FS1

From Exampleproblems

Find the Fourier series for $|x|\,$ , $- \pi < x < \pi \,$ .

Following the rules from the link above,

$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(nx) + b_n \sin(nx))\,$

$a_0 = \frac{1}{\pi} \int_{-\pi}^\pi |x|\, dx = \frac{2}{\pi}\int_0^\pi x \,dx = \pi\,$

$a_n = \frac{1}{\pi} \int_{-\pi}^\pi |x| \cos(nx)\, dx = \frac{2}{\pi}\int_0^\pi x \cos(nx) \,dx\,$

$= \frac{2}{\pi}\left(\left[\frac{x \sin(nx)}{n}\right]_0^\pi - \int_0^\pi \frac{\sin(nx)}{n}\, dx \right) = \frac{2}{\pi} \left[\frac{\cos(nx)}{n^2}\right]_0^\pi\,$

$= \frac{2 (-1)^n}{\pi n^2} - \frac{2}{\pi n^2} = \frac{ 2((-1)^n-1) }{\pi n^2}\,$ .

$b_n = \frac{1}{\pi} \int_{-\pi}^\pi |x| \sin(nx)\,dx = \frac{1}{\pi}\int_{-\pi}^{0} -x \sin(nx)\, dx + \frac{1}{\pi} \int_{0}^{\pi} x \sin(nx)\, dx\,$

$= \frac{1}{\pi}\left[\left[\frac{x\cos(nx)}{n}\right]_{-\pi}^0 - \int_{-\pi}^0 \frac{\cos(nx)}{n}\,dx\right] +\frac{1}{n}\left[\left[\frac{-x \cos(nx)}{n}\right]_0^\pi + \int_0^\pi \frac{\cos(nx)}{n}\,dx\right]\,$

$= \frac{1}{\pi} \left[ \frac{\pi (-1)^n}{n} - \left[\frac{\sin(nx)}{n^2}\right]_{-\pi}^0\right] + \frac{1}{\pi}\left[\frac{-\pi(-1)^n}{n} + \left[\frac{\sin(nx)}{n^2}\right]_0^\pi\right]\,$

$= \frac{1}{\pi} \frac{\pi (-1)^n}{n} + \frac{1}{\pi}\frac{-\pi(-1)^n}{n} = 0\,$ .

So,

$|x| = \frac{\pi}{2} + \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n-1}{n^2} \cos(nx) = \frac{\pi}{2} + \frac{2}{\pi} \sum_{n=0}^\infty \frac{-2}{(2n+1)^2} \cos\left((2n+1)x\right)\,$