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I'm trying to compute the Fourier series of $$f(x)=\frac{1}{2-\cos(x)}$$ on the interval $[0, 2\pi]$. It is an even function, so I need to determine the $a_n$ coefficients. They are given by the following integral:

$$a_n=\frac{1}{\pi}\int_{0}^{2\pi}\frac{\cos(nx/2)}{2-\cos(x)}dx.$$

However, for some reason I can't compute this integral and am thinking I might be doing something wrong. Can anyone help?

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You can use Chebyshev polynomials to write the numerator and denominator as polynomials in $\cos\frac x2$, and then use long polynomial division to separate the fraction into a polynomial in $\cos\frac x2$ (easily integrated) and a remainder term. The remainder will need additional tricks, but at least they will be the same for every $n$. –  Henning Makholm Dec 15 '12 at 18:26
    
That's well beyond the scope of the class I'm taking. Perhaps the question was given in error. Is there no elementary way to do this? Did I set up the integral correctly? –  Alex Dec 15 '12 at 18:27
    
Yes, the integral looks correct (modulo possible errors in the normalization factor, which I don't care enough to look up). What makes you think it is beyond the scope of your course. If "Chebyshev polynomials" looks scary, just think of it as applying the angle-addition formulas from basic trig to $\cos\frac x2$ repeatedly until you reach the desired multiple. –  Henning Makholm Dec 15 '12 at 18:35
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2 Answers

up vote 6 down vote accepted

Let $\alpha = 2-\sqrt{3}$. For $|z|=1$ we have the following Laurent expansion:

$$ \frac{1}{2-\frac{z+z^{-1}}{2}} =\frac{-2z}{z^2-4z+1} = \frac{\alpha}{\sqrt{3}z(1-\frac{\alpha}{z})} + \frac{1}{\sqrt{3}(1-\alpha z)}=\\ \frac{1}{\sqrt{3}}\left(\dotsc +\frac{\alpha^2}{z^2} +\frac{\alpha}{z}+1 + \alpha z + \alpha^2 z^2 + \dotsc \right) $$

Set $z = e^{ i x}$ to get the required Fourier transform:

$$ \frac{1}{2-\cos(x)} = \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}} \sum_{n=1}^\infty (2-\sqrt{3})^n \cos(n x). $$

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It gives you the requested Fourier expansion and a fortiori the value of all these integrals. –  WimC Dec 15 '12 at 19:16
    
@Alex Look closely at the result. It's exactly the cosine series you want. –  Potato Dec 15 '12 at 19:21
    
If you don't mind, can you explain the logic behind this a little bit? Why did you set alpha to that? What is the first line? The $\frac{1}{2-\dots}$. –  Alex Dec 15 '12 at 21:18
    
@Alex $\alpha$ and $\alpha^{-1}$ are the roots of $z^2-4z+1$. That's why $\alpha$ appears in the partial fractions decomposition. The first expression is based on a fact that can be applied in many cases, namely that $$\cos(x) = \frac{z+z^{-1}}{2}$$ where $z = e^{ix}$. The last equality is not always relevant for computations and simply $|z|=1$ suffices. At a higher level, this is an example that for analytic periodic functions Fourier transform, $z$-transform and Laurent expansions are all just different manifestations of the same underlying structure. –  WimC Dec 16 '12 at 7:35
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With a general $n$ in the integral formulas, Mathematica didn't finish the integration after more than a minute, which indicates something nontrivial is going on.

However, for particular $n$ values, it finishes very quickly.

f[x_] = 1/(2 - Cos[x]);
L = 2 Pi;
a[n_] := If[n == 0, 2/L*Integrate[f[x], {x, 0, L}], 
   2/L*Integrate[f[x] Cos[n Pi x/L], {x, 0, L}]]
Table[a[n], {n, 0, 10}]

Mathematica graphics

FC[x_, n_] := 1/2*a[0] + Sum[a[i] Cos[i Pi x/L], {i, 1, n}]; 
Plot[Evaluate[{f[x], FC[x, 5]}], {x, 0, L}, 
 PlotStyle -> {{Thick, Black}, {Thick, Blue}}, ImageSize -> 400]

Mathematica graphics

Hope that helps.

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