[{"type":"thinking","thinking":"Zeta23's ZetaBounds.lean works with the B₁ sawtooth integral representation (ZetaSum_aux lemmas: ζ(s) = Σ₁^N n^{−s} + N^{1−s}/(s−1) + N^{−s}/2 + s∫_N^∞ (⌊x⌋+1/2−x)/x^{s+1} dx — this is the CLASSICAL representation valid for σ > 0... used for ZetaUpperBnd on σ near 1).\n\nFor σ > −1, there's the extended representation: ζ(s) = 1/(s−1) + 1/2 + s/2·∫₁^∞ B̃₂(x)/x^{s+2}... hmm the B₂ extension: integrating the B₁ integral once more: ∫₁^∞ B̃₁(x)x^{−s−1}dx = [B̃₂(x)/2·x^{−s−1}]₁^∞ + (s+1)/2∫₁^∞ B̃₂(x)x^{−s−2}dx = −B₂(0)/2 + (s+1)/2∫B̃₂x^{−s−2} (B̃₂ periodic, B̃₂(0) = B₂ = 1/6; boundary at ∞ vanishes for σ > −1...). So |ζ(s)| for σ ∈ [−1/2+, ...] hmm wait — the second integral ∫₁^∞ B̃₂(x) x^{−s−2} dx converges absolutely for σ > −1: |∫| ≤ (s+1)/2·(1/6)/(σ+1)... \n\nSo the EXTENDED representation gives |ζ(s)| ≤ poly(|s|) for σ > −1... and iterating: for σ > −4, integrate 4 times. Each IBP adds a factor (s+k). This is a concrete, Γ-free route but requires:\n- The B₁ representation as starting point: exists in Zeta23 (ZetaSum_aux1: Σ_a^b... and the infinite-tail version? Zeta0EqZeta used in ZetaUpperBnd — \"ζ(s) = Σ₁^N + N^{1−s}/(1−s) + N^{−s}/2 + s∫_N^∞...\" with `Zeta0EqZeta` — that's for σ > 0 hmm — wait is it? ZetaUpperBnd used σ ≥ 1−A/log|t|. Let me check the exact statement of the representation — is it valid for σ > −1? The B₁ integral ∫_N^∞ (⌊x⌋+1/2−x)x^{−s−1}dx converges for σ > 0 (|integrand| ≤ x^{−σ−1}). For σ ≤ 0 it diverges. So the B₁ rep covers σ > 0 only. To go below, IBP extensions.\n\nHmm — how much work is the IBP chain in Lean? Each step: intervalIntegral integration by parts of B̃_k(x)·x^{−s−k} — needs: B̃_k antiderivative relations (B̃_{k+1}' = (k+1)B̃_k·... periodic Bernoulli functions — Zeta23 has B1 only. Building B2, B3, B4 periodic functions + their integrals... This is its own mini-project (~1 session).\n\nWAIT. Better idea — **do we even need σ < 0?** Let me recheck the geometry requirement. The R-bound theorem range: σ ∈ [−5/2, 3], |t| ≥ 10 (D-036 §1). The DISC for the Davenport argument: must contain the application points in its INTERIOR region. What if we use a disc centered at s₀ = 2 + it with RADIUS chosen so the disc stays in σ > 0, but the BOUND is only delivered for σ ∈ [1/2, 3]? Then handle σ ∈ [−5/2, 1/2] by the FE reflection of the R-bound itself?? The FE for logDeriv: ζ′/ζ(s) = (log χ₀)′(s) − ζ′/ζ(1−s). For σ ∈ [−5/2, 1/2]: 1−s has re ∈ [1/2, 7/2] — where the R-bound (right version) applies... but wait — (log χ₀)′(s) = log 2 + log π + (π/2)cot(πs/2) − ψ(1−s)·... hmm χ₀(s) = 2^s π^{s−1} sin(πs/2)Γ(1−s): logDeriv = log 2 + log π + (π/2)cot(πs/2) − ψ(1−s). The ψ(1−s) term: re(1−s) ∈ [1/2, 7/2], |im(1−s)| = |t| ≥ 10: digamma bound |ψ(1−s)| ≤ C log|t| — Zeta23 has digamma strip bounds (norm_digamma_le — where? Y.lean used `norm_digamma_le hw hw'` and `StirlingVert.digamma_stirling`) ✓. cot: |cot(πs/2)| ≤ 2 for |t| ≥ 10 ✓ elementary.