[{"type":"thinking","thinking":"Approved plan:\n1. D-039: prove the digamma sub-bound (upgrade R-004's H2 link to proved).\n2. Then the valley theorem attack.\n3. (User handles TeX install separately.)\n\nD-039 task: prove |ψ(w) − log w + 1/(2w)| ≤ 1/|w|² for Re w ≥ −3/2, |w| ≥ 5 (the D-036 digamma remainder bound used in the R(s) assembly). Actually the exact statement from D-036 §1: \"Digamma: ψ(w) = log w − 1/(2w) + r(w), |r(w)| ≤ 1/|w|² for Re w ≥ −3/2, |w| ≥ 5 (E-021 verifies numerically)\".\n\nHow to prove it rigorously?\n\nStandard approach: Binet's second formula or the integral representation. For Re w > 0:\nψ(w) = log w − 1/(2w) − ∫₀^∞ (1/(e^t−1) − 1/t + 1/2) e^{−wt} dt / ... Let me recall Binet's second formula:\nlog Γ(w) = (w−½)log w − w + ½log(2π) + 2∫₀^∞ arctan(t/w)/(e^{2πt}−1) dt.\nAnd ψ(w) = log w − 1/(2w) − 2∫₀^∞ t/(t²+w²)·1/(e^{2πt}−1) dt — that's Binet's formula for ψ: \nψ(w) = log w − 1/(2w) − 2∫₀^∞ t/((t²+w²)(e^{2πt}−1)) dt, valid for Re w > 0.\n\nSo r(w) = −2∫₀^∞ t/((t²+w²)(e^{2πt}−1)) dt for Re w > 0. Bound: |r(w)| ≤ 2∫₀^∞ t/(|t²+w²|(e^{2πt}−1)) dt. For Re w > 0: |t²+w²| ≥ ... if w = u+iv with u ≥ 0: |t²+w²| ≥ t²·sin-ish... For |arg w| ≤ π/2−ε... There's a classical bound: |r(w)| ≤ 1/(12|w|²)·sec²(θ/2)-type (θ = arg w). \n\nBut our range is Re w ≥ −3/2 (w = s/2 with s on the contour σ₀ = −5/2: w = −5/4 + it/2) — OUTSIDE Re w > 0. Use the recurrence to shift into Re > 0: ψ(w) = ψ(w+2) − 1/w − 1/(w+1). Then r(w) = ψ(w) − log w + 1/(2w) = [ψ(w+2) − log(w+2) + 1/(2(w+2))] + [log(w+2) − log w] − 1/(2(w+2)) − 1/w − 1/(w+1) + 1/(2w)... = r(w+2) + log(1+2/w) − 1/w − 1/(w+1) − 1/(2(w+2)) + 1/(2w)... let me just carefully derive and bound each piece for |w| ≥ 5, Re w ≥ −3/2 (so Re(w+2) ≥ 1/2 > 0 ✓ Binet applies).\n\nr(w) = ψ(w) − log w + 1/(2w)\nψ(w) = ψ(w+2) − 1/w − 1/(w+1)\nψ(w+2) = log(w+2) − 1/(2(w+2)) + r(w+2)\nSo r(w) = log(w+2) − log w − 1/(2(w+2)) − 1/w − 1/(w+1) + 1/(2w) + r(w+2)\n= log(1 + 2/w) − 1/(2(w+2)) − 1/(2w) − 1/(w+1) + r(w+2).\n\nThe elementary part E(w) := log(1+2/w) − 1/(2(w+2)) − 1/(2w) − 1/(w+1). Expand for large w: log(1+2/w) = 2/w − 2/w² + 8/(3w³) − 4/w⁴ + ...; 1/(2(w+2)) = 1/(2w)·1/(1+2/w) = 1/(2w) − 1/w² + 2/w³ − 4/w⁴...; 1/(2w); 1/(w+1) = 1/w − 1/w² + 1/w³ − 1/w⁴...\nE = [2/w − 2/w² + 8/(3w³) − 4/w⁴ + 32/(5w⁵)...] − [1/(2w) − 1/w² + 2/w³ − 4/w⁴ + ...] − 1/(2w) − [1/w − 1/w² + 1/w³ − 1/w⁴ + ...]