Gaussian quadrature — Examples

Since $\mathrm{deg}({\mathrm{\ell }}_i^2)=2$, exactness [F2] gives $Q_2({\mathrm{\ell }}_i^2)=\int {\mathrm{\ell }}_i^{2}d\mu$. Expanding the finite rule sum [F1] and using the cardinal values [F3] gives $Q_2({\mathrm{\ell }}_i^2)=w_i$, hence $\int {\mathrm{\ell }}_i^{2}d\mu =w_i$.

The formula for ${\mathrm{\ell }}_1$ gives ${\mathrm{\ell }}_1(x{)}^2=(x^2-2x_{2}x+x_2^2)\mathrm{/}(x_1-x_2{)}^2$. Integrating term by term using [F4] proves the displayed expression for $w_1$. Replacing the omitted node $x_2$ by $x_1$ gives the displayed expression for $w_2$. Their strict positivity is [F2].

Since each coordinate has two distinct nodes by [F5], the sixteen coordinate tuples are distinct: two different tuples differ in a coordinate whose node entries are distinct.

The node-count statement of [F3] gives $2\cdot 2\cdot 2\cdot 2=16$. For each coordinate degree ${\alpha }_j\le 3$, [F4] evaluates the corresponding one-dimensional sum as its moment. Distributing the finite sums in [F1] therefore gives the product of those four moments, which is [F2]. Taking the specified exponents proves the displayed calculation and the general coordinatewise exactness claim.

By [F10], every relative Borel set is $B\cap [-1,1]$ for ambient Borel $B$. The interval is closed, hence ambient Borel, so this intersection is ambient Borel and is Lebesgue measurable by [F1]. Restricting Lebesgue measure to this sigma-algebra therefore defines the asserted positive measure; disjoint relative Borel unions are the same ambient measurable unions, so countable additivity is inherited. For $x\in [-1,1]$, [F9] gives an $\epsilon >0$ such that $(x-\epsilon ,x+\epsilon )\cap [-1,1]$ lies in any chosen relative open neighborhood $U$ of $x$. Choose a closed interval $I$ of positive length inside that intersection, using a one-sided interval when $x$ is an endpoint. Then $I$ is relative Borel and [F2] gives $\mu (I)>0$. Since $U=I\bigsqcup (U\setminus I)$ and both sets are relative Borel, countable additivity and nonnegativity from [F1] imply $\mu (U)\ge \mu (I)>0$. Thus every $x$ lies in the relative support, which has at least three points.

The functions $1,x,x^2,x^3$ are continuous, hence Riemann integrable by [F3]. Their primitives $x,x^2\mathrm{/}2,x^3\mathrm{/}3,x^4\mathrm{/}4$, evaluated at $-1$ and $1$, give respectively $2,0,2\mathrm{/}3,0$. The Riemann–Lebesgue equality in [F3] gives the stated moment values.

The monic polynomials $1$ and $X$ are orthogonal in the required degrees: $\int xd\mu =0$. The monic polynomial $X^2-1\mathrm{/}3$ is orthogonal to $1$ because ${\mu }_2-(1\mathrm{/}3){\mu }_0=0$, and to $X$ because ${\mu }_3-(1\mathrm{/}3){\mu }_1=0$. Uniqueness [F4] identifies $p_0,p_1,p_2$ as stated. By [F11], $\sqrt{3}>0$ exists uniquely. The identity $X^2-1\mathrm{/}3=(X-1\mathrm{/}\sqrt{3})(X+1\mathrm{/}\sqrt{3})$ gives exactly the two roots $\pm 1\mathrm{/}\sqrt{3}$.

Here $h_0={\mu }_0=2$, $h_1=\int x^{2}d\mu =2\mathrm{/}3$, and both recurrence diagonal entries are zero by the odd moments. Hence ${\beta }_1=h_1\mathrm{/}{h}_0=1\mathrm{/}3$, and [F5] gives the displayed $J_2$. Direct multiplication shows its two displayed unit eigenvectors have eigenvalues $-1\mathrm{/}\sqrt{3}$ and $1\mathrm{/}\sqrt{3}$; [F6] identifies those as the Gaussian nodes.

For either unit eigenvector, the first-coordinate square is $1\mathrm{/}2$. Applying [F7] with ${\mu }_0=2$ gives each weight as $1$. Exactness through degree three is [F8], proving the stated rule.