IDC402 Non Linear Dynamics

2022-01-11

Alternate way of viewing: Bernoulli Shift.

$x_n=0.a_i$

Where $a_i$ is the binary expansion.

Then, $x_{n+1}=(x_n«1)$ and then set first digit to 0.

Hence, after m steps, $x_m$ is in leading order. Hence, any error in order $2^m$ becomes of leading order in $m$ steps

Periodic Orbit of the Map

Order $p$ cycle if $x_0,\dots ,x_{p-1}$ is an orbit.

Period 1

Also called fixed point.

$$F(x)=x$$

First one is $0$

Second one is $2-2x=x⟹x=2/3$

Period 2

$$F^2(x)=x$$

Again, plotting $F^2$,

Which has 4 solutions.

2 of these are trivial, which are in the 1 orbit anyway, $0,\frac{2}{3}$

The new ones are $x=\frac{2}{5},\frac{4}{5}$

Hence all solutions are

$$x=\{0,\frac{2}{3},\frac{2}{5},\frac{4}{5}\}$$

Periods of Shift Map

Period 1

$$x_{n+1}=x_n$$

$x=0.\overline{0},0.\overline{1}$

Period 2

$$x_{n+2}=x_n$$

$x=0.\overline{01},0.\overline{10}$

Period P

$x=0.\overline{\text{any sequence of P digits}}$

Hence, there are $2^P$ distinct sequences hence so many $x_0$ which have period $P$.