a question on real-analytic functions
David Bernier
that f(r) is rational for every rational r in the domain of definition.
Some examples: x|-> 1/x
x|-> P(x), P in Q[x],
x|-> 9/23* (1/(x^2 + 1))
The most general form I can come up with is:
- P in Q[x],
- T in Q[x], such that T has no zeros inside ]0, 1[
- Let f(x) = P(x)/T(x) .
That's all I could think of. I chose the domain of definition ]0, 1[ to
be small, for example compared to the Reals .
So I'm curious to know whether any other such functions f exist.
BTW: The set of all such f forms a Q-algebra, it seems.
David Bernier
David C. Ullrich
need f be a rational rational function?
I bet there are other examples - if nobody comes up with
an elementary closed-from example I bet you could construct
a power series with irrational coefficients that takes rational
values at all the rationals.
If you can't do that in a nice way I bet you could
do that by moderately subtle brute force. Maybe something
like this:
Enumerate the rationals q1, ... .
Choose irrationals a0, a1 such that a0 + a1*q1 is rational.
Fix a0 and a1 at this point, and also declare that whatever
the other a's turn out to be, they will satisfy
a_n + a_{n+1}*q1 = 0
for every even number n > 0. (Also vow that |a_n|<=1/n!
for all n, so our power series will give an entire function.)
Now choose a_2,...a_5 consistent with the constraints so
far, so that the sum of the first six terms of the series
is rational at q2. Add constraints on blocks of _four_
consectutive terms (starting with a7,...a10) so that
the sum of those groups of four terms will vanish at
q2. Etc.
I'm not claiming that exactly what I said here works,
but I don't see why something very much like it should
not work. The idea is that at each qn you add constraints
that _start_ at places farther and farther out in the
series, so ech a_n is required to satisfy only finitely
many conditions...
[Um, I think it's clear that a power series with
irrational coefficients does give an "other" example:
If f = P/T with P and T in Q[x] then the fact that
t*f = P shows that the coefficients in the power series
for f satisfy a recurrence with rational coefficients
and initial conditions, right?]
David C. Ullrich
Robert Israel
David C. Ullrich <ull...@math.okstate.edu> wrote:
>Well, it's a cute question: If f is rational on the rationals
>need f be a rational rational function?
>I bet there are other examples - if nobody comes up with
>an elementary closed-from example I bet you could construct
>a power series with irrational coefficients that takes rational
>values at all the rationals.
Yes. This has come up before. See for example my posting of 29 March
1998 on the subject Re: Conjecture in algebraic geometry.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Dave L. Renfro
[sci.math 13 Mar 2002 23:43:19 -0800]
http://mathforum.org/epigone/sci.math/zumrizom
wrote
In this same thread Robert Israel mentioned an earlier post of his:
http://mathforum.org/epigone/sci.math/khoustelwhand/6fkids$cjq$1...@nntp.ucs.ubc.ca
http://groups.google.com/groups?selm=6fkids%24cjq%241%40nntp.ucs.ubc.ca
Israel showed that there are c many analytic functions that
map the rationals to the rationals and then points out that there
are only countably many functions of the type you've described.
In case anyone has a further interest in this, I thought I
mention a few papers related to this question.
[1] Philip Franklin, "Analytic transformations of everywhere dense
point sets", Trans. Amer. Math. Soc. 27 (1925), 91-100.
[Announced in Bull. Amer. Math. Soc. 30 (1924), 482.]
[JFM 51.0166.01]
Theorem I (p. 94): "For any two enumerable linear point sets,
each everywhere dense on an open interval,
an analytic function can be found which maps
the two intervals on one another, and effects
a one to one correspondence between the point
sets."
[[ NOTE: I don't believe that any of the following later references
mention Franklin's paper ]]
[2] Paul Erdos, "Some unsolved problems", Michigan Math. J. 4 (1957),
291-300. [MR 20 #5157; Zbl 81.00102]
Problem #24 (p. 197): "Does there exist an entire function f, not
of the form f(x) = a + bz, such that the
number f(x) is rational or irrational
according to as x is rational or irrational?
More generally, if A and B are two
denumerable, dense sets, does there exist
an entire function which maps A onto B?"
[3] Fred Gross, "Research Problem #19", Bull. Amer. Math. Soc. 71
(1965), 853.
