 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/11i/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

p_{- 1} = 1, q_{- 1} = 0, p_{0} = a_{0}, q_{0} = 1,

and, for

n \geq 1

p_{n} = a_{n} p_{n - 1} + p_{n - 2}, q_{n} = a_{n} q_{n - 1} + q_{n - 2} .

M (a) = (a 1 10)

, these recurrences are equivalent to the continued-fraction matrix identity

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

The associated Möbius transformation sends a positive tail

β

θ_{n} = [a_{0}, \dots, a_{n}, β] = \frac{β p _{n} + p _{n - 1}}{β q _{n} + q _{n - 1}} .

Since

det M (a_{j}) = - 1

, taking determinants gives

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

Moreover,

θ_{n} - \frac{p _{n}}{q _{n}} = \frac{p _{n - 1} q _{n} - p _{n} q _{n - 1}}{q _{n} ( β q _{n} + q _{n - 1} )},

whereas

θ_{n} - \frac{p _{n - 1}}{q _{n - 1}} = \frac{β ( p _{n} q _{n - 1} - p _{n - 1} q _{n} )}{q _{n - 1} ( β q _{n} + q _{n - 1} )} .

Their signs are opposite, so

θ_{n}

lies between the two adjacent convergents. For

n = 0

, the second endpoint is interpreted as

p_{- 1} / q_{- 1} = \infty

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