LemmaStatement: AI-generatedProof: AI-generated

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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every fixed positive squared logarithmic density satisfies the budget

Statement

Fix $c > 0$ and $K > 0$ . For $d \geq c (\log n)^{2}$ , set $α = (K + 1) / \sqrt{d}$ and $β = K / \sqrt{d} + \sqrt{12 \log n / d}$ . Eventually $0 < α \leq 1 / 24$ , $0 < β < 1 / 4$ , and $π / (96 α) > \log n / \log (1 / (2 β) - 1) + 1$ . Consequently the criterion inequality in the inverse-sine-budget lemma holds for every $0 < a \leq α$ and $0 < b \leq β$ .

Facts & Assumptions

Given: The data and hypotheses in the Statement, with $n$ tending to infinity through natural numbers and $d$ ranging over $d \geq c (\log n)^{2}$ .

Goal: Establish the conclusions uniformly over this range of $d$ .

[F1]

The elementary inverse-sine budget is An elementary sufficient budget for the synchronization criterion; eventual assertions and uniform comparisons have the meaning in Asymptotic quantifiers and uniform parameter ranges.

[F2]

The natural logarithm has domain $(0, \infty)$ (The natural logarithm as the inverse of the exponential function). It is strictly increasing, is onto $R$ , satisfies $\log 1 = 0$ , and obeys the product and quotient laws (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F3]

Every nonnegative real number has a unique nonnegative square root, whose square is that number (Square roots exist: a unique √a ≥ 0 with (√a)² = a; the positives are {x² : x ≠ 0}). Order comparisons and product identities for these roots follow by squaring nonnegative quantities in the real ordered field.

[F4]

The constant $π$ is positive (Pi as twice the smallest positive zero of cosine). Every real number has an integer part (Integer part: for every real x there is exactly one integer m with m ≤ x < m + 1), so a natural-number cutoff can exceed any specified positive real cutoff.

Proof

1.1givenF1F2F3F4algebra

By [F2], for any real $M$ there is $R > 0$ with $\log R = M$ . For natural $n > R$ , monotonicity gives $\log n > M$ ; [F4] supplies a natural-number cutoff beyond $R$ . Thus $ℓ = \log n$ eventually exceeds every real constant. Applying the same argument to $ℓ$ shows that $\log ℓ$ eventually exceeds every real constant as well. All logarithms below consequently have positive arguments for sufficiently large $n$ . Put $C = (K + \sqrt{12}) / \sqrt{c} > 0$ and $A = π \sqrt{c} / (96 (K + 1)) > 0$ . For $ℓ \geq 1$ and every allowed $d$ , squaring nonnegative quantities in [F3] gives $\sqrt{d} \geq \sqrt{c} ℓ$ , so $0 < α \leq (K + 1) / (\sqrt{c} ℓ)$ and $0 < β \leq K / (\sqrt{c} ℓ) + \sqrt{12} / (\sqrt{c} \sqrt{ℓ}) \leq C / \sqrt{ℓ}$ . These estimates are uniform in $d$ . In particular $α \leq 1 / 24$ when $ℓ \geq 24 (K + 1) / \sqrt{c}$ , and $β < 1 / 4$ when $\sqrt{ℓ} > 4 C$ .

2.1F2F3step 1.1algebra

For $\sqrt{ℓ} > 4 C$ , put $q = 1 / (2 β) - 1$ . Step 1.1 gives $q \geq \sqrt{ℓ} / (2 C) - 1 \geq \sqrt{ℓ} / (4 C) > 1$ . The log laws imply $\log \sqrt{ℓ} = (\log ℓ) / 2$ , since $(\sqrt{ℓ})^{2} = ℓ$ . Hence $\log q \geq (\log ℓ) / 2 - \log (4 C)$ . Requiring $\log ℓ > \max {0, 4 \log (4 C)}$ ensures $\log q \geq (\log ℓ) / 4 > 0$ . Consequently $\log n / \log q + 1 \leq 4 ℓ / \log ℓ + 1$ , while $π / (96 α) \geq A ℓ$ .

3.1F1F2step 1.1step 2.1algebra∎

Choose $n$ sufficiently large that all preceding conditions hold, $n > 6$ , $\log ℓ > 8 / A$ and $ℓ > 2 / A$ ; step 1.1 establishes a cutoff independent of $d$ . Then $4 ℓ / \log ℓ + 1 < A ℓ / 2 + A ℓ / 2 = A ℓ \leq π / (96 α)$ , proving the budget. If $0 < a \leq α$ and $0 < b \leq β$ , then $a \leq 1 / 24$ , $b < 1 / 4$ , and $1 / (2 b) - 1 \geq q > 1$ . Monotonicity of the logarithm, together with $\log n > 0$ , gives $\log n / \log (1 / (2 b) - 1) + 1 \leq \log n / \log q + 1 < π / (96 α) \leq π / (96 a)$ . The inverse-sine-budget lemma now yields its criterion inequality for every such $a, b$ .

Depends on

Used by · 1

Earlier thresholds with explicit source quantifiers and a fixed-scale deduction

Dependency tree · 2 levels

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