7

Motivation:
I was initially asked to compute the sum
$$ \sum_{k=1}^n \cos(k) $$ I was able to find a closed-form expression for it. This led me to a more general question: is it possible to find a closed form for
$$ \sum_{k=1}^n \cos^r(k) $$ where $r,n \in \mathbb{N}$?

My attempt:
I attempted to derive a closed-form expression for the case where $r$ is even. I suspect that the odd case to be solvable using a similar method (if it worked for the evens), but I haven’t worked it out yet. Here's what I did for the even case, assuming $r$ is even and that $r/2=m$:

$$\sum\limits_{k=1}^n\cos^{2m}(k)$$ $$=2^{-2m}\sum\limits_{k=1}^n\left(e^{ik}+e^{ik}\right)^{2m}$$ $$=2^{-2m}\sum\limits_{k=1}^n\sum_{j=0}^{2m}\dbinom{2m}{j}e^{i(2m-2j)k}$$ $$=2^{-2m}\sum\limits_{k=1}^n\left(\dbinom{2m}{m}+2\sum_{j=0}^{m-1}\dbinom{2m}{j}\cos((2m-2j)k)\right)$$ $$=\frac{\dbinom{2m}{m}}{2^m}n+2^{-{2m-1}}\sum_{j=0}^{m-1}\dbinom{2m}{j}\sum\limits_{k=1}^n\cos((2m-2j)k)$$ $$=\frac{\dbinom{2m}{m}}{2^{2m}}n+ 2^{-2m}\sum_{j=0}^{m-1}\dbinom{2m}{j}\left(1+\frac{\sin((2n+1)(m-j)}{\sin(m-j)}\right)$$ $$= \frac{\dbinom{2m}{m}}{2^{2m}}n+2^{-2m} \left(\frac{2^m-\binom{2m}{m}}{2}\right)+ 2^{-2m}\sum_{j=0}^{m-1}\dbinom{2m}{j}\left(\frac{\sin((2n+1)(m-j))}{\sin(m-j)}\right)$$ $$=\frac{1}{2}+\frac{\dbinom{2m}{m}}{2^{2m}}(n-1/2)+2^{-2m}\sum_{j=0}^{m-1}\dbinom{2m}{j}\left(\frac{\sin((2n+1)(m-j))}{\sin(m-j)}\right)$$

I couldn't find a closed form for the last sum $$ \sum_{j=0}^{m-1} \binom{2m}{j} \cdot \frac{\sin\left((2n+1)(m-j)\right)}{\sin(m-j)} $$

Rócherz
  • 4,241
pie
  • 8,483
  • How do you go from $\cos^{2m}k$ to $(e^{ik}e^{-ik})^m/2^m?$ It should have $2m$ in the exponents. – Thomas Andrews May 02 '25 at 23:21
  • @ThomasAndrews it is a typo – pie May 02 '25 at 23:22
  • 2
    Also, it is generally bad when taking a case $m$ even and call it $2m,$ since it confuses matters. It is apt to get errors like this. Pick a new variable , $m=2p$ or something. – Thomas Andrews May 02 '25 at 23:25
  • @ThomasAndrews I changed m tp r – pie May 02 '25 at 23:32
  • $$\sum_{k=1}^{n} \cos(kx) = {\csc\left(\frac{x}{2}\right) \sin\left(\frac{nx}{2}\right) \cos\left(\frac{1}{2} (n + 1) x\right)}{}\ \sum_{k=1}^{n} \cos^2(kx) = \frac{1}{2} \left( \csc(x) \sin(nx) \cos\left(\frac{(n + 1)x}{2}\right) + n \right) $$ – Mason May 03 '25 at 00:38
  • A question about summations of periodic functions should include a Fourier series tag. – Mason May 03 '25 at 00:41
  • I think all the tools are here – Mason May 03 '25 at 00:59

1 Answers1

14

$$\begin{align} \sum_{k=1}^{n}\cos^r(k) &=\sum_{k=1}^{n}\left(\frac{e^{ik}+e^{-ik}}{2}\right)^r\\ &=\frac{1}{2^r}\sum_{k=1}^{n}\sum_{m=0}^{r}\binom{r}{m}e^{mik}e^{(m-r)ik}\\ &=\frac{1}{2^r}\sum_{m=0}^{r}\binom{r}{m}\sum_{k=1}^{n}e^{(2m-r)ik}\\ &=\begin{cases}\frac{1}{2^r}\sum\limits_{m=0}^{r}\binom{r}{m}e^{(2m-r)i}\cdot\frac{1-e^{(2m-r)in}}{1-e^{(2m-r)i}}&\text{if}\,r\,\text{is odd}\\\frac{1}{2^r}\sum\limits_{\substack{m=0\\2m\neq r}}^{r}\binom{r}{m}e^{(2m-r)i}\cdot\frac{1-e^{(2m-r)in}}{1-e^{(2m-r)i}}+\frac{n}{2^r}\binom{r}{r/2}&\text{otherwise}\end{cases} \end{align}$$ Now we take real part of the summand: $$\Re\left(e^{(2m-r)i}\cdot\frac{1-e^{(2m-r)in}}{1-e^{(2m-r)i}}\right)=\csc\frac{2m-r}{2}\sin\frac{n(2m-r)}{2}\cos\frac{(n+1)(2m-r)}{2}$$ Thus $$\boxed{\sum_{k=1}^{n}\cos^r(k)=\begin{cases}\frac{1}{2^r}\sum\limits_{m=0}^{r}\binom{r}{m}\csc\frac{2m-r}{2}\sin\frac{n(2m-r)}{2}\cos\frac{(n+1)(2m-r)}{2}&\text{if}\,r\,\text{is odd}\\\frac{1}{2^r}\sum\limits_{\substack{m=0\\2m\neq r}}^{r}\binom{r}{m}\csc\frac{2m-r}{2}\sin\frac{n(2m-r)}{2}\cos\frac{(n+1)(2m-r)}{2}+\frac{n}{2^r}\binom{r}{r/2}&\text{otherwise}\end{cases}}$$

Thinh Dinh
  • 8,233