# Fourier series of |sin x|

1. Sep 24, 2008

### Narcol2000

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

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

The answer however gives

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

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: Sep 24, 2008
2. Sep 24, 2008

### mathman

You mad a mistake in calculating an. There is no n in the numerator. It should be (-1)n-1-1. Then all the terms for odd n are 0 and the even terms remain. (I suggest you redo the calculation).

3. Sep 24, 2008

### Narcol2000

yeah i found the mistake,
I get the numerator to be
$$2((-1)^{n-1} - 1)$$

which does lead to the right answer.

thx for your help.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook