How to show that recurrence coeficients can be written as $a²_n=\frac{<P_n,P_n>}{<P_{n-1},P_ {n-1}>},\; b_n= \frac{<x P_n,P_n>}{<P_n,P_n>}$

by Mario Gonzalez   Last Updated June 13, 2019 12:20 PM

can someone help me.

We have $$\mu \in M(R)$$ is a probability measure, then orthogonal polynomials

$$p_n(z) = p_n(z;\mu) = k_n z^n + \gamma_n z^{n-1}$$

satisfy ahree-term recurrence relation $$x p_n(x) = a_{n+1} p_{n+1}(x) + b_n p_n(x) + a_n p_{n-1}(x),\;\;(1)$$ with inicial conditions $$p_{-1}=0$$ and $$p_0(x) =1.$$

The recurrence coefficients are given by $$a_n = \int x p_{n-1}(x) p_n(x) d \mu(x) =\frac{k_{n-1}}{k_n}>0$$ $$b_n = \int x p_n²(x) d \mu(x) =\frac{\gamma_n}{k_n}-\frac{\gamma_{n+1}}{k_{n+1}} \in R$$

The monic polynimials $$P_n(x)= p_n(x;\mu)/k_n,$$ satisfy the three term recurrence relation $$P_{n+1}(x) = (x-b_n) P_n(x) - a²_n P_{n-1}(x)$$

The question is:

How to show that the recurrence coefficients in (1) can be written as $$a²_n=\frac{}{},\; b_n= \frac{}{}$$

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