Past exam of the approximation theory and finite element methods course of the University of Oxford / 2008 / A / 2

Use the standard convention that the orthogonal polynomials are nonzero. Their exact degrees then follow from the stated assumptions by induction: ${\varphi }_0$ is a nonzero constant, and if ${\varphi }_0,\dots ,{\varphi }_{r-1}$ span $P_{r-1}$, a nonzero ${\varphi }_r\in P_r$ orthogonal to all of them cannot lie in $P_{r-1}$.

Let $t_1,\dots ,t_m$ be the distinct sign-changing roots of ${\varphi }_r$ in $(0,1)$, and put $q(x)={\prod }_{j=1}^m(x-t_j)$. Then ${\varphi }_{r}q$ has constant sign between and across all its roots, and is not identically zero. Because $w>0$ in the interior, ${\int }_0^{1}w{\varphi }_{r}q$ is nonzero. If $m<r$, orthogonality to $P_{r-1}$ would make this integral zero. Thus $m\ge r$.

A degree-$r$ polynomial cannot have more than $r$ roots counted with multiplicity, so all $r$ roots are real, simple, and strictly inside $(0,1)$. This is the standard zeros of orthogonal polynomials argument, with the positive weight ensuring strict noncancellation.

The displayed recurrence requires monic normalization, which is not explicitly stated in the preamble. Arbitrary independent rescaling of orthogonal polynomials preserves orthogonality but changes the recurrence coefficients. Choose their monic representatives for this and the following calculations.

Expand $x{\varphi }_n$ in the orthogonal polynomial basis:The leading coefficient of ${\varphi }_{n+1}$ is one. For $j\le n-2$, $(x{\varphi }_n,{\varphi }_j)=({\varphi }_n,x{\varphi }_j)=0$, because the second factor has degree below $n$. Thus only $j=n,n-1$ remain. Inner products giveThe last equality uses $x{\varphi }_{n-1}-{\varphi }_n\in P_{n-1}$. Rearranging yields exactly the three-term recurrence for monic orthogonal polynomials

For weight $x^2$, the moments are ${\int }_0^1{x}^{j+2}dx=1/(j+3)$. Monic orthogonalization gives ${\varphi }_0=1$ and ${\varphi }_1=x-(1/4)/(1/3)=x-3/4$.

For ${\varphi }_2=x^2+ax+b$, orthogonality to $1,x$ requiresSolving gives $a=-4/3$, $b=2/5$. HenceThese are the nodes for the two-node Gaussian quadrature with quadratic weight.

Exactness on constants and linear functions givesWith the nodes just computed, the solution isTo prove cubic exactness rather than merely moment matching, divide a polynomial $u$ of degree at most three by the monic ${\varphi }_2$:The weighted integral of $q_1{\varphi }_2$ vanishes by orthogonality, and its two nodal values vanish because the nodes are roots of ${\varphi }_2$. The remaining $r_1$ is integrated exactly by the two equations above. This proves Gaussian quadrature exactness through degree three.