looking for...
max01000011
infinitely differentiable at EVERY point in R
analytic at NO point in R
my mind is blank
thx
max
Zdislav V. Kovarik
max01000011 <pxa...@SPAMinwind.it> wrote:
:... a function f:R->R
This has been around several times. Search through past
contributions would be much easier if the authors made their
subject lines more story-telling, such as
C-infinity non-analytic
instead of your
looking for...
This being said, here is one such function:
sum[n=0 to infinity] cos(2^n*x)/n!
Cheers, ZVK(Slavek)
Zdislav V. Kovarik
sum[n=0 to infinity] sin(2^n*x)/n!
(see my other response)
max01000011
"Zdislav V. Kovarik" <kov...@mcmail.cis.McMaster.CA> ha scritto nel
messaggio news:ab6fej$4...@mcmail.cis.mcmaster.ca...
> In article <t4zB8.17451$US3.4...@twister1.libero.it>,
> max01000011 <pxa...@SPAMinwind.it> wrote:
> :... a function f:R->R
> :infinitely differentiable at EVERY point in R
> :analytic at NO point in R
> :my mind is blank
> :thx
> :max
> :
> This has been around several times. Search through past
> contributions would be much easier if the authors made their
> subject lines more story-telling, such as
>
> C-infinity non-analytic
>
> instead of your
>
> looking for...
note taken. thanks for yous wise suggestion :)
> This being said, here is one such function:
>
> sum[n=0 to infinity] cos(2^n*x)/n!
ok. a proof?
max01000011
proof? is it simple?
thx
max
max01000011
"Zdislav V. Kovarik" <kov...@mcmail.cis.McMaster.CA> ha scritto nel
messaggio news:ab6fj0$o...@mcmail.cis.mcmaster.ca...
mhhh... do you know of a natural way to "generalise" the canonocal example
of a f C-oo on R and not analytic in 0, namely exp(-1/x^2) ?
Denis Feldmann
"max01000011" <pxa...@SPAMinwind.it> a écrit dans le message news:
WCzB8.17735$US3.4...@twister1.libero.it...
Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
f(x-q_n)/2^n), where q_n is an enumeration of Q...
>
>
Denis Feldmann
"max01000011" <pxa...@SPAMinwind.it> a écrit dans le message news:
WCzB8.17735$US3.4...@twister1.libero.it...
>
Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
Dave L. Renfro
[sci.math 6 May 2002 13:44:51 -0400]
http://mathforum.org/epigone/sci.math/jarplunswerm
wrote (in part):
> This has been around several times. Search through past
> contributions would be much easier if the authors made their
> subject lines more story-telling, such as
>
> C-infinity non-analytic
>
> instead of your
>
> looking for...
I didn't even read this thread earlier because I didn't
think it would be anything I'd be interested in. It just
so happens that I have some stuff written up about nowhere
analytic C-infinity functions. It's part of a LaTex file
manuscript I have (an English translation someone did for
me, and which I typed up and added a lot of additional notes to,
of V. S. Fedorov's 1987 Thesis (under A. M. Olevskii) "On
intersections of continuous functions with families of smooth
functions"), so I'll post the C-infinity comments later this
evening or tomorrow sometime -- I first need to de-TeX and reformat
the text so that it'll be suitable for posting.
Since I'm sortta a Baire category freak, it'll be slanted towards
that end. [Most C-infinity functions are nowhere analytic in the
sense of Baire category when you use the standard norm on C-infinity
(or any norm that generates the topology of uniform convergence for
all orders of derivatives on compact sets).]
Dave L. Renfro
David C. Ullrich
Well, it's infinitely differentiable, because if you
formally take the k-th derivative term by term you get
f^(k) ~ sum[n=1 to infinity] 2^(kn) S(2^n*x)/n!
where S is either cos, -sin, -cos or sin, depending on k,
and that series converges absolutely.
To show it's nowhere analytic involves a trick or two.
