Find the closed form of $\sum_{k=1}^{n}\cos^{2m+1}\left(\frac{(2k-1)\pi}{2n+1}\right)$

Question

Add it: Let $m,n$ be postive integers. Find the closed form of

$$f=\sum_{k=1}^{n} \cos^{2m+1}{\left(\dfrac{(2k-1)\pi}{2n+1}\right)}$$

for $m, n \in \mathbb{N}^{+}$.

Maybe use Euler

\begin{align} 2\cos{x} &=e^{ix}+e^{-ix} \\ \dfrac{\pi}{2n+1} &=x \end{align}

then

\begin{align} f &= \dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}(w^{(2k-1)}+w^{-(2k-1)})^{2m+1} \\ &=\dfrac{1}{2^{2m+1}}\sum_{k=1}^{n}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{-(2k-1)i}w^{((2m+1)-i)(2k-1)}\\ &=\dfrac{1}{(2w)^{2m+1}}\sum_{i=0}^{2m+1}\binom{2m+1}{i}w^{2i}\sum_{k=1}^{n}w^{(4m-4i+2)k} \end{align}

where $w=e^{ix}$.

Answer 1

The function $\frac{(2n+1)/z}{z^{2n+1}-1}$ has residue $1$ at each $2n+1^\text{st}$ root of unity. So we need to account for the residues at $0$ and $\infty$.

$$\newcommand{\Res}{\operatorname*{Res}} \begin{align} \sum_{k=1}^n\cos^{2m+1}\left(\pi\,\frac{2k-1}{2n+1}\right) &=\frac12-\frac12\sum_{k=0}^{2n}\cos^{2m+1}\left(2\pi\,\frac{k}{2n+1}\right)\\ &=\frac12+\frac12\Res_{z=0}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\ &-\frac12\Res_{z=\infty}\left(\frac1{2^{2m+1}}\frac{(2n+1)/z}{z^{2n+1}-1}\left(z+\frac1z\right)^{2m+1}\right)\\ &=\frac12-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=0}^\infty z^{(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\ &-\frac{2n+1}{2^{2m+2}}\left[z^0\right]\left(\sum_{k=1}^\infty z^{-(2n+1)k}\left(z+\frac1z\right)^{2m+1}\right)\\ &=\frac12-\frac{2n+1}{2^{2m+1}}\sum_{k=0}^m\binom{2m+1}{k}[\,2n+1\,|\,2m+1-2k\,] \end{align} $$ where $[\dots]$ are Iverson Brackets.

Note that for $m\lt n$, the sum is $\frac12$ because $2n+1$ cannot divide $2m+1-2k$.