Chapter 11.3, Problem 7E

Interpretation Introduction

Interpretation:

To construct the fractals von Koch snowflake curve, taking an equilateral triangle as ${\text{S}}_{\text{0}}$ and doing the von Koch procedure on each of its three sides.

1. To show that ${\text{S}}_{\text{1}}$ looks like a David star.

2. To draw ${\text{S}}_{\text{2}}{\text{ and S}}_{\text{3}}$.

3. To show that the curve ${\text{S = S}}_{\infty }$ has an infinite arc length.

4. The area of the region bounded by $\text{S}$ is to be found.

5. To find the similarity dimension of S.

Concept Introduction:

• Fractals are the complex geometric shapes with a fine structure at arbitrarily small scales.

• The similarity dimension d is defined as:

  $\text{d =}\frac{\text{ln m}}{\text{ln r}}$, where the values of m and r are determined by inspection.

• The procedure of constructing the Koch snowflake is by deleting the middle third of ${\text{S}}_{\text{0}}$ and replacing it with the other side of the equilateral triangle. This way, ${\text{S}}_{\text{n}}$ is obtained by applying this procedure on ${\text{S}}_{\text{n-1}}$

Answer to Problem 7E

Solution:

1. It is shown that ${\text{S}}_{\text{1}}$ looks like a David star.

2. ${\text{S}}_{\text{2}}{\text{ and S}}_{\text{3}}$ is drawn.

3. It is shown that ${\text{S = S}}_{\infty }$ have an infinite arc length.

4. The area of the region surrounded by $\text{S}$ is $\frac{2\sqrt{3}}{5}$.

5. The similarity dimension of the Koch snowflake is $\text{d =}\frac{\text{ln}\left(\text{4}\right)}{\text{ln}\left(\text{3}\right)}$.

Explanation of Solution

1. The von Koch snowflake curve can be constructed by the computer:

   

2. 

3. The arc length of ${\text{S}}_{\text{0}}\text{= 3}$; each iteration arc length is multiplied by $\frac{4}{3}$.

   Therefore, for ${\text{S = S}}_{\infty }$:

   The arc length is $=3{\left(\frac{4}{3}\right)}^{\infty }=\infty$

4. ${\text{S}}_{\text{0}}$ has a side length $\text{= 1}$, and the area is $\frac{\sqrt{3}}{4}$. In every iteration, a new equilateral triangle is added to for each side having the side length as $\frac{1}{3}$.

   Therefore:

   The side length of ${\text{S}}_n={\left(\frac{1}{3}\right)}^n$

   The number of sides of ${\text{S}}_{\text{n}}=3{(4)}^{\text{n}}$

   The area added in each iteration: ${\text{S}}_{n-1}{\text{to S}}_{\text{n}}=\left({\text{Number of sides of S}}_{n-1}\right)\frac{\sqrt{3}}{4}{\left(\frac{1}{3}\left({\text{Side length of S}}_{n-1}\right)\right)}^2$

   $\begin{array}{l}=3{\left(4\right)}^{n-1}\frac{\sqrt{3}}{4}{\left(\frac{1}{3}{\left(\frac{1}{3}\right)}^{n-1}\right)}^2\\ =3{\left(4\right)}^{n-1}\frac{\sqrt{3}}{4}{\left({\left(\frac{1}{3}\right)}^n\right)}^2\\ =\frac{3{\left(4\right)}^{n-1}\sqrt{3}}{4\left(3^{2n}\right)}=\frac{3\sqrt{3}}{16}{\left(\frac{4}{9}\right)}^{\text{n}}\end{array}$

   Add the area of ${\text{S}}_{\text{0}}$ and the area added after each iteration:

   $\begin{array}{l}\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{16}{\displaystyle \sum _{n=1}^{\infty }{\left(\frac{4}{9}\right)}^{\text{n}}}=\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{16}\frac{\frac{4}{9}}{1-\frac{4}{9}}\\ =\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{16}\frac{4}{5}=\frac{\sqrt{3}}{4}\left(1+\frac{3}{5}\right)\\ =\frac{2\sqrt{3}}{5}\end{array}$

   Thus, the area of the region enclosed by $\text{S}$ is $\frac{2\sqrt{3}}{5}$

5. The Koch snowflakes are constructed by using three Koch curves. To recreate the Koch curve, each Koch curve would require four copies of itself scaled by the factor $\frac{1}{3}$.

   Thus, the Koch snowflake can be constructed by taking four complete copies of itself and scaling each of them by factor $\frac{1}{3}$, after which each scaled Koch snowflake is cut into thirds and placed onto the full-size Koch snowflake.

   Therefore, the similarity dimension d of the Koch snowflake is given by:

   $4=3^d$

   Rearrange the above expression:-

   $\text{d =}\frac{\text{ln}\left(\text{4}\right)}{\text{ln}\left(\text{3}\right)}$

   Thus, the similarity dimension of the Koch snowflake is $\text{d =}\frac{\text{ln}\left(\text{4}\right)}{\text{ln}\left(\text{3}\right)}$.