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Is there a faster approach for finding the Fourier series of $$\sin(x/2)~,\cos(x/2)~~~\text{etc}$$ other than the usual approach?

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Fourier series generally apply to period function a period $2\pi$. These are not quite so periodic. You could define though a function equal to these on the interval $[0,2\pi]$, and define for periodicity outside that interval. –  Thomas Andrews Mar 7 '13 at 2:21
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The usual approach is pretty fast, isn't it? You have to calculate integrals like $\int\sin nx\sin(x/2)\,dx$, and you can use trig identities to transform that product of sines into a simple sum, which sum is easy to integrate. –  Gerry Myerson Mar 7 '13 at 2:29
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1 Answer

I'll give a non-standard derivation for $\cos (x/2)$ on $[-\pi,\pi]$. Faster or not, you tell.

The function being even, we work with cosines. Observe that if $$f(x)\sim \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx\tag1$$ then $$f''(x)\sim -\sum_{n=1}^\infty n^2 a_n \cos nx\tag2$$ I am writing $\sim $ instead of $=$ here to avoid getting arrested by the convergence police. Naturally, we want to plug (1) and (2) into $f''= (-1/4)f$ and equate coefficients. This does not work.

The problem is, $f''$ is not exactly $(-1/4)f$. The values of $f'$ at the endpoints don't match, and this discontinuity contributes a point mass to $f''$. The correct formula is $$f''=(-1/4)f+\delta_{\pi}+\delta_{-\pi}\tag3$$ It's easy to find the coefficients for the delta function: $$\delta_\pi +\delta_{-\pi}= \frac{c_0}{2}+ \sum_{n=1}^\infty c_n\cos nx, \quad c_n= (-1)^n\frac{2}{\pi}\tag4$$ because "integrating" against a delta function amounts to evaluation. Now plug into (3). $$-n^2 a_n = (-1/4)a_n + c_n\tag5$$
which yields $$ a_n = \frac{-c_n}{n^2-1/4}\tag6$$ with $c_n$ as in (4).

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+1 for the convergence police. –  nbubis Mar 7 '13 at 6:48
    
Convergence police is a good term. Also, why not substitute the number $\tfrac{1}{2}$ in $\cos{(x/2)}$ with a general $k$. –  NiftyKitty95 Mar 7 '13 at 8:36
    
@Nick One could do that, although the easy case of integer $k$ must be treated separately. I "leave this as an exercise for the interested reader." –  user53153 Mar 7 '13 at 12:44
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