Compute $sum_{ngeq1}frac{1}{pi^2+n^2}$ by expanding $e^{pi x}$ in its Fourier series The 2019...

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Compute $sum_{ngeq1}frac{1}{pi^2+n^2}$ by expanding $e^{pi x}$ in its Fourier series



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4












$begingroup$



Compute $$sum_{ngeq1}frac{1}{pi^2+n^2}.$$



by expanding $e^{pi x}$ in its Fourier series.




So I calculated that



$$c_n=frac{1}{2pi}int_{-pi}^{pi}e^{pi x} e^{-inx} dx=frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)},$$



which is correct. Now,



$$f(x)=e^{pi x}=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}e^{inx}.$$



At $x=0$ we have



$$1=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}=frac{sinh(pi^2)}{pi}sum_{ninmathbb{Z}}frac{(-1)^n}{pi-in}.$$



Taking out the case $n=0$ and splitting the sum by $n rightarrow -n$ in one of them we get



begin{align}
1&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+sum_{ngeq1}frac{(-1)^n}{pi-in} + sum_{ngeq1}frac{(-1)^{-n}}{pi+in}right)\
&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+2pisum_{ngeq1}frac{(-1)^n}{pi^2+n^2}right).
end{align}



So I almost have the desired sum here, only thing bothering me is the $(-1)^n$. How do I proceed?



Note that I'm not interested in alternate solutions. I already know at least 2 other methods.






summation power-series fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
    $endgroup$
    – mrtaurho
    Mar 22 at 21:07










  • $begingroup$
    Fixed it just before you wrote that comment :P My apologies!
    $endgroup$
    – Parseval
    Mar 22 at 21:07












  • $begingroup$
    It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
    $endgroup$
    – mrtaurho
    Mar 22 at 21:08






  • 3




    $begingroup$
    Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
    $endgroup$
    – Parseval
    Mar 22 at 21:09








  • 2




    $begingroup$
    Try evaluating at $x=pi$ instead? That should cancel away the minuses.
    $endgroup$
    – Semiclassical
    Mar 22 at 21:20
















4












$begingroup$



Compute $$sum_{ngeq1}frac{1}{pi^2+n^2}.$$



by expanding $e^{pi x}$ in its Fourier series.




So I calculated that



$$c_n=frac{1}{2pi}int_{-pi}^{pi}e^{pi x} e^{-inx} dx=frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)},$$



which is correct. Now,



$$f(x)=e^{pi x}=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}e^{inx}.$$



At $x=0$ we have



$$1=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}=frac{sinh(pi^2)}{pi}sum_{ninmathbb{Z}}frac{(-1)^n}{pi-in}.$$



Taking out the case $n=0$ and splitting the sum by $n rightarrow -n$ in one of them we get



begin{align}
1&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+sum_{ngeq1}frac{(-1)^n}{pi-in} + sum_{ngeq1}frac{(-1)^{-n}}{pi+in}right)\
&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+2pisum_{ngeq1}frac{(-1)^n}{pi^2+n^2}right).
end{align}



So I almost have the desired sum here, only thing bothering me is the $(-1)^n$. How do I proceed?



Note that I'm not interested in alternate solutions. I already know at least 2 other methods.






summation power-series fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
    $endgroup$
    – mrtaurho
    Mar 22 at 21:07










  • $begingroup$
    Fixed it just before you wrote that comment :P My apologies!
    $endgroup$
    – Parseval
    Mar 22 at 21:07












  • $begingroup$
    It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
    $endgroup$
    – mrtaurho
    Mar 22 at 21:08






  • 3




    $begingroup$
    Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
    $endgroup$
    – Parseval
    Mar 22 at 21:09








  • 2




    $begingroup$
    Try evaluating at $x=pi$ instead? That should cancel away the minuses.
    $endgroup$
    – Semiclassical
    Mar 22 at 21:20














4












4








4


1



$begingroup$



Compute $$sum_{ngeq1}frac{1}{pi^2+n^2}.$$



by expanding $e^{pi x}$ in its Fourier series.




