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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Clenshaw–Curtis and extrapolation quadrature

1 · Prerequisites

2 · Summary

Clenshaw–Curtis quadrature is defined by integrating the Chebyshev–Lobatto interpolant, with explicit DCT-I endpoint conventions and a midpoint case at degree zero. It is exact through degree n. For m=(a+b)/2, d=(b−a)/2 and g(t)=f(m+dt), the approximation estimate is ∣∫abf−Qn,[a,b](f)∣≤(b−a)(1+ΛnL)En(g), where En(g) is the best uniform degree-n error on [−1,1] and ΛnL is the Lobatto interpolation norm. Affine pullback identifies En(g) with the best degree-n error of f on [a,b]. The factor (b−a) follows by integrating the interpolation residual over the physical interval; on [−1,1] the factor is 2. The source contract's shorter [−1,1] expression without that factor is not used.

The page also defines a finite nested Romberg table. A finite Euler–Maclaurin formula supplies a uniform even-power expansion, and a separate induction proves the cancellations under that expansion hypothesis. An observed error ratio alone does not establish the hypothesis or the resulting order.

For a finite quadrature functional annihilating polynomials through degree r−1, the Peano kernel gives an integral remainder and an L1 bound. The nine-node closed Newton–Cotes rule is verified exactly through degree nine; its negative weights give a sample-perturbation norm larger than its signed weight sum. The even-extension DFT algorithm inherits only the computed-DFT status in its supplier contract and makes no new stability or accuracy claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite Romberg table from nested composite trapezoid levels

Definition

Input a real function f on a nondegenerate closed interval [a,b] and an integer L≥0. The output is the finite triangular table (Rℓ,j)0≤j≤ℓ≤L and its top entry RL,L.

  1. For each ℓ=0,…,L, set Nℓ:=2ℓ∈N using natural exponentiation of Exponentiation of natural numbers, mn, and its agreement with the integer power in R. Since 20=1 and 2ℓ+1=2ℓ⋅2, induction gives Nℓ≥1, as required by Composite trapezoidal rule. Set hℓ:=(b−a)/Nℓ and Rℓ,0:=TNℓ(f) using its composite trapezoidal value. Natural powers in real scalar expressions are understood through their canonical real images.
  2. For j=1,…,L, and then for each ℓ=j,…,L, set Rℓ,j:=Rℓ,j−1+Rℓ,j−1−Rℓ−1,j−14j−1.

For j≥1, claim 3 of Monotonicity of x↦xn and of n↦an gives 4j≥4>1, using the natural-to-real power identity of Exponentiation of natural numbers, mn, and its agreement with the integer power in R; thus every displayed denominator 4j−1 is the positive real 4j−1 and is invertible by The reals form a field. This definition returns only the stated finite table; it makes no accuracy claim for nonsmooth inputs or finite-precision arithmetic.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Bernoulli polynomials by a normalized coefficient recursion

Definition

Let ι:N→R be the canonical natural map of The canonical natural ι(n)=n⋅1F of a field. Define polynomials Bm∈R[X] recursively. Put B0(X)=1. If m≥1 and Bm−1(X)=∑q=0m−1βm−1,qXq, set Bm(X):=ι(m)∑q=0m−1βm−1,q(Xq+1ι(q+1)−1ι(q+1)ι(q+2)). Every sum is finite. The denominator factors are positive by Canonical naturals are positive and strictly increasing, hence invertible in R by The reals form a field. Write Bm:=Bm(0) for the Bernoulli number of index m.

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Type-I cosine transform for Chebyshev–Lobatto data

Definition

Let n∈N with n≥1 and let y=(y0,…,yn)∈Rn+1. In real-valued formulas, each natural index denotes its canonical real image (The canonical natural ι(n)=n⋅1F of a field). Since n>0 by Canonical naturals are positive and strictly increasing, division by n is defined in R (The reals form a field). The type-I discrete cosine transform, in the Lobatto convention used here, is the vector X=(X0,…,Xn) defined by Xk:=y0+(−1)kyn+2∑j=1n−1yjcos⁡ ⁣(jkπn),0≤k≤n. Here (−1)k is the natural power of the real number −1 from Integer powers am, and the sum is the finite sum of Finite sums and finite products, by recursion, empty when n=1. The data yj are ordered from the right endpoint toward the left endpoint at nodes cos⁡(jπ/n). This definition fixes the unnormalised DCT-I convention; the transform is not defined at n=0.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Closed Newton–Cotes rule on eight equal panels

Definition

For j=0,…,8, set xj=−1+j/4 and wj:=114175(989,5888,−928,10496,−4540,10496,−928,5888,989)j. For any real-valued function f on [−1,1], define the closed nine-node rule Q8(f):=∑j=08wjf(xj), where the sum is finite. The nodes are equally spaced and include both endpoints; the subsequent weight and exactness lemma verifies that these coefficients are the integrated Lagrange-cardinal weights.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Peano kernel of a finite quadrature functional

Definition

Let a<b, let x1,…,xs∈[a,b] be a finite indexed list (repetitions are allowed), and let w1,…,ws∈R. For every Riemann-integrable f:[a,b]→R, define the quadrature value and residual by Q(f):=∑i=1swif(xi),L(f):=∫abf(x) dx−Q(f). Every formal real polynomial's evaluation function is continuous by 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 and integrable by A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion. In the following exactness predicate, L(p) means L applied to that evaluation function. For a natural r≥1, say that Q is exact through degree r−1 if L(p)=0 for the zero polynomial and every nonzero formal real polynomial p with degree at most r−1 (Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials). Here r−1 is the natural predecessor specified by The factorial n! and the falling factorial nk‾, defined by recursion in N. For any natural q, the factorial q! in a real-valued formula denotes its canonical real image under The canonical natural ι(n)=n⋅1F of a field. This denominator is nonzero: q!≠0 by The factorial n! and the falling factorial nk‾, defined by recursion in N, every nonzero natural is a successor by Every nonzero natural number is a successor and therefore is at least 1 in the natural order Order on the natural numbers, and its canonical real image is positive by Canonical naturals are positive and strictly increasing, hence invertible by The reals form a field. For t∈[a,b] define gr,t(x):={(x−t)r−1(r−1)!,x>t,0,x≤t,Kr(t):=L(gr,t). The power is the natural power of Integer powers am. For r≥2, gr,t is polynomial and continuous on each side of t by 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; the polynomial branch extends to t with value 0 because r−1≥1, agreeing with the zero branch. Thus it is continuous on [a,b] and Riemann-integrable by A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion. For r=1 it is the indicator of (t,b]; if t=b it is zero, and if t<b, for each ε>0 choose s∈(t,b) with s−t<ε and use the partition with breakpoints a,t,s,b after deleting repeated endpoints. The indicator is constant on all panels except [t,s], where its oscillation is 1, so its upper-minus-lower sum is at most s−t<ε; Riemann's criterion: a bounded f on [a,b] is Darboux integrable if and only if for every real ε>0 there is a partition P with U(f,P)−L(f,P)<ε makes it integrable. At r=1 the denominator is 0!=1. Once K1 is shown integrable, changing its values at the finitely many quadrature nodes does not change its integral by Changing an integrable function at finitely many points changes neither its integrability nor its integral.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

The Chebyshev recurrence gives degree at most its index

Statement

For every integer n≥0, the coefficient list of the first-kind Chebyshev polynomial Tn from Chebyshev polynomials of the first and second kinds by their three-term recurrences has no nonzero coefficient above index n. In particular, deg⁡Tn≤n.

