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Re: Re: Re: Infinite sum of gaussians
This is indeed why the argument fails. What confused me was failing to
notice the obvious fact that Abs[w* E^(-(z-w)^2/2)] does not -> 0 as
w->Infinity in the complex plane, although of course it does ->0 for
real w, in particulalar for integers. So of course there is no doubt
that the infinite sums exist and define complex analytic function on
the entire complex plane.
Andrzej
On 17 Apr 2005, at 20:27, Andrzej Kozlowski wrote:
> 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_]= . 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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