From: gerry@mpce.mq.edu.au (Gerry Myerson)
Newsgroups: sci.math
Subject: Re: continued fraction
Date: Wed, 07 Oct 1998 10:03:36 +1100
In article ,
edgar@math.ohio-state.edu (G. A. Edgar) wrote:
=> In article <361A465B.246A@dds.nl>, Johan Bosman wrote:
=>
=> > Is there a way to determine a formula for the continued fraction of an
=> > arbitrary 3rd-degree algebraic number?
=>
=> I believe none is known. Not even for a single 3rd-degree number.
Agreed. There are, however, clever ways of computing partial quotients
using the minimal equation (as opposed to first calculating a decimal
expansion). See, e.g., Enrico Bombieri & Alfred J van der Poorten,
Continued fractions of algebraic numbers, in Bosma & van der Poorten,
eds, Computational Algebra and Number Theory, Kluwer, 1995, pp 137--152.
Gerry Myerson (gerry@mpce.mq.edu.au)
==============================================================================
From: "Arthur L. Rubin" <216-5888@mcimail.com>
Newsgroups: sci.math,alt.gobment.lones
Subject: Re: continued fraction
Date: Mon, 12 Oct 1998 08:06:28 +0700
Ronald Bruck wrote:
> So we don't know what the set of real numbers whose continued-fraction
> decomposition is bounded, looks like. Can we say what the set of reals
> with only 1's and 2's appearing in the continued-fraction decomposition
> looks like? Surely it's dense; is it first-category or second? What's
> its measure? (Zero?) What are its topological properties?
I believe it resembles the Cantor set; (probably) measure 0, perfect, i
somorphic to 2^omega considered as an ordered topological space in
itself.
(This is completely ignoring the rationals which only have 1 and 2 in
their
representation.)
--
Arthur L. Rubin 216-5888@mcimail.com
==============================================================================
From: kramsay@aol.com (KRamsay)
Newsgroups: sci.math
Subject: Re: continued fraction
Date: 12 Oct 1998 17:53:44 GMT
: Can we say what the set of reals
: with only 1's and 2's appearing in the continued-fraction decomposition
: looks like? Surely it's dense;
No, it has no elements in (1/3,1/4) for instance. (If you had allowed
for finitely many exceptional partial quotients, you would get a dense
set of course.)
Its complement is dense, because in any open interval there is a
subinterval consisting of the numbers whose continued fraction starts
out with a given string [c0;c1,c2,...,cn] and among those there is a
subinterval consisting of numbers whose next partial quotient is 3.
: is it first-category or second?
Since every interval contains an open interval not in the set, the
set is nowhere dense, hence "very much" of first category.
: What's its measure? (Zero?)
Zero. As someone else noted, there's a theorem saying that the set of
numbers where the partial quotients don't obey a certain statistical
distribution is zero.
:What are its topological properties?
Closed and totally disconnected. The smallest element is 1; the next
one is [1,2,1,2,1,2,...]=(1+sqrt(3))/2. The rationals in the set are
dense in it, but each rational is an isolated point, as all numbers
sufficiently close to a given rational have a large partial quotient
either one or two steps after position of the last coefficient of the
rational. The irrational elements are a perfect set.
It has cardinality c; we can associate in a 1-1 manner to each number
between 0 and 1 with binary expansion .b0b1b2... the number
[b0+1,b1+1,b2+1,...].
: So we don't know what the set of real numbers whose continued-fraction
: decomposition is bounded, looks like.
It contains all rationals and quadratic irrationals.
It's the union over n>1 of the set of numbers whose partial quotients
are bounded by n. Since each term of the union is nowhere dense and
has measure zero, it has measure zero and is of first category. By
Baire's theorem, it's complement is dense.
That x has bounded partial quotients is equivalent to the existence of
epsilon>0 such that for all rationals p/q we have |x-p/q|>epsilon/q^2.
Thus the set is invariant under translation by rationals, which shows
more explicitly that both it and its complement are dense.
Keith Ramsay "Thou Shalt not hunt statistical significance with
kramsay@aol.com a shotgun." --Michael Driscoll's 1st commandment
==============================================================================
From: gerry@mpce.mq.edu.au (Gerry Myerson)
Newsgroups: sci.math
Subject: Re: continued fraction
Date: Mon, 12 Oct 1998 10:58:23 +1100
In article <6vquam$bqt$1@math.usc.edu>, bruck@math.usc.edu (Ronald Bruck) wrote:
> I should think that what he meant is pretty clear. No, it's NOT that
> the "partial quotients" are unbounded, because they CONVERGE to the
> number (that's true of ANY real number). What he means is that the set of
> a[i] in the continued fraction decomposition are unbounded.