Facts & Assumptions

Given: The recursively defined first-kind Chebyshev polynomials.

[A1]

n ranges over nonnegative integer indices.

[F1]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: T0=1, T1=X, Tk+2=2XTk+1−Tk, and polynomial multiplication uses coefficient convolution.

[F2]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials: a nonzero formal polynomial is a finite coefficient list whose degree is its largest nonzero coefficient index; the zero polynomial has no degree.

[F3]

The principle of mathematical induction: induction on natural indices is valid.

Proof

1.1givenA1F1basealgebra

Define P(k) to assert that Tk has no nonzero coefficient above k and Tk+1 has none above k+1. At k=0, this holds because T0=1 and T1=X by [F1].

1.2A1F1F2IHalgebra

Assume P(k). By [F1], the coefficient of XTk+1 at index q is the coefficient of Tk+1 at index q−1 (zero when q=0), so it vanishes for q>k+2. The coefficient of Tk vanishes for those indices as well; hence Tk+2=2XTk+1−Tk has no coefficient above k+2. Together with the second assertion of P(k), this is P(k+1).

2.1givenA1F3step 1.1step 1.2inductiondischarge-induction

By [F3], steps 1.1 and 1.2 prove P(k) for every natural k, so every Tn has no nonzero coefficient above n.

3.1givenF1F2F3step 2.1inductiondischarge-inductionalgebra∎

By the convolution definition in [F1] and finite distributivity, evaluation at 1 respects polynomial sums and products. Evaluating the recurrence at 1 therefore gives T0(1)=T1(1)=1 and Tk+2(1)=2Tk+1(1)−Tk(1). Induction yields Tn(1)=1 for every natural n, so no Tn is the zero polynomial. Its degree is therefore defined by [F2], and the support bound in step 2.1 gives deg⁡Tn≤n.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

The Chebyshev recurrence agrees with cosine multiple angles

Statement

For every integer n≥0 and every real θ, the first-kind Chebyshev polynomial from Chebyshev polynomials of the first and second kinds by their three-term recurrences satisfies Tn(cos⁡θ)=cos⁡(nθ).

Facts & Assumptions

Given: An integer n≥0 and a real angle θ.

[A1]

n is a nonnegative integer and θ∈R.

[F1]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: T0=1, T1=X, and Tk+2=2XTk+1−Tk for every k≥0.

[F2]

The addition formulas for sine and cosine: cos⁡((k+2)θ)=2cos⁡θcos⁡((k+1)θ)−cos⁡(kθ) for every integer k≥0 and real θ.

[F3]

The principle of mathematical induction: an identity true at the initial indices and preserved by the stated recurrence holds for every natural index.

Proof

1.1givenA1F1basealgebra

For the fixed real θ in [A1], define P(k) to assert the identity at both indices k and k+1. At k=0, the identities for T0 and T1 give P(0).

1.2givenA1F1F2IHalgebra

Fix θ∈R as in [A1] and suppose P(k). By the polynomial recurrence [F1], Tk+2(cos⁡θ)=2cos⁡θ Tk+1(cos⁡θ)−Tk(cos⁡θ). Substituting the two induction hypotheses and then [F2] gives Tk+2(cos⁡θ)=cos⁡((k+2)θ), so P(k+1) holds.

2.1A1F3step 1.1step 1.2inductiondischarge-induction∎

By [F3], steps 1.1 and 1.2 prove P(k) for every natural k; its first clause is the identity for every n in [A1].

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Derivative, endpoint, parity, and boundedness identities for Bernoulli polynomials

Statement

For the polynomials of Bernoulli polynomials by a normalized coefficient recursion and every natural m≥1, Bm′(x)=ι(m)Bm−1(x),∫01Bm(x) dx=0. In particular, B0(x)=1 and B1(x)=x−12. Consequently Bm(1)=Bm(0) for m≥2. The reflection identity Bm(1−x)=(−1)mBm(x) holds for every m≥0, so Bm(0)=0 for odd m≥3. Each Bm is bounded on [0,1].

Facts & Assumptions

Given: The recursive coefficient definition of Bm and an integer index m.

[A1]

m is a natural index; derivative and mean identities are asserted for m≥1.

[F1]

Bernoulli polynomials by a normalized coefficient recursion: if Bm−1(x)=∑q=0m−1βm−1,qxq, then Bm(x)=ι(m)∑q=0m−1βm−1,q(xq+1/ι(q+1)−1/(ι(q+1)ι(q+2))).

[F3]
[F6]

Monotone change of variable for Riemann-integrable functions: the decreasing affine map x↦1−x preserves the Riemann integral after multiplication by its absolute derivative 1.

[F7]

The principle of mathematical induction: induction over the natural polynomial index is valid.

Proof

1.1givenA1F1F2F3algebra

For the nonnegative natural index m of [A1], the case m=1 gives B1(x)=x−12 by [F1]. For every m≥1, [F1]–[F2] give Bm′(x)=ι(m)∑q=0m−1βm−1,qxq=ι(m)Bm−1(x). Integrating each monomial in [F1] over [0,1], the integral of xq+1/ι(q+1) is 1/(ι(q+1)ι(q+2)), which cancels the subtracted constant term; hence ∫01Bm=0.

1.2givenF1basealgebra

The reflection formula holds at m=0 because B0(x)=1 by [F1], so B0(1−x)=B0(x)=(−1)0B0(x).

2.1givenF1F3step 1.1algebra

For m≥2, [F3] and step 1.1 applied to Bm−1 give Bm(1)−Bm(0)=ι(m)∫01Bm−1=0. Thus the endpoints agree for every such m.

