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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}]
and
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[-(-k)^2/2],{k,-Infinity,Infinity}]],100]
2.5066282880429055448306790538639603781474512715189099785077187561072857
447639\
10390142584776971960969
N[FullSimplify[Sum[Exp[-(30 - k)^2/2],
{k, -Infinity, Infinity}]], 100]
various messages
-2.0771591956161771304`2.107209964269863*^-48
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
FullSimplify[f[0]]
0
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
derivative:
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]])
N[%]
-2.2333485739105708*^-14
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
>
> http://functions.wolfram.com/EllipticFunctions/EllipticTheta3/06/01/
>
> using the second q-series formula.
>
>
> Daniel Lichtblau
> Wolfram Research
>
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