+1}$ in $1+x\leq e^x$. – Zir Jan 15 '17 at 2:43
Take the log of all sides, so we just need to prove that
$$n\ln n-n+1\le\ln(n! )\le (n+1)\ln n-n+1$$
Then, it is easy to see that
$$\ln(n! )=\sum_{k=1}^n\ln(k)$$
And that is the right Riemann sum to the following integral:
$$\sum_{k=1}^n\ln(k)\ge\int_1^n\ln(x)\ dx=n\ln n-n+1$$
So we have proven the left side.
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https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
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# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values. Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
|
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined".
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
| 1,660,658,483,000,000,000
|
text/html
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crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00167.warc.gz
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|
# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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Can anyone explain, what am I understanding wrong. Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, .
|
With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there. Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
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text/html
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crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00167.warc.gz
| 510,181,825
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|
# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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You can even plot an histogram on the samples to see the (unnormalized) posterior. In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value. Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g.
|
Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution. b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others). b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3).
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
| 1,660,658,483,000,000,000
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# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice. Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
|
[1] Edit: not true, see @Xi'an's comment. Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant.
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
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|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00167.warc.gz
| 510,181,825
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# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly.
|
Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand.
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
| 1,660,658,483,000,000,000
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text/html
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crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00167.warc.gz
| 510,181,825
| 68,042
|
# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
|
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a). The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
|
https://stats.stackexchange.com/questions/138644/confusion-in-gibbs-sampling
| 1,660,658,483,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00167.warc.gz
| 510,181,825
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|
# Confusion in Gibbs sampling
I am self-studying Gibbs sampling from a book. The book introduces metropolis hastings algortihm to generate representative values from a posterior distribution. So we know $p(D | \theta) p(\theta)$ but not in the normalized form. So we generate those values.
Now when it introduces Gibbs sampling it quotes,
Gibbs sampling is especially useful when the complete joint posterior, $p(θ_i |D)$, cannot be analytically determined and cannot be directly sampled, but all the conditional distributions, $p(θ_i |\{θ_{j \ne i}\}, D)$, can be determined and directly sampled.
Suppose we have two parameters $\theta_1$ and $\theta_2$ so for determining $p(θ_1|θ_2, D)$, we need $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$, and now I am all the more confused. First it says
when we can't analytically determine the posterior
and then uses $\frac{p(θ_1, θ_2|D)}{p(θ_2|D)}$ , where we don't know yet the posterior distribution, because that is not "analytically determined". Can anyone explain, what am I understanding wrong.
Since I'm not sure where are you stuck at, I'll try multiple shots:
Explanation 1: The thing is that you only need the form of the unnormalized posterior, and that is why it's enough if you can get: $$p(\theta_1 | \theta_2, D) \propto p(\theta_1, \theta_2 | D)$$ The normalizing constant is not interesting, this is very common in bayesian statistics, . With Gibbs sampling, Metropolis-Hasting or any other Monte Carlo method, what you are doing is drawing samples from this posterior. That is, the more density around a point, the more samples you'll get there.
Then, once you have enough samples from this posterior distribution, you know that the normalized density at some point $x$ is the proportion of samples that felt at that point.
You can even plot an histogram on the samples to see the (unnormalized) posterior.
In other words, if I give you the samples $1,3,4,5,1.....,3,4,16,1$ and I tell you these are samples from a density function, you know to compute the probability of every value.
Explanation 2: If you observe the analytical form of your unnormalized posterior (you always know it [1]), two things can happen:
a) It has the shape of some known distribution (e.g. Gaussian): then you can get the normalize posterior since you know the normalizing constant of a gaussian distribution.
b) It has an ugly form that corresponds to no familiar distribution: then you can always sample with Metropolis-Hastings (there are others).
b.1) M-H is not the most efficient of the methods (you reject a lot of samples, usually more than 2/3). If the posterior was ugly but the conditionals of the individual variables are pretty (known distributions) then you can do Gibbs sampling by sampling for one single variable at a time.
Explanation 3 If you use conjugate priors for the individual variables, the denominator of their conditional probability will be always nice and familiar, and you will know what the normalizing constant in the denominator is. This is why Gibbs sampling is so popular when the joint probability is ugly but the conditional probabilities are nice.
Maybe this thread, and specially the answer with puppies, helps you: Why Normalizing Factor is Required in Bayes Theorem?
[1] Edit: not true, see @Xi'an's comment.
Update (example)
Imagine you have:
$$P(\theta_1, \theta_2 | D) = \frac{ p(D , \theta_1, \theta_2)} {\int_{\theta_1, \theta_2} p(D , \theta_1, \theta_2) \text{d} \theta_1, \theta_2} \propto p(D , \theta_1, \theta_2)$$
If the joint probability is complicated, then you can't know the normalization constant. Sometimes, if it does not contain things like large $\sum$ or $\prod$ that would make it painful to compute, you can even plot the posterior. In this case you would have some 2-D plot with axes $\theta_1$ and $\theta_2$. Yet, you plot is right up to a missing constant. Sampling algorithms say "ok, I don't know what the normalization factor is, but if I might draw samples from this function in such a way that, if $p(D, \theta_1=x_1, \theta_2=x_2)$ is two times $p(D, \theta_1=x_3, \theta_2=x_4)$, then then I should get the sample $(x_1, x_2)$ twice as much as $(x_3, x_4)$"
Gibbs sampling does this by sampling every variable separatedly. Imagine $\theta_1$ is a mean $\mu$ and that its conditional probability is (forget about the $\sigma$'s, imagine we know them): $$p(\mu | D) = \frac{ \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) } {\int \mathcal{N}(D | \mu, \sigma_d) \mathcal{N}(\mu, \sigma) \text{d} \mu}$$ The product of two normals is another normal with new parameters (see conjugate priors and keep that table always at hand. Even memorize the ones you end up using the most). You do the multiplication, you drop everything that does not depend on $\mu$ into a constant $K$ and you get something that you can express as: $$p(\mu | D) = K \exp\left(\frac{1}{a}(\mu - b)^2\right)$$ It has the functional form of a Gaussian. Therefore since you know it is a density, $K$ must be the normalizing factor of $\mathcal{N}(b, a)$.
Thus, your posterior is a Gaussian distribution with posterior parameters (b,a).
The short version is that the product of the prior and the likelihood have the functional form of a familiar distribution (it actually has the form of the prior if you chose conjugates), then you know how to integrate it. For instance, the integral of the $\exp(...)$ element of a normal distribution, that is, a normal without its normalizing factor, is the inverse of its normalizing factor.
• in Explanation 2, it is not always the case that you know the analytical form of your unnormalized posterior. Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals. ( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1). I'll add an example to the answer if it's not clear yet. Feb 22, 2015 at 10:42
• Yes @Alberto please include an example. Feb 22, 2015 at 11:00
• Yes, so that the whole (1/marginal_like)*\int likelihood*prior = 1. The example will arrive in a few minutes :) Feb 22, 2015 at 11:49
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Take for instance the case of latent variables: the unnormalized posterior may involve an intractable integral. Feb 21, 2015 at 19:54
• So what I understand is, that when we apply $\frac{p(θ_1,θ_2|D)}{p(θ_2|D)}$ here even if we don't know the normalized posterior, but analytically it is $k∗likelihood∗prior$ but when we plug it in the formula for finding the conditional probability that constant will cancel out from both the denominator and numerator and we will get the conditionals.
|
( Even when we don't know the full posterior with the normalizing constant ). Is that right ? Feb 22, 2015 at 8:59
• It's not that they cancel out, it's that if you have some likelihood * prior / marginal_likelihood which is e.g $\exp(C*(x-\mu)^2)/marginal_like$ you know, since the posterior is a density, that marginal_like is the constant factor of your normal distribution (so that the whole thing integrates to 1).
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https://math.stackexchange.com/questions/2792471/linearization-of-system-of-odes-around-operating-point-transfer-functions-and
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# Linearization of System of ODEs around Operating Point / Transfer Functions and State Space
I have this system of ODEs and I'm trying to get a linearized version of it around the "operating point" $\overline{x}_1 = 1$
$$\left\{\begin{matrix} \ddot{x_1}(t)+2\dot{x_1}(t)+2x_1^2(t)-2\dot{x_2}(t)=0 \\ 2\ddot{x_2}(t)+2\dot{x_2}(t)-2\dot{x_1}(t)=f(t) \end{matrix}\right.$$
So I define a small perturbation $\delta x_1$, $\delta x_2$ and $\delta f$ around the operating point $\overline{x}_1$, $\overline{x}_2$ and $\overline{f}$
$$\delta x_1 = x_1 - \overline{x}_1 \Rightarrow \dot{x_1} = \dot{\delta x_1} \Rightarrow \ddot{x_1} = \ddot{\delta x_1}$$
$$\delta x_2 = x_2 - \overline{x}_2 \Rightarrow \dot{x_2} = \dot{\delta x_2} \Rightarrow \ddot{x_2} = \ddot{\delta x_2}$$
$$\delta f = f - \overline{f}$$
I use Taylor polynomial to linearize $x_1^2(t)$ around $\overline{x}_1=1$ as
$$x_1^2 \approx \overline{x}_1^2 + 2\overline{x}_1 \delta x_1 = 1 + 2\delta x_1$$
I replace all in the original equations:
$$\left\{\begin{matrix}\delta\ddot{x_1}(t)+2\delta\dot{x_1}(t)+2\left [1+2\delta x_1(t) \right ] - 2 \delta \dot{x_2}(t)=0 \\ 2\delta \ddot{x_2}(t)+2\delta \dot{x_2}(t)-2\delta \dot{x_1}(t)=\overline{f}+\delta f(t) \end{matrix}\right.$$
This system is "linear", but not homogeneous, because it has constant terms $2$ and $\overline{f}$. In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow.
How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$
By the way, I got this systems of ODEs from this physical system:
• How did you determine that $x_1=1$ is the operating point? You will need a non-constant $\bar f$, as with a constant one you get $x_1=0$ as equilibrium point, just from physical considerations. Note that $$\frac{d}{dt}\left[\frac12 \dot x_1(t)^2+\dot x_2(t)^2+\frac23x_1(t)^3\right]=f(t)\dot x_2(t)-2(\dot x_1(t)-\dot x_2(t))^2,$$ where the last term continuously loses energy, leading to $x_1 \to 0$. You will need a very specific $f$ to continuously replace that lost energy. Commented May 23, 2018 at 7:24
• This is a problem from a textbook. It specifically ask to linearize about $x_1=1$. Thank you Commented May 23, 2018 at 7:35
When linearising a non-linear system of the form $\dot{x} = g(x,f)$ at an operating point $\bar{x}$ and $\bar{f}$ requires that $g(\bar{x},\bar{f})=0$. Since $\bar{x}_1$ is given and $g(x,f)$ is not a function of $x_2$, then $g(\bar{x},\bar{f})=0$ only has a solution when $x_2$ is omitted from the state space vector, so $x$ only contains $x_1$, $\dot{x}_1$ and $\dot{x}_2$ and no $x_2$. So $\bar{\dot{x}}_2$ can then be a non-zero constant, which can be chosen such that $g(\bar{x},\bar{f})=0$ can be satisfied.
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In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow. How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$
By the way, I got this systems of ODEs from this physical system:
• How did you determine that $x_1=1$ is the operating point?
|
You will need a non-constant $\bar f$, as with a constant one you get $x_1=0$ as equilibrium point, just from physical considerations. Note that $$\frac{d}{dt}\left[\frac12 \dot x_1(t)^2+\dot x_2(t)^2+\frac23x_1(t)^3\right]=f(t)\dot x_2(t)-2(\dot x_1(t)-\dot x_2(t))^2,$$ where the last term continuously loses energy, leading to $x_1 \to 0$. You will need a very specific $f$ to continuously replace that lost energy.
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http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
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Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
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I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
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Very interesting! – goblin Apr 18 '15 at 3:08
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# Can we extend the definition of a homomorphism to binary relations? This is going to be quite a long post. The actual questions will be at the end of it in section "Questions." INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra.
|
It seems more difficult to me because there seems not to be a unique obvious way of doing it. I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
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Very interesting! – goblin Apr 18 '15 at 3:08
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Both fail to satisfy a property I thought they should satisfy. I will explain this later. ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$.
|
Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition. Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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|
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| 195,675,263
| 23,186
|
# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
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Very interesting! – goblin Apr 18 '15 at 3:08
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$\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism. Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition. Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
|
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism.