2.2givenA1F2F3F4F6F7step 1.1IHinductionalgebra

Let m≥1 and assume Bm−1(1−x)=(−1)m−1Bm−1(x) for all x. By [F2], [F4], and step 1.1, the derivative of Dm(x):=(−1)mBm(1−x)−Bm(x) is Dm′(x)=(−1)m+1ι(m)Bm−1(1−x)−ι(m)Bm−1(x)=0, using the induction hypothesis. Hence Dm is constant by [F3]; its integral on [0,1] is zero because step 1.1 gives ∫01Bm=0 and [F6] gives ∫01Bm(1−x) dx=∫01Bm(x) dx. A constant with zero integral on an interval of length one is zero, so reflection holds at m.

3.1A1F7step 1.2step 2.2induction

The base case in step 1.2 and the induction step in step 2.2 prove the reflection identity for every natural m in [A1] by [F7].

4.1givenA1F5step 2.1step 3.1inductiondischarge-inductionalgebra∎

For the index m in [A1], if it is odd and at least 3, step 3.1 gives Bm(1)=−Bm(0) while step 2.1 gives Bm(1)=Bm(0); therefore Bm(0)=0. Finally each Bm is a polynomial and hence continuous, so [F5] bounds it on the compact interval [0,1].

LemmaStatement: Literature-sourcedProof: AI-generatedOpen item page →

Exact integrals of Chebyshev polynomials

Statement

For every integer k≥0, let μk:=∫−11Tk(x) dx, where Tk is the first-kind Chebyshev polynomial. Then μk={2,k=0,0,k odd,21−k2,k≥2 even. Equivalently, μ2r=2/(1−4r2) for every integer r≥0.

Facts & Assumptions

Given: An integer k≥0 and the polynomial Tk of Chebyshev polynomials of the first and second kinds by their three-term recurrences.

[A1]

k is a nonnegative integer.

[F1]

The Chebyshev recurrence agrees with cosine multiple angles: Tk(cos⁡θ)=cos⁡(kθ) for every real θ.

[F2]

Monotone change of variable for Riemann-integrable functions: a monotone differentiable bijection with integrable derivative changes a Riemann integral by the absolute derivative.

[F3]

Signs, monotonicity intervals, and ranges of sine and cosine and The derivatives of sine and cosine are cosine and minus sine: cos⁡ decreases on [0,π], sin⁡θ≥0 there, and (cos⁡θ)′=−sin⁡θ.

[F5]

The addition formulas for sine and cosine: sin⁡ucos⁡v=12(sin⁡(u+v)+sin⁡(u−v)).

[F7]

Quarter-turn values and shifts by pi/2 and pi and The principle of mathematical induction: induction using the shift by π gives cos⁡(mπ)=(−1)m for natural m≥0.

Proof

1.1givenA1F1F2F3F4algebra

Fix the integer k in [A1]. The map θ↦cos⁡θ is a decreasing differentiable bijection from [0,π] to [−1,1] with derivative −sin⁡θ; [F2]–[F4] verify the change-of-variable hypotheses. Therefore [F1]–[F4] give μk=∫0πTk(cos⁡θ)sin⁡θ dθ=∫0πcos⁡(kθ)sin⁡θ dθ.

2.1givenA1F5F6F7step 1.1algebra

For the nonnegative integer k in [A1], the case k=0 gives μ0=∫0πsin⁡θ dθ=2 by step 1.1. For k≥1, [F5] gives sin⁡θcos⁡(kθ)=12(sin⁡((k+1)θ)−sin⁡((k−1)θ)). [F6]–[F7] evaluate each nonzero-frequency sine integral; the m=0 term is identically zero.

3.1givenF6F7step 2.1algebra∎

If k is odd, both positive integers among k−1 and k+1 are even, so [F6]–[F7] and step 2.1 give μk=0. If k≥2 is even, both k−1 and k+1 are odd, so they give μk=1/(k+1)−1/(k−1)=2/(1−k2). Together with the k=0 case, this proves the formula.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Weighted cosine orthogonality on the Lobatto grid

Statement

For n∈N with n≥1 and k,ℓ∈N with 0≤k,ℓ≤n, define Skℓ:=12cos⁡(0)+∑j=1n−1cos⁡ ⁣(jkπn)cos⁡ ⁣(jℓπn)+12(−1)k+ℓ. Then Skℓ=n when k=ℓ=0 or k=ℓ=n, Skℓ=n/2 when k=ℓ is strictly between 0 and n, and Skℓ=0 when k≠ℓ. The half-endpoint convention is the one used by the DCT-I of Type-I cosine transform for Chebyshev–Lobatto data.

Facts & Assumptions

Given: A natural n≥1 and naturals 0≤k,ℓ≤n.

[A1]

The grid size is n+1 with n≥1, and 0≤k,ℓ≤n in the natural-number order.

[F1]

Type-I cosine transform for Chebyshev–Lobatto data fixes the Lobatto angles jkπ/n, the canonical real-image convention, and the half-weighted endpoint convention.

[F2]

The addition formulas for sine and cosine and The subtraction formulas for sine and cosine give the sine-difference telescoping identity and the cosine product-to-sum identity.

[F3]

Quarter-turn values and shifts by pi/2 and pi gives the shifts by π/2 and the quarter-turn values. Iterating its shift identities by induction on the natural m gives sin⁡(mπ/2)=0 for even m, ∣sin⁡(mπ/2)∣=1 for odd m, and cos⁡(mπ/2)=0 for odd m (The principle of mathematical induction).

[F4]

Signs, monotonicity intervals, and ranges of sine and cosine gives strict increase of sine on [0,π/2] and strict decrease on [π/2,π]; with sin⁡0=sin⁡π=0 from Quarter-turn values and shifts by pi/2 and pi, this implies sin⁡θ>0 for 0<θ<π.

Proof

1.1givenA1F1F2F4algebra

For the n in [A1], and any natural frequency m with 1≤m<2n, put α:=mπ/n and Cm:=∑j=0n−1cos⁡(jα). Positivity of canonical naturals gives n>0, and 1≤m<2n gives 0<α/2<π; hence sin⁡(α/2)>0 by [F4]. The half-weighted endpoints are the DCT-I convention fixed by [F1]. By [F2], 2sin⁡(α/2)Cm=∑j=0n−1(sin⁡((j+12)α)−sin⁡((j−12)α))=sin⁡((n−12)α)+sin⁡(α/2)=2sin⁡(nα/2)cos⁡((n−1)α/2). Since sin⁡(α/2)>0, this gives Cm=sin⁡(mπ/2)cos⁡((n−1)mπ/(2n))/sin⁡(mπ/(2n)).

