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Clenshaw–Curtis and extrapolation quadrature — Examples

1 · Prerequisites

2 · Summary

The examples show why endpoint smoothness matters for Romberg extrapolation, compute the three-point Clenshaw–Curtis weights and its exact quadratic evaluation, and verify the sharp sample-error amplification of the nine-point closed Newton–Cotes rule. The last example keeps deterministic Peano integration error separate from perturbations in the supplied samples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

The square-root endpoint does not have Romberg's even-power order

Example

For $f (x) = \sqrt{x}$ on $[0, 1]$ , let $T_{N}$ be the composite trapezoidal value with mesh $h = 1 / N$ , and set $S_{N} : = T_{2 N} + (T_{2 N} - T_{N}) / 3 = (4 T_{2 N} - T_{N}) / 3$ , the first-column Richardson value formed from the $N$ - and $2 N$ -panel levels. For $N = 2^{k}$ with $k \in N$ , this is $R_{k + 1, 1}$ of Finite Romberg table from nested composite trapezoid levels whenever its level cap $L$ satisfies $L \geq k + 1$ . For every integer $N \geq 1$ , $\frac{8 \sqrt{2} - 11}{144} h^{3 / 2} \leq \int_{0}^{1} f (x) d x - S_{N} .$ The coefficient is positive, so this error is not $O (h^{2})$ as $N \to \infty$ .

Facts & Assumptions

Given: $N \geq 1$ , $h = 1 / N$ , and $f (x) = \sqrt{x}$ on $[0, 1]$ .

[A1]

$N$ is a positive integer and $h = 1 / N$ .

[F1]

Composite trapezoidal rule: $T_{N}$ is the sum of panel trapezoids on the uniform mesh; its individual panel values add to $T_{N}$ .

[F2]

Second-derivative error bound for the composite trapezoidal rule: on a panel of width $h$ where $f \in C^{2}$ , the absolute difference between its integral and its trapezoid is at most $h^{3} \sup ∣ f^{''} ∣ / 12$ .

[F3]

Finite Romberg table from nested composite trapezoid levels: its first-column recurrence is $T_{2 N} + (T_{2 N} - T_{N}) / 3$ when applied to the adjacent $N$ - and $2 N$ -panel levels; for dyadic $N$ this is the corresponding table entry.

[F4]

Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Rational powers $a^{r}$ of a positive base, Continuity and derivatives of positive-base real powers, The exponential definition of real powers agrees with the existing rational powers, A function differentiable at $c$ is continuous at $c$ , and Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ : $\sqrt{x}$ is the nonnegative square root, is differentiable and hence continuous on $(0, \infty)$ , and is continuous at $0$ by $0 \leq x < ε^{2} \Rightarrow 0 \leq \sqrt{x} < ε$ .

[F5]

Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, and The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : for $u \in (0, 1]$ , substituting $x = t^{2}$ on $[0, \sqrt{u}]$ gives $\int_{0}^{u} \sqrt{x} d x = \int_{0}^{\sqrt{u}} 2 t^{2} d t = 2 u \sqrt{u} / 3$ .

[F6]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ : the integral and the sum of panel trapezoids decompose over the uniform mesh.

[F7]

Continuity and derivatives of positive-base real powers, The exponential definition of real powers agrees with the existing rational powers, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , and Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: on each closed panel with endpoints $i h$ and $(i + 1) h$ , for $i \geq 1$ , $f \in C^{2}$ and $f^{''} (x) = - 1 / (4 x \sqrt{x})$ ; the displayed derivative formulas are continuous there because the root is positive and continuous and the denominator does not vanish.

[F8]

Romberg cancels a proved finite even-power expansion: the cancellation conclusion requires the finite even-power expansion hypothesis stated there; it does not assert second-order accuracy for an endpoint-singular input without that expansion.

Proof

1.1givenA1F1F4F5F6algebra

Continuity from [F4] permits the substitution [F5] on $[0, 1]$ and on the first panel $[0, h]$ . Thus $\int_{0}^{1} \sqrt{x} d x = 2 / 3$ and the first-panel integral is $2 h \sqrt{h} / 3$ , while its trapezoid is $h \sqrt{h} / 2$ ; its excess is therefore $h \sqrt{h} / 6 = h^{3 / 2} / 6$ . For every panel, the chord joining endpoint values lies below $\sqrt{x}$ : for $u, v \geq 0$ and $0 \leq λ \leq 1$ , the square of $(1 - λ) \sqrt{u} + λ \sqrt{v}$ is at most $(1 - λ) u + λ v$ , since the difference is $λ (1 - λ) (\sqrt{u} - \sqrt{v})^{2}$ . Thus each panel integral is at least its trapezoid and $e_{N} : = 2 / 3 - T_{N} \geq h^{3 / 2} / 6$ .

