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Re: Infinite sum of gaussians

• To: mathgroup at smc.vnet.net
• Subject: [mg56024] Re: [mg55955] Infinite sum of gaussians
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 13 Apr 2005 01:11:13 -0400 (EDT)
• References: <200504120926.FAA27573@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

On 12 Apr 2005, at 18:26, Valeri Astanoff wrote:

> Dear group,
>
> Could anyone prove or disprove this equality, at least for z being real
> :
>
> Sum[Exp[-((z - k)^2/2)], {k, -Infinity, Infinity}] ==
>  Sqrt[2*Pi] + Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] - Sqrt[2*Pi])
>
> ?
>
>
> Valeri
>
>
>

Here is a sketch of an attempted proof. It is only a sketch (I do not 
have much free time to spend on this) and I have not checked all the 
details but I am pretty sure it can be made rigorous..

First, suppose that z is not just real but an integer. In this case 
Mathematica gives:


FullSimplify[Sum[Exp[-((z - k)^2/2)], {k, -Infinity, Infinity}] ==
    Sqrt[2*Pi] + Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] - 
Sqrt[2*Pi]), z $B":(B Integers]


Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] == 
EllipticTheta[3, 0, 1/Sqrt[E]]

Now, this has to be true, because since z is an integer the sum 
Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}]  is the same as 
the sum
Sum[E^((-(1/2))*s^2), {s, -Infinity, Infinity}] and for this 
Mathematica gives


Sum[E^((-(1/2))*s^2), {s, -Infinity, Infinity}]

EllipticTheta[3, 0, 1/Sqrt[E]]

So the identity is true for integer z.

Now consider the function

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])

We would like to prove that it is equal to Sqrt[2 Pi] for every z. We 
know the function is equal to 2Pi for every integer z. (In fact it is 
easy to show that it is an even periodic function with period 1).
We can also show that it's derivative at every integer 0 is zero. For 
example, we have


D[f[z], z]


2*Pi*(-Sqrt[2*Pi] + EllipticTheta[3, 0, 1/Sqrt[E]])*
    Sin[2*Pi*z] + Sum[(k - z)/E^((1/2)*(z - k)^2),
    {k, -Infinity, Infinity}]

If z is an integer Sin[2*Pi*z] is certianly 0. Also Sum[(k - 
z)/E^((1/2)*(z - k)^2),{k, -Infinity, Infinity}] is also obviously 0.


In fact Mathematica gives:


FullSimplify[D[f[z], z], z $B":(B Integers]


Sum[(k - z)/E^((1/2)*(z - k)^2), {k, -Infinity, Infinity}]

which clearly is 0 if z is an integer.

  For the second derivative we get:

D[f[z], {z, 2}]

4*Pi^2*(-Sqrt[2*Pi] + EllipticTheta[3, 0, 1/Sqrt[E]])*
    Cos[2*Pi*z] + Sum[(k - z)^2/E^((1/2)*(z - k)^2) -
     E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}]

which again can be shown to be 0 for integer z by the above methods.

I am pretty sure that a similar argument should show that all the 
derivatives vanish.

But since the function is analytic everywhere it's values are 
determined by it's value and the values of all its derivatives at just 
a single point. So it should be 2Pi everywhere.

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/

• References:
  • Infinite sum of gaussians
    • From: "Valeri Astanoff" <astanoff@yahoo.fr>