1.2givenF2F3algebra

If m is even, the numerator factor sin⁡(mπ/2) is 0 by [F3], so Cm=0 for 1≤m<2n. If m is odd, write (n−1)mπ/(2n)=mπ/2−mπ/(2n); the addition formulas and cos⁡(mπ/2)=0 give cos⁡((n−1)mπ/(2n))=sin⁡(mπ/2)sin⁡(mπ/(2n)), hence Cm=1. For m=0 and m=2n, direct evaluation gives Cm=n.

2.1step 1.2algebra

Define Dm:=12+∑j=1n−1cos⁡(jmπ/n)+12(−1)m. Since Cm=1+∑j=1n−1cos⁡(jmπ/n), step 1.2 gives Dm=0 for 1≤m<2n and D0=D2n=n.

3.1givenA1F1F2step 2.1algebra∎

For k,ℓ in [A1], [F2] makes each product in Skℓ one half the sum of the cosine terms with natural frequency indices ∣k−ℓ∣ and k+ℓ, so Skℓ=12(D∣k−ℓ∣+Dk+ℓ). If k≠ℓ, then 1≤∣k−ℓ∣<2n and 1≤k+ℓ<2n except the impossible distinct-endpoint case k+ℓ=2n, so both terms vanish by step 2.1. If k=ℓ=0 or k=ℓ=n, both indices are 0 or 2n, giving Skℓ=n. If k=ℓ is interior, the indices are 0 and 2k with 1≤2k<2n, giving Skℓ=n/2. These cases prove the statement.

TheoremStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Peano representation and L1 error bound

Statement

Let a<b, let x1,…,xs∈[a,b] and w1,…,ws∈R define the finite quadrature Q and residual L of Peano kernel of a finite quadrature functional, and let r∈N with r≥1. Assume L vanishes on the evaluation functions of the zero polynomial and every nonzero formal polynomial of degree at most r−1, and let f∈Cr([a,b]). For the kernel Kr(t)=L(gr,t) of Peano kernel of a finite quadrature functional, L(f)=∫abKr(t)f(r)(t) dt,∣L(f)∣≤(∫ab∣Kr(t)∣ dt)∥f(r)∥∞.

Facts & Assumptions

Given: A nondegenerate interval [a,b], a finite real weighted quadrature exact on Pr−1, an integer r≥1, and f∈Cr([a,b]).

[A1]

a<b, r∈N with r≥1, L vanishes on the evaluation functions of the zero polynomial and every nonzero formal polynomial of degree at most r−1, and f has continuous derivatives through order r.

[F1]

Peano kernel of a finite quadrature functional: L(f)=∫abf−∑iwif(xi) and Kr(t)=L(gr,t), where gr,t(x)=(x−t)+r−1/(r−1)! for r≥2 and g1,t(x)=1{x>t}; the factorial image is positive and invertible.

[F2]

Higher derivatives and the classes Ck and C∞: f∈Cr([a,b]) has continuous derivatives through order r, including one-sided endpoint derivatives.

[F3]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): for a continuous derivative, integration over [a,x] equals the endpoint increment of a primitive.

[F5]

A continuous function on a closed rectangle has repeated Riemann integrals in every coordinate order, all equal to its multiple integral: a continuous function on a closed rectangle has equal repeated Riemann integrals in either coordinate order.

[F7]

A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion: continuous functions on closed intervals are Riemann-integrable.

[F8]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous function on a compact metric rectangle is uniformly continuous.

[F9]

Changing an integrable function at finitely many points changes neither its integrability nor its integral: changing finitely many values preserves Riemann integrability and leaves the integral unchanged.

Proof

1.1givenF2F3F4inductionalgebra

We first establish Taylor's integral remainder by induction on the natural order q (The principle of mathematical induction). For r=1, [F3] gives f(x)=f(a)+∫axf′(t) dt. If the remainder formula holds at order q≥1, put u(t):=(x−t)q/q!. The product rule and FTC give ∫axu′(t)f(q)(t) dt=[u(t)f(q)(t)]ax−∫axu(t)f(q+1)(t) dt, and u′(t)=−(x−t)q−1/(q−1)!. Rearranging yields ∫ax(x−t)q−1(q−1)!f(q)(t) dt=(x−a)qq!f(q)(a)+∫ax(x−t)qq!f(q+1)(t) dt. Thus induction yields f(x)=∑j=0r−1f(j)(a)(x−a)j/j!+∫ax(x−t)r−1f(r)(t)/(r−1)! dt.

2.1givenA1F1F6step 1.1

Extend the last integral to [a,b] by setting its kernel to zero when t≥x. With Pr−1(x):=∑j=0r−1f(j)(a)(x−a)j/j!, the result is f(x)=Pr−1(x)+∫abgr,t(x)f(r)(t) dt. Since Pr−1 is zero or has degree at most r−1, [A1] gives L(Pr−1)=0.

3.1givenF1F2F5F6F7F8step 2.1

Suppose first that r≥2. The function (x,t)↦gr,t(x)f(r)(t) is continuous on [a,b]2 because r−1≥1 and f(r) is continuous. By [F5]–[F7], applying the integral part of L to step 2.1 and interchanging the two rectangle integrals gives ∫ab(∫abgr,t(x) dx)f(r)(t) dt. Each finite evaluation term of Q commutes with the t-integral by [F6]. Therefore L(f)=∫abL(gr,t)f(r)(t) dt=∫abKr(t)f(r)(t) dt. The function (x,t)↦gr,t(x) is uniformly continuous on the compact square by [F8], so its x-integral and its finite weighted evaluations vary continuously with t; hence Kr is continuous and integrable.

3.2givenF1F3F4F6F7F9F10step 2.1algebra

If r=1, exactness on constants gives ∑iwi=b−a. For t outside the finite node set, [F1] gives K1(t)=(b−t)−∑xi>twi. It is bounded and affine between consecutive nodes. Extend each affine piece continuously to its closed subinterval; the resulting pieces are integrable by [F7] and add to an integrable piecewise function by [F6]. The kernel differs from that function at only finitely many nodes, so it is integrable and has the same integral by [F9]. Since f′ is continuous, [F10] makes K1f′ integrable; changing the kernel values at the finite node set does not change that product integral by [F9]. By [F3]–[F4] and integration by parts, ∫ab(b−t)f′(t) dt=∫abf(t) dt−(b−a)f(a), while ∑iwi∫axif′(t) dt=Q(f)−(b−a)f(a). Subtracting gives ∫abK1(t)f′(t) dt=L(f).

4.1givenF2F10F11step 3.1step 3.2algebra∎

By [F2] and [F11], f(r) is bounded on [a,b], so ∥f(r)∥∞<∞. In either case, the kernel is integrable and ∣Kr∣ is integrable by [F10]. Apply monotonicity in [F10] to −∣Kr∣∥f(r)∥∞≤Krf(r)≤∣Kr∣∥f(r)∥∞, and use the representation in step 3.1 or 3.2. This gives the asserted bound.

