convergence on the boundary ?
amy666
given a taylor series and its radius of convergence ( thus no special cases where the radius is undecidable )
for which elements on the boundary do we have convergence ?
i want to learn about this !
an idea :
let the circle be mapped to the interval [0,2pi] by letting the angle from the origin intersect with the circle.
and for this interval [0,2pi] define a function f(theta)
such that
f(theta) = 1 if we have convergence
f(theta) = 0 if we have divergence
if f(theta) can be given by a fourier series ,
how can we compute it from the given taylor series ?
probably a FAQ ?
regards
tommy1729
Dave L. Renfro
> convergence on the boundary of the radius of
> convergence of a taylor series.
>
> given a taylor series and its radius of convergence
> ( thus no special cases where the radius is undecidable )
>
> for which elements on the boundary do we have convergence ?
The convergence set has to be F_sigma_delta, since
the (pointwise) convergence set for any sequence
of continuous functions is F_sigma_delta.
Herzog and Piranian (together) proved in 1949 that
any F_sigma subset of |z| = 1 can be the convergence set
of some power series with radius of convergence 1. Lukasenko
proved in 1978 that some G_delta subsets of |z| = 1 cannot
be the convergence set of any power series with radius of
convergence 1. For a fairly elementary survey of the problem
of characterizing the convergence set for a power series
in C (complex numbers), see Thomas W. Korner, "The
behavior of power series on their circle of convergence"
[pp. 56-94 in "Banach Spaces, Harmonic Analysis, and
Probability Theory", Springer Lecture Notes in Mathematics
#995, Springer-Verlag, 1983]. This is a beautifully written
paper that contains detailed proofs of virtually everything
and is pitched at the level of a beginning graduate student
in math.
Dave L. Renfro
amy666
thanks.
but i dont have access to books.
i wanted it discussed here on sci.math.
regards
tommy1729
Bill
"amy666" <tomm...@hotmail.com> wrote in message
news:32552154.1228860634...@nitrogen.mathforum.org...
> convergence on the boundary of the radius of convergence of a taylor
> series.
>
> given a taylor series and its radius of convergence ( thus no special
> cases where the radius is undecidable )
>
> for which elements on the boundary do we have convergence ?
>
>
> i want to learn about this !
>
> an idea :
>
> let the circle be mapped to the interval [0,2pi] by letting the angle from
> the origin intersect with the circle.
>
> and for this interval [0,2pi] define a function f(theta)
>
> such that
>
> f(theta) = 1 if we have convergence
> f(theta) = 0 if we have divergence
>
I think a relevant idea is that the Lebesgue measure of { theta: f(theta)
= 1} is 0 if the function lies in the Hardy space H^2(D). If the function
lies in the Dirichlet space, then the logarithmic capacity of { theta:
f(theta) = 1} is 0. The latter is a much stronger statement. Here is a nice
book:
P. Kahane R. Salem,Ensembles Parfaits et Séries Trigonométriques, Hermann,
Paris, 1963.
I know it is written in French, but I learned enough of the French language
to read most of it, so I can assure you that it is readable even if you
don't know any French at all. With regard to your other post, I don't think
you can expect the same level of exposition in a newsgroup that is available
in a book. In a newsgroup community, I am very greatful when anyone offers
me any sort of helpful advice or guides me towards the information I am
seeking.
Bill
Bill
"Bill" <Bill_...@comcast.net> wrote in message
news:ghn0j...@news7.newsguy.com...
>
> "amy666" <tomm...@hotmail.com> wrote in message
> news:32552154.1228860634...@nitrogen.mathforum.org...
>> convergence on the boundary of the radius of convergence of a taylor
>> series.
>>
>> given a taylor series and its radius of convergence ( thus no special
>> cases where the radius is undecidable )
>>
>> for which elements on the boundary do we have convergence ?
>>
>>
>> i want to learn about this !
>>
>> an idea :
>>
>> let the circle be mapped to the interval [0,2pi] by letting the angle
>> from the origin intersect with the circle.