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http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
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Very interesting! – goblin Apr 18 '15 at 3:08
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Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
|
• other algebraic structures? (2) Has anything similar to the definitions I'm giving been tried in literature? (3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.) -
Instead of relations I think it is a better idea to work with spans, or correspondences.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
| 1,467,040,698,000,000,000
|
text/html
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crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00172-ip-10-164-35-72.ec2.internal.warc.gz
| 195,675,263
| 23,186
|
# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets.
|
In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity). Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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If you wanted to, though, use the following. Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$. For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$. -
I know very little about category theory. Could you explain what a morphism between two binary relations is?
|
– user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it?
|
– user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
| 4,081
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A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie. Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e.
|
using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$. As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
| 1,467,040,698,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-26/segments/1466783396029.85/warc/CC-MAIN-20160624154956-00172-ip-10-164-35-72.ec2.internal.warc.gz
| 195,675,263
| 23,186
|
# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
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Very interesting! – goblin Apr 18 '15 at 3:08
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Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims. Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986).
|
It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard. The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York. • Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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Masson, Paris, 1984.) I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful. However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said).
|
If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
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http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
|
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions. A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
|
http://math.stackexchange.com/questions/148715/can-we-extend-the-definition-of-a-homomorphism-to-binary-relations
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| 195,675,263
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# Can we extend the definition of a homomorphism to binary relations?
This is going to be quite a long post. The actual questions will be at the end of it in section "Questions."
INTRODUCTION
After receiving an answer to this question about extending the definition of a continuous function to binary relations, I started thinking about doing the same with homomorphisms in abstract algebra. It seems more difficult to me because there seems not to be a unique obvious way of doing it.
I tried a modest task first: to do this for semigroups (so for groups too in particular). Ring or algebra homomorphisms seem more difficult to generalize. I have ended up with two definitions for semigroup homomorphisms, but I'm not sure if they're of any use. Both fail to satisfy a property I thought they should satisfy. I will explain this later.
ATTEMPTED DEFINITIONS
Let $S,T$ be semigroups. A function $\varphi:S\to T$ is a homomorphism if it satisfies $$\varphi(ab)=\varphi(a)\varphi(b).\tag1$$
If $\rho\subseteq S\times T$ is a binary relation, the following also makes sense:
$$\rho(ab)=\rho(a)\rho(b),\tag2$$
but now $\rho(ab),\rho(a),\rho(b)$ are not elements of $T$, but of $2^T$. Yet $(2)$ makes sense, because $2^T$ also has a structure of a semigroup, with the operation defined by $$AB=\{ab\,|\,a\in A,\,b\in B\}.$$
This gives an idea for a definition.
Definition 1. $\rho$ is element-wise homomorphic iff for any $a,b\in S,$ we have $$\rho(ab)=\rho(a)\rho(b).$$
But there's another idea. Every function $\varphi:S\to T$ can be extended to a function $\overline\varphi : 2^S\to 2^T$ as follows. $$\overline\varphi(A)=\varphi(A),$$ where $\varphi(A)$ denotes the image of $A$ under $\varphi.$ Obviously, $$\overline\varphi(\{a\})=\{\varphi(a)\}.$$
Now let $\varphi$ be a homomorphism.
Fact 1. $\overline\varphi$ is a semigroup homomorphism from $2^S$ to $2^T$.
Proof. Let $A,B\subseteq S.$ Then
$$\begin{eqnarray} \overline\varphi(AB)&=&\{\varphi(ab)\,|\, a\in A,\,b\in B\}\\ &=&\{\varphi(a)\varphi(b)\,|\,a\in A,\,b\in B\}\\ &=&\{xy\,|\,x\in\varphi(A),\,y\in\varphi(B)\}\\ &=&\{xy\,|\,x\in\overline\varphi(A),\,y\in\overline\varphi(B)\}\\ &=&\overline\varphi(A)\overline\varphi(B). \end{eqnarray}$$
Also, it is obvious that if $\overline\varphi$ is a homomorphism, then $\varphi$ is a homomorphism too, because it essentially a restriction of $\overline\varphi$ to the subsemigroup $S$ of the semigroup $2^S.$ This gives us
Fact 2. $\varphi:S\to T$ is a homomorphism iff $\overline\varphi:2^S\to 2^T$ is a homomorphism.
Now $\rho$ also induces a function $\overline\rho:2^S\to 2^T$ by $$\overline\rho(A)=\rho(A),$$ where $\rho(A)$ denotes the image of $A$ under $\rho.$ This gives another idea for a definition.
Definition 2. $\rho$ is set-wise homomorphic iff $\overline\rho:2^S\to 2^T$ is a semigroup homomorphism.
WHY I THINK IT DOESN'T WORK
I think a good definition of a homomorphic relation $\rho$ should give that $\rho^{-1}$ is also homomorphic. This would be a generalization of the fact that a bijective homomorphism is an isomorphism. (That is, we don't need to check whether the inverse function is a homomorphism.) Both definitions fail here. Let $\varphi:\mathbb Z\to\mathbb Z$ be the trivial homomorphism. Then $$\overline{\left(\varphi^{-1}\right)}(\{1\}+\{-1\})=\overline{\left(\varphi^{-1}\right)}(\{0\})=\mathbb Z.$$
But $$\overline{\left(\varphi^{-1}\right)}(\{1\})=\varnothing$$ and $$\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing,$$ so
$$\overline{\left(\varphi^{-1}\right)}(\{1\})\overline{\left(\varphi^{-1}\right)}(\{-1\})=\varnothing.$$
QUESTIONS
(1) Is there a standard definition of a "homomorphic binary relation" for
• semigroups?
• other algebraic structures?
(2) Has anything similar to the definitions I'm giving been tried in literature?
(3) When defining a "homomorphic binary relation", what could make one defintion more sensible than another? (This is imprecise of course, but I want to ask it in case someone has a precise answer.)
-
Instead of relations I think it is a better idea to work with spans, or correspondences. A span between two objects $a$ and $b$ in a category $C$ is a diagram of the form $a \leftarrow c \to b$. Spans admit a notion of composition described for example at the nLab in any category with pullbacks (which includes in particular any algebraic category), giving a category $\text{Span}(C)$. If $a$ and $b$ are sets and $R$ a relation between them, then letting $c = \{ (x, y) : x \in a, y \in b, xRy \}$ and thinking of the obvious projections $c \to a, c \to b$ we see that the category of relations embeds into the category of spans of sets. In general every category $C$ embeds into $\text{Span}(C)$ (just take $c = a$ and the map $c \to a$ to be the identity).
Recall that a relation between two sets $S, T$ can be described as a subobject of $S \times T$, but this is just some special kind of morphism $R \to S \times T$ and in full generality there's no reason to impose any additional requirements on this morphism. If you wanted to, though, use the following.
Definition: A relational homomorphism between two algebraic structures $R, S$ is a substructure of $R \times S$.
For example a relational homomorphism between two semigroups is a subsemigroup of $R \times S$.
-
I know very little about category theory. Could you explain what a morphism between two binary relations is? – user23211 May 23 '12 at 12:45
@ymar: the binary relations are the morphisms. – Qiaochu Yuan May 23 '12 at 12:45
Ah, so "the category of relations" is the category in which sets are objects and relations are morphisms? – user23211 May 23 '12 at 12:47
@ymar: yep. ${}{}{}$ – Qiaochu Yuan May 23 '12 at 12:48
Thanks a lot, I'd never think of such a definition myself. And it actually works here that "$\rho$ homomorphic implies $\rho^{-1}$ homomorphic". So, by this definition, a congruence on a semigroup $S$ for example is a relational endomorphism of $S$, right? It seems that the structure of the semigroup of such relational endomorphisms would be extremely difficult to understand in the general case, wouldn't it? – user23211 May 23 '12 at 14:01
This is a question by bygone asker/user, who apparently was satisfied with answer it got, but for the sake of not leaving the question half-answered (from my perspective), I'll point out that a notion strictly on semigroups does appear in Howie's Fundamentals of Semigroup Theory in an exercise on p. 42. I'll reproduce it here with Howie's notation, who writes function and relations to the right. If $\rho$ is a relation on $A\times B$ (i.e. subset thereof), you can have a notion of "mapping" along the lines of a (union-of-outputs) multi-valued function by
$$a\rho = \{b\in B:(a,b)\in \rho\}$$
If $S$ and $T$ are semigroups, define a relational morphism $\mu$ from $S$ to $T$ as satisfying both the following
RM1: $[\forall a\in S]\, a\mu\neq \emptyset$
RM2: $[\forall a,b\in S]\, (a\mu)(b\mu) \subseteq (ab)\mu$.
Furthermore, this relational morphism is said injective if it also satisfies
RM3: $[\forall a,b\in S]\, a\mu\cap b\mu \neq \emptyset \Rightarrow a\mu = b\mu$.
It is left as exercise to the reader to show that every relational morphism is a subsemigroup of the direct product $S\times T$ (seen as a semigroup).
A closer look at the definitions above sees that only RM2 actually has something to do with the semigroups operations; RM1 and RM3 only deal with properties of the multivalued-function-like application of $\rho$, which by itself does not get baptized in any way by Howie.
Anyhow, Howie goes on to define divisibility of semigroups in the usual manner, i.e. using run-of-the-mill function-based [homo]morphim as: $S$ divides $T$ if there exists a subsemigroup $U$ of $T$ and a [run-of-the-mill, function-based] [homo]morphism $\psi$ from $U$ onto $S$; so, equivalently, $S$ is a quotient subsemigroup of $T$.
As the final point of the exercise (relating the two notions of morphism), you are asked to show that $S$ divides $T$ if and only if there exists an injective relational morphism from $S$ to $T$.
Alas from this I conclude that the notion of relational morphism doesn't appear particularly fruitful since all we used it for is to say something we could say just as nicely with function-based [homo]morphims.
Howie later notes on p. 44:
The idea of a relational morphism appears in Eilenberg (1976) within the chapters written by Tilson, and is much used in applications of semigroups to language theory—see Pin (1986). It might have been more logical to call it a morphic relation, but the term 'relational morphism' is now standard.
The more detailed refs cited are:
• Eilenberg, S. (1976). Automata, languages and machines, Vol. B. Academic Press, New York.
• Pin, J.-E. (1986). Varieties of formal languages, North Oxford Academic Publishers, London, (A translation of Varietes de lagages formels. Masson, Paris, 1984.)
I haven't looked at those references in-depth to see why/how this notion of relational morphism is fruitful.
However, Pin says that any relational morphims as above factorizes through its "inverse projections", i.e. if $\mu$ is a relational morphism from $S$ to $T$, then the graph of $\mu$, i.e. $\{(a,b) : b\in a\mu\}$ is a subsemigroup of $S \times T$ (as already said). If we denote this subsemigroup by $R$, then there exist [function-based] morphism $\alpha : R \to S$ and $\beta : R \to T$, such that $\alpha$ is surjective and $\mu = \alpha^{-1}\beta$, where $\alpha$ is inverted as a relation. Given that this decomposition is possible for every relational morphism, Pin says:
Thus, if one has trouble to think in terms of relational morphims, one can always go back to usual morphisms.
Tilson introduced some additional terminology and has some insightful observations. He calls a relation that satisfies RM1 a fully defined relation. As we've already noted RM3 can be applied to relations in general which can be called injective. If a relation is injective, then its inverse is a partial function. Furthermore if an injective relation is fully defined then its inverse is a surjective partial function.
Relations satisfying RM1 have also been called left-total and the injective ones have also been called left-unique by Kilp, Knauer and Mikhalev. Or another way of putting it is that RM1+RM3 are the injective multi-valued functions.
A deeper result involving relational morphisms appears in Lambek's paper "The Butterfly and the Serpent", which I really enjoyed reading.
First I'll recast the above definitions in Lambek's terminology. Lambek defines a homomorphic relation to be a binary relation $\rho \subseteq A\times B$ whose graph is a subalgebra of the direct product. If $\rho$ additionally meets RM1, Lambek calls it universally defined, which he expresses using the equivalent definition $1_A \subseteq \rho^\vee\rho$, where $\rho^\vee$ is the inverse relation; Lambek writes relations in infix notation with respect to their arguments and composes them right-to-left: $c (\sigma\rho) a$ iff $[\exists b]\, c\sigma b \wedge b\rho a$. Note that unlike the other authors mentioned, Lambek does not include RM1 in his definition of homomorphic relation, so Lambek's definition is the only one mentioned here that coincides with the one that @Qiaochu Yuan gave in his answer above.