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Finite Euler–Maclaurin formula for the composite trapezoid

Statement

Let a<b, let N≥1 and p≥1 be integers, put h=(b−a)/N and xi=a+ih, and let f∈C2p([a,b]) in the sense of Higher derivatives and the classes Ck and C∞. Write Bm(x) for Bernoulli polynomials by a normalized coefficient recursion and Bm:=Bm(0). For ui(x):=(x−xi)/h on [xi,xi+1], the composite trapezoidal value TN(f) of Composite trapezoidal rule satisfies TN(f)=∫abf(x) dx+∑r=1pB2rh2r(2r)!(f(2r−1)(b)−f(2r−1)(a))−h2p(2p)!∑i=0N−1∫xixi+1B2p(ui(x))f(2p)(x) dx. In particular, with M2p:=max⁡0≤u≤1∣B2p(u)∣, the absolute value of the final remainder is at most (b−a)M2ph2p∥f(2p)∥∞/(2p)!.

Facts & Assumptions

Given: A nondegenerate interval [a,b], integers N,p≥1, the uniform mesh, and f∈C2p([a,b]).

[A1]

The parameters satisfy a<b, N≥1, p≥1, h=(b−a)/N>0, and f has continuous derivatives through order 2p on [a,b].

[F1]

Composite trapezoidal rule: TN(f)=h(f(a)/2+∑i=1N−1f(xi)+f(b)/2).

[F2]

Bernoulli polynomials by a normalized coefficient recursion and Derivative, endpoint, parity, and boundedness identities for Bernoulli polynomials: B1(u)=u−12, Bm′=mBm−1, Bm(1)=Bm(0) for m≥2, and Bm(0)=0 for odd m≥3.

[F3]

Higher derivatives and the classes Ck and C∞: C2p([a,b]) means the derivatives through order 2p are continuous, with one-sided endpoint derivatives.

[F6]

The factorial n! and the falling factorial nk‾, defined by recursion in N: (k)! is the positive product of the first k positive integer real images.

Proof

1.1givenA1F1F2F5F6algebra

On a panel [xi,xi+1], put u=(x−xi)/h. Integration by parts using B1(u)=u−12 gives h∫xixi+1B1(u)f′(x) dx=h2(f(xi)+f(xi+1))−∫xixi+1f(x) dx. Summing this identity over the N panels and using [F1] and [F5] yields TN(f)−∫abf=J1, where Jk:=hkk!∑i=0N−1∫xixi+1Bk(ui(x))f(k)(x) dx for 1≤k≤2p.

1.2givenA1F2F3F4F5algebra

For 1≤k<2p, differentiate Bk+1(ui(x)) to get (k+1)Bk(ui(x))/h. Integrating by parts on each panel therefore gives Jk=hk+1(k+1)!∑i=0N−1[Bk+1(ui(x))f(k)(x)]xixi+1−Jk+1. Because Bk+1(1)=Bk+1(0) by [F2], the interior endpoint terms telescope, leaving hk+1Bk+1(k+1)!(f(k)(b)−f(k)(a)) as the boundary term.

2.1givenstep 1.2F2algebra

Write bq:=Bqhqq!(f(q−1)(b)−f(q−1)(a)) for q≥2. If k+1 is odd and at least 3, [F2] makes that boundary term zero, so step 1.2 gives Jk=−Jk+1. If k+1=2r is even, step 1.2 gives J2r−1=b2r−J2r. For each r<p, the odd-index recurrence at k=2r gives J2r=−J2r+1; hence J2r−1=b2r+J2r+1. Finite induction on r=1,…,p−1 (The principle of mathematical induction) iterates these identities; followed by the final even-index recurrence, it gives J1=∑r=1pB2rh2r(2r)!(f(2r−1)(b)−f(2r−1)(a))−J2p.

3.1givenA1F2F5F6F7step 1.1step 2.1algebra∎

The polynomial B2p is continuous by [F2], so [F7] ensures M2p=max⁡[0,1]∣B2p∣ exists and is finite. By the definition of J2p and the bound ∣B2p(u)∣≤M2p on each unit cell, ∣J2p∣≤h2pM2p∥f(2p)∥∞∑i(xi+1−xi)/(2p)!=(b−a)M2ph2p∥f(2p)∥∞/(2p)!. Substituting step 2.1 into step 1.1 proves the stated formula and remainder bound.

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Even-power expansion of trapezoid levels with a uniform remainder

Statement

Let a<b, let j∈N with j≥1, and let f∈C2j+2([a,b]). For every integer N≥1, set h=(b−a)/N, I=∫abf, and cr:=B2r(2r)!(f(2r−1)(b)−f(2r−1)(a)),1≤r≤j+1, using the Bernoulli numbers of Bernoulli polynomials by a normalized coefficient recursion. Then TN(f)=I+∑r=1jcrh2r+εN,∣εN∣≤Cjh2j+2, where the level-independent finite constant Cj:=∣cj+1∣+(b−a)M2j+2(2j+2)!∥f(2j+2)∥∞,M2j+2:=max⁡0≤u≤1∣B2j+2(u)∣, does not depend on N. Thus the remainder bound is uniform over every set of dyadic levels.

Facts & Assumptions

Given: a<b, an integer j≥1, f∈C2j+2([a,b]), and an integer trapezoid panel count N≥1.

[F1]

Finite Euler–Maclaurin formula for the composite trapezoid: for every p≥1, the composite trapezoid equals the integral, the first p endpoint Bernoulli corrections, and a remainder bounded by (b−a)M2ph2p∥f(2p)∥∞/(2p)!.

[F2]

Higher derivatives and the classes Ck and C∞: C2j+2 supplies all continuous endpoint derivatives through order 2j+2.

[F3]

Composite trapezoidal rule: TN(f) uses the uniform mesh of width h=(b−a)/N.

Proof

1.1givenF1F2F3

Apply [F1] with p=j+1. The formula has the displayed coefficients c1,…,cj+1 and a remainder ρN satisfying ∣ρN∣≤(b−a)M2j+2h2j+2∥f(2j+2)∥∞/(2j+2)!.

2.1givenF1F2F4step 1.1algebra∎

By [F2] and [F4], each derivative norm in cj+1 and Cj is finite; the Bernoulli maximum exists by the Euler–Maclaurin theorem's construction. Put εN:=cj+1h2j+2+ρN. The triangle inequality gives ∣εN∣≤(∣cj+1∣+(b−a)M2j+2∥f(2j+2)∥∞/(2j+2)!)h2j+2=Cjh2j+2. Every term in Cj depends only on f,a,b,j, not on N, so this is the claimed uniform remainder.

