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just Liouville's Theorem and an understanding of isolated singularities and uniform convergence. It follows quite easily from one of the most mind blowing formulas in all of math: The complex sine function satisfies | just Liouville's Theorem and an understanding of isolated singularities and uniform convergence. It follows quite easily from one of the most mind blowing formulas in all of math: The complex sine function satisfies | ||
| − | <math>\frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2.</math> | + | <math>\frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}.</math> |
--[[User:Bell|Steve Bell]] 14:02, 21 January 2011 (UTC) | --[[User:Bell|Steve Bell]] 14:02, 21 January 2011 (UTC) | ||
[[ Discussion HW2 ECE301 Spring2011|Back to Discussion HW2 ECE301 Spring2011]] | [[ Discussion HW2 ECE301 Spring2011|Back to Discussion HW2 ECE301 Spring2011]] | ||
Revision as of 10:03, 21 January 2011
How to Evaluate a Complicated Sum
Does somebody know how to calculate the sum
$ \sum_{k=-\infty}^\infty \frac{1}{1+k^2} ? \ $
Wolfram said answer is π * coth(π). is there any easier way to do that? Yimin. Jan 20
Discussion/help
Big hint: See page 189 of the classic book, "Complex Analysis," by Lars Ahlfors and think about letting z=i in formula (11). After that, it will be like falling off a log ... head first.--Steve Bell 12:32, 21 January 2011 (UTC)
Ok, I've got more time now. Here's that famous formula, (11) on p. 189:
$ \pi\cot \pi z =\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2} $
where z is a complex number. The cotangent function is the complex cotangent given by
$ \cot z=\frac{\cos z}{\sin z}=i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}. $
You don't have to know much complex analysis to cook up this formula... just Liouville's Theorem and an understanding of isolated singularities and uniform convergence. It follows quite easily from one of the most mind blowing formulas in all of math: The complex sine function satisfies
$ \frac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty\frac{1}{(z-n)^2}. $
--Steve Bell 14:02, 21 January 2011 (UTC)