Suppose it's analytic in an interval of length L.
Choose N so 2^(-N)*2*Pi < L. Write
f(x) = sum[n=0 to infinity] cos(2^n*x)/n!
= sum{n=0 to N-1] + sum[n=N to infinity]
= I(x) + II(x).
Now I is certainly analytic on the entire line. And
II is analytic in an interval of length L, but if
you look at it you see II is periodic with period
2^(-N)*2*Pi; since 2^(-N)*2*Pi < L it follows that
II is analytic on the entire line. So f is analytic
on the entire line.
In particular f is analytic at x = 0. But that
can't be:
f^(4*k)(0) = sum[n=1 to infinity] 16^(k*n)/n! = exp(16^k),
which shows that a power series with coefficients f^(k)(0)/k!
has radius of convergence 0.
>
David C. Ullrich
David C. Ullrich
I believe that this is probably an example of the required
type, but proving that might be tricky. Seems simpler to
show that the example ZK gave is nowhere analytic (as I
did just now in my reply in the _first_ thread on this
topic...)
Maybe proving this example works is easy(???)
David C. Ullrich
David C. Ullrich
wrote:
[...]
>
>Since I'm sortta a Baire category freak, it'll be slanted towards
>that end. [Most C-infinity functions are nowhere analytic in the
>sense of Baire category when you use the standard norm on C-infinity
>(or any norm that generates the topology of uniform convergence for
>all orders of derivatives on compact sets).]
Surely you don't actually mean "norm".
(If f_n -> f in this topology then it's clear that f_n' -> f'
as well, so the derivative is a continuous map from C-infinity to
itself. So if C-infinity were normable the derivative would be
a bounded map, but exponentials show it isn't.)
>Dave L. Renfro
David C. Ullrich
The World Wide Wade
"Denis Feldmann" <denis.f...@wanadoo.fr> wrote:
> > mhhh... do you know of a natural way to "generalise" the canonocal example
> > of a f C-oo on R and not analytic in 0, namely exp(-1/x^2) ?
>
> Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
> f(x-q_n)/2^n), where q_n is an enumeration of Q...
Mmmm ... and the proof? (C^oo on R is easy, lack of analyticity seems
harder).
--WWW.
Denis Feldmann
"The World Wide Wade" <wrame...@attbi.remove13.com> a écrit dans le
message news: wrameyxiii-A92A9...@news.attbi.com...
Baire theorem?
>
> --WWW.
G. A. Edgar
<pxa...@SPAMinwind.it> wrote:
The Fabius function is an interesting example of this.
I have a picture at <http://www.math.ohio-state.edu/~edgar/selfdiff/>
Reference:
J. Fabius,
"A probabilistic example of a nowhere-analytic $C^\infty$-function"
Zeit. Wahrscheinlichkietstheorie Verw. Geb. 5 (1966) 173--174
This example is in [0,1], but there is a natural way to extend
it to the whole line to retain the property everywhere.
An exposition is in
K. Stromberg
Probability for Analysts
Chapman \& Hall
New York
1994
--
G. A. Edgar http://math.ohio-state.edu/~edgar/
David C. Ullrich
Um???
>>
>> --WWW.
>
>
David C. Ullrich
Dave L. Renfro
http://mathforum.org/epigone/sci.math/gaxcherlyel
wrote
>> Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
>> f(x-q_n)/2^n), where q_n is an enumeration of Q...
>
> Mmmm ... and the proof? (C^oo on R is easy, lack of
> analyticity seems harder).
The points of non-analyticity form a closed set, so it suffices
to prove that f isn't analytic at each rational number. You can
prove this by showing that the Taylor expansion about each rational
number is the zero function (and hence it doesn't converge to f in
any neighborhood of that rational number), and presumably this
can be done by term-by-term differentiations, using the chain
rule and what know about the derivatives of exp(-1/x^2) at x=0.