[[ Gross apparently asks essentially the same question, but I
have not actually looked at this reference. Maurer claims
that his result answers negatively the question Gross asked,
but see Gross' comment at the end of the Zbl review of
Maurer's paper (URL given in [4] below). ]]
[4] Ward D. Maurer, "Conformal equivalence of countable dense sets",
Proc. Amer. Math. Soc. 18 (1967), 269-270.
[MR 35 #6829; Zbl 189.36204]
http://www.emis.de/cgi-bin/Zarchive?an=0189.36204
Theorem: "Let A and B be two countable dense sets in the complex
plane. Then there exists an entire function taking A
onto B."
[5] Karl F. Barth and Walter J. Schneider, "Entire functions
mapping countable dense subsets of the reals onto each other
monotonically", J. London Math. Soc. (2) 2 (1970), 620-626.
[MR 42 #4729; Zbl 201.09203]
http://www.emis.de/cgi-bin/Zarchive?an=0201.09203
MR (by W. D. Maurer): "This paper provides further information
about a problem due to Erdos concerning the
existence of functions mapping arbitrary
countable dense sets onto others. The reviewer
[Proc. Amer. Math. Soc. 18 (1967), 269-270;
MR 35 #6829] showed that, given any two
countable dense subsets of the complex plane,
there exists an entire function taking one of
these onto the other. In this paper we are
talking about the line, not the plane, but
the conclusion is stronger: the function takes
points of the first countable dense set, and
only such points, into points of the second
one. The argument is very tedious and, in fact,
is not completely carried out; only the case
in which both of the given sets are the
rationals is completed (the function involved
in this case, however, is transcendental, as
it is in the general case). As an application,
the authors prove that strictly monotone
increasing entire functions may assume any
sequence of values whatsoever at all positive
integers, just so long as these values
themselves form an increasing sequence."
[6] Stuart Rankin and Daihachiro Sato, "Entire functions mapping
countable dense subsets of the reals onto each other
monotonically", Bull. Aust. Math. Soc. 10 (1974), 67-70.
[MR 49 #10883; Zbl 275.30020]
MR (by T. Kovari): "By modifying a technique of W. D. Maurer,
the authors give a new proof of the following
result of K. F. Barth and W. J. Schneider: Let
A and B be countable dense subsets of the real
line; then there exists a transcendental entire
function f such that the restriction of f to the
real line is a real homeomorphism, and f(A) = B."
[7] Emil A. Corena, "Problem #10361", Amer. Math. Monthly, 101
(1994), 175.
"Do there exist nonlinear C^1 functions f:R --> R such
that for any rational x, f(x) is also rational and for
any irrational x, f(x) is also irrational?"
Dave L. Renfro
Dave L. Renfro
[sci.math 13 Mar 2002 23:43:19 -0800]
http://mathforum.org/epigone/sci.math/zumrizom
wrote (in part and while taking a break from watching the
Fritsche/Ullrich blood-bath that's currently taking place):
> In this same thread Robert Israel mentioned an earlier post of his:
>
> http://mathforum.org/epigone/sci.math/khoustelwhand/6fkids$cjq$1...@nntp.ucs.ubc.ca
> http://groups.google.com/groups?selm=6fkids%24cjq%241%40nntp.ucs.ubc.ca
>
> Israel showed that there are c many analytic functions that
> map the rationals to the rationals and then points out that there
> are only countably many functions of the type you've described.
Hummm...It's clear there are only countably many. However, after
looking again, it appears that Israel didn't actually SAY there
are countably many . . .
Dave L. Renfro
Denis Feldmann
"Dave L. Renfro" <renf...@cmich.edu> a écrit dans le message news:
28ae5e5e.0203...@posting.google.com...
> David Bernier <ez...@yahoo.com>
> [sci.math 13 Mar 2002 23:43:19 -0800]
> http://mathforum.org/epigone/sci.math/zumrizom
>
> wrote (in part and while taking a break from watching the
> Fritsche/Ullrich blood-bath that's currently taking place):
>
> > In this same thread Robert Israel mentioned an earlier post of his:
> >
> >
http://mathforum.org/epigone/sci.math/khoustelwhand/6fkids$cjq$1...@nntp.ucs.ub
c.ca
> > http://groups.google.com/groups?selm=6fkids%24cjq%241%40nntp.ucs.ubc.ca
> >
> > Israel showed that there are c many analytic functions that
> > map the rationals to the rationals and then points out that there
> > are only countably many functions of the type you've described.
