Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Zeta horizontal logarithmic derivative comparison

Statement

There are absolute d>0,C>0, with d<c0, such that for ∣t∣≥3 and σ≥1−d/log⁡(∣t∣+2), ∣ζ′(σ+it)ζ(σ+it)∣≤Clog⁡(∣t∣+2).

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Riemann zeta classical zero free region: There is an absolute c0>0 such that ζ has no zeros in σ≥1−c0/log⁡(∣t∣+2). The pole at s=1 is not a zero.

[F2]

A local formula for the logarithmic derivative of zeta: Uniformly for −1≤σ≤2 and s=σ+it≠1 whose ordinate is not that of a nontrivial zero, ζ′ζ(s)=−1s−1+∑ρ: ∣t−ℑρ∣<11s−ρ+O(log⁡(∣t∣+2)), where zeros occur with multiplicity. For ∣t∣≥2 the pole term is absorbed into the error, giving the usual large-height local formula.

[F3]

Zeta logarithmic derivative zero bound: Write s=σ+it and let ρ range over nontrivial zeta zeros with multiplicity. With the Hadamard constant B, ζ′ζ(s)=B+∑ρ(1s−ρ+1ρ)−1s−1+log⁡π2−Γ′(1+s/2)2Γ(1+s/2). This is a meromorphic identity, using convergent genus-one terms. Uniformly for −1≤σ≤2, ∣t∣≥3 and ζ(s)≠0, Re⁡ζ′ζ(s)=∑ρRe⁡1s−ρ−12log⁡∣t∣+O(1), and this real series is absolutely convergent.

[F4]

The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1: For ℜs>1, if ζ(s):=∑n≥1n−s, then −ζ′(s)ζ(s)=∑n≥1Λ(n)n−s.

Proof

1.1

Put L=log⁡(∣t∣+2), s1=1+L−1+it. The Euler series and the real-axis simple-pole expansion give ∣ζ′/ζ(s1)∣≤−ζ′/ζ(1+L−1)=O(L). The same comparison holds for every σ≥1+L−1; for σ≥2 it is even bounded by the convergent series at two. The real-part formula now gives ∑ρRe⁡(1/(s1−ρ))=O(L), all summands being positive.

F3F4
1.2

For ∣Im⁡ρ−t∣≤1, the region theorem implies 1−Re⁡ρ≥c0/(KL) with an absolute K, since log⁡(∣Im⁡ρ∣+2)≤KL. Choose d<c0/(2K). For 1−d/L≤σ≤1+1/L, the positive real parts of s−ρ and s1−ρ are comparable, hence ∣s−ρ∣≥c∣s1−ρ∣. Consequently ∣1/(s−ρ)−1/(s1−ρ)∣≤C/(L∣s1−ρ∣2)≤CRe⁡(1/(s1−ρ)).

F1algebra
2.1

For ordinates off the zero ordinates, subtract the two local logarithmic-derivative formulas. Their pole terms are bounded at these heights. Sum the preceding comparison over the common local zero set and use its positive-sum bound to obtain O(L). The constants do not depend on the distance of t from an ordinate. Taking limits from non-ordinates extends the bound to all t, because the entire horizontal segment is zero-free. Together with the Euler-series range this proves the assertion.

F2step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

18 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