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Exact weights and degree-nine exactness of the eight-panel rule

Statement

For the nodes xj=−1+j/4 and rule Q8 of Closed Newton–Cotes rule on eight equal panels, the weights are the integrals of the degree-eight Lagrange cardinal polynomials. Interpreting polynomial inputs by their evaluation functions, the rule is exact for the zero polynomial and every nonzero formal polynomial of degree at most 9.

Facts & Assumptions

Given: The nine equally spaced nodes xj=−1+j/4, their weights wj, and the finite rule Q8.

[F1]

Closed Newton–Cotes rule on eight equal panels: Q8(f)=∑j=08wjf(xj) with the displayed symmetric rational weights.

[F2]

Cardinal-polynomial interpolation and Cardinal interpolation functions are polynomials of the stated degrees: each cardinal polynomial ℓj has degree 8, equals 1 at xj, equals 0 at the other eight nodes, and is ∏m≠j(x−xm)/(xj−xm); every finite cardinal sum is zero or has degree at most 8.

[F3]

Existence and uniqueness of the cardinal-polynomial interpolant and A nonzero real polynomial of degree n has no more than n distinct real roots: the degree-at-most-eight interpolant at nine distinct nodes is unique; equivalently, two such polynomials agreeing at all nine nodes are equal.

[F4]

Monotone change of variable for Riemann-integrable functions: under t=4(x+1), ∫−11ℓj(x) dx=14∫08ℓj((t−4)/4) dt.

Proof

1.1givenF2F4algebra

Put Pj(t)=∏0≤m≤8, m≠j(t−m) and Dj=∏m≠j(j−m). Substituting t=4(x+1) into [F2] cancels the eight factors 1/4 between numerator and denominator, giving ℓj((t−4)/4)=Pj(t)/Dj. For j=0,1,2,3,4, the pairs (Dj;coefficients of Pj(t) from t0 upward) are (40320;[40320,−109584,118124,−67284,22449,−4536,546,−36,1]), (−5040;[0,−40320,69264,−48860,18424,−4025,511,−35,1]), (1440;[0,−20160,44712,−36706,15289,−3580,478,−34,1]), (−720;[0,−13440,32048,−28692,12864,−3195,447,−33,1]), and (576;[0,−10080,24876,−23312,10993,−2864,418,−32,1]). These coefficients follow by multiplying the eight factors, and each denominator is the product of the corresponding integer differences.

2.1givenF1F2F4F5step 1.1algebra

Integrating each coefficient list by [F5] gives ∫08Pj(t)/Dj dt=(3956,23552,−3712,41984,−18160)j/14175 for j=0,1,2,3,4. The factor 1/4 in [F4] therefore gives ∫−11ℓj(x) dx=(989,5888,−928,10496,−4540)j/14175, exactly the first five weights in [F1]. Reflection x↦−x sends node xj to x8−j and, by the product formula in [F2], sends ℓj to ℓ8−j; the substitution preserves the integral. Thus the remaining four cardinal integrals equal the corresponding symmetric weights in [F1].

3.1givenF1F2F3F5step 2.1algebra

For the zero polynomial and every nonzero formal polynomial p of degree at most 8, its cardinal interpolant ∑j=08p(xj)ℓj is the evaluation of a formal polynomial of degree at most 8 or is zero, and it agrees with the evaluation function of p at all nine nodes by [F2]. Uniqueness [F3] makes the two evaluation functions equal. Integrating this finite identity and using step 2.1 and [F5] yields ∫−11p=∑jp(xj)∫−11ℓj=Q8(p).

4.1givenF1F5step 3.1algebra∎

The nodes and weights in [F1] occur in equal pairs under x↦−x, so Q8(x9)=0 by cancellation of the paired terms and the zero central term. The integral of the odd polynomial x9 over [−1,1] is zero by the same substitution. The zero polynomial, and each nonzero formal polynomial of degree at most 9, has the form p+cX9 for a zero polynomial or nonzero formal polynomial p of degree at most 8 and a scalar c; step 3.1 and linearity therefore prove exactness through degree 9.

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Romberg cancels a proved finite even-power expansion

Statement

Let a<b, let m≥1 and L≥m be integers, and let f∈C2m+2([a,b]). Write hℓ=(b−a)2−ℓ and let (Rℓ,j) be the table from Finite Romberg table from nested composite trapezoid levels. If the trapezoid expansion has coefficients c1,…,cm as in Even-power expansion of trapezoid levels with a uniform remainder, then the recurrence cancels all first m even powers and there is a finite Cm,L such that ∣RL,m−∫abf(x) dx∣≤Cm,LhL2m+2. No error-order conclusion is made from a ratio of observed consecutive trapezoid errors.

Facts & Assumptions

Given: a<b, integers m≥1, L≥m, f∈C2m+2([a,b]), and its finite Romberg table.

[F1]

Finite Romberg table from nested composite trapezoid levels: Rℓ,0=T2ℓ(f) and Rℓ,j=Rℓ,j−1+(Rℓ,j−1−Rℓ−1,j−1)/(4j−1) for 1≤j≤ℓ.

[F2]

Even-power expansion of trapezoid levels with a uniform remainder: for every level ℓ, Rℓ,0=I+∑r=1mcrhℓ2r+εℓ,0 with ∣εℓ,0∣≤C0hℓ2m+2, where I=∫abf and C0 is independent of ℓ.

[F3]

Composite trapezoidal rule: hℓ−1=2hℓ and the initial table entries are the nested trapezoid values on these levels.

[F4]

The principle of mathematical induction: induction on the finite extrapolation-column index is valid.

Proof

1.1givenF1F2baseinduction

By [F1]–[F2], column zero has the expansion Rℓ,0=I+∑r=1mcrhℓ2r+εℓ,0 uniformly for ℓ≥0. We prove by induction on the column j that, for every ℓ≥j, Rℓ,j=I+∑r=j+1mdr,jhℓ2r+εℓ,j,∣εℓ,j∣≤Cjhℓ2m+2, with coefficients dr,j independent of ℓ. For j=0, this is [F2] with dr,0=cr.

2.1givenF1F3step 1.1IHinductionalgebra

Assume the induction hypothesis for column j−1 with 1≤j≤m. Then Rℓ,j−1 has terms indexed r=j,…,m and Rℓ−1,j−1 has the same coefficients with hℓ−12r=4rhℓ2r by [F3]. Substitution in [F1] multiplies the coefficient of hℓ2r by 1+1−4r4j−1=4j−4r4j−1. For r=j this factor is zero, so the jth error power cancels; the remaining powers have indices r>j.

