       Re: Infinite sum of gaussians

• To: mathgroup at smc.vnet.net
• Subject: [mg56051] Re: [mg55955] Infinite sum of gaussians
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 14 Apr 2005 08:55:05 -0400 (EDT)
• References: <20050413104220.92517.qmail@web25306.mail.ukl.yahoo.com> <b447027f5c7a94966c215b491cf6f04a@mimuw.edu.pl> <91fdbb6be3f4178ff4167ae0d9132c85@mimuw.edu.pl>
• Sender: owner-wri-mathgroup at wolfram.com

```I think the function is clearly analytic but my mistake was to assume
that one can differentiate these double infinite series term by term.
Curiously Mathematica seems to do it automatically!

Andrzej

On 13 Apr 2005, at 21:41, Andrzej Kozlowski wrote:

> Hm... I am beginning to have doubts. First of all:
>
>
> N[Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] -
>      Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] -
>        Sqrt[2*Pi]) == Sqrt[2*Pi] /. z -> 1/2,100]
>
> False
>
> This indeed suggests that something may be wrong with my proof.
> However, the only possibility is the assumption that
> Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] is an analytic
> function in the complex plane, since I have no doubt that the function
> defined by the difference between  the LHS and the RHS is 0 at z=0 and
> so are all its derivatives. If the function is analytic in the whole
> complex plane than everything is proved. So presumably the double
> infinite sum can't be an analytic function?
> I'll have to to think about it more when I find a bit of time, unless
> someone else can clarify this ;-)
>
> Andrzej Kozlowski
>
>
> On 13 Apr 2005, at 19:50, Andrzej Kozlowski wrote:
>
>>
>> Yes, but this is not equivalent to your statement. Your statement was
>>
>> Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] -
>>    Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] -
>>      Sqrt[2*Pi]) == Sqrt[2*Pi]
>>
>> If you try this at z=1/2
>>
>>
>>
>> N[Sum[E^((-(1/2))*(z - k)^2), {k, -Infinity, Infinity}] -
>>      Cos[2*Pi*z]*(EllipticTheta[3, 0, 1/Sqrt[E]] -
>>        Sqrt[2*Pi]) == Sqrt[2*Pi] /. z -> 1/2]
>>
>>
>> True
>>
>> I do not have access to sci.math.num-analysis but I am convinced my
>> proof is correct.
>>
>> Andrzej
>>
>> On 13 Apr 2005, at 19:42, Valeri Astanoff wrote:
>>
>>> If I trust mathematica's numerics with z=1/2, the sum of gaussians
>>> can't be a constant
>>> nor the sinusoid I posted :
>>>
>>>
>>> {Sum[Exp[-(z-k)^2/2], {k, -Infinity, Infinity}],
>>>   Sqrt[2*Pi]} /. z -> 1/2 // N[#,9]&
>>>
>>>  {2.50662826,
>>>   2.50662827}
>>>
>>>
>>>
>>> { Sum[Exp[-(z-k)^2/2], {k, -Infinity, Infinity}],
>>>    Sqrt[2*Pi]+Cos[2*Pi*z]*(EllipticTheta[3,0,1/Sqrt[E]]-Sqrt[2*Pi])}
>>> /. z -> 1/2 //  N[#,35]&
>>>
>>> {2.5066282612190954600008515157581306,
>>>   2.5066282612190954600008515157581301}
>>>
>>>
>>>  If you have a minute, have a look at the refutation
>>>  posted by James Van Buskirk on sci.math.num-analysis.
>>>
>>> Regards,
>>>
>>> v.a.
>>>
>>>
>>>
>>>
>>>  Découvrez le nouveau Yahoo! Mail : 250 Mo d'espace de stockage pour
>>> vos mails !
>>> Créez votre Yahoo! Mail
>

```

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