And finally we get to non-trivial facts. Lambek gives the following equivalent characterizations of a Maltsev variety (of which examples are plentiful, as you probably know, including groups, quasigroups, Heyting algebras, complemented lattices):
• (the usual one) there exists a ternary term $f\, xyz$ such that $f\, xyy = x$ and $f\, yyz = z$
• (what Maltsev proved) any two congruences permute; equivalently their join in the lattice of congruences is equal to their composition, so $\rho\sigma=\sigma\rho=\rho\sqcup\sigma$
• (finally we get to) any homomorphic relation is difunctional, meaning that $\rho\rho^\vee\rho = \rho$. This property can be restated as $\rho$ being a regular element using the usual notion from semigroup theory, although Lambek expresses his displeasure with this latter terminology.
• (one more) any reflexive homomorphic relation is a congruence relation.
Homomorphic relations are perhaps not that exciting by themselves, but a derived notion surely gets interesting. Define a subcongruence to be a homomorphic relation from $A$ to $A$ that is symmetric ($\rho^\vee \subseteq \rho$) and transitive ($\rho\rho \subseteq \rho$), but not necessarily reflexive ($1_A \subseteq \rho$).
The notion of subcongruence allows a nice characterization of a generalization of Maltsev varieties called Grousat varieties, which can be defined by any of the following equivalent statements:
• there exist two a ternary terms $f\, xyz$ and $g\, xyz$ such that $f\, xyy = x$, $f\, yyz = g\, yzz$, and $g\, yyz = z$
• any two congruences 3-permute, meaning that $\rho\sigma\rho=\sigma\rho\sigma=\rho\sqcup\sigma$, i.e that equality is also their join in the lattice of congruences
• any homomorphic relation satisfies $\rho\rho^\vee\rho\rho^\vee = \rho\rho^\vee$, which is saying that $\rho\rho^\vee$ is an idempotent.
The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem.
Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it. Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
Anyway, Goursat's Theorem:
If $\rho$ is any homomorphic relation from $A$ to $B$ such that $\rho^\vee\rho$ and $\rho\rho^\vee$ are subcongruences of $A$ and $B$ respectively, in particular if $\rho$ is any homomorphic relation whatsoever between two algebras in a Goursat variety, then $\mathrm{Dom}\,\rho\rho^\vee / \rho\rho^\vee \simeq \mathrm{Dom}\, \rho^\vee\rho / \rho^\vee\rho$.
Since long MathJaX posts get horribly slow to edit (some $O(n^2)$ algorithm at work probably), I'll stop here by just mentioning that Lambek shows that one can derive the Snake Lemma from Goursat's Theorem as well.
-
Very interesting! – goblin Apr 18 '15 at 3:08
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The punchline of Lambek's paper is that one can state a general version of Grousat's theorem in a Grousat variety. Furthermore, in the variety of groups, one recovers Zassenhaus' butterfly lemma as a consequence of this formulation of Grousat's theorem. Before I end this (rather long post!) with Grousat's theorem, we need one more definition, which alas Lambek only gives as a formula and doesn't put any name to it.
|
Given a subcongruence $\sigma$ on $A$, define $\mathrm{Dom}\,\sigma = \{a\in A : a \sigma a\}$. Since $\sigma$ is not necessarily reflexive, it makes good sense to consider this set. I suppose this notation can be read as "domain" of $\sigma$, but that seems a little misleading given $A$; perhaps reading it as the [sub]diagonal of $\sigma$ makes more sense.
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https://math.stackexchange.com/questions/633757/order-of-conjugate-of-an-element-given-the-order-of-its-conjugate?noredirect=1
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# Order of conjugate of an element given the order of its conjugate
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...
How to find this answer? Sorry for my bad question, but I need this for my study.
• Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32
$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.
Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.
For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.
Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.
Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!)
If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.
• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08
Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.
• I still don't get it. – Yagami Jan 10 '14 at 14:58
In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.
CONCLUSION: $o(a)=o(bab^{-1})$.
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If $a$ has order $6$, then the order of $bab^{-1}$ is...
How to find this answer? Sorry for my bad question, but I need this for my study. • Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32
$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer.
|
But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$. Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little. For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.
|
https://math.stackexchange.com/questions/633757/order-of-conjugate-of-an-element-given-the-order-of-its-conjugate?noredirect=1
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# Order of conjugate of an element given the order of its conjugate
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...
How to find this answer? Sorry for my bad question, but I need this for my study.
• Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32
$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.
Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.
For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.
Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.
Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!)
If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.
• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08
Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.
• I still don't get it. – Yagami Jan 10 '14 at 14:58
In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.
CONCLUSION: $o(a)=o(bab^{-1})$.
| 911
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Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order. Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms.
|
Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!) If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.
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https://math.stackexchange.com/questions/633757/order-of-conjugate-of-an-element-given-the-order-of-its-conjugate?noredirect=1
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# Order of conjugate of an element given the order of its conjugate
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...
How to find this answer? Sorry for my bad question, but I need this for my study.
• Hint: conjugation is an automorphism – dani_s Jan 10 '14 at 14:27
• @dani_s: that would be too much... it is a basic question and you are proposing to look at the automorphism.. Not a good idea i guess.. – user87543 Jan 10 '14 at 15:32
$$|bab^{-1}|=k\to (bab^{-1})^k=e_G$$ and $k$ is the least positive integer. But $e_G=(bab^{-1})^k=ba^kb^{-1}$ so $a^k=e_G$ so $6\le k$. Obviously, $k\le 6$ (Why?) so $k=6$.
Two good pieces of advice are already out here that prove the problem directly, but I'd like to decompose and remix them a little.
For a group $G$ and any $g\in G$, the map $x\mapsto gxg^{-1}$ is actually a group automorphism (self-isomorphism). This is a good exercise to prove if you haven't already proven it.
Intuitively, given an isomorphism $\phi$, $\phi(G)$ looks just like $G$, and $\phi(g)$ has the same group theoretic properties as $g$. (This includes order.) This motivates you to show that $g^n=1$ iff $\phi(g)^n=1$, and so (for minimal choice of $n$) they share the same order.
Here's a slightly more general statement for $\phi$'s that aren't necessarily isomorphisms. Let $\phi:G\to H$ be a group homomorphism of finite groups. Then for each $g\in G$, the order of $\phi(g)$ divides the order of $g$. (Try to prove this!)
If $\phi$ is an isomorphism, then so is $\phi^{-1}$, and so the order of $\phi(g)$ divides the order of $g$, and the order of $\phi^{-1}(\phi(g))=g$ divides the order of $\phi(g)$, and thus they're equal.
• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08
Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$. Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$.
• I still don't get it. – Yagami Jan 10 '14 at 14:58
In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough). Easy to see that if $k=n$ then $(bab^{-1})^n=ba^nb^{-1}=beb^{-1}=e$, hence $k=n$.
CONCLUSION: $o(a)=o(bab^{-1})$.
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• @Andreas It seems this question (and variants) are destined to be prototypical examples of an abstract duplicate (e.g. recall the recent question). In fact, even the comments are becoming duplicate! – Bill Dubuque Jan 12 '14 at 17:56
• @BillDubuque, an optimistic view of the fact that the comments are becoming duplicates is that we are reaching a consensus on a canonical form for answers and comments ;-) – Andreas Caranti Jan 12 '14 at 18:08
Note that $(bab^{-1})^2 = bab^{-1}bab^{-1} = ba^2b^{-1}$.
|
Similarly $(bab^{-1})^n = ba^nb^{-1}$ for any $n$. When will $ba^nb^{-1} = 1$ using the information about $a$? Then you just have to check to see that $ba^mb^{-1} \not = 1$ for any $1 \leq m < n$. • I still don't get it. – Yagami Jan 10 '14 at 14:58
In general, let $o(a)=n$ and $o(bab^{-1})=k$, then $(bab^{-1})^k=ba^kb^{-1}=e$, by Cancellation Law in group, we can get $a^k=e$, since $o(a)=n$, then $k \geq n$ (in fact we can get $n|k$, but in this proof $k \geq n$ is enough).
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https://gamedev.stackexchange.com/questions/138165/how-can-i-move-and-rotate-an-object-in-an-infinity-or-figure-8-trajectory/138167
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# How can I move and rotate an object in an “infinity” or “figure 8” trajectory?
I know that the easiest way to move an object with the figure 8 trajectory is:
x = cos(t);
y = sin(2*t) / 2;
but I just don't know how to rotate it, lets says with a new variable r as rotation, how do I merge it into the above formula ? can anyone please advise me on what is the simplest and cheapest way/formula to move and rotate the figure 8 trajectory ?
## 1 Answer
The object should point in the direction of the derivative, which is [-sin(t), cos(2t)]. Its angle is atan2(cos(2t), -sin(t)).
Edit: OP is apparently asking how to rotate the "trajectory," not the object itself.
To rotate the figure, choose an angle, θ, in radians, that you'd like the trajectory to be rotated. The position along this rotated figure is:
x = cos(θ) * cos(t) - sin(θ) * sin(2t)/2
y = sin(θ) * cos(t) + cos(θ) * sin(2t)/2
• so how would I modify the formula to get a rotated figure of 8 ? – user1998844 Mar 3 '17 at 18:24
• That is a completely different question than the one I answered. I'll edit my answer with a solution to this question. – Drew Cummins Mar 3 '17 at 18:31
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can anyone please advise me on what is the simplest and cheapest way/formula to move and rotate the figure 8 trajectory ? ## 1 Answer
The object should point in the direction of the derivative, which is [-sin(t), cos(2t)]. Its angle is atan2(cos(2t), -sin(t)). Edit: OP is apparently asking how to rotate the "trajectory," not the object itself. To rotate the figure, choose an angle, θ, in radians, that you'd like the trajectory to be rotated.
|
The position along this rotated figure is:
x = cos(θ) * cos(t) - sin(θ) * sin(2t)/2
y = sin(θ) * cos(t) + cos(θ) * sin(2t)/2
• so how would I modify the formula to get a rotated figure of 8 ? – user1998844 Mar 3 '17 at 18:24
• That is a completely different question than the one I answered. I'll edit my answer with a solution to this question. – Drew Cummins Mar 3 '17 at 18:31
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https://engineering.stackexchange.com/questions/54395/how-much-force-is-needed-to-break-off-the-stick
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# How much force is needed to break off the stick
Let's consider the following figure
The grey box contains a blue stick which is fixed. The blue stick has a length of $$a+b+c$$ and two diameters $$f,h$$. The diameter $$h$$ describes the part $$b$$ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $$F$$ is pushing against the withe plane like in the picture. How much force is needed to break off the stick in part $$b$$?
• How much effort have you applied to try to obtain a proposed solution? Commented Feb 26, 2023 at 20:03
• I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane. Commented Feb 26, 2023 at 20:26
• I just have some knowledge about bending sticks and not about stuff like in my picture. Commented Feb 26, 2023 at 20:43
• Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. Commented Feb 26, 2023 at 20:57
• Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? Commented Feb 26, 2023 at 21:18
## 1 Answer
We assume the distance from F to the hinge to be
$$X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis. When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B.
$$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$
we assume the stick will break at yield stress and ignore 2nd hardening, or if we have it we plug it.
$$\sigma_y=\frac{MC}{I_{Beam}}=\frac{(F*x)(e+f/2)}{bh^3/12+bh(e+f/2)^2}$$ $$F*X=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{e+f/2}$$
$$F=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{(e*f/2)*(a+b+c+d/2)}$$
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# How much force is needed to break off the stick
Let's consider the following figure
The grey box contains a blue stick which is fixed. The blue stick has a length of $$a+b+c$$ and two diameters $$f,h$$. The diameter $$h$$ describes the part $$b$$ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $$F$$ is pushing against the withe plane like in the picture.
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How much force is needed to break off the stick in part $$b$$? • How much effort have you applied to try to obtain a proposed solution? Commented Feb 26, 2023 at 20:03
• I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane.
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https://engineering.stackexchange.com/questions/54395/how-much-force-is-needed-to-break-off-the-stick
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# How much force is needed to break off the stick
Let's consider the following figure
The grey box contains a blue stick which is fixed. The blue stick has a length of $$a+b+c$$ and two diameters $$f,h$$. The diameter $$h$$ describes the part $$b$$ of the stick. The stick is fixed in the plane but the plane is not connected to the grey box. A force $$F$$ is pushing against the withe plane like in the picture. How much force is needed to break off the stick in part $$b$$?
