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











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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 $xin [0,1)$.





  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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    up vote
    4
    down vote

    favorite












    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 $xin [0,1)$.





    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?












    share|cite|improve this question
























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      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 $xin [0,1)$.





      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?












      share|cite|improve this question













      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 $xin [0,1)$.





      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?









      dynamical-systems ergodic-theory chaos-theory






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      asked Nov 23 at 8:31









      sleeve chen

      2,95341851




      2,95341851






















          2 Answers
          2






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          I think it means:





          1. The binary representation of the number $fleft(xright)$ 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.

          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 $fleft(xright)$ 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.







          share|cite|improve this answer




























            up vote
            3
            down vote













            Let $x=0,b_1ldots b_n$, with $b_iin{0,1}$, be the representation of $x$ in base $2$.
            Then $2x=b_1,b_2ldots b_n$ hence $f(x)=0,b_2ldots b_n$, 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_1ldots b_n\
            f(x) & 0,b_2ldots b_n\
            f^{circ 2}(x) & 0,b_3ldots 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).






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              5
              down vote



              accepted










              I think it means:





              1. The binary representation of the number $fleft(xright)$ 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.

              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 $fleft(xright)$ 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.







              share|cite|improve this answer

























                up vote
                5
                down vote



                accepted










                I think it means:





                1. The binary representation of the number $fleft(xright)$ 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.

                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 $fleft(xright)$ 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.







                share|cite|improve this answer























                  up vote
                  5
                  down vote



                  accepted







                  up vote
                  5
                  down vote



                  accepted






                  I think it means:





                  1. The binary representation of the number $fleft(xright)$ 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.

                  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 $fleft(xright)$ 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.







                  share|cite|improve this answer












                  I think it means:





                  1. The binary representation of the number $fleft(xright)$ 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.

                  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 $fleft(xright)$ 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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                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 9:06









                  Sam Streeter

                  1,481417




                  1,481417






















                      up vote
                      3
                      down vote













                      Let $x=0,b_1ldots b_n$, with $b_iin{0,1}$, be the representation of $x$ in base $2$.
                      Then $2x=b_1,b_2ldots b_n$ hence $f(x)=0,b_2ldots b_n$, 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_1ldots b_n\
                      f(x) & 0,b_2ldots b_n\
                      f^{circ 2}(x) & 0,b_3ldots 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).






                      share|cite|improve this answer

























                        up vote
                        3
                        down vote













                        Let $x=0,b_1ldots b_n$, with $b_iin{0,1}$, be the representation of $x$ in base $2$.
                        Then $2x=b_1,b_2ldots b_n$ hence $f(x)=0,b_2ldots b_n$, 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_1ldots b_n\
                        f(x) & 0,b_2ldots b_n\
                        f^{circ 2}(x) & 0,b_3ldots 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).






                        share|cite|improve this answer























                          up vote
                          3
                          down vote










                          up vote
                          3
                          down vote









                          Let $x=0,b_1ldots b_n$, with $b_iin{0,1}$, be the representation of $x$ in base $2$.
                          Then $2x=b_1,b_2ldots b_n$ hence $f(x)=0,b_2ldots b_n$, 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_1ldots b_n\
                          f(x) & 0,b_2ldots b_n\
                          f^{circ 2}(x) & 0,b_3ldots 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).






                          share|cite|improve this answer












                          Let $x=0,b_1ldots b_n$, with $b_iin{0,1}$, be the representation of $x$ in base $2$.
                          Then $2x=b_1,b_2ldots b_n$ hence $f(x)=0,b_2ldots b_n$, 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_1ldots b_n\
                          f(x) & 0,b_2ldots b_n\
                          f^{circ 2}(x) & 0,b_3ldots 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).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 23 at 9:06









                          Fabio Lucchini

                          7,77311326




                          7,77311326






























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