A Cantor set of specified dimension

mark

Let $d \in (0,1)$. Find a generalized Cantor set whose dimension is $d$.

theoldernoah

An IFS for our generalized Cantor set is [imath] \left\{\frac{1}{r}x,\frac{1}{r}x + \frac{1-r}{r}\right\} [/imath].

Therefore our scaling factor is simply just [imath] \frac{1}{r} [/imath]. If we now solve for [imath] d [/imath] we will find our dimension.

[dmath] \left(\frac{1}{r}\right)^d+\left(\frac{1}{r}\right)^d=1 \Rightarrow 2\left(\frac{1}{r}\right)^d=1 \Rightarrow \left(\frac{1}{r}\right)^d=\frac{1}{2} \Rightarrow r=\frac{1}{\sqrt[d]{\frac{1}{2}}} \Rightarrow r=\sqrt[d]{2}.[/dmath]

mark

@theoldernoah You know $d$ - it's the target dimension. You want to know, "what value of $r$ will yield the target dimension? So you should solve for $r$ in terms of $d$, rather than the other way around.

Rex

so, we're using $d = \frac{log3}{log2}$ , right? which gives us $r = \frac{1}{2^{\frac{1}{d}}}$ which I'm told is something like r = 0.645760. Is that right?

eric

To correct: $D =\frac{log(number\ of\ pieces)}{log(magnification\ factor)}$

for example, we can verify.
Let $D= log(2)/log(3) \simeq .6309297536$ , so $D\in (0,1)$. then $r=2^d$ solving for r we get we could get .6457601172... which r is between 0 and 1.

$r = \frac{1}{2^{\frac{1}{d}}}$,

if $D= log(2)/log(3) \simeq .6309297536$ and $r = \frac{1}{2^{\frac{1}{d}}}$ then
$r= \frac{1}{3}$

mark

@Rex and @eric - Not quite. $d$ is known but unspecified. So you solve for $r$ in terms of $d$ and leave it as is. Thus, once you get to
$$r = \frac{1}{2^{\frac{1}{d}}},$$
you're done!

Cornelius

This is just aesthetics.
$$\Bigg( \frac{1}{r}\Bigg)^{d} + \Bigg( \frac{1}{r} \Bigg)^{d}= 2\Bigg( \frac{1}{r}\Bigg)^{d}=1$$

Note that $$\Bigg( \frac{1}{r}\Bigg)^{d}=\frac{1^{d}}{r^{d}}=\frac{1}{r^{d}}$$

So $$2\Bigg( \frac{1}{r}\Bigg)^{d}=\frac{2}{r^{d}}=1 \Rightarrow r^{d}=2 \Rightarrow r=\sqrt[d]{2}$$