Consider $$ p_{k+1} \zeta_{k+1}^2 + q_{k+1}\zeta_{k+1} + r_{k+1} = 0 $$
where
$$p_{k+1} = p_0P_k^2 + q_0P_kQ_k + r_0Q_k^2$$ $$q_{k+1} = 2p_0P_kP_{k-1} + q_0 ( P_kQ_{k-1} + Q_kP_{k-1} ) + 2r_0Q_kQ_{k-1}$$ $$r_{k+1} = p_0P_{k-1}^2 + q_0 P_{k-1}Q_{k-1} + r_0Q_{k-1}^2 = p_k $$
Given that $p_k \neq 0$ for any given $k$
Now the discriminant of this equation turns out to be: $$\Delta = (q_0^2 - 4p_0r_0 )( P_kQ_{k-1} - P_{k-1}Q_k)^2 $$
Specifically, $$C_k = \frac{ P_k } { Q_k } $$ denote the $k-th$ convergent of a simple continued fraction, so the identity $$P_kQ_{k-1}+Q_kP_{k-1} = \pm 1 $$ may be useful for our calculation.
I noticed that the first bracketed term of $\Delta$ was simply when $k=-1$, so is there some sort of easy way of achieving that expression without any intense algebra?
Kind Regards,
Let's see, for you, $\alpha = P_k,\gamma = Q_k, $ $\beta = P_{k-1},\delta = Q_{k-1}, $ $$ p = \left( \begin{array}{cc} P_k & P_{k-1} \\ Q_k & Q_{k-1} \end{array} \right) $$
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We have a binary form $\langle A,B,C \rangle$ meaning $f(x,y) = A x^2 + B xy + C y^2.$ We create the Hessian matrix $$ h = \left( \begin{array}{cc} 2A & B \\ B & 2 C \end{array} \right) $$ with discriminant $$ \Delta = B^2 - 4 AC. $$ To get back to the triple of coefficients we halve the diagonal entries but keep one of the off diagonal entries as is, for $B.$
Given a matrix
$$ p = \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$ we calculate the symmetric matrix $p^T h p$ to give a new one, $$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2 \rangle $$ Note that, about the discriminant $\Delta,$ we automatically have the new discriminant being $\Delta \det^2 p.$
Given $$ Q = \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta^2 \\ \alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right), $$ there is a typographical error on page 23 of Magnus, actually $\det Q = (\alpha \delta - \beta \gamma)^3.$
If we now write the triple $(A,B,C)$ as a row vector, we find $$ (A,B,C) \left( \begin{array}{ccc} \alpha^2 & 2 \alpha \beta & \beta^2 \\ \alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\ \gamma^2 & 2 \gamma \delta & \delta^2 \end{array} \right) = $$ $$ ( A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2) $$ Compare our earlier $ p^T h p =$ $$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2 \rangle $$