3.1givenF1step 2.1IHalgebra

The new remainder is εℓ,j=εℓ,j−1+(εℓ,j−1−εℓ−1,j−1)/(4j−1). Thus ∣εℓ,j∣≤Cj−1(1+1+22m+24j−1)hℓ2m+2, so a finite level-independent Cj exists under the inductive remainder bound.

4.1F2F4step 1.1step 2.1step 3.1inductiondischarge-inductionalgebra∎

Step 1.1 supplies the base column, and steps 2.1 and 3.1 advance the expansion and its uniform remainder from column j−1 to column j. By [F4], induction therefore gives the asserted expansion at every column through m. At j=m the coefficient sum is empty, so RL,m=I+εL,m, and step 3.1 bounds its remainder by CmhL2m+2.

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Peano remainder bound for the eight-panel rule

Statement

If f∈C10[−1,1], then the rule Q8 of Closed Newton–Cotes rule on eight equal panels satisfies ∣∫−11f(x) dx−Q8(f)∣≤(∫−11∣K10(t)∣ dt)∥f(10)∥∞, where K10(t)=(1−t)1010!−∑j=08wj(xj−t)+99!,xj=−1+j/4, and u+:=max⁡(u,0).

Facts & Assumptions

Given: A function f∈C10[−1,1] and the nine-node quadrature Q8.

[A1]

f has continuous derivatives through order 10 on [−1,1].

[F1]

Closed Newton–Cotes rule on eight equal panels: Q8(f)=∑j=08wjf(xj) on the nodes xj=−1+j/4.

[F2]

Exact weights and degree-nine exactness of the eight-panel rule: Q8 is exact for the zero polynomial and every nonzero polynomial of degree at most 9.

[F3]

Peano kernel of a finite quadrature functional: for r=10, K10(t)=L(g10,t), where g10,t(x)=(x−t)+9/9! and L(u)=∫−11u−Q8(u).

[F4]

Peano representation and L1 error bound: exactness through degree 9 and f∈C10 imply ∣L(f)∣≤(∫−11∣K10(t)∣ dt)∥f(10)∥∞.

Proof

1.1givenF1F3algebra

By [F1] and [F3], K10(t)=∫−11(x−t)+9/9! dx−∑j=08wj(xj−t)+9/9!. For −1≤t≤1, the integral is ∫t1(x−t)9/9! dx=(1−t)10/10!, by the power antiderivative and endpoint evaluation. This gives the displayed kernel.

2.1givenA1F2F4step 1.1∎

The exactness condition of [F4] holds by [F2], and its smoothness condition is [A1]. Applying [F4] to this Q8 and substituting the kernel from step 1.1 proves the bound.

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DCT-I coefficients of the Chebyshev–Lobatto interpolant

Statement

Let n∈N with n≥1, and let g∈C[−1,1]. Put yj=g(cos⁡(jπ/n)) for 0≤j≤n, and let Xk be the DCT-I of Type-I cosine transform for Chebyshev–Lobatto data. Define a0:=X02n,an:=Xn2n,ak:=Xkn(1≤k<n). Then pn(t):=∑k=0nakTk(t) is the unique polynomial of degree at most n interpolating g at the Chebyshev–Lobatto nodes cos⁡(jπ/n); in particular pn=InLg in the convention of Best uniform polynomial error and Chebyshev interpolation operators.

Facts & Assumptions

Given: An integer n≥1 and a continuous g:[−1,1]→R.

[F1]

Type-I cosine transform for Chebyshev–Lobatto data: Xk=y0+(−1)kyn+2∑j=1n−1yjcos⁡(jkπ/n).

[F2]

Weighted cosine orthogonality on the Lobatto grid: the half-endpoint weighted cosine products sum to n at matching endpoint frequencies, n/2 at matching interior frequencies, and 0 at distinct frequencies.

[F3]

The Chebyshev recurrence agrees with cosine multiple angles: Tk(cos⁡θ)=cos⁡(kθ) for every real θ.

[F4]

Best uniform polynomial error and Chebyshev interpolation operators: the Lobatto nodes are xjL,n=cos⁡(jπ/n) and InLg is the Lagrange-cardinal interpolant at those nodes.

[F5]

Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum: InLg is a polynomial of degree at most n and has values InLg(xjL,n)=g(xjL,n) at every node.

[F6]

Cardinal-polynomial interpolation and Existence and uniqueness of the cardinal-polynomial interpolant: at n+1 strictly increasing distinct real nodes, the degree-at-most-n Lagrange interpolant is the unique polynomial with the given node values.

[F7]

The Chebyshev recurrence gives degree at most its index: each Tk has degree at most k.

Proof

1.1givenF1F3algebra

Use the half-endpoint sum notation ∑j=0n′vj:=v0/2+∑j=1n−1vj+vn/2. By [F1], Xk=2∑j=0n′yjcos⁡(jkπ/n). The recurrence-angle identity [F3] gives Tk(cos⁡(jπ/n))=cos⁡(jkπ/n), so the definition of ak then gives pn(cos⁡(jπ/n))=1n∑k=0n′Xkcos⁡(jkπ/n).

2.1givenF1F2step 1.1algebra

Substitute the expression for Xk from step 1.1 into the displayed reconstruction sum and interchange the two finite sums. By [F2], every frequency m≠j contributes zero. If j is an endpoint, its outer half-weight and the diagonal sum n contribute (1/n)⋅2⋅(1/2)n yj=yj; if j is interior, the diagonal sum n/2 contributes (1/n)⋅2⋅(n/2)yj=yj. Thus pn(cos⁡(jπ/n))=yj for each j.

3.1givenF4F5F6F7step 2.1algebra∎

By [F7], pn has degree at most n. The interpolant InLg has the same degree bound and node values by [F4]–[F5]. Reorder the common node set increasingly; it remains n+1 distinct nodes because cos⁡ is strictly decreasing on [0,π]. Apply the uniqueness statement [F6] to the two polynomials with the same values from step 2.1. Hence pn=InLg.

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Clenshaw–Curtis quadrature with the midpoint zero-degree case

Definition

Let a<b, n∈N, m=(a+b)/2, d=(b−a)/2, and f∈C[a,b]. For n≥1, set g(t):=f(m+dt) on [−1,1], let yj=g(cos⁡(jπ/n)), and form a0,…,an from the DCT-I coefficients in DCT-I coefficients of the Chebyshev–Lobatto interpolant. With the Chebyshev moments μk of Exact integrals of Chebyshev polynomials, define Qn,[a,b](f):=d∑k=0nakμk. For n=0, define instead Q0,[a,b](f):=(b−a)f(m), the midpoint rule. This is the explicit endpoint and zero-degree convention for Clenshaw–Curtis quadrature. The sums and coefficients are exact real quantities; finite-precision evaluation is a separate algorithm.

