Mixing
Wrong value of sum using fourier series
$begingroup$
I have the following $2 pi $ periodic function which is $t$ for $0leq t <pi$ and $0$ for $pileq t<2pi$
I'm asked to find its complex fourier series representation. So I calculate
$c_k=frac{1}{2pi}int_{0}^{pi}te^{-ikt}dt$ which results in $frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})$ and $c_0=frac{1}{2pi}int_{0}^{pi}tdt=frac{pi}{4}$. So our complex fourier series is
$f(t)=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})e^{ikt}=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{(-1)^k-1}{k^2})e^{ikt}$
Then I am asked to find the value of the fourier series at $t=pi$.
In their solution, they simply say that the answer is $frac{pi}{2}$
But I don't see how they got there.
Plugging in $pi$, I get:
$f(pi)=0=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{-1}{pi k^2}*(-1)$ for $k$ odd. Now, if I solve, I get $frac{-pi^2}{4}$
So I don't see what I did wrong and how they got $frac{pi}{2}$.
Thanks for your help !
real-analysis sequences-and-series complex-analysis fourier-analysis fourier-series
asked Feb 2 at 13:09
PoujhPoujh
6111516
$endgroup$
add a comment |
$begingroup$
I have the following $2 pi $ periodic function which is $t$ for $0leq t <pi$ and $0$ for $pileq t<2pi$
I'm asked to find its complex fourier series representation. So I calculate
$c_k=frac{1}{2pi}int_{0}^{pi}te^{-ikt}dt$ which results in $frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})$ and $c_0=frac{1}{2pi}int_{0}^{pi}tdt=frac{pi}{4}$. So our complex fourier series is
$f(t)=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})e^{ikt}=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{(-1)^k-1}{k^2})e^{ikt}$
Then I am asked to find the value of the fourier series at $t=pi$.
In their solution, they simply say that the answer is $frac{pi}{2}$
But I don't see how they got there.
Plugging in $pi$, I get:
$f(pi)=0=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{-1}{pi k^2}*(-1)$ for $k$ odd. Now, if I solve, I get $frac{-pi^2}{4}$
So I don't see what I did wrong and how they got $frac{pi}{2}$.
Thanks for your help !
real-analysis sequences-and-series complex-analysis fourier-analysis fourier-series
asked Feb 2 at 13:09
PoujhPoujh
6111516
$endgroup$
1
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42
add a comment |
$begingroup$
I have the following $2 pi $ periodic function which is $t$ for $0leq t <pi$ and $0$ for $pileq t<2pi$
I'm asked to find its complex fourier series representation. So I calculate
$c_k=frac{1}{2pi}int_{0}^{pi}te^{-ikt}dt$ which results in $frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})$ and $c_0=frac{1}{2pi}int_{0}^{pi}tdt=frac{pi}{4}$. So our complex fourier series is
$f(t)=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})e^{ikt}=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{(-1)^k-1}{k^2})e^{ikt}$
Then I am asked to find the value of the fourier series at $t=pi$.
In their solution, they simply say that the answer is $frac{pi}{2}$
But I don't see how they got there.
Plugging in $pi$, I get:
$f(pi)=0=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{-1}{pi k^2}*(-1)$ for $k$ odd. Now, if I solve, I get $frac{-pi^2}{4}$
So I don't see what I did wrong and how they got $frac{pi}{2}$.
Thanks for your help !
real-analysis sequences-and-series complex-analysis fourier-analysis fourier-series
asked Feb 2 at 13:09
PoujhPoujh
6111516
$endgroup$
I have the following $2 pi $ periodic function which is $t$ for $0leq t <pi$ and $0$ for $pileq t<2pi$
I'm asked to find its complex fourier series representation. So I calculate
$c_k=frac{1}{2pi}int_{0}^{pi}te^{-ikt}dt$ which results in $frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})$ and $c_0=frac{1}{2pi}int_{0}^{pi}tdt=frac{pi}{4}$. So our complex fourier series is
$f(t)=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{pi(-1)^k}{-ik}+frac{(-1)^k-1}{k^2})e^{ikt}=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{1}{2pi} (frac{(-1)^k-1}{k^2})e^{ikt}$
Then I am asked to find the value of the fourier series at $t=pi$.
