2

I am looking for a nice(er) proof of the following identity: $$\sum_{k=1}^{n-1} (-1)^k \cos^{2j}\frac{\pi k}{2n} = -\frac{1}{2}, ~ 0<j<n,~j,n\in \mathbb{Z}^+.$$ I have been able to prove it using the following strategy. Express the cosine in terms of the complex exponential, obtaining the summand $(-1)^k\left(1/2+e^{i\pi k/n}/4+e^{-i\pi k/n}/4\right)^j$. Use the trinomial expansion to expand the bracket, then sum over $k$ which is now a geometric series. In particular, I find $$-\sum_{p+q\leq j} {j\choose p~q} \frac{1}{2^{j-p-q}}\frac{1}{4^p}\frac{1}{4^q} \frac{(-1)^{n+p-q}+e^{i(p-q)\pi/n}}{1+e^{i(p-q)\pi/n}}=-\sum_{p+q\leq j} {j\choose p~q} \frac{1}{2^{j-p-q}}\frac{1}{4^p}\frac{1}{4^q}\times\begin{cases}1, ~ n+p-q=\text{even}\\0, ~ n+p-q=\text{odd}\end{cases},$$ where the odd case follows as the result would be imaginary. The trinomial then reconstitutes to $$-(1/2+1/4+1/4)^j/2 \pm (1/2-1/4-1/4)^j/2=-1/2,$$ where the choice of sign depends on the parity of $n$.

Is there some deeper interpretation of this identity, and/or a slicker way to prove it? There is some clear overlap with Sum of Fourth Powers of cosine series has closed form solution., epecially the answer given there by Claude Leibovici.

lucubrative
  • 21

1 Answers1

2

I would do a series of simplifications, to reduce to sums-of-roots-of-unity. First of all, $$\sum_{k=1}^{n-1}(-1)^k\cos^{2j}\frac{\pi k}{2n}=S\implies\sum_{k=0}^{2n-1}(-1)^k\cos^{2j}\frac{\pi k}{2n}=2S+1$$ (this is $\sum_{k=0}^{2n-1}a_k$ with $a_k=a_{2n-k}$, $a_n=0$ and $a_0=1$). Next, to get rid of $(-1)^k$, use $$\sum_{k=0}^{2n-1}(-1)^k a_k=2\sum_{k=0}^{n-1}a_{2k}-\sum_{k=0}^{2n-1}a_k;$$ thus, $2S+1=2S_{n,j}-S_{2n,j}$ where $S_{n,j}=\sum_{k=0}^{n-1}\cos^{2j}(k\pi/n)$.

Now introduce the $n$-th root of unity $\omega=\exp(2i\pi/n)$. By the binomial formula, $$2^{2j}S_{n,j}=\sum_{k=0}^{n-1}\omega^{-kj}(1+\omega^k)^{2j}=\sum_{r=0}^{2j}\binom{2j}{r}\sum_{k=0}^{n-1}\omega^{k(r-j)}.$$ The inner sum is $0$ unless $n\mid(r-j)$. If $0<j<n$, the latter holds if and only if $r=j$.

Finally, we get $S_{n,j}=\frac{n}{2^{2j}}\binom{2j}{j}$ for $0<j<n$, and $2S+1=0$, hence $S=-1/2$.

metamorphy
  • 39,111