43.09.01 · numerical-analysis / 09-numerical-quadrature

Newton-Cotes rules and their error via the Peano kernel

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Anchor (Master): Davis-Rabinowitz 1984 *Methods of Numerical Integration* 2e (Academic Press) Ch. 2-4 (interpolatory and Newton-Cotes rules, the Peano kernel theory, the negative-weight instability of high-order rules); Brass-Petras 2011 *Quadrature Theory* (AMS) Ch. 1-2 (the error functional, Peano kernels, definiteness and best constants); Krylov 1962 *Approximate Calculation of Integrals* (Macmillan) Ch. 6-7 (Newton-Cotes coefficients and convergence)

Intuition Beginner

You often need the area under a curve between two points, but the function may be one you cannot integrate by hand, or you may only know its values at a few inputs. The oldest workaround is to replace the awkward curve by a simple one — a polynomial that agrees with it at a handful of sample points — and integrate the simple one instead. Because a polynomial's area is a quick sum, you get a fast estimate of the true area. This unit is about that idea, the standard recipes it produces, and how to say precisely how wrong each recipe is.

The recipes all share one shape. You pick a few sample inputs, read off the function's value at each, multiply each value by a fixed number called its weight, and add up the products. The weights depend only on where the sample points sit, not on the function, so you compute them once and reuse them forever. A rule with two endpoints gives the trapezium rule, which replaces the curve by a straight line and measures the area of the resulting trapezoid. A rule with two endpoints and the midpoint gives Simpson's rule, which fits a parabola instead.

A pleasant accident makes some of these recipes better than they look. A rule built to be exact for straight lines might, by a lucky symmetry, also be exact for cubics. Simpson's rule is the famous case: it is designed around a parabola yet integrates any cubic with no error at all. The largest degree of polynomial a rule integrates exactly is its badge of quality, and chasing a higher badge is what separates a crude rule from a sharp one.

The catch comes when you get greedy. Adding more evenly spaced sample points should buy more accuracy, and for a while it does. But past a certain point the weights start turning negative and swinging to large sizes, so small errors in the function values get amplified instead of averaged out. That is why the simple high-point recipes are abandoned in favour of either chopping the interval into many small pieces or placing the points more cleverly — the subjects of the next two units.

Visual Beginner

Picture the curve over an interval and the simple shape that stands in for it. For the trapezium rule the stand-in is the straight line joining the two endpoint values, and the estimate is the area of the trapezoid under that line. For Simpson's rule the stand-in is the parabola through the two endpoints and the midpoint, and the estimate is the area under that parabola. The table below lists the three basic rules, the simple shape each one integrates, and the badge — the highest polynomial degree the rule gets exactly right.

rule	sample points	stand-in shape	highest degree exact
midpoint	midpoint only	flat bar	$1$
trapezium	two endpoints	straight line	$1$
Simpson	two endpoints and midpoint	parabola	$3$

Each row reads the same way: the stand-in shape is a low-degree polynomial, but the badge in the last column can be higher than the shape's own degree. The midpoint and Simpson rules each win a bonus degree from the left-right symmetry of their sample points, which is why Simpson, built on a parabola, is exact for every cubic.

Worked example Beginner

Let us estimate the area under $f (x) = x^{3}$ from $0$ to $2$ with Simpson's rule, then compare with the exact answer.

Step 1. Find the exact area. The area under $x^{3}$ from $0$ to $2$ is the value of $x^{4} \div 4$ at the endpoints: $2^{4} \div 4 - 0 = 16 \div 4 = 4$ . So the true area is $4$ .

Step 2. Set up Simpson's rule. It samples the two endpoints $0$ and $2$ and the midpoint $1$ . The rule is: take the width of the interval, divide by $6$ , and multiply by the combination "value at the left end, plus four times the value at the midpoint, plus value at the right end". The width is $2$ , so the prefactor is $2 \div 6 = 1 \div 3$ .

Step 3. Read off the three values. At $x = 0$ , $f = 0$ . At $x = 1$ , $f = 1$ . At $x = 2$ , $f = 8$ . The combination is $0 + 4 \times 1 + 8 = 12$ .

Step 4. Assemble the estimate. Multiply the prefactor by the combination: $(1 \div 3) \times 12 = 4$ . So Simpson's rule gives exactly $4$ .

Step 5. Compare. The estimate $4$ equals the true area $4$ — no error at all, even though $x^{3}$ is not a parabola.

What this tells us: Simpson's rule integrates the cube perfectly because its sample points are symmetric about the midpoint, and that symmetry makes the rule exact for every cubic. The bonus degree is not luck you hope for case by case; it is built into the rule.

Check your understanding Beginner

Formal definition Intermediate+

Throughout, $f : [a, b] \to R$ , and $x_{0} < x_{1} < \dots < x_{n}$ are $n + 1$ distinct nodes in $[a, b]$ ; $Π_{k}$ is the space of real polynomials of degree at most $k$ , $L_{i}$ the Lagrange cardinal polynomials on the nodes from 43.08.01, and $π_{n + 1} (x) = \prod_{i = 0}^{n} (x - x_{i})$ the node polynomial of 43.08.02. The notation $f \in C^{m} [a, b]$ means $f$ has continuous derivatives through order $m$ . For a real number $u$ , the truncated power is $u_{+} = max (u, 0)$ , and $u_{+}^{k} = (u_{+})^{k}$ .

