Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 6D / b

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.