A_{n-1}(z) \dots A_0 (z)$$ for $n=1,2,\dots$. We have, by , that $${\label}{2.8} 
T_{mp+b} = T_b (T_p)^m$$ suggesting that $T_p$ plays a basic role. By and , $${\label}{2.9} 
\det (T_p) =1$$

A fundamental quantity is the discriminant $${\label}{2.10} 
\Delta(z) = {\text{\rm{Tr}}}(T_p(z))$$ By , we have $${\label}{2.10a} 
T_n(z) = \begin{pmatrix} p_n(z) & q_{n-1}(z) \\
p_{n-1}(z) & q_{n-2}(z) \end{pmatrix}$$ where $q_n(z)$ is a polynomial of degree $n$ that is essentially the polynomial of the second kind (the normalization is not the standard one but involves an extra $a_p$).

By and , $T_p(z)$ has eigenvalues $${\label}{2.11} 
\Gamma_\pm (z) = {\frac}{\Delta(z)}{2} \pm \sqrt{\bigl(\tfrac{\Delta (z)}{2}\bigr)^2 -1}$$ In a moment, we will define branch cuts in such a way that on all of ${{\mathbb{C}}}\backslash 
\text{cuts}$, $${\label}{2.12x} 
{\lvert\Gamma_+(z)\rvert} > {\lvert\Gamma_- (z)\rvert}$$ so implies the Lyapunov exponent is given by $${\label}{2.12}
\lim_{n\to\infty} \, {\frac}{1}{n} \, \log \|T_n(z)\| = {\frac}{1}{p} \, \log {\lvert\Gamma_+(z)\rvert} 
\equiv \gamma(z)$$

means ${\lvert\Gamma_+\rvert}={\lvert\Gamma_-\rvert}$ if and only if $\Delta(z)\in 
[-2,2]$, and one shows that this only happens if $z$ is real. Moreover, if $\Delta(z)\in 
(-2,2)$, then $\Delta'(x)\neq 0$. Thus, for $x$ very negative, $(-1)^p \Delta(x) >0$ and solutions of $(-1)^p \Delta(x)=\pm 2$ alternate as $+2,-2,-2,+2,+2,-2,-2,\dots$, which we label as $${\label}{2.13} 
\alpha_1 < \beta_1 \leq \alpha_2 < \beta_2 