# Fourier series of |sin x|

I'm trying to find the fourier series of |sin x| between -pi and pi.

I've got it down to:

$$a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} |sin x| cos (nx) dx$$

which i wrote as:

$$a_n = \frac{2}{\pi}\int^{\pi}_0 sin x cos (nx) dx$$

writing

$$sin x cos (nx) = \frac{1}{2} (sin (n+1)x - sin (n-1)x)$$

I eventually get

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

giving

$$f(x) = \frac{2}{\pi} + \sum^{\infty}_{n=1} \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)} cos(nx)$$

$$f(x) = \frac{2}{\pi} + \frac{4}{\pi}\sum^{\infty}_{n=1} \frac{cos (2nx)}{4n^2 - 1}$$

I don't see how they arrive at this... if anyone can let me know where i've gone wrong or if i'm missing something :S

Last edited:

mathman
$$2((-1)^{n-1} - 1)$$