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I'm stuck on a seemingly simple problem: What is the fourier Series for $\sin^k(x/2)$? I've tried Mathematica with no luck.

Thanks for your help!

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1 Answer 1

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Use $$ \sin^n\theta = \frac{2}{2^n} \sum_{k=0}^{\frac{n-1}{2}} (-1)^{(\frac{n-1}{2}-k)} \binom{n}{k} \sin{((n-2k)\theta)} \tag{$n= {}$odd} $$ $$ \sin^n\theta = \frac{1}{2^n} \binom{n}{\frac{n}{2}} + \frac{2}{2^n} \sum_{k=0}^{\frac{n}{2}-1} (-1)^{(\frac{n}{2}-k)} \binom{n}{k} \cos{((n-2k)\theta)} \tag{$n={}$even} $$ and $$ \mathcal{F}_x\sin(kx/2)=i\sqrt{2\pi}(\delta(k-2\omega)+\delta(k+2\omega)). $$ Oh you just want the Fourier Series. So (odd) and (even) should be enough.

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