20091106, 01:07  #1 
"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 20091106 at 01:13 
20091106, 03:23  #2 
"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 20091106 at 03:26 
20091106, 04:54  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9390_{10} Posts 
Wolfram Alpha says:
http://mathworld.wolfram.com/JacobiThetaFunctions.html The answer is somewhere there. 
20091106, 05:59  #4 
"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!

20091106, 06:29  #5 
"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

20091106, 07:18  #6  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·5·313 Posts 
Quote:
with And, apparently, this follows from the Poisson sum formula Last fiddled with by Batalov on 20091106 at 07:25 Reason: TeX is hard! :) 

20091106, 09:47  #7  
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
Quote:
I think you might find the whole thread "Mysterious Connection" in Puzzles instructive. Quote:
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". 

20091106, 10:13  #8  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
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 

20091106, 12:46  #9  
"Lucan"
Dec 2006
England
194A_{16} Posts 
Quote:
David 

20091107, 07:42  #10  
"Kyle"
Feb 2005
Somewhere near M52..
3×5×61 Posts 
Quote:
You're right. I'm not going to share my mistake... it is rather embarrassing! 

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