In working on a physics problem, I've come across sums of trigonometric functions of the following form:
$$S(n,L) = -4^{n}+2^{2n}\sum_{k=0}^{L}\left[\cos\left(\frac{k\pi}{L+1}\right)\right]^{2n}$$
where $L$ and $n$ are positive integers. These sums aren't particularly recognizable to me, so if an expert is familiar with sums of these types it would be helpful to get their comments. In any case here is what I've tried to see how they behave:
To begin understanding these sums, I've tried evaluating them as a function of $n$ at fixed $L$. The first few results are $S(n,1) = 0, S(n,2) = 2, S(n,3) = 2^{n+1}$. For larger $L$, numerical experimentation indicates that the result is significantly more complicated. Can anyone shed light on possible method of attacking these kinds of sums?
Let $\theta=\pi/(L+1)$. Using the fact that $$ \cos(k\theta)=\frac{e^{ik\theta}+e^{-ik\theta}}{2}, $$ we get $$ \begin{align} 2^{2n}\sum_{k=0}^L[\cos(k\theta)]^{2n} &=2^{2n}\sum_{k=0}^L\left(\frac{e^{ik\theta}+e^{-ik\theta}}{2}\right)^{2n} \\&=2^{2n}\sum_{k=0}^L\left(\frac{e^{ik\theta}+e^{-ik\theta}}{2}\right)^{2n} \\&=\sum_{k=0}^L \sum_{r=0}^{2n} \binom{2n}r (e^{ik\theta})^r(e^{-ik\theta})^{2n-r} \\&=\sum_{r=0}^{2n} \binom{2n}r\sum_{k=0}^L ( e^{2i\theta(r-n)} )^k. \end{align} $$ Let us now zoom in on the inner sum $\sum_{k=0}^L ( e^{2i\theta(r-n)} )^k,$ which is equal to $$ \sum_{k=0}^L ( e^{i2\pi (r-n)/(L+1)} )^k. $$ This is a finite geometric series of the form $1 + \alpha + \alpha^2 + \dots + \alpha^L$, where $\alpha = e^{i2\pi (r-n)/(L+1)}$. If $\alpha\neq 1$, then the sum is equal to $$\frac{1-\alpha^{L+1}}{1-\alpha}$$ However, since $\alpha^{L+1}=e^{i2\pi(r-n)}=1$, the geometric series sums to zero in the case $\alpha\neq 1$.
All that remains are the terms for which $\alpha=1$. This occurs if and only if $(r-n)/(L+1)$ is an integer. Clearly, if $\alpha=1$, then the geometric series evaluates to $1+\alpha+\dots+\alpha^L=1+1+\dots+1=L+1$.
To summarize, we see that the only time the sum $\sum_{k=0}^L ( e^{i2\pi (r-n)/(L+1)} )^k$ is nonzero is when $r=n+m(L+1)$ for some integer $m\in \mathbb Z$, in which case the sum is equal to $L+1$. It follows that the entire summation over $r$ from $0$ to $2n$ is equal to $$ \sum_{r=0}^{2n} \binom{2n}r\sum_{k=0}^L ( e^{2i\theta(r-n)} )^k =(L+1)\sum_{m=-\lfloor 2n/(L+1)\rfloor}^{\lfloor 2n/(L+1)\rfloor} \binom{2n}{n+m(L+1)} $$ In words, you sum is equal to $(L+1)$ times a certain sum of binomial coefficients. Specifically, you look at the $(2n)^\text{th}$ row of Pascal's triangle, and you add up every $(L+1)^\text{st}$ term, starting with the central term of $\binom{2n}n$.