Re: Re: Re: Infinite sum of gaussians

*To*: mathgroup at smc.vnet.net*Subject*: [mg56177] Re: [mg56173] Re: Re: Infinite sum of gaussians*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 18 Apr 2005 03:08:35 -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>*Sender*: owner-wri-mathgroup at wolfram.com

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>

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**Re: Re: Infinite sum of gaussians**

**Re: Re: Re: Infinite sum of gaussians**