Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 19F / Solution

Use the weighted inner productHere the interval is nondegenerate and the indicated integrals are finite, as required for the given orthogonal polynomials. Strict positivity follows because $w>0$ and a nonzero polynomial cannot vanish on an interval. Since each $Q_j$ is monic of degree $j$, $Q_0,\dots ,Q_{n+1}$ form a basis of the polynomials of degree at most $n+1$. Comparing leading coefficients, expandBy orthogonality, $c_j=⟨x{Q}_n,Q_j⟩/h_j$. For $j\le n-2$, symmetry of multiplication by $x$ givesbecause $x{Q}_j$ has degree at most $n-1$ and is a linear combination of $Q_0,\dots ,Q_{n-1}$. Only the last two terms remain. Their coefficients areFor the second equality, $x{Q}_{n-1}-Q_n$ has degree at most $n-1$, so its inner product with $Q_n$ vanishes. Rearranging proves the three-term recurrence for monic orthogonal polynomials:For $n=0$ this reads $Q_1=x-a_0$ because $Q_{-1}=0$ and $Q_0=1$. The coefficient $b_0$ is therefore immaterial and is not determined by the polynomials. The usual convention is $b_0=0$; if a positive $b_0$ is desired, any positive value gives the identical recurrence. The strictly positive norm-ratio formula is the assertion for $n\ge 1$, not a division by the norm of $Q_{-1}$.