ExampleConstruction: AI-generatedVerification: 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.

Freezing a varying slow waveform leaves a first-order coupling error

Example

Consider $x^{'} = 1$ , $z^{'} = x$ , $(x_{0}, z_{0}) = (0, 0)$ on $[0, T]$ , with $T > 0$ and $N \geq 1$ equal macro steps $H = T / N$ . The exact solution is $x (t) = t$ , $z (t) = t^{2} / 2$ . The frozen-slow method Frozen-slow explicit Euler with fast microsteps has $x_{n} = n H$ and $z_{N} = T^{2} / 2 - T H / 2$ , so its terminal $z$ error is $T H / 2$ . For any integer $m \geq 1$ , set $h = H / m$ and instead update each midpoint-sampled block by $z_{n, 0} = z_{n}$ and $z_{n, j + 1} = z_{n, j} + h (x_{n} + (j + \frac{1}{2}) h)$ for $0 \leq j < m$ . Then one macro update adds exactly $H x_{n} + H^{2} / 2$ , and this alternative gives the exact solution at every macro node for this model. This is a model-specific diagnostic, not a claim about any unverified symmetric composition.

Facts & Assumptions

Given: The model, initial values, $T > 0$ , and integer $N \geq 1$ in the Example.

[F1]

The frozen-slow algorithm sets $x^{+} = x + H f (x, z)$ and uses the original $x$ during its fast updates $z_{j + 1} = z_{j} + h g (x, z_{j})$ (Frozen-slow explicit Euler with fast microsteps).

[A1]

Here $f (x, z) = 1$ , $g (x, z) = x$ , and $H = T / N$ .

Proof

1.1F1A1inductionalgebra

Direct integration gives $x (t) = t$ and $z (t) = t^{2} / 2$ . By [F1] and [A1], induction on $n$ gives $x_{n} = n H$ ; because $g (x_{n}, z) = x_{n}$ is constant across a frozen block, $z_{n + 1} = z_{n} + H x_{n}$ . Starting with $z_{0} = 0$ , summation gives $z_{N} = H^{2} \sum_{n = 0}^{N - 1} n = H^{2} N (N - 1) / 2 = T^{2} / 2 - T H / 2$ , so the terminal error is $T H / 2 > 0$ .

2.1A1step 1.1algebra∎

Under midpoint sampling, the fast block increment is $\sum_{j = 0}^{m - 1} h (x_{n} + (j + \frac{1}{2}) h) = H x_{n} + h^{2} \sum_{j = 0}^{m - 1} (j + \frac{1}{2}) = H x_{n} + H^{2} / 2$ , since $m h = H$ and $\sum_{j = 0}^{m - 1} (j + \frac{1}{2}) = m^{2} / 2$ . If $z_{n} = (n H)^{2} / 2$ , this increment gives $z_{n + 1} = ((n + 1) H)^{2} / 2$ ; with $z_{0} = 0$ , induction proves exactness at every macro node.

Depends on

Frozen-slow explicit Euler with fast microsteps

Used by

Nothing in the library uses this result yet.

Dependency tree · one level

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