\n\nAnd the zero-sum: R(s) = ζ′/ζ(s) − Σ_{|γ−t|<1} 1/(s−ρ). Via FE: ζ′/ζ(s) = (log χ₀)′(s) − ζ′/ζ(1−s). And ζ′/ζ(1−s) = R(1−s) + Σ_{|γ'+t|<1} 1/(1−s−ρ')... zeros of ζ(1−·) at ordinate −t: the zeros ρ' with |γ' − (−t)| < 1 = zeros 1−ρ with |−γ+t|<1 → γ near t. So:\nR(s) = (log χ₀)′(s) − R(1−s) − Σ_{|γ−t|<1}1/(1−s−(1−ρ)) − Σ_{|γ−t|<1}1/(s−ρ)\nNow 1/(1−s−(1−ρ)) = 1/(ρ−s) = −1/(s−ρ). So −Σ 1/(ρ−s)·(−1)... let me redo: −ζ′/ζ(1−s) = −R(1−s) − Σ_{|γ−t|<1} 1/((1−s)−ρ') where ρ' ranges over zeros near Im = −t. Substitute ρ' = 1−ρ̄?? Hmm — zeros of ζ at w near 1−s's height: ζ(1−s) zeros are at 1−s = ρ₀ where ρ₀ zeros of ζ: s = 1−ρ₀: for s = σ+it with σ ∈ [−5/2,1/2], t ≥ 10: ρ₀ = 1−s has im = −t: zeros with im ≈ −t: ρ₀ = 1−ρ where ρ has im ≈ t (by the FE symmetry of zeros: zeros at 1−s ⟺ ...). Zero set symmetry: ρ zero ⟹ 1−ρ̄? The nontrivial zeros are symmetric under ρ ↦ 1−ρ (FE) AND ρ ↦ ρ̄ (conjugate). Zeros with im ≈ −t: γ' ≈ −t: write ρ' = 1−ρ with ρ a zero: im(1−ρ) = −γ ≈ −t ⟺ γ ≈ t ✓ (FE symmetry maps zeros to zeros: if ρ is a zero, so is 1−ρ ✓ riemannZeta_one_sub: ζ(1−ρ) = χ...ζ(ρ)·... — yes 1−ρ is a zero when ρ is, for nontrivial zeros... careful with the FE factor nonvanishing — for nontrivial zeros ρ (0<re<1): ζ(1−ρ) = χ(1−ρ)ζ(ρ)·... hmm actually ζ(1−s) = χ(1−s)ζ(s): at s = ρ nontrivial zero: ζ(1−ρ) = χ(1−ρ)·0 = 0 ✓ if χ finite there ✓).\nSo Σ_{ρ': |γ' +t|<1} 1/(1−s−ρ') = Σ_{ρ: |γ−t|<1} 1/(1−s−(1−ρ)) = Σ 1/(ρ−s) = −Σ 1/(s−ρ).\nThen: R(s) = (log χ₀)'(s) − R(1−s) + Σ_{|γ−t|<1}1/(s−ρ) − Σ_{|γ−t|<1}1/(s−ρ) = (log χ₀)'(s) − R(1−s). \n**THE ZERO SUMS CANCEL EXACTLY!** So:\n|R(s)| ≤ |(log χ₀)'(s)| + |R(1−s)| ≤ C log|t| + (A log(|t|+2) + B₀).\n\nTHIS IS BEAUTIFUL — the FE-reflection of the R-bound cancels the zero sums identically. So the full plan becomes:\n\n**ii′-5 plan (revised route b′):**\n1. **Right-region R-bound** (σ ∈ [1/2, 7/2], |t| ≥ 10) via the Davenport disc argument with the disc B(2+it, R) staying in σ > 0 hmm — wait: to cover σ ∈ [1/2, 7/2] in the inner region: center 2+it, inner radius 3/2+δ... covering σ = 1/2 needs |s−s₀| ≤ 3/2 interior; covering σ = 7/2 needs |s−s₀| ≤ 3/2. So inner radius = 3/2+δ₁, zero-collection radius slightly bigger, outer radius R = 3/2 + 2δ... with R < 2 the outer disc stays σ > 0!! e.g. δ = 1/4: outer R = 2: dips to σ = 0 exactly — need R < 2 STRICT: R = 15/8: leftmost σ = 1/8 > 0 ✓✓. So the whole disc stays in σ ≥ 1/8 > 0 and |ζ| there: split σ ≥ 1/8 into [1/8, ...] hmm — the interior strip σ ∈ [1/8, 1−A/log|t|) still needs a bound!! ARGH — the critical-strip-interior bound again!