Definition (interpolatory quadrature rule). A quadrature rule on the nodes $x_{0}, \dots, x_{n}$ is a functional $Q (f) = \sum_{i = 0}^{n} w_{i} f (x_{i})$ approximating $\int_{a}^{b} f (x) d x$ . The rule is interpolatory if its weights are the integrals of the cardinal polynomials, $$ w_i = \int_a^b L_i(x),dx, \qquad i = 0, \dots, n, $$ equivalently if $Q (f) = \int_{a}^{b} p_{n} (x) d x$ where $p_{n} \in Π_{n}$ is the interpolant of $f$ at the nodes. The error functional is $E (f) = \int_{a}^{b} f (x) d x - Q (f)$ , a linear functional in $f$ .

Definition (degree of exactness). The degree of exactness (or degree of precision) of a rule $Q$ is the largest integer $d$ such that $E (p) = 0$ for every $p \in Π_{d}$ , while $E (q) \neq = 0$ for some $q \in Π_{d + 1}$ . An interpolatory rule on $n + 1$ nodes has $d \geq n$ , since it integrates the interpolant exactly and the interpolant of any $p \in Π_{n}$ is $p$ itself.

Definition (closed Newton-Cotes rules). The closed Newton-Cotes rule of order $n$ uses the $n + 1$ equispaced nodes $x_{i} = a + ih$ , $h = (b - a) / n$ , including both endpoints. The first three are $$ \text{(trapezium, } n=1\text{)}\quad Q(f) = \tfrac{h}{2}\big(f_0 + f_1\big), \qquad \text{(Simpson, } n=2\text{)}\quad Q(f) = \tfrac{h}{3}\big(f_0 + 4f_1 + f_2\big), $$ $$ \text{(Simpson's } 3/8, \ n=3\text{)}\quad Q(f) = \tfrac{3h}{8}\big(f_0 + 3f_1 + 3f_2 + f_3\big), $$ with $f_{i} = f (x_{i})$ . The open midpoint rule ( $n = 0$ interior node) is $Q (f) = (b - a) f (\frac{a + b}{2})$ . Their degrees of exactness are $1$ , $3$ , $3$ , and $1$ respectively: the trapezium and Simpson's $3/8$ achieve $d = n$ , while the midpoint and Simpson rules gain one extra degree from the symmetry of their nodes about the interval midpoint.

Definition (Peano kernel). Let $E$ be a linear functional that annihilates $Π_{r - 1}$ , i.e. $E (p) = 0$ for all $p \in Π_{r - 1}$ , and that is continuous on $C^{r} [a, b]$ . Its Peano kernel of order $r$ is $$ K_r(t) = \frac{1}{(r-1)!},E_x\big[(x - t)+^{,r-1}\big], $$ the functional applied to the truncated power $x \mapsto (x-t)+^{r-1} $w i t h$ t $h e l df i x e d . T h esy mb o l s$ \int $,$ \prod $,$ \sum $,$ \partial $, t h e t r u n c a t e d p o w er$ u_+^k $, an d t h e n o d e p o l y n o mia l$ \pi_{n+1}$ are recorded in _meta/NOTATION.md.

Counterexamples to common slips Intermediate+

"The degree of exactness of a rule on $n + 1$ nodes is exactly $n$ ." It is at least $n$ , and is strictly larger when the nodes are symmetric and $n$ is even: Simpson's rule on $3$ nodes has degree $3$ , not $2$ . The error functional kills the next monomial by an odd-symmetry argument, not by the interpolation count alone.
"A nonnegative integrand forces a nonnegative quadrature estimate." Only when all weights are nonnegative. High-order Newton-Cotes rules have negative weights, so $Q (f)$ can be negative for a strictly positive $f$ , and the rule amplifies data error rather than averaging it.
"The Peano kernel needs $f$ to be a polynomial." The kernel representation $E (f) = \int K_{r} (t) f^{(r)} (t) d t$ holds for every $f \in C^{r} [a, b]$ ; the polynomial-annihilation hypothesis is on the functional $E$ , supplying the exactness that makes the kernel well-defined, not on the integrand.
"A one-signed Peano kernel is automatic once $E$ annihilates $Π_{r - 1}$ ." Definiteness is an extra fact to be checked rule by rule. The trapezium and Simpson kernels happen to be one-signed, which is why their errors take the clean mean-value form; a general rule's kernel can change sign, and then only the integral bound $∣ E (f) ∣ \leq ∥ f^{(r)} ∥_{\infty} \int ∣ K_{r} ∣$ survives.

Key theorem with proof Intermediate+

The signature result is the Peano kernel theorem: a linear functional that integrates polynomials up to degree $r - 1$ exactly has its error written as a single integral of the $r$ -th derivative of $f$ against a fixed kernel determined by the functional alone. This separates the error into a function-dependent part — the derivative $f^{(r)}$ — and a rule-dependent part — the kernel $K_{r}$ — exactly as the interpolation error of 43.08.02 split into a derivative and the node polynomial. When the kernel keeps one sign, the mean value theorem collapses the integral to a single derivative evaluation, recovering the textbook error terms. This follows Süli-Mayers ^{[Süli-Mayers §7.4]}.