\n= (2/w − 1/(2w) − 1/(2w) − 1/w) + (−2 + 1 + 1)/w² + (8/3 − 2 − 1)/w³ + (−4 + 4 + 1)/w⁴ + ...\n= 0 + 0/w² + (8/3 − 3)/w³ + 1/w⁴ + ... = (−1/3)/w³ + 1/w⁴ + O(1/w⁵).\nSo E(w) ~ −1/(3w³) — the leading cancellation is exact (as expected — the recurrence preserves the asymptotic). So r(w) = r(w+2) + E(w) with E(w) = O(1/w³).\n\nBound: |E(w)| for |w| ≥ 5, Re w ≥ −3/2: need explicit. Write E(w) = log(1+2/w) − 2/w + [2/w − 1/(2w) − 1/(2(w+2)) − 1/(w+1)]. Hmm let me just bound each: |log(1+u) − u| ≤ |u|²/2·(1/(1−|u|)) for |u| < 1: with u = 2/w, |u| ≤ 2/5 = 0.4: |log(1+2/w) − 2/w| ≤ (2/|w|)²/2·1/(1−0.4) = (2/w²)·(5/3) = 10/(3w²)... these crude bounds give |E| ≤ O(1/w²) not the needed O(1/w³) directly — need the cancellation. Better: E(w) = ∫-form or Taylor with remainder: E is analytic in 1/w; E(w) = −1/(3w³) + R₄ with |R₄| ≤ C/|w|⁴: compute the exact series remainder: E(w) = f(1/w) with f analytic in |u| ≤ 0.4 (u = 1/w): f(u) = log(1+2u) − 1/(2(1+2u))·u − u/2 − u/(1+u). Then f(u) = Σ c_k u^k with c_0=c_1=c_2=0, c_3 = −1/3. |f(u) + u³/3| ≤ (max |f^{(4)}|/24)|u|⁴ — bound via Cauchy: on |u| = 0.45: max |f|: compute: |log(1+2u)| ≤ −log(1−0.9) = 2.3; |u/(2(1+2u))| ≤ 0.45/(2·0.1) = 2.25; |u|/2 = 0.225; |u/(1+u)| ≤ 0.45/0.55 = 0.82: total ≤ 5.6 + 1/3. Cauchy: |c_4| ≤ M/r⁴... |f(u)+u³/3| ≤ M|u|⁴/(r³(r−|u|)) with r = 0.45, |u| ≤ 0.4: M ≈ 6: ≤ 6·|u|⁴/(0.45³·0.05) = 6/(0.0045) = 1316·|u|⁴ — crude but explicit: |E(w)| ≤ 1/(3|w|³) + 1316/|w|⁴. At |w|=5: 1316/625 = 2.1 — vs 1/(3·125) = 0.0027 — the crude Cauchy bound is useless at |w|=5. Need sharper — just bound E(w) DIRECTLY (it's an explicit function!): for |w| ≥ 5, Re w ≥ −3/2, bound each term numerically-analytically... Actually simplest rigorous route: E(w) is explicit elementary; compute max_{|w|≥5, Re w≥−3/2} |E(w)|·|w|³ analytically-ish: |E(w)|·|w|³: as |w|→∞: → 1/3. The max on the domain: could be near w = −3/2±5i... or near the boundary |w|=5 with Re w = −3/2. We can bound: |E(w)| ≤ sup... For a clean proof: evaluate |E| on a grid + derivative bound (interval methods) — that's a small computation, same spirit as E-023's M₀. OR use a crude-but-sufficient bound: we only need |r(w)| ≤ C/|w|² with SOME explicit C (D-036 used 1/|w|² but the R-constant A=2/B₀=2 has slack; a C=2 or 3 is fine — N₀ barely moves).