Im working on the interlacing properties of zeros of orthogonal polynomials $p_n(x)$ - proved by Gabor Szegö (Orthogonal Polynomials - Theorem 3.3.2 ) The Theorem says: Let $x_1 < … <x_n$ be the zeroes of $p_n(x)$, $x_0 = a, x_{n+1} =b$. Then each interval $[x_\theta, x_{\theta+1} ]$, $\theta = 0,1,…,n$ contains exactly one zero of $p_{n+1}(x)$.
I understand most of the proof, but there is a little part, which is not clear to me. So the proof says if $\psi =x_n$ is the greatest zero of $p_n(x)$ , then $p_n'(\psi) > 0$.
I dont see why this follows, i thought it depends on the degree of the polynomial if $p'_n(\psi)$ is smaller or bigger than zero.
Thanks for helping!
Since the coefficient of $x^n$ in $p_n(x)$ is positive (by definition), we have $p_n(x) \to +\infty$ as $x \to +\infty$. So the last time that the graph of $p_n$ crosses the $x$-axis, it must be heading upwards.