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Re: Re: Re: Re: Infinite sum of gaussians
Andrzej Kozlowski wrote:
> I now believe the equation is not true on mathematical grounds.
James Van Buskirk and Dave Rusin show on sci.math.num-analysis why it is
not correct. In essence they gave a detailed argument of the sort Carl
Woll outlined.
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.
What I was trying to indicate, not very clearly, is that for a sum that
is not evaluated in closed form, N[Sum[...,prec]] might or might not
give a reliable result. One is relying on the behavior of NValues for
Sum to make appropriate use of the prec argument. Even NSum may not be
reliable (it is not, in this case; I'm not sure if that constitutes a
bug or feature). But the messages low accuracy of the result in both
cases gives warning that something may be amiss.
> 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
The sum is analytic. Here are two ways to see this. One is to recognize
the closed forms as indicated in previous replies by myself and Maxim
Rytin. The other is a standard proof: Each summand is analytic, so it
suffices to show that, for any fixed z, the sum converges in a
neighborhood of z. But that's pretty clear from the rapid decay in
Exp[-k^2] as k ranges over the integers.
As for the small quantities in the difference between actual and
originally proposed values, that appears to be fairly interesting. There
was some discussion of it in the sci.math.num-analysis thread.
http://groups-beta.google.com/group/sci.math.num-analysis/browse_frm/thread/97b7b0e07e14abbe/e8174dd813b6cf59?tvc=1#e8174dd813b6cf59
If nothing else this makes for a really nice counterexample. Of
somehing. (Maybe of several things.)
Daniel Lichtblau
Wolfram Research
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