In their solution, they simply say that the answer is $frac{pi}{2}$
But I don't see how they got there.
Plugging in $pi$, I get:
$f(pi)=0=frac{pi}{4}+sum_{-infty, kneq 0}^{infty}frac{-1}{pi k^2}*(-1)$ for $k$ odd. Now, if I solve, I get $frac{-pi^2}{4}$
So I don't see what I did wrong and how they got $frac{pi}{2}$.
Thanks for your help !
real-analysis sequences-and-series complex-analysis fourier-analysis fourier-series
real-analysis sequences-and-series complex-analysis fourier-analysis fourier-series
asked Feb 2 at 13:09
PoujhPoujh
6111516
asked Feb 2 at 13:09
PoujhPoujh
6111516
asked Feb 2 at 13:09
PoujhPoujh
6111516
asked Feb 2 at 13:09
PoujhPoujh
6111516
asked Feb 2 at 13:09
PoujhPoujh
6111516
6111516
1
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42
add a comment |
1
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42
1
1
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $pi$ from below and $0$ from above, or $frac{pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($frac t2$ on $(-pi,pi)$). We can't drop them until we choose $pi$ as the point to evaluate at - because there, $e^{ikpi}=e^{-ikpi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $pi$ - that's a correct expression. And I get $frac{pi}{4}+frac{pi}{4}=frac{pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
add a comment |
$begingroup$
Your series serms correct, but I am at a loss as to how you got $(-pi^2)/4$.
If you know the reciprocal square sum
$Sigma_{n=1}^{infty}(1/n^2)=pi^2/6$
then you can subtract one quarter of that:
$Sigma_{n=1}^{infty}(1/(2n)^2)=pi^2/24$
and thus
$Sigma_{n=1}^{infty}(1/n^2)|_{ntext{ odd}}=pi^2/8$.
Plug that into your sum, remembering that there is also a division by $pi$, and the correct solution should come out.
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097272%2fwrong-value-of-sum-using-fourier-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $pi$ from below and $0$ from above, or $frac{pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($frac t2$ on $(-pi,pi)$). We can't drop them until we choose $pi$ as the point to evaluate at - because there, $e^{ikpi}=e^{-ikpi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $pi$ - that's a correct expression. And I get $frac{pi}{4}+frac{pi}{4}=frac{pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
add a comment |
$begingroup$
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $pi$ from below and $0$ from above, or $frac{pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($frac t2$ on $(-pi,pi)$). We can't drop them until we choose $pi$ as the point to evaluate at - because there, $e^{ikpi}=e^{-ikpi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $pi$ - that's a correct expression. And I get $frac{pi}{4}+frac{pi}{4}=frac{pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
$endgroup$
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
add a comment |
$begingroup$
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $pi$ from below and $0$ from above, or $frac{pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($frac t2$ on $(-pi,pi)$). We can't drop them until we choose $pi$ as the point to evaluate at - because there, $e^{ikpi}=e^{-ikpi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $pi$ - that's a correct expression. And I get $frac{pi}{4}+frac{pi}{4}=frac{pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
$endgroup$
It's a standard result that the Fourier series evaluated at a jump discontinuity of the (otherwise nice) function is the average of the left and right limits. So, here, that should be the average of $pi$ from below and $0$ from above, or $frac{pi}{2}$. The solution you mention checks out.
So, where did your calculation go wrong?
You've got the right coefficients.
That series for $f(t)$ - the $frac1k$ terms don't go away just yet. Those correspond to the odd part of $f$ ($frac t2$ on $(-pi,pi)$). We can't drop them until we choose $pi$ as the point to evaluate at - because there, $e^{ikpi}=e^{-ikpi}$ and terms on opposite sides of zero cancel for a principal value sum of zero.
