Newsgroups: sci.math
From: thomasc@lobachevskii.geom.umn.edu (Thomas Colthurst)
Subject: Re: A Fractal Dimension
Date: Thu, 6 Aug 1992 10:32:17 GMT

In article <1992Aug6.001108.17231@Princeton.EDU> amlogan@phoenix.Princeton.EDU (Adam M. Logan) writes:
>Let S be the set of reals in [0, 1) such that, when they are written in 
>a continued fraction, only ones and twos appear as partial denominators.
>What is the Hausdorff dimension of S?

Quick Answer:

0.642954...


Long Answer:

Consider the two functions f(x) = 1/(1+x) & g(x) = 1/(2+x) acting on 
[0,1].  It is clear that we can get all the continued fractions with ones
and twos as their partial denominators by considering the forward orbits of
0 under all possible combinations of these maps.  The functions f & g are
contractive on (0,1] -- in particular, f' = -1/(1+x)^2 < 1 and
g' = -1/(2+x)^2 < 1/4.  We can thus think of {f,g} as an iterated function
system, and use IFS machinery.  {f,g} is disconnected -- f([0,1]) = [1/2,1],
g([0,1]) = [1/3,1/2], g([1/3,1]) = [1/3,3/7], f([1/3,1]) = [1/2,3/4], so 
the Hausdorff dimension is equal to the fractal dimension which is equal
to the solution d of |s_f|^d + |s_g|^d = 1, where s_f and s_g are the
scaling factors of f and g.  But wait, you say, f & g 's scaling factors
depend on x.  No matter.  We simply take the x (on our attractor) which
maximizes the solution d to (1/(1+x)^2)^d  + (1/(2+x)^2)^d = 1; locally
around that point, the dimension will be d, and since it is the largest,
it will dominate that of the entire set.

For the range we are interested in, d is a decreasing function of x,
so we seek the smallest value on the attractor (the smallest number
representable by an infinite continued fraction containing only ones
and twos as partial denominators).  This turns out to be the fixed point
of g of f, and is approximately 0.366025.  Solving for d gives 0.642954...

This, of course, all generalizes.  If we wish to find the Hausdorff
dimension of all continued fractions with partial denominators
restricted to p_1, p_2, .. p_n, then we simply have to solve
/sum (1/(p_i+x)^2)^d = 1 for x = 1/(p_n+1/(p_1+x)) [assuming p_1 is
the least and p_n is the greatest allowed partial denominator].

-Thomas C

==============================================================================

Newsgroups: sci.math
From: shallit@graceland.uwaterloo.ca (Jeffrey Shallit)
Subject: Re: A Fractal Dimension
Date: Thu, 6 Aug 1992 11:56:35 GMT

In article <1992Aug6.001108.17231@Princeton.EDU> amlogan@phoenix.Princeton.EDU (Adam M. Logan) writes:
>Let S be the set of reals in [0, 1) such that, when they are written in 
>a continued fraction, only ones and twos appear as partial denominators.
>What is the Hausdorff dimension of S?

.53128049 < dim(E2) < .53128051

For this, see

@article{Good:1941,
	author = "I. J. Good",
	title = "The fractional dimensional theory of continued fractions",
	journal = "Proc. Cambridge Phil. Soc.",
	volume = 37,
	year = 1941,
	pages = "199-228",
	note = "(Corrigenda:  {\bf 105} (1989), 607)"}

@incollection{Bumby:1985,
	author = "R. T. Bumby",
	title = "Hausdorff dimension of sets arising in number theory",
	booktitle = "Number Theory",
	editor = "D. V. Chudnovsky and G. V. Chudnovsky and H. Cohn and
M. B. Nathanson",
	publisher = SV,
	series = LNIM,
	volume = 1135,
	year = 1985,
	note = "MR 87a\#11074",
	pages = "1-8"}

@article{Hensley:1989b,
	author = "D. Hensley",
	title = "The {Hausdorff} dimensions of some continued fraction
{Cantor} sets",
	journal = JNT,
	volume = 33,
	year = 1989,
	pages = "182-198"}


JNT = "Journal of Number Theory".

You might also see my survey article, Real Numbers with Bounded Partial Quotients,
which has just appeared in the latest issue of L'Enseignement Math.

Jeff Shallit