[It would follow from local uniform convergence of the n'th order
differentiated series, for each n = 1, 2, ..., which seems evident
since the 2^(-n) factor sufficiently damps out the terms for the
Weierstrass M-test to be applied to the n'th order differentiated
series (whose terms have the form of a rational function of x
times exp(-1/x^2) times 2^(-n)).]
Dave L. Renfro
Dave L. Renfro
[sci.math Mon, 06 May 2002 23:28:09 GMT]
http://mathforum.org/epigone/sci.math/jarplunswerm
wrote
>> sense of Baire category when you use the standard norm on C-infinity
>> (or any norm that generates the topology of uniform convergence for
>> all orders of derivatives on compact sets).]
>
> Dave L. Renfro
>
> Surely you don't actually mean "norm".
>
> (If f_n -> f in this topology then it's clear that f_n' -> f'
> as well, so the derivative is a continuous map from C-infinity to
> itself. So if C-infinity were normable the derivative would be
> a bounded map, but exponentials show it isn't.)
Ahh, a non-normable metric on a vector space (besides the
L^p spaces for 0 < p < 1).
Hummm...I guess this is what all that semi-norm stuff was about
in Rudin's "Functional Analysis" (Section 1.44).
Dave L. Renfro [slowly getting back up, tail between legs . . .]
max01000011
"Dave L. Renfro" <renf...@cmich.edu> ha scritto nel messaggio
news:28ae5e5e.02050...@posting.google.com...
> The World Wide Wade <wrame...@attbi.remove13.com>
> [sci.math Tue, 07 May 2002 01:49:00 GMT]
> http://mathforum.org/epigone/sci.math/gaxcherlyel
>
> wrote
>
> >> Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
> >> f(x-q_n)/2^n), where q_n is an enumeration of Q...
> >
> > Mmmm ... and the proof? (C^oo on R is easy, lack of
> > analyticity seems harder).
>
> The points of non-analyticity form a closed set,
how come?
Chip Eastham
Since the "singularities" are dense in R, the function would be
singular at every point of R. I think that's a basic result of
complex variables, limit point of singularities is a singularity.
-- chip
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu>
Yeah, I think so. A countable collection of seminorms defines a
metric, but it almost never defines a normed space. (heh - we
leave it to you to define a topology on the class of, say,
separable metrizable topological vector spaces so that this
is true in the sense of category...)
The simplest/most intuitive example of a non-normable metrizable
topological vector space is probably C(R) (with uniform convergence
on compact sets). If the topology were given by a norm ||.||
then that semi-norm stuff would show that there exists a compact
K such that ||f|| <= c||f||_K, and hence uniform convergence on K
would imply uniform convergence on every compact subset of R.
>Dave L. Renfro [slowly getting back up, tail between legs . . .]
If you're not actually putting us on then skip the next
paragraph:
Surely you're putting us on, and saying "norm" instead of
"metric" was just an online slip. I imagine that everyone
thinks metric -> norm at some point, but they realize that's
wrong a little later (and since you mention L^p you evidently
did realize it was wrong long ago...)
David C. Ullrich
David C. Ullrich
wrote:
>The World Wide Wade <wrame...@attbi.remove13.com>
>[sci.math Tue, 07 May 2002 01:49:00 GMT]
>http://mathforum.org/epigone/sci.math/gaxcherlyel
>
>wrote
>
>>> Yes. Let f(x)=exp(-1/x^2) for x>0, 0 for x <=0, and g(x) =sum(
>>> f(x-q_n)/2^n), where q_n is an enumeration of Q...
>>
>> Mmmm ... and the proof? (C^oo on R is easy, lack of
>> analyticity seems harder).
>
>The points of non-analyticity form a closed set, so it suffices
>to prove that f isn't analytic at each rational number.
Yes...
