Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 2F (Topics in Analysis)

For each node define its Lagrange interpolation cardinal polynomial ${\ell }_i(t)={\prod }_{j\ne i}(t-x_j)/(x_i-x_j)$. Its square has degree $2n-2<2n$ and nodal values ${\ell }_i(x_j{)}^2={\delta }_{ij}$. Exactness therefore gives $a_i={\int }_{-1}^1{\ell }_i(t{)}^{2}dt$. The polynomial is nonzero, so continuity makes this integral strictly positive. Thus $\boxed{a_i>0}$ for every node.

Apply exactness to the constant polynomial one. Its integral is two and its value at every node is one, so $\boxed{\sum _{i=1}^n{a}_i=2}$.

Put $P(t)={\prod }_i(t-x_i)$. For every polynomial $q$ of degree at most $n-1$, $Pq$ has degree at most $2n-1$ and vanishes at all nodes; thus ${\int }_{-1}^{1}Pq=0$. The monic degree-$n$ orthogonal polynomial for this integral is unique: the difference of two such monic polynomials has degree below $n$ and is orthogonal to itself, hence is zero. Consequently $P$ is the monic normalization of the degree-$n$ Legendre polynomial, proving the given nodes are exactly its roots. Together the three results characterize this Gaussian quadrature rule.