Composite quadrature

A single-panel rule approximates a whole interval by one low-degree polynomial. A composite rule partitions the interval and applies that same low-degree construction locally. Accuracy improves by refining the mesh instead of raising a global interpolation degree.

Use the convention in lecture note Section 5.6: $n$ is the number of small intervals and there are $n+1$ grid nodes:

Errors are always $I-Q$. We derive global constants from the single-panel theorem, keeping the grid width distinct from the Simpson panel width [1][1] S. Rojas, “Lecture Notes on Computational Mathematics,” 2025. Course lecture note distributed with MTH2051; local source course-lecture-notes.pdf., [2][2] M. U. School of Mathematics, “MTH2051: Introduction to Computational Mathematics: Course Lecture Notes and Study Workbooks,” 2026. Monash University course materials and study workbooks, Semester 2, 2026., [3][3] R. L. Burden and J. D. Faires, Numerical Analysis, 9th ed. Brooks/Cole, Cengage Learning, 2011..

Three composite rules

From local to global error

Trapezoidal and midpoint rules have global order two, whereas Simpson has global order four. A local $O(h^3)$ error is summed over order $1/h$ panels and becomes $O(h^2)$. Local and global orders cannot be interchanged. These orders describe bounds; a vanishing leading coefficient can give a higher observed order or exactness for a particular function.

Let

Replacing the remainder derivative by $B_r$ gives explicit bounds. A sufficient Simpson condition for tolerance $\mathrm{tol}>0$ is

Choose the smallest positive even integer at least $((b-a)^5B_4/(180\mathrm{tol}))^{1/4}$. This requires an actual derivative bound; agreement between two computed meshes alone does not provide the same guarantee.

Convexity, bounds and cost

When $f''\geq0$, the error signs give $M_n\leq I\leq T_n$. Convexity gives a direct proof too: on each panel the graph lies below its endpoint chord, so its integral lies below the trapezoid. The average of symmetric function values about the midpoint is at least the midpoint value; integrate that inequality to obtain the midpoint lower bound.

A direct evaluation requires $n+1$ function calls for trapezoidal, $n$ for midpoint, and $n+1$ for Simpson. Simpson requires an even $n$. Comparing equal mesh counts and equal evaluation costs are distinct comparisons. Error constants involve derivatives of different orders, so there is no universal ranking for all integrands.

These are algebraic identities for arbitrary sampled values and need no smoothness. Their fourth-order error interpretation still requires the earlier hypotheses. The last identity is also the first Richardson extrapolation of trapezoidal values.

Observed order and failed smoothness

If the error has a nonzero leading term $I-Q(h)=Ch^p+o(h^p)$, then

Divide numerator and denominator by $h^p$ and take limits. The nonzero $C$ ensures a nonzero denominator eventually. A bound $O(h^p)$ alone does not imply that the ratio tends to $2^p$.

For $f(x)=\sqrt{x}$, endpoint derivatives are unbounded, so neither the usual second-order nor fourth-order theorem applies. Avoiding endpoint samples does not restore smoothness. In fact its composite midpoint absolute error is $\Theta(h^{3/2})$: the first panel error is $(2/3-1/\sqrt2)h^{3/2}$, and concavity makes all panel errors negative. For panel $i\geq1$, the local second-derivative bound gives magnitude at most $h^{3/2}/(96i^{3/2})$. Summing the convergent series gives an upper bound of the same order, while the nonzero first panel gives a lower bound.

Experiment and executable example

Begin with the exponential to inspect error ratios, then switch to the square root to see smoothness fail. The narrow-peak example shows how small differences on coarse meshes can simply mean that the important region has not been sampled.

This implementation uses the same weights as the experiment, with $n$ counting small intervals:

def composite_simpson(f, a, b, n):
    if not isinstance(n, int) or not a < b or n < 2 or n % 2:
        raise ValueError("require a < b and positive even n")
    h = (b - a) / n
    total = f(a) + f(b)
    for i in range(1, n):
        total += (4 if i % 2 else 2) * f(a + i * h)
    return h * total / 3

For $x^4$ on $[0,1]$, the error is exactly $-2h^4/15$, since the fourth derivative is constantly $24$. This example simultaneously checks weights, error sign, and the definition of mesh width.