- #1
Narcol2000
- 25
- 0
I'm trying to find the Fourier series of |sin x| between -pi and pi.
I've got it down to:
[tex]
a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} |sin x| cos (nx) dx
[/tex]
which i wrote as:
[tex]
a_n = \frac{2}{\pi}\int^{\pi}_0 sin x cos (nx) dx
[/tex]
writing
[tex]
sin x cos (nx) = \frac{1}{2} (sin (n+1)x - sin (n-1)x)
[/tex]
I eventually get
[tex]
a_n = \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)}
[/tex]
giving
[tex]
f(x) = \frac{2}{\pi} + \sum^{\infty}_{n=1} \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)} cos(nx)
[/tex]
The answer however gives
[tex]
f(x) = \frac{2}{\pi} + \frac{4}{\pi}\sum^{\infty}_{n=1} \frac{cos (2nx)}{4n^2 - 1}
[/tex]
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
I've got it down to:
[tex]
a_n = \frac{1}{\pi} \int^{\pi}_{-\pi} |sin x| cos (nx) dx
[/tex]
which i wrote as:
[tex]
a_n = \frac{2}{\pi}\int^{\pi}_0 sin x cos (nx) dx
[/tex]
writing
[tex]
sin x cos (nx) = \frac{1}{2} (sin (n+1)x - sin (n-1)x)
[/tex]
I eventually get
[tex]
a_n = \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)}
[/tex]
giving
[tex]
f(x) = \frac{2}{\pi} + \sum^{\infty}_{n=1} \frac{2(n(-1)^n - 1)}{\pi(n^2 - 1)} cos(nx)
[/tex]
The answer however gives
[tex]
f(x) = \frac{2}{\pi} + \frac{4}{\pi}\sum^{\infty}_{n=1} \frac{cos (2nx)}{4n^2 - 1}
[/tex]
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
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