Compute the Fourier series and prove the equality

Question

Compute the Fourier series for the function f(x) = |x| on the interval [-π, π]. Use the result and theorem 2.3

to show that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{π^2}{8}$$

My Solution: I found the Fourier series but I don't know how to prove the above equality. $$f(x)= \frac{π}{2} + \frac{2}{π}\sum_{n=1}^\infty \frac{(-1)^n-1}{n^2}cos(nx)$$

Answer 1

$f(x)=|x|\sim \frac{π}{2} + \frac{2}{π}\sum\limits_{n=1}^\infty \frac{(-1)^n-1}{n^2} \cos(nx)=\frac{π}{2}-\frac{4}{\pi}\left(\sum\limits_{k=0}^\infty \frac{1}{(2k+1)^2} \cos(nx)\right)$

what happens when you set $x=0$ ?

Answer 2

This answer is from the book 'Higher Mathematics':

The fuction $f(x)=|x|$ is even, that means that it's Fourier series does not have sines. The coeficient $a_0$ is equal to:

$a_0=\frac2{\pi}\int^{\pi}_0|x|dx=\frac2{\pi}\int^{\pi}_0 x dx=\pi$

For n\neq 0 we get:

$a_n=\frac2{\pi}\int^{\pi}_0 x \cos nx dx=\frac 2{\pi}x\frac{\sin nx}n\big|^{\pi}_0-\frac 2{n\pi} \int^{\pi}_0 \sin nx dx=2\frac{\cos nx -1}{n^2\pi}$

that is:

$a_{2a}=0, a_{2k-1}=-\frac 4{(2k-1)^2\pi}; (k=1, 2, 3,\cdot\cdot\cdot)$

Hence the Fourier series of the function $f(x)-|x|$ will be:

$\frac{\pi}2-\frac 4{\pi}\big(\frac{\cos x}{1^2}+\frac{\cos 3x}{3^2}+\frac{\cos 5x}{5^2}+\cdot\cdot\cdot\frac{\cos (2n-1)x}{(2n-1)^2}+\cdot\cdot\cdot\big)\space\space\space\space\space (1)$

Function $f(x)=|x|$ satisfies the hypotheses of theorem of the Fourier series of a continuous function.Hence the series (1) is everywhere convergent. Its sum is equal to $|x|$ for any value of x inside the interval$(-\pi, \pi)$. What is more , since the function $f(x)=|x|$ is even , the sum of its Fourier series is $f(x)$ at the extremities of the interval $(\pi, \pi)$ as well. Indeed , for an even function we have $F(-\pi)=f(\pi)$ such that the arithmetic mean between the values $f(-\pi)$ and $f(\pi)$ coincides with each one of these values. Thus we have:

$|x|=\frac{\pi}2-\frac 4{\pi}\big(\frac{\cos x}{1^2}+\frac{\cos 3x}{3^2}+\cdot\cdot\cdot \big); (-\pi \leq x \leq \pi)$$\space\space\space\space (2)$

In particular, substituting into (2) one of the values $x=\pm\pi$ or $x=0$, we find that:

$\frac 1{1^2}+\frac 1{3^2}+\frac 1{5^2}+\cdot\cdot\cdot=\frac {\pi^2} 8$