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Is it possible to compute the Fourier Series of $f(x)$? It seems that it would admit a Fourier cosine representation because it is even (by looking at the graph), but the periodicity is a little strange. Is this even possible?

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Yes, try this on WolframAlpha. You might want to compare the result to $Sin(x)$ by removing the Abs. Regards. – Amzoti Dec 15 '12 at 1:36
On what interval do you want it? Perhaps $\[0,2\pi]\,$ ? This is an even function, so its Fourier series shouldn't be that hard to evaluate... – DonAntonio Dec 15 '12 at 1:37
I think $[0, 2\pi]$. – Alex Dec 15 '12 at 1:40
It seems that it'll just be the Fourier series of $sin(x)$ on $[0,2\pi]$ and the Fourier series of $-sin(x)$ on $[-2\pi,0]$. Can anyone confirm this? – Alex Dec 15 '12 at 3:12
Why would you do that? The function is even so it's the same in both intervals you mention: no need to work twice – DonAntonio Dec 15 '12 at 20:28
up vote 5 down vote accepted

$$c_0:=\frac{1}{\pi}\int_0^{2\pi}\sin x\,dx=\left.-\frac{1}{\pi}\,(\cos x)\right|_0^{2\pi}=0$$

$$\forall\,\,n\neq \pm \,1\,\,,\,c_n:=\frac{1}{2\pi}\int_0^{2\pi}\sin x\,e^{-inx}\,dx=\frac{1}{4\pi i}\int_0^{2\pi}\left(e^{ix}-e^{-ix}\right)e^{-inx}\,dx=$$

$$=\frac{1}{4\pi i}\int_0^{2\pi}\left(e^{ix(1-n)}-e^{-ix(1+n)}\right)dx=\frac{1}{4\pi i}\left[\frac{1}{i(1-n)}e^{ix(1-n)}+\frac{1}{i(1+n)}e^{ix(1+n)}\right]_0^{2\pi}=$$

$$\frac{1}{4\pi i}\left[\frac{1}{i(1-n)}\left(e^{2\pi i(1-n)}-1\right)+\frac{1}{i(1+n)}\left(e^{2\pi i(1+n)}-1\right)\right]=0$$

$$c_{-1}=\frac{1}{2\pi}\int_0^{2\pi}\sin x\,e^{ix}\,dx=\frac{1}{4\pi i}\int_0^{2\pi}\left(e^{2ix}-1\right)dx=\frac{1}{4\pi i}\left(-2\pi\right)=\frac{i}{2}$$

$$c_1=\frac{1}{2\pi}\int_0^{2\pi}\sin x\,e^{-ix}\,dx=\frac{1}{4\pi i}\int_0^{2\pi}\left(1-e^{-2ix}\right)dx=\frac{1}{4\pi i}2\pi=-\frac{i}{2}$$

Thus, in a pretty expected and even boring way, we get:

$$\sin x=c_{-1}e^{-ix}+c_1e^{ix}=\frac{i}{2}\left(e^{-ix}-e^{ix}\right)=\frac{e^{ix}-e^{-ix}}{2i}$$

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