I have a dynamical system model whose equilibria depend on the solution of the following finite sum:
\begin{align} \sum_{j\neq i}^n\frac{\sin(\theta_j-\theta_i)}{\left(1-\cos(\theta_j-\theta_i)\right)^{3/2}}=0 \end{align}
I think I'm ok to suppose $\theta_i=0$ and run the sum from $j=1$ to $j=n-1$. Is this alright? Also, can I jettison the denominator altogether and focus on the sine terms? The solutions should be confined to the domain $(\theta_j-\theta_i)\in[0,2\pi]\:\forall\: j$. Im sure they are related to the roots of unity but haven't worked that out. Maybe there is some clever way to solve this equation once and for all?
\begin{align} \sum_{j=1}^{n-1}\sin\left(\frac{2\pi j}{n}\right) = 0 \end{align}
Does this exhaust the solutions? Assume that it does for a moment. This constraint on the angles in the system then leads finally to an additional trig sum that has some interesting properties. It appears that if the $l|k$ then the $k^{th}$ term contains the $l^{th}$ term with an additional factor. viz:
\begin{align} \frac{1}{2\sqrt{2}}\sum_{j=1}^{n-1}\frac{1}{\sqrt{1-\cos(\frac{2\pi j}{n})}} = \frac{1}{4}\sum_{j=1}^{n-1}\csc\left(\frac{\pi j}{n}\right) \end{align}
The first 12 terms:
\begin{align} &0,\:\frac{1}{4}\:,\frac{1}{\sqrt{3}}\:,\frac{1}{4}+\frac{1}{\sqrt{2}}\:,\sqrt{1+\frac{2}{\sqrt{5}}}\:,\frac{5}{4}+\frac{1}{\sqrt{3}},\\[2mm] &\frac{1}{\sqrt{2 \left(1+\sin \left(\frac{\pi}{14}\right)\right)}}+\frac{1}{\sqrt{2-2 \sin \left(\frac{3 \pi }{14}\right)}}+\frac{1}{\sqrt{2\left(1+\cos \left(\frac{\pi}{7}\right)\right)}},\\[2mm] &\frac{1}{4}+\frac{1}{\sqrt{2}}+\sqrt{2+\sqrt{2}},\\[2mm] &\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{2-2 \sin \left(\frac{\pi }{18}\right)}}+\frac{1}{\sqrt{2 \left(1+\cos \left(\frac{\pi}{9}\right)\right)}}+\frac{1}{2} \csc \left(\frac{\pi}{9}\right),\frac{1}{4}+\sqrt{5}+\sqrt{1+\frac{2}{\sqrt{5}}},\\[2mm] &\frac{1}{\sqrt{2 \left(1+\sin\left(\frac{\pi }{22}\right)\right)}}+\frac{1}{\sqrt{2-2 \sin \left(\frac{3 \pi}{22}\right)}}+\frac{1}{\sqrt{2 \left(1+\sin \left(\frac{5 \pi}{22}\right)\right)}}+\frac{1}{\sqrt{2 \left(1+\cos \left(\frac{\pi}{11}\right)\right)}}+\frac{1}{2} \csc \left(\frac{\pi}{11}\right),\\[2mm] &\frac{5}{4}+\sqrt{6}+\sqrt{\frac{5}{6}+\sqrt{\frac{2}{3}}} \end{align}
You see for example 6 is divisible by 2 and 3 and the 6th term is equal to the 2nd term plus the 3rd term plus 1. This sequence must be related the the symmetry point group $C_{nh}$ and its subgroups but how?