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I've been learning Fourier sequence for a while and now I'm stuck. I have a function $$f(x)=x^2,\quad x\in[-\pi, \pi ]$$ and once evaluating this to Fourier we get: $$f(x)=\frac{\pi^2}{3} + 4\sum_{n=1}^{\infty} \frac{(-1)^{-n}}{n^2}\,\cos{(nx)}$$

Now it says: with the aid of the Fourier sequence (above) calculate the following sequence: $$\sum_{n=1}^{\infty} \frac{1}{n^4}$$

How do I tackle such problem, I've goggled everything but haven't found anything similar.
Any help on this?

Jack D'Aurizio
  • 353,855

1 Answers1

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By applying Parseval's theorem we get

$$ \int_{-\pi}^{\pi}(x^2)^2\,dx =2\pi\cdot\frac{\pi^4}{9}+16\pi \sum_{n\geq 1}\frac{1}{n^4}\tag{1} $$ but the LHS of $(1)$ equals $\frac{2\pi^5}{5}$, hence by rearranging we get: $$ \zeta(4)=\sum_{n\geq 1}\frac{1}{n^4}=\frac{1}{16\pi}\left(\frac{2\pi^5}{5}-\frac{2\pi^5}{9}\right)=\color{red}{\frac{\pi^4}{90}}.\tag{2}$$ Have also a look at this historical thread.

Jack D'Aurizio
  • 353,855
  • Thank you Sir, the answer that you got is listed within my answers! Can you suggest any other series apart from 1/n^4 to practice, I see that this is one is very popular. – Eugen Sunic Sep 08 '16 at 15:28
  • @wesewx, see http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90?noredirect=1&lq=1 – xpaul Sep 08 '16 at 21:05