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 Incoming links: Periodic points of the doubling map

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 32A / b / ii / Solution Created 2026-09-24 Updated 2026-09-25

By part (i),

F^{n} (x) = x

exactly when

(2^{n} - 1) x \in Z .

Thus the periodic points of the doubling map fixed by

F^{n}

are

x = \frac{j}{2 ^{n} - 1}, j = 0, \dots, 2^{n} - 2,

and their number is

2^{n} - 1 .

For

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.

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