Plugging in $pi$ - that's a correct expression. And I get $frac{pi}{4}+frac{pi}{4}=frac{pi}{2}$ for it. Your error is somewhere in that last step, for which you didn't show any work.
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
answered Feb 2 at 13:40
jmerryjmerry
17.1k11633
17.1k11633
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
add a comment |
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Well, in the last step, I put the $frac{pi}{4}$ on the other side and then solved for $sum frac{1}{k^2}$, k odd. I think that's the problem
$endgroup$
– Poujh
Feb 2 at 13:54
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
$begingroup$
Just a question, to get the second $frac{pi}{4}$, you seem to have evaluated $ sum_{0}^{infty} frac{1}{(2m+1)^2}$ whose value is $ frac{pi^2}{8}$ then used the fact that we have $sum_{-infty}^{infty}=2 sum_{0}^{infty}$. Do you think they supposed we should know the value of that sum ?
$endgroup$
– Poujh
Feb 2 at 13:57
1
1
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
$begingroup$
It does seem rather necessary for this one. Of course, we can easily derive it from the well-known sum of all the reciprocals of squares.
$endgroup$
– jmerry
Feb 2 at 14:02
add a comment |
$begingroup$
Your series serms correct, but I am at a loss as to how you got $(-pi^2)/4$.
If you know the reciprocal square sum
$Sigma_{n=1}^{infty}(1/n^2)=pi^2/6$
then you can subtract one quarter of that:
$Sigma_{n=1}^{infty}(1/(2n)^2)=pi^2/24$
and thus
$Sigma_{n=1}^{infty}(1/n^2)|_{ntext{ odd}}=pi^2/8$.
Plug that into your sum, remembering that there is also a division by $pi$, and the correct solution should come out.
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
Your series serms correct, but I am at a loss as to how you got $(-pi^2)/4$.
If you know the reciprocal square sum
$Sigma_{n=1}^{infty}(1/n^2)=pi^2/6$
then you can subtract one quarter of that:
$Sigma_{n=1}^{infty}(1/(2n)^2)=pi^2/24$
and thus
$Sigma_{n=1}^{infty}(1/n^2)|_{ntext{ odd}}=pi^2/8$.
Plug that into your sum, remembering that there is also a division by $pi$, and the correct solution should come out.
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
Your series serms correct, but I am at a loss as to how you got $(-pi^2)/4$.
If you know the reciprocal square sum
$Sigma_{n=1}^{infty}(1/n^2)=pi^2/6$
then you can subtract one quarter of that:
$Sigma_{n=1}^{infty}(1/(2n)^2)=pi^2/24$
and thus
$Sigma_{n=1}^{infty}(1/n^2)|_{ntext{ odd}}=pi^2/8$.
Plug that into your sum, remembering that there is also a division by $pi$, and the correct solution should come out.
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
Your series serms correct, but I am at a loss as to how you got $(-pi^2)/4$.
If you know the reciprocal square sum
$Sigma_{n=1}^{infty}(1/n^2)=pi^2/6$
then you can subtract one quarter of that:
$Sigma_{n=1}^{infty}(1/(2n)^2)=pi^2/24$
and thus
$Sigma_{n=1}^{infty}(1/n^2)|_{ntext{ odd}}=pi^2/8$.
Plug that into your sum, remembering that there is also a division by $pi$, and the correct solution should come out.
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
answered Feb 2 at 13:39
Oscar LanziOscar Lanzi
13.6k12136
13.6k12136
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097272%2fwrong-value-of-sum-using-fourier-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
The value will definitely be $frac{pi}{2}$ as it will be the average value at the jump in $f(t)$ from $pi$ to 0 when $t = pi$.
$endgroup$
– Paul
Feb 2 at 13:42