Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 11I (Number Theory)

Setand, for $n\ge 1$, defineWriting $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$, these recurrences are equivalent to the continued-fraction matrix identityThe associated Möbius transformation sends a positive tail $\beta$ to the continued-fraction tail formula

Since $\det M(a_j)=-1$, taking determinants givesMoreover,whereasTheir signs are opposite, so ${\theta }_n$ lies between the two adjacent convergents. For $n=0$, the second endpoint is interpreted as $p_{-1}/q_{-1}=\infty$.

Let ${\beta }_{n+1}>0$ be the complete quotient after $a_n$. Part (a) writes the irrational number $\theta$ asso $\theta$ lies between the adjacent continued fraction convergents $p_n/q_n$ and $p_{n-1}/q_{n-1}$. Their separation isBecause all $a_j$ are positive integers, the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ implies $q_n\to \infty$. Hence this separation tends to zero, and the squeeze theorem gives

The continued-fraction matrix for the original finite continued fraction isEach $M(a_j)$ is a symmetric matrix, so reversing the order gives the transpose:Reading off the columns proves the reversal identity for a finite continued fraction:These are respectively the $n$th and $(n-1)$th convergents of the reversed continued fraction.

Pure periodicity means that $\theta$ reappears as the tail after one period. The continued-fraction tail formula therefore givesAfter cross-multiplication,

Now letBy part (c), the matrix for one reversed period isso its positive fixed point $\eta$ satisfiesPutting $\eta =-1/X$ and multiplying by $-X^2$ turns this equation into $f(X)=0$. The product of the two roots of $f$ is $-p_{n-1}/q_n<0$; because $\theta >0$, the other root ${\theta }^{\prime }$ is negative. Consequently $-1/{\theta }^{\prime }>0$ is the positive fixed point selected by convergence of the reversed continued fraction. The reversed-period identity for a purely periodic continued fraction is therefore