The a[i] *are* the partial quotients; the things that converge to the
number are the *convergents*, usually denoted p[i]/q[i].
> (Perhaps there is a more precise version of the conjecture,
> which says that a[i] --> infinity? I'd be interested in hearing more!)
For almost all x, and for all a, the frequency with which a appears as a
partial quotient in the continued fraction expansion of x is
log base 2 of ( (a + 1)^2 / (a + 1)^2 - 1 ).
"Almost all x" means all but a set of measure zero. So, for example, almost
all x have about 41.5% of their partial quotients 1. Of course, "frequency"
is to be understood as a limit as the number of partial quotients considered
goes to infinity.
Whether the exceptional set of measure zero contains none, some, or all of
the algebraic irrationals of degree 3 and up is unknown (so far as I know).
I think most of us would vote for "none" but of course mathematical truth
is not determined by majority rule.
> So we don't know what the set of real numbers whose continued-fraction
> decomposition is bounded, looks like. Can we say what the set of reals
> with only 1's and 2's appearing in the continued-fraction decomposition
> looks like? Surely it's dense; is it first-category or second? What's
> its measure? (Zero?) What are its topological properties?
I don't know, off-hand, but I do know people have looked at this kind of
question, and I can tell you where to look if you really want to know;
Reviews in Number Theory. I think the 1-2 set has Hausdorff dimension
strictly between 0 and 1.
I've removed alt.gobment.lones from the newsgroup list because I choose not
to post to newsgroups I know nothing about.
Gerry Myerson (gerry@mpce.mq.edu.au)
==============================================================================
From: bumby@lagrange.rutgers.edu (Richard Bumby)
Newsgroups: sci.math,alt.gobment.lones
Subject: Re: continued fraction
Date: 13 Oct 1998 18:52:16 -0400
bruck@math.usc.edu (Ronald Bruck) writes:
>...
>I should think that what he meant is pretty clear. No, it's NOT that
>the "partial quotients" are unbounded, because they CONVERGE to the
>number (that's true of ANY real number). What he means is that the set of
>a[i] in the continued fraction decomposition are unbounded. That does NOT
>mean that it converges to infinity, because it doesn't prevent low numbers
>from recurring. (Perhaps there is a more precise version of the conjecture,
>which says that a[i] --> infinity? I'd be interested in hearing more!)
"Partial quotient" is the name for the a[i]. The rational numbers
converging to the number represented, that you get by truncating the
continued fraction, are usually called "convergents".
Lang and Trotter did (and interpreted) calculations of continued
fractions of algebraic numbers, including the cube root of 2. The
paper appeared in J. reine angew. Math. 255 (1972), 112--134 (MR46
#558 or Reviews in Number Theory Y65-201). The calculations show the
expected distribution of many small and few large partial quotients.
This is "expected" in the sense that for almost all numbers, the
density of k with a[k]>=n is log(1+1/n)/log(2), and it is believed
that algebraic numbers of degree greater than 2 do not have "special"
continued fractions.
>...
>So we don't know what the set of real numbers whose continued-fraction
>decomposition is bounded, looks like. Can we say what the set of reals
>with only 1's and 2's appearing in the continued-fraction decomposition
>looks like? Surely it's dense; is it first-category or second? What's
>its measure? (Zero?) What are its topological properties?
In the report of the New York Number Theory Seminar from 1983-4
(Springer Lecture Notes 1135), I have an article that includes a
description of a calculation showing that this set has Hausdorff
dimension between 0.5312 and 0.5314. (I have a better computer now, so
if there is enough demand, I can calculate the result more
accurately.)
The set is *clearly* nowhere dense. Including a_0 among the partial
quotients being restricted to {1,2} gives a set whose smallest element
has continued fraction [_1,2_] and largest element [_2,1_] (where the _
are used to mark a periodic block). Those having a_0=1 lie in the
interval from [_1,2_] to [1,_1,2_], and those with a_0=2 lie in the
interval from [2,_2,1_] to [_2,1_]. This leads to a Cantor set
construction of the set, but one in which the size and location of the
segment removed from each interval depends on a parameter representing the
history of the construction leading to the interval.
--
R. T. Bumby ** Rutgers Math || Amer. Math. Monthly Problems Editor 1992--1996
bumby@math.rutgers.edu ||
Telephone: [USA] 732-445-0277 (full-time message line) FAX 732-445-5530