LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-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.

Explicit tail for the periodized Gaussian image sum

Statement

For $τ > 0$ , integer $L \geq 1$ , and $x \in [0, 1]$ , put $ϕ_{τ, L} (x) = \sum_{n = - L}^{L} \exp ⁣ (- \frac{π (x + n)^{2}}{τ}), R_{τ, L} = \frac{2 \exp (- π L^{2} / τ)}{1 - \exp (- π L / τ)} .$ Then $0 \leq ϕ_{τ} (x) - ϕ_{τ, L} (x) \leq R_{τ, L}$ , and the denominator defining $R_{τ, L}$ is positive.

Facts & Assumptions

Given: $τ > 0$ , $L \geq 1$ , and $x \in [0, 1]$ .

[F1]

Periodized Gaussian gridding kernel and Local uniform convergence of the periodized Gaussian: $ϕ_{τ} (x)$ is the convergent sum over all integer images.

[F2]

Real elementary functions and their domains: $π = 4 \arctan (1) \geq 2 > 0$ , because $(1 + t^{2})^{- 1} \geq 1 / 2$ on $[0, 1]$ ; the exponential power series gives $\exp (0) = 1$ .

[F3]

The exponential function is strictly increasing and The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $0 < \exp (- π L / τ) < 1$ .

[F4]

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges: for $0 < q < 1$ , $\sum_{h \geq 0} q^{h} = 1 / (1 - q)$ and $\sum_{h \geq 1} q^{h} = q / (1 - q)$ .

Proof

1.1givenF1F2F3F4algebra

Let $q = \exp (- π L / τ) \in (0, 1)$ by [F2]–[F3]. For positive omitted indices write $n = L + h$ with $h \geq 1$ . Since $0 \leq x \leq 1$ , $(x + n)^{2} \geq n^{2} = (L + h)^{2} \geq L^{2} + L h$ , so their sum is at most $\exp (- π L^{2} / τ) \sum_{h \geq 1} q^{h}$ . By [F4] this is at most $\exp (- π L^{2} / τ) / (1 - q)$ .

1.2givenF1F2F3F4algebra

For negative omitted indices write $n = - (L + 1 + h)$ with $h \geq 0$ . Then $∣ x + n ∣ = L + 1 + h - x \geq L + h$ , and $(L + h)^{2} \geq L^{2} + L h$ . Their sum is at most $\exp (- π L^{2} / τ) \sum_{h \geq 0} q^{h} = \exp (- π L^{2} / τ) / (1 - q)$ by [F4].

2.1step 1.1step 1.2F1F3algebra∎

The two omitted tails are nonnegative and exhaust the indices outside $[- L, L]$ . Adding steps 1.1 and 1.2 gives the claimed remainder $R_{τ, L}$ ; [F3] also proves its denominator is positive.

Depends on

Used by

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Sources

Jens Keiner, Stefan Kunis, and Antje Vollrath, The Nonequispaced FFT and its Applications, Part I, slides 44–50 (standard reference, not scraped)
Leslie Greengard and June-Yub Lee, Accelerating the Nonuniform Fast Fourier Transform (standard reference, not scraped)