So I calculated that



$$c_n=frac{1}{2pi}int_{-pi}^{pi}e^{pi x} e^{-inx} dx=frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)},$$



which is correct. Now,



$$f(x)=e^{pi x}=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}e^{inx}.$$



At $x=0$ we have



$$1=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}=frac{sinh(pi^2)}{pi}sum_{ninmathbb{Z}}frac{(-1)^n}{pi-in}.$$



Taking out the case $n=0$ and splitting the sum by $n rightarrow -n$ in one of them we get



begin{align}
1&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+sum_{ngeq1}frac{(-1)^n}{pi-in} + sum_{ngeq1}frac{(-1)^{-n}}{pi+in}right)\
&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+2pisum_{ngeq1}frac{(-1)^n}{pi^2+n^2}right).
end{align}



So I almost have the desired sum here, only thing bothering me is the $(-1)^n$. How do I proceed?



Note that I'm not interested in alternate solutions. I already know at least 2 other methods.






summation power-series fourier-series







share|cite|improve this question











$endgroup$





Compute $$sum_{ngeq1}frac{1}{pi^2+n^2}.$$



by expanding $e^{pi x}$ in its Fourier series.




So I calculated that



$$c_n=frac{1}{2pi}int_{-pi}^{pi}e^{pi x} e^{-inx} dx=frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)},$$



which is correct. Now,



$$f(x)=e^{pi x}=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}e^{inx}.$$



At $x=0$ we have



$$1=sum_{nin{mathbb{Z}}}frac{sinh{(pi^2)}(-1)^n}{pi(pi-in)}=frac{sinh(pi^2)}{pi}sum_{ninmathbb{Z}}frac{(-1)^n}{pi-in}.$$



Taking out the case $n=0$ and splitting the sum by $n rightarrow -n$ in one of them we get



begin{align}
1&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+sum_{ngeq1}frac{(-1)^n}{pi-in} + sum_{ngeq1}frac{(-1)^{-n}}{pi+in}right)\
&=frac{sinh(pi^2)}{pi}left(frac{1}{pi}+2pisum_{ngeq1}frac{(-1)^n}{pi^2+n^2}right).
end{align}



So I almost have the desired sum here, only thing bothering me is the $(-1)^n$. How do I proceed?



Note that I'm not interested in alternate solutions. I already know at least 2 other methods.






summation power-series fourier-series




summation power-series fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 23:45









Andrews

1,2962423




1,2962423










asked Mar 22 at 20:35









ParsevalParseval

3,0741719




3,0741719












  • $begingroup$
    There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
    $endgroup$
    – mrtaurho
    Mar 22 at 21:07










  • $begingroup$
    Fixed it just before you wrote that comment :P My apologies!
    $endgroup$
    – Parseval
    Mar 22 at 21:07












  • $begingroup$
    It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
    $endgroup$
    – mrtaurho
    Mar 22 at 21:08






  • 3




    $begingroup$
    Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
    $endgroup$
    – Parseval
    Mar 22 at 21:09








  • 2




    $begingroup$
    Try evaluating at $x=pi$ instead? That should cancel away the minuses.
    $endgroup$
    – Semiclassical
    Mar 22 at 21:20


















  • $begingroup$
    There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
    $endgroup$
    – mrtaurho
    Mar 22 at 21:07










  • $begingroup$
    Fixed it just before you wrote that comment :P My apologies!
    $endgroup$
    – Parseval
    Mar 22 at 21:07












  • $begingroup$
    It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
    $endgroup$
    – mrtaurho
    Mar 22 at 21:08






  • 3




    $begingroup$
    Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
    $endgroup$
    – Parseval
    Mar 22 at 21:09








  • 2




    $begingroup$
    Try evaluating at $x=pi$ instead? That should cancel away the minuses.
    $endgroup$
    – Semiclassical
    Mar 22 at 21:20
















$begingroup$
There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
$endgroup$
– mrtaurho
Mar 22 at 21:07




$begingroup$
There is a typo. The computed coefficient should be $$c_n=frac{sinh(pi^2)}pifrac{(-1)^color{red}{n}}{pi-in}$$ It is of minor matter since you used the right coefficient later on but I was pretty confused in the first place by this typo.
$endgroup$
– mrtaurho
Mar 22 at 21:07












$begingroup$
Fixed it just before you wrote that comment :P My apologies!
$endgroup$
– Parseval
Mar 22 at 21:07






