Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 19C / b / Solution

Inductively, $p_n$ is monic of degree $n$, because the leading term of $p_{n+1}$ comes only from $x{p}_n$. Assume $p_0,\dots ,p_n$ are mutually orthogonal. The definitions giveUsing the preceding recurrence in the formgivesso $⟨p_{n+1},p_{n-1}⟩=0$ by the definition of ${\beta }_n$. If $j\le n-2$, then $x{p}_j$ lies in the span of $p_{j-1},p_j,p_{j+1}$, all orthogonal to $p_n$, and the remaining recurrence terms are also orthogonal to $p_j$. Hence $⟨p_{n+1},p_j⟩=0$. Induction proves that the recurrence defines monic orthogonal polynomials.