

Title: A summation of series probalem Post by comehome1981 on Feb 4^{th}, 2009, 10:14am Someone know the value of the following series: sum^{infty}_{k=infty} k^2 * exp(k^2/2) , k is an integer I did it as follows: (1) sum^{N}_{k=N} k^2 * exp(k^2/2) = sum^{N}_{k=N}{ residues of pi*cot(pi*z) f(z) at integers N, N+1, .... , 0, 1, ... , N } (2) int_{C_N} (pi*cot(pi*z ) f(z) ) dz = 2pi*i sum^{N}_{k=N}{ residues of pi*cot(pi*z ) f(z) at integers N, N+1, .... , 0, 1, ... , N } ( By the Residue Theorem ) where C_N be a square with vertices at (N+1/2) x (+ 1 + i ) (3) lim_{N>infty} int_C_N (pi* cot(pi*z ) f(z) ) dz = 2pi*i (int_{infty}^{infty} x^2 exp(x^2/2) dx) I have question about the equality in (3) Why the two integral are equal?? Thanks so much for responses. 

Title: Re: A summation of series probalem Post by Eigenray on Feb 4^{th}, 2009, 3:59pm on 02/04/09 at 10:14:30, comehome1981 wrote:
They're not. If they were we would have http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifn^{2} e^{n^2/2} = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx^{2}e^{x^2/2}dx but the LHS is 2.506627759, while the RHS is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} = 2.506628275. One problem with your approach is that the integrals of z^{2} cot(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifz) e^{z^2/2}, over the sides of the square, do not converge. You would want to keep Im z relatively small to avoid e^{z^2/2} blowing up. Your sum can be written in terms of the theta function http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gife^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifn^2 t} as 1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif'(1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif)). I don't know if there is a closed form for this. Where did you get this problem? 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 5^{th}, 2009, 11:22am Eigenray, thanks for your reply. well, From my work, I just come up with this series from normal distribution somehow. they are really closed in your numerical values, is it possible just because of the error coming from computer? and actually they are equal? Or if the following are equal sum^{infty}_{k=infty} k^2 * exp(k^2/2)= sum^{infty}_{k=infty} exp(k^2/2) I need if this two are equal or not? Anyone knows?? 

Title: Re: A summation of series probalem Post by Eigenray on Feb 5^{th}, 2009, 12:21pm Unless there is a bug in Mathematica, Maple and PARI/GP they are not equal. It is easy to approximate the sum: let f(x) = x^{2} e^{x^2/2}. Then f is decreasing for x > http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2. Therefore by considering Riemann sums, E_{N} = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{k>N} f(k) < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{N}^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif} f(x)dx. < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif_{N}^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif} x^{2} e^{x N/2} dx = 2e^{N^2/2} (8+4N^{2}+N^{4})/N^{3}, which is < .4*10^{9} for N=7. This means that the sum is approximated to within 10^{9} by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif_{k=7}^{7} f(k) ~ 2.506627759. On the other hand, the integral is well known to be http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} ~ 2.506628275. 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 6^{th}, 2009, 7:33am Thanks. That helps a lot. I think they are not equal in (3). But is the following ture? sum^{infty}_{k=infty} k^2 * exp(k^2/2)= sum^{infty}_{k=infty} exp(k^2/2) I have been trying using program to calculate, and they are not equal. Just wonder if there is a closed form for above equation. 