• How much effort have you applied to try to obtain a proposed solution? Commented Feb 26, 2023 at 20:03
• I don't have an idea how I could solve this because I had never a mechanical problem with a notch. What I also can say is that I see two different ways how this plane could move: One way would be striaght downward if the force is close to the notch or the plane is rotated if the force comes from the outer part of the plane. Commented Feb 26, 2023 at 20:26
• I just have some knowledge about bending sticks and not about stuff like in my picture. Commented Feb 26, 2023 at 20:43
• Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. Commented Feb 26, 2023 at 20:57
• Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? Commented Feb 26, 2023 at 21:18
## 1 Answer
We assume the distance from F to the hinge to be
$$X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis. When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B.
$$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$
we assume the stick will break at yield stress and ignore 2nd hardening, or if we have it we plug it.
$$\sigma_y=\frac{MC}{I_{Beam}}=\frac{(F*x)(e+f/2)}{bh^3/12+bh(e+f/2)^2}$$ $$F*X=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{e+f/2}$$
$$F=\frac{\sigma y*(bh^3/12+bh(e+f/2)^2)}{(e*f/2)*(a+b+c+d/2)}$$
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Commented Feb 26, 2023 at 20:43
• Apply your knowledge about bending sticks to try to solve the problem. We should like to see how far that takes you. Commented Feb 26, 2023 at 20:57
• Does the white plane slide against the grey plane or does it tilt ie pivot at the lower left corner? Commented Feb 26, 2023 at 21:18
## 1 Answer
We assume the distance from F to the hinge to be
$$X_F=a+b+c+d/2$$ We calculate the equivalent I of the cantilever beam, with the parallel axis.
|
When it bends it will rotate about a point at the lower corner of the gray support, call it point A. Let's annotate the thickness of the bar, B. $$I_{Beam} =I_{stick}+ A_{stick}*Y^2_{stick}$$ $$I_{stick}= bh^3/12$$ $$I_{Beam}=bh^3/12+bh(e+f/2)^2$$
we assume the stick will break at yield stress and ignore 2nd hardening, or if we have it we plug it.
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https://stats.stackexchange.com/questions/592820/how-can-i-find-the-expectation-value-to-this-problem
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# How can I find the expectation value to this problem?
At a wedding reception on an evening the representative of the host is taking it as an occasion to exercise and explain a classical analytic problem. specifically, he insists that he would start serving the food only when the first table, which is arranged for 12 guests to dine together, has guests born in every twelve months of the year. assume that any given guest is equally likely to be born in any of the twelve months of the year, and that new guests were arriving at every two minutes then. what is the expected waiting time of the first arriving guest before the food gets served eventually?
Since this looked like a Coupon Collector's problem variation, my initial approach was to determine the sum of the expected value of each guests of unique birth months.
X ~ FS(p) [First Success Distribution]
X = time needed until food gets served
$$E[X] = E[X1] + E[X2] + ... + E[X12]$$
$$=> E[X] = 12/12 + 12/11 + ... + 12/1$$
However, this is where i ran into problem, since I don't know how to handle the arrival at every two minutes in my equation. Should I just multiply by 2? Or am i missing something very obvious or basic trivia? Help will be appreciated.
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# How can I find the expectation value to this problem? At a wedding reception on an evening the representative of the host is taking it as an occasion to exercise and explain a classical analytic problem. specifically, he insists that he would start serving the food only when the first table, which is arranged for 12 guests to dine together, has guests born in every twelve months of the year.
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assume that any given guest is equally likely to be born in any of the twelve months of the year, and that new guests were arriving at every two minutes then. what is the expected waiting time of the first arriving guest before the food gets served eventually? Since this looked like a Coupon Collector's problem variation, my initial approach was to determine the sum of the expected value of each guests of unique birth months.
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https://math.stackexchange.com/questions/3003033/show-lim-x-to-x-0-fxx-x-0-0-when-f-mathbbr-subset-mathbbr
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# Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing.
Show $$\lim_{x \to x_0^+} f(x)(x-x_0) =0$$ when $$f(\mathbb{R}) \subset \mathbb{R}^+$$ & monotone increasing.
Try
I need to show,
$$\forall \epsilon >0, \exists \delta >0 : x \in (x_0, x_0 + \delta) \Rightarrow |f(x) (x-x_0)| < \epsilon$$
I think I could find some upper bound $$M >0$$ such that $$|f(x) (x-x_0)| \le M |x - x_0|$$.
Let $$M = f(x_0 + \epsilon)$$, and let $$\delta = \frac{\epsilon}{\max \{2M, 2 \}}$$, then clearly $$f(x) \le f(x_0 + \epsilon) = M$$
But I'm not sure $$|f(x) (x-x_0)| \le M |x - x_0|$$.
Any hint about how I should proceed?
Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $$x\leq x_0$$.
Use $$M=f(x_0+1)$$ and cosider $$\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$$.
Fix $$\varepsilon>0$$. Let $$M=f(x_0+1)$$ and choose $$\delta=\mathrm{min}\{1,\frac{\varepsilon}{M}\}$$. For each $$x\in(x_0,x_0+\delta)$$, $$|f(x)|\leq M$$ since $$f$$ is strictly increasing. Thus, $$|f(x)(x-x_0)|\leq M|x-x_0|.
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# Show $\lim_{x \to x_0^+} f(x)(x-x_0) =0$ when $f(\mathbb{R}) \subset \mathbb{R}^+$ & monotone increasing. Show $$\lim_{x \to x_0^+} f(x)(x-x_0) =0$$ when $$f(\mathbb{R}) \subset \mathbb{R}^+$$ & monotone increasing. Try
I need to show,
$$\forall \epsilon >0, \exists \delta >0 : x \in (x_0, x_0 + \delta) \Rightarrow |f(x) (x-x_0)| < \epsilon$$
I think I could find some upper bound $$M >0$$ such that $$|f(x) (x-x_0)| \le M |x - x_0|$$.
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Let $$M = f(x_0 + \epsilon)$$, and let $$\delta = \frac{\epsilon}{\max \{2M, 2 \}}$$, then clearly $$f(x) \le f(x_0 + \epsilon) = M$$
But I'm not sure $$|f(x) (x-x_0)| \le M |x - x_0|$$. Any hint about how I should proceed? Hint: Observe \begin{align} |f(x)(x-x_0)|\leq |f(x_0)||x-x_0| \end{align} for all $$x\leq x_0$$. Use $$M=f(x_0+1)$$ and cosider $$\delta=\min\{\frac{1}{2},\frac{\epsilon}{2M}\}$$.
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http://math.stackexchange.com/questions/607103/prove-sin2-theta-cos2-theta-1/607121
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine. -
By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem?
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– user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
-
By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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– 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram).
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From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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– Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity.
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– Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles.
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It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it?
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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– Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering?
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– Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative. Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer. $(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!}
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- \ldots$ and $\cos(x)$ is its derivative. Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer. As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem.
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We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse. Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics.
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http://math.stackexchange.com/questions/607103/prove-sin2-theta-cos2-theta-1/607121
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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Archaelogists have discovered it inscribred in stones in excavations in Babylon! -
This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!)
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the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$. • Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
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This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially. -
$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say.
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– Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere.
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# Prove $\sin^2\theta + \cos^2\theta = 1$
How do you prove the following trigonometric identity: $$\sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
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By definition? This really depends on how the functions are defined to begin with. – egreg Dec 14 '13 at 22:37
Do you know pythagorean theorem? – user63181 Dec 14 '13 at 22:40
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? – Nick Dec 14 '13 at 23:10
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. – 1015 Dec 14 '13 at 23:19
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. – Nick Dec 14 '13 at 23:52
Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$
$\phantom{X}$
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The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. – Nick Apr 12 '14 at 16:59
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$
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Sweet.${}{{}{}}$ – Git Gud Dec 14 '13 at 22:48
Bravo! Wow, this is priceless. – Nick Dec 14 '13 at 22:53
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. – Leox Dec 15 '13 at 0:08
@Leox Series definition. There are other alternatives too. – Git Gud Dec 15 '13 at 0:23
Excellent! ${}{}{}{}+ 1$ – amWhy Dec 15 '13 at 12:56
Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:
$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$
Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}
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See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. – Nick Dec 14 '13 at 22:47
Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem, $${AC}^2+{BC}^2 = {AB}^2$$ Dividing by $AB^2$, \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align}
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This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. – Silverfish Jan 1 '15 at 18:36
@Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. – Nick Jan 2 '15 at 7:38
I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. – Silverfish Jan 2 '15 at 9:09
@Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ – Nick Jan 2 '15 at 9:38
In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get
$$2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$
So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.
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This is exactly the solution given by Sami Ben Romdhane above! – Idris Dec 19 '14 at 12:54
$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$
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You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. – egreg Dec 14 '13 at 22:53
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? – Nick Dec 14 '13 at 23:07
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. – egreg Dec 14 '13 at 23:24
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? – Nick Dec 14 '13 at 23:36
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… – Dylan Yott Dec 14 '13 at 23:50
If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:
$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.
Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.
$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.
Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.
As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)
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hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. – Nick Dec 14 '13 at 23:23
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. – Dylan Yott Dec 14 '13 at 23:31
Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.
Now, from the Pyth. theorem
$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$
By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!
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This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. – R R Dec 14 '13 at 23:57
We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:
• Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
• Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
• Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.
This makes \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}
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Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:
Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$
To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.
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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:
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Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. – Joao Oct 22 '14 at 5:51
On one hand, $$\int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x),$$ On the other hand, $$\int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2},$$ Hence, by subtraction, we will have that, $$0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2}$$ or, equivalently, $$\sin^2(x)+\cos^2(x)=1.$$ I have not seen this proof elsewhere. It is fun.
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Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. – Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.
Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives
$$A(\theta)^n = A(n \theta) \approx I$$
where $I$ is the identity matrix. By taking determinants, we get
$$(\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$
for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that
$$\cos(\theta)^2 + \sin(\theta)^2 = 1$$
To get the details right, we select sequences of integers $a_n, b_n$ with
$$\lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0$$ $$\lim_{n \to \infty} a_n = +\infty$$
e.g. this can be done by continued fractions. Then,
$$\lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$
and consequently
$$\lim_{n \to \infty} (\det A(\theta))^{a_n} = 1$$
Another method in the same vein is to also define vectors
$$v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right)$$
so that $A(\theta) v(\varphi) = v(\theta + \varphi)$
However, we know that for every angle $\theta$:
$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1$$
In particular, we have
$$\frac{1}{2} \leq \| v(\theta) \| \leq 2$$
and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.
By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.
However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.
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– Bookend Oct 9 '15 at 11:58
Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$
We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$
-
Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$.
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It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection. Define the matrix
$$A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right)$$
By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.
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https://math.stackexchange.com/questions/1762036/why-cant-you-count-up-to-aleph-null
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# Why can't you count up to aleph null?
Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:
1. $\aleph_0$ is the cardinality of the natural numbers
2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.
But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$.
Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$. But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above.
This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.
• Can you count to $\aleph_0$?. I am not even going to start to see if I can.,
– user328032
Apr 28, 2016 at 1:19
• It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. Apr 28, 2016 at 1:20
• @CameronWilliams Yes, but then what would be the last element of the set? Apr 28, 2016 at 1:21
• I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. Apr 28, 2016 at 1:22
• @Timtech That's the thing. There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. Apr 28, 2016 at 1:22
This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number. It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers.
If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.
• Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets. Since $\aleph_0$ is a cardinal number, it is also a set. Apr 28, 2016 at 22:01
$$\{1,2,\ldots,\text{ an infinite list of numbers },\ldots , \aleph_0\}$$
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# Why can't you count up to aleph null? Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:
1. $\aleph_0$ is the cardinality of the natural numbers
2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.
|
But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$. Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$.
|
https://math.stackexchange.com/questions/1762036/why-cant-you-count-up-to-aleph-null
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# Why can't you count up to aleph null?
Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:
1. $\aleph_0$ is the cardinality of the natural numbers
2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.
But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$.
Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$. But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above.
This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.
• Can you count to $\aleph_0$?. I am not even going to start to see if I can.,
– user328032
Apr 28, 2016 at 1:19
• It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. Apr 28, 2016 at 1:20
• @CameronWilliams Yes, but then what would be the last element of the set? Apr 28, 2016 at 1:21
• I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. Apr 28, 2016 at 1:22
• @Timtech That's the thing. There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. Apr 28, 2016 at 1:22
This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number. It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers.
If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.
• Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets. Since $\aleph_0$ is a cardinal number, it is also a set. Apr 28, 2016 at 22:01
$$\{1,2,\ldots,\text{ an infinite list of numbers },\ldots , \aleph_0\}$$
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But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above. This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.
|
• Can you count to $\aleph_0$?. I am not even going to start to see if I can.,
– user328032
Apr 28, 2016 at 1:19
• It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. Apr 28, 2016 at 1:20
• @CameronWilliams Yes, but then what would be the last element of the set?
|
https://math.stackexchange.com/questions/1762036/why-cant-you-count-up-to-aleph-null
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| 417,244,395
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|
# Why can't you count up to aleph null?
Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:
1. $\aleph_0$ is the cardinality of the natural numbers
2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.
