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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?

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2 Answers 2

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I think it means:

  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.
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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).

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