>You can
>prove this by showing that the Taylor expansion about each rational
>number is the zero function
Not for any reason I can see. The Taylor expansion of f_n about
q_n is 0, if f_n(x) = f(x-q_n), but there are other terms in
the definition of g. (The f_m with q_m > q_n are ok, but there
exist m with q_m < q_n...)
Like I said to Denis, I suspect that this is in fact an example
of the required type, but I don't see how to prove it. Using
the example ZK gave seems much easier (I posted a proof under
"looking for..." that that example behaves as advertised.)
>(and hence it doesn't converge to f in
>any neighborhood of that rational number), and presumably this
>can be done by term-by-term differentiations, using the chain
>rule and what know about the derivatives of exp(-1/x^2) at x=0.
>[It would follow from local uniform convergence of the n'th order
>differentiated series, for each n = 1, 2, ..., which seems evident
>since the 2^(-n) factor sufficiently damps out the terms for the
>Weierstrass M-test to be applied to the n'th order differentiated
>series (whose terms have the form of a rational function of x
>times exp(-1/x^2) times 2^(-n)).]
Not sure I followed that - is this an explanation for why
the Taylor series about q_n is 0?
David C. Ullrich
wrote:
How do you know that q_n is a singularity (by which I take it you
mean that g is not analytic at q_n)? That seems likely - how do
you prove it?
The problem is this: Let's say f_n(x) = f(x-q_n)/2^n. Then q_n
is certainly a singularity of f_n; the Taylor series for f_n
does not converge to f_n in any neighborhood of q_n. How do
you know that the power series for g does not converge to g
in any neighborhood of q_n? There are all those other terms;
how do you know that it does not happen that the other terms
make the power series converge? Seems unlikely.
>-- chip
David C. Ullrich
The World Wide Wade
renf...@cmich.edu (Dave L. Renfro) wrote:
> You can
> prove this by showing that the Taylor expansion about each rational
> number is the zero function (and hence it doesn't converge to f in
> any neighborhood of that rational number), and presumably this
> can be done by term-by-term differentiations
If all Taylor coefficents vanish on a dense set, then all derivatives
vanish on a dense set, hence the function is 0 everywhere. (In fact it's
easy to see this function and its derivative is positive everywhere => the
Taylor expansion is never the zero function.)
--WWW.
Dave L. Renfro
[sci.math Tue, 07 May 2002 14:03:38 GMT]
http://mathforum.org/epigone/sci.math/gaxcherlyel
wrote (in part, in response to a post of mine):
>> You can
>> prove this by showing that the Taylor expansion about each
>> rational number is the zero function
>
> Not for any reason I can see. The Taylor expansion of f_n about
> q_n is 0, if f_n(x) = f(x-q_n), but there are other terms in
> the definition of g. (The f_m with q_m > q_n are ok, but there
> exist m with q_m < q_n...)
>
> Like I said to Denis, I suspect that this is in fact an example
> of the required type, but I don't see how to prove it. Using
> the example ZK gave seems much easier (I posted a proof under
> "looking for..." that that example behaves as advertised.)
Well, darn it, aren't they supposed to be orthogonal or
something? [Yes, I know this is nonsense. But that's how
I feel right now . . .]
>> (and hence it doesn't converge to f in
>> any neighborhood of that rational number), and presumably this
>> can be done by term-by-term differentiations, using the chain
>> rule and what know about the derivatives of exp(-1/x^2) at x=0.
>> [It would follow from local uniform convergence of the n'th order
>> differentiated series, for each n = 1, 2, ..., which seems evident
>> since the 2^(-n) factor sufficiently damps out the terms for the
>> Weierstrass M-test to be applied to the n'th order differentiated
>> series (whose terms have the form of a rational function of x
>> times exp(-1/x^2) times 2^(-n)).]
>
> Not sure I followed that - is this an explanation for why
> the Taylor series about q_n is 0?
I'm not sure of much right now. I think I've been working on too
many different things the past couple of days to keep anything
straight.