But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$.
Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$. But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above.
This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.
• Can you count to $\aleph_0$?. I am not even going to start to see if I can.,
– user328032
Apr 28, 2016 at 1:19
• It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. Apr 28, 2016 at 1:20
• @CameronWilliams Yes, but then what would be the last element of the set? Apr 28, 2016 at 1:21
• I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. Apr 28, 2016 at 1:22
• @Timtech That's the thing. There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. Apr 28, 2016 at 1:22
This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number. It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers.
If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.
• Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets. Since $\aleph_0$ is a cardinal number, it is also a set. Apr 28, 2016 at 22:01
$$\{1,2,\ldots,\text{ an infinite list of numbers },\ldots , \aleph_0\}$$
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Apr 28, 2016 at 1:21
• I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. Apr 28, 2016 at 1:22
• @Timtech That's the thing.
|
There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. Apr 28, 2016 at 1:22
This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number.
|
https://math.stackexchange.com/questions/1762036/why-cant-you-count-up-to-aleph-null
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crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00443.warc.gz
| 417,244,395
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|
# Why can't you count up to aleph null?
Recently I learned about the infinite cardinal $\aleph_0$, and stumbled upon a seeming contradiction. Here are my assumptions based on what I learned:
1. $\aleph_0$ is the cardinality of the natural numbers
2. $\aleph_0$ is larger than all finite numbers, and thus cannot be reached simply by counting up from 1.
But then I started wondering: the cardinality of the set $\{1\}$ is $1$, the cardinality of the set $\{1, 2\}$ is $2$, the cardinality of the set $\{1, 2, 3\}$ is 3, and so on. So I drew the conclusion that the cardinality of the set $\{1, 2, \ldots n\}$ is $n$.
Based on this conclusion, if the cardinality of the natural numbers is $\aleph_0$, then the set of natural numbers could be denoted as $\{1, 2, \ldots \aleph_0\}$. But such a set implies that $\aleph_0$ can be reached by counting up from $1$, which contradicts my assumption #2 above.
This question has been bugging me for a while now... I'm not sure where I've made a mistake in my reasoning or if I have even used the correct mathematical terms/question title/tags to describe it, but I'd sure appreciate your help.
• Can you count to $\aleph_0$?. I am not even going to start to see if I can.,
– user328032
Apr 28, 2016 at 1:19
• It seems to me that you want this to be an ordered set, but it does not really make sense to tack on $\aleph_0$ to the end in the way that you want. Apr 28, 2016 at 1:20
• @CameronWilliams Yes, but then what would be the last element of the set? Apr 28, 2016 at 1:21
• I can count up to $\aleph_0$. Just give me $\aleph_0$ seconds added to my life and I hope I will be able to be patient enough to do this... Countable doesn't mean you can count to it, it just means it contains the whole numbers excluding all the rational decimals between them. Apr 28, 2016 at 1:22
• @Timtech That's the thing. There isn't a "last" element here. There is a maximal element, but not a last. Last implies that you can reach that element in finitely many steps. "Last" is somewhat of a colloquialism. Apr 28, 2016 at 1:22
This is a good example where intuition about a pattern breaks down; what is true of finite sets is not true of infinite sets in general. The natural numbers $\textit{cannot}$ be denoted by the set $A=\{1,2,...,\aleph_0\}$ as the set $\aleph_0$ is not a natural number. It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers.
If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.
• Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets. Since $\aleph_0$ is a cardinal number, it is also a set. Apr 28, 2016 at 22:01
$$\{1,2,\ldots,\text{ an infinite list of numbers },\ldots , \aleph_0\}$$
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It is true that the cardinality of $A$ is $\aleph_0$ (a good exercise), but it contains more than just natural numbers. If $\aleph_0$ were a natural number then, as you point out, we would have a contradiction. However $\aleph_0$ is the $\textit{cardinality}$ of the natural numbers, and not a natural number itself. By definition, $\aleph_0$ is the least ordinal number with which the set $\omega$ of natural numbers may be put into bijection.
|
• Both... In $ZFC$ $\textit{everything}$ is a set, but more explicitly, the definition of cardinal numbers I know is this: Let $A$ be a set. Then the cardinal number of $A$ is the least ordinal $\kappa$ such that there exists a bijection $f: \kappa \to A$. Now by definition, ordinals are transitive sets that are well-ordered by $"\in"$, and since cardinals are in particular ordinals, they are sets.
|
https://blender.stackexchange.com/questions/90357/plotting-3d-surfaces
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# Plotting 3D Surfaces
Building on from my previous question, how can I plot something more complex than a sphere (and not as easy to split into a set of parametric equations)?
Say, for example
$$2x^2+y^2-5z^2+z-7x=16$$
Graphed on Desmos (use the z-slider to change the z value) for those who cannot as easily visualize the 3D surface.
• You could potentially use Python script to build a mesh by layers using successive values of Z. Sep 16, 2017 at 9:41
• Welcome to the site :) It looks like you have three separate accounts. If you would care to register one of them fully, we can merge them together and you'll be able to edit your question, accept answers, and do all those other fun things. Thanks Sep 16, 2017 at 22:24
## Discrete approach with Python
A possible approch that avoids solving the equation would consist in interating through all possible points in the 3D space, check whether or not in each one of them the equation is verified, and if it is create a vertex there.
As in the 3D space there are infinite points, we must discretize it and limits the boundaries of the investigation domain. For the sake of semplicity, I choose to consider, for example, only integer numbers from -100 to +100 for each axis. Here's the code for such iteration:
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
Then, for each vector we'll calculate the left member of the equation and check if it's equal to the right member of the equation previusly defined. As we discretized the domain, it's best to give to the result a bit of "tollerance" in verifying the equation (that's why I'm using the "+/-toll" trick upon the right member verification).
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
Here's a possible code with all the lines needed to create an object in the scene:
import bpy
import bmesh
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
goal=16 #right member of the equation
toll=5 #threshold
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Examples of usage:
function=2*x**2+y**2-5*z**2+z-7*x
Note: as you can see, it takes a while to iterate through all the 8 000 000 positions...
This script only provides some of the vertices due to the approximation, but, with a proper resolution, you can obtain a good representation of the surface.
The meshing of the vertices into a surface needs it's own approach. This "brute force plotting" I'm proposing does not go well with the "on the fly" definitions of faces. The set of point is not structured in any way. You'll need something to mesh the pointcloud.
Here's another version of the code where you can better control the domain size and resolution using floats to inspect the region closer to the origin in high definition:
import bpy
import bmesh
max=7
min=-7
prec=10 #precision of the calc (1=1 units, 10 = 0.1 units, 100 = 0.01 units...)
tol=0.5 #tollerance
def f(x, y, z): return 2*x**2+y**2-5*z**2+z-7*x -16 #the function = 0
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("XYZ Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
for x in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for y in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for z in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
if(abs(f(x,y,z))<tol):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Here's a possible result:
You can for example export points to Meshlab and take advantage of the built in tools. Here's an example using the previusly obtained set of points.
As far as I know, there isn't an add-on or method that can graph more complex mathematical functions in the way you want. As an alternative, you can put your equation into nodes and then render it with a Volume Scatter shader.
Begin with the default cube. Add a new material, and add a Texture Coordinate node (Shift + A > Input > Texture Coordinate) and Separate XYZ node (Shift + A > Converter > Separate XYZ):
Then, create your equation using Math nodes (Shift + A > Converter > Math). Here, I created your example equation:
Now, add a Color Mix node (Shift + A > Color > MixRGB) and set it to Multiply. Plug the texture coordinates in to one slot, and a Value node in to the other. This will scale our window:
Finally, multiply the output by 10 to increase the value, then plug it into the Density of a Volume Scatter and a Volume Absorption shader added together:
This will nicely display the function when rendered, though the shading isn't ideal:
click for larger image
To optimize rendering, change the following settings:
Although it doesn't give you a workable mesh, this can definitely give you a renderable graph.
We can also improve the look of it by coloring it:
click for larger image
• Need 50 rep to comment... @ScottMilner I was hoping there was a way to preview in viewport directly (not viewport shading: rendered, but viewport shading: solid). The less CPU power, the better I am actually looking only to visualize the shape (so quality is not important. Quantity is)
– math
Sep 16, 2017 at 5:25
The equation in the form in the question is a quadric surface. By completing the square we can translate the axis and reduce to form
(x-7/4)² - 49/16 + 1/2y² - 5/2(z - 1/10)² + 5/200 - 8 = 0
or
X² + 0.5Y² - 2.5Z² - - 11.0375 = 0
A sample of the surface created by iterating through z (an xy plane) then finding the 0, 1 or 2 roots for each y using the quadratic formula to solve for x. Notice the missing faces where x=0. Can run thru solving for y and have missing faces around y=0. Overlaying the two gives the shape, but not a pretty mesh.
I've put in both the original equation f(x, y, z) and the translated form f2(x, y, z).
import bpy
import bmesh
from math import sqrt
def f(x, y, z):
return 2 * x * x + y * y - 5 * z * z + z -7 * x -16
# f transformed to have origin (0, 0, 0)
def f2(x, y, z):
x * x + 0.5 * y * y - 2.5 * z * z - 11.0375
axis = -b / (2 * a)
check = b * b - 4 * a * c
if abs(check) < 0.0001:
return True, axis, axis
if check < 0:
return False, None, None
dist = sqrt(check) / (2 * a)
return True, axis - dist, axis + dist
bm = bmesh.new()
dom = [m for m in range(-10, 11)]
use_f2 = False # use f2()
planes = []
# coefficents of x
if use_f2:
a, b = 1, 0 # f2()
else:
a, b = 2, -7 # f()
for z in dom:
# for each xy plane
neg_line = []
pos_line = []
for y in dom:
# for each line in plane
if use_f2:
# constant f2()
c = 0.5 * y * y - 2.5 * z * z -11.0375
else:
# constant f()
c = y * y - 5 * z * z + z -16
# find roots
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
neg_root = bm.verts.new((neg_root, y, z))
pos_root = bm.verts.new((pos_root, y, z))
pos_line.append(pos_root)
neg_line.append(neg_root)
planes.append([neg_line, pos_line])
# skin it
if len(planes) > 1:
for i in range(len(planes) - 1):
n0, p0 = planes[i]
n1, p1 = planes[i + 1]
# pos
for j in range(len(n0) - 1):
verts = [n0[j], n1[j], n1[j+1], n0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
verts = [p0[j], p1[j], p1[j+1], p0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
# make an object
scene = bpy.context.scene
bm.to_mesh(mesh)
scene.objects.active = ob
ob.scale *= 0.01
To solve for y for f(x, y, z) edit
a, b = 1, 0
c = 2 * x * x - 5 * z * z + z -7 * x - 16
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
#print(neg_root, pos_root)
neg_root = bm.verts.new((x, neg_root, z))
pos_root = bm.verts.new((x, pos_root, z))
Result after running in both x and y and joining
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Say, for example
$$2x^2+y^2-5z^2+z-7x=16$$
Graphed on Desmos (use the z-slider to change the z value) for those who cannot as easily visualize the 3D surface. • You could potentially use Python script to build a mesh by layers using successive values of Z. Sep 16, 2017 at 9:41
• Welcome to the site :) It looks like you have three separate accounts.
|
If you would care to register one of them fully, we can merge them together and you'll be able to edit your question, accept answers, and do all those other fun things. Thanks Sep 16, 2017 at 22:24
## Discrete approach with Python
A possible approch that avoids solving the equation would consist in interating through all possible points in the 3D space, check whether or not in each one of them the equation is verified, and if it is create a vertex there.
|
https://blender.stackexchange.com/questions/90357/plotting-3d-surfaces
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crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00116.warc.gz
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|
# Plotting 3D Surfaces
Building on from my previous question, how can I plot something more complex than a sphere (and not as easy to split into a set of parametric equations)?
Say, for example
$$2x^2+y^2-5z^2+z-7x=16$$
Graphed on Desmos (use the z-slider to change the z value) for those who cannot as easily visualize the 3D surface.
• You could potentially use Python script to build a mesh by layers using successive values of Z. Sep 16, 2017 at 9:41
• Welcome to the site :) It looks like you have three separate accounts. If you would care to register one of them fully, we can merge them together and you'll be able to edit your question, accept answers, and do all those other fun things. Thanks Sep 16, 2017 at 22:24
## Discrete approach with Python
A possible approch that avoids solving the equation would consist in interating through all possible points in the 3D space, check whether or not in each one of them the equation is verified, and if it is create a vertex there.
