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Re: Infinite sum of gaussians

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56026] Re: [mg55955] Infinite sum of gaussians
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 14 Apr 2005 08:54:18 -0400 (EDT)
  • References: <200504120926.FAA27573@smc.vnet.net> <714d7a167f299bc9aef477a98aea382e@mimuw.edu.pl>
  • Sender: owner-wri-mathgroup at wolfram.com

Of course it is enough to show that all the derivatives of f[z] vanish 
at 0, which is obvious without using Mathematica.
Matheamtica gives:


FullSimplify[D[f[z], {z, 3}] /. z -> 0]


Sum[k^3/E^(k^2/2) - (3*k)/E^(k^2/2),
   {k, -Infinity, Infinity}]

Strangely Mathematica can't see that this is 0 but if we apply N we get:


N[%]

0.

Andrzej


On 13 Apr 2005, at 12:23, Andrzej Kozlowski wrote:

> 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/
>


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