| Summary: | 博士 === 淡江大學 === 數學學系 === 85 === \par The present study comprises two major parts. The condition
that the simple curve constructed based on a symmetric Cantor
set is a fractal is discussed inthe first part. Let $a$ be the
ratio (${{\rm the~ residual~ length}\over {\rm the~ original~
length}}$) in each step of the construction. We show that if $a
> 1/4$, then the simple curve obtained is a fractal;otherwise,
it is not. The Hausdorff dimension of the fractal thus obtained
canbe evaluated. The Hausdorff dimension of the set of continued
fractions is estimated in the second part of this study. A
simple, yet efficient, algorithmwhich is similar to the
procedure adopted in the construction of a Cantor set is
proposed for the estimation of the lower bound of the Hausdorff
dimension ofcontinued fractions. For the set $\{\xi~ :~ a_n\ge
a^{b^n}~{\rm for~all}~ n\},$ the present algorithm provides the
greatest lower bound.{\bf Theorem} {\it If $a_1 > {1/4},$ then
$C$ is a fractal; otherwiseit is not a fractal}.\par Let $r$ be
a positive integer. Let us define two sets of integers depending
on $r$ by$$\eqalign {A_n =& \{ r+n+2k-1~~\vert~ k~{\rm is~ an~
interger~ and~} 0\le k\le n-1\}\cr {\rm and}~~~~~~~~~&\crB_n =&
\{ r+n+2k~~\vert~ k~{\rm is~ an~ integer~ and ~} 0\le k\le n-2
\},~~\forall n>1.\cr}$$Also, $A_1=\{r\}$ and $B_1=\phi.$ Let
$\vert\cdot\vert$ denote the cardinalnumber; thus $\vert
A_n\vert=n$ and $\vert B_n\vert =n-1.$\vskip .5cm\par {\bf
Theorem } {\itLet $T = \{\xi : a_n\in A_n~ {\rm for~ all}~ n\}$.
Then } $\dim T = 1/2.$\bigskip \vskip .5cm\par {\bf Definition }
Let us define two sets $A_n = \{ n^b, (n+2)^b, \cdots,(3n-2)
^b\}$ and$B_n = \{ x : n^b < x < (3n-2)^b\ {\rm and}\ x\not\in
A_n\},$ $\forall n\ge 1$.\vskip .5 cm\par{\bf Theorem } {\it Let
$D = \{\xi : a_i\in (n^b)\ {\rm and}\ a_i\ge i^b\ {\rm for~
all}\ i\}$.Then} $$\dim D \ge {1\over 2b}.$$\bigskip \vskip
.5cm\par Let us define two sets $A_n = \{ [a^{b^n}]+n+1,
[a^{b^n}]+n+3, \cdots, 3[a^{b^n}]+n+1\}$ and$B_n = \{ [a^{b^n}]+
n+2, [a^{b^n}]+n+4, \cdots,3[a^{b^n}]+n \}$ for all $n\ge
1$,where [$x$] denote the largest integer $m$ with $m\le x$.The
numbers of sets $A_n$ and $B_n$ are $[a^{b^n}]+1$ and
$[a^{b^n}],$ respectively,for all $n\ge 1$. \vskip .5 cm\par{\bf
Theorem } {\it Fix $a, b>1$.Let $S = \{\xi : a_n\ge a^{b^n} for\ all\ n\}$.Then we have} $$\hbox{dim}S \ge {1\over {b+1}}.$$
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