As in the 3D space there are infinite points, we must discretize it and limits the boundaries of the investigation domain. For the sake of semplicity, I choose to consider, for example, only integer numbers from -100 to +100 for each axis. Here's the code for such iteration:
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
Then, for each vector we'll calculate the left member of the equation and check if it's equal to the right member of the equation previusly defined. As we discretized the domain, it's best to give to the result a bit of "tollerance" in verifying the equation (that's why I'm using the "+/-toll" trick upon the right member verification).
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
Here's a possible code with all the lines needed to create an object in the scene:
import bpy
import bmesh
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
goal=16 #right member of the equation
toll=5 #threshold
for x in range(-100,100):
for y in range(-100,100):
for z in range(-100,100):
function=2*x**2+y**2-5*z**2+z-7*x #function here
if((function<(goal+toll))and(function>(goal-toll))):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Examples of usage:
function=2*x**2+y**2-5*z**2+z-7*x
Note: as you can see, it takes a while to iterate through all the 8 000 000 positions...
This script only provides some of the vertices due to the approximation, but, with a proper resolution, you can obtain a good representation of the surface.
The meshing of the vertices into a surface needs it's own approach. This "brute force plotting" I'm proposing does not go well with the "on the fly" definitions of faces. The set of point is not structured in any way. You'll need something to mesh the pointcloud.
Here's another version of the code where you can better control the domain size and resolution using floats to inspect the region closer to the origin in high definition:
import bpy
import bmesh
max=7
min=-7
prec=10 #precision of the calc (1=1 units, 10 = 0.1 units, 100 = 0.01 units...)
tol=0.5 #tollerance
def f(x, y, z): return 2*x**2+y**2-5*z**2+z-7*x -16 #the function = 0
mesh = bpy.data.meshes.new("mesh")
obj = bpy.data.objects.new("XYZ Function", mesh)
scene = bpy.context.scene
scene.objects.active = obj
obj.select = True
mesh = bpy.context.object.data
bm = bmesh.new()
for x in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for y in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
for z in [float(j)/prec for j in range(min*prec, max*prec+1,1)]:
if(abs(f(x,y,z))<tol):
bm.verts.new([x,y,z])
bm.to_mesh(mesh)
bm.free()
Here's a possible result:
You can for example export points to Meshlab and take advantage of the built in tools. Here's an example using the previusly obtained set of points.
As far as I know, there isn't an add-on or method that can graph more complex mathematical functions in the way you want. As an alternative, you can put your equation into nodes and then render it with a Volume Scatter shader.
Begin with the default cube. Add a new material, and add a Texture Coordinate node (Shift + A > Input > Texture Coordinate) and Separate XYZ node (Shift + A > Converter > Separate XYZ):
Then, create your equation using Math nodes (Shift + A > Converter > Math). Here, I created your example equation:
Now, add a Color Mix node (Shift + A > Color > MixRGB) and set it to Multiply. Plug the texture coordinates in to one slot, and a Value node in to the other. This will scale our window:
Finally, multiply the output by 10 to increase the value, then plug it into the Density of a Volume Scatter and a Volume Absorption shader added together:
This will nicely display the function when rendered, though the shading isn't ideal:
click for larger image
To optimize rendering, change the following settings:
Although it doesn't give you a workable mesh, this can definitely give you a renderable graph.
We can also improve the look of it by coloring it:
click for larger image
• Need 50 rep to comment... @ScottMilner I was hoping there was a way to preview in viewport directly (not viewport shading: rendered, but viewport shading: solid). The less CPU power, the better I am actually looking only to visualize the shape (so quality is not important. Quantity is)
– math
Sep 16, 2017 at 5:25
The equation in the form in the question is a quadric surface. By completing the square we can translate the axis and reduce to form
(x-7/4)² - 49/16 + 1/2y² - 5/2(z - 1/10)² + 5/200 - 8 = 0
or
X² + 0.5Y² - 2.5Z² - - 11.0375 = 0
A sample of the surface created by iterating through z (an xy plane) then finding the 0, 1 or 2 roots for each y using the quadratic formula to solve for x. Notice the missing faces where x=0. Can run thru solving for y and have missing faces around y=0. Overlaying the two gives the shape, but not a pretty mesh.
I've put in both the original equation f(x, y, z) and the translated form f2(x, y, z).
import bpy
import bmesh
from math import sqrt
def f(x, y, z):
return 2 * x * x + y * y - 5 * z * z + z -7 * x -16
# f transformed to have origin (0, 0, 0)
def f2(x, y, z):
x * x + 0.5 * y * y - 2.5 * z * z - 11.0375
axis = -b / (2 * a)
check = b * b - 4 * a * c
if abs(check) < 0.0001:
return True, axis, axis
if check < 0:
return False, None, None
dist = sqrt(check) / (2 * a)
return True, axis - dist, axis + dist
bm = bmesh.new()
dom = [m for m in range(-10, 11)]
use_f2 = False # use f2()
planes = []
# coefficents of x
if use_f2:
a, b = 1, 0 # f2()
else:
a, b = 2, -7 # f()
for z in dom:
# for each xy plane
neg_line = []
pos_line = []
for y in dom:
# for each line in plane
if use_f2:
# constant f2()
c = 0.5 * y * y - 2.5 * z * z -11.0375
else:
# constant f()
c = y * y - 5 * z * z + z -16
# find roots
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
neg_root = bm.verts.new((neg_root, y, z))
pos_root = bm.verts.new((pos_root, y, z))
pos_line.append(pos_root)
neg_line.append(neg_root)
planes.append([neg_line, pos_line])
# skin it
if len(planes) > 1:
for i in range(len(planes) - 1):
n0, p0 = planes[i]
n1, p1 = planes[i + 1]
# pos
for j in range(len(n0) - 1):
verts = [n0[j], n1[j], n1[j+1], n0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
verts = [p0[j], p1[j], p1[j+1], p0[j + 1]]
if None not in verts:
f = bm.faces.new(verts)
# make an object
scene = bpy.context.scene
bm.to_mesh(mesh)
scene.objects.active = ob
ob.scale *= 0.01
To solve for y for f(x, y, z) edit
a, b = 1, 0
c = 2 * x * x - 5 * z * z + z -7 * x - 16
has_solution, neg_root, pos_root = quad_root(a, b, c)
if has_solution:
#print(neg_root, pos_root)
neg_root = bm.verts.new((x, neg_root, z))
pos_root = bm.verts.new((x, pos_root, z))
Result after running in both x and y and joining
| 2,340
| 8,011
|
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| 3.59375
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.896606
|
This will scale our window:
Finally, multiply the output by 10 to increase the value, then plug it into the Density of a Volume Scatter and a Volume Absorption shader added together:
This will nicely display the function when rendered, though the shading isn't ideal:
click for larger image
To optimize rendering, change the following settings:
Although it doesn't give you a workable mesh, this can definitely give you a renderable graph.
|
We can also improve the look of it by coloring it:
click for larger image
• Need 50 rep to comment... @ScottMilner I was hoping there was a way to preview in viewport directly (not viewport shading: rendered, but viewport shading: solid). The less CPU power, the better I am actually looking only to visualize the shape (so quality is not important.
|
https://stats.stackexchange.com/questions/591229/generating-random-variable-which-has-a-power-distribution-of-box-and-tiao-1962
| 1,709,588,368,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00885.warc.gz
| 552,287,119
| 41,580
|
# Generating random variable which has a power distribution of Box and Tiao (1962)
Box and Tiao (Biometrika 1962) use a distribution whose density has the following form: $$f(x; \mu, \sigma, \alpha) = \omega \exp\left\{ -\frac{1}{2} \Big\vert\frac{x-\mu}{\sigma}\Big\vert^{\frac{2}{(1+\alpha)}} \right\},$$ where $$\omega^{-1} = [\Gamma(g(\alpha)]\,2^{g(\alpha)}\sigma$$ is the normalizing constant with $$g(\alpha) = \frac{3}{2} + \frac{\alpha}{2},$$ $$\sigma \gt 0,$$ and $$-1 \lt \alpha \lt 1$$.
When $$\alpha=0$$ this reduces to the normal distribution; when $$\alpha=1$$ it reduces to the double exponential (Laplace) distribution, and when $$\alpha \to -1^{+}$$ it tends to a uniform distribution.
How can I generate random numbers from this distribution for any such value of $$\alpha$$?
Box & Tiao refer to this as a "convenient class of power distributions," referencing Diananda (1949), Box (1953), and Turner (1960).
Because $$\mu$$ and $$\sigma$$ just establish a unit of measurement and the absolute value reflects values around the origin, the basic density is proportional to $$\exp(-z^p/2)$$ where $$p = 2/(1+\alpha)$$ and $$z \ge 0.$$ Changing variables to $$y = z^p$$ for $$0\lt p \lt \infty$$ changes the probability element to
$$\exp(-z^p/2)\mathrm{d}z \to \exp(-y/2) \mathrm{d}\left(y^{1/p}\right) = \frac{1}{p}y^{1/p - 1}e^{-y/2}\mathrm{d}y.$$
Since $$p = 2/(1+\alpha),$$ this is proportional to a scaled Gamma$$(1/p)$$ = Gamma$$((1+\alpha)/2)$$ density, also known as a Chi-squared$$(1+\alpha)$$ density.
Thus, to generate a value from such a distribution, undo all these transformations in reverse order:
Generate a value $$Y$$ from a Chi-squared$$(1+\alpha)$$ distribution, raise it to the $$2/(1+\alpha)$$ power, randomly negate it (with probability $$1/2$$), multiply by $$\sigma,$$ and add $$\mu.$$
This R code exhibits one such implementation. n is the number of independent values to draw.
rf <- function(n, mu, sigma, alpha) {
y <- rchisq(n, 1 + alpha) # A chi-squared variate
u <- sample(c(-1,1), n, replace = TRUE) # Random sign change
y^((1 + alpha)/2) * u * sigma + mu
}
Here are some examples of values generated in this fashion (100,000 of each) along with graphs of $$f.$$
Generating Chi-squared variates with parameter $$1+\alpha$$ near zero is problematic. You can see this code works for $$1+\alpha = 0.1$$ (bottom left), but watch out when it gets much smaller than this:
The spike and gap in the middle should not be there.
The problem lies with floating point arithmetic: even double precision does not suffice. By this point, though, the uniform distribution looks like a good approximation.
### Appendix
This R code produced the plots. It uses the showtext library to access a Google font for the axis numbers and labels. Few of these fonts, if any, support Greek or math characters, so I had to use the default font for the plot titles (using mtext). Otherwise, everything is done with the base R plotting functions hist and curve. Don't be concerned about the relatively large simulation size: the total computation time is far less than one second to generate these 400,000 variates.
library(showtext)
showtext_auto()
#
# Density calculation.
#
f <- function(x, mu, sigma, alpha)
exp(-1/2 * abs((x - mu) / sigma) ^ (2 / (1 + alpha)))
C <- function(mu, sigma, alpha, ...)
integrate(\(x) f(x, mu, sigma, alpha), -Inf, Inf, ...)\$value
#
# Specify the distributions to plot.
#
Parameters <- list(list(mu = 0, sigma = 1, alpha = 0),
list(mu = 10, sigma = 2, alpha = 1/2),
list(mu = 0, sigma = 3, alpha = -0.9),
list(mu = 0, sigma = 4, alpha = 0.99))
#
# Generate the samples and plot summaries of them.
#
n.sim <- 1e5 # Sample size per plot
set.seed(17) # For reproducibility
pars <- par(mfrow = c(2, 2), mai = c(1/2, 3/4, 3/8, 1/8)) # Shrink the margins
for (parameters in Parameters)
with(parameters, {
x <- rf(n.sim, mu, sigma, alpha)
hist(x, freq = FALSE, breaks = 100, family = "Informal",
xlab = "", main = "", col = gray(0.9), border = gray(0.7))
mtext(bquote(list(mu==.(mu), sigma==.(sigma), alpha==.(alpha))),
cex = 1.25, side = 3, line = 0)
omega <- 1 / C(mu, sigma, alpha) # Compute the normalizing constant
curve(omega * f(x, mu, sigma, alpha), add = TRUE, lwd = 2, col = "Red")
})
par(pars)
• That's some clean code...
– Zen
Oct 5, 2022 at 17:20
• Nice and very fastly delivered:) answer.
