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

Question

Consider the following function $f(x) = \frac{1-a^2}{1+a^2-2a \cos x}$. I'm looking for its represention of this kind: $\sum_{\mathbb{Z}} a_k e^{ikx}$

So we can rewrite function as $f(x) = \frac{1}{1+a^2}\frac{1-a^2}{1-\frac{2a \cos x}{1+a^2}}$ And consider it is as a geometric sequence, with $q=\frac{2a \cos x}{1+a^2} = \frac{a (e^{ix}+e^{-ix})}{1+a^2}$

So the series probally look like: $\sum_{\mathbb{Z}} \left(\frac{a}{1+a^2}\right)^{|k|} e^{ikx}$. Is it correct? Does it work for complex a?

Answer 1

You are dealing with the well-known Fourier series of the Poisson kernel. For any $0\leq r <1$,

$$ P_r(\theta) = \sum_{n\in\mathbb{Z}} r^{|n|}e^{in\theta} = \frac{1-r^2}{1-2r\cos\theta +r^2} = \operatorname{Re}\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right)$$ and if $r$ does not belong to the $[0,1)$ interval it is enough to apply the above result to $-r$ or $\pm\frac{1}{r}$.
For any complex $r$ such that $|r|<1$, $\sum_{n\in\mathbb{N}}r^n e^{in\theta}$ is simply a geometric series.

Answer 2

$$ \begin{align} f(x)&=\frac{1-\alpha}{1-2\alpha cosx+\alpha^2}\\ &=\frac{2-\alpha(e^{ix}+e^{-ix})}{1+\alpha^2-\alpha(e^{ix}+e^{-ix})}-1\\&=\frac{(1-\alpha e^{ix})+(1-\alpha e^{-ix})}{(1-\alpha e^{ix})(1-\alpha e^{-ix})}-1\\ &=\frac{1}{(1-\alpha e^{ix})}+\frac{1}{(1-\alpha e^{-ix})}-1\\ &=\sum_{k=0}^\infty{\alpha^ke^{ixk}}+\sum_{k=0}^\infty{\alpha^ke^{-ixk}}-1\\ &=-1+2\sum_{k=0}^\infty{\alpha^kcos(kx)} \end{align} $$