# Cotangent summation (proof)

How to sum up this thing, i tried it with complex number getting nowhere so please help me with this,$$\sum_{k=0}^{n-1}\cot\left(x+\frac{k\pi}{n}\right)=n\cot(nx)$$

Another chance is given by the residue theorem. $\cot(\pi z)$ is a meromorphic function with simple poles at the integers, with residue $\frac{1}{\pi}$. It follows that: $$\cot(\pi z) = \frac{1}{\pi}\sum_{m\in\mathbb{Z}}\frac{1}{z-m} = \frac{1}{\pi z}+\frac{2}{\pi}\sum_{m\geq 1}\frac{z}{z^2-m^2}.\tag{1}$$ On the other hand, due to $(1)$ both $$\sum_{k=0}^{n-1}\cot\left(x+\frac{\pi k}{n}\right)\quad\text{and}\quad n\cot(n x)$$ are meromorphic functions with the same set of singularities (simple poles) and the same residues at these poles. In particular, $n \cot(nx)$ is the logarithmic derivative of $\sin(nx)$ and $\cot\left(x+\frac{\pi k}{n}\right)$ is the logarithmic derivative of $\sin\left(x+\frac{\pi k}{n}\right)$. The identity: $$\prod_{k=0}^{n-1}\sin\left(x+\frac{\pi k}{n}\right) = 2^{1-n}\sin(nx) \tag{2}$$ is well-known and can be proved in a similar spirit.

• Very much like the look of this. A small question - couldn't the same comment on poles and residues also be made of, say $n\cot(nx)+e^x$? This reminds me of the Herglotz trick, which is actually used to establish the sum you mention. Surely once you have the sum expansion, the proof just reduces to checking that the series expansion of both sides would be identical? – πr8 Jan 10 '16 at 19:09
• @πr8: there is a point one needs to prove, i.e. that in the RHS of $(1)$ there are no contributes given by holomorphic functions like $e^z$. However, that is a simple consequence of the fact that $\cot$ is the logarithmic derivative of $\sin$, and the Weierstrass product of $\sin$ is well-known. – Jack D'Aurizio Jan 10 '16 at 20:10
• I was once able to prove the nontrivial equivalent of $(2)$ for $x=0,$ namely that $$\prod_{k=1}^{n-1}\sin\bigg(\frac kn~\pi\bigg)=\frac n{2^{n-1}},$$ but even that I forgot how to do. As for extending the formula to $x\neq0,$ I haven't got the slightest clue. – Lucian Jan 11 '16 at 5:49

$$\cot(nx)=\dfrac1{\tan(nx)}=\dfrac{1-\binom n2\tan^2x+\cdots}{\binom n1\tan x-\binom n3\tan^3x+\cdots}=\dfrac{\cot^nx-\binom n2\cot^{n-2}x+\cdots}{\binom n1\cot^{n-1}x-\binom n3\cot^{n-3}x+\cdots}(\text{multiplying the N & D by }\cot^nx)$$

If $\cot(nx)=\cot(nA)\iff\tan nx=\tan nA, nx=nA+m\pi$ where $m$ is any integer

$x=A+\dfrac{m\pi}n$ where $m\equiv0,1,\cdots,n-2,n-1\pmod n$

So, the roots of $$\cot^nx-\binom n1\cot nA\cdot\cot^{n-1}x-\binom n2\cot^{n-2}x+\cdots=0$$ are $\cot\left(A+\dfrac{m\pi}n\right)$ where $m\equiv0,1,\cdots,n-2,n-1\pmod n$

$$\implies\sum_{m=0}^{n-1}\cot\left(A+\dfrac{m\pi}n\right)=\binom n1\cot nA=n\cot nx$$

Hint: Consider the argument of $$\prod_{k=0}^{n-1}\left(1+i\cot\left(x+\frac{k\pi}{n}\right)\right)$$ in two different ways.

• please tell your method – Maverick Jan 17 '16 at 2:23