How do I start solving this question, what are the steps?
a) Find the Fourier series for the function $f(x) = x^4$ on the interval $[−π, π]$.
b) Hence prove that $$1-\frac{1}{2^4}+\frac{1}{3^4}-\frac{1}{4^4}+.....=\frac{7\pi ^4}{720}$$
using
$$\sum _{n=1}^{\infty }\left(\frac{\left(-1\right)^{n-1}}{n^2}\right)=\frac{\pi ^2}{12}$$
On
We need to calculate the coefficients $a_n$ of the cosine series. To do so, we will determine the integral
$$\begin{align} I(a)&=\int_{-\pi}^\pi x^4\cos (ax)\,dx\\\\ &=\text{Re}\left(\frac{d^4}{da^4}\int_{-\pi}^\pi e^{iax}\,dx\right)\\\\ &=\text{Re}\left(\frac{d^4}{da^4}\frac{2\sin (\pi a)}{a}\right)\\\\ &=\frac{d^4}{da^4}\frac{2\sin (\pi a)}{a}\\\\ &=\frac{2\sin(\pi a)}{a^5}\left((\pi a)^4-12(\pi a)^2+24\right)+\frac{2\cos(\pi a)}{a^5}\left(4(\pi a)^3-24(\pi a)\right) \end{align}$$
Setting $a=n$ gives
$$\begin{align} \int_{-\pi}^\pi x^4\cos (nx)\,dx&=(-1)^n\frac{8(\pi n)^3-48(\pi n)}{n^5}\\\\ &=(-1)^n \frac{8\pi^3}{n^2}-(-1)^n \frac{48\pi}{n^4} \end{align}$$
Therefore, we have
$$x^4=\pi^4/5+\sum_{n=1}^\infty \left((-1)^n \frac{8\pi^2}{n^2}-(-1)^n \frac{48}{n^4}\right)\,\cos(nx) \tag 1$$
Setting $x=0$ in $(1)$ yields
$$0=\pi^4/5+\sum_{n=1}^\infty \left((-1)^n \frac{8\pi^2}{n^2}-(-1)^n \frac{48}{n^4}\right) \tag 2$$
whereupon solving $(2)$ for the series of interest reveals that
$$\begin{align} \sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n^4}&=\frac1{48}\left(-\pi^4/5+\sum_{n=1}^\infty (-1)^{n-1} \frac{8\pi^2}{n^2}\right) \\\\ &=\frac1{48}\left(-\pi^4/5+8\pi^4/12\right) \\\\ &=\frac{7\pi^4}{20} \end{align}$$
as was to be shown!
The function $f(x)=x^4$ is even, so you only need cosines. Fourier coefficient is $\int_{-\pi}^\pi x^4\cos(nx)dx$, which can be computed by integration by parts. Once you are done, just compute $f(1)$.