Mathematica 9 is now available
Services & Resources / Wolfram Forums
MathGroup Archive
*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: Re: Re: Infinite sum of gaussians

I now believe the equation is not true on mathematical grounds. But I  
would not trust any numerical verifications of it. To see why consider  
the following two series.

Sum[Exp[-(30 - k)^2/2], {k, -Infinity, Infinity}]


Sum[Exp[-(-k)^2/2], {k, -Infinity, Infinity}]

Now as k runs through integer values from -Infinity to +Infinity -(20 -  
k)^2/2 and -(-k)^2/2 must run though precisely the same set of values,  
so since the series are absolutely convergent they should be equal.  
However, Mathematica gives:



N[FullSimplify[Sum[Exp[-(30 - k)^2/2],
     {k, -Infinity, Infinity}]], 100]

various messages


so one hundred digits of precision is insufficient to show that these  
two values are the same. This problem appears to be very ill posed and  
therefore I do not think numerical arguments are convincing.  
Nevertheless I think the identity is not satisfied.  This can be best  
proved by an argument involving Fourier series mentioned in Carl Woll's  
posting. However, I would like to return again to my original argument  
to try to understand where I went wrong. Consider again the function

f[z_] := 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]

we certainly have



What sort of function is f? Well, it is clearly not complex analytic.  
In fact the sum Sum[E^((-(1/2))*(z - k)^2),
      {k, -Infinity, Infinity}] cannot converge for all complex z, since  
E^((-(1/2))*(z - k)^2) is certainly complex analytic as a function of  
k, and one can prove that if g is a complex analytic function in the  
entire complex plane we must have Sum[g[k],{k,-Infinity,Infinity}]==0.  
So f is not defined for all complex values of z. But it is defined for  
all real values and the function so obtained is real analytic. I think  
I can prove that but I admit I have not considered this carefully. But  
if it is a real analytic function it is also determined by its value at  
just one point and the valus of the derivatives at that point. Note  
also that the function f has obviously period 1.
So let's consider again what happens that the point 0. We know that the  
f itself takes the value 0 there.

Mathematica also returns:

FullSimplify[D[f[x], {x, 3}] /. x -> 0]

Sum[k^3/E^(k^2/2) - (3*k)/E^(k^2/2),
   {k, -Infinity, Infinity}]

This is clearly zero, and so are all the odd derivatives. What about  
the even ones. Well, I believe now i was wrong to say that they are 0  
but I think they are extremely small. Let's look again at the second  

FullSimplify[D[f[x], {x, 2}] /. x -> 0]

Sum[k^2/E^(k^2/2) - E^(-(k^2/2)), {k, -Infinity,
     Infinity}] + 4*Pi^2*(-Sqrt[2*Pi] +
     EllipticTheta[3, 0, 1/Sqrt[E]])



I think this really is an extremly small number rather than 0. If this  
is indeed so and if the function is really real analytic, as I believe  
than we can see what happens. The function is 0 for integer values of  
x. For non-integer x it can be expressed as a power series in odd  
powers in  x-Floor[x], with extremely small coefficients. So the values  
of f remain always very close to zero, to an extent that is impossible  
to reliably determine by numerical means.

Andrzej Kozlowski

On 15 Apr 2005, at 17:47, Daniel Lichtblau wrote:

> *This message was transferred with a trial version of CommuniGate(tm)  
> Pro*
> Valeri Astanoff wrote:
>> Andrzej,
>> I'm not a mathematician, just an engineer, and what puzzles me is this
>> :
>> In[1]:=
>> { 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]&
>> Out[1]=
>> {2.5066282612190954600008515157581306,
>>  2.5066282612190954600008515157581301}
>> If I trust mathematica, this should induce to think
>> the equality is false.
>> Anyway, thanks for the time you spent.
>> Regards,
>> v.a.
> If I am seeing those numbers correctly, they differ only in the last
> digit. So I would be strongly induced to think the equality is true (at
> least for z=1/2), and that the numeric evaluation of
> Sum[E^(-(1/2-k)^2/2), {k,-Infinity,Infinity}]
> at 35 digits has done something bordering on magic.
> By the way, here is an identity. That sum of Gaussians is
> Exp[-z^2/2]*EllipticTheta[3,z/(2*I),1/Sqrt[E]]
> This follows from
> using the second q-series formula.
> Daniel Lichtblau
> Wolfram Research

  • Prev by Date: Re: removing subelements
  • Next by Date: Re: Eigensystem Bug? OS-X
  • Previous by thread: Re: Re: Infinite sum of gaussians
  • Next by thread: Typesetting Mathematica code