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.

PCHIP has third-order value accuracy on uniform nonstationary grids

Statement

Let a<b, let f be C3 on an open interval containing [a,b], and suppose ∣f′(x)∣≥c>0 on [a,b]. For each integer N≥2, put h=(b−a)/N>0 and xj=a+jh, j=0,…,N, and let Ph be the PCHIP interpolant to the values f(xj). There are constants C,h0>0, depending only on f, c, and [a,b], such that for every such grid with h<h0, max⁡x∈[a,b]∣Ph(x)−f(x)∣≤Ch3. The assertion is limited to uniform grids and intervals on which the derivative stays bounded away from zero; it makes no third-order claim at an extremum.

Facts & Assumptions

Given: Reals a<b, the stated C3 function and derivative bound on [a,b], an integer N≥2, and the uniform grid with h=(b−a)/N>0.

[F1]

PCHIP slope selection: PCHIP uses the prescribed nodal values and selected slopes in the cubic Hermite formula.

[F2]

PCHIP slope selection has no zero denominator: those slopes and pieces are finite and well-defined.

[F3]

Piecewise cubic Hermite interpolant: on a cell the Hermite formula is the linear combination of H00,H01,H10,H11 with value and slope data.

[F4]

Interpolation, C1 matching, and cubic reproduction for Hermite pieces: each Hermite piece interpolates its endpoint values and slopes; the resulting interpolant is C1 on the closed interval (with one-sided endpoint derivatives); and each piece reproduces any cubic polynomial from its endpoint values and derivatives.

[F5]

PCHIP slopes are second-order accurate on uniform nonstationary grids: for all sufficiently fine uniform grids, every PCHIP nodal slope differs from f′(xj) by at most Csh2 for a constant Cs independent of the cell.

[F6]

Taylor's theorem with an integral remainder: the degree-two Taylor polynomial of a C3 function has remainder at most Mh3/6 on a cell of width h when ∣f′′′∣≤M there. Applied to f′ with degree one, the same theorem bounds its remainder by Mh2/2.

[F7]

Continuity and derivative on a real interval: C3 on an open interval means the derivatives through order three exist and are continuous there.

[F8]

A continuous real function on a compact subset of R is bounded: a continuous real function on a compact subset of R is bounded there.

[F9]

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

[F10]

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 and minimum.

[F11]

Real and complex modulus product and triangle inequalities: the real triangle inequality; applying it twice gives ∣∣u∣−∣v∣∣≤∣u−v∣.

Proof

1.1F6F7F8F9given

Since a<b and h=(b−a)/N>0, the grid is admissible and [a,b] is compact by [F9]. By [F7] and [F8], choose M≥0 with ∣f′′′∣≤M on [a,b]. Fix a cell [xi,xi+h] and let T2 be the degree-two Taylor polynomial of f at xi. By [F6], the residual r=f−T2 satisfies r(xi)=r′(xi)=0 and, for every x∈[xi,xi+h], ∣r(x)∣≤Mh3/6; at the right endpoint it also satisfies ∣r(xi+h)∣≤Mh3/6 and ∣r′(xi+h)∣≤Mh2/2. The derivative bound is [F6] applied to f′ with degree one.

1.2F1F2F3F5givenalgebra

Let H be the exact-derivative Hermite piece for f on this cell. The PCHIP piece Ph and H have identical endpoint values. By [F3], their difference is h(mi−f′(xi))H10(t)+h(mi+1−f′(xi+h))H11(t). Since [F2] guarantees these PCHIP slopes are defined and ∣H10(t)∣=t(1−t)2≤1, ∣H11(t)∣≤1, [F5] gives ∣Ph(x)−H(x)∣≤2Csh3 for every x in the cell.

2.1F3F4step 1.1algebra

For this exact-derivative Hermite piece H, linearity of the formula [F3] and reproduction [F4] give H=T2+Hr, where Hr(t)=r(xi+h)H01(t)+hr′(xi+h)H11(t). On 0≤t≤1, ∣H01(t)∣=t2(3−2t)≤1 and ∣H11(t)∣=t2(1−t)≤1, so ∣Hr(t)∣≤(1/6+1/2)Mh3=2Mh3/3. Together with ∣r(x)∣≤Mh3/6 from step 1.1, this yields ∣H(x)−f(x)∣≤5Mh3/6 for every x in the cell.

3.1step 2.1step 1.2F1F4F7F9F10F11algebra∎

Combining steps 2.1 and 1.2 gives ∣Ph(x)−f(x)∣≤(5M/6+2Cs)h3 for every x∈[a,b], since the cells cover the interval. The constants do not depend on i or N; put C=5M/6+2Cs. To justify the maximum in the statement, admissibility of the grid for [F1] gives x0<⋯<xN, so a<b. By [F4], Ph is C1 and hence continuous on [a,b]; by [F7], f is continuous there. Set E(x)=∣Ph(x)−f(x)∣. For x,z∈[a,b], applying the triangle inequality twice gives ∣∣Ph(x)−f(x)∣−∣Ph(z)−f(z)∣∣≤∣(Ph(x)−f(x))−(Ph(z)−f(z))∣≤∣Ph(x)−Ph(z)∣+∣f(x)−f(z)∣, so E is continuous on [a,b]. By [F9], [a,b] is compact, and it is nonempty; [F10] therefore gives a point where E attains its maximum. The pointwise bound just proved bounds that maximum by Ch3.

Depends on

Used by

Dependency tree · two levels

48 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