The value of \zeta(4) with fourier series

Question

Given a function as follows :

$f(x) = \pi^2 - x ^2$ on $|x|<\pi$ and $f(x+2\pi)=f(x)$

I did find its Fourier expansion

$f(x)=\frac{2}{3}\pi^2$+$\sum_{n \geq 1}\frac{4}{n^2}(-1)^{n+1}cosx$

And by putting $x=\pi$, I got the zeta of 2 , $\zeta(2) = \sum_{n \geq 1}\frac{1}{n^2} = \frac{\pi^2}{6}$

I d like to find the zeta of 4,$\zeta(4) = \frac{\pi^4}{90}$ with fourier expansion. If possible, what is the method ? If this function $f(x)$ is Not proper, What shlould i use another function? Please, advise me.

I do guess that it may be used Square Error

Answer 1

Using Parseval's identity, one has: $$\|f\|^2=\sum_{n=-\infty}^{+\infty}|c_n(f)|^2.$$ Besides, one has: $$\|f\|^2=\frac{1}{2\pi}\int_{-\pi}^\pi |f(t)|^2\mathrm{d}t=\frac{8\pi^4}{15}.$$ Moreover, one has: $$\sum_{n=-\infty}^{+\infty}|c_n(f)|^2=\frac{4\pi^4}{9}+8\zeta(4).$$ Finally, one has: $$\zeta(4)=\frac{\pi^4}{90}.$$

Remark. I used that $c_0(f)=a_0(f)$ and $c_{\pm n}(f)=\frac{1}{2}(a_n(f)\pm ib_n(f)).$

Answer 2

$f(x) = \pi^4-x^4$, $f(x+2 \pi)=f(x)$

$$f(x) = \sum_{n=0}^{\infty} a_n \cos{n x} $$

$$a_0 = \frac1{2 \pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) = \frac{4 \pi^4}{5}$$

$$a_n = \frac1{\pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) \cos{n x} = (-1)^{n+1} \frac{8 (n^2 \pi^2-6)}{n^4}$$

Thus,

$$f(x) = \frac{4 \pi^4}{5} - 8 \sum_{n=1}^{\infty} (-1)^{n} \frac{n^2 \pi^2-6}{n^4} \cos{n x}$$

$$f(\pi) = 0 \implies \frac{4 \pi^4}{5} - 8 \pi^2 \sum_{n=1}^{\infty} \frac1{n^2} + 48 \sum_{n=1}^{\infty} \frac1{n^4} = 0$$

Using

$$\sum_{n=1}^{\infty} \frac1{n^2} = \frac{\pi^2}{6} $$

We may deduce that

$$\sum_{n=1}^{\infty} \frac1{n^4} = \frac{\pi^4}{90} $$