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Define a function $f(x) =(2\cos(\pi x))^{10} $$f\in L^{1}$ so it's one-period. I would like to calculate the Fourier coefficient $\hat{f}(2)$.

So we get $\displaystyle\hat{f}(n)=\int_{0}^{1}e^{-2\pi inx}(2\cos(\pi x))^{10}dx$ , $n\in \mathbb{Z}$

I know that we can write $\cos(\pi x)^{10}$ = $((e^{i\pi x}+e^{-i\pi x})/2)^{10}$ and with the binomial formula we get

$$\sum_{k=0}^{10}\binom{10}{k}(e^{i\pi x})^{10-k}(e^{-i\pi x})^{k}$$

So the Fourier coefficient $\displaystyle \hat{f}(n) = \int_{0}^{1}e^{-2\pi inx}2^{10}\sum_{k=0}^{10}\binom{10}{k}(e^{i\pi x})^{10-k}(e^{-i\pi x})^{k}dx$ , where $n\in \mathbb{Z}$

I have no idea how to easily integrate this one. I appreciate any help.

karvens
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lisa
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  • Maybe you should try $e^{-2πinx}=\cos(2nπx)-i \sin(2nπx)$. As you only need the secound coefficient, you can do n=2 and try to use trigonometric identities. You could then try cos(t)=t and do the integral. – mlainz Oct 04 '14 at 15:27
  • According to mathematica, the result is not very beautiful: $\frac{\left(-4096+53760 \pi ^2-231168 \pi ^4+375040 \pi ^6-193536 \pi ^8+14175 \pi ^{10}\right) \sin (4)}{128 \left(-64+880 \pi ^2-4092 \pi ^4+7645 \pi ^6-5269 \pi ^8+900 \pi ^{10}\right)}$ – mlainz Oct 04 '14 at 15:36
  • @mlainz: Integrate[Exp[-4Pi I x](2Cos[Pi x])^10,{x,0,1}] returns $120$ – robjohn Oct 05 '14 at 03:27

1 Answers1

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$$ \begin{align} \int_0^1e^{-4\pi ix}(2\cos(\pi x))^{10}\,\mathrm{d}x &=\int_0^1e^{-4\pi ix}\left(e^{i\pi x}+e^{-i\pi x}\right)^{10}\,\mathrm{d}x\\ &=\int_0^1e^{-4\pi ix}\binom{10}{3}e^{7\pi ix}e^{-3\pi ix}\,\mathrm{d}x\\ &=\binom{10}{3} \end{align} $$ All the other terms in the binomial expansion of $\left(e^{i\pi x}+e^{-i\pi x}\right)^{10}$ are multiples of $e^{2k\pi ix}$ for $k\ne2$, and therefore, integrate to $0$ against $e^{-4\pi ix}$.

robjohn
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