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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!

malilini
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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.

Hans Lundmark
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  • I think i messed something up. Im concentrating especially on Krawtchouk Polynomials - that's why i was confused by the degree. So the proof is about orthogonal polynomials in general. Can i use this theorem somehow to prove the interlacing properties for Krawtchouk Polynomials? – malilini Apr 17 '22 at 19:53
  • Can't you just change the sign of the polynomials if needed? That doesn't affect the orthogonality. – Hans Lundmark Apr 17 '22 at 20:57
  • I thought so. But if im not completely wrong a Krawtchouk polynomial with odd degree will be heading downwards after the last zero. Thats why im confused about the proof. – malilini Apr 17 '22 at 22:27
  • So just change the sign of the odd-degree polynomials then. – Hans Lundmark Apr 18 '22 at 06:12
  • Something Iam missing... the next step in the proof says, that $K_{n+1}(b) > 0$. Why do i know that? – malilini Apr 18 '22 at 14:13
  • It's already been proved that all zeros of every $p_i(x)$ lie in the open interval $(a,b)$. And $p_{n+1}(x)$ tends to $+\infty$ as $x \to +\infty$, so it has to be positive already when it leaves that interval at $x=b$, otherwise it would have a zero outside the interval. – Hans Lundmark Apr 18 '22 at 14:48
  • Thanks a lot for helping. My problem is just, that i cannot transfer it to a Krawtchouk polyniomial, because it can tend fo $+- \infty$ – malilini Apr 19 '22 at 10:23
  • I don't think I understand what your problem is. Can't you just let $p_0=K_0$, $p_1=-K_1$, $p_2=K_2$, $p_3=-K_3$, and so on? Then $p_n$ has the same zeros as $K_n$ for each $n$, and the highest coefficient of $p_n$ is positive, and the polynomials $p_n$ are orthogonal. So you should be able to apply the theorem to $p_n$ to get the conclusion that you want for $p_n$, and hence for $K_n$. – Hans Lundmark Apr 19 '22 at 10:57