Oh, well, for what it's worth, here's an example that I got
from Eric Bedford years ago. He said it was floating around
Princeton back when he was a student there --->>>
SUM(n=0 to infinity) of 2^(-2^n) * exp{ -1 / [sin(x*2^n)]^2 }.
Maybe the super dampening factor makes the verification easier,
but I'm going to pass on saying anything more and get back to
grading my last class's exams, finishing the "translation" from
LaTeX to text of an essay on nowhere analytic C-infinity functions
(history of, Baire category results, etc.), finish photocopying
something that's due back at our interlibrary loan office today,
finish an e-mail to a student about what they missed on the exam,
and lots of other fun things.
Dave L. Renfro
Dave L. Renfro
[sci.math Tue, 07 May 2002 13:54:50 GMT]
http://mathforum.org/epigone/sci.math/jarplunswerm
wrote (in part):
> Surely you're putting us on, and saying "norm" instead of
> "metric" was just an online slip. I imagine that everyone
> thinks metric -> norm at some point, but they realize that's
> wrong a little later (and since you mention L^p you evidently
> did realize it was wrong long ago...)
Yeah, I know they're different: inner product ==> norm ==> metric
in the appropriate context, with none of these reversible; the
L^p spaces suffice to give examples. [Another way for the second
one is to observe that any metric defined from a norm is translation
invariant, and then find a metric on a vector space that isn't
translation invariant.]
Unfortunately, it was actually incorrect in a manuscript that I
lifted it from (not intended for publication), something that I'll
be posting later today (I hope). Oh well . . .
Maybe I can turn the events of today to my advantage --->>>
** A new large number for my large numbers list -- # mistakes I've made.
** A concrete example of asymptotic equivalence -- # posts with errors
vs. number of posts.
** Problems that I can use on analysis tests -- find the error(s)
in the following argument.
** An example of a high Cantor-Bendixson rank (for Bill
Taylor) -- number of re-posts needed to reach a perfect kernel.
Dave L. Renfro
Chip Eastham
For all the other f_k(x) = f(x - q_k)/2^n, k != n, the terms are
analytic at q_n, with a radius of convergence up to the nearest
singularity.
The Taylor series for f_n around q_n is all zeroes, and so does
converge (just not to f_n). So in fact the Taylor series for f
around q_n suffers from the accumulation of other "singularities"
at q_n and fails to converge in any open interval around q_n.
regards, chip
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu>
>(*) SUM(n=0 to infinity) of 2^(-2^n) * exp{ -1 / [sin(x*2^n)]^2 }.
>
>Maybe the super dampening factor makes the verification easier,
Possibly. Or it could be the extreme amount of symmetry. (Consider
the argument I gave for sum[n=0 to infinity] cos(2^n*x)/n!; there
I verified it was not analytic at 0 by explicitly calculating the
derivatives at the origin, something you could do for (*)
but not for the example above - also there the fact that
sum[n=N to infinity] sin(2^n*x)/n! has a small period helped
a lot in showing that non-analyticity at the origin implied
nowhere analytic - you could also do that with (*).)
>but I'm going to pass on saying anything more and get back to
>grading my last class's exams, finishing the "translation" from
>LaTeX to text of an essay on nowhere analytic C-infinity functions
>(history of, Baire category results, etc.), finish photocopying
>something that's due back at our interlibrary loan office today,
>finish an e-mail to a student about what they missed on the exam,
>and lots of other fun things.
>
>Dave L. Renfro
David C. Ullrich
David C. Ullrich
wrote:
In fact f_k is analytic at q_n, with a radius of convergence
of exactly |q_n - q_k|. And since there are q_k arbitrarily
close to q_n that means that we don't know anything about the
_sum_ of the f_k being analytic at q_n (or if we do you certainly
haven't proved it.)
If the f_k were all analytic in some _fixed_ neighborhood of q_n
then maybe one could argue this way.
