Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

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 
>> vos mails !
>> Créez votre Yahoo! Mail


  • Prev by Date: Re: Algebra in Mathematica
  • Next by Date: Re: Re: Numerical Optimization involving equation solving
  • Previous by thread: Re: Infinite sum of gaussians
  • Next by thread: Re: Infinite sum of gaussians