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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

Set
$$
p_{-1}=1,\quad q_{-1}=0,
\qquad
p_0=a_0,\quad q_0=1,
$$
and, for $n\geq1$, define
$$
p_n=a_np_{n-1}+p_{n-2},
\qquad
q_n=a_nq_{n-1}+q_{n-2}.
$$
Writing $M(a)=\left(\begin{smallmatrix}a&1\\1&0\end{smallmatrix}\right)$, these recurrences are equivalent to the <continued-fraction matrix> identity
$$
M(a_0)\cdots M(a_n)
=\begin{pmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{pmatrix}.
$$
The associated <Möbius transformation> sends a positive tail $\beta$ to the <continued-fraction tail formula>
$$
\boxed{\theta_n=[a_0,\ldots,a_n,\beta]
=\frac{\beta p_n+p_{n-1}}{\beta q_n+q_{n-1}}.}
$$

Since $\det M(a_j)=-1$, taking determinants gives
$$
\boxed{p_nq_{n-1}-p_{n-1}q_n=(-1)^{n+1}=(-1)^{n-1}.}
$$
Moreover,
$$
\theta_n-\frac{p_n}{q_n}
=\frac{p_{n-1}q_n-p_nq_{n-1}}
{q_n(\beta q_n+q_{n-1})},
$$
whereas
$$
\theta_n-\frac{p_{n-1}}{q_{n-1}}
=\frac{\beta(p_nq_{n-1}-p_{n-1}q_n)}
{q_{n-1}(\beta q_n+q_{n-1})}.
$$
Their signs are opposite, so $\theta_n$ lies between the two adjacent convergents. For $n=0$, the second endpoint is interpreted as $p_{-1}/q_{-1}=\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 11I / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 11I / a / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 11I / a / Solution

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