Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 19C (Numerical Analysis)

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.

Inductively, $p_n$ is monic of degree $n$, because the leading term of $p_{n+1}$ comes only from $x{p}_n$. Assume $p_0,\dots ,p_n$ are mutually orthogonal. The definitions giveUsing the preceding recurrence in the formgivesso $⟨p_{n+1},p_{n-1}⟩=0$ by the definition of ${\beta }_n$. If $j\le n-2$, then $x{p}_j$ lies in the span of $p_{j-1},p_j,p_{j+1}$, all orthogonal to $p_n$, and the remaining recurrence terms are also orthogonal to $p_j$. Hence $⟨p_{n+1},p_j⟩=0$. Induction proves that the recurrence defines monic orthogonal polynomials.

The probabilists' Hermite polynomials satisfywhich follow by differentiating the Rodrigues formula. Thusso ${\alpha }_n=0$ and ${\beta }_n=n$.

For the squared norm, substitute the Rodrigues formula and integrate by parts $n$ times. All boundary terms vanish under the Gaussian weight, and ${\mathrm{He}}_n^{(n)}=n!$, giving

Let $x_1,\dots ,x_N$ be the distinct real roots of ${\mathrm{He}}_N$. The Gauss-Hermite quadrature rule iswhere $w_j$ is the weighted integral of the $j$th Lagrange interpolation polynomial. Equivalently,The orthogonality of ${\mathrm{He}}_N$ makes the rule exact for every polynomial of degree at most