\n\nHmm wait — do we actually have a problem at σ ∈ [1/8, 0.78]? |ζ(σ+it)| for σ ∈ [1/8, 1/2]-ish, |t| ≥ 10. From the B₁-integral representation (valid σ > 0!): ζ(s) = 1/(s−1) + 1/2 + s∫₁^∞ B̃₁(x)x^{−s−1}dx (the ZetaSum_aux1 with N = 1, tail form): |ζ(s)| ≤ 1/|s−1| + 1/2 + |s|·(1/2)∫₁^∞ x^{−σ−1}dx = 1/|s−1| + 1/2 + |s|/(2σ). For σ ≥ 1/8: |s|/(2σ) ≤ 4|s| ≤ 4(|t|+4) — POLYNOMIAL (linear in |t|)!! The B₁ representation bound: |ζ(s)| ≤ 1/|s−1| + 1/2 + |s|/(2σ) — valid for ALL σ > 0! That covers σ ∈ [1/8, ...] entirely — no gap!!\n\nLet me double-check the representation: for σ > 0: ζ(s) = 1/(s−1) + 1/2 + s∫₁^∞ B̃₁(x)x^{−s−1}dx where B̃₁(x) = x−⌊x⌋−1/2 (sawtooth, |B̃₁| ≤ 1/2 — Zeta23 `abs_B1_le_half` ✓). ∫₁^∞ |B̃₁|x^{−σ−1} ≤ (1/2)∫₁^∞ x^{−σ−1}dx = 1/(2σ) ✓. So |ζ(s)| ≤ 1/|s−1| + 1/2 + |s|/(2σ) — for σ ≥ 1/8, |t| ≥ 10: ≤ 1/9 + 1/2 + 4·|s| ≤ 0.7 + 4(|t|+3)... ✓ |ζ| ≤ C|t| on the disc. THAT'S THE PIECE. And it EXISTS in Zeta23 (the ZetaSum_aux1 representation + abs_B1_le_half) — need to check the exact available form (is there a \"ζ = ...\" equation lemma for σ > 0?).\n\nSo the full ii′-5 route (b′′) — NO Stirling, NO FE-|ζ|-bound, NO left-strip work:\n\n**A. |ζ| bound on the disc B(2+it, 15/8)** (all points have σ ≥ 1/8): via the B₁-integral representation: |ζ(s)| ≤ 1/|s−1| + 1/2 + |s|/(2σ) ≤ C·|t|. (Zeta23 pieces exist.)\n\n**B. Davenport disc argument** for σ ∈ [1/2, 3]-ish (inner radius 3/2+δ around 2+it):\n   - zeros within |γ−t| < 1: all have |ρ − s₀| ≤ √((2)²+1) < 3 hmm: ρ = β+iγ, 0<β<1: |ρ−s₀|² = (2−β)² + (t−γ)² ≤ 4+1 = 5 < (3/2+...)... √5 ≈ 2.236 — hmm the inner radius 3/2+δ < √5 — the near zeros are NOT all inside the inner disc! Let me redo the radii: zeros to collect: |γ−t| < 1 ⟹ |ρ−s₀| ≤ √5 ≈ 2.24. Application points: σ ∈ [1/2, 3]: |s−s₀| ≤ 3/2. So: collection radius r_c = √5 ≈ 2.24, inner/app radius r_a = 3/2 + small, outer radius R < 2 (to stay σ > 0)... but r_c = 2.24 > R = 2!! The collected zeros must be INSIDE the outer disc (to factor them out of g) — √5 > 2 breaks the geometry. Hmm!\n\n   Wait — but the zeros with β < 1 have |2−β| > 1: |ρ−s₀| ≥ 2−β > 1... the zeros sit at distance ∈ (1, √5) from s₀. For the Blaschke factoring g(s) = ζ(s)/∏(s−ρₖ): on the OUTER circle |s−s₀| = R: need |s−ρₖ| bounded BELOW (so that 1/∏|s−ρₖ| doesn't blow up): |s−ρₖ| ≥ R − |ρₖ−s₀| ≥ R − √5 — NEGATIVE for R < √5!! The standard Davenport uses R significantly larger than the collection radius. If R must exceed √5 ≈ 2.24, the disc dips to σ = 2 − R ≤ −0.24 < 0. DAMN.