# Dynamic system $f(x) = 2x$ mod $1$

I am reading the following paper:

ergodic theory of chaos and strange attractors, by J.-P. Eckmann (can be easily downloaded)

My question is from an example on p.620 (bottom left):

Consider a dynamical system $$f(x) = 2x \ \ \ \text{mod } 1$$ for $$x\in [0,1)$$.

1. The paper says this dynamical system is "left-shift with leading digit truncation" in binary notation. What is this?

2. "Repeated measurements in time will yield eventually all binary digits of the initial point." what does this mean?

1. The binary representation of the number $$f\left(x\right)$$ is the same as the binary representation of $$x$$ shifted one place to the left and with the leading digit cut off. For example, in binary notation, we have $$\frac{3}{4} = 0.11$$, and $$f\left(\frac{3}{4}\right) = \frac{1}{2} = 0.1$$.
2. Reading the example as in the paper, you didn't mention that "measurement" here really means measurement of whether a number is greater or less than $$\frac{1}{2}$$. By repeatedly applying $$f\left(x\right)$$ and measuring the outcome each time, we can ascertain for each binary digit of $$x$$ whether it is $$0$$ or $$1$$, hence we can determine all binary digits of $$x$$ in this way.
Let $$x=0,b_1\ldots b_n$$, with $$b_i\in\{0,1\}$$, be the representation of $$x$$ in base $$2$$. Then $$2x=b_1,b_2\ldots b_n$$ hence $$f(x)=0,b_2\ldots b_n$$, that's $$f(x)$$ is obtained from $$x$$ by remove left-handed binary digit as asserted in 1. Consequently, the orbit of $$x$$ is: $$\begin{array} x & 0,b_1\ldots b_n\\ f(x) & 0,b_2\ldots b_n\\ f^{\circ 2}(x) & 0,b_3\ldots b_n\\ \end{array}$$ Consequently, the first digit of the sequence $$x, f(x), f^{\circ 2}(x), \ldots$$ gives the sequence of binary digits of $$x$$ (statement 2).