$begingroup$
Fixed it just before you wrote that comment :P My apologies!
$endgroup$
– Parseval
Mar 22 at 21:07














$begingroup$
It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
$endgroup$
– mrtaurho
Mar 22 at 21:08




$begingroup$
It is still there ;) Within your computation of $c_n$ aswell as your definition of $f(x)$
$endgroup$
– mrtaurho
Mar 22 at 21:08




3




3




$begingroup$
Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
$endgroup$
– Parseval
Mar 22 at 21:09






$begingroup$
Yes, It was on two places! Only bad thing with copy paste is that it also applies to errors and typos.
$endgroup$
– Parseval
Mar 22 at 21:09






2




2




$begingroup$
Try evaluating at $x=pi$ instead? That should cancel away the minuses.
$endgroup$
– Semiclassical
Mar 22 at 21:20




$begingroup$
Try evaluating at $x=pi$ instead? That should cancel away the minuses.
$endgroup$
– Semiclassical
Mar 22 at 21:20










1 Answer
1






active

oldest

votes


















2












$begingroup$

Just to put in into an answer rather than into a comment I have to say that I am almost sure the only, and most likely intended way is to evaluate the series at $x=pi$, as pointed out by Semiclassical. This is yet alone the only possibilty to get rid of the $(-1)^n$ since this factor has nothing to do with the original function but more with underlying structure of the Fourier Coefficients.



A typical approach to find oscillating series, such as the one you derived from the non-oscillating one, is by playing the good old even odd cancelation game $($here similiar is done with the Riemann Zeta Function and its relative, the Dirichlet Eta Function$)$. However, applying this method to the given problem, and ignoring the constant factors for a moment, we get the following



$$underbrace{sum_{nge1}frac{(-1)^n}{pi^2+n^2}}_{=S_1}=underbrace{sum_{nge1}frac1{pi^2+n^2}}_{=S}-underbrace{2sum_{nge1}frac1{pi^2+(2n)^2}}_{=S_2}$$



Our main goal is to find $S$ wherease $S_1$ can be computed by your given formula. What remains to find is an expression for $S_2$. And from hereon we cannot get any further by only taking the Fourier Series Expansion of $f(x)=e^{pi x}$, at least as far as I can tell. One could use the expansion of $F(x)=e^{ax}$, for a positive real number $a$, instead and it is quite funny following this ansatz since it is way off what I think was intended by this question. Computing $c_n$ for $F(x)$ we obtain



$$F(x)=frac{sinh(api)}pisum_{ninmathbb Z}frac{color{red}{(-1)^n}}{a-in}e^{-inx}$$



Well, there is it again, the oscillating minus sign. However, note that for $a=fracpi2$ we are actually able to deduce an expression for $S_2$; by evaluating at $x=pi$ ... Moreover I am certainly sure that this is not the expected method, even though it works afterall.




Long story short: to deal with oscillating minus signs is easily done by splitting the sum up into even and odd parts but sometimes it makes the problem even more complicated than it actually is.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
    $endgroup$
    – Parseval
    Mar 22 at 22:29










  • $begingroup$
    @Parseval Always glad to help!
    $endgroup$
    – mrtaurho
    Mar 22 at 22:31












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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Just to put in into an answer rather than into a comment I have to say that I am almost sure the only, and most likely intended way is to evaluate the series at $x=pi$, as pointed out by Semiclassical. This is yet alone the only possibilty to get rid of the $(-1)^n$ since this factor has nothing to do with the original function but more with underlying structure of the Fourier Coefficients.



A typical approach to find oscillating series, such as the one you derived from the non-oscillating one, is by playing the good old even odd cancelation game $($here similiar is done with the Riemann Zeta Function and its relative, the Dirichlet Eta Function$)$. However, applying this method to the given problem, and ignoring the constant factors for a moment, we get the following



$$underbrace{sum_{nge1}frac{(-1)^n}{pi^2+n^2}}_{=S_1}=underbrace{sum_{nge1}frac1{pi^2+n^2}}_{=S}-underbrace{2sum_{nge1}frac1{pi^2+(2n)^2}}_{=S_2}$$