\n\n   Fix: use zeros with |γ−t| < 1/2 only?... the R-bound's sum is over |γ−t| < 1 — the sum is FIXED by the theorem statement. Hmm. Options: (i) shrink the application region to σ ∈ [3/2, 3] and reflect more? The FE-reflection needs R(1−s) for re(1−s) ∈ [1/2, 7/2] — full right region. Circular.\n   (ii) Choose the collection radius as the actual max zero distance √((2−β_min)²+1)... β can approach 0 (zeros near the left edge β → 0+): |ρ−s₀| → √5. Can't shrink.\n   (iii) Use an ELLIPSE/rectangle instead of a disc? Borel–Carathéodory is a disc theorem; there are annulus versions... complicated.\n   (iv) Accept the dip: let the outer disc dip to σ = −1/2 (R = 5/2): then need |ζ| poly bound on σ ∈ [−1/2, 0]: via ONE IBP extension of the B₁ rep: ζ(s) = 1/(s−1) + 1/2 − s/2·[B̃₂ term]... the B₂ extension: ∫B̃₁x^{−s−1} = (1/2)B₂(0)·(−1)... hmm let me redo: ∫₁^∞ B̃₁(x)x^{−s−1}dx — IBP with B̃₁ = B̃₂'/2 (B̃₂' = 2B̃₁ a.e.): = [B̃₂(x)/2·x^{−s−1}]₁^∞ + (s+1)/2·∫₁^∞ B̃₂(x)x^{−s−2}dx = −B₂(0)/2·1 + (s+1)/2·∫₁^∞ B̃₂ x^{−s−2}dx (boundary at 1: B̃₂(1) = B₂(0) = 1/6: [F]₁^∞ = 0 − (1/6)/2·1 = −1/12). So ζ(s) = 1/(s−1) + 1/2 + s(−1/12 + (s+1)/2∫₁^∞ B̃₂ x^{−s−2}dx): the integral converges for σ > −1: |∫| ≤ (1/6)/(σ+1)·... |B̃₂| ≤ 1/6: |∫| ≤ (1/6)/(σ+1). So |ζ(s)| ≤ 1/|s−1| + 1/2 + |s|/12 + |s(s+1)|/12·1/(σ+1) — for σ ≥ −1/2: 1/(σ+1) ≤ 2: |ζ| ≤ C|s|² ✓ polynomial! Covers σ ∈ [−1/2, 0] with ONE extra IBP.\n\n   So geometry: R = 5/2 (dips to σ = −1/2), |ζ| ≤ C(1+|t|)² on the whole disc via B₁-rep (σ > 0 part) + B₂-rep (σ ≥ −1/2 part). Collection radius √5 ≈ 2.236 < R = 5/2 ✓; clearance on outer circle: R − √5 ≥ 0.264 — hmm smallish; use R = 11/4 = 2.75: dips to σ = −3/4; B₂-rep needs σ > −1 ✓ covers; clearance 2.75−2.236 = 0.51 ✓. Or R = 3: dips to −1 — B₂ rep boundary... use R = 3 with B₃-rep (σ > −2)? Each IBP level gives one more unit. Let me not fix radii yet — design doc will pin them. Actually — D-064's sketch said \"radius ~ 9/2\" — with R = 9/2 dipping to σ = −5/2: that needs the B₄-rep (σ > −3) hmm or more. Finer point: LARGER R gives better clearance (R − √5 bigger) but needs more IBP levels (σ > −R+2). R = 4: dips to −2: needs σ > −2 hmm B₃ rep (σ > −2 ✓ with three IBP levels?). Wait let me recount: B₁ rep: σ > 0. B₂ rep: σ > −1. B₃ rep: σ > −2. So R = 4 needs B₄ hmm: σ = −2 boundary: B₃ rep covers σ > −2: at σ = −2 exactly the integral borderline diverges; R = 4 dips to σ = −2 ON the circle — need the bound ON the circle including σ = −2: use R = 4 with B₄ rep (σ > −3) ✓ covers σ ≥ −2 comfortably. Meh — these are constants; the design will pin them.