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From: Maury Barbato on 31 Oct 2009 03:45
TCL wrote:

>
> "Maury Barbato" <mauriziobarbato(a)aruba.it> wrote in
> message
> news:106944995.141200.1256921300699.JavaMail.root(a)gall
> ium.mathforum.org...
> > <snipped>
> > I don't see how to prove that this lim sup is
> infinity.
> > For any even integer n, you obtain
> >
> > |f^(n)(0)|/n! = sum_{k=0 to inf}
> (2^(n*k))/((n!)*(k!))
> >
> > So?
>
> This is exp(2^n)/n!.
> Now apply ratio test.
>
> TCL
>
>

Ok, ok, ... I was quite tired ...
From: Maury Barbato on 31 Oct 2009 07:27
David C. Ullrich wrote:

> On Thu, 29 Oct 2009 15:26:40 EDT, Maury Barbato
> <mauriziobarbato(a)aruba.it> wrote:
>
> >Hello,
> >let a:N->Q a bijection, and let f:R->R be the
> >function
> >
> > e^[-(1/x)] if x > 0
> >f(x) =
> > 0 if x <= 0
> >
> >
> >Define g:R->R as follows
> >
> >g(x) = sum_{n=0 to inf} [2^(-n)]*f(x - a(n))
> >
> >g is well defined and it's easy to prove, using
> >the theorem on the derivative of a function series,
> >that g is C^(inf).
> >I think, anyhow, that g is nowhere analytic, but
> >I can't give a rigorous proof. Do you have some
> idea?
>
> As Zladislav points out this does't seem to be
> immediately
> obvious. I'd be very surprised if it were false...
>

Hmm, I was quite sure that there should be a simple
proof, but I'm changing my mind ... trying again
and again, I can't find a proof!

I found the function g above in the first edition
of the book "Geometry of Manifolds" by Bishop and
Crittenden. The authors, maybe taking into consideration
the complexity of their example, replaced g with the
following function

f(x)=sum_{n=0 to inf} [2^(-(2^n))]*exp[-(csc((2^n)*x))^2]

It's easy to see that f is well defined and C^(inf).
Since all the derivates of exp(-1/(x^2)) are zero in
x=0, you obtain, using repeatedly the chain rule

f^(k)(0)=0 for every k in N.

So f is not analytic in x=0, and, being periodic
with period pi, it's not analytic in every x=k*pi,
with k in Z. Dropping the first n+1 terms of the
series, you obtain a function, with period
pi/(2^(k+1)), which is not analytic in x=0. Noting
that the sum of the first n+1 terms of the series that
defines f is analytic in (0,pi/(2^n)), you obtain that
f is not analytic in x=k*pi/(2^(n+1)), for every k in
Z.

> >Thank you very much for your attention.
> >My Best Regards,
> >Maury Barbato
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)

Maury Barbato
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