Our main goal is to find $S$ wherease $S_1$ can be computed by your given formula. What remains to find is an expression for $S_2$. And from hereon we cannot get any further by only taking the Fourier Series Expansion of $f(x)=e^{pi x}$, at least as far as I can tell. One could use the expansion of $F(x)=e^{ax}$, for a positive real number $a$, instead and it is quite funny following this ansatz since it is way off what I think was intended by this question. Computing $c_n$ for $F(x)$ we obtain



$$F(x)=frac{sinh(api)}pisum_{ninmathbb Z}frac{color{red}{(-1)^n}}{a-in}e^{-inx}$$



Well, there is it again, the oscillating minus sign. However, note that for $a=fracpi2$ we are actually able to deduce an expression for $S_2$; by evaluating at $x=pi$ ... Moreover I am certainly sure that this is not the expected method, even though it works afterall.




Long story short: to deal with oscillating minus signs is easily done by splitting the sum up into even and odd parts but sometimes it makes the problem even more complicated than it actually is.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
    $endgroup$
    – Parseval
    Mar 22 at 22:29










  • $begingroup$
    @Parseval Always glad to help!
    $endgroup$
    – mrtaurho
    Mar 22 at 22:31
















2












$begingroup$

Just to put in into an answer rather than into a comment I have to say that I am almost sure the only, and most likely intended way is to evaluate the series at $x=pi$, as pointed out by Semiclassical. This is yet alone the only possibilty to get rid of the $(-1)^n$ since this factor has nothing to do with the original function but more with underlying structure of the Fourier Coefficients.



A typical approach to find oscillating series, such as the one you derived from the non-oscillating one, is by playing the good old even odd cancelation game $($here similiar is done with the Riemann Zeta Function and its relative, the Dirichlet Eta Function$)$. However, applying this method to the given problem, and ignoring the constant factors for a moment, we get the following



$$underbrace{sum_{nge1}frac{(-1)^n}{pi^2+n^2}}_{=S_1}=underbrace{sum_{nge1}frac1{pi^2+n^2}}_{=S}-underbrace{2sum_{nge1}frac1{pi^2+(2n)^2}}_{=S_2}$$



Our main goal is to find $S$ wherease $S_1$ can be computed by your given formula. What remains to find is an expression for $S_2$. And from hereon we cannot get any further by only taking the Fourier Series Expansion of $f(x)=e^{pi x}$, at least as far as I can tell. One could use the expansion of $F(x)=e^{ax}$, for a positive real number $a$, instead and it is quite funny following this ansatz since it is way off what I think was intended by this question. Computing $c_n$ for $F(x)$ we obtain



$$F(x)=frac{sinh(api)}pisum_{ninmathbb Z}frac{color{red}{(-1)^n}}{a-in}e^{-inx}$$



Well, there is it again, the oscillating minus sign. However, note that for $a=fracpi2$ we are actually able to deduce an expression for $S_2$; by evaluating at $x=pi$ ... Moreover I am certainly sure that this is not the expected method, even though it works afterall.




Long story short: to deal with oscillating minus signs is easily done by splitting the sum up into even and odd parts but sometimes it makes the problem even more complicated than it actually is.







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
    $endgroup$
    – Parseval
    Mar 22 at 22:29










  • $begingroup$
    @Parseval Always glad to help!
    $endgroup$
    – mrtaurho
    Mar 22 at 22:31














2












2








2





$begingroup$

Just to put in into an answer rather than into a comment I have to say that I am almost sure the only, and most likely intended way is to evaluate the series at $x=pi$, as pointed out by Semiclassical. This is yet alone the only possibilty to get rid of the $(-1)^n$ since this factor has nothing to do with the original function but more with underlying structure of the Fourier Coefficients.



