Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 17B (Numerical Analysis Rb)

Let $Q=Q_{n+1}$. By polynomial division, every $q\in {\mathcal{P}}_{n+1+k}$ has a unique decompositionEvery node is a zero of $Q$, so $I_n(Qs)=0$. The assumed degree-$n$ exactness also gives $I_n(r)=I(r)$. HenceTherefore $I_n(q)=I(q)$ for every $q\in {\mathcal{P}}_{n+1+k}$ if and only iffor every $s\in {\mathcal{P}}_k$. This proves the nodal-polynomial criterion for quadrature exactness.

Suppose that an $(n+1)$-node quadrature rule were exact on ${\mathcal{P}}_{2n+2}$. The nodal polynomial $Q_{n+1}$ has degree $n+1$, so $Q_{n+1}^2\in {\mathcal{P}}_{2n+2}$. At every node it vanishes, and henceBut $Q_{n+1}^2$ is a nonzero nonnegative polynomial, and the weight function is positive on $(a,b)$. Thereforea contradiction. This is the degree ceiling for quadrature exactness.

For each node $x_i^{(n)}$, letbe its Lagrange interpolation polynomial. It has degree $n$ and satisfies ${\ell }_i(x_j^{(n)})={\delta }_{ij}$. Since ${\ell }_i^2\in {\mathcal{P}}_{2n}$, exactness givesThe integrand is nonnegative and is positive except at finitely many points, soThis proves the positivity of quadrature weights from degree 2n exactness.

Fix a continuous function $f$ and $\epsilon >0$. By the Weierstrass approximation theorem, choose a polynomial $p\in {\mathcal{P}}_m$ such thatFor every $n\ge m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial giveConsequentlySince $\epsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the convergence of positive quadrature rules.