mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2009-11-06, 01:07   #1
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default 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
davieddy is offline   Reply With Quote
Old 2009-11-06, 03:23   #2
Primeinator
 
Primeinator's Avatar
 
"Kyle"
Feb 2005
Somewhere near M52..

3×5×61 Posts
Default

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
Primeinator is offline   Reply With Quote
Old 2009-11-06, 04:54   #3
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

941610 Posts
Default

Wolfram Alpha says:
 <br />
\sum e^{-2 k^2 \pi^2} = \theta_3(0,e^{-2 \pi^2})<br />
http://mathworld.wolfram.com/JacobiThetaFunctions.html
The answer is somewhere there.
Batalov is offline   Reply With Quote
Old 2009-11-06, 05:59   #4
philmoore
 
philmoore's Avatar
 
"Phil"
Sep 2002
Tracktown, U.S.A.

3·373 Posts
Default

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!
philmoore is offline   Reply With Quote
Old 2009-11-06, 06:29   #5
Primeinator
 
Primeinator's Avatar
 
"Kyle"
Feb 2005
Somewhere near M52..

3×5×61 Posts
Default

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
Primeinator is offline   Reply With Quote
Old 2009-11-06, 07:18   #6
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

24C816 Posts
Default

Quote:
Originally Posted by Batalov View Post
Wolfram Alpha says:
 <br />
\sum e^{-2 k^2 \pi^2} = \theta_3(0,e^{-2 \pi^2})<br />
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.:
 <br />
\frac {\theta_3(0,e^{-\pi x})} {\theta_3(0,e^{-\pi/x})} = \frac {1} {\sqrt x}<br />
with <br />
x=2 \pi<br />

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! :-)
Batalov is offline   Reply With Quote
Old 2009-11-06, 09:47   #7
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Quote:
Originally Posted by Primeinator View Post
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 View Post
P.S. I now scrolled through it enough and (43) is q.e.d.:
 <br />
\frac {\theta_3(0,e^{-\pi x})} {\theta_3(0,e^{-\pi/x})} = \frac {1} {\sqrt x}<br />
with <br />
x=2 \pi<br />

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".
davieddy is offline   Reply With Quote
Old 2009-11-06, 10:13   #8
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2·3·13·83 Posts
Default

Quote:
Originally Posted by Primeinator View Post
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
davieddy is offline   Reply With Quote
Old 2009-11-06, 12:46   #9
davieddy
 
davieddy's Avatar
 
"Lucan"
Dec 2006
England

2×3×13×83 Posts
Default

Quote:
Originally Posted by philmoore View Post
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
davieddy is offline   Reply With Quote
Old 2009-11-07, 07:42   #10
Primeinator
 
Primeinator's Avatar
 
"Kyle"
Feb 2005
Somewhere near M52..

3·5·61 Posts
Default

Quote:
Originally Posted by davieddy View Post
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 View Post
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!
Primeinator is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
FATAL ERROR, can somebody explain? Nils Hardware 11 2012-07-21 18:06
Can anyone explain 'iterations' for factoring? petrw1 PrimeNet 8 2007-08-11 18:28
Could someone please explain my blurb? jasong jasong 5 2007-07-19 00:43
Could someone explain how the Fermat factoring programs work? jasong Information & Answers 3 2006-09-12 02:25
Equality of quadratic equations dsouza123 Math 2 2004-07-30 09:03

All times are UTC. The time now is 08:45.

Thu May 6 08:45:06 UTC 2021 up 28 days, 3:25, 0 users, load averages: 1.28, 1.46, 1.54

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.