Re: Re: Re: Infinite sum of gaussians

*To*: mathgroup at smc.vnet.net*Subject*: [mg56175] Re: [mg56173] Re: Re: Infinite sum of gaussians*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 18 Apr 2005 03:08:32 -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> <200504170707.DAA08493@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 17 Apr 2005, at 16:07, Maxim wrote: >> >> > > 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. Yes, you are right, and in fact that is what I thought at first. But after Carl Woll's message I realized that I could easily prove the following: if f[z] is everywhere complex analytic and Abs[z f[z]]->0 as z->Infinity then Sum[f[z],{z,-Infinity,Infinity}]==0. I thought that this shows that that the above sum can't be convergent everywhere, but I have had not time to think about it for more than a few minutes at a time so I am probably not missing something even now. > > 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. > Of course but nobody ever said that real analytic is the same as C^Ininity. Real analitic means that the Taylor series converges everhwehre and is equal to the value of the function. This is all that was needed in this case anyway. 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>

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

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

**Re: Re: Infinite sum of gaussians***From:*Maxim <ab_def@prontomail.com>