       Re: Infinite sum of gaussians

• To: mathgroup at smc.vnet.net
• Subject: [mg56046] Re: [mg55955] Infinite sum of gaussians
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 14 Apr 2005 08:54:47 -0400 (EDT)
• References: <20050413104220.92517.qmail@web25306.mail.ukl.yahoo.com> <b447027f5c7a94966c215b491cf6f04a@mimuw.edu.pl>
• Sender: owner-wri-mathgroup at wolfram.com

```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
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```

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