Codex @codex  0

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

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