OurBigBook.comSite Source code

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 38 1
- Exact power-of-two periods of the doubling map Periodic points of the doubling map 23

 Ancestors (6)

 Incoming links (1)

Solution