The cantor set

💼 Case: consider the tent map

$$ x_{n+1}=\left \{ \begin{matrix} r x_n & \text{if} & x_n \le \frac 12 \\ r(1-x_n) & \text{if} & x_n >\frac 12 \end{matrix} \right . $$

1. If we consider $r=2$

   All iterates of the map from the interval $[0,1]$ remain bounded within the interval $[0,1]$

1. If we consider $r=3$ lets try and find the points within $[0,1]$ do not escape to $-\infin$

   • Points that escape after one iteration have $f(x_0)>1$

     $f(x)=1$ for $x=1/2,2/3$

     if $x_0 \in (\frac 13, \frac 23)$ : $x_0$ escapes after 1 iteration

• After two iterations we must have ${f(x_0)\in (\frac 12 ,\frac 23)}$

  this happens for ${x_0\in \left ( \frac 19 , \frac 29 \right )} \text{ and } {x_0\in \left ( \frac 79 , \frac 89 \right ) }$

• After three iterations we can repeat the previous the values which we get are part of the Cantor set

We can construct the cantor set as following

1. Remove 1 piece of length $\frac 13$
2. Remove 2 pieces of length $\frac 13 ^2$
3. Remove 3 pieces of length $\frac 12^3$
4. At step $n$ we remove $2^{n-1}$ of length $\frac 13^n$

In total what we have removed is

$$ \small\begin{aligned} \sum^\infin_{n=1}\frac{2^{n-1}}{3^n}=\frac 13 \sum^\infin_{n=0}\left ( \frac{2}{3}\right )^n = \frac 13 \frac{1}{1-\frac 23}=1 \end{aligned} $$

After $n$ steps we have $2^n$ pieces of length $\frac 13^n$ so that the length $\frac 23^n \to 0$ as $n\to \infin$

Properties:

• There is structure on all length scales
• The Cantor set is self-similar
• The dimension of the Cantor set is not an integer
• The Cantor set has zero length

Von Koch curve

If we look at the length of the curve we have

$$ \begin{aligned} L_1&=\frac 43 L_0 \\ L_2&=\left ( \frac{4}{3} \right )^2 L_0 \\ &\; \; \vdots \\ L_\infin&=\infin

\end{aligned} $$

Total length of the Von Koch curve is infinite

• 🗒️ Note: on area

The arclength between any two points in Euclidian space of the curve is infinite

So we cant express the positions of points using a 1D grid but 2D is redundant so it has non-integer dimensions between 1 and 2

Self similarity

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/949b372f-a131-4642-a322-a16a5b775eec/Self_similarity.gif" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/949b372f-a131-4642-a322-a16a5b775eec/Self_similarity.gif" width="40px" />

Self similarity: Two objects are similar if they have the same shape, regardless of their size

</aside>

Similarity transformations:

• Scaling

• Rotation

• Translation

• A function $f(x)$ is scale-invariant if its shape is preserved under rescaling of the variable $x$

  $$ f(\lambda x)=f(x)g(\lambda) $$

  where $\lambda$ is the rescaling factor and $g$ is a function of $\lambda$

  🗒️ Note: here $f(x)$ looks the same regardless of the scale on which we look at it

⚽ Goal: lets try to remove this abstract $g(\lambda)$ function

• If we set $x=1$ on $f(\lambda x)=f(x)g(\lambda)$ we get

  $$ \begin{aligned} g(\lambda )&=\frac{f(\lambda)}{f(1)} \\ \Rightarrow \qquad f(\lambda x)&=f(x) \frac{f(\lambda)}{f(1)} \end{aligned} $$

• Now if we differentiate with respect to $\lambda$ we get

  $$ xf'(\lambda x)=\frac{f(x)}{f(1)}f'(\lambda) $$