= Solution
Pure periodicity means that $\theta$ reappears as the tail after one period. The <continued-fraction tail formula> therefore gives
$$
\theta=[a_0,\ldots,a_n,\theta]
=\frac{p_n\theta+p_{n-1}}{q_n\theta+q_{n-1}}.
$$
After cross-multiplication,
$$
\boxed{f(\theta)
=q_n\theta^2+(q_{n-1}-p_n)\theta-p_{n-1}=0.}
$$
Now let
$$
\eta=[\overline{a_n,a_{n-1},\ldots,a_0}].
$$
By part (c), the matrix for one reversed period is
$$
\begin{pmatrix}p_n&q_n\\p_{n-1}&q_{n-1}\end{pmatrix},
$$
so its positive fixed point $\eta$ satisfies
$$
p_{n-1}\eta^2+(q_{n-1}-p_n)\eta-q_n=0.
$$
Putting $\eta=-1/X$ 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 $\theta'$ is negative. Consequently $-1/\theta'>0$ 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
$$
\boxed{-\frac1{\theta'}=[\overline{a_n,a_{n-1},\ldots,a_0}].}
$$
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