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=How to Evaluate a Complicated Sum= | =How to Evaluate a Complicated Sum= | ||
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<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> | ||
| − | Notice that one over sine squared is the cosecant squared. Any formula you know for the real trig functions also holds in the complex case. The | + | Notice that one over sine squared is the cosecant squared. Any formula you know for the real trig functions also holds in the complex case. The antiderivative of cosecant squared is minus the cotangent function. The idea is to antidifferentiate the famous formula term by term and combine the plus and minus n terms to get the cotangent formula. |
See page 189 of the classic book, "Complex Analysis," by Lars Ahlfors for the whole story. | See page 189 of the classic book, "Complex Analysis," by Lars Ahlfors for the whole story. | ||
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(I can't resist adding here that writing the formula in Wolfram alpha probably is the easiest way to evaluate the sum!) | (I can't resist adding here that writing the formula in Wolfram alpha probably is the easiest way to evaluate the sum!) | ||
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| + | [[2011 Spring MA 530 Bell|Go to the MA530 Spring 2011]] | ||
| − | + | [[Discussion HW2 ECE301 Spring2011|Back to Discussion HW2 ECE301 Spring 2011]] | |
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| − | [[ Discussion HW2 ECE301 Spring2011|Back to Discussion HW2 ECE301 | + | |
Latest revision as of 06:35, 21 March 2013
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
Here is an answer from Steve Bell, the instructor of MA 530, Complex Analysis for first year grad students in the Math PhD program. You can understand this material after a solid course on complex variables like MA 425 or MA 525. The answer is based on a famous formula for the complex cotangent function,
$ \cot z=\frac{\cos z}{\sin z}=i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}. $
The formula is
$ \pi\cot \pi z =\frac{1}{z}+\sum_{n=1}^\infty\frac{2z}{z^2-n^2} $
where z is a complex number. If you let z=i in this formula, you will find your answer.
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. The formula 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}. $
Notice that one over sine squared is the cosecant squared. Any formula you know for the real trig functions also holds in the complex case. The antiderivative of cosecant squared is minus the cotangent function. The idea is to antidifferentiate the famous formula term by term and combine the plus and minus n terms to get the cotangent formula.
See page 189 of the classic book, "Complex Analysis," by Lars Ahlfors for the whole story.
--Steve Bell 14:02, 21 January 2011 (UTC)
(I can't resist adding here that writing the formula in Wolfram alpha probably is the easiest way to evaluate the sum!)
