Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 17B (Numerical Analysis)

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.

Expansion of $\det (xI-A_n)$ along the last row giveswith $D_0=1$, $D_1=x-{\alpha }_1$. This is exactly the recurrence defining $P_n$, so induction gives $P_n=\det (xI-A_n)$.

The three-term recurrence theorem with the displayed coefficients identifies $P_n$ with the monic orthogonal polynomial $p_n$. Part (b) says its zeros are precisely the eigenvalues of the real symmetric Jacobi matrix $A_n$, and part (a) says those zeros are distinct and all belong to $(-1,1)$.