How to iteratively construct a fractal with dimension d = ln(a)/ ln(b)?

Question

I know that if the dimension of a fractal is $\ln(a)/\ln(b)$ then $a$ is the number of copies needed to construct a larger version of it, and $b$ is the magnification factor.

However I'm not sure where to go from here.

Answer 1

As Mark noted, you need $a$ parts scaled by a factor of $b$.

Let's look at a simple example with $a=4$ and $b=5$. You start with the segment $[0,1]$, lets call this $C_0$. You scale it by a factor of $b=5$, and get the segment $[0,1/5]$. Now you take four copies of that, and place them inside the first segment, e.g. $$ [0,1/5], \quad [1/4,1/4+1/5]\quad [1/2,1/2+1/5]\quad [3/4,3/4+1/5] $$ Let's call the union of these four sets $C_1$.

Now we repeat this process. At every step we define $C_i$ by four copies of $C_{i-1}$ scaled by 1/5, mathematically $$ 1/5 C_{i-1} \cup \left( 1/4+1/5 C_{i-1} \right)\cup \left(1/2+1/5 C_{i-1} \right) \cup \left(3/4 +1/5 C_{i-1}\right) $$

You get a sequence of sets $C_i$.

As they are all contained within each other ($C_i \subset C_{i-1}$), we can take their intersection and get the desired set, i.e. $\bigcap C_i$ solves your problem.

Two comments:

Here we have $\ln(a)/\ln(b)<1$, hence we can fit this set in the 1 dimensional real space. For larger dimensions, you need to start with larger dimensional sets; the unit cube in any dimension is usually a good set to start.

Here we defined a sequence with $C_i \subset C_{i-1}$. This is not necessary (but something weaker is). If it does not hold, you need however a more complicated limit set than just the intersection of the $C_i$.