– Yves
Oct 5, 2022 at 17:32
• Beautiful solution. Thank you! Oct 6, 2022 at 18:19
• @whuber: Can you please show us how you generated the lovely plots? Oct 6, 2022 at 18:26
• @user67724 Done.
– whuber
Oct 6, 2022 at 19:05
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| 3.703125
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CC-MAIN-2024-10
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latest
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# Generating random variable which has a power distribution of Box and Tiao (1962)
Box and Tiao (Biometrika 1962) use a distribution whose density has the following form: $$f(x; \mu, \sigma, \alpha) = \omega \exp\left\{ -\frac{1}{2} \Big\vert\frac{x-\mu}{\sigma}\Big\vert^{\frac{2}{(1+\alpha)}} \right\},$$ where $$\omega^{-1} = [\Gamma(g(\alpha)]\,2^{g(\alpha)}\sigma$$ is the normalizing constant with $$g(\alpha) = \frac{3}{2} + \frac{\alpha}{2},$$ $$\sigma \gt 0,$$ and $$-1 \lt \alpha \lt 1$$.
|
When $$\alpha=0$$ this reduces to the normal distribution; when $$\alpha=1$$ it reduces to the double exponential (Laplace) distribution, and when $$\alpha \to -1^{+}$$ it tends to a uniform distribution. How can I generate random numbers from this distribution for any such value of $$\alpha$$? Box & Tiao refer to this as a "convenient class of power distributions," referencing Diananda (1949), Box (1953), and Turner (1960).
|
https://stats.stackexchange.com/questions/591229/generating-random-variable-which-has-a-power-distribution-of-box-and-tiao-1962
| 1,709,588,368,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00885.warc.gz
| 552,287,119
| 41,580
|
# Generating random variable which has a power distribution of Box and Tiao (1962)
Box and Tiao (Biometrika 1962) use a distribution whose density has the following form: $$f(x; \mu, \sigma, \alpha) = \omega \exp\left\{ -\frac{1}{2} \Big\vert\frac{x-\mu}{\sigma}\Big\vert^{\frac{2}{(1+\alpha)}} \right\},$$ where $$\omega^{-1} = [\Gamma(g(\alpha)]\,2^{g(\alpha)}\sigma$$ is the normalizing constant with $$g(\alpha) = \frac{3}{2} + \frac{\alpha}{2},$$ $$\sigma \gt 0,$$ and $$-1 \lt \alpha \lt 1$$.
When $$\alpha=0$$ this reduces to the normal distribution; when $$\alpha=1$$ it reduces to the double exponential (Laplace) distribution, and when $$\alpha \to -1^{+}$$ it tends to a uniform distribution.
How can I generate random numbers from this distribution for any such value of $$\alpha$$?
Box & Tiao refer to this as a "convenient class of power distributions," referencing Diananda (1949), Box (1953), and Turner (1960).
Because $$\mu$$ and $$\sigma$$ just establish a unit of measurement and the absolute value reflects values around the origin, the basic density is proportional to $$\exp(-z^p/2)$$ where $$p = 2/(1+\alpha)$$ and $$z \ge 0.$$ Changing variables to $$y = z^p$$ for $$0\lt p \lt \infty$$ changes the probability element to
$$\exp(-z^p/2)\mathrm{d}z \to \exp(-y/2) \mathrm{d}\left(y^{1/p}\right) = \frac{1}{p}y^{1/p - 1}e^{-y/2}\mathrm{d}y.$$
Since $$p = 2/(1+\alpha),$$ this is proportional to a scaled Gamma$$(1/p)$$ = Gamma$$((1+\alpha)/2)$$ density, also known as a Chi-squared$$(1+\alpha)$$ density.
Thus, to generate a value from such a distribution, undo all these transformations in reverse order:
Generate a value $$Y$$ from a Chi-squared$$(1+\alpha)$$ distribution, raise it to the $$2/(1+\alpha)$$ power, randomly negate it (with probability $$1/2$$), multiply by $$\sigma,$$ and add $$\mu.$$
This R code exhibits one such implementation. n is the number of independent values to draw.
rf <- function(n, mu, sigma, alpha) {
y <- rchisq(n, 1 + alpha) # A chi-squared variate
u <- sample(c(-1,1), n, replace = TRUE) # Random sign change
y^((1 + alpha)/2) * u * sigma + mu
}
Here are some examples of values generated in this fashion (100,000 of each) along with graphs of $$f.$$
Generating Chi-squared variates with parameter $$1+\alpha$$ near zero is problematic. You can see this code works for $$1+\alpha = 0.1$$ (bottom left), but watch out when it gets much smaller than this:
The spike and gap in the middle should not be there.
The problem lies with floating point arithmetic: even double precision does not suffice. By this point, though, the uniform distribution looks like a good approximation.
### Appendix
This R code produced the plots. It uses the showtext library to access a Google font for the axis numbers and labels. Few of these fonts, if any, support Greek or math characters, so I had to use the default font for the plot titles (using mtext). Otherwise, everything is done with the base R plotting functions hist and curve. Don't be concerned about the relatively large simulation size: the total computation time is far less than one second to generate these 400,000 variates.
library(showtext)
showtext_auto()
#
# Density calculation.
#
f <- function(x, mu, sigma, alpha)
exp(-1/2 * abs((x - mu) / sigma) ^ (2 / (1 + alpha)))
C <- function(mu, sigma, alpha, ...)
integrate(\(x) f(x, mu, sigma, alpha), -Inf, Inf, ...)\$value
#
# Specify the distributions to plot.
#
Parameters <- list(list(mu = 0, sigma = 1, alpha = 0),
list(mu = 10, sigma = 2, alpha = 1/2),
list(mu = 0, sigma = 3, alpha = -0.9),
list(mu = 0, sigma = 4, alpha = 0.99))
#
# Generate the samples and plot summaries of them.
#
n.sim <- 1e5 # Sample size per plot
set.seed(17) # For reproducibility
pars <- par(mfrow = c(2, 2), mai = c(1/2, 3/4, 3/8, 1/8)) # Shrink the margins
for (parameters in Parameters)
with(parameters, {
x <- rf(n.sim, mu, sigma, alpha)
hist(x, freq = FALSE, breaks = 100, family = "Informal",
xlab = "", main = "", col = gray(0.9), border = gray(0.7))
mtext(bquote(list(mu==.(mu), sigma==.(sigma), alpha==.(alpha))),
cex = 1.25, side = 3, line = 0)
omega <- 1 / C(mu, sigma, alpha) # Compute the normalizing constant
curve(omega * f(x, mu, sigma, alpha), add = TRUE, lwd = 2, col = "Red")
})
par(pars)
• That's some clean code...
– Zen
Oct 5, 2022 at 17:20
• Nice and very fastly delivered:) answer.
– Yves
Oct 5, 2022 at 17:32
• Beautiful solution. Thank you! Oct 6, 2022 at 18:19
• @whuber: Can you please show us how you generated the lovely plots? Oct 6, 2022 at 18:26
• @user67724 Done.
– whuber
Oct 6, 2022 at 19:05
| 1,414
| 4,604
|
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| 3.703125
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.750842
|
Thus, to generate a value from such a distribution, undo all these transformations in reverse order:
Generate a value $$Y$$ from a Chi-squared$$(1+\alpha)$$ distribution, raise it to the $$2/(1+\alpha)$$ power, randomly negate it (with probability $$1/2$$), multiply by $$\sigma,$$ and add $$\mu.$$
This R code exhibits one such implementation. n is the number of independent values to draw.
|
rf <- function(n, mu, sigma, alpha) {
y <- rchisq(n, 1 + alpha) # A chi-squared variate
u <- sample(c(-1,1), n, replace = TRUE) # Random sign change
y^((1 + alpha)/2) * u * sigma + mu
}
Here are some examples of values generated in this fashion (100,000 of each) along with graphs of $$f.$$
Generating Chi-squared variates with parameter $$1+\alpha$$ near zero is problematic.
|
https://stats.stackexchange.com/questions/591229/generating-random-variable-which-has-a-power-distribution-of-box-and-tiao-1962
| 1,709,588,368,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00885.warc.gz
| 552,287,119
| 41,580
|
# Generating random variable which has a power distribution of Box and Tiao (1962)
Box and Tiao (Biometrika 1962) use a distribution whose density has the following form: $$f(x; \mu, \sigma, \alpha) = \omega \exp\left\{ -\frac{1}{2} \Big\vert\frac{x-\mu}{\sigma}\Big\vert^{\frac{2}{(1+\alpha)}} \right\},$$ where $$\omega^{-1} = [\Gamma(g(\alpha)]\,2^{g(\alpha)}\sigma$$ is the normalizing constant with $$g(\alpha) = \frac{3}{2} + \frac{\alpha}{2},$$ $$\sigma \gt 0,$$ and $$-1 \lt \alpha \lt 1$$.
When $$\alpha=0$$ this reduces to the normal distribution; when $$\alpha=1$$ it reduces to the double exponential (Laplace) distribution, and when $$\alpha \to -1^{+}$$ it tends to a uniform distribution.
How can I generate random numbers from this distribution for any such value of $$\alpha$$?
Box & Tiao refer to this as a "convenient class of power distributions," referencing Diananda (1949), Box (1953), and Turner (1960).
Because $$\mu$$ and $$\sigma$$ just establish a unit of measurement and the absolute value reflects values around the origin, the basic density is proportional to $$\exp(-z^p/2)$$ where $$p = 2/(1+\alpha)$$ and $$z \ge 0.$$ Changing variables to $$y = z^p$$ for $$0\lt p \lt \infty$$ changes the probability element to
$$\exp(-z^p/2)\mathrm{d}z \to \exp(-y/2) \mathrm{d}\left(y^{1/p}\right) = \frac{1}{p}y^{1/p - 1}e^{-y/2}\mathrm{d}y.$$
Since $$p = 2/(1+\alpha),$$ this is proportional to a scaled Gamma$$(1/p)$$ = Gamma$$((1+\alpha)/2)$$ density, also known as a Chi-squared$$(1+\alpha)$$ density.
Thus, to generate a value from such a distribution, undo all these transformations in reverse order:
Generate a value $$Y$$ from a Chi-squared$$(1+\alpha)$$ distribution, raise it to the $$2/(1+\alpha)$$ power, randomly negate it (with probability $$1/2$$), multiply by $$\sigma,$$ and add $$\mu.$$
This R code exhibits one such implementation. n is the number of independent values to draw.
rf <- function(n, mu, sigma, alpha) {
y <- rchisq(n, 1 + alpha) # A chi-squared variate
u <- sample(c(-1,1), n, replace = TRUE) # Random sign change
y^((1 + alpha)/2) * u * sigma + mu
}
Here are some examples of values generated in this fashion (100,000 of each) along with graphs of $$f.$$
Generating Chi-squared variates with parameter $$1+\alpha$$ near zero is problematic. You can see this code works for $$1+\alpha = 0.1$$ (bottom left), but watch out when it gets much smaller than this:
The spike and gap in the middle should not be there.
The problem lies with floating point arithmetic: even double precision does not suffice. By this point, though, the uniform distribution looks like a good approximation.
### Appendix
This R code produced the plots. It uses the showtext library to access a Google font for the axis numbers and labels. Few of these fonts, if any, support Greek or math characters, so I had to use the default font for the plot titles (using mtext). Otherwise, everything is done with the base R plotting functions hist and curve. Don't be concerned about the relatively large simulation size: the total computation time is far less than one second to generate these 400,000 variates.
library(showtext)
showtext_auto()
#
# Density calculation.
#
f <- function(x, mu, sigma, alpha)
exp(-1/2 * abs((x - mu) / sigma) ^ (2 / (1 + alpha)))
C <- function(mu, sigma, alpha, ...)
integrate(\(x) f(x, mu, sigma, alpha), -Inf, Inf, ...)\$value
#
# Specify the distributions to plot.
#
Parameters <- list(list(mu = 0, sigma = 1, alpha = 0),
list(mu = 10, sigma = 2, alpha = 1/2),
list(mu = 0, sigma = 3, alpha = -0.9),
list(mu = 0, sigma = 4, alpha = 0.99))
#
# Generate the samples and plot summaries of them.
#
n.sim <- 1e5 # Sample size per plot
set.seed(17) # For reproducibility
pars <- par(mfrow = c(2, 2), mai = c(1/2, 3/4, 3/8, 1/8)) # Shrink the margins
for (parameters in Parameters)
with(parameters, {
x <- rf(n.sim, mu, sigma, alpha)
hist(x, freq = FALSE, breaks = 100, family = "Informal",
xlab = "", main = "", col = gray(0.9), border = gray(0.7))
mtext(bquote(list(mu==.(mu), sigma==.(sigma), alpha==.(alpha))),
cex = 1.25, side = 3, line = 0)
omega <- 1 / C(mu, sigma, alpha) # Compute the normalizing constant
curve(omega * f(x, mu, sigma, alpha), add = TRUE, lwd = 2, col = "Red")
})
par(pars)
• That's some clean code...
