Riley Quinn

2022-03-16

Can someone prove that $\sum _{i=1}^{89}{\mathrm{sin}}^{2n}\left(\frac{\pi }{180}i\right)$ is a dyadic rational for all positive integers n, or find counterexample?

IdodaHekbed7mx

$\sum _{j=0}^{k-1}{\mathrm{sin}}^{2n}\left(\frac{j\pi }{k}\right)=\sum _{j=0}^{k-1}{\left(\frac{{e}^{\frac{ji\pi }{k}}-{e}^{-\frac{ji\pi }{k}}}{2i}\right)}^{2n}$
$=\frac{1}{{\left(2i\right)}^{2n}}\sum _{j=0}^{k-1}\sum _{m=0}^{2n}\left(\begin{array}{c}2n\\ m\end{array}\right){\left({e}^{\frac{ji\pi }{k}}\right)}^{m}{\left(-{e}^{-\frac{ji\pi }{k}}\right)}^{2n-m}$
$=\frac{1}{{\left(-4\right)}^{n}}\sum _{m=0}^{2n}\left(\begin{array}{c}2n\\ m\end{array}\right){\left(-1\right)}^{m}\sum _{j=0}^{k-1}{e}^{\frac{j\left(2m-2n\right)i\pi }{k}}$
That last sum (over j) is a sum of roots of unity - specifically, if $d=GCD\left(m-n,k\right)$, the last sum is the sum of the $\frac{k}{d}$ roots of unity d times. So the sum is equal to 0, except when $d=k$, in which case all the terms are 1 and the sum is k. This implies that the summand of the sum over m is an integer, so the entire expression is an integer divided by ${4}^{n}$.
In the case that $n, we get the simple expression
$\frac{1}{{\left(-4\right)}^{n}}\left(\begin{array}{c}2n\\ n\end{array}\right){\left(-1\right)}^{n}k=\frac{1}{{4}^{n}}\left(\begin{array}{c}2n\\ n\end{array}\right)k$
For the original problem I posed, we have
$2\sum _{j=1}^{89}{\mathrm{sin}}^{2n}\left(\frac{\pi i}{180}\right)+1$
$={\mathrm{sin}}^{2n}\left(0\right)+\sum _{j=1}^{89}{\mathrm{sin}}^{2n}\left(\frac{\pi i}{180}\right)+{\mathrm{sin}}^{2n}\left(\frac{90\pi }{180}\right)+\sum _{j=91}^{179}{\mathrm{sin}}^{2n}\left(\frac{\pi i}{180}\right)$
$=\sum _{j=0}^{179}{\mathrm{sin}}^{2n}\left(\frac{\pi i}{180}\right)$
and we have proven that the last sum is a dyadic ratioanl, so $\sum _{j=1}^{89}{\mathrm{sin}}^{2n}\left(\frac{\pi i}{180}\right)$ is a dyadic rational as well.

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