Question

See next comment on fourier series:

Answer 1

Find the Fourier series of abs[x]

\[f=\frac{a_0}{2}+\sum_{n=1}^{\infty}[a_ncos(nx)+b_nsin(nx)]\]\[a_0=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx\]\[a_n=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)cos(nx)dx\]\[b_n=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)sin(nx)dx\]

I found:\[a_0=0\]\[b_n=0\] Since they are odd functions integrated on symmetric bounds.

But for bn:
\[b_n=\frac{1}{\pi}\int\limits_{-\pi}^{\pi}|x|\cos(nx)dx=\frac{2}{\pi}(\frac{|x|}{n}\sin(nx)+\frac{1}{n^2}\cos(nx))|_{0}^{\pi}\]
I got simplified to:
\[\frac{2(-1)^n-2}{\pi n^2}\]

So:\[\sum_{n=1}^{\infty}\frac{2(-1)^n-2}{\pi n^2}\cos(nx)\]

But this is wrong, notice in attached plot the series is shifted down too far.

Answer 2

I solve it again.

Answer 3

Already saw my mistake hehe,
\[a_0=\pi\]

Answer 4

I calculated it as pi then forgot thinking it was 0...