>The Taylor series for f_n around q_n is all zeroes, and so does
>converge (just not to f_n). So in fact the Taylor series for f
>around q_n suffers from the accumulation of other "singularities"
>at q_n and fails to converge in any open interval around q_n.
>
>regards, chip
David C. Ullrich
David C. Ullrich
wrote:
Another way to say what the problem is: A sum that involves
sums of things analytic except for singularities with all
the properties you've used here, but the sum _is_ analytic.
This is going to be an uncountable sum, for simplicity.
For every x define h_x by h_x(x) = 1, h_x(t) = 0 for t <> x.
Then h = sum_x h_x is analytic everywhere (because it is
identically 1.) That's in spite of the fact that h_x is
analytic except for a singularity at x, so the singularities
of h_x definitely accumulate at 0, etc.
Dave L. Renfro
[sci.math 6 May 2002 13:47:12 -0400]
http://mathforum.org/epigone/sci.math/gaxcherlyel
Zdislav V. Kovarik began the thread "C-infinity nohwere analytic,
was Re: looking for..." a few days ago. I contributed a couple
of times on May 7 (one of my posts had several errors) and I also
mentioned that I was writing an essay on nowhere analytic
C-infinity functions that I'd post in sci.math. [It turns out I
was sick May 7 and May 8 (didn't realize how bad until May 8).
Maybe this is why so many of my May 7 posts had errors . . . ??]
Anyway, I've tried posting the essay to sci.math via google.com
twice in the past couple of days and neither time has it showed
up that I can tell -- neither of the first two attempts are at
http://groups.google.com/groups?oi=djq&as_ugroup=sci.math
or at
http://mathforum.org/epigone/sci.math
So then I posted the essay via the Math Forum, and you can find
it here --->>>
ESSAY ON NOWHERE ANALYTIC C-INFINITY FUNCTIONS
http://mathforum.org/epigone/sci.math/brengunsul
If you have any interest in this topic, I encourage you to
look at what I've written. It's fairly thorough, probably more
so than anything you can find in English -- even in print.
I discuss the early history of examples, Zahorski's complete
characterization for the (C)-points and (P)-points of a function
[sometimes called Type I and Type II singularities (perhaps not
respectively)], various Baire category proofs that most
C-infinity functions are nowhere analytic, and I even toss
out some research problems that appear promising (along with
my suggestions for how one might begin; these are things I
worked on a few years ago, not for very long however, which
it's appearing more and more likely that I'll never have a
chance to get back to). In fact, I'd rate this essay as among
my top four essays, the others being
HISTORICAL ESSAY ON CONTINUITY OF DERIVATIVES
http://mathforum.org/epigone/sci.math/pringquibroi
HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS
http://mathforum.org/epigone/sci.math/dwillydwul
ESSAY ON NON-DIFFERENTIABILITY POINTS OF MONOTONE FUNCTIONS
http://mathforum.org/epigone/sci.math/queewirand
The reason I'm announcing my essay with this post is that lately
(since this past January, in fact) posts made via the Math
Forum don't seem to be showing up anywhere else. Of course,
if my attempts at posting to google.com aren't working, then
this present post also might not be seen by anyone. However,
I'm hoping that maybe things will be different if I post in
reply to someone rather than by beginning a new thread. Plus,
I'm posting this from my home computer rather than at work,
although that shouldn't make a difference when posting via
google groups.
Dave L. Renfro
David C. Ullrich
wrote:
[...]
>
>The reason I'm announcing my essay with this post is that lately
>(since this past January, in fact) posts made via the Math
>Forum don't seem to be showing up anywhere else. Of course,
>if my attempts at posting to google.com aren't working, then
>this present post also might not be seen by anyone.
Here we see this post - haven't seen any of those mathforum
posts, and the other recent posts from google haven't appeared.
>However,
>I'm hoping that maybe things will be different if I post in
>reply to someone rather than by beginning a new thread. Plus,
>I'm posting this from my home computer rather than at work,
>although that shouldn't make a difference when posting via
>google groups.
