mersenneforum.org > Math Can anyone explain/prove this equality?
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 2009-11-06, 01:07 #1 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Can anyone explain/prove this equality? (1/sqrt(2pi))(1 + 2*(e^(-1²/2)+e^(-2²/2)+e^(-3²/2)+...)) = 1 + 2*(e^(-2pi²1²)+e^(-2pi²2²)+e^(-2pi²3²)+...) David Last fiddled with by davieddy on 2009-11-06 at 01:13
 2009-11-06, 03:23 #2 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 3·5·61 Posts First attempt was to convert to infinite series, test for convergence using the ratio test... left hand side yields limit as n goes to infinity of ((n^2 + 2n + 1)/n^2)) and the right hand side yields limit as n goes to infinity of ((n^2)/(n^2 + 2n + 1))...which gives an answer of 1 on both sides = test is inconclusive I will try again later with a different test, though I think this may be the first step in the approach to take (infinite series). If I am an idiot, please call me out on it. I defer to your superior mathematical knowledge as mine is limited to ODE and three semesters of calculus and it has been a little while since I have had these classes. I may be missing something obvious as to why this isn't a good route to go... Edit: Perhaps the Integral Test.... I will try it when I get time. Last fiddled with by Primeinator on 2009-11-06 at 03:26
 2009-11-06, 04:54 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 939010 Posts Wolfram Alpha says: $ \sum e^{-2 k^2 \pi^2} = \theta_3(0,e^{-2 \pi^2})$ http://mathworld.wolfram.com/JacobiThetaFunctions.html The answer is somewhere there.
 2009-11-06, 05:59 #4 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3·373 Posts Thanks for the reference, Serge. I recognized it as a theta function identity, but that has always been one of the more obscure (to me) aspects of the theory of the Riemann zeta function. I hope to understand it better at some point!
 2009-11-06, 06:29 #5 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 3×5×61 Posts I'll let the pros take it from here. The left hand side of the equation passes the integral test nicely (the calculator was nice for integrating e^((-x^2)/2) from 1 to infinity and converges to around .4. However, the right hand side fails the integral test. Apparently more math than I have had is needed to solve this
2009-11-06, 07:18   #6
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2·3·5·313 Posts

Quote:
 Originally Posted by Batalov Wolfram Alpha says: $ \sum e^{-2 k^2 \pi^2} = \theta_3(0,e^{-2 \pi^2})$ http://mathworld.wolfram.com/JacobiThetaFunctions.html The answer is somewhere there.
P.S. I now scrolled through it enough and (43) is q.e.d.:
$
\frac {\theta_3(0,e^{-\pi x})} {\theta_3(0,e^{-\pi/x})} = \frac {1} {\sqrt x}
$

with $
x=2 \pi
$

And, apparently, this follows from the Poisson sum formula

Last fiddled with by Batalov on 2009-11-06 at 07:25 Reason: TeX is hard! :-)

2009-11-06, 09:47   #7
davieddy

"Lucan"
Dec 2006
England

11001010010102 Posts

Quote:
 Originally Posted by Primeinator I'll let the pros take it from here. The left hand side of the equation passes the integral test nicely (the calculator was nice for integrating e^((-x^2)/2) from 1 to infinity and converges to around .4. However, the right hand side fails the integral test. Apparently more math than I have had is needed to solve this
http://mersenneforum.org/showpost.ph...2&postcount=26

I think you might find the whole thread "Mysterious Connection"
in Puzzles instructive.

Quote:
 Originally Posted by Batalov P.S. I now scrolled through it enough and (43) is q.e.d.: $ \frac {\theta_3(0,e^{-\pi x})} {\theta_3(0,e^{-\pi/x})} = \frac {1} {\sqrt x}$ with $ x=2 \pi$ And, apparently, this follows from the Poisson sum formula
THX
That link (equation 5) hits the nail on the head, and seems to
endorse the sentiment I expressed in the post I linked above:
"I suspect our friend Monsieur Fourier may be involved".

2009-11-06, 10:13   #8
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by Primeinator First attempt was to convert to infinite series, test for convergence using the ratio test... left hand side yields limit as n goes to infinity of ((n^2 + 2n + 1)/n^2)) and the right hand side yields limit as n goes to infinity of ((n^2)/(n^2 + 2n + 1))...which gives an answer of 1 on both sides = test is inconclusive
No.

The ratio is e-(n+1)[sup]2/2[/sup] /e-n[sup]2/2[/sup]

=e-(2n+1)/2 which goes to zero like hell in a handcart

David

2009-11-06, 12:46   #9
davieddy

"Lucan"
Dec 2006
England

194A16 Posts

Quote:
 Originally Posted by philmoore Thanks for the reference, Serge. I recognized it as a theta function identity, but that has always been one of the more obscure (to me) aspects of the theory of the Riemann zeta function. I hope to understand it better at some point!
No I wasn't sure it wasn't a homework problem, but I am now

David

2009-11-07, 07:42   #10
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

3×5×61 Posts

Quote:
 Originally Posted by davieddy http://mersenneforum.org/showpost.ph...2&postcount=26 I think you might find the whole thread "Mysterious Connection" in Puzzles instructive.
Thank you.

Quote:
 Originally Posted by davieddy No. The ratio is e-(n+1)[sup]2/2[/sup] /e-n[sup]2/2[/sup] =e-(2n+1)/2 which goes to zero like hell in a handcart David
You're right. I'm not going to share my mistake... it is rather embarrassing!

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