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How can we get or prove that the 'fractal dimension' of the Cantor set is $\log_{3} (2)$?

I know how to prove by evaluating the poles of $f(s)= \sum \limits_{n=1}^{\infty} 2^{n-1} 3^{-sn}$, and then I take the real pat of the poles which is $\log _{3} (2)$ with complex period $ \frac{2\pi i}{\log 3}$.

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This is a standard argument. A simple google search turned up: – JavaMan Nov 8 '11 at 21:02
It seems you are evaluating the complex dimensions of this fractal (in the sense of Lapidus). Is that what you mean by "fractal dimension"?? – GEdgar Nov 9 '11 at 1:51
up vote 6 down vote accepted

Given an $r>0$, let $N_r(X)$ be the smallest number of balls of radius $r$ that it takes to cover $X$. The fractal dimension of $X$ is $$ \inf\{d:\limsup_{r\to0}\;r^dN_r(x)=0\}\tag{1} $$ Considering the "middle-thirds" construction of $C$, $N_r(C)=2^k$ when $r=3^{-k}$. Using $(1)$, we want to find the infimum of $d$ so that $\limsup\limits_{k\to\infty}\; 3^{-kd}2^k=0$, and that is $d=\frac{\log(2)}{\log(3)}$.

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