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From: Dave L. Renfro on 30 Oct 2009 09:47
Zdislav V. Kovarik wrote (in part):

> (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
From: Maury Barbato on 30 Oct 2009 08:47
Zdislav V. Kovarik wrote:

> A correction of Cauchy-Hadamard formula below.
>
> On Thu, 29 Oct 2009, Zdislav V. Kovarik wrote:
>
> >
> >
> > On Thu, 29 Oct 2009, 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?
> > >
> > > Thank you very much for your attention.
> > > My Best Regards,
> > > Maury Barbato
> >
> > (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!)
> >
> Correction: 1/R = lim sup (abs (f^(n)(0))/n!)^(1/n)
>
> My apologies.
>

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?

> > 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).
> >
From: TCL on 30 Oct 2009 14:09

"Maury Barbato" <mauriziobarbato(a)aruba.it> wrote in message
news:106944995.141200.1256921300699.JavaMail.root(a)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


From: W^3 on 30 Oct 2009 14:23
In article
<106944995.141200.1256921300699.JavaMail.root(a)gallium.mathforum.org>,
Maury Barbato <mauriziobarbato(a)aruba.it> wrote:

> Zdislav V. Kovarik wrote:
>
> > A correction of Cauchy-Hadamard formula below.
> >
> > On Thu, 29 Oct 2009, Zdislav V. Kovarik wrote:
> >
> > >
> > >
> > > On Thu, 29 Oct 2009, 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?
> > > >
> > > > Thank you very much for your attention.
> > > > My Best Regards,
> > > > Maury Barbato
> > >
> > > (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!)
> > >
> > Correction: 1/R = lim sup (abs (f^(n)(0))/n!)^(1/n)
> >
> > My apologies.
> >
>
> 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?

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.

> > > 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).
> > >
From: W^3 on 30 Oct 2009 19:22
In article <aderamey.addw-E29479.11232330102009(a)News.Individual.NET>,
W^3 <aderamey.addw(a)comcast.net> wrote:

> In article
> <106944995.141200.1256921300699.JavaMail.root(a)gallium.mathforum.org>,
> Maury Barbato <mauriziobarbato(a)aruba.it> wrote:
>
> > Zdislav V. Kovarik wrote:
> >
> > > A correction of Cauchy-Hadamard formula below.
> > >
> > > On Thu, 29 Oct 2009, Zdislav V. Kovarik wrote:
> > >
> > > >
> > > >
> > > > On Thu, 29 Oct 2009, 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?
> > > > >
> > > > > Thank you very much for your attention.
> > > > > My Best Regards,
> > > > > Maury Barbato
> > > >
> > > > (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!)
> > > >
> > > Correction: 1/R = lim sup (abs (f^(n)(0))/n!)^(1/n)
> > >
> > > My apologies.
> > >
> >
> > 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?
>
> 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.

That < should have been >.

> > > > 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).
> > > >
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