\n\nPlan for D-039:\n1. Binet formula for Re w > 0: r(w) = −2∫₀^∞ t/((t²+w²)(e^{2πt}−1))dt. Bound for Re w ≥ 1/2, |w| ≥ 3 (say): |t²+w²| ≥ t² + Re(w²)... for Re w > 0: |t² + w²|² = (t²+u²−v²)² + 4u²v² ≥ ... standard: |t²+w²| ≥ t² + |w|²·cos(2θ)-ish... For |arg w| < π/4: |t²+w²| ≥ (t²+|w|²)/√2-ish. Our w = s/2 with s = σ+it on the contour, σ ∈ [−5/2, 4], |t| ≥ 10: w = u+iv with |v| ≥ 5 — |arg w| near ±π/2 — the bound |t²+w²| ≥ 2|w|t·... : |t²+w²| = |w−it||w+it| ≥ |w|·|... let me use: |t²+w²| ≥ |Im(t²+w²)| = 2|uv| — with |v| ≥ 5 and u ≥ 1/2 (after shift): ≥ 2·(1/2)·5 = 5. Hmm weak. Better: |t²+w²| = |w+it||w−it| ≥ ... for real t: |w ± it| = |v ± t·(∓i)... u² + (v±t)²... |w+it|² = u² + (t+v)² hmm w+it = u + i(v+t): |w+it| ≥ u. And |w−it| ≥ u. Product ≥ u²?? too weak. Product: |t²+w²| = |w−it||w+it|: as t varies, min over t of |w−it||w+it|: at t = ±v·... the product = |t² + w²|: t² + w² = t² + u² − v² + 2iuv: modulus² = (t²+u²−v²)² + 4u²v²: minimized at t² = v²−u² (if v > u): min value = 4u²v²: so |t²+w²| ≥ 2u|v|. With u ≥ 1/2 (shifted w+2 has Re ≥ −3/2+2 = 1/2) and |v| ≥ 5: ≥ 2·(1/2)·|v| = |v| ≥ 5. So |r(w+2)| ≤ 2∫₀^∞ t/( (2u|v|)·(e^{2πt}−1) )dt = (1/(u|v|))·∫ t/(e^{2πt}−1)dt = (1/(u|v|))·(1/24)·... ∫₀^∞ t/(e^{2πt}−1)dt = 1/(4π²)·π²/6·... = 1/24. So |r(w+2)| ≤ 2·(1/(2u|v|))·(1/24)·... let me redo: |r(w+2)| ≤ 2∫ t/(|t²+(w+2)²|(e^{2πt}−1))dt ≤ 2∫ t/(2U|V|(e^{2πt}−1))dt where W = w+2 = U+iV: = (1/(U|V|))·(1/24) = 1/(24 U|V|). With U ≥ 1/2: ≤ 1/(12|V|). And |V| ≥ 5. So |r(w+2)| ≤ 1/(12·(1/2)·... ≤ 1/(12·|V|)... For |w| ≥ 5: |V| = |Im w| ≥ ... w on contour: |Im w| = |t|/2 ≥ 5 ✓. And |V| ≥ |w| − 2 − |Re w|... |V| ≥ |t|/2 ≥ 5 ≥ |w|/2-ish (|w| ≤ |t|/2+2: for |t| ≥ 10, |w| ≤ |t|/2 + 2 ≤ |t|/2·1.4... |V| = |t|/2 ≥ |w|/1.4·... fine: |r(w+2)| ≤ 1/(12·|V|)·... ≤ c/|w| with explicit c. But we need 1/|w|² decay! 1/(12|V|) is only O(1/|w|) — NOT sufficient for D-036's r(w) ≤ 1/|w|² claim!\n\nHmm wait — the true asymptotic r(w) ~ −1/(12w²)·... ψ(w) = log w − 1/(2w) − 1/(12w²) + O(1/w⁴): so r(w) ≈ −1/(12w²) — the integral bound should give |w|⁻² decay: |r(w+2)| ≤ 2∫ t/(|t²+W²|(e^{2πt}−1))dt: I bounded |t²+W²| ≥ 2U|V| (independent of t!) losing the t-dependence. Better: |t²+W²| ≥ max(t²·something, ...): |t²+W²| ≥ (t² + |W|²·cos(2θ)) for θ = arg W ≤ π/4... our W has |arg| near π/2 — fails. Alternative split: for t ≤ |V|/2: |t²+W²| ≥ |2UV| ≥ |V| (U≥1/2)... and for t large: t². Hmm — but the integral ∫t/(e^{2πt}−1) concentrates at t ~ 1/(2π) — SMALL t — so the relevant bound is |t²+W²| at t ~ 0.16: |t²+W²| ≥ |2UV| = |V|·(2U) ≥ |V| — so r ≈ ∫ small-t t/(|V|(e^{2πt}−1)) ~ 1/(24|V|) — but the TRUE r ~ 1/(12|W|²) ~ 1/(12|V|²): the bound 1/(24|V|) is loose by |V| — because at small t, t²+W² ≈ W² + t²: |W²+t²| ≥ |W|²·... for t ≪ |V|: W² + t² = (U²−V²+t²) + 2iUV: with |V| ≥ 5, t ≤ 1: |W²+t²| ≥ V² − U² − t² ≥ |V|²(1 − ...)