= Solution
Let $Q=Q_{n+1}$. By <polynomial division>, every $q\in\mathcal P_{n+1+k}$ has a unique decomposition
$$
q=Qs+r,
\qquad s\in\mathcal P_k,\qquad r\in\mathcal P_n.
$$
Every node is a zero of $Q$, so $I_n(Qs)=0$. The assumed degree-$n$ exactness also gives $I_n(r)=I(r)$. Hence
$$
I_n(q)=I(r),
\qquad
I(q)=I(Qs)+I(r).
$$
Therefore $I_n(q)=I(q)$ for every $q\in\mathcal P_{n+1+k}$ if and only if
$$
I(Qs)=\langle Q,s\rangle=0
$$
for every $s\in\mathcal P_k$. This proves the <nodal-polynomial criterion for quadrature exactness>.
Back to article page
