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
Order $p$ cycle if ${x_0,\dots,x_{p-1}}$ is an orbit.
Also called fixed point.
\[F(x) = x\]First one is $0$
Second one is $2-2x = x \implies x = 2/3$
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
\[x = \left\{0, \frac{2}{3}, \frac{2}{5}, \frac{4}{5}\right\}\]$x = {0.\bar{0}, 0.\bar 1}$
$x = {0.\overline{01}, 0.\overline{10}}$
$x = {0.\overline{\text{any sequence of P digits}}}$
Hence, there are $2^P$ distinct sequences hence so many $x_0$ which have period $P$.