CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)

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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.

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Distinct pure frequencies differing by a reciprocal-lattice shift have identical samples

Statement refuted

Distinct pure frequencies produce distinct sample sequences on the lattice $h Z$ : if $f_{1} (x) = e^{2 π i ξ x}$ and $f_{2} (x) = e^{2 π i η x}$ with $η \neq ξ$ , then $f_{1} (h k) \neq f_{2} (h k)$ for some $k \in Z$ .

Facts & Assumptions

Given: Countable Choice, a sampling spacing $h > 0$ , a frequency $ξ \in R$ and a nonzero integer $m$ , with $f_{1} (x) = e^{2 π i ξ x}$ and $f_{2} (x) = e^{2 π i (ξ + m / h) x}$ for $x \in R$ (The complex exponential by its power series, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

$e^{z + w} = e^{z} e^{w}$ for all complex $z, w$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[F2]

$e^{z} = 1$ if and only if $z \in 2 π i Z$ , and $e^{i π} + 1 = 0$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Counterexample

1.1F1F2givenalgebra

For every $k \in Z$ the addition law [F1] gives $f_{2} (h k) = e^{2 π i ξ h k} e^{2 π i (m / h) h k} = f_{1} (h k) e^{2 π i m k}$ , and $e^{2 π i m k} = 1$ because $m k \in Z$ and $2 π i m k \in 2 π i Z$ [F2]. Hence $f_{2}$ and $f_{1}$ have identical samples on $h Z$ .

2.1step 1.1F1F2givenalgebra∎

The two functions are distinct: at $x = h / (2 m)$ , which is a real number because $m \neq 0$ , one has $f_{2} (x) f_{1} (x)^{- 1} = e^{2 π i (m / h) (h / (2 m))} = e^{i π} = - 1$ by [F1] and [F2], and $- 1 \neq 1$ , so $f_{2} \neq f_{1}$ . Together with step 1.1 this refutes the displayed statement: the frequencies $ξ$ and $ξ + m / h$ are distinct, yet every procedure reading only the samples on $h Z$ sees the same data. The witness consists of pure frequencies of constant modulus one, which are bounded but not square-integrable on $R$ ; it therefore does not contradict the $L^{2}$ reconstruction theorem of the A page.

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Sources

Richard S. Laugesen, Harmonic Analysis Lecture Notes (arXiv:0903.3845) (standard reference, not scraped)
Lior Silberman, Fourier series and the Poisson summation formula (Math 604/613 notes, UBC) (standard reference, not scraped)