>>
>> and for this interval [0,2pi] define a function f(theta)
>>
>> such that
>>
>> f(theta) = 1 if we have convergence
>> f(theta) = 0 if we have divergence
>>
>
> I think a relevant idea is that the Lebesgue measure of { theta:
> f(theta) = 0} is 0 if the function lies in the Hardy space H^2(D). If the
> function lies in the Dirichlet space, then the logarithmic capacity of {
> theta: f(theta) = 0} is 0. The latter is a much stronger statement. Here
William Elliot
> convergence on the boundary of the radius of convergence of a taylor series.
>
> given a taylor series and its radius of convergence ( thus no special
> cases where the radius is undecidable )
>
> for which elements on the boundary do we have convergence ?
> i want to learn about this !
>
> an idea :
>
For blatantly flaunting deliberate illiteracy,
he deserves Capital Punishment:
The first person singular pronoun 'I' is capitalized.
Sentences begin with a capital letter.
Proper names are capitalized.
David C. Ullrich
wrote:
>convergence on the boundary of the radius of convergence of a taylor series.
>
>given a taylor series and its radius of convergence ( thus no special cases where the radius is undecidable )
>
>for which elements on the boundary do we have convergence ?
This is a long story. I don't think that a complete answer is known.
(That is, a complete answer to the question of what subsets of the
boundary can be the set of convergence for some power series.)
Renfro has provided two classical references and a third that
I wasn't aware of - given that the third is by Korner I tend
to believe him when he says it's an excellent survey.
Hmm, come to think of it it's not just a long story, it's
any one of several long stories, depending on exactly
what you mean by the question. Are you asking about
the set on the boundary where the series converges,
or the set on the boundary where the function defined
by the series has a radial limit, or some other sort of
limit, or what?
>
>i want to learn about this !
Then you're going to need to get access to books somehow.
>an idea :
>
>let the circle be mapped to the interval [0,2pi] by letting the angle from the origin intersect with the circle.
>
>and for this interval [0,2pi] define a function f(theta)
>
>such that
>
>f(theta) = 1 if we have convergence
>f(theta) = 0 if we have divergence
>
>if f(theta) can be given by a fourier series ,
>
>how can we compute it from the given taylor series ?
I seriously doubt that there's any reasonably simple way
to do this.
It's also very hard to see why one would care, because
the Fourier series for f does not determine f anyway.
The set of convergence can for example be a set of
measure 0, in which case that Fourier series is going
to be simply 0, in which case it says nothing at all
about the set of convergence.
>
>probably a FAQ ?
>
>
>regards
>
>tommy1729
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
wrote:
>
>"amy666" <tomm...@hotmail.com> wrote in message
>news:32552154.1228860634...@nitrogen.mathforum.org...
>> convergence on the boundary of the radius of convergence of a taylor
>> series.
>>
>> given a taylor series and its radius of convergence ( thus no special
>> cases where the radius is undecidable )
>>
>> for which elements on the boundary do we have convergence ?
>>
>>
>> i want to learn about this !
>>
>> an idea :
>>
>> let the circle be mapped to the interval [0,2pi] by letting the angle from
>> the origin intersect with the circle.
>>
>> and for this interval [0,2pi] define a function f(theta)
>>
>> such that
>>
>> f(theta) = 1 if we have convergence
>> f(theta) = 0 if we have divergence
>>
>
>I think a relevant idea is that the Lebesgue measure of { theta: f(theta)
>= 1} is 0 if the function lies in the Hardy space H^2(D).
Of course, _if_, as I suspect, "convergence" means convergence
of the series, then it's also worth mentioning that this is a
hugely deep theorem (as opposed to the corresponding statement
about the set where the function has a radial limit).
Unless of course I'm being stupid again and H^2 is much
easier than L^2 here, but I doubt that.
