Re: Re: Re: Infinite sum of gaussians

```On 17 Apr 2005, at 16:07, Maxim wrote:
>>
>>
>
> This is wrong on several points. In fact, Sum[E^(-(z - k)^2/2), {k,
> -Infinity, Infinity}] is analytic everywhere in the complex plane.
> Since
> we already know that this sum is equal to Sqrt[2*Pi]*EllipticTheta[3,
> Pi*z, E^(-2*Pi^2)], all its properties, including analyticity, follow
> from
> the properties of EllipticTheta.
>
> Actually, since the sum of E^(-(z - k)^2/2) is very well-behaved (the
> terms decay faster than, say, E^(-k^2/4))
>  it is trivial to prove the
> uniform convergence in z and therefore the validity of the termwise
> differentiation as well as analyticity directly. The fact that the
> series
> is double infinite is of no importance; we can always rewrite it as two
> series from 1 to +Infinity.

Yes, you are right, and in fact that is what I thought at first.  But
after Carl Woll's message I realized  that I could easily prove the
following:  if f[z] is everywhere complex analytic  and Abs[z f[z]]->0
as z->Infinity then Sum[f[z],{z,-Infinity,Infinity}]==0. I thought that
this shows that that the above sum can't be convergent everywhere, but
I have had not time to think about it for more than a few minutes at a
time so I am probably not missing something even now.

>
> Also it's not correct that a real infinitely differentiable function
> can
> be defined by its value and the values of its derivatives at a point.
> If
> we take f[z] == E^-z^-2 for z != 0 and f[0] == 0, then all the (real)
> derivatives at 0 vanish.
>

Of course but nobody ever said that real analytic is the same as
C^Ininity. Real analitic means that the Taylor series converges
everhwehre and is equal to the value of the function. This is all that
was needed in this case anyway.

Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/

```

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