Periodic points of the doubling map

The fixed points of

F^{n}

are

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

Thus

F^{n}

has

2^{n} - 1

fixed points.

Table of contents
- Exact power-of-two periods of the doubling map Periodic points of the doubling map

Exact power-of-two periods of the doubling map

For

n = 2^{k}

, every proper period dividing

n

divides

n /2

. Hence the number of points of exact period

2^{k}

is

(2^{2^{k}} - 1) - (2^{2^{k - 1}} - 1) = 2^{2^{k}} - 2^{2^{k - 1}} .

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

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