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        general >> complex analysis >> A summation of series probalem
        
(Message started by: comehome1981 on Feb 4^th, 2009, 10:14am)

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:

Why the two integral are equal??

They're not. If they were we would have
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gifn² e^{-n^2/2} = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx²e^{-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 z² 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) = x² 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} x² e^{-x N/2} dx
= 2e^{-N^2/2} (8+4N²+N⁴)/N³,
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=-7⁷ 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) = x² 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} (1-4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²t²) exp(-2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²t²),
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(1-4http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²)exp(-2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²) + 2(1-16http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²)exp(-8http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²) + 2(1-36http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²)exp(-18http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif²) + ... ]
~ 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 f₀ to be any function you want (so long as both sides converge) and then there is a unique constant c such that f(x) = f₀(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:

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 ??

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:

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/~wwuYaBBImages/symbols/pm.gif1.)

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:

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 x³ = - (-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

Title: Re: A summation of series probalem
Post by comehome1981 on Feb 7^th, 2009, 1:19pm

yes I got it , thanks