Re: Re: Infinite sum of gaussians

*To*: mathgroup at smc.vnet.net*Subject*: [mg56173] Re: Re: Infinite sum of gaussians*From*: Maxim <ab_def at prontomail.com>*Date*: Sun, 17 Apr 2005 03:07:39 -0400 (EDT)*References*: <200504120926.FAA27573@smc.vnet.net><d3ibr6$9un$1@smc.vnet.net> <200504141254.IAA28085@smc.vnet.net> <200504150847.EAA11453@smc.vnet.net> <d3qheo$oeh$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Sat, 16 Apr 2005 08:13:12 +0000 (UTC), Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > 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 > This is wrong on several points. In fact, Sum[E^(-(z - k)^2/2), {k, -Infinity, Infinity}] is analytic everywhere in the complex plane. Since we already know that this sum is equal to Sqrt[2*Pi]*EllipticTheta[3, Pi*z, E^(-2*Pi^2)], all its properties, including analyticity, follow from the properties of EllipticTheta. Actually, since the sum of E^(-(z - k)^2/2) is very well-behaved (the terms decay faster than, say, E^(-k^2/4)), it is trivial to prove the uniform convergence in z and therefore the validity of the termwise differentiation as well as analyticity directly. The fact that the series is double infinite is of no importance; we can always rewrite it as two series from 1 to +Infinity. Also it's not correct that a real infinitely differentiable function can be defined by its value and the values of its derivatives at a point. If we take f[z] == E^-z^-2 for z != 0 and f[0] == 0, then all the (real) derivatives at 0 vanish. As to using N[Sum[...], prec], there is probably a minor inconsistence in the semantics: if N has to call NSum, then it cannot guarantee that all (or even any) of the returned digits will be correct, unlike, say, N[Pi, prec]. WorkingPrecision in NSum, NIntegrate, NDSolve, etc. only determines the precision used at each computation step; while it is useful for keeping track of precision (for example, may detect the loss of digits), in general it cannot give a reliable estimate of the error in the final result. I believe it is better to just use NSum, where we have more control over the settings: for example, NSum[Exp[-(30 - k)^2/2], {k, -Infinity, Infinity}, NSumTerms -> 50, WorkingPrecision -> 100] gives the result with 100 digits of precision, that is, no digits are lost at all. This is hardly surprising, since the sum converges very rapidly. (Using N[Sum[...]] is just out of the question: thus, the correct value of f''[0] is around -10^-32, not -10^-14). In principle a rigorous proof can be numerical; if we take z == 1/2 and expand EllipticTheta[3, 0, 1/Sqrt[E]] into a series, the identity becomes Sum[E^(-(1/2 - k)^2/2) + E^(-k^2/2), {k, -Infinity, Infinity}] == 2*Sqrt[2*Pi]. Now In[1]:= Select[Range[100], Sum[E^(-(1/2 - k)^2/2) + E^(-k^2/2), {k, -#, #}] > 2*Sqrt[2*Pi]&, 1] Out[1]= {12} This means that the left-hand side is already greater than 2*Sqrt[2*Pi] if we take the sum from -12 to 12; since all the terms are positive, the identity cannot be true. The point is that Less/Greater use significance arithmetic; if we evaluate In[2]:= N[Sum[E^(-(1/2 - k)^2/2) + E^(-k^2/2), {k, -12, 12}] - 2*Sqrt[2*Pi], 10] Out[2]= 3.95515733259688519694588`10.*^-34 then Mathematica tells us that the accuracy of the result is greater than 43 and so the absolute error is less than 10^-43, therefore the result is verifiably different from zero. So this can be a proof, but only barring the defects of Mathematica's significance arithmetic. Also it's possible to use Interval. Maxim Rytin m.r at inbox.ru

**Follow-Ups**:**Re: Re: Re: Infinite sum of gaussians***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**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>