Theorem (Peano kernel representation). Let $E$ be a linear functional on $C^{r} [a, b]$ , continuous in the sense that it commutes with the integral remainder below, and suppose $E (p) = 0$ for all $p \in Π_{r - 1}$ . Then for every $f \in C^{r} [a, b]$ , $$ E(f) = \int_a^b K_r(t), f^{(r)}(t),dt, \qquad K_r(t) = \frac{1}{(r-1)!},E_x\big[(x-t)_+^{,r-1}\big]. $$ If moreover $K_{r}$ does not change sign on $[a, b]$ , there exists $η \in (a, b)$ with $$ E(f) = f^{(r)}(\eta)\int_a^b K_r(t),dt = \frac{E(x^r)}{r!},f^{(r)}(\eta). $$

Proof. Fix $f \in C^{r} [a, b]$ . Taylor's theorem with the integral form of the remainder 02.05.02, expanded about $a$ , gives $$ f(x) = \underbrace{\sum_{j=0}^{r-1} \frac{f^{(j)}(a)}{j!}(x-a)^j}{=:,p(x),\in,\Pi{r-1}} ;+; \frac{1}{(r-1)!}\int_a^x (x-t)^{r-1} f^{(r)}(t),dt . $$ Extend the inner integral to the full interval using the truncated power: since $(x - t)^{r - 1} = (x - t)_{+}^{r - 1}$ for $t \leq x$ and the truncated power vanishes for $t > x$ , the remainder equals $\frac{1}{( r - 1 )!} \int_{a}^{b} (x - t)_{+}^{r - 1} f^{(r)} (t) d t$ . Apply $E$ to both sides. By hypothesis $E (p) = 0$ because $p \in Π_{r - 1}$ , so $E$ annihilates the polynomial part and acts only on the remainder: $$ E(f) = E_x!\left[\frac{1}{(r-1)!}\int_a^b (x-t)+^{r-1} f^{(r)}(t),dt\right]. $$ Exchanging the order of $E_{x}$ and the integration over $t$ — legitimate because the remainder is jointly continuous and $E$ is built from integration and finite point-evaluations, both of which commute with the $t$ -integral by Fubini — moves $E_{x}$ inside: $$ E(f) = \int_a^b \frac{1}{(r-1)!},E_x\big[(x-t)+^{r-1}\big], f^{(r)}(t),dt = \int_a^b K_r(t),f^{(r)}(t),dt . $$ This is the representation. For the mean-value form, suppose $K_{r} (t) \geq 0$ throughout (the case $K_{r} \leq 0$ is identical with signs reversed). Then $K_{r}$ is a nonnegative weight against which $f^{(r)}$ is integrated, and since $f^{(r)}$ is continuous, the weighted mean value theorem for integrals supplies $η \in (a, b)$ with $\int_{a}^{b} K_{r} (t) f^{(r)} (t) d t = f^{(r)} (η) \int_{a}^{b} K_{r} (t) d t$ . Finally, applying the representation to the monomial $f (x) = x^{r}$ , whose $r$ -th derivative is the constant $r!$ , gives $E (x^{r}) = r! \int_{a}^{b} K_{r} (t) d t$ , so $\int_{a}^{b} K_{r} = E (x^{r}) / r!$ , and the mean-value form follows. $□$

Corollary (trapezium and Simpson error terms). On a single interval $[a, b]$ with $h = b - a$ , the trapezium rule has degree of exactness $1$ , Peano kernel $K_{2} (t) = \frac{1}{2} (t - a) (t - b) \leq 0$ of order $r = 2$ , and error $$ E(f) = -\frac{(b-a)^3}{12},f''(\eta), \qquad \eta \in (a,b). $$ Simpson's rule has degree of exactness $3$ , a one-signed Peano kernel of order $r = 4$ , and error $E (f) = - \frac{( b - a ) ^{5}}{2880} f^{(4)} (η)$ . The negative sign and the explicit constant both come straight from $\int K_{r}$ : the kernel's fixed sign is what licenses the single-point derivative evaluation.

Bridge. The Peano kernel theorem is the foundational reason quadrature error has the same architecture as interpolation error: both factor into a high derivative of $f$ , which the analyst cannot touch, and a fixed object determined by the rule — the node polynomial there, the kernel here. This is exactly the integrated form of the interpolation remainder, since integrating $f - p_{n} = f^{(n + 1)} (ξ) π_{n + 1} / (n + 1)!$ over $[a, b]$ produces an error functional annihilating $Π_{n}$ , whose Peano kernel repackages $π_{n + 1}$ into a one-signed weight. The result builds toward the composite and Gauss rules: it appears again in 43.09.02, where summing single-interval kernels across a partition gives the global $O (h^{2})$ and $O (h^{4})$ errors, and in 43.09.03, where maximising the degree of exactness past $n$ is the design principle of Gauss quadrature. The central insight is that exactness on $Π_{r - 1}$ is equivalent to the error depending only on $f^{(r)}$ , which generalises the single mean value theorem of 02.05.02 from one secant to an $r$ -th-order weighted average and is dual to the divided-difference reading of the interpolation remainder. Putting these together, the sign of the kernel is the whole question of whether a rule has a clean mean-value error or only an integral bound.

Exercises Intermediate+

Exercise 5 (medium, symbolic).

Compute the trapezium Peano kernel directly from the definition $K_{2} (t) = E_{x} [(x - t)_{+}] /1!$ on $[a, b]$ , and confirm it is $\leq 0$ .

Hint

$E (g) = \int_{a}^{b} g d x - \frac{h}{2} (g (a) + g (b))$ . Apply it to $g (x) = (x - t)_{+}$ for fixed $t \in (a, b)$ : the integral is $\int_{t}^{b} (x - t) d x = (b - t)^{2} /2$ , and the endpoint terms use $(a - t)_{+} = 0$ , $(b - t)_{+} = b - t$ .

