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Re: Infinite sum of gaussians
*To*: mathgroup at smc.vnet.net
*Subject*: [mg56053] Re: [mg55955] Infinite sum of gaussians
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Thu, 14 Apr 2005 08:55:15 -0400 (EDT)
*References*: <20050413142142.78605.qmail@web25305.mail.ukl.yahoo.com>
*Sender*: owner-wri-mathgroup at wolfram.com
On 13 Apr 2005, at 23:21, Valeri Astanoff wrote:
> Seems that only odd derivatives are null... Maybe this could explain?
>
Yes, you are right. Checking this again i see that Mathematica returns
f[z_]:=Sum[E^((-(1/2))*(z- k)^2), {k, -Infinity, Infinity}] -
Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] -
Sqrt[2*Pi]) - Sqrt[2*Pi]
D[f[z],{z,2}]/.z->0
4*Pi^2*(EllipticTheta[3, 0, 1/Sqrt[E]] - Sqrt[2*Pi]) +
Sum[k^2/E^(k^2/2) - E^(-(k^2/2)),
{k, -Infinity, Infinity}]
Which now definitely does not look like zero.
In fact I was too much in a hurry and too convinced that the result
must be true (and if it were true this would have to be zero), so I
declared it "obvious" without careful checking. This demonstrates how
one can always "prove' any result one beleives to be true.
Of course the fact that the odd derivatives are zero follows simply
form the fact that we have an analytic even function (of period 1). So
maybe after all the this termwise differentiation of a double infinite
series is valid, and my mistake was simply to believe that the even
derivatives were 0 whiteout careful checking.
Andrzej
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