Gaussian Quadrature

The trapezoid and Simpson rules use equally spaced points and choose only the weights. Gauss had a better idea: choose the points too. With $n$ points there are $2n$ numbers to choose, so a rule can be made exact for all polynomials of degree up to $2n - 1$, which is what interpolation at $n$ fixed points could never do.

Key idea

Two-point Gauss rule: $\int_{-1}^{1}f(x)\,dx \approx f\left(-\frac{1}{\sqrt{3}}\right) + f\left(\frac{1}{\sqrt{3}}\right)$, exact for every cubic. In general the $n$-point Gauss–Legendre rule is exact up to degree $2n - 1$ (three points: nodes $0, \pm\sqrt{3/5}$, weights $\frac{8}{9}, \frac{5}{9}, \frac{5}{9}$). On $[a, b]$, substitute $x = \frac{a + b}{2} + \frac{b - a}{2}t$ and multiply by $\frac{b - a}{2}$.

Why it works

Ask for $w_1f(x_1) + w_2f(x_2)$ to equal $\int_{-1}^{1}f$ for $f = 1, x, x^{2}, x^{3}$: $$w_1 + w_2 = 2,\quad w_1x_1 + w_2x_2 = 0,\quad w_1x_1^{2} + w_2x_2^{2} = \tfrac{2}{3},\quad w_1x_1^{3} + w_2x_2^{3} = 0.$$ By symmetry try $x_2 = -x_1$, $w_1 = w_2 = 1$: the odd equations hold automatically, and $2x_1^{2} = \frac{2}{3}$ gives $x_1 = \frac{1}{\sqrt{3}}$. For the interval $[a, b]$, the substitution maps $[-1, 1]$ onto $[a, b]$ and $dx = \frac{b - a}{2}\,dt$, which is where the extra factor comes from.

Worked example: Exact for a cubic

Apply the 2-point rule to $\int_{-1}^{1}(x^{3} + x^{2} + 1)\,dx$.

At $\pm\frac{1}{\sqrt{3}}$ the $x^{3}$ terms cancel, and $x^{2} = \frac{1}{3}$ at both.
Sum: $2\left(\frac{1}{3} + 1\right) = \frac{8}{3}$.
Exact: $0 + \frac{2}{3} + 2 = \frac{8}{3}$. ✓

Answer: $\frac{8}{3}$

Worked example: On another interval

Estimate $\int_0^1 e^{x}\,dx$ with the 2-point rule.

Nodes: $\frac{1}{2} \pm \frac{1}{2\sqrt{3}} \approx 0.2113249$ and $0.7886751$.
Weights: $\frac{b - a}{2} = \frac{1}{2}$ each.
$\frac{1}{2}\left(e^{0.2113249} + e^{0.7886751}\right) \approx 1.717896$. The exact value is $1.718282$: error $3.9 \times 10^{-4}$.
Simpson's rule with 3 points has error $5.8 \times 10^{-4}$: two well-placed points beat three equally spaced ones.

Answer: $\approx 1.717896$

Worked example: One degree too many

What does the 2-point rule give for $\int_{-1}^{1}x^{4}\,dx$?

$\left(\frac{1}{\sqrt{3}}\right)^{4} \times 2 = \frac{2}{9}$, but the integral is $\frac{2}{5}$.
Error $\frac{2}{9} - \frac{2}{5} = -\frac{8}{45}$: degree 4 is beyond $2n - 1 = 3$.

Answer: $\frac{2}{9}$, error $-\frac{8}{45}$

Where do the nodes come from in general? They are the zeros of the Legendre polynomial $P_n$, the polynomials that are orthogonal on $[-1, 1]$; that orthogonality is exactly what makes the rule exact up to degree $2n - 1$. The weights are all positive, which keeps the rule stable. The price: the nodes for $n$ and $n + 1$ points share nothing (apart from $0$), so increasing $n$ throws away all earlier function values. Practical codes therefore use a fixed $n$ on many panels (composite Gauss), or pairs of rules built to share nodes (Gauss–Kronrod), which also give an error estimate.

The two Gauss nodes for [0, 1], at 0.211 and 0.789: not at the ends, and not at the middle.

Gauss–Legendre rules. The nodes are the zeros of the Legendre polynomials.
points $n$	nodes on $[-1, 1]$	weights	exact up to degree
1	$0$	$2$	1
2	$\pm\frac{1}{\sqrt{3}}$	$1, 1$	3
3	$0, \pm\sqrt{\frac{3}{5}}$	$\frac{8}{9}, \frac{5}{9}, \frac{5}{9}$	5

Common mistake

Moving the nodes to $[a, b]$ but forgetting to multiply by $\frac{b - a}{2}$. Check with $f = 1$: the rule must give $b - a$, the length of the interval. Without the factor it gives $2$ (the length of $[-1, 1]$) for every interval.

Gaussian Quadrature

Practice

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Sources and further reading