find $\sum_{k=1}^{13}\sin^{2n}\left({\frac{k\pi}{13}}\right)$?

Question

Prove that $\sum_{k=1}^{13}\sin^{10}\left({\frac{k\pi}{13}}\right)=\frac{819}{256}$.

In general, how to find $\sum_{k=1}^{13}\sin^{2n}\left({\frac{k\pi}{13}}\right)$?

Thanks in advance.

Answer 1

Suppose we seek to evaluate

$$S = \sum_{k=1}^{m-1} \sin^{2q}(k\pi/m) = \sum_{k=0}^{m-1} \sin^{2q}(2\pi k/2/m).$$

Introducing $\zeta_k = \exp(2\pi i k/2/m)$ (root of unity) we get

$$S = \sum_{k=0}^{m-1} \frac{1}{(2i)^{2q}} (\zeta_k - 1/\zeta_k)^{2q}.$$

We also have

$$\sum_{k=m}^{2m-1} \frac{1}{(2i)^{2q}} (\zeta_k - 1/\zeta_k)^{2q} \\ = \sum_{k=0}^{m-1} \frac{1}{(2i)^{2q}} (\zeta_k \exp(2\pi i m/2/m) - 1/\zeta_k/\exp(2\pi i m/2/m))^{2q} \\ = \sum_{k=0}^{m-1} \frac{1}{(2i)^{2q}} (-\zeta_k + 1/\zeta_k)^{2q} \\ = \sum_{k=0}^{m-1} \frac{1}{(2i)^{2q}} (\zeta_k - 1/\zeta_k)^{2q} = S.$$

We conclude that

$$S = \frac{1}{2} \sum_{k=0}^{2m-1} \frac{1}{(2i)^{2q}} (\zeta_k - 1/\zeta_k)^{2q}.$$

Introducing

$$f(z) = \frac{(-1)^q}{2^{2q+1}} \left(z-\frac{1}{z}\right)^{2q} \frac{2mz^{2m-1}}{z^{2m}-1} \\ = \frac{(-1)^q}{2^{2q+1}} \frac{(z^2-1)^{2q}}{z^{2q}} \frac{2mz^{2m-1}}{z^{2m}-1}$$

we then have

$$S = \sum_{k=0}^{2m-1} \mathrm{Res}_{z=\zeta_k} f(z).$$

Observe that the term $(z^2-1)^{2q}$ cancels the poles at $\pm 1$ produced by $z^{2m}-1$ which however is perfectly acceptable as they correspond to $\zeta_0 = 1$ and $\zeta_m = -1$ where $\zeta_k-1/\zeta_k$ is zero as well.

Residues sum to zero so we obtain

$$S + \mathrm{Res}_{z=0} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$

Now for the residue at zero we see that when $2q-1\lt 2m-1$ or $q\lt m$ the residue is zero. Otherwise we get

$$\frac{(-1)^q}{2^{2q+1}} [z^{2q-2m}] (z^2-1)^{2q} \frac{2m}{z^{2m}-1} \\ = \frac{(-1)^q}{2^{2q+1}} [z^{2q}] (z^2-1)^{2q} \frac{2m z^{2m}}{z^{2m}-1} \\ = -2m \frac{(-1)^q}{2^{2q+1}} \sum_{p=0}^q {2q\choose p} (-1)^{2q-p} [z^{2q-2p}] \frac{z^{2m}}{1-z^{2m}}.$$

We must have $p=q-lm$ where $l\ge 1.$ This yields

$$-2m \frac{1}{2^{2q+1}} \sum_{l=1}^{\lfloor q/m\rfloor} {2q\choose q-lm} (-1)^{lm}.$$

This is correct even when $q\lt m.$

Continuing with the residue at infinity we find

$$\mathrm{Res}_{z=\infty} f(z) = - \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z) \\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{(-1)^q}{2^{2q+1}} \frac{(1/z^2-1)^{2q}}{1/z^{2q}} \frac{2m/z^{2m-1}}{1/z^{2m}-1} \\ = - \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{(-1)^q}{2^{2q+1}} \frac{(1-z^2)^{2q}}{z^{2q}} \frac{2mz}{1-z^{2m}} \\ = - \mathrm{Res}_{z=0} \frac{(-1)^q}{2^{2q+1}} \frac{(z^2-1)^{2q}}{z^{2q+1}} \frac{2m}{1-z^{2m}}.$$

This is the same as the first residue at zero except now $l$ starts at $l=0$ and we obtain

$$-2m \frac{1}{2^{2q+1}} \sum_{l=0}^{\lfloor q/m\rfloor} {2q\choose q-lm} (-1)^{lm}.$$

Joining the two pieces we finally have

$$\bbox[5px,border:2px solid #00A000]{ m \frac{1}{2^{2q}} {2q\choose q} + m \frac{1}{2^{2q-1}} \sum_{l=1}^{\lfloor q/m\rfloor} {2q\choose q-lm} (-1)^{lm}.}$$