2.1givenA1F2F6F7step 1.1algebra

For panels indexed by $i = 1, \dots, N - 1$ , [F2] and [F7] bound each nonnegative panel excess by $h^{3} / (48 i h \sqrt{i h}) = h^{3 / 2} / (48 i^{3 / 2})$ , using the rational-power laws in Laws of rational exponents. The finite sum satisfies $\sum_{i = 1}^{N - 1} i^{- 3 / 2} \leq 3$ : for $N = 1$ it is empty; for $N = 2$ it is $1$ ; and for $N \geq 3$ , if $i - 1 \leq x \leq i$ then $x^{3 / 2} \leq i^{3 / 2}$ by Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ . If $x < i$ , inverse order Inverses of positives are positive, and reciprocation reverses order gives $x^{- 3 / 2} > i^{- 3 / 2}$ , and if $x = i$ equality holds. Integral monotonicity If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ gives $i^{- 3 / 2} \leq \int_{i - 1}^{i} x^{- 3 / 2} d x$ for $2 \leq i \leq N - 1$ . Summing and using Truncated integrals of rational powers yields $\sum_{i = 1}^{N - 1} i^{- 3 / 2} \leq 1 + \int_{1}^{N - 1} x^{- 3 / 2} d x \leq 3$ . Adding the first-panel excess $h^{3 / 2} / 6$ gives $e_{N} \leq (1 / 6 + 3 / 48) h^{3 / 2} = 11 h^{3 / 2} / 48$ . Hence $h^{3 / 2} / 6 \leq e_{N} \leq 11 h^{3 / 2} / 48.$

3.1givenF3F8step 2.1algebra∎

The function is not $C^{1}$ at $0$ , because its right difference quotient is $\sqrt{x} / x = 1 / \sqrt{x}$ and is unbounded as $x ↓ 0$ ; in particular it does not meet the smoothness assumptions behind the finite even-power expansion used for [F8]. Nevertheless the one-step value $S_{N}$ is defined. Writing $e_{2 N} = 2 / 3 - T_{2 N}$ and using the preceding bounds at mesh $h / 2$ and $h$ gives $\int_{0}^{1} f - S_{N} = \frac{4 e_{2 N} - e_{N}}{3} \geq \frac{4 (h / 2)^{3 / 2} / 6 - 11 h^{3 / 2} / 48}{3} = \frac{8 \sqrt{2} - 11}{144} h^{3 / 2} .$ The constant is positive since $(8 \sqrt{2})^{2} = 128 > 121 = 11^{2}$ . Dividing by $h^{2}$ makes the lower bound grow as a positive constant times $h^{- 1 / 2}$ , so the error cannot be $O (h^{2})$ . This illustrates why [F8] cannot be applied without its expansion hypothesis.

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Three-point Clenshaw–Curtis integration of a quadratic

Example

For the three-point rule ( $n = 2$ ) on $[- 1, 1]$ , the nodes are $1, 0, - 1$ and the quadrature weights in that order are $1 / 3, 4 / 3, 1 / 3$ . Hence it integrates $1$ , $x$ , and $x^{2}$ exactly, with respective values $2$ , $0$ , and $2 / 3$ .

Facts & Assumptions

Given: The degree $n = 2$ Clenshaw–Curtis rule on $[- 1, 1]$ and the sample data of the polynomials $1, x, x^{2}$ at its three nodes.

[F1]

Type-I cosine transform for Chebyshev–Lobatto data: for $n = 2$ , $X_{k} = y_{0} + (- 1)^{k} y_{2} + 2 y_{1} \cos (k π / 2)$ .

[F2]

DCT-I coefficients of the Chebyshev–Lobatto interpolant: $a_{0} = X_{0} / 4$ , $a_{1} = X_{1} / 2$ , and $a_{2} = X_{2} / 4$ .

[F3]

Exact integrals of Chebyshev polynomials: $μ_{0} = 2$ , $μ_{1} = 0$ , and $μ_{2} = - 2 / 3$ .

[F4]

Clenshaw–Curtis quadrature with the midpoint zero-degree case: $Q_{2, [- 1, 1]} (f) = \sum_{k = 0}^{2} a_{k} μ_{k}$ since $d = 1$ .

[F5]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: $T_{0} (x) = 1$ , $T_{1} (x) = x$ , and $T_{2} (x) = 2 x^{2} - 1$ .