·... V² − U² − t²: V² ≥ 25, U² ≤ (2.5)²... wait U = Re W = Re w + 2 ∈ [−1/2, 6]: U² ≤ 36 hmm for σ up to 4: U up to 6. |W²|² = (U²−V²)² + 4U²V² = (U²+V²)² = |W|⁴ — so |W²| = |W|². And |W² + t²| ≥ |W|² − t² ≥ |W|²(1 − t²/|W|²) ≥ |W|²(1−t²/25·...). So for t ≤ 1, |W| ≥ 7: |W²+t²| ≥ |W|²(1 − 1/49) ≥ 0.98|W|². Then split the integral at t=1: \n∫₀¹ t/(|W²+t²|(e^{2πt}−1))dt ≤ 1/(0.98|W|²)·∫₀¹ t/(e^{2πt}−1)dt ≤ 1/(0.98|W|²)·(1/24)·... \n∫₁^∞ t/(t²·(e^{2πt}−1))dt ≤ ∫₁^∞ 1/(t e^{2πt}−...)dt — tiny: ∫₁^∞ t/(t²(e^{2πt}−1))dt ≤ ∫₁^∞ e^{-2πt}·(1/(t(1−e^{−2πt})))dt ≤ 2∫₁^∞ e^{-2πt}/t dt ≤ 2e^{-2π}/(2π) ≈ 3e-4·... negligible.\nSo |r(W)| ≤ 2[1/(0.98|W|²·24) + 3e-4·...] hmm: the 2 prefactor: |r(W)| ≤ 2∫₀^∞ t/(|t²+W²|(e^{2πt}−1))dt ≤ 2[1/(0.98|W|²)·(1/24) + ε] with ε ≤ ∫₁^∞ t/(t²(e^{2πt}−1))dt ≤ ~1e-3. For |W| ≥ 7: ε ≤ 1e-3 ≤ 0.05/|W|² (since |W|² ≥ 49: 0.05/49 = 1e-3 ✓). So |r(W)| ≤ 2[1/(23.5|W|²) + 0.05/|W|²] ≤ 2·(0.0426+0.05)/|W|² = 0.185/|W|². \n\nWait but this needs U² + V² = |W|² ≥ 49 and the lower bound |W²+t²| ≥ |W|²−t²: |W²+t²|² = |W|⁴ + 2t² Re(W²) + t⁴: if Re(W²) = U²−V² < 0 (our case: V large), then |W²+t²|² = |W|⁴ − 2t²(V²−U²) + t⁴ ≥ |W|⁴ − 2t²|W|² + t⁴ = (|W|²−t²)² ✓ so |W²+t²| ≥ |W|²−t² for t ≤ |W| ✓. So |r(W)| ≤ 0.185/|W|² for Re W ≥ 1/2, |W| ≥ 7. Then r(w) = r(w+2) + E(w): |r(w)| ≤ 0.185/|w+2|² + |E(w)|. With |w| ≥ 5: |w+2| ≥ |w|−2 ≥ 3... need |w+2| ≥ 7?? no — the r(W) bound needs |W| ≥ 7: |w+2| ≥ 7 fails for |w| = 5..7. Adjust: use the bound for |W| ≥ 5 (recompute constants: 1−t²/25 = 0.96: |r(W)| ≤ 2[1/(23.04·... )]... fine, constants shift slightly. Or shift by 3 instead of 2 for small |w|. Detail.\n\nAnd E(w): |E(w)| = |log(1+2/w) − 1/(2(w+2)) − 1/(2w) − 1/(w+1)| — bound via Taylor: E(w) = −1/(3w³) + O(w⁻⁴): |E(w)| ≤ C_E/|w|³ with C_E explicit via a careful series-bound: E(w) = f(1/w), f(u) = log(1+2u) − u/2 − u/(2(1+2u)) − u/(1+u): f analytic on |u| < 1/2: Cauchy bound on |u| = 0.49: M = max|f| ≤ |log(1+2u)| + |u|/2 + |u/(2(1+2u))| + |u/(1+u)|: |log(1+2u)| ≤ −log(1−2·0.49) = −log(0.02) = 3.9; |u|/2 ≤ 0.25; |u/(2(1+2u))|: |1+2u| ≥ 1−0.98 = 0.02: ≤ 0.49/0.04 = 12.25; |u/(1+u)| ≤ 