Re: Re: Re: Infinite sum of gaussians

*To*: mathgroup at smc.vnet.net*Subject*: [mg56133] Re: [mg56077] Re: [mg56032] Re: Infinite sum of gaussians*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 16 Apr 2005 03:52:28 -0400 (EDT)*References*: <200504120926.FAA27573@smc.vnet.net><d3ibr6$9un$1@smc.vnet.net> <200504141254.IAA28085@smc.vnet.net> <200504150847.EAA11453@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I now believe the equation is not true on mathematical grounds. But I would not trust any numerical verifications of it. To see why consider the following two series. Sum[Exp[-(30 - k)^2/2], {k, -Infinity, Infinity}] and Sum[Exp[-(-k)^2/2], {k, -Infinity, Infinity}] Now as k runs through integer values from -Infinity to +Infinity -(20 - k)^2/2 and -(-k)^2/2 must run though precisely the same set of values, so since the series are absolutely convergent they should be equal. However, Mathematica gives: N[FullSimplify[Sum[Exp[-(-k)^2/2],{k,-Infinity,Infinity}]],100] 2.5066282880429055448306790538639603781474512715189099785077187561072857 447639\ 10390142584776971960969 N[FullSimplify[Sum[Exp[-(30 - k)^2/2], {k, -Infinity, Infinity}]], 100] various messages -2.0771591956161771304`2.107209964269863*^-48 so one hundred digits of precision is insufficient to show that these two values are the same. This problem appears to be very ill posed and therefore I do not think numerical arguments are convincing. Nevertheless I think the identity is not satisfied. This can be best proved by an argument involving Fourier series mentioned in Carl Woll's posting. However, I would like to return again to my original argument to try to understand where I went wrong. Consider again 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])-Sqrt[2 Pi] we certainly have FullSimplify[f[0]] 0 What sort of function is f? Well, it is clearly not complex analytic. In fact the sum Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] cannot converge for all complex z, since E^((-(1/2))*(z - k)^2) is certainly complex analytic as a function of k, and one can prove that if g is a complex analytic function in the entire complex plane we must have Sum[g[k],{k,-Infinity,Infinity}]==0. So f is not defined for all complex values of z. But it is defined for all real values and the function so obtained is real analytic. I think I can prove that but I admit I have not considered this carefully. But if it is a real analytic function it is also determined by its value at just one point and the valus of the derivatives at that point. Note also that the function f has obviously period 1. So let's consider again what happens that the point 0. We know that the f itself takes the value 0 there. Mathematica also returns: FullSimplify[D[f[x], {x, 3}] /. x -> 0] Sum[k^3/E^(k^2/2) - (3*k)/E^(k^2/2), {k, -Infinity, Infinity}] This is clearly zero, and so are all the odd derivatives. What about the even ones. Well, I believe now i was wrong to say that they are 0 but I think they are extremely small. Let's look again at the second derivative: FullSimplify[D[f[x], {x, 2}] /. x -> 0] Sum[k^2/E^(k^2/2) - E^(-(k^2/2)), {k, -Infinity, Infinity}] + 4*Pi^2*(-Sqrt[2*Pi] + EllipticTheta[3, 0, 1/Sqrt[E]]) N[%] -2.2333485739105708*^-14 I think this really is an extremly small number rather than 0. If this is indeed so and if the function is really real analytic, as I believe than we can see what happens. The function is 0 for integer values of x. For non-integer x it can be expressed as a power series in odd powers in x-Floor[x], with extremely small coefficients. So the values of f remain always very close to zero, to an extent that is impossible to reliably determine by numerical means. Andrzej Kozlowski On 15 Apr 2005, at 17:47, Daniel Lichtblau wrote: > *This message was transferred with a trial version of CommuniGate(tm) > Pro* > Valeri Astanoff wrote: >> Andrzej, >> >> I'm not a mathematician, just an engineer, and what puzzles me is this >> : >> >> In[1]:= >> { 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 -> >> 1/2 // >> N[#,35]& >> >> Out[1]= >> {2.5066282612190954600008515157581306, >> 2.5066282612190954600008515157581301} >> >> If I trust mathematica, this should induce to think >> the equality is false. >> >> Anyway, thanks for the time you spent. >> >> >> Regards, >> >> v.a. > > If I am seeing those numbers correctly, they differ only in the last > digit. So I would be strongly induced to think the equality is true (at > least for z=1/2), and that the numeric evaluation of > Sum[E^(-(1/2-k)^2/2), {k,-Infinity,Infinity}] > at 35 digits has done something bordering on magic. > > By the way, here is an identity. That sum of Gaussians is > > Exp[-z^2/2]*EllipticTheta[3,z/(2*I),1/Sqrt[E]] > > This follows from > > http://functions.wolfram.com/EllipticFunctions/EllipticTheta3/06/01/ > > using the second q-series formula. > > > Daniel Lichtblau > Wolfram Research >

**Follow-Ups**:**Re: Re: Re: Re: Infinite sum of gaussians***From:*Daniel Lichtblau <danl@wolfram.com>

**Typesetting Mathematica code***From:*"J. McKenzie Alexander" <jalex@lse.ac.uk>

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

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

**Re: Re: Infinite sum of gaussians***From:*Daniel Lichtblau <danl@wolfram.com>