OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 33B / b / i / Solution

Let

I_{0} = [x_{0}, x_{1}], I_{1} = [x_{1}, x_{2}], I_{2} = [x_{2}, x_{3}] .

The covering relations are

I_{0} ⟶ I_{1}, I_{1} ⟶ I_{2}, I_{2} ⟶ I_{0}, I_{1}, I_{2} .

Thus the covering graph has adjacency matrix

A = 001101011 .

The two closed length-three walks

I_{2} \to I_{2} \to I_{2} \to I_{2}, I_{2} \to I_{0} \to I_{1} \to I_{2}

give a horseshoe from two closed covering walks for

F^{3}

, so

F

must be chaotic. Moreover,

tr A = 1, tr (A^{3}) = 7.

By counting cycles in an interval covering graph, the number of primitive length-three cyclic itineraries is

\frac{7 - 1}{3} = 2.

Hence

F

must have at least, and the connect-the-dots realization shows that it need have only,

2

distinct 3-cycles.

Solved by gpt-5.6-sol high.

 Ancestors (12)