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I need to find the complex Fourier series of this function, and I'm having problems calculating these integers:

$$|a|<1$$

$$x\in [-\pi,\pi]$$

$$f(x)=\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}$$

$$a_0=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}dx$$

$$b_n=\int_{-\pi}^{\pi}\frac{1-a\cos(x)}{1-2a\cos(x)+a^2}\sin(nx)dx$$

Cameron Buie
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Deiota
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1 Answers1

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Note: The following answers the question before it was edited and changed.

Let $z = e^{ix}$ (so $|z|=1$) then

$$ f(x) = \frac{1-a^2}{(1-a z)(1-a z^{-1})} = \frac{a}{z-a} + \frac{1}{1- a z}. $$

As a function of $z$ this has the following Laurent expansion on the unit circle:

$$ \frac{a}{z}\sum_{k=0}^{\infty} \frac{a^k}{z^k} + \sum_{k=0}^{\infty} a^k z^k = 1 + \sum_{k=1}^{\infty}a^k(z^k + z^{-k}) = 1 + 2 \sum_{k=1}^{\infty} a^k \cos(k x) $$

This gives the Fouries series of $f$ without calculating any integral.

The same can be done for the latest form of $f$.

$$ f(x) = \frac{2-a (z + z^{-1})}{2(1-a z)(1 - a z^{-1})} = \frac{a}{2 (z-a)}+\frac{1}{2 (1 - a z)}+\frac{1}{2} $$

The Fourier series is now apparent given the previous result:

$$ f(x) = 1 + \sum_{k=1}^{\infty} a^k \cos(k x) $$

WimC
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  • That's pretty cool but I don't understand how you did it... – Deiota Apr 27 '13 at 13:55
  • What is the Laurent expansion of this. $$ f(x) = \frac{2-a (z + z^{-1})}{2(1-a z)(1 - a z^{-1})} = \frac{a}{2 (z-a)}+\frac{1}{2 (1 - a z)}+\frac{1}{2}$$ – Deiota Apr 30 '13 at 12:40
  • @Amccds Do you know the geometric series? Since $|z|=1$ and $|a|<1$ that suffices to expand this function as Laurent (power) series in $z$. – WimC Apr 30 '13 at 13:32
  • I'm asking as you told me to the first exercise, to get the fourier complex serie... – Deiota Apr 30 '13 at 13:39
  • @winC can you please do the Laurent expansion on the unit circle like you did for the first case so then we get to the complex fourier serie? – Deiota Apr 30 '13 at 19:35
  • @Amccds Divide the first series by $2$ and add $\tfrac{1}{2}$ to get the second series. – WimC May 01 '13 at 05:01
  • @WinC The right result my book says the sum should start from 0 , the second result should be "0+sum from 0 too..." – Deiota May 01 '13 at 10:45
  • Solved thanks a lot – Deiota May 02 '13 at 16:52
  • Wow using the unit circle like that. – mathreadler Aug 15 '16 at 14:59