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/ >

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