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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.

Opposite real and reciprocal cutoff trends in the Ewald parameter

Example

Fix the unit cell, a neutral dipole with nonzero total absolute charge $Q$ , and real and reciprocal cutoffs $R, K_{c} \geq 1$ . The real-space potential-tail bound decreases strictly with $α$ , while the reciprocal-space potential-tail bound increases strictly with $α$ . Thus choosing $α$ at fixed cutoffs and target tolerance balances two opposing certified contributions.

Facts & Assumptions

Given: A fixed neutral configuration with $Q > 0$ , fixed cutoffs $R, K_{c} \geq 1$ , and a positive parameter $α$ .

[A1]

The Ewald formulas use Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The real and reciprocal potential-tail bounds are $B_{R}^{U} = Q C_{R} (\sqrt{π} α)^{- 1} T_{3} (α^{2}, ⌊ R ⌋)$ and $B_{K}^{U} = 4 π Q C_{K} V^{- 1} T_{3} ((4 α^{2})^{- 1}, ⌊ K_{c} ⌋)$ (Explicit real and reciprocal Ewald potential, energy, and force tail bounds).

[F2]

$T_{3} (β, M) = \sum_{m = M}^{\infty} (m + 1)^{3} e^{- β m^{2}}$ is finite for $β > 0$ , and each term is positive (Uniform lattice-shell counts and explicit Gaussian tail controllers).

Verification

technique · direct

1.1givenA1F1F2construct

For $α > 0$ set $R (α) = \frac{Q C_{R}}{\sqrt{π} α} \sum_{m = M_{R}}^{\infty} (m + 1)^{3} e^{- α^{2} m^{2}}$ and $K (α) = \frac{4 π Q C_{K}}{V} \sum_{m = M_{K}}^{\infty} (m + 1)^{3} e^{- m^{2} / (4 α^{2})}$ , where $M_{R} = ⌊ R ⌋$ and $M_{K} = ⌊ K_{c} ⌋$ . These are exactly the two bounds in [F1], and they are finite by [F2].

1.2F1F2algebra

If $0 < α_{1} < α_{2}$ , then for every $m \geq M_{R} \geq 1$ , $e^{- α_{2}^{2} m^{2}} / α_{2} < e^{- α_{1}^{2} m^{2}} / α_{1}$ because both positive factors strictly decrease. The positive convergent sum therefore gives $R (α_{2}) < R (α_{1})$ .

2.1F1F2algebra∎

For every $m \geq M_{K} \geq 1$ , $e^{- m^{2} / (4 α_{2}^{2})} > e^{- m^{2} / (4 α_{1}^{2})}$ , since $α_{2} > α_{1}$ makes the negative exponent less negative. Summing positive terms gives $K (α_{2}) > K (α_{1})$ . Hence a fixed-tolerance certificate must balance the real and reciprocal tails; the theorem does not infer an optimal parameter from FFT work or RMS data.

Depends on

Used by

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Dependency tree · two levels

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