band edges where $\Gamma_+ -\Gamma_- 
\to 0$. For later purpose, we note that $\langle \binom{0}{1}, (T_p(z) - \Gamma_- 
\boldsymbol{1}) \binom{1}{0}\rangle = \langle \binom{0}{1}, T_p(z) \binom{1}{0} 
\rangle$ has a finite limit at such points. Later we will be looking at $$\begin{aligned}
a(z) (\Gamma_+^m - \Gamma_-^m) &= \biggl\langle\binom{0}{1}, T_p(z)\binom{1}{0} 
\bigg\rangle \, {\frac}{\Gamma_+^m - \Gamma_-^m}{\Gamma_+ - \Gamma_-} \\ 
&  \to \bigg\langle \binom{0}{1}, T_p(z) \binom{1}{0} \bigg\rangle m\Gamma_+^{m-1} \end{aligned}$$ if $\Gamma_+ - \Gamma_- \to 0$. This is zero if and only if $\langle \binom{0}{1}, 
T_p(z) \binom{1}{0}\rangle =0$, that is, if and only if the edge of the band is a Dirichlet data point.

and imply $${\label}{2.32} 
T_{mp}(z) = T_p(z)^m = \Gamma_+^m P_+ + \Gamma_-^m P_-$$ so $${\label}{2.33} 
T_{mp}(z)\binom{0}{1} = [a(z)(\Gamma_+^m - \Gamma_-^m)]\binom{0}{1} + 
[b(z) \Gamma_+^m + (1-b(z)) \Gamma_-^m)] \binom{1}{0}$$ Thus, by for $b\geq 0$, $$\begin{aligned}
P_{mp+b-1} &= \bigg\langle \binom{0}{1}, T_b T_{mp}\binom{1}{0}\bigg\rangle {\label}{2.37a} \\
&= [(\Gamma_+^m - \Gamma_-^m)a(z)] q_{b-2}(z) + [b(z) \Gamma_+^m + (1-b(z)) 
\Gamma_-^m)] p_{b-1}(z) {\label}{2.34} \end{aligned}$$ where $${\label}{2.35} 
q_{-2}(z) \equiv 1 \qquad q_{-1}(z) \equiv 0$$

For $b=0$, $p_{b-1}\equiv 0$ and $q_{b-2}=1$, so $${\label}{2.36} 
p_{mp-1}(z) = (\Gamma_+^m - \Gamma_-^m) a(z)$$ Its zeros are thus points where $a(z)=0$ or where $\Gamma_+^m = \Gamma_-^m$, except that at branch points, $a(z)$ can have a pole which can cancel 