This immediately implies that once one proves that the density of zeros exists, that it is given by $dk$.

5\. It is remarkable that this result is new, given that it is so elegant and its proof so simple! I think this is because the OP community most often focuses on measures and doesn’t think so much about the recursion parameters and the Schrödinger operator community doesn’t usually think of zeros of $P_n$.

[E2.2]{} Let $b_n\equiv 0$, $a_n \equiv {\frac}12$ which has period $p=1$. It is well-known in this case that the $P_n$ are essentially Chebyshev polynomials of the second kind, that is, $${\label}{2.19} 
P_n (\cos\theta) = {\frac}{1}{2^n}\, {\frac}{\sin(n+1)\theta}{\sin\theta}$$ Thus $P_{m-1}$ has zeros at points where $${\label}{2.20x} 
\theta ={\frac}{j\pi}{m} \qquad j=1, \dots, m-1$$ (the zeros at $\theta=0$ and $\theta=\pi$ are cancelled by the $\sin(\theta)$). $k(x)=\pi -\arccos (x)$ and is . We see that Theorem \[T2.1\] generalizes the obvious result on the zeros of the Chebyshev polynomials of the second kind.

By , zeros of $P_{mp-1}$ are precisely points where the $12$ matrix element of $T_{mp}$ vanishes, that is, points where $\binom{1}{0}$ is an eigenvector of $T_{mp}$. That is, zeros of $P_{mp-1}$ are Dirichlet points for this period $mp$ problem.

When 