Box-counting dimension for a self-affine set

mark
Wed, 15 Nov 2017 15:13:47 +0000
Quiz or Test Prep

Consider the following IFS of four Affine functions:

[dmath]
\begin{align}
f_1(\vec{x}) &= \left(
\begin{array}{cc}
\frac{1}{4} & 0 \\
0 & \frac{1}{2}
\end{array}
\right) \cdot \vec{x} \\
f_2(\vec{x}) &= \left(
\begin{array}{cc}
\frac{1}{4} & 0 \\
0 & \frac{1}{2}
\end{array}
\right) \cdot \vec{x} + \left(
\begin{array}{c}
\frac{1}{4} \\
\frac{1}{2} \
\end{array}
\right)\\
f_3(\vec{x}) &= \left(
\begin{array}{cc}
\frac{1}{4} & 0 \\
0 & \frac{1}{2}
\end{array}
\right) \cdot \vec{x} + \left(
\begin{array}{c}
\frac{1}{2} \\
0
\end{array}
\right) \\
f_4(\vec{x}) &= \left(
\begin{array}{cc}
\frac{1}{4} & 0 \\
0 & \frac{1}{2}
\end{array}
\right) \cdot \vec{x} + \left(
\begin{array}{c}
\frac{3}{4} \\
\frac{1}{2}
\end{array}
\right)
\end{align}
[/dmath]

Several iterations of this IFS applied to the unit square are shown below:

In particular, note that the image of the square is not another square - it's a rectangle that's twice as tall as it is wide. After 5 steps away from the square, we see a good approximation to the limiting attractor, which we denote $E$.

Show that $n^{\text{th}}$ cover of $E$ generated by this process consists of $4^n$ rectangles of width $1/4^n$ and height $1/2^n$.

How many squares of side length $1/4^n$ must each of the rectangles in the previous part be decomposed into to cover the attractor with squares?

Use the squares from the previous part to compute $N_{4^{-k}}(E)$. Don't count squares that intersect $E$ only in their boundaries.

Compute the box counting dimension of $E$.