LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pass

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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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A monotone-range Taylor remainder for small Boys arguments

Statement

Let $m, N \in N_{0}$ , $T \geq 0$ , and $N + 2 \geq T$ . Then $∣ F_{m} (T) - \sum_{k = 0}^{N} \frac{(- T)^{k}}{k! (2 m + 2 k + 1)} ∣ \leq \frac{T^{N + 1}}{(N + 1)! (2 m + 2 N + 3)} .$ In particular, this is a valid finite Taylor certificate whenever $0 \leq T \leq 1$ .

Facts & Assumptions

Given: $F_{m}$ and its absolutely convergent series as in Boys functions and lower incomplete-gamma notation and Boys integral, power series, recurrence, and incomplete-gamma representation, with $m, N \geq 0$ , $T \geq 0$ , and $N + 2 \geq T$ .

[F1]

Boys integral, power series, recurrence, and incomplete-gamma representation: $F_{m} (T) = \sum_{k \geq 0} (- T)^{k} / [k! (2 m + 2 k + 1)]$ .

Proof

1.1givenF1algebra

Set $a_{k} = T^{k} / [k! (2 m + 2 k + 1)]$ . For $T > 0$ , $\frac{a_{k + 1}}{a_{k}} = \frac{T}{k + 1} \frac{2 m + 2 k + 1}{2 m + 2 k + 3} \leq \frac{T}{k + 1} .$ For every $k \geq N + 1$ , $k + 1 \geq N + 2 \geq T$ , and the second factor is strictly less than one, so the positive terms decrease from the first omitted term onward.

2.1givenF1step 1.1algebra∎

The tail in [F1] after term $N$ is alternating with decreasing positive magnitudes by step 1.1; pairing consecutive terms shows its absolute value is at most $a_{N + 1}$ . That value is the displayed bound. If $T = 0$ , the tail and bound are both zero.

Depends on

Used by

Dependency tree · two levels

13 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

Trygve Helgaker, Molecular Integral Evaluation, 10th Sostrup Summer School lecture notes (2008) (standard reference, not scraped)
L. E. McMurchie and E. R. Davidson, One and Two Electron Integrals over Cartesian Gaussian Functions, UCRL-13791 (1976) (standard reference, not scraped)