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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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