Answer

For $t \in (a, b)$ , $\int_{a}^{b} (x - t)_{+} d x = \int_{t}^{b} (x - t) d x = \frac{( b - t ) ^{2}}{2}$ . The rule term is $\frac{h}{2} ((a - t)_{+} + (b - t)_{+}) = \frac{h}{2} (0 + (b - t)) = \frac{( b - a ) ( b - t )}{2}$ . So $K_{2} (t) = \frac{( b - t ) ^{2}}{2} - \frac{( b - a ) ( b - t )}{2} = \frac{( b - t )}{2} [(b - t) - (b - a)] = \frac{( b - t ) ( a - t )}{2} = \frac{1}{2} (t - a) (t - b)$ . On $(a, b)$ the factors $(t - a) > 0$ and $(t - b) < 0$ , so $K_{2} (t) \leq 0$ . Integrating, $\int_{a}^{b} K_{2} = - h^{3} /12$ , recovering $E (f) = - h^{3} f^{''} (η) /12$ . Rubric: full credit for the integral, the rule term, the factored kernel, and the sign.

Exercise 7 (hard, symbolic).

Prove that any quadrature rule with all weights $w_{i} > 0$ satisfies $\sum_{i} ∣ w_{i} ∣ = \sum_{i} w_{i} = b - a$ , and explain why a rule with a negative weight has $\sum_{i} ∣ w_{i} ∣ > b - a$ .

Hint

Every interpolatory rule is exact for the constant $f \equiv 1$ , which forces $\sum_{i} w_{i} = \int_{a}^{b} 1 d x = b - a$ . Compare $\sum ∣ w_{i} ∣$ with $\sum w_{i}$ when some weight is negative.

Answer

An interpolatory rule integrates $f \equiv 1 \in Π_{0}$ exactly, so $\sum_{i} w_{i} = \int_{a}^{b} 1 d x = b - a$ . If every $w_{i} > 0$ then $∣ w_{i} ∣ = w_{i}$ and $\sum_{i} ∣ w_{i} ∣ = \sum_{i} w_{i} = b - a$ . If some weight is negative, write $S_{+} = \sum_{w_{i} > 0} w_{i}$ and $S_{-} = \sum_{w_{i} < 0} ∣ w_{i} ∣ > 0$ ; then $\sum_{i} w_{i} = S_{+} - S_{-} = b - a$ , while $\sum_{i} ∣ w_{i} ∣ = S_{+} + S_{-} = (b - a) + 2 S_{-} > b - a$ . The quantity $\sum ∣ w_{i} ∣$ is the operator norm of $Q$ acting on data with sup-norm $1$ , so a negative weight strictly inflates the amplification of input error above the minimal value $b - a$ . Rubric: full credit for the constant-exactness step, the split into positive and negative parts, and the strict inequality.

Exercise 8 (hard, symbolic).

Show that the midpoint rule $Q (f) = (b - a) f (\frac{a + b}{2})$ has degree of exactness $1$ , with order- $2$ Peano kernel one-signed, and derive its error $E (f) = \frac{( b - a ) ^{3}}{24} f^{''} (η)$ .

Hint

Check $E (1) = E (x) = 0$ by symmetry but $E (x^{2}) \neq = 0$ . For the kernel, center at $m = (a + b) /2$ and compute $K_{2} (t) = E_{x} [(x - t)_{+}]$ . The sign of the constant is opposite to the trapezium rule.

Answer

Take $[a, b] = [- c, c]$ with $m = 0$ , $h = 2 c$ . Then $E (1) = 2 c - 2 c \cdot 1 = 0$ and $E (x) = 0 - 2 c \cdot 0 = 0$ by symmetry, so degree of exactness $\geq 1$ ; but $E (x^{2}) = \int_{- c}^{c} x^{2} - 2 c \cdot 0 = \frac{2 c ^{3}}{3} \neq = 0$ , pinning it at $1$ . The order- $2$ kernel is $K_{2} (t) = \int_{- c}^{c} (x - t)_{+} d x - 2 c (0 - t)_{+} = \frac{( c - t ) ^{2}}{2} - 2 c (- t)_{+}$ . For $t \geq 0$ this is $\frac{( c - t ) ^{2}}{2} \geq 0$ ; for $t < 0$ it is $\frac{( c - t ) ^{2}}{2} - 2 c (- t) = \frac{( c + t ) ^{2}}{2} \geq 0$ after simplification, so $K_{2} \geq 0$ throughout — one-signed, opposite to the trapezium. Then $\int K_{2} = \int_{- c}^{c} \frac{( c - ∣ t ∣ ) ^{2}}{2} d t = \frac{c ^{3}}{3} = \frac{( b - a ) ^{3}}{24}$ , giving $E (f) = f^{''} (η) (b - a)^{3} /24$ . Rubric: full credit for the exactness check, the sign of the kernel on both halves, and the constant $1/24$ .

Advanced results Master

The single-interval rules fix the local picture; the master layer turns the Peano representation into a precise theory of when a rule's error is sign-definite, how the negative-weight instability of high-order Newton-Cotes follows from the kernel sign structure, and why the entire equispaced family must be abandoned past moderate order.

Theorem 1 (Peano kernel, general order and best constant). Let $E$ be a continuous linear functional on $C^{r} [a, b]$ annihilating $Π_{r - 1}$ . Then $E (f) = \int_{a}^{b} K_{r} (t) f^{(r)} (t) d t$ with $K_{r} (t) = E_{x} [(x - t)_{+}^{r - 1}] / (r - 1)!$ , and the best possible bound is $∣ E (f) ∣ \leq (\int_{a}^{b} ∣ K_{r} (t) ∣ d t) ∥ f^{(r)} ∥_{\infty}$ , with equality approached by functions whose $r$ -th derivative matches $sign K_{r}$ . When $K_{r}$ is one-signed the bound sharpens to the mean-value identity $E (f) = f^{(r)} (η) \int_{a}^{b} K_{r}$ , and the error constant is $c_{r} = \int_{a}^{b} K_{r} = E (x^{r}) / r!$ ^{[Brass-Petras Ch. 2]}. The order $r$ at which one applies the theorem is the degree of exactness plus one; raising $r$ trades a higher derivative of $f$ for a smaller kernel, the quantitative form of the exactness-versus-smoothness exchange.