>
> Hummm...It's clear there are only countably many.
Clear?? I could have bet there was almost as many of them as permutations of
the rationals, i.e. 2^aleph_0...
Dave L. Renfro
[sci.math Sat, 16 Mar 2002 00:46:47 +0100]
http://mathforum.org/epigone/sci.math/zumrizom
wrote (in response to my response to my post):
***Renfro (from initial post)
>>> Israel showed that there are c many analytic functions that
>>> map the rationals to the rationals and then points out that
>>> there are only countably many functions of the type you've
>>> described.
***Renfro (from follow-up post)
>> Hummm...It's clear there are only countably many.
***Feldmann
> Clear?? I could have bet there was almost as many of them as
> permutations of the rationals, i.e. 2^aleph_0...
***Renfro (from follow-up post)
>> However, after looking again, it appears that Israel didn't
>> actually SAY there are countably many . . .
***Israel (from his March 29, 1998 post, not in Feldmann's post)
And these functions have the cardinality of the continuum.
So L is a lot bigger than the rational functions with rational
coefficients.
I don't see how counting permutations of the rational numbers
has anything to do with counting rational functions with rational
number coefficients. Shouldn't we be counting finite sequences of
rational numbers, or am I missing something?
The only point to my follow-up post was that my comment "... points
out that there are only countably many functions of the type ..."
isn't quite accurate. I must have read between Israel's lines for
what he was obviously implying and then forgot that this was
"between the lines" and stated that he actually said it.
Dave L. Renfro
Denis Feldmann
> Denis Feldmann <denis.f...@wanadoo.fr>
> [sci.math Sat, 16 Mar 2002 00:46:47 +0100]
> http://mathforum.org/epigone/sci.math/zumrizom
>
> wrote (in response to my response to my post):
>
> ***Renfro (from initial post)
> >>> Israel showed that there are c many analytic functions that
> >>> map the rationals to the rationals and then points out that
> >>> there are only countably many functions of the type you've
> >>> described.
>
> ***Renfro (from follow-up post)
> >> Hummm...It's clear there are only countably many.
>
> ***Feldmann
> > Clear?? I could have bet there was almost as many of them as
> > permutations of the rationals, i.e. 2^aleph_0...
>
> ***Renfro (from follow-up post)
> >> However, after looking again, it appears that Israel didn't
> >> actually SAY there are countably many . . .
>
> ***Israel (from his March 29, 1998 post, not in Feldmann's post)
> And these functions have the cardinality of the continuum.
> So L is a lot bigger than the rational functions with rational
> coefficients.
Yes, I see now I mixed up the attributions, sorry. What I meant was what R.
Israel said: "these functions= those who maps Q to Q", have the continuum
cardinality, and so are more numerous that "those functions = the rational
ones", which are (obviously :-)) denumerable
>
> I don't see how counting permutations of the rational numbers
> has anything to do with counting rational functions with rational
> number coefficients.
Shouldn't we be counting finite sequences of
> rational numbers, or am I missing something?
No, you are of course right,and this was my argument too; we were simply
not speaking of the same functions.
>
> The only point to my follow-up post was that my comment "... points
> out that there are only countably many functions of the type ..."
> isn't quite accurate. I must have read between Israel's lines for
> what he was obviously implying and then forgot that this was
> "between the lines" and stated that he actually said it.
>
Ok.
> Dave L. Renfro
Pertti Lounesto
>"these functions= those who maps Q to Q", have the continuum
>cardinality, and so are more numerous that "those functions
>= the rational ones", which are (obviously :-)) denumerable
>
Could you give an example of a an analytic function R -> R
that maps Q to Q, but is not rational?
Denis Feldmann
"Pertti Lounesto" <plou...@pp.htv.fi> a écrit dans le message news:
3C935DBD...@pp.htv.fi...
This was done by Robert Israel some times ago. IIRC, one possible idea is
to enumerate the rationals , define P_n(x)=(x-q_i)(x-q_2)....(x-q_n), then
put f(x)= sum(P_n(x)/n!) (or something similar, to assure uniform
convergence, and f analytic).
>