A typical approach to find oscillating series, such as the one you derived from the non-oscillating one, is by playing the good old even odd cancelation game $($here similiar is done with the Riemann Zeta Function and its relative, the Dirichlet Eta Function$)$. However, applying this method to the given problem, and ignoring the constant factors for a moment, we get the following



$$underbrace{sum_{nge1}frac{(-1)^n}{pi^2+n^2}}_{=S_1}=underbrace{sum_{nge1}frac1{pi^2+n^2}}_{=S}-underbrace{2sum_{nge1}frac1{pi^2+(2n)^2}}_{=S_2}$$



Our main goal is to find $S$ wherease $S_1$ can be computed by your given formula. What remains to find is an expression for $S_2$. And from hereon we cannot get any further by only taking the Fourier Series Expansion of $f(x)=e^{pi x}$, at least as far as I can tell. One could use the expansion of $F(x)=e^{ax}$, for a positive real number $a$, instead and it is quite funny following this ansatz since it is way off what I think was intended by this question. Computing $c_n$ for $F(x)$ we obtain



$$F(x)=frac{sinh(api)}pisum_{ninmathbb Z}frac{color{red}{(-1)^n}}{a-in}e^{-inx}$$



Well, there is it again, the oscillating minus sign. However, note that for $a=fracpi2$ we are actually able to deduce an expression for $S_2$; by evaluating at $x=pi$ ... Moreover I am certainly sure that this is not the expected method, even though it works afterall.




Long story short: to deal with oscillating minus signs is easily done by splitting the sum up into even and odd parts but sometimes it makes the problem even more complicated than it actually is.







share|cite|improve this answer









$endgroup$



Just to put in into an answer rather than into a comment I have to say that I am almost sure the only, and most likely intended way is to evaluate the series at $x=pi$, as pointed out by Semiclassical. This is yet alone the only possibilty to get rid of the $(-1)^n$ since this factor has nothing to do with the original function but more with underlying structure of the Fourier Coefficients.



A typical approach to find oscillating series, such as the one you derived from the non-oscillating one, is by playing the good old even odd cancelation game $($here similiar is done with the Riemann Zeta Function and its relative, the Dirichlet Eta Function$)$. However, applying this method to the given problem, and ignoring the constant factors for a moment, we get the following



$$underbrace{sum_{nge1}frac{(-1)^n}{pi^2+n^2}}_{=S_1}=underbrace{sum_{nge1}frac1{pi^2+n^2}}_{=S}-underbrace{2sum_{nge1}frac1{pi^2+(2n)^2}}_{=S_2}$$



Our main goal is to find $S$ wherease $S_1$ can be computed by your given formula. What remains to find is an expression for $S_2$. And from hereon we cannot get any further by only taking the Fourier Series Expansion of $f(x)=e^{pi x}$, at least as far as I can tell. One could use the expansion of $F(x)=e^{ax}$, for a positive real number $a$, instead and it is quite funny following this ansatz since it is way off what I think was intended by this question. Computing $c_n$ for $F(x)$ we obtain



$$F(x)=frac{sinh(api)}pisum_{ninmathbb Z}frac{color{red}{(-1)^n}}{a-in}e^{-inx}$$



Well, there is it again, the oscillating minus sign. However, note that for $a=fracpi2$ we are actually able to deduce an expression for $S_2$; by evaluating at $x=pi$ ... Moreover I am certainly sure that this is not the expected method, even though it works afterall.




Long story short: to deal with oscillating minus signs is easily done by splitting the sum up into even and odd parts but sometimes it makes the problem even more complicated than it actually is.








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answered Mar 22 at 22:00









mrtaurhomrtaurho

6,15271641




6,15271641








  • 1




    $begingroup$
    Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
    $endgroup$
    – Parseval
    Mar 22 at 22:29










  • $begingroup$
    @Parseval Always glad to help!
    $endgroup$
    – mrtaurho
    Mar 22 at 22:31














  • 1




    $begingroup$
    Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
    $endgroup$
    – Parseval
    Mar 22 at 22:29










  • $begingroup$
    @Parseval Always glad to help!
    $endgroup$
    – mrtaurho
    Mar 22 at 22:31








1




1




$begingroup$
Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
$endgroup$
– Parseval
Mar 22 at 22:29




$begingroup$
Thanks for this. You're absolutely right, it is not the expected method. Its much easier using Parsevals equation (not mine) or evaluating at $x=pi.$ However I ended up having a discussion with a classmate who thought it was impossible to do it by $x=0$ and now I can rub this answer in his face. Thanks!
$endgroup$
– Parseval
Mar 22 at 22:29












$begingroup$
@Parseval Always glad to help!
$endgroup$
– mrtaurho
Mar 22 at 22:31




$begingroup$
@Parseval Always glad to help!
$endgroup$
– mrtaurho
Mar 22 at 22:31


















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