Theorem 2 (definiteness of the classical rules). The trapezium rule is negative definite of order $2$ : its kernel $K_{2} (t) = \frac{1}{2} (t - a) (t - b) \leq 0$ , error $- (b - a)^{3} f^{''} (η) /12$ . The midpoint rule is positive definite of order $2$ , error $+ (b - a)^{3} f^{''} (η) /24$ — half the magnitude and opposite sign, so the two bracket the true integral for convex $f$ and their weighted average $\frac{1}{3} (midpoint) + \frac{2}{3}$ rescaled gives Simpson. Simpson's rule is negative definite of order $4$ , error $- (b - a)^{5} f^{(4)} (η) /2880$ ^{[Davis-Rabinowitz Ch. 2]}. Definiteness fails for high-order Newton-Cotes: the kernel of the $n = 8$ closed rule changes sign, so no single-derivative mean-value form exists and only the integral bound $∣ E ∣ \leq ∥ f^{(r)} ∥_{\infty} \int ∣ K_{r} ∣$ remains, with $\int ∣ K_{r} ∣ ≫ ∣ \int K_{r} ∣$ .

Theorem 3 (negative weights and divergence of high-order Newton-Cotes). For the closed Newton-Cotes rule on $n + 1$ equispaced nodes, the weights $w_{i}^{(n)}$ are rational and, beginning at $n = 8$ , some are negative; the sum of absolute weights satisfies $\sum_{i} ∣ w_{i}^{(n)} ∣ \to \infty$ as $n \to \infty$ , growing like $2^{n} / (n lo g n)$ ^{[Davis-Rabinowitz Ch. 2]}. By the Pólya-Steklov theorem a sequence of quadrature rules converges $Q_{n} (f) \to \int_{a}^{b} f$ for every $f \in C [a, b]$ if and only if it is exact on a dense polynomial set and $sup_{n} \sum_{i} ∣ w_{i}^{(n)} ∣$ is finite; the unbounded absolute-weight sum makes high-order Newton-Cotes fail the second condition, so there exist continuous $f$ for which $Q_{n} (f)$ diverges. The mechanism mirrors the Runge phenomenon of 43.08.02 exactly: the same equispaced node polynomial whose sup norm swells near the endpoints produces the oscillating cardinal-polynomial integrals that are the negative weights.

Theorem 4 (the two repairs). Two structurally different repairs restore convergence. The composite strategy fixes the order at $1$ or $2$ and subdivides: applying the trapezium or Simpson rule on each of $N$ subintervals of width $h = (b - a) / N$ and summing gives error $O (h^{2})$ or $O (h^{4})$ with all weights positive and uniformly bounded, the subject of 43.09.02; the Euler-Maclaurin expansion there is the full asymptotic series whose leading term is the single-interval Peano constant. The Gauss strategy keeps a single interval but abandons equispacing, placing $n + 1$ nodes at the roots of the degree- $(n + 1)$ orthogonal polynomial to raise the degree of exactness from $n$ to $2 n + 1$ while keeping all weights positive, the subject of 43.09.03. Both repairs are dictated by the Peano picture: composite rules shrink the kernel by shrinking the interval; Gauss rules shrink the error by raising the order $r$ past what any interpolatory equispaced rule can reach.

Synthesis. The Peano kernel theorem is the foundational reason quadrature error is one theorem seen at different orders: the error of any rule exact on $Π_{r - 1}$ is a single integral of $f^{(r)}$ against a fixed kernel, so analysing a rule reduces to analysing the one function $K_{r}$ . The central insight is that the sign of the kernel decides everything — a one-signed kernel gives the mean-value error $f^{(r)} (η) \int K_{r}$ and positive weights, while a sign-changing kernel gives only an integral bound and the negative weights that destroy convergence; this is exactly the line between the usable low-order rules and the discarded high-order Newton-Cotes family. Putting these together, the negative-weight instability is dual to the Runge phenomenon of 43.08.02: both are the endpoint swelling of the equispaced node polynomial, read once as interpolation divergence and once as quadrature divergence, with the same cure of clustering nodes or chopping the interval. The bridge is that raising exactness past the interpolation count $n$ demands either many small intervals (composite, 43.09.02) or non-equispaced nodes (Gauss, 43.09.03), and the Peano constant $\int K_{r}$ is the quantity each repair drives to zero. This generalises the single mean value theorem of 02.05.02 into an $r$ -th-order weighted-mean statement, and is the integrated shadow of the interpolation remainder of 43.08.02, one node polynomial governing both how well $p_{n}$ approximates $f$ and how well $\int p_{n}$ approximates $\int f$ .

Full proof set Master

Proposition 1 (interpolatory rules have degree of exactness $\geq n$ ). An interpolatory quadrature rule on $n + 1$ distinct nodes integrates every $p \in Π_{n}$ exactly.

