Intuitive meaning of fractal dimension.

Question

I'm studying M. Barnsley's book 'Fractals Everywhere', but I'm stuck in the chapter 'Fractal Dimension'.

Suppose $(X, d)$ is a complete metric space and let $A \in \mathcal{H}(X)$ be a nonempty compact subset of $X$. Write $\mathcal{N}(A, \varepsilon)$ for the smallest amount of closed balls with radius $\varepsilon$ needed to cover $A$. Barnsley states the following:

The intuitive idea behind fractal dimension is that the set $A$ has fractal dimension $D$ if $\mathcal{N}(A, \varepsilon) \approx C \varepsilon^{-D}$ for some positive constant $C$, where $f(x) \approx g(x)$ if $\lim_{\varepsilon \to 0} \frac{\ln(f(x))}{\ln(g(x))}=1$.

I don't understand the intuition behind this definition. Can you explain this a little bit better?

Answer 1

(This is more for future reader's than for you)

This is just a really fancy formula for counting how many boxes, squares, can cover a particular fractal. An example will clarify everything...

Take the humble square. It has an area of $L^D$ where $L$ is the length of a side and D is the dimensionality of the square, which is 2. Now if you want to find out how many of these boxes are needed to cover a square, you need to know additional information. What's the side length of one of these smaller squares? Well denote its side length by $\epsilon$. Now you just divide the area of the square by the area of a box. This gives the number of small boxes needed to cover the square. $$N={{L^D} \over {{\epsilon}^D}}$$ But we want to change $\epsilon$ not the area of the square, so just denote the area by a constant $C$. $$N=C \cdot {\epsilon}^{-D}$$ take the logarithm of both sides $$\ln(N)=\ln(C \cdot {\epsilon}^{-D})$$ apply the power rule and multiplication rule $$\ln(N)=\ln(C)+D \cdot \ln({S})$$ where $S={1 \over {\epsilon}}$ $$\ln(N)-\ln(C)=D \cdot \ln(S)$$ $$D={{\ln(N)-\ln(C)} \over {\ln(S)}}$$ substitute for the case of the square and see that it works, then move on to fractals like the Sierpinski triangle and Koch snowflake.

Answer 2

Imagine a Sierpinski triangle of side length $1$. And imagine one of its Sierpinski sub-triangles of side length $1/2^n$. If it helps, think about one of the sub-triangles of side length $1/4$.

Now suppose $\epsilon$ just the right size so your disc of radius $\epsilon$ just barely covers such a s sub-triangle. You need $3^n$ such discs to cover the whole thing. (In the $1/4$ case, there are $9$ sub-triangles of that size that comprise the full triangle. So you need $9=3^2$ such discs.)

So $\mathcal{N}(A, 1/4) =9$. More generally, $\mathcal{N}(A, 1/2^n) =3^n$. So there is an exponential relationship between $\epsilon$ and $\mathcal{N}$. You can rewrite this as $\mathcal{N}(A, \epsilon) =\epsilon^{\log_{1/2}(3)}=\epsilon^{-\ln(3)/\ln(2)}$. The fact that the $\ln(3)/\ln(2)$ doesn't depend on $\epsilon$ is what is neat, and makes that quantity worthy of being given a name: the fractal dimension of the Sierpinski triangle.

Added later: Oh, and the Sierpinski triangle is a "perfect" fractal because of the equality in the relationship $\mathcal{N}(A, 1/2^n) =3^n$. If you had some sort of shape that was less "perfect", but still had the asymptotic behavior defined here, it gets to have a fractal dimension too.