 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-3/2g/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 2G / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Introduce

p_{- 1} = 1

,

q_{- 1} = 0

,

p_{0} = a_{0}

and

q_{0} = 1

. Composition of the Möbius transformations

z \mapsto a_{j} + 1/ z

represents a finite continued fraction. Their matrices give

(p_{n} q_{n} p_{n - 1} q_{n - 1}) = (a_{0} 1 10) (a_{1} 1 10) \dots (a_{n} 1 10) .

Indeed, multiplying the last factor gives the linear recurrences

p_{n} = a_{n} p_{n - 1} + p_{n - 2}

and

q_{n} = a_{n} q_{n - 1} + q_{n - 2}

, which reproduce the continued fraction convergents of the continued fraction. Each factor has determinant

- 1

. Taking determinants therefore yields

p_{n} q_{n - 1} - q_{n} p_{n - 1} = (- 1)^{n + 1} .

This identity also proves that the numerator and denominator in the first column are coprime, so no unnoticed cancellation changes the displayed matrix.

 Read the full article

 New to topics? Read the docs here!