MHB Simplifying the Summation Identity Using Complex Numbers

[anemone]

Hi,

I have been trying to solve this difficult problem for some time and I thought of at least two ways to prove it but to no avail...the second method that I thought of was to employ binomial expansion on the denominator and that did lead me to the result where it only has x terms in my final proof, but it did not lead to the desired result (and I should be able to tell beforehand that I wouldn't get anywhere near to the desired form of the result by expanding the expression on the denominator too)...I am really at my wit's end and really very mad at myself now and I'd be grateful if someone could throw some light on to the problem...thanks in advance.:)Let $$-1<x<1$$, show that $$\sum_{k=0}^{6} {\frac{1-x^2}{1-2x\cos (\frac{2\pi k}{7})+x^2}}=\frac{7\left(1+x^7 \right)}{1-x^7}$$.

 

[Opalg]

Let $$-1<x<1$$, show that $$\sum_{k=0}^{6} {\frac{1-x^2}{1-2x\cos (\frac{2\pi k}{7})+x^2}}=\frac{7\left(1+x^7 \right)}{1-x^7}$$.

This problem cries out for the use of complex numbers. I'll prove a slightly more general result.Let $\omega = e^{2\pi i/n}$, with complex conjugate $\overline{\omega} = \omega^{-1}$. The $n$th roots of unity are $\omega^k\ (0\leqslant k\leqslant n-1)$, and $\displaystyle 1-x^n = \prod_{k=0}^{n-1}(1-\omega^k x)$. It follows that there must be a partial-fraction decomposition of the form $$\frac n{1-x^n} = \sum_{k=0}^{n-1}\,\frac{s_k}{1-\omega^k x}.$$ To find the coefficients $s_j$, multiply both sides by $1-\omega^jx$ to get $$\frac {n(1-\omega^jx)}{1-x^n} = s_j + (1-\omega^jx)f(x)$$ for some function $f(x)$ that is continuous at $\omega^{-j}.$ Then $$s_j = \lim_{x\to\omega^{-j}}(s_j + (1-\omega^jx)f(x)) = \lim_{x\to\omega^{-j}}\frac {n(1-\omega^jx)}{1-x^n} = \lim_{x\to\omega^{-j}}\frac {-n\omega^j}{-nx^{n-1}} = 1,$$ (using l'Hôpital's rule to evaluate the limit). Therefore $$\frac n{1-x^n} = \sum_{k=0}^{n-1}\,\frac1{1-\omega^k x}.$$ Multiply that by 2, and use the facts that $\overline{\omega}^k = \omega^{n-k}$ and $\omega^k + \overline{\omega}^k = 2\cos\bigl(\frac{2k\pi}n\bigr)$, to get $$\frac {2n}{1-x^n} = \sum_{k=0}^{n-1}\,\biggl(\frac1{1-\omega^k x} + \frac1{1-\overline{\omega}^k x}\biggr) = \sum_{k=0}^{n-1}\,\frac{2-2x\cos\bigl(\frac{2k\pi}n\bigr)}{1-2x\cos\bigl(\frac{2k\pi}n\bigr) + x^2}.$$ Finally, subtract $n$ from both sides to get $$\frac{n(1+x^n)}{1-x^n} = \frac {2n}{1-x^n} - n = \sum_{k=0}^{n-1}\,\biggl(\frac{2-2x\cos\bigl(\frac{2k\pi}n\bigr)}{1-2x\cos\bigl(\frac{2k\pi}n\bigr) + x^2} - 1\biggr) = \sum_{k=0}^{n-1}\,\frac{1-x^2}{1-2x\cos\bigl(\frac{2k\pi}n\bigr) + x^2}.$$