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Re: Re: Re: Infinite sum of gaussians

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/
>
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