Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
How statement and proof provenance work

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

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • 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.

Composite Simpson error bound for a four-times continuously differentiable function

Statement

Let a<b, let n≥2 be an even integer, and put h=(b−a)/n. If f∈C4([a,b]), set M4=max⁡x∈[a,b]∣f(4)(x)∣. Then the value QS(f;a,b,n) of Composite Simpson rule on an even uniform grid satisfies

∣∫abf(x) dx−QS(f;a,b,n)∣≤(b−a)h4180M4.

The integral is the Riemann integral. This is an exact-arithmetic quadrature bound; evaluation and summation roundoff are not included.

Facts & Assumptions

Given: Reals a<b, an even integer n≥2, h=(b−a)/n, and f∈C4([a,b]).

[F1]

Higher derivatives and the classes Ck and C∞: C4([a,b]) makes f(4) continuous on the closed interval.

[F2]

Heine-Borel by bisection: every closed bounded interval [a,b] is compact: every closed bounded interval [a,b] is compact.

[F3]

Extreme value theorem: a continuous real function on a nonempty compact subset of R attains a greatest and a least value: a continuous real function on a nonempty compact subset of R attains its maximum.

[F4]

Composite Simpson rule on an even uniform grid: the composite value is the sum, over j<n/2, of h3(f(x2j)+4f(x2j+1)+f(x2j+2)).

[F5]

Peano kernel identity for one Simpson panel: on a panel of length L, the Simpson error is ∫Kf(4), where K≤0 and ∫∣K∣=L5/2880.

[F7]

Triangle inequality for finite sums: the absolute value of a finite sum is at most the sum of the absolute values.

[F8]

If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a): pointwise order between integrable functions on an increasing interval gives the same order between their integrals.

Proof

1.1givenF1F2F3F4F5F8F9construct

By [F1], f(4) is continuous. The interval is compact by [F2], so [F3] gives M4=max⁡[a,b]∣f(4)∣<∞. For each j=0,…,n/2−1, put u=x2j and L=2h. The local rule from [F4] is S(f;u,u+L); by [F5], its error is Ej=∫uu+LKu,u+Lf(4) and ∫uu+L∣Ku,u+L∣=(2h)5/2880. Since Ku,u+L≤0, its absolute value equals its negative, hence is integrable by [F5] and [F9]. The integrable functions satisfy −M4∣Ku,u+L∣≤Ku,u+Lf(4)≤M4∣Ku,u+L∣. Monotonicity [F8] and scalar linearity [F9] therefore give ∣Ej∣≤M4∫uu+L∣Ku,u+L∣=M4(2h)5/2880.

2.1step 1.1F4F6algebra

Additivity [F6] gives ∫abf as the sum of the local panel integrals. Summing the local Simpson expressions in [F4], each interior even-indexed value appears in two adjacent blocks and each odd-indexed value appears once; hence their sum is exactly QS(f;a,b,n). Therefore the global error is ∑j=0n/2−1Ej.

3.1step 1.1step 2.1F7algebra∎

The finite-sum triangle inequality [F7] yields ∣∑jEj∣≤(n/2)M4(2h)5/2880=nM4h5/180. Using nh=b−a gives the stated bound.

Depends on

Used by

Dependency tree · two levels

61 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources