Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 32A (Dynamical Systems)

A map is Devaney chaos when:

Write a point as a binary expansionApart from the harmless choice of expansion at dyadic rationals, the binary shift representation of the doubling map is

Every open interval contains a binary cylinder specified by a finite initial word. Given cylinders $U$ and $V$, choose a binary sequence beginning with the word for $U$ and place the word for $V$ after it. A suitable iterate shifts the second word to the front, proving topological transitivity.

Given any cylinder, repeat its defining word forever. The resulting point is periodic and lies in that cylinder, so periodic points are dense.

Finally, given $x$ and any neighbourhood, choose $N$ so large that changing only digits after the first $N$ stays inside that neighbourhood. Choose the later tail so that after $N$ shifts it is either $0$ or $3/4$, whichever is farther from $F^N(x)$. The separation is at least $3/8$, so any smaller fixed constant, for example $\delta =1/4$, proves sensitivity. Hence the doubling map is chaotic in Devaney's sense.

Concatenate every finite binary word, ordered first by length and then lexicographically:Every finite binary block occurs in this expansion. Since iterating $F$ shifts digits to the left, its orbit enters every binary cylinder and is therefore dense. The digit string is not eventually periodic, so it gives a concrete nonperiodic chaotic orbit.

A fixed point satisfiesso $x$ is an integer. In $[0,1)$ the only possibility isEquivalently, a binary sequence fixed by the left shift is constant; $0.000\dots {}_2=0$, while $0.111\dots {}_2=1$ represents the excluded endpoint, which is the same circle point.

Points whose period divides two solveso, after removing the fixed point zero, the unique two-cycle is

Points whose period divides four are $j/15$ for $j=0,\dots ,14$. Removing the three points fixed by $F^2$, namely $0,5/15,10/15$, leaves twelve points of exact period four. They form the three orbitsandStarting at another point on one displayed orbit gives the corresponding cyclic rotation.

The result follows by induction. It is true for $n=1$. IfthenTherefore

By part (i), $F^n(x)=x$ exactly whenThus the periodic points of the doubling map fixed by $F^n$ areand their number isFor $n=2$ this gives three points: the fixed point and the two points in the two-cycle. For $n=4$ it gives fifteen: those three points and the twelve points in the three four-cycles found above.

Take $n=2^k$. Every proper divisor of $n$ divides $n/2=2^{k-1}$, so the points fixed by $F^n$ that do not have exact period $n$ are precisely those fixed by $F^{n/2}$. Part (ii) therefore givesBecause different starting points are counted separately, no division by the period is made. This is the exact power-of-two periods of the doubling map.