A Nowhere Analytic Function
Maury Barbato
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?
Thank you very much for your attention.
My Best Regards,
Maury Barbato
Timothy Murphy
> 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.
--
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
Robert Israel
> 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.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Zdislav V. Kovarik
(There is always a risk that infinite summation may "iron out" the lack of
analyticity.)
If you settle for another example of a C^(inf) nowhere analytic function
with an easy proof, take
f(x) = sum(n=0 to infinity) cos(2^n * x)/n!
It is easy to calculate its derivatives at x=0, and when you apply
Cauchy-Hadamard formula for radius of convergence, it will come out as
zero.
1/R = lim sup (abs (f^(n)(0))/n!)
Now, f is (2*pi) periodic, so integer multiples of 2*pi are points of
non-analyticity.
If you drop finitely many initial terms of the series, the period shortens
by powers of 2, so you get lack of analyticity at all binary fractions of
2*pi and their integer multiples. That is a dense subset, and you are
done.
(Inspired by Weierstrass's example of an everywhere continuous, nowhere
differentiable function.)
Cheers, ZVK(Slavek).
Zdislav V. Kovarik
Correction: 1/R = lim sup (abs (f^(n)(0))/n!)^(1/n)
My apologies.
Maury Barbato
Is exp(-1/x) not bounded on (0,inf)? Ehm?!?
Am I drunk tonight?
Robert Israel
> Timothy Murphy <gayl...@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.
Maury Barbato
> 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 ...
David C. Ullrich
<gayl...@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.)
David C. Ullrich
<maurizi...@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
Dave L. Renfro
> (There is always a risk that infinite summation may "iron out"
> the lack of analyticity.)
For those interested, some examples in which various types
of singularities in the partial sums of infinite series of
functions are proved to be NOT "ironed out" in the limit
can be found in the following, both of which are freely
available on the internet.
Israel Halperin, "Discontinuous functions with the Darboux
property", Canadian Mathematical Bulletin 2 (1959), 111-118.
[See p. 13.]
http://books.google.com/books?id=-st_B62xKbYC&pg=PA111
Ernest William Hobson, "The theory of Functions of a Real
Variable and the Theory of Fourier's Series", 1st edition
Cambridge University Press, 1907.
[See Sections 421-424 (pp. 607-620).]
http://books.google.com/books?id=PxgPAAAAIAAJ
Dave L. Renfro
Maury Barbato
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?
TCL
"Maury Barbato" <maurizi...@aruba.it> wrote in message
news:106944995.141200.12569...@gallium.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
W^3
<106944995.141200.12569...@gallium.mathforum.org>,
Maury Barbato <maurizi...@aruba.it> wrote:
The last sum = e^(2^n)/n! < e^(2^n)/n^n. The nth root of the last
expression = e^(2^n/n)/n -> oo.
W^3
W^3 <aderam...@comcast.net> wrote:
That < should have been >.
Maury Barbato
>
> "Maury Barbato" <maurizi...@aruba.it> wrote in
> message
> news:106944995.141200.1256921300699.JavaMail.root@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 ...
Maury Barbato
> On Thu, 29 Oct 2009 15:26:40 EDT, Maury Barbato
> <maurizi...@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