Re: Re: Re: Infinite sum of gaussians

*To*: mathgroup at smc.vnet.net*Subject*: [mg56178] Re: [mg56173] Re: Re: Infinite sum of gaussians*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 18 Apr 2005 03:08:36 -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> <f4784ed34f283d5fa142445ede5b13f4@mimuw.edu.pl> <80d088e4c619e7f9e019ad446d9c0c35@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I forgot to add that the function f must (of course) decay sufficientlyy fast at infinity. This may be the reason why the argument fails. Andrzej On 17 Apr 2005, at 19:16, Andrzej Kozlowski wrote: > Well, I still can't see what is wrong. Perhaps I have not been > sleeping enough. Maybe someone can help me resolve this puzzle: > > Suppose f is a function analytic in the enitre compelx plain. Conisder > the function g[z_]:= Pi*f[z]*Cot[Pi*z]. This has poles at integer z > with residues equal to 1. Now consider a large square with sides > parallel to the axes not passing through any integer real numbers and > integrate g[z] over it. By Cauchy's theorem the integral is > Sum[f[i],{i,-n,n}] where the interval from -n to n is the largest such > interval inside the square over which we are integrating. Now make > n->Infinity. This seems to show that Sum[f[i],{i,_Infinity, > Infinity}]==0. Now, for a fixed z apply this to f[w_]= E^(-(z-w)^2/2) > . This certainly analityic but we know Sum[E^((z-i)^2/2),{i,_Infinity, > Infinity}]=!=0. > So what has gone wrong here? > > Andrzej > > > On 17 Apr 2005, at 18:32, Andrzej Kozlowski wrote: > >> >> 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>