Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge 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.

A conditional binomial error bound for Parareal

Statement

Let xn+1=Fn(xn) be the sequential fine trajectory and let Unk be the exact-arithmetic table from Finite-budget Parareal correction sweeps. Let D⊆X contain every xn and every iterate used in the requested sweeps. Suppose constants a,b,M≥0 satisfy, for every 0≤n<N and u,v∈D, ∥Gn(u)−Gn(v)∥≤a∥u−v∥,∥(Fn−Gn)(u)−(Fn−Gn)(v)∥≤b∥u−v∥, and ∥Un0−xn∥≤M for all 0≤n≤N. Use (nk) for the number of k-element subsets of an n-element set, set it to 0 for k>n, and put A=max⁡{1,a}. Then for every node 0≤n≤N and completed sweep k≥0, ∥Unk−xn∥≤M(nk)Anbk, with b0=1. The hypotheses include all comparison points; invariance of D under Fn,Gn alone does not ensure that corrected iterates stay in D.

For a fixed horizon T=NH, suppose a≤1+CaH, b≤CbHq+1 for constants Ca,Cb≥0 and integer q≥1, and one M works uniformly as N varies. Then, for each fixed k≤N, ∥UNk−xN∥≤M(CbT)kexp⁡(CaT)Hqk. Thus the exact-solve error is O(Hqk) at each fixed iteration k under these uniform assumptions. Neither estimate includes finite-solver error; that effect requires a separate update-defect analysis.

Facts & Assumptions

Given: The exact Parareal recurrence, its sequential fine trajectory, a common comparison set, and the stated Lipschitz and initial-error bounds.

[F1]

Fine and coarse propagators on a common macro grid: the sequential fine trajectory is xn+1=Fn(xn) from x0.

[F2]

Finite-budget Parareal correction sweeps: U0k=x0 and Un+1k+1=Gn(Unk+1)+Fn(Unk)−Gn(Unk).

[F3]

A norm on a real vector space, the induced metric, and the dictionary with the metric axioms: the norm satisfies ∥u+v∥≤∥u∥+∥v∥.

[F5]

The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y): exp⁡(u+v)=exp⁡(u)exp⁡(v) for real u,v.

[F6]

The real exponential function and the number e by a power series: for every real x, exp⁡(x) is defined by the convergent series ∑n=0∞xn/n!.

[A1]

All fine and coarse evaluations are exact and defined, all comparison points lie in D, the two Lipschitz inequalities hold on D, and the coarse initial-row errors are at most M.

[A2]

In the fixed-horizon branch, H=T/N, T>0, a≤1+CaH, b≤CbHq+1, q≥1, and the envelope M is independent of N.

Proof

1.1givenF1F2F3A1algebra

Put enk=Unk−xn. Subtract xn+1=Fn(xn) from the recurrence in [F2] and add and subtract Gn(xn); this gives en+1k+1=Gn(Unk+1)−Gn(xn)+(Fn−Gn)(Unk)−(Fn−Gn)(xn). By [A1] and the triangle inequality in [F3], En+1k+1≤aEnk+1+bEnk, where Enk=∥enk∥.

1.2F2A1induction

At node 0, E0k=0 for every k by [F2]. At sweep k=0, [A1] gives En0≤M≤MAn. These are the initial cases for induction on the node, simultaneously over all completed sweeps.

2.1step 1.1step 1.2A1inductionalgebra

Assume the bound holds at node n. For the level-zero row at node n+1, [A1] gives En+10≤M≤MAn+1. For each k≥0, [step 1.1] gives En+1k+1≤MAnbk+1(a(nk+1)+(nk))≤MAn+1bk+1((nk+1)+(nk)). The last inequality uses a≤A and 1≤A. Partitioning the (k+1)-element subsets of an (n+1)-element set according to whether they contain its final element gives (nk+1)+(nk)=(n+1k+1), including the zero convention. Thus the bound holds at node n+1, and induction proves the first estimate.

3.1A2F4F5F6step 2.1constructalgebra∎

In the fixed-horizon branch, A≤1+CaH≤exp⁡(CaH) by [A2], [F4], and the definition of exp⁡ in [F6]. Repeated use of [F5] gives AN≤exp⁡(CaNH)=exp⁡(CaT). Each k-element subset of an N-element set has a unique increasing k-tuple, so (Nk)≤Nk. Substituting these bounds and b≤CbHq+1 into the first estimate gives ∥UNk−xN∥≤MNkexp⁡(CaT)(CbHq+1)k=M(CbT)kexp⁡(CaT)Hqk. For fixed T,k and uniform M, this is O(Hqk).

Depends on

Used by

Dependency tree · two levels

38 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