Trigonometric Identities and Fourier Series

Question

I have the series:

$$2+\sum_{m=1}^n 4(-1)^m\cos(m\pi x)$$

Here, $x\in (-1,1)$.

I need to show that this equals some fraction with only cosine terms and $n$ (no $m$).

Just looking for some pointers on where to start as I haven't the faintest clue!

Answer 1

Hint: $$ (-1)^m\cos(m\pi x)=\frac{e^{im\pi (x+1)}+e^{-im\pi (x+1)}}{2}=\frac{z^m+z^{-m}}{2} $$ where $z\neq 1\neq z^{-1}$.

Answer 2

Use

$$\cos{(m \pi x)} = \Re{[e^{i m \pi x}]}$$

and evaluate the subsequent geometrical series.

$$\begin{align}\sum_{m=1}^{n} (-1)^m e^{i m \pi x}&= \frac{e^{i \pi (x+1)}-e^{i \pi (n+1) (x+1)}}{1+e^{i \pi x}}\\\end{align}$$

Take real parts:

$$\implies \sum_{m=1}^{n} (-1)^m \cos{(\pi m x)} = -\frac{1}{2} + (-1)^n \frac{\cos{[(n+1/2) \pi x]}}{\cos{(\pi x/2)}}$$

$$2+\sum_{m=1}^n 4(-1)^m\cos(m\pi x) = (-1)^n \frac{4 \cos{[(n+1/2) \pi x]}}{\cos{(\pi x/2)}}$$