This follows on from this post.

Recall the Doubling Mapping D:[0,1)\rightarrow [0,1) given by:

\displaystyle D(x)=\begin{cases} 2x & \text{ if }x<1/2 \\ 2x-1 & \text{ if }x\geq 1/2 \end{cases}

At the end of the last post we showed that this dynamical system displays sensitivity to initial conditions. Now we show that it displays topological mixing (a chaotic orbit) and density of periodic points.

First we must talk about periodic points.

Periodic Points

Consider, for example, the initial state \displaystyle x_0=\frac{1}{9}. The orbit of x_0 is given by:

\displaystyle \text{orb}(x_0)=\left\{\frac{1}{9},\frac29,\frac49,\frac89,\frac79,\frac59,\frac19,\frac29,\dots\right\}

Here we see \frac19 repeats itself and so gets ‘stuck’ in a repeating pattern:


The orbit of x_0=1/9.

The orbit of any fraction, e.g. \displaystyle x_0=\frac{4}{243}, must be periodic, because \displaystyle D\left(\frac{i}{243}\right) is either equal to \displaystyle \frac{2i}{243} of \displaystyle \frac{2i-243}{243} and so the orbit consists only of states of the form:

\displaystyle \frac{i}{243},

and there are only 243 of these and so after 244 iterations, some state must be repeated and so we get locked into a periodic cycle.

If we accept the following:


A fraction \frac{p}{q} has a recurring binary expansion:

\displaystyle \frac{p}{q}=0.b_1\dots b_m\overline{a_1a_2\dots a_n}_2,

then this is another way to see that fractions are (eventually) periodic. Take for example,

\displaystyle x_0=0.101,101,101,101,\dots_2=0.\overline{101}_2=\frac{5}{7}.

Then, recalling D chops off the first binary digit:

\displaystyle x_1=D(x_0)=0.01,101,101,101,\dots_2,

\displaystyle x_2=D(x_1)=0.1,101,101,101,\dots_2,

\displaystyle x_3=D(x_2)=0.101,101,101,\dots_2=0.\overline{101}_2=x_0,

we see that if x_0 is recurring (after a point), then it is (eventually) periodic.


The (eventually) periodic points of D are those x_0\in [0,1) with a recurring binary expansion (aka the fractions).


Show that

\displaystyle 0.\overline{101}_2=\frac{5}{7}.


A Chaotic Orbit

From the above, and the previous post, we know to find an initial state that has a chaotic orbit (never repeating) it cannot be recurring. We also want it to get close to every possible point. Consider the following:

  • one digit binaries: 0.0_2,0.1_2
  • two digit binaries: 0.00_2,0.01_2,0.10_2,0.11_2
  • three digit binaries: 0.000_2,0.001_2,0.010_2,0.011_2,0.100_2,0.101_2,0.110_2,0.111_2.

Make a state that agrees to these to one, two, three binary places (recall that numbers are close when they agree to a number of binary/decimal places)


Now this state has a terminating binary expansion and so is a fraction. However note it’s orbit gets close to everything:


This orbit gets close to everything: draw a horizontal line and some iterate is close to it… but is eventually fixed at x=0.

If instead we keep the pattern going, with all the four digit binaries, the five digit binaries, etc., etc., we get an initial state that is not periodic and gets close to every possible state in [0,1), because it eventually agrees to every state to an arbitrary number of binary digits.

Such an initial state has a chaotic orbit.

Density of Periodic Points

Have we got close to every state a periodic state. Yes! Take any x\in [0,1), for example


Note that x is not a fraction. Then the initial state x_0=\overline{0.01101}_2 is periodic, and close to x (because they agree to five binary places).

This is the density of periodic points and the mega-sensitivity to initial conditions mentioned in the previous post. I believe that \sqrt{2} -1 has a chaotic orbit, yet it is very close to the periodic point x_0=\overline{0.01101}_2 which has very different behaviour:


\sqrt{2}-1\approx 0.\overline{01101}_2, but their orbits have very different behaviour: chaotic (I think) in red vs periodic in green.

Therefore D:[0,1)\rightarrow [0,1) is a chaotic mapping.

Prediction in the Presence of Chaos

In the real world, measurement come with an error. Suppose, for example, that we measure many aspects of the weather and feed them into our computer models that predict temperature.

Now the problem is, if we think the temperature today is 15.5, and measurement error of 0.1 is present, and 15.6, 15.5 and 15.4 have very different orbits, what temperature should be put into our computer model? The answer is we should put all of them in! We should put in an ensemble of initial states. So rather than T_0=15.5 we should look at perhaps 100 starting values:

T_0=15.450,15.451,15.452,\dots,15.500,\dots, 15.550.


Here we see the orbits of ten initial states close to T_0=15.5. Note that after three days there is already quite a large variance between the predicted temperatures. After five days there isn’t really any pattern at all. The best we can do is predict, perhaps T_1\approx 13, T_2\approx 18 and T_3\approx 7. After this it is a lottery.