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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?

Viuツ
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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 $n\neq 2k.$

Thomas Andrews
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    A little more direct than re-indexing and simplifying the sum is just replacing $z$ with $z^{-1}$ (so that the only check is instead $z^{-1} + (z^{-1})^{-1} = z + z^{-1}$). – Brian Moehring Jun 17 '23 at 22:59
  • As I was writing down the solution now, I was wondering what you mean with "adding"? How you get from: ${z^{n-2k}}$ to your solution? – Viuツ Jun 21 '23 at 12:36
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    $$2\cos^n x=\cos^n x+\cos^n x=\frac1{2^n}\sum_{k=0}^n\binom nk z^{n-2k}+\frac1{2^n}\sum_{k=0}^{n}\binom nk z^{-(n-2k)}$$ Where the first term is the term you got, while the second term is the $j$ case, with $j$ renamed $k.$ @Viuツ – Thomas Andrews Jun 21 '23 at 14:06