I am unable to get $2^{-|k|}$ as the Fourier coefficients of $\frac {3}{5-4\cos(x)}$ on $[0,2\pi]$ Kindly give me some clue as how to get this value $2^{-|k|}$. i am using the formula to find coefficients:
$f_k$ = $ \frac {1}{2\pi} $ $\int_{0}^{2\pi}\frac {3}{5-4\cos(x)}e^{-ikx} dx$ but since $\frac {3}{5-4\cos(x)}$ is infinitely differentiable so i am not getting the required one.
Thanks
Assuming we can represent
$$\frac{3}{5-4 \cos{x}} = \sum_{n=-\infty}^{\infty} c_n \, e^{i n x} $$
then $$c_n = \frac{3}{2 \pi} \int_{-\pi}^{\pi} dx \, \frac{e^{i n x}}{5-4 \cos{x}} $$
We may tackle this integral using, e.g., the residue theorem. Here, let $e^{i x}=z$; then the integral is
$$i \frac{3}{2 \pi} \oint_{|z|=1} dz \frac{z^n}{2 z^2-5 z+2} $$
Assuming $n \ge 0$, the poles of the integrand are at $z_{\pm} = (5 \pm 3)/4$. The only pole inside the unit circle is at $z_-=1/2$. By the residue theorem, the integral is
$$i \frac{3}{2 \pi} (i 2 \pi) \frac{(1/2)^n}{4 (1/2)-5} = 2^{-n} $$
It should be clear that the result should be the same for $n \lt 0$. Thus, for all $n$,