Chapter 11.4, Problem 8E

Interpretation Introduction

Interpretation:

Redo the previous questionexcept adding an element of randomness to the process: to generate ${\text{S}}_{\text{n+1}}$ from ${\text{S}}_{\text{n}}$, flip a coin. If the result is heads, delete the second quarter of every interval in ${\text{S}}_{\text{n}}$; if it is tails, delete the third quarter. The limiting set is an example of a random fractal.

1. Can you find the box dimension of this set? Does this question even make sense? In other words, might the answer depend on a particular sequence of heads and tails that happen to come up?

2. Suppose if tail comes up, we delete the first quarter. Could this make a difference? For instance, what if we had a long string of tails?

Concept Introduction:

• Let $\text{S}$ be a subset of the $\text{D-}$ dimensional Euclidean space, and let $\text{N}\left(\text{ε}\right)$ be the minimum number of the $\text{D-}$ dimensional cubes of side e needed to cover $\text{S}$. How does $\text{N}\left(\text{ε}\right)$ depend on $\text{ε}$? For the smooth curve of length $\text{L}$, $\text{N}\left(\text{ε}\right)\propto \frac{\text{A}}{{\text{ε}}^{\text{2}}}$, the key observation is that the dimension of the set equals the exponent d in the power law $\text{N}\left(\text{ε}\right)\propto \frac{\text{1}}{{\text{ε}}^{\text{d}}}$.

• This power law also holds for most fractal sets $\text{S}$, except that d is no longer an integer, where d is the dimension called the capacity of the box.

• Consider a fractal that divides the closed unit interval $\left[\text{0,1}\right]$ into four quarters.

Answer to Problem 8E

Solution:

1. The box dimension exists for both cases and equals to: $\frac{\text{ln 3}}{\text{ln 4}}$ .

2. If we had a long string of tails, then there would be no effect of the box dimension.

Explanation of Solution

Consider a fractal that divides the closed unit interval $\left[\text{0,1}\right]$ into four quarters.

Consider the set ${\text{S}}_{\text{1}}$ that is produced after deleting the open second quarter from left that generates ${\text{S}}_{\text{n+1}}$ from ${\text{S}}_{\text{n}}$ by deleting the second quarter of each of the intervals in ${\text{S}}_{\text{n}}$.

1. Consider the formula given below for the fractal:

   $\text{N}\left(\text{ε}\right){\text{= 3}}^{\text{n}}\text{, }\frac{\text{1}}{\text{ε}}{\text{= 4}}^{\text{n}}$

   Case-1: When the result is heads, delete the second quarter of every interval ${\text{S}}_{\text{n}}$.

   Then, there will be no change in $\text{N}\left(\text{ε}\right)$ and $\text{ε}$. So:-

   $\text{N}\left(\text{ε}\right){\text{= 3}}^{\text{n}}\text{, }\frac{\text{1}}{\text{ε}}{\text{= 4}}^{\text{n}}$

   Now, use the below mentioned formula to find the box dimension $\text{d}$:-

   $\text{d =}\underset{n\to \infty }{\text{lim}}\left(\frac{\text{lnN}\left(\text{ε}\right)}{\frac{\text{1}}{\text{ε}}}\right)$

   Now, substitute the values $\text{N}\left(\text{ε}\right){\text{= 3}}^{\text{n}}\text{, }\frac{\text{1}}{\text{ε}}{\text{= 4}}^{\text{n}}$ in the above box dimension formula. Then:-

   $\text{d =}\underset{n\to \infty }{\text{lim}}\left(\frac{\text{lnN}\left(\text{ε}\right)}{\frac{\text{1}}{\text{ε}}}\right)$

   $\text{=}\underset{n\to \infty }{\lim }\left(\frac{{\text{ln3}}^{\text{n}}}{{\text{ln4}}^{\text{n}}}\right)$

   $\text{=}\frac{\text{n}\underset{n\to \infty }{\lim }\left(\text{ln 3}\right)}{\text{n}\underset{n\to \infty }{\lim }\left(\text{ln 4}\right)}$

   $\text{=}\frac{\text{ln 3}}{\text{ln 4}}$

   Thus, the required box dimension $\text{d}$ is: $\text{=}\frac{\text{ln 3}}{\text{ln 4}}$.

   Case-$2$: When the result is tails, delete the third quarter of every interval ${\text{S}}_{\text{n}}$.

   Then, there will be no change in $\text{N}\left(\text{ε}\right)$ and $\text{ε}$. So:-

   $\text{N}\left(\text{ε}\right){\text{= 3}}^{\text{n}}\text{, }\frac{\text{1}}{\text{ε}}{\text{= 4}}^{\text{n}}$

   Now, use the below given formula to find the box dimension $\text{d}$:-

   $\text{d =}\underset{n\to \infty }{\text{lim}}\left(\frac{\text{lnN}\left(\text{ε}\right)}{\frac{\text{1}}{\text{ε}}}\right)$

   Now, substitute the values $\text{N}\left(\text{ε}\right){\text{= 3}}^{\text{n}}\text{, }\frac{\text{1}}{\text{ε}}{\text{= 4}}^{\text{n}}$ in the above box dimension formula. Then:-

   $\text{d =}\underset{n\to \infty }{\text{lim}}\left(\frac{\text{lnN}\left(\text{ε}\right)}{\frac{\text{1}}{\text{ε}}}\right)$

   $\text{=}\underset{n\to \infty }{\lim }\left(\frac{{\text{ln3}}^{\text{n}}}{{\text{ln4}}^{\text{n}}}\right)$

   $\text{=}\frac{\text{n}\underset{n\to \infty }{\lim }\left(\text{ln 3}\right)}{\text{n}\underset{n\to \infty }{\lim }\left(\text{ln 4}\right)}$

   $\text{=}\frac{\text{ln 3}}{\text{ln 4}}$

   Thus, the required box dimension $\text{d}$ is: $\text{d =}\frac{\text{ln 3}}{\text{ln 4}}$ .

   Now, see case-$1$ and case-$2$. Both the conditions have the same box dimension.

   Hence, box dimension exists for both the cases and equals to $\frac{\text{ln 3}}{\text{ln 4}}$.

2. Consider that if tails come up, delete the first quarter.

   When tails show up, delete the first quarter, which means that there will be no effect of the values of $\text{N}\left(\text{ε}\right)$ and $\text{ε}$.

   This means that the box dimension will be the same.

   Hence, there will be no difference if tails come up and delete the first quarter.

   Since the box dimension does not depend upon either head or tails, if we had a long string of tails, then there would be no effect of the box dimension.