The Hausdorff Dimensions of Geometrical Fractals

• Main Authors: LIANG, JYH-CHING, 梁志慶
• Other Authors: TSENG SHIOJENN
• Format: Others
• Language: zh-TW
• Published: 1997
• Online Access: http://ndltd.ncl.edu.tw/handle/15854281806928742326

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} $$\hbox{dim}S \ge {1\over {b+1}}.$$