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Re: Re: Re: Infinite sum of gaussians
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/
>>
>
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