Proof. Let $p \in Π_{n}$ . Its interpolant at the $n + 1$ nodes is $p$ itself, by the uniqueness half of the unisolvence theorem of 43.08.01, since $p \in Π_{n}$ agrees with itself at the nodes and is the unique such polynomial of degree $\leq n$ . The interpolatory rule is defined by $Q (p) = \int_{a}^{b} (interpolant of p) = \int_{a}^{b} p$ , so $E (p) = \int_{a}^{b} p - Q (p) = 0$ . Hence the degree of exactness is at least $n$ . $□$

Proposition 2 (one extra degree for symmetric rules with even $n$ ). If the nodes are symmetric about the interval midpoint $m = (a + b) /2$ and $n$ is even, the interpolatory rule has degree of exactness $\geq n + 1$ .

Proof. Translate so $m = 0$ , i.e. $[a, b] = [- c, c]$ with nodes symmetric: $x_{i} = - x_{n - i}$ . Symmetric nodes give symmetric cardinal polynomials $L_{i} (- x) = L_{n - i} (x)$ , hence symmetric weights $w_{i} = w_{n - i}$ . The error functional $E (f) = \int_{- c}^{c} f - \sum_{i} w_{i} f (x_{i})$ then annihilates every odd function: for odd $g$ , $\int_{- c}^{c} g = 0$ and $\sum_{i} w_{i} g (x_{i}) = \sum_{i} w_{n - i} g (- x_{n - i}) = - \sum_{i} w_{i} g (x_{i})$ , forcing $\sum_{i} w_{i} g (x_{i}) = 0$ . The monomial $x^{n + 1}$ is odd when $n$ is even, so $E (x^{n + 1}) = 0$ ; combined with exactness on $Π_{n}$ (Proposition 1), the degree of exactness is $\geq n + 1$ . For Simpson ( $n = 2$ ) this gives degree $3$ . $□$

Proposition 3 (Peano kernel representation). For $E$ continuous on $C^{r} [a, b]$ annihilating $Π_{r - 1}$ and $f \in C^{r} [a, b]$ , $E (f) = \int_{a}^{b} K_{r} (t) f^{(r)} (t) d t$ with $K_{r} (t) = E_{x} [(x - t)_{+}^{r - 1}] / (r - 1)!$ .

Proof. Taylor's theorem with integral remainder about $a$ writes $f (x) = p (x) + R (x)$ with $p \in Π_{r - 1}$ and $R (x) = \frac{1}{( r - 1 )!} \int_{a}^{x} (x - t)^{r - 1} f^{(r)} (t) d t$ . Replacing $(x - t)^{r - 1}$ by $(x - t)_{+}^{r - 1}$ extends the upper limit to $b$ without changing the value, since the integrand vanishes for $t > x$ . Apply $E$ : linearity and $E (p) = 0$ give $E (f) = E (R) = E_{x} [\frac{1}{( r - 1 )!} \int_{a}^{b} (x - t)_{+}^{r - 1} f^{(r)} (t) d t]$ . The functional $E$ is a finite combination of an integral over $x$ and point evaluations in $x$ ; each commutes with the $t$ -integral by the Fubini-Tonelli theorem applied to the jointly continuous integrand $(x, t) \mapsto (x - t)_{+}^{r - 1} f^{(r)} (t)$ on the compact square. Interchanging yields $E (f) = \int_{a}^{b} \frac{1}{( r - 1 )!} E_{x} [(x - t)_{+}^{r - 1}] f^{(r)} (t) d t = \int_{a}^{b} K_{r} (t) f^{(r)} (t) d t$ . $□$

Proposition 4 (sign-definite kernel gives the mean-value error and the error constant). If $K_{r}$ from Proposition 3 does not change sign, there is $η \in (a, b)$ with $E (f) = f^{(r)} (η) \int_{a}^{b} K_{r}$ , and $\int_{a}^{b} K_{r} = E (x^{r}) / r!$ .

Proof. Assume $K_{r} \geq 0$ (otherwise apply the argument to $- E$ ). Since $f^{(r)} \in C [a, b]$ attains its min $m$ and max $M$ , and $K_{r} \geq 0$ , $m \int K_{r} \leq \int K_{r} f^{(r)} \leq M \int K_{r}$ . If $\int K_{r} = 0$ then $E (f) = 0$ for all $f$ and any $η$ works; otherwise dividing by $\int K_{r} > 0$ places the quotient $E (f) / \int K_{r}$ in $[m, M]$ , and the intermediate value theorem applied to the continuous $f^{(r)}$ supplies $η \in (a, b)$ with $f^{(r)} (η) = E (f) / \int K_{r}$ , i.e. $E (f) = f^{(r)} (η) \int K_{r}$ . For the constant, apply Proposition 3 to $f (x) = x^{r}$ : then $f^{(r)} \equiv r!$ , so $E (x^{r}) = \int_{a}^{b} K_{r} (t) r! d t = r! \int_{a}^{b} K_{r}$ , whence $\int_{a}^{b} K_{r} = E (x^{r}) / r!$ . $□$

Proposition 5 (operator norm and the convergence criterion). The quantity $\sum_{i} ∣ w_{i} ∣$ is the operator norm of the rule $Q$ on $(C [a, b], ∥ \cdot ∥_{\infty})$ , and a family ${Q_{n}}$ exact on a dense polynomial set converges for all $f \in C [a, b]$ iff $sup_{n} \sum_{i} ∣ w_{i}^{(n)} ∣ < \infty$ .

