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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?

Américo Tavares
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  • @VFG: Did you calculate the Fourier coefficients? Can you see any similarity to $\zeta(2)$? How can you manipulate the Fourier series? – AD - Stop Putin - Aug 27 '10 at 20:41
  • AD: Yes. One of the difficulties I have is dealing with a complex rather than a trigonometric Fourier series. – Américo Tavares Aug 27 '10 at 20:56
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    I believe you are supposed to be using Parseval's identity. – Qiaochu Yuan Aug 27 '10 at 21:16
  • Qiaochu Yuan: That is a good hint, thanks. Anyhow could you please elaborate a little bit? – Américo Tavares Aug 27 '10 at 21:37
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    I 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 – Riley E Jun 02 '11 at 12:55
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    @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). – Riley E Jun 02 '11 at 13:04
  • @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

    $$\pi ^{2p}=\frac{2\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }a_{n,2p}\cdot \cos n\pi .$$

    ...

     – Américo Tavares Jun 02 '11 at 14:58
  • See also http://math.stackexchange.com/questions/374221/fourier-series-of-fx-x and http://math.stackexchange.com/a/324158 – Martin Sleziak Nov 11 '15 at 14:30
  • @MartinSleziak Thanks for both links, in particular the first one! – Américo Tavares Nov 11 '15 at 21:31

3 Answers3

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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.

Aryabhata
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  • @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. – Aryabhata Aug 27 '10 at 23:29
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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.

night owl
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  • It seems strange that you have included Aryabhatta's answer word for word within your own. – Jonas Meyer Jun 02 '11 at 07:36
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    @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. – night owl Jun 02 '11 at 07:42
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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.