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From: Maury Barbato on 29 Oct 2009 13:16
Robert Israel wrote:

> Timothy Murphy <gayleard(a)eircom.net> writes:
>
> > Maury Barbato 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?
> >
> > I don't see why your series is convergent, say for
> x = 0,
> > since you might have a(n) < 0 and very small
> > for arbitrarily large n.
>
> The problem is not when a(n) is small, but when it is
> large.
> I think he means to use a bounded f, say replace
> exp(-1/x) by exp(-1/x)/(1 + exp(-1/x)). Then the
> 2^(-n) ensures
> that the series converges uniformly.
> --

Is exp(-1/x) not bounded on (0,inf)? Ehm?!?
Am I drunk tonight?

> Robert Israel
> israel(a)math.MyUniversitysInitials.ca
> Department of Mathematics
> http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver,
> BC, Canada
From: Robert Israel on 30 Oct 2009 00:08
Robert Israel <israel(a)math.MyUniversitysInitials.ca> writes:

> Timothy Murphy <gayleard(a)eircom.net> writes:
>
> > Maury Barbato 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?
> >
> > I don't see why your series is convergent, say for x = 0,
> > since you might have a(n) < 0 and very small
> > for arbitrarily large n.
>
> The problem is not when a(n) is small, but when it is large.

Sorry, forget that. Must be lack of sleep.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Maury Barbato on 30 Oct 2009 03:46
Timothy Murphy wrote:

> Maury Barbato 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?
>
> I don't see why your series is convergent, say for x
> = 0,
> since you might have a(n) < 0 and very small
> for arbitrarily large n.
>

Ehm?! I defined f:R->R as

f(x) = exp(-1/x) if x > 0

f(x) = 0 if x <= 0

This is a bounded function, so ...

>
> --
> Timothy Murphy
> e-mail: gayleard /at/ eircom.net
> tel: +353-86-2336090, +353-1-2842366
> s-mail: School of Mathematics, Trinity College,
> Dublin 2, Ireland
From: David C. Ullrich on 30 Oct 2009 07:51
On Thu, 29 Oct 2009 19:40:54 +0000, Timothy Murphy
<gayleard(a)eircom.net> wrote:

>Maury Barbato 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?
>
>I don't see why your series is convergent, say for x = 0,
>since you might have a(n) < 0 and very small
>for arbitrarily large n.

Look again, a little more carefully.




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.)
From: David C. Ullrich on 30 Oct 2009 07:59
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...

>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.)
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