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Old 2005-12-07, 19:47   #1
Crook
 
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Nov 2004

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Default Why zeta(-2n)=0 ?

I know this wouldn't even deserve to be posted, but could someone explain me why zeta(-2n)=0 ? Regards.
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Old 2005-12-08, 02:34   #2
John Renze
 
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Quote:
Originally Posted by Crook
I know this wouldn't even deserve to be posted, but could someone explain me why zeta(-2n)=0 ? Regards.
It's hard to answer this without knowing your mathematical level, but here is one way to look at it.

Start with one version of the functional equation for the zeta function:

zeta(1-s) == 2 (2pi)^(-s) cos(s*pi/2)gamma(s)zeta(s)

See Equation 10 at http://eww2lx.wri.wolfram.com/RiemannZetaFunction.html or many other sources for this equation.

Substitute s = 2n+1 where n is a positive integer to get

zeta(-2n) == 2 (2pi)^(-s) cos(n*pi + 1/2)(2n+1)!*zeta(2n+1)

cos(n*pi + 1/2) == 0 and zeta(2n+1) is finite, so you get

zeta(-2n) == 0

Hope this helps,
John
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