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 $xin [0,1)$.
The paper says this dynamical system is "left-shift with leading digit truncation" in binary notation. What is this?
"Repeated measurements in time will yield eventually all binary digits of the initial point." what does this mean?
dynamical-systems binary ergodic-theory chaos-theory
edited Dec 6 '18 at 22:07
Adam
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
add a comment |
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 $xin [0,1)$.
The paper says this dynamical system is "left-shift with leading digit truncation" in binary notation. What is this?
"Repeated measurements in time will yield eventually all binary digits of the initial point." what does this mean?
dynamical-systems binary ergodic-theory chaos-theory
edited Dec 6 '18 at 22:07
Adam
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
add a comment |
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 $xin [0,1)$.
The paper says this dynamical system is "left-shift with leading digit truncation" in binary notation. What is this?
"Repeated measurements in time will yield eventually all binary digits of the initial point." what does this mean?
dynamical-systems binary ergodic-theory chaos-theory
edited Dec 6 '18 at 22:07
Adam
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
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 $xin [0,1)$.
The paper says this dynamical system is "left-shift with leading digit truncation" in binary notation. What is this?
"Repeated measurements in time will yield eventually all binary digits of the initial point." what does this mean?
dynamical-systems binary ergodic-theory chaos-theory
dynamical-systems binary ergodic-theory chaos-theory
edited Dec 6 '18 at 22:07
Adam
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
edited Dec 6 '18 at 22:07
Adam
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
edited Dec 6 '18 at 22:07
Adam
1,1951919
edited Dec 6 '18 at 22:07
Adam
1,1951919
edited Dec 6 '18 at 22:07
Adam
1,1951919
1,1951919
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
asked Nov 23 '18 at 8:31
sleeve chensleeve chen
3,06641852
3,06641852
add a comment |
add a comment |
2 Answers
2
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oldest
votes
I think it means:
- 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.
- 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.
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
add a comment |
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).
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I think it means:
- 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.
- 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.
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
add a comment |
I think it means:
- 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.
- 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.
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
add a comment |
I think it means:
- 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.
- 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.
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
I think it means:
- 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 $fleft(frac{3}{4}right) = frac{1}{2} = 0.1$.
- 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.
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
answered Nov 23 '18 at 9:06
Sam StreeterSam Streeter
1,485417
1,485417
add a comment |
add a comment |
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).
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
add a comment |
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).
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
add a comment |
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).
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
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).
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
answered Nov 23 '18 at 9:06
Fabio LucchiniFabio Lucchini
7,81311426
7,81311426
add a comment |
add a comment |
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