Calculating the Fourier Series of |sin x|

[Narcol2000]

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

I've got it down to:

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

which i wrote as:

<br /> a_n = \frac{2}{\pi}\int^{\pi}_0 sin x cos (nx) dx<br />

writing

<br /> sin x cos (nx) = \frac{1}{2} (sin (n+1)x - sin (n-1)x)<br />

I eventually get

<br /> a_n = \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)}<br />

giving

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

The answer however gives

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

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

 

[mathman]

You mad a mistake in calculating a_n. 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).

 

[Narcol2000]

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

which does lead to the right answer.

thx for your help.