\n\n   Actually hold on — maybe cleaner: choose R such that the dip is an integer-friendly level. Whatever — D-067 will pin: candidate R = 4, B₄-level rep (σ > −3). The IBP chain: build periodic Bernoulli B̃₂, B̃₃, B̃₄ with antiderivative relations and bounds |B̃ₖ| ≤ ... — hmm, ~half a session of build. Alternatively — is there a Mathlib/Zeta23 lemma giving the extended representations directly? Zeta23's EulerMaclaurin.lean has sum_eq_integral_add_integral_deriv (the EM formula) — the IBP chain for ζ specifically isn't there (only B₁-level used). Mathlib: `riemannZeta` representations...? There's the LSeries/Hurwitz stuff. Probably fastest to build the B̃ₖ chain directly.\n\nHmm, wait — actually, one more idea to avoid the IBP chain entirely: **bound |ζ| on the left arc via the FE applied with the digamma... no. OR use the asymmetric contour: make the zero-collection not the full |γ−t|<1 but prove the R-bound with a DIFFERENT sum?** No — the sum shape is fixed by ii′-6's needs (the partial-fraction near-sum from the R(s) definition).\n\nHmm, alternatively — change the REFERENCE POINT for the disc: center at s₀ = 2 + it but use a disc of radius R large and bound |ζ| on the left arc via... the IBP chain is genuinely needed. Fine — it's honest work, Γ-free, and each level is mechanical.\n\nActually — WAIT. One more: maybe don't factor zeros at all — the FromPNTPlus StrongPNTPrefix might contain a COMPLETE Davenport-style lemma already! It has BorelCaratheodoryDeriv, LogOfAnalyticFunction, ZerosBound — these are the pieces of the classical proof of the \"partial fraction bound\" for −ζ′/ζ used in PNT! MediumPNT.lean:752: `‖ζ'(σ+tI)/ζ(σ+tI)‖ ≤ ...` for σ ∈ Ici (1−A/log^9) — and 2291: logzetabnd ‖ζ'/ζ‖ ≤ C·log⁹|t| — the PNT-level bound on ζ'/ζ! Let me look at what StrongPNTPrefix/MediumPNT actually prove about ζ'/ζ — there might be a lemma like \"‖ζ'/ζ(s) − Σ_{|γ−t|<1} 1/(s−ρ)‖ ≤ C log²t\" already there (the classical Ch. 15/17 statement used for the zero-free region!). The zero-free region proof via \"−Re ζ′/ζ ≤ ... − 1/(s−ρ₀)\" uses EXACTLY this kind of partial-fraction + bound. Let me read StrongPNTPrefix.lean's theorem list and MediumPNT's ζ'/ζ bounds.","signature":""},{"type":"text","text":"The geometry issue: covering σ down to −5/2 forces the disc past σ = 0. But before designing around it — StrongPNTPrefix/MediumPNT may already contain the Davenport-style ζ′/ζ bound (PNT zero-free region proofs need exactly this). Reading their theorem lists."},{"type":"tool_use","id":"Bash_167","name":"Bash","input":{"command":"cd D:/Python/Quantum/zeta-23-lean && grep -n \"theorem\\|lemma\" Zeta23/FromPNTPlus/StrongPNTPrefix.lean | head -40","description":"List StrongPNTPrefix theorems"},"extra_content":null}]