Let $f(x) := |\cos \frac x 2 |$ for $x \in \mathbb R$. Show by using Eulers formula that the $n$'th Fourier coefficient $c_n$.

Question

Let $f(x) := |\cos \frac x 2 |$ for $x \in \mathbb R$.

Show by using Eulers formula that the $n$'th Fourier coefficient $$c_n = \frac 1 {2\pi} \frac {(-1)^{n-1}} {n^2-\frac 1 4}$$ with respect to the ortonormal basis $\{e^{inx}\}$ if $c_n = (f,e^{inx}) = \frac 1 {2\pi} \int^{\pi}_{-\pi} f(x) e^{-inx} dx$.

I know Eulers formula for $\cos$ is $\cos(x) = \frac {e^{ix} + e^{-ix}} 2$. But expanding $\cos \frac x 2$ using this I don't get the proper formula for $c_n$.

Can someone help me out ?

Answer 1

$$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}|\cos \frac{x}{2}|e^{-inx}dx=\frac{1}{\pi}\int_{0}^{\pi}\cos \frac{x}{2}\cos nx dx\\=\frac{1}{2\pi}\int_{0}^{\pi}[\cos(n+1/2)x+\cos (n-1/2)x]dx\\=\frac{1}{2\pi}\left[\frac{\sin(n\pi+\pi/2)}{n+1/2}-\frac{\sin(n\pi-\pi/2)}{n-1/2}\right]\\=\frac{1}{2\pi}(-1)^{n-1}\left[\frac{1}{n-1/2}-\frac{1}{n+1/2}\right]\\=\frac{1}{2\pi}\frac{(-1)^{n-1}}{n^2-\frac{1}{4}}$$ Euler's formula comes into action at the last step of the first line in the answer above.