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Trigonometry Prove the summation identity.

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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:mad: at myself now and I'd be grateful if someone could throw some light on to the problem...thanks in advance.:)


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

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let \(\displaystyle -1<x<1\), show that \(\displaystyle \sum_{k=0}^{6} {\frac{1-x^2}{1-2x\cos (\frac{2\pi k}{7})+x^2}}=\frac{7\left(1+x^7 \right)}{1-x^7}\).
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 $$\frac{n(1+x^n)}{1-x^n} = \frac {2n}{1-x^n} - n = \sum_{k=0}^{n-1}\,\biggl(\frac{2-2x\cos\bigl(\frac{2k\pi}n\bigr)}{1-2x\cos\bigl(\frac{2k\pi}n\bigr) + x^2} - 1\biggr) = \sum_{k=0}^{n-1}\,\frac{1-x^2}{1-2x\cos\bigl(\frac{2k\pi}n\bigr) + x^2}.$$
 
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