I know how to prove $\zeta (2)=\pi ^{2}/6$ by using the trigonometric Fourier series expansion of $x^{2}/4$. How can one prove the same result using the complex Fourier series of $f(x)=x$ for $0\leq x\leq 1$? Any suggestion?
Hint on how to prove $\zeta ( 2) =\pi ^{2}/6$ using the complex Fourier series of $f(x)=x$
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0@VFG: Did you calculate the Fourier coefficients? Can you see any similarity to $\zeta(2)$? How can you manipulate the Fourier series? – 2010-08-27
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0AD: Yes. One of the difficulties I have is dealing with a complex rather than a trigonometric Fourier series. – 2010-08-27
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2I believe you are supposed to be using Parseval's identity. – 2010-08-27
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0Qiaochu Yuan: That is a good hint, thanks. Anyhow could you please elaborate a little bit? – 2010-08-27
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1I just want to point out this process can be generalized to give you $\zeta(2n)$ for all $n\in\mathbb{N}$. In this case dealing with $[0,1]$ is simpler than $[-\pi,\pi]$, so $f(z)=\sum_{n=-\infty}^{\infty}c_ne^{2\pi inz}$ where $c_n=\int_0^1 f(z)e^{-2\pi inz}dz$. The coefficients of $f(x)=x^{2n}$ will give you $\zeta(2n)$. However, if you instead use the Bernoulli polynomials, the integration by parts turns out much nicer (the $uv|_0^1$ terms all go away). http://en.wikipedia.org/wiki/Bernoulli_polynomials – 2011-06-02
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1@Americo Tavares: The Fourier coefficients of $f(x)=x$ actually only involve $\frac{1}{n}$, so it doesn't quite give you $\zeta(2)$. However, Parseval's Theorem says $\int_0^1 |f(z)|^2dz=\sum_{n=-\infty}^{\infty}|c_n|^2$. The left hand side is easy enough to evaluate, and the right hand side will give you the $\frac{1}{n^2}$ you need (note that $c_0=0$, so the sum doesn't blow up on you). – 2011-06-02
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0@Riley E: Thanks a lot for your explanations. A few weeks ago I computed the trigonometric Fourier series for $f(x)=x^{2p},x\in \left[ -\pi ,\pi \right] $ $$\frac{2\pi ^{2p}}{2p+1}+\sum_{n=1}^{\infty }a_{n,2p}\cdot \cos nx,$$ where $$a_{n,2p}=\frac{2}{\pi }\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.$$ For $x=\pi $, it gives $$\pi ^{2p}=\frac{2\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }a_{n,2p}\cdot \cos n\pi .$$ ... – 2011-06-02
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0See also http://math.stackexchange.com/questions/374221/fourier-series-of-fx-x and http://math.stackexchange.com/a/324158 – 2015-11-11
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0@MartinSleziak Thanks for both links, in particular the first one! – 2015-11-11
3 Answers
Use the definition:
Say $f$ is defined on $[-\pi, \pi]$.
If $f(z) = \sum_{-\infty}^{\infty} {c_{n} e^{inz}}$
then
$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}{f(z)e^{-inz}} dz$
If you put $f(z) = z$, can you work out what $c_{n}$ turns out to be?
To integrate, you can try integration by parts.
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0@VFG: I am curious though, was this homework? It is ok to ask I believe, as long as you say it is homework and show what you have tried so far. – 2010-08-27
Extending off from Aryabhatta answer:
For our situation: We have $f(x)=x ~~~{\text{ for }} 0\leq x\leq 1$
$2L=1,\Rightarrow L=\frac{1}{2}$
So restating we have:
$f(x) = \displaystyle\sum_{n=-\infty}^{\infty} {c_{n} e^{inx}}, \text{ where }c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{f(x)e^{-inx}} ~\mathrm{d}x,~~~~~~n=0,~\pm 1,~\pm 2, \cdots~ $
$ \Rightarrow~~ c_{n} = \displaystyle\frac{1}{2\pi}\int_{-\frac{1}{2}}^{\frac{1}{2}}{xe^{-inx}}~\mathrm{d}x $
After integrating the complex Fourier coefficient we see that we get the following:
$\Rightarrow~~~~\displaystyle c_n=i\left(\frac{\cos(\frac{n}{2})}{2\pi n}-\frac{\sin(\frac{n}{2})}{\pi n^2}\right),~~~\text{for }n \in \mathbb{R}$
Lastly plugging back $c_n$ into $f(x)$ we then get our desired result for $n=0,~\pm 1,~\pm 2, \cdots~$.
Please update if you see any mistakes with any of the work. It has been quite some time since I work with Fourier Series and went off from my head. Feel free to edit mistakes as necessary if willing.
Thanks.
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0It seems strange that you have included Aryabhatta's answer word for word within your own. – 2011-06-02
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0@Jonas: His really was not needed to be restated, I was just putting so people could follow along without having to scroll back and forth between the two. His was just for some generic interval as I was trying to ask the question at hand. Could remove if seems strange. That is why I added the note above before proceeding. – 2011-06-02
This is not related but you would like to see this article: A Short Proof of ζ (2) = π2/6 T.H. Marshall American Math monthly April 2010.