Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 32A / b

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.