Questions on doubling map D(x)

Question

For the doubling map D on [0, 1): $$D(x)=\begin{cases} 2x & 0\leq x< \frac{1}{2} \\ 2x-1 & \frac{1}{2}\leq x <1 \end{cases}$$

1. List all points whose orbits end up landing on 0 and are thereby eventually fixed.
2. Let x ∈ [0, 1) and suppose that x is given in binary form as $a_0, a_1, a_2 ...$ where each $a_j$ is either 0 or 1. First give a formula for the binary representation of D(x). Then explain why this causes orbits of D generated by a computer to end up eventually fixed at 0.

I have no idea how to solve this can anyone provide any hints?

Answer 1

Can you guess how the binary representation behaves under the doubling map? Recall that $x \in [0,1)$ has binary representation $a_1,a_2,a_3,\dots$ (I drop the $a_0$ term for a more convenient notation) if and only if $$ x = \sum_{k \geq 1} a_k 2^{-k}.$$ Note that $x < 1/2$ if and only if$^*$ $a_1 = 0$ (and, hence, $x \geq 1/2$ if and only if $a_1 = 1$). Thus, if $x = (a_1,a_2,\dots)$, then what is $D(x)$ in binary? This can then be used to tackle both questions. Try it yourself first and only check the spoiler below if you get stuck.

$^*$Here, I implicitly assume that we take the finite representation if both a finite and an infinite one exist (as is the case for $x=1/2$).

Solution: Plugging the binary representation into the formula for the doubling map shows that $D(a_1,a_2,\dots) = (a_2,a_3,\dots)$. So, the doubling map shifts the sequence to the left, always dropping the first entry. In particular, if a point $x$ admits a finite binary representation, i.e. where $a_k = 0$ for all $k \geq n$, then $D^n(x) = 0$. Conversely, if $x$ does not admit a finite binary representation, then it admits an infinite binary representation in which there is a non-zero entry $a_{n_k}$ with $n_k \geq k$ for all $k \in \mathbb{N}$. In this case, $D^{n_k}(x) = (a_{n_k},a_{n_k+1},\dots) \ne 0$ for all $k \in \mathbb{N}$, so $x$ does not eventually get mapped to 0. Lastly, of course, a computer always works with finite representations.