Proof. For $∥ f ∥_{\infty} \leq 1$ , $∣ Q (f) ∣ = ∣ \sum_{i} w_{i} f (x_{i}) ∣ \leq \sum_{i} ∣ w_{i} ∣$ , so $∥ Q ∥ \leq \sum_{i} ∣ w_{i} ∣$ ; choosing continuous $f$ with $∥ f ∥_{\infty} = 1$ and $f (x_{i}) = sign w_{i}$ gives $Q (f) = \sum_{i} ∣ w_{i} ∣$ , so $∥ Q ∥ = \sum_{i} ∣ w_{i} ∣$ . For convergence, the functionals $Q_{n} - \int_{a}^{b}$ are bounded on $C [a, b]$ with norms controlled by $\sum_{i} ∣ w_{i}^{(n)} ∣ + (b - a)$ ; if $sup_{n} \sum_{i} ∣ w_{i}^{(n)} ∣ < \infty$ they are uniformly bounded, and since they vanish on the dense polynomial subspace, the Banach-Steinhaus theorem gives $Q_{n} (f) \to \int f$ for every $f$ . Conversely, if $sup_{n} \sum_{i} ∣ w_{i}^{(n)} ∣ = \infty$ , the uniform boundedness principle produces an $f \in C [a, b]$ with $sup_{n} ∣ Q_{n} (f) ∣ = \infty$ , so convergence fails. This is the Pólya-Steklov theorem; the divergence of high-order Newton-Cotes follows from $\sum_{i} ∣ w_{i}^{(n)} ∣ \to \infty$ . $□$

Connections Master

The quadrature weights $w_{i} = \int_{a}^{b} L_{i}$ are integrals of the Lagrange cardinal polynomials of 43.08.01, and the rule is exact on $Π_{n}$ because it integrates the interpolant; the Newton-form and barycentric machinery of that unit supplies the cardinal polynomials whose integrals are tabulated as the Newton-Cotes coefficients, and the same node polynomial $π_{n + 1}$ that controls interpolation error reappears as the kernel-generating object here.
The Peano kernel representation is the integrated form of the interpolation error theorem of 43.08.02: integrating $f - p_{n} = f^{(n + 1)} (ξ) π_{n + 1} / (n + 1)!$ over $[a, b]$ yields an error functional annihilating $Π_{n}$ , and the negative-weight divergence of high-order equispaced Newton-Cotes is the quadrature face of the Runge phenomenon, driven by the identical endpoint swelling of the equispaced node polynomial that drives interpolation divergence.
The composite trapezium and Simpson rules of 43.09.02 sum the single-interval Peano errors across a partition to obtain the global $O (h^{2})$ and $O (h^{4})$ convergence, and the Euler-Maclaurin formula developed there is the full asymptotic expansion whose leading coefficient is the single-interval error constant $\int K_{r} = E (x^{r}) / r!$ computed in this unit; Romberg integration extrapolates that expansion.
Gauss quadrature in 43.09.03 is the response to the degree-of-exactness ceiling found here: by placing nodes at the roots of the degree- $(n + 1)$ orthogonal polynomial of 43.08.05 rather than at equispaced points, it raises the exactness from $n$ to $2 n + 1$ and keeps all weights positive, escaping the negative-weight instability that the Peano kernel sign analysis of this unit exposes.
The mean-value collapse of the sign-definite kernel rests on the same weighted mean value theorem for integrals that underlies the Lagrange remainder of 02.05.02; the Peano representation generalises the single mean value theorem to an $r$ -th-order statement, recovering the Taylor remainder as the special case where the functional is point evaluation minus its Taylor polynomial.

Historical & philosophical context Master

Interpolatory integration by equispaced ordinates was systematised by Isaac Newton in the Methodus differentialis (circulated 1676, published 1711) and tabulated for low orders by Roger Cotes, whose posthumous Harmonia mensurarum (1722) gave the coefficients for the rules through eleven points; the family has carried both names since. Thomas Simpson published the three-point parabolic rule in his Mathematical Dissertations (1743), though the rule was known to Cavalieri and Gregory earlier, another of the subject's standard misattributions ^{[Newton 1711]}.

The systematic error theory is twentieth-century. Giuseppe Peano gave the kernel representation of the remainder of a quadrature formula in a 1913 note in the Rendiconti of the Accademia dei Lincei, expressing the error of any linear functional vanishing on polynomials up to a fixed degree as a definite integral of a derivative of the integrand against a fixed kernel obtained by applying the functional to the truncated power ^{[Peano 1913]}. The divergence of high-order Newton-Cotes was established through the Pólya-Steklov convergence criterion: George Pólya in 1933 and Vladimir Steklov independently characterised convergence of quadrature sequences by uniform boundedness of the absolute-weight sums, explaining why the negative weights that first appear at the eight-point closed rule are fatal. The definiteness theory and best-constant analysis were developed by the mid-century quadrature school surveyed by Philip Davis and Philip Rabinowitz and given its modern functional-analytic form by Helmut Brass and Knut Petras.

Bibliography Master

@book{sulimayers2003,
  author    = {S\"{u}li, Endre and Mayers, David F.},
  title     = {An Introduction to Numerical Analysis},
  publisher = {Cambridge University Press},
  year      = {2003}
}

@book{davisrabinowitz1984,
  author    = {Davis, Philip J. and Rabinowitz, Philip},
  title     = {Methods of Numerical Integration},
  edition   = {2},
  publisher = {Academic Press},
  year      = {1984}
}

@book{brasspetras2011,
  author    = {Brass, Helmut and Petras, Knut},
  title     = {Quadrature Theory: The Theory of Numerical Integration on a Compact Interval},
  series    = {Mathematical Surveys and Monographs},
  volume    = {178},
  publisher = {American Mathematical Society},
  year      = {2011}
}