If you're posting from home you must have an ISP(?) - if
not you can get unlimited internet access from various
ISP's (eg earthlink) for a little over $20/month. Your
ISP should have a news server; posting to an actual
news server is the most reliable way I've found to post
things to usenet.
Dave L. Renfro
[sci.math Fri, 10 May 2002 13:45:26 GMT]
http://mathforum.org/epigone/sci.math/gaxcherlyel
wrote (in part):
> If you're posting from home you must have an ISP(?) - if
> not you can get unlimited internet access from various
> ISP's (eg earthlink) for a little over $20/month. Your
> ISP should have a news server; posting to an actual
> news server is the most reliable way I've found to post
> things to usenet.
I did this back in Fall 1999 but I couldn't get it to
produce posts with MY NAME ("author" always came out as
"dlrenfro"). I didn't work very hard at trying to fix this
because I always put my name at the end of every post (for
search purposes) and I'm always too busy with other things
to bother. [[I don't "magically" come by all the details that
show up in my posts (and my still modest number of publications,
although a book I've been working on for several years will make
anything you've seen by me trivial in comparison), as I'm sure
you know. For instance, I have copies of all but a couple of the
items listed in the bibliography of my nowhere analytic C-infinity
essay. Several of these I obtained from the authors themselves.]]
But then my posts began appearing ONLY in my local server's
newsgroup, or whatever, and that did it for me. I decided that
since I mostly read the posts via the Math Forum anyway (because
it was often at my office during office hours . . .), I'd just
post through them. I'd been doing this until late January or
early February of this year when I began realizing that no one
was seeing my posts anymore. I don't know why, but for some reason
posts made through the Math Forum haven't been getting past the
Math Forum site. [Yes, I've written to them. So has Leroy Quet, who
was also posting through the Math Forum site. We both began posting
via google groups as a result of this problem, which is apparently
continuing. (I bet you haven't seen any posts by G.E. Ivey lately.
He's still posting via the Math Forum, and none of his posts have
been showing up at the google groups archive.)] Anyway, I don't know
why my essay on nowhere analytic C-infinity functions never got
posted when I sent it via google groups. Maybe it was too long.
Maybe there's a problem with starting a new thread (but I've started
new threads in recent weeks with no problem).
But enough of this. I'm off to the library now. This means that I
might be posting some new interesting tidbits later today that I
come across, such as these recent items --->>>
More numerical coincidences
http://mathforum.org/epigone/sci.math/clehpheucrun
exponentially rational numbers
http://mathforum.org/epigone/sci.math/khingkraldsler
Obtaining prime numbers from exponential towers
http://mathforum.org/epigone/sci.math/slitrayquau
Dave L. Renfro
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu>
>[sci.math Fri, 10 May 2002 13:45:26 GMT]
>http://mathforum.org/epigone/sci.math/gaxcherlyel
>
>wrote (in part):
>
>> If you're posting from home you must have an ISP(?) - if
>> not you can get unlimited internet access from various
>> ISP's (eg earthlink) for a little over $20/month. Your
>> ISP should have a news server; posting to an actual
>> news server is the most reliable way I've found to post
>> things to usenet.
>
>I did this back in Fall 1999 [...]
>
>But then my posts began appearing ONLY in my local server's
>newsgroup, or whatever, and that did it for me. I decided that
>since I mostly read the posts via the Math Forum anyway (because
>it was often at my office during office hours . . .), I'd just
>post through them. I'd been doing this until late January or
>early February of this year when I began realizing that no one
>was seeing my posts anymore. I don't know why, but for some reason
>posts made through the Math Forum haven't been getting past the
>Math Forum site.
Sounds like the same problem in both situations (same symptoms,
anyway). The difference is if you're paying that $20/month they
have an incentive to fix it when you complain.
David C. Ullrich