My task is to find the real Fourier series for $ f(x)=\cos(x)^n$.
What I did so far: I set $z = e^{ix}$ and $cos(x) = \frac{z+z^{-1}}{2}$
With that I came to to following: $$cos(x)^n = \frac{1}{2^n}(z+z^{-1})^n$$
Now I only need to find a series representation of this. Which turned out harder than I thought. I came to something like this: $$\frac{1}{2^n}(z+z^{-1})^n=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}z^{n-k}z^{-k}$$
But now I struggle to simplify this to some. Can someone help me simplify this to some sort of cosine equation? Or am I missing a point?
Replacing $k$ with $n-j,$ you also get:
$$\cos^n(x)=\frac1{2^n}\sum_{j=0}^{n}\binom n{n-j}z^{n-2(n-j)}$$
But $\binom n{n-j}=\binom nj,$ and $n-2(n-j)=2j-n=-(n-2j).$
So adding, we get:
$$2\cos^n x=\frac1{2^n}\sum_{k=0}^{n}\binom nk \left(z^{n-2k}+z^{-(n-2k)}\right)=\frac1{2^n}\sum_{k=0}^{n}\binom nk 2\cos((n-2k)x)$$
Now, there are different cases, whether $n=2k$ or $n\neq 2k.$