– Zen
Oct 5, 2022 at 17:20
• Nice and very fastly delivered:) answer.
– Yves
Oct 5, 2022 at 17:32
• Beautiful solution. Thank you! Oct 6, 2022 at 18:19
• @whuber: Can you please show us how you generated the lovely plots? Oct 6, 2022 at 18:26
• @user67724 Done.
– whuber
Oct 6, 2022 at 19:05
| 1,414
| 4,604
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.750842
|
You can see this code works for $$1+\alpha = 0.1$$ (bottom left), but watch out when it gets much smaller than this:
The spike and gap in the middle should not be there. The problem lies with floating point arithmetic: even double precision does not suffice. By this point, though, the uniform distribution looks like a good approximation. ### Appendix
This R code produced the plots.
|
It uses the showtext library to access a Google font for the axis numbers and labels. Few of these fonts, if any, support Greek or math characters, so I had to use the default font for the plot titles (using mtext). Otherwise, everything is done with the base R plotting functions hist and curve. Don't be concerned about the relatively large simulation size: the total computation time is far less than one second to generate these 400,000 variates.
|
https://stats.stackexchange.com/questions/591229/generating-random-variable-which-has-a-power-distribution-of-box-and-tiao-1962
| 1,709,588,368,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00885.warc.gz
| 552,287,119
| 41,580
|
# Generating random variable which has a power distribution of Box and Tiao (1962)
Box and Tiao (Biometrika 1962) use a distribution whose density has the following form: $$f(x; \mu, \sigma, \alpha) = \omega \exp\left\{ -\frac{1}{2} \Big\vert\frac{x-\mu}{\sigma}\Big\vert^{\frac{2}{(1+\alpha)}} \right\},$$ where $$\omega^{-1} = [\Gamma(g(\alpha)]\,2^{g(\alpha)}\sigma$$ is the normalizing constant with $$g(\alpha) = \frac{3}{2} + \frac{\alpha}{2},$$ $$\sigma \gt 0,$$ and $$-1 \lt \alpha \lt 1$$.
When $$\alpha=0$$ this reduces to the normal distribution; when $$\alpha=1$$ it reduces to the double exponential (Laplace) distribution, and when $$\alpha \to -1^{+}$$ it tends to a uniform distribution.
How can I generate random numbers from this distribution for any such value of $$\alpha$$?
Box & Tiao refer to this as a "convenient class of power distributions," referencing Diananda (1949), Box (1953), and Turner (1960).
Because $$\mu$$ and $$\sigma$$ just establish a unit of measurement and the absolute value reflects values around the origin, the basic density is proportional to $$\exp(-z^p/2)$$ where $$p = 2/(1+\alpha)$$ and $$z \ge 0.$$ Changing variables to $$y = z^p$$ for $$0\lt p \lt \infty$$ changes the probability element to
$$\exp(-z^p/2)\mathrm{d}z \to \exp(-y/2) \mathrm{d}\left(y^{1/p}\right) = \frac{1}{p}y^{1/p - 1}e^{-y/2}\mathrm{d}y.$$
Since $$p = 2/(1+\alpha),$$ this is proportional to a scaled Gamma$$(1/p)$$ = Gamma$$((1+\alpha)/2)$$ density, also known as a Chi-squared$$(1+\alpha)$$ density.
Thus, to generate a value from such a distribution, undo all these transformations in reverse order:
Generate a value $$Y$$ from a Chi-squared$$(1+\alpha)$$ distribution, raise it to the $$2/(1+\alpha)$$ power, randomly negate it (with probability $$1/2$$), multiply by $$\sigma,$$ and add $$\mu.$$
This R code exhibits one such implementation. n is the number of independent values to draw.
rf <- function(n, mu, sigma, alpha) {
y <- rchisq(n, 1 + alpha) # A chi-squared variate
u <- sample(c(-1,1), n, replace = TRUE) # Random sign change
y^((1 + alpha)/2) * u * sigma + mu
}
Here are some examples of values generated in this fashion (100,000 of each) along with graphs of $$f.$$
Generating Chi-squared variates with parameter $$1+\alpha$$ near zero is problematic. You can see this code works for $$1+\alpha = 0.1$$ (bottom left), but watch out when it gets much smaller than this:
The spike and gap in the middle should not be there.
The problem lies with floating point arithmetic: even double precision does not suffice. By this point, though, the uniform distribution looks like a good approximation.
### Appendix
This R code produced the plots. It uses the showtext library to access a Google font for the axis numbers and labels. Few of these fonts, if any, support Greek or math characters, so I had to use the default font for the plot titles (using mtext). Otherwise, everything is done with the base R plotting functions hist and curve. Don't be concerned about the relatively large simulation size: the total computation time is far less than one second to generate these 400,000 variates.
library(showtext)
showtext_auto()
#
# Density calculation.
#
f <- function(x, mu, sigma, alpha)
exp(-1/2 * abs((x - mu) / sigma) ^ (2 / (1 + alpha)))
C <- function(mu, sigma, alpha, ...)
integrate(\(x) f(x, mu, sigma, alpha), -Inf, Inf, ...)\$value
#
# Specify the distributions to plot.
#
Parameters <- list(list(mu = 0, sigma = 1, alpha = 0),
list(mu = 10, sigma = 2, alpha = 1/2),
list(mu = 0, sigma = 3, alpha = -0.9),
list(mu = 0, sigma = 4, alpha = 0.99))
#
# Generate the samples and plot summaries of them.
#
n.sim <- 1e5 # Sample size per plot
set.seed(17) # For reproducibility
pars <- par(mfrow = c(2, 2), mai = c(1/2, 3/4, 3/8, 1/8)) # Shrink the margins
for (parameters in Parameters)
with(parameters, {
x <- rf(n.sim, mu, sigma, alpha)
hist(x, freq = FALSE, breaks = 100, family = "Informal",
xlab = "", main = "", col = gray(0.9), border = gray(0.7))
mtext(bquote(list(mu==.(mu), sigma==.(sigma), alpha==.(alpha))),
cex = 1.25, side = 3, line = 0)
omega <- 1 / C(mu, sigma, alpha) # Compute the normalizing constant
curve(omega * f(x, mu, sigma, alpha), add = TRUE, lwd = 2, col = "Red")
})
par(pars)
• That's some clean code...
– Zen
Oct 5, 2022 at 17:20
• Nice and very fastly delivered:) answer.
– Yves
Oct 5, 2022 at 17:32
• Beautiful solution. Thank you! Oct 6, 2022 at 18:19
• @whuber: Can you please show us how you generated the lovely plots? Oct 6, 2022 at 18:26
• @user67724 Done.
– whuber
Oct 6, 2022 at 19:05
| 1,414
| 4,604
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 30, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.703125
| 4
|
CC-MAIN-2024-10
|
latest
|
en
| 0.750842
|
#
f <- function(x, mu, sigma, alpha)
exp(-1/2 * abs((x - mu) / sigma) ^ (2 / (1 + alpha)))
C <- function(mu, sigma, alpha, ...)
integrate(\(x) f(x, mu, sigma, alpha), -Inf, Inf, ...)\$value
#
# Specify the distributions to plot. #
Parameters <- list(list(mu = 0, sigma = 1, alpha = 0),
list(mu = 10, sigma = 2, alpha = 1/2),
list(mu = 0, sigma = 3, alpha = -0.9),
list(mu = 0, sigma = 4, alpha = 0.99))
#
# Generate the samples and plot summaries of them.
|
#
n.sim <- 1e5 # Sample size per plot
set.seed(17) # For reproducibility
pars <- par(mfrow = c(2, 2), mai = c(1/2, 3/4, 3/8, 1/8)) # Shrink the margins
for (parameters in Parameters)
with(parameters, {
x <- rf(n.sim, mu, sigma, alpha)
hist(x, freq = FALSE, breaks = 100, family = "Informal",
xlab = "", main = "", col = gray(0.9), border = gray(0.7))
mtext(bquote(list(mu==.
|
https://math.stackexchange.com/questions/1184338/gibbs-phenomenon-and-fourier-series
| 1,571,861,776,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570987836295.98/warc/CC-MAIN-20191023201520-20191023225020-00216.warc.gz
| 581,563,792
| 32,428
|
# Gibbs Phenomenon and Fourier Series
a) Show the partial sum $$S = \frac{4}{\pi} \sum_{n=1}^N \frac{\sin((2n-1)t)}{2n-1}$$ which may also be written as $$\frac{2}{\pi}\int_0^x\frac{\sin(2Nt)}{\sin(t)}dt$$ has extrema at $x= \frac{m\pi}{2N}$
where $m$ is any positive integer except m=2kN, k also integer.
Solution: The derivative of $$S = \frac{2}{\pi}\frac{\sin(2Nx)}{\sin(x)} = 0$$ $$\text{where }\sin(2Nx)=0,$$ $\sin(x)$ cannot equal zero.
$\sin(x) = 0$ where $x$ is a multiple of $\pi$.
Therefore, $$\sin(2Nx)=0$$ where $x=\frac{m\pi}{2N}$
however $\sin(x)$ cannot equal zero, $$\sin(\frac{m\pi}{2N})\neq 0$$ som is any positive integer except $m=2kN$, $k$ also integer. Is this complete?!
b) Consider the first extrema to the right of the discontinuity, located at $x=\frac{\pi}{2N}$. By considering a suitable small angle formula show that the value of the sum at this point $$S(\frac{\pi}{2N})≈\frac{2}{\pi}\int_0^{\pi} \frac{\sin(u)}{u}du$$
Solution: I'm not sure which small angle formula i'm meant to considering?! Taylor series of sin? or how to consider it? I see that $$S(\frac{\pi}{2N}) = \frac{4}{\pi} (\sin(\frac{\pi}{2N})+\frac{\sin(\frac{3\pi}{2N})}{3}+\frac{\frac{\sin(5\pi)}{2N}}{5}+\ldots)$$ $$=\frac{2}{\pi}(\frac{\pi}{N}(\frac{\sin(\frac{\pi}{2N})}{\frac{\pi}{2N}}+\frac{\sin(\frac{3\pi}{2N})}{\frac{3\pi}{2N}}+\frac{\sin(\frac{5\pi}{2N})}{\frac{5\pi}{2N}}+\cdots)$$
c) and by getting a computer to evaluate this numerically show that $$S(\frac{\pi}{2N})≈1.1790$$ independently of $N$.
Not really sure how I could show this?
Hence comment on the accuracy of Fourier series at discontinuities (also known as Gibbs phenomenon). Given that the error at $\frac{π}{2N}$ is nearly constant explain why the Fourier Convergence theorem is, or is not, valid for this problem?
Where a function has a jump discontinuity, the fourier series will overshoot as it approaches the discontinuity. As the number of terms in the fouler series increases, the amount of overshoot will converge to a constant percentage (around 17.9) of the amount of the jump
could someone explain this to me please? -sorry for the long winded question!
In this answer, it is shown that the overshoot, on each side, is approximately $$\frac1\pi\int_0^\pi\frac{\sin(t)}{t}\mathrm{d}t-\frac12=0.089489872236$$ of the total jump. Thus, the overshoot you mention is twice that.
| 799
| 2,373
|
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| 4.0625
| 4
|
CC-MAIN-2019-43
|
latest
|
en
| 0.841029
|
Solution: The derivative of $$S = \frac{2}{\pi}\frac{\sin(2Nx)}{\sin(x)} = 0$$ $$\text{where }\sin(2Nx)=0,$$ $\sin(x)$ cannot equal zero. $\sin(x) = 0$ where $x$ is a multiple of $\pi$. Therefore, $$\sin(2Nx)=0$$ where $x=\frac{m\pi}{2N}$
however $\sin(x)$ cannot equal zero, $$\sin(\frac{m\pi}{2N})\neq 0$$ som is any positive integer except $m=2kN$, $k$ also integer.
|
Is this complete?! b) Consider the first extrema to the right of the discontinuity, located at $x=\frac{\pi}{2N}$. By considering a suitable small angle formula show that the value of the sum at this point $$S(\frac{\pi}{2N})≈\frac{2}{\pi}\int_0^{\pi} \frac{\sin(u)}{u}du$$
Solution: I'm not sure which small angle formula i'm meant to considering?!
|
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