@article{peano1913,
  author  = {Peano, Giuseppe},
  title   = {Resto nelle formule di quadratura espresso con un integrale definito},
  journal = {Atti della Reale Accademia dei Lincei, Rendiconti},
  volume  = {22},
  year    = {1913},
  pages   = {562--569}
}

@book{krylov1962,
  author    = {Krylov, Vladimir I.},
  title     = {Approximate Calculation of Integrals},
  publisher = {Macmillan},
  year      = {1962}
}

@article{polya1933,
  author  = {P\'{o}lya, George},
  title   = {\"{U}ber die Konvergenz von Quadraturverfahren},
  journal = {Mathematische Zeitschrift},
  volume  = {37},
  year    = {1933},
  pages   = {264--286}
}

Prerequisites

43.08.01
43.08.02
02.05.02

Tier anchors

beginner: Süli-Mayers 2003 *An Introduction to Numerical Analysis* (Cambridge) §7.1-7.3 (interpolatory quadrature, the trapezium and Simpson rules and their error terms, at the level of a first numerical-methods course); Burden-Faires 2011 *Numerical Analysis* 9e (Brooks/Cole) §4.3 for the Newton-Cotes weight bookkeeping on worked nodes
intermediate: Süli-Mayers 2003 *An Introduction to Numerical Analysis* (Cambridge) §7.1-7.5 (interpolatory quadrature, degree of exactness, the closed Newton-Cotes family, the trapezium and Simpson error terms, and the Peano kernel theorem for the error functional); Stoer-Bulirsch 2002 *Introduction to Numerical Analysis* 3e (Springer) §3.1 (Newton-Cotes formulae and the Peano representation of the remainder)
master: Davis-Rabinowitz 1984 *Methods of Numerical Integration* 2e (Academic Press) Ch. 2-4 (interpolatory and Newton-Cotes rules, the Peano kernel theory, the negative-weight instability of high-order rules); Brass-Petras 2011 *Quadrature Theory* (AMS) Ch. 1-2 (the error functional, Peano kernels, definiteness and best constants); Krylov 1962 *Approximate Calculation of Integrals* (Macmillan) Ch. 6-7 (Newton-Cotes coefficients and convergence)

References

Süli, E. & Mayers, D. F. — An Introduction to Numerical Analysis · Cambridge University Press (2003). Chapter 7 ('Numerical integration -- I') builds quadrature by integrating the interpolating polynomial: §7.2 defines the interpolatory rule int_a^b f approx sum_i w_i f(x_i) with weights w_i = int_a^b L_i(x) dx the integrals of the Lagrange cardinal polynomials, and proves such a rule integrates every polynomial of degree at most n exactly. §7.3 specialises to the closed Newton-Cotes family on equally spaced nodes -- the trapezium rule with error -(b-a)^3/12 f''(eta), Simpson's rule with error -(b-a)^5/2880 f^{(4)}(eta) -- and notes the gain of one extra degree of exactness for even n by symmetry. §7.4-7.5 develop the Peano kernel theorem: a linear functional E that annihilates Pi_k has the representation E(f) = int K(t) f^{(k+1)}(t) dt with Peano kernel K(t) = E_x[(x-t)_+^k]/k!, and when K is one-signed the mean-value form E(f) = f^{(k+1)}(eta) int K(t) dt holds; the chapter also records that high-order Newton-Cotes rules acquire negative weights and lose practical stability.
Davis, P. J. & Rabinowitz, P. — Methods of Numerical Integration · Academic Press, 2nd edition (1984). Chapters 2-4 give the comprehensive interpolatory-quadrature account: the Newton-Cotes weights as integrals of cardinal polynomials and their closed-form rational values, the Peano kernel representation E(f)=int K(t)f^{(r)}(t)dt with K(t)=E_x[(x-t)_+^{r-1}]/(r-1)!, the definiteness criterion and the resulting mean-value error forms, and the table of Newton-Cotes coefficients showing the first negative weight appearing at the eight-point (n=8) closed rule, with the sum of absolute weights growing without bound so that the rules amplify rounding error and fail to converge for general continuous integrands -- the structural reason high-order equispaced Newton-Cotes is abandoned in favour of composite rules and Gauss quadrature.
Brass, H. & Petras, K. — Quadrature Theory: The Theory of Numerical Integration on a Compact Interval · American Mathematical Society, Mathematical Surveys and Monographs 178 (2011). Chapters 1-2 treat the error functional E of a quadrature rule as a bounded linear functional on a Sobolev-type space, prove the Peano kernel theorem in the general form E(f)=int_a^b K_r(t) f^{(r)}(t) dt for E annihilating Pi_{r-1}, with K_r(t)=E_x[(x-t)_+^{r-1}]/(r-1)!, establish that a definite (one-signed) kernel yields the best-constant error bound |E(f)| <= |int K_r| sup|f^{(r)}| with equality form E(f)=f^{(r)}(eta) int K_r, and characterise when classical rules are definite; the negative-weight phenomenon and its consequence for the operator norm sum|w_i| of high-order Newton-Cotes are derived from the kernel sign structure.
Peano, G. — Resto nelle formule di quadratura espresso con un integrale definito · Atti della Reale Accademia dei Lincei, Rendiconti (5) 22 (1913), 562-569. The original note expressing the remainder of a quadrature formula -- and more generally of a linear functional vanishing on polynomials up to a given degree -- as a definite integral of a fixed kernel against a derivative of the integrand, K(t) obtained by applying the functional to the truncated power (x-t)_+^k; this is the result now universally called the Peano kernel theorem.

Estimated time

beginner: 20m
intermediate: 50m
master: 90m