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How to prove this identity for $2\le n\in\Bbb N^+, a , b > 0$? $$ \sum_{k=0}^{n-1}\frac1{a^2\cos^2\left(\frac kn⋅π\right)+b^2\sin^2\left(\frac kn⋅π\right)}=\frac{n}{ab}⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n} $$

First I apply double angle formula $$\frac1{a^2\cos^2\left(\frac kn⋅π\right)+b^2\sin^2\left(\frac kn⋅π\right)}=\frac1{\frac{a^2-b^2}2\cos(2k\pi/n)+\frac{a^2+b^2}2}$$ Similar to Finite Series $\sum_{k=1}^{n-1}\frac1{1-\cos(\frac{2k\pi}{n})}$ and Complex Partial Fraction Decomposition

I get this identity from Riemann sum of this integral: \begin{align*} \int_0^π\frac{\mathrm{{}d}t}{a^2\cos^2t+b^2\sin^2t}&=\lim_{n→∞}\fracπn\sum_{k=0}^{n-1}\frac1{a^2\cos^2\left(\frac kn⋅π\right)+b^2\sin^2\left(\frac kn⋅π\right)}\\ &=\lim_{n→∞}\fracπ{ab}⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}\\ &=\fracπ{a b} \end{align*} It looks like the sum of gravitational potential from points $(a\cos(\frac kn\cdot\pi),b\sin(\frac kn\cdot\pi))$ on an ellipse.

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1 Answers1

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Let $ζ=e^{2i\pi/n}$, the identity is equivalent to $$\sum_{k=0}^{n-1}\frac1{\frac{a^2-b^2}2\frac{\zeta^k+\zeta^{-k}}2+\frac{a^2+b^2}2} =\frac{n}{ab}⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}$$ Partial fractions, $$\frac1{\frac{a^2-b^2}2\frac{z+z^{-1}}2+\frac{a^2+b^2}2}=\frac{{b - a}}{{a b \left((a + b) z + a - b\right)}} + \frac{{a + b}}{{a b \left((a - b) z^{-1} + a + b\right)}} $$ multiply the identity by $ab$, let $z=\zeta^k$, $$\sum_{k=0}^{n-1}\left(\frac{{b - a}}{{(a + b) z + a - b}} + \frac{{a + b}}{{(a - b) z^{-1}+ a + b}}\right)=n⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}$$ since $\{z^{-1}:k=0,\dots,n-1\}=\{z:k=0,\dots,n-1\}$, we can replace $z^{-1}$ with $z$ in the second term $$\tag1\label1\sum_{k=0}^{n-1}\left(\frac{{b - a}}{{(a + b) z + a - b}} + \frac{{a + b}}{{(a - b) z+ a + b}}\right)=n⋅\frac{(a+b)^n+(b-a)^n}{(a+b)^n-(b-a)^n}$$ Let $w=\frac1{\frac{a + b}{b - a}z- 1}$ then $(\frac1w+1)^n=(\frac{a + b}{b - a}z)^n=(\frac{a + b}{b - a})^n\implies(1+w)^n-w^n(\frac{a + b}{b - a})^n=0$,

so for $k=0,\dots,n-1$ all $w$ are the roots of the polynomial $f(x)=(1+x)^n-x^n(\frac{a + b}{b - a})^n$

so $$ (1+x)^n-x^n(\frac{a + b}{b - a})^n=\left(1-(\frac{a + b}{b - a})^n\right)\prod_{k=0}^{n-1}(x-w) $$ equating coefficient of $x^{n-1}$ $$\sum_{k=0}^{n-1} w=\frac n{(\frac{a + b}{b - a})^n-1}$$ substitute $w=\frac{{b - a}}{{(a + b) z + a - b}}$ $$\tag2\label2 \sum_{k=0}^{n-1}\frac{b-a}{(a + b)z+a-b}=\frac n{(\frac{a + b}{b - a})^n-1}$$ replace $a$ with $-a$ and multiply by $-1$ $$\tag3\label3 \sum_{k=0}^{n-1}\frac{a+b}{(a-b)z+a+b}=\frac n{1-(\frac{-a + b}{a+b})^n}$$ adding \eqref{2} and \eqref{3} $$ \sum_{k=0}^{n-1}\left(\frac{b-a}{(a + b)z+a-b}+\frac{a+b}{(a-b)z+a+b}\right)=\frac n{(\frac{a + b}{b - a})^n-1}+\frac n{1-(\frac{-a + b}{a+b})^n}$$ This simplifies to \eqref{1}

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