>If the function
>lies in the Dirichlet space, then the logarithmic capacity of { theta:
>f(theta) = 1} is 0. The latter is a much stronger statement. Here is a nice
>book:
>
>P. Kahane R. Salem,Ensembles Parfaits et Séries Trigonométriques, Hermann,
>Paris, 1963.
>
>I know it is written in French, but I learned enough of the French language
>to read most of it, so I can assure you that it is readable even if you
>don't know any French at all. With regard to your other post, I don't think
>you can expect the same level of exposition in a newsgroup that is available
>in a book. In a newsgroup community, I am very greatful when anyone offers
>me any sort of helpful advice or guides me towards the information I am
>seeking.
>
>Bill
>
>
>
>> if f(theta) can be given by a fourier series ,
>>
>> how can we compute it from the given taylor series ?
>>
>>
>> probably a FAQ ?
>>
>>
>> regards
>>
>> tommy1729
>
David C. Ullrich
David C. Ullrich
Duh, of course it's not - it's clear that a.e. convergence
for H^2 implies a.e convergence for L^2.
Dave L. Renfro
> Renfro has provided two classical references
> and a third that I wasn't aware of - given that
> the third is by Korner I tend to believe him
> when he says it's an excellent survey.
I should mention that one of the drawbacks to
Korner's survey is that he gives almost no references
or historical comments (maybe none at all -- it's
been several years since I last looked at it and
don't remember now). My guess is his survey was
lightly edited from class/seminar notes, and maybe
the preface says something to this effect (again,
I don't remember). As for giving super-detailed and
complete proofs, his survey is a little like C. A.
Rogers' book "Hausdorff Measures", except of course
Rogers' book gives a lot of references.
Dave L. Renfro
Bill
>>> an idea :
>>>
>>> let the circle be mapped to the interval [0,2pi] by letting the angle from
>>> the origin intersect with the circle.
>>>
>>> and for this interval [0,2pi] define a function f(theta)
>>>
>>> such that
>>>
>>> f(theta) = 1 if we have convergence
>>> f(theta) = 0 if we have divergence
>>>
>> I think a relevant idea is that the Lebesgue measure of { theta: f(theta)
>> = 0} is 0 if the function lies in the Hardy space H^2(D).
>
> Of course, _if_, as I suspect, "convergence" means convergence
> of the series, then it's also worth mentioning that this is a
> hugely deep theorem (as opposed to the corresponding statement
> about the set where the function has a radial limit).
>
I confess I was thinking about radial (and non-tangential limits).
I suspect that the answers sought lie in Antoni Zygmund's treatise,
Trigonometric Series.
Bill
amy666
in general : no , f is not determined by the fourier series.
therefore i wrote : " if f can be given by a fourier series "
IF
how do we determine if its possible to give f by its fourier series , given the taylor series and radius of the original function ?
> The set of convergence can for example be a set of
> measure 0, in which case that Fourier series is going
> to be simply 0, in which case it says nothing at all
> about the set of convergence.
>
> >
> >probably a FAQ ?
> >
> >
> >regards
> >
> >tommy1729
>
> 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.)
regards
tommy1729
Bill
Keep in mind that the original function f may only be defined almost
everywhere, so in many cases
the Fourier Series for f is a better behaved function which may even be
defined at more points than
the original function. In a nutshell, the better the behavior of the Fourier
coefficients (which at once implies
that f is in L^1 or L^2), the better the behavior of the Fourier Series for
f. Here, "better" could can translated
to "the Fourier coefficent a_n doesn't grow too fast as n gets large".
Bill
David C. Ullrich
wrote:
_No_ function is determined by its Fourier series.
That's why I said "does not determine". Given any
function f that _has_ a Fourier series there is always
_another_ function with the _same_ Fourier series.
Obviously.
>therefore i wrote : " if f can be given by a fourier series "
>
>IF
>
>
>how do we determine if its possible to give f by its fourier series , given the taylor series and radius of the original function ?
Exactly what do you mean when you say "give f by its fourier series"?
I can think of several different things that might mean.