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Polynomial exactness and best-approximation error for Clenshaw–Curtis

Statement

Let a<b, n∈N, and f∈C[a,b]. Put m=(a+b)/2, d=(b−a)/2, and g(t)=f(m+dt) on [−1,1]. Let En(g) and ΛnL be the best uniform polynomial error and the Chebyshev–Lobatto interpolation norm from Best uniform polynomial error and Chebyshev interpolation operators. Interpreting polynomial inputs by their evaluation functions, the rule Qn,[a,b] of Clenshaw–Curtis quadrature with the midpoint zero-degree case is exact for the zero polynomial and every nonzero formal polynomial of degree at most n, and ∣∫abf(x) dx−Qn,[a,b](f)∣≤(b−a)(1+ΛnL)En(g). Because affine pullback preserves the supremum norm and polynomial degree, En(g) is also the best degree-n uniform error of f on [a,b].

Facts & Assumptions

Given: a<b, n∈N, and f∈C[a,b].

[A1]

n∈N, and the normalized function is g(t)=f(m+dt) with m=(a+b)/2 and d=(b−a)/2>0.

[F1]

Clenshaw–Curtis quadrature with the midpoint zero-degree case: for n≥1, Qn,[a,b](f)=d∑k=0nakμk with the DCT-I coefficients and Chebyshev moments; at n=0 it is (b−a)f(m).

[F2]

DCT-I coefficients of the Chebyshev–Lobatto interpolant: for n≥1, ∑k=0nakTk=InLg.

[F3]

Exact integrals of Chebyshev polynomials: μk=∫−11Tk(t) dt for every k≥0.

[F4]

Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum: InLp=p for every p∈Pn, including n=0.

[F5]

Chebyshev interpolation error is controlled by best polynomial error: ∥g−InLg∥∞≤(1+ΛnL)En(g) for every n≥0.

[F6]

Best uniform polynomial error and Chebyshev interpolation operators: En(g)=inf⁡p∈Pn∥g−p∥∞, ΛnL is the operator norm of InL, and I0Lg=g(0).

[F7]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials and The binomial theorem in R: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k: if q(X)=∑k=0nckXk and α,β∈R, then q(α+βt)=∑k=0nck∑j=0kι(kj)αk−jβjtj. Regrouping these finite sums gives a coefficient list with no index above n; since d=(b−a)/2>0, this applies to both affine maps t↦m+dt and x↦(x−m)/d.

[F9]

Monotone change of variable for Riemann-integrable functions: for the increasing affine bijection t↦m+dt, ∫abP(x) dx=d∫−11P(m+dt) dt.

Proof

1.1givenF1F2F3F9F10algebra

For n≥1, let pn(t)=∑k=0nakTk(t). By [F1]–[F3] and [F10], Qn,[a,b](f)=d∑kak∫−11Tk(t) dt=d∫−11pn(t) dt. The affine substitution [F9] makes this ∫abP(x) dx for P(x)=pn((x−m)/d), and [F2] identifies P with the interpolant pullback. For n=0, [F1] is the integral of the constant polynomial P(x)=f(m) by [F9]. Thus Qn,[a,b](f)=∫abP(x) dx in every case.

2.1givenF4F7step 1.1algebra

Let q be the zero polynomial or a nonzero formal polynomial of degree at most n, interpreted by its evaluation function on [a,b], and put q~(t)=q(m+dt). By [F7], q~∈Pn. The interpolation reproduction in [F4] gives InLq~=q~, so step 1.1 yields Qn,[a,b](q)=∫abq(x) dx. This proves polynomial exactness.

2.2givenF5F8F10step 1.1algebra

Set P(x)=InLg((x−m)/d) as in step 1.1. The pullback bijection gives ∥f−P∥∞,[a,b]=∥g−InLg∥∞,[−1,1]. By [F8], ∣∫ab(f−P)∣≤(b−a)∥f−P∥∞. Combining this with Q(f)=∫P from step 1.1 and [F5] proves the displayed bound.

3.1givenF6F7algebra∎

For each p∈Pn on [−1,1], the pullback x↦p((x−m)/d) is in Pn on [a,b] by [F7] and has the same supremum error against f as p has against g. Conversely, every degree-at-most-n polynomial on [a,b] pulls back to one on [−1,1]. The two infima defining best uniform error are therefore equal.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Computed Clenshaw–Curtis from an even-extension DFT

Definition

Input real numbers a<b, a natural n≥0, sampled values y0,…,yn, and the phase/FFT controls for Arbitrary-length mixed-radix and Bluestein DFT: a phase tolerance 0<τ≤1/4 (required to be exact rational when n≥1), integer dyadic exponent limits emin⁡≤emax⁡, and integers p0≥2, pmax⁡≥p0, J≥1, and dmax⁡≥0. Set m=(a+b)/2 and d=(b−a)/2. For an underlying f∈C[a,b], admissible samples satisfy y0=f(m) when n=0 and yj=f(m+dcos⁡(jπ/n)) for 0≤j≤n when n≥1, using the canonical real images of The canonical natural ι(n)=n⋅1F of a field. When n≥1, also require every yj to lie in the representable real set F of the same floating-point model instance used by the called DFT; this ensures its complex input coordinates satisfy that procedure's input condition. No representability condition is needed for n=0, when no DFT call is made. The output is an approximate real quadrature value or the first status returned by the called DFT procedure.

  1. If n=0, return (b−a)y0, the midpoint convention of Clenshaw–Curtis quadrature with the midpoint zero-degree case.
  2. If n≥1, form z∈C2n by zj:=yj+0i for 0≤j≤n and z2n−j:=yj+0i for 1≤j<n. This is the even extension (y0,…,yn,yn−1,…,y1), with real samples embedded as complex numbers with zero imaginary part. Call Arbitrary-length mixed-radix and Bluestein DFT on z with length 2n and the same controls (τ,emin⁡,emax⁡,p0,pmax⁡,J,dmax⁡). If it returns a status instead of a vector, return that status.
  3. Otherwise, take the real parts Xk of output entries 0≤k≤n, form the DCT-I coefficients a0=X0/(2n), an=Xn/(2n), and ak=Xk/n for 1≤k<n as in Type-I cosine transform for Chebyshev–Lobatto data, and return d∑k=0nakμk with the exact moments μk in Clenshaw–Curtis quadrature with the midpoint zero-degree case.

The postprocessing is exact real arithmetic applied to the approximate DFT output. The procedure makes no end-to-end floating-point error or stability claim; it only propagates the DFT routine's declared failure statuses.

5 · Examples, counterexamples and false statements

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