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Topic: A summation of series probalem (Read 25751 times) 

comehome1981
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A summation of series probalem
« on: Feb 4^{th}, 2009, 10:14am » 
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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.

« Last Edit: Feb 4^{th}, 2009, 10:16am by comehome1981 » 
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Eigenray
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Re: A summation of series probalem
« Reply #1 on: Feb 4^{th}, 2009, 3:59pm » 
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on Feb 4^{th}, 2009, 10:14am, comehome1981 wrote: Why the two integral are equal?? 
 They're not. If they were we would have n^{2} e^{n^2/2} = x^{2}e^{x^2/2}dx but the LHS is 2.506627759, while the RHS is {2} = 2.506628275. One problem with your approach is that the integrals of z^{2} cot(z) 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 (t) = e^{n^2 t} as 1/ '(1/(2)). I don't know if there is a closed form for this. Where did you get this problem?


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comehome1981
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Re: A summation of series probalem
« Reply #2 on: Feb 5^{th}, 2009, 11:22am » 
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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??

« Last Edit: Feb 5^{th}, 2009, 11:27am by comehome1981 » 
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Eigenray
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Re: A summation of series probalem
« Reply #3 on: Feb 5^{th}, 2009, 12:21pm » 
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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 > 2. Therefore by considering Riemann sums, E_{N} = _{k>N} f(k) < _{N}^{} f(x)dx. < _{N}^{} 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 _{k=7}^{7} f(k) ~ 2.506627759. On the other hand, the integral is well known to be {2} ~ 2.506628275.


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comehome1981
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Re: A summation of series probalem
« Reply #4 on: Feb 6^{th}, 2009, 7:33am » 
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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.

« Last Edit: Feb 6^{th}, 2009, 7:34am by comehome1981 » 
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Eigenray
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Re: A summation of series probalem
« Reply #5 on: Feb 6^{th}, 2009, 1:05pm » 
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No they are not equal. But there is a reason that the sums are so close to the corresponding integrals. e^{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 (1/t) = t (t). So e^{k^2/2} = (1/(2)) = {2} (2) But (2) = 1 + 2exp(2^2) + 2exp(8^2) + 2exp(18^2) + ... ~ 1 + 5*10^{9} + 10^{34} + 10^{77} + ... = 1.0000000053505759821... is very close to 1. So e^{k^2/2} differs from e^{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) = f(x) e^{2i x t} dx = {2} (14^{2}t^{2}) exp(2^{2}t^{2}), so by Poisson summation, f(n) = F(n) = {2} [ 1 + 2(14^{2})exp(2^{2}) + 2(116^{2})exp(8^{2}) + 2(136^{2})exp(18^{2}) + ... ] ~ {2} [ 1  2*10^{7}  2*10^{32}  5*10^{75}  ... ] = {2} * 0.99999979411830293505...

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


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comehome1981
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Re: A summation of series probalem
« Reply #7 on: Feb 6^{th}, 2009, 2:04pm » 
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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)?


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towr
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Re: A summation of series probalem
« Reply #8 on: Feb 6^{th}, 2009, 2:42pm » 
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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)

« Last Edit: Feb 6^{th}, 2009, 2:42pm by towr » 
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Eigenray
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Re: A summation of series probalem
« Reply #9 on: Feb 6^{th}, 2009, 4:22pm » 
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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 0 (again, as long the series converge), and then just solve for f(0). (Or with 0 replaced by any integer, other than 1.) 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.

« Last Edit: Feb 6^{th}, 2009, 4:24pm by Eigenray » 
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comehome1981
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Re: A summation of series probalem
« Reply #10 on: Feb 7^{th}, 2009, 12:09pm » 
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on Feb 6^{th}, 2009, 2:42pm, towr wrote: any function where f(0)=0 and f(x)=f(x) (so for example, f(x)=x or f(x)=sin(x) ) 
 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 ??


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comehome1981
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Re: A summation of series probalem
« Reply #11 on: Feb 7^{th}, 2009, 12:33pm » 
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on Feb 6^{th}, 2009, 4:22pm, Eigenray wrote: Or you can pick f(n) to be whatever you want for n 0 (again, as long the series converge), and then just solve for f(0). (Or with 0 replaced by any integer, other than 1.) 
 I think this is easier. Thanks


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towr
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Re: A summation of series probalem
« Reply #12 on: Feb 7^{th}, 2009, 1:05pm » 
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on Feb 7^{th}, 2009, 12:09pm, comehome1981 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 ?? 
 Because x^{3} =  (x)^{3}, so for the first series the positive and negative k cancel eachother out. And for the second series x = (x), so there also the negative and positive k cancel eachother out. So both series are zero, and zero equals zero. 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


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comehome1981
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Re: A summation of series probalem
« Reply #13 on: Feb 7^{th}, 2009, 1:19pm » 
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yes I got it , thanks


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