I'm studying signal processing. I've found the associated Fourier Series for a message $m(t)$ = $t^2$ over the interval $[-1, 1]$ with period $T = 2$.
However, I'm then asked to verify that
$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}.$$
The particular Fourier series not withstanding, this seems to be a common question framing in my text, to follow finding a Fourier Series with this verification, and I'm not quite sure what it is asking, or if it relates to my previous result at all. I thought I might be able to find it by equating the result to the energy integral, i.e.
$E_g = \frac{1}{2\pi} \int m(t)^2 dt$
and therefore not need to use the Fourier Series I've found at all, but the result is just similar.
The Fourier series here is $$ x^2={1\over 3}+\sum_{n=1}^\infty {4(-1)^n\over n^2 \pi^2}\cos(n\pi x), \quad -1<x<1. $$
By Parseval's Identity, if $$ f(x)={1\over 2}a_0+\sum_{n=1}^\infty [a_n\cos(n\pi x/\ell)+b_n\sin(n\pi x/\ell)], \quad -\ell<x<\ell, $$ then $$ \|f\|^2=\ell\left({1\over 2}a_0^2+\sum_{n=1}^\infty (a_n^2+b_n^2)\right), $$ where the norm here is the $L^2$ norm on $-1<x<1$.
In this context, this leads to \begin{align} \|x^2\|^2&={1\over 2}\left({2\over 3}\right)^2+\sum_{n=1}^\infty \left({4(-1)^n\over n^2 \pi^2}\right)^2\\ \int_{-1}^1 (x^2)^2\,dx&={2\over 9}+\sum_{n=1}^\infty {16\over n^4 \pi^4}\\ {2\over 5}&={2\over 9}+\sum_{n=1}^\infty {16\over n^4 \pi^4}\\ {8\over 45}\cdot{\pi^4\over 16}&=\sum_{n=1}^\infty {1\over n^4}\\ {\pi^4\over 90}&=\sum_{n=1}^\infty {1\over n^4}. \end{align}