7
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

(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.

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.