CounterexampleConstruction: 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.

The absolute-value extension gives a zero complex-step quotient despite its real derivative

Statement refuted

For every open $U \subseteq C$ , every $f : U \to C$ , every open real interval $I$ with $x \in I \subseteq U \cap R$ on which $f$ is real-valued and differentiable at $x$ , and every $h > 0$ with $x + i h \in U$ , the quotient $Im f (x + i h) / h$ equals $(f ∣_{I})^{'} (x)$ .

Facts & Assumptions

Given: $U = C$ , $f (z) = ∣ z ∣$ , $x = 1$ , $I = (0, 2)$ , and any $h > 0$ .

[F1]

Real and imaginary parts, complex conjugation, and modulus: if $z = a + b i$ , then $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ and $Im z = b$ .

[F2]

The complex-step derivative estimator and its component-error model: the notation $C_{h} f (x) = Im f (x + i h) / h$ is defined there for holomorphic $f$ that is real-valued on a real interval around $x$ ; the counterexample tests the same quotient formula without holomorphy.

[F3]

Complex-step differentiation is second-order for holomorphic functions real on the real axis: its derivative-error conclusion assumes $f$ is holomorphic near the vertical segment and real on a real interval around $x$ .

Counterexample

Given: The function, point and step in the Facts & Assumptions.

1.1givenF1algebra

On a real neighborhood of $1$ , $f (t) = ∣ t ∣ = t$ , so the ordinary derivative of the real restriction at $1$ is $1$ .

1.2givenF1algebra

By [F1], $f (1 + i h) = \sqrt{1 + h^{2}}$ is real. Thus the quotient in the Statement is $0$ for every $h > 0$ , which is not the real derivative $1$ .

2.1givenF1F2F3step 1.2algebra∎

The complex difference quotient along real increments $s$ is $(∣ 1 + s ∣ - 1) / s = 1$ for $∣ s ∣ < 1$ and $s \neq 0$ . Along imaginary increments it is $(\sqrt{1 + s^{2}} - 1) / (i s) = - i s / (\sqrt{1 + s^{2}} + 1) \to 0$ . The limits differ, so $f$ is not complex differentiable at $1$ and the formally defined estimator in [F2] and guarantee in [F3] do not apply; step 1.2 instead refutes the unqualified quotient claim.

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Sources

Nicholas J. Higham, What Is the Complex Step Approximation? (standard reference, not scraped)
Mike Giles, Monte Carlo methods, Lecture 6: Complex Variable Trick (standard reference, not scraped)
R. Boudjemaa, M. G. Cox, A. B. Forbes, and P. M. Harris, Automatic Differentiation Techniques and their Application in Metrology, NPL Report CMSC 26/03 (standard reference, not scraped)