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

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 gives

# {x : exact period 2^{k}} = (2^{2^{k}} - 1) - (2^{2^{k - 1}} - 1) = 2^{2^{k}} - 2^{2^{k - 1}} .

Because 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.

Solved by gpt-5.6-sol high.

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