Proof

1.1givenF1F2F4algebra

For sample values $(y_{0}, y_{1}, y_{2})$ at $(1, 0, - 1)$ , [F1] and the exact values $\cos (0) = 1$ , $\cos (π / 2) = 0$ , $\cos (π) = - 1$ give $X_{0} = y_{0} + 2 y_{1} + y_{2}$ , $X_{1} = y_{0} - y_{2}$ , and $X_{2} = y_{0} - 2 y_{1} + y_{2}$ . Substituting [F2] into [F4] gives $Q_{2} (f) = \frac{y_{0} + 2 y_{1} + y_{2}}{2} - \frac{y_{0} - 2 y_{1} + y_{2}}{6} = \frac{y_{0}}{3} + \frac{4 y_{1}}{3} + \frac{y_{2}}{3} .$

2.1givenF3F5step 1.1algebra∎

For $1$ , the sample vector is $(1, 1, 1)$ and the rule gives $2$ . For $x$ , the vector is $(1, 0, - 1)$ and the rule gives $0$ . For $x^{2}$ , it is $(1, 0, 1)$ and the rule gives $2 / 3$ . Direct integration gives these same values, respectively $2$ , $0$ , and $[x^{3} / 3]_{- 1}^{1} = 2 / 3$ . Equivalently, [F5] gives $x^{2} = (T_{0} + T_{2}) / 2$ , and [F3] gives integral $μ_{0} / 2 + μ_{2} / 2 = 2 / 3$ . Thus all three evaluations agree with the exact integrals.

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Negative weights amplify sample perturbations in the nine-point Newton–Cotes rule

Example

For a sample vector $s = (s_{0}, \dots, s_{8}) \in R^{9}$ , write ${\hat{Q}}_{8} (s) : = \sum_{j = 0}^{8} w_{j} s_{j}$ . If a perturbed sample vector is $s + δ$ with $∣ δ_{j} ∣ \leq ε$ , then $∣ {\hat{Q}}_{8} (s + δ) - {\hat{Q}}_{8} (s) ∣ \leq \frac{13714}{4725} ε .$ This constant is sharp and exceeds $2$ , the signed sum of the weights and the total weight of any positive rule exact on constants over $[- 1, 1]$ . Independently, for $f \in C^{10} [- 1, 1]$ the quadrature error satisfies the Peano bound $(\int_{- 1}^{1} ∣ K_{10} (t) ∣ d t) ∥ f^{(10)} ∥_{\infty}$ of Peano remainder bound for the eight-panel rule.

Facts & Assumptions

Given: A sample vector $s \in R^{9}$ , the nine weights of $Q_{8}$ , and real perturbations $δ_{0}, \dots, δ_{8}$ satisfying $∣ δ_{j} ∣ \leq ε$ .

[A1]

$ε \geq 0$ and $∣ δ_{j} ∣ \leq ε$ for each node.

[F1]

Closed Newton–Cotes rule on eight equal panels: the rule has weights with numerators $(989, 5888, - 928, 10496, - 4540, 10496, - 928, 5888, 989)$ and common denominator $14175$ .

[F2]

Exact weights and degree-nine exactness of the eight-panel rule: these coefficients are the integrated cardinal weights and the rule is exact through degree $9$ .

[F3]

Peano remainder bound for the eight-panel rule: for $f \in C^{10}$ , its integral error is bounded by $∥ K_{10} ∥_{1} ∥ f^{(10)} ∥_{\infty}$ with the stated explicit kernel.

Proof

1.1givenA1F1algebra

The positive weight numerators sum to $34746$ and the absolute values of the negative numerators sum to $928 + 4540 + 928 = 6396$ . Thus $\sum_{j} ∣ w_{j} ∣ = (34746 + 6396) / 14175 = 41142 / 14175 = 13714 / 4725$ , while $\sum_{j} w_{j} = 2$ . The triangle inequality and [A1] give $∣ {\hat{Q}}_{8} (s + δ) - {\hat{Q}}_{8} (s) ∣ = ∣ \sum_{j} w_{j} δ_{j} ∣ \leq ε \sum_{j} ∣ w_{j} ∣$ . Equality holds by taking $δ_{j} = ε$ when $w_{j} > 0$ and $δ_{j} = - ε$ when $w_{j} < 0$ (with either value when $w_{j} = 0$ ), so the bound is sharp.

2.1givenF2F3step 1.1∎

For any separately supplied $f \in C^{10} [- 1, 1]$ , [F2] and [F3] bound its deterministic integration error as stated. That smooth-function bound does not remove the sample-perturbation amplification in step 1.1.

Clenshaw–Curtis and extrapolation quadrature — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The square-root endpoint does not have Romberg's even-power order

Example

Facts & Assumptions

Proof

Three-point Clenshaw–Curtis integration of a quadratic

Example

Facts & Assumptions

Proof

Negative weights amplify sample perturbations in the nine-point Newton–Cotes rule

Example

Facts & Assumptions

Proof

Sources