Title: Re: A summation of series probalem Post by Eigenray on Feb 6^{th}, 2009, 1:05pm No they are not equal. But there is a reason that the sums are so close to the corresponding integrals. http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gife^{k^2/2} = 1 + 2e^{1/2} + 2e^{4/2} + 2e^{9/2} + ... ~ 1 + 1.2 + 0.27 + 0.022 + 0.00067 + 0.0000075 + ... converges rapidly enough, but we can make it converge even faster. Using Poisson summation, one can show that the theta function satisfies http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(1/t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gift http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(t). So http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gife^{k^2/2} = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(1/(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif)) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) But http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/theta.gif(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif) = 1 + 2exp(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^2) + 2exp(8http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^2) + 2exp(18http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^2) + ... ~ 1 + 5*10^{9} + 10^{34} + 10^{77} + ... = 1.0000000053505759821... is very close to 1. So http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gife^{k^2/2} differs from http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gife^{x^2/2}dx by around 10^{8}. Similarly, let f(x) = x^{2} e^{x^2/2}. Then the Fourier transform of f is F(t) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.giff(x) e^{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi x t} dx = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} (14http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2}t^{2}) exp(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2}t^{2}), so by Poisson summation, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.giff(n) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifF(n) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} [ 1 + 2(14http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2})exp(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2}) + 2(116http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2})exp(8http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2}) + 2(136http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2})exp(18http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif^{2}) + ... ] ~ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} [ 1  2*10^{7}  2*10^{32}  5*10^{75}  ... ] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif} * 0.99999979411830293505... 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 6^{th}, 2009, 1:13pm Ok, now it is sure that they are not equal. A bad news for me. Anyway, really thanks for those explanation. 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 6^{th}, 2009, 2:04pm Sorry for another question: Wonder if there exists a function f(x) s.t. sum^{infty}_{k=infty} x^2 * exp(x^2/2)*f(x)= sum^{infty}_{k=infty} exp(x^2/2)*f(x) is it possible to know what is the form of f(x)? 

Title: Re: A summation of series probalem Post by towr on Feb 6^{th}, 2009, 2:42pm f(x)=0 is the most obvious candidate. And further any function where f(0)=0 and f(x)=f(x) (so for example, f(x)=x or f(x)=sin(x) ) Aside from functions where both series are 0, I don't think there are any solutions. (But the ones above are probably not the only ones that accomplish this) 

Title: Re: A summation of series probalem Post by Eigenray on Feb 6^{th}, 2009, 4:22pm You can pick f_{0} to be any function you want (so long as both sides converge) and then there is a unique constant c such that f(x) = f_{0}(x) + c works. Or you can pick f(n) to be whatever you want for n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0 (again, as long the series converge), and then just solve for f(0). (Or with 0 replaced by any integer, other than http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1.) And of course any linear combination of solutions is also a solution. The space of all functions is infinite dimensional so one equation won't narrow it down much. But whether there is any "natural" such function I don't know. 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 7^{th}, 2009, 12:09pm on 02/06/09 at 14:42:03, towr wrote:
why f(x)=x will do? sum^{infty}_{x=infty} x^2 * exp(x^2/2) *x= sum^{infty}_{x=infty} exp(x^2/2) *x ?? 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 7^{th}, 2009, 12:33pm on 02/06/09 at 16:22:51, Eigenray wrote:
I think this is easier. Thanks 

Title: Re: A summation of series probalem Post by towr on Feb 7^{th}, 2009, 1:05pm on 02/07/09 at 12:09:08, comehome1981 wrote:
It goes for any pair of odd and even functions. sum^{infty}_{x=infty} x^2 * exp(x^2/2) *x = sum^{infty}_{x=infty} x^3 * exp(x^2/2) = sum^{1}_{x=infty} x^3 * exp(x^2/2) + 0^2*exp(0) + sum^{infty}_{x=1} x^2 * exp(x^2/2) = sum^{infty}_{x=1} (x)^3 * exp((x)^2/2) + sum^{infty}_{x=1} x^2 * exp(x^2/2) = sum^{infty}_{x=1} x^3 * exp(x^2/2) + sum^{infty}_{x=1} x^2 * exp(x^2/2) = sum^{infty}_{x=1} x^3 * exp(x^2/2) + sum^{infty}_{x=1} x^2 * exp(x^2/2) = 0 

Title: Re: A summation of series probalem Post by comehome1981 on Feb 7^{th}, 2009, 1:19pm yes I got it , thanks 

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