0.49/0.51 = 0.96: M ≤ 17.4. Hmm the 1/(1+2u) near u = −0.5 blows the Cauchy radius. Use r = 0.4: |1+2u| ≥ 0.2: |u/(2(1+2u))| ≤ 0.4/0.4 = 1; |log(1+2u)| ≤ −log(0.2) = 1.61; M ≤ 1.61+0.2+1+0.78 = 3.6. Then f(u) = −u³/3 + Σ_{k≥4} c_k u^k with |c_k| ≤ M/r^k = 3.6/0.4^k: |f(u)+u³/3| ≤ 3.6·Σ_{k≥4}(|u|/0.4)^k = 3.6·(|u|/0.4)⁴/(1−|u|/0.4): for |u| ≤ 0.2 (|w| ≥ 5): ≤ 3.6·(0.5)⁴/0.5·|u|⁴·... = 3.6·0.0625/0.5 = 0.45·|u|⁴/0.4⁴?? let me redo: Σ_{k≥4}(|u|/0.4)^k = (|u|/0.4)⁴/(1−|u|/0.4): |u| ≤ 0.2: = (0.5)⁴/0.5 = 0.125: so |f+u³/3| ≤ 3.6·0.125 = 0.45 — dimensionless ✓ meaning |E(w) + 1/(3w³)| ≤ 0.45/|w|⁴. At |w| = 5: 0.45/625 = 7.2e-4. So |E(w)| ≤ 1/(3|w|³) + 0.45/|w|⁴ ≤ (1/3·0.2 + 0.45/25)/|w|²·... = (0.0667+0.018)/|w|²·... ≤ 0.085/|w|²·(1/|w|)... hmm: |E| ≤ |w|⁻³(1/3 + 0.45/5) = 0.423/|w|³ ≤ 0.423/(5·|w|²) = 0.085/|w|².\n\nTotal: |r(w)| ≤ 0.185/|w+2|² + 0.085/|w|². With |w+2| ≥ |w|−2 ≥ 3/5|w| (|w| ≥ 5): 0.185·(25/9)/|w|² = 0.514/|w|². Total ≤ (0.514+0.085)/|w|² ≈ 0.6/|w|². So |r(w)| ≤ 0.6/|w|² for |w| ≥ 5, Re w ≥ −3/2 — UNDER the D-036 claim of 1/|w|² ✓ with margin.\n\nWait — need to double-check the constraint |arg| / Re W: the r(W) bound used |W²+t²| ≥ |W|²−t² which holds always (shown above). And Re W ≥ 1/2 for Binet validity (Re W > 0). Our W = w+2, Re w ≥ −3/2 ⟹ Re W ≥ 1/2 ✓. Also w on the OTHER contour lines: σ ∈ [−5/2, 4] all covered by Re w ≥ −5/4 ≥ −3/2 ✓. And |Im w| ≥ 5 (|t| ≥ 10) so |w| ≥ 5 ✓.\n\nAlso — hmm, Binet's formula validity: the second Binet formula for ψ: ψ(w) = log w − 1/(2w) − 2∫₀^∞ t/((t²+w²)(e^{2πt}−1))dt for Re w > 0 ✓ (standard). \n\nSo the proof works: |r(w)| ≤ 0.6/|w|² for |w| ≥ 5, Re w ≥ −3/2. This upgrades H2's digamma piece to proved (with the D-036 bound 1/|w|² holding with slack). The rest of H2 (the R-bound assembly A=2/B₀=2) was already traced analytically in D-036 §1 — with the digamma piece now proved, the whole R-bound becomes PROVED.\n\nLet me double check the R-assembly uses: D-036 §1: \"Digamma: ψ(w) = log w − 1/(2w) + r(w), |r(w)| ≤ 1/|w|² for Re w ≥ −3/2, |w| ≥ 5\". Our proof gives 0.6/|w|² ≤ 1/|w|² ✓.\n\nOne caveat: the asymptotic sign: r(w) ≈ −1/(12w²): ψ(w) = log w − 1/(2w) − 1/(12w²) + 1/(120w⁴) − ... so r(w) = −1/(12w²) + O(w⁻⁴): |r| ~ 1/(12|w|²) = 0.083/|w|² — our bound 0.6/|w|² is ~7× loose — fine for a first explicit proof.\n\nAlso should double-check my E(w) formula: r(w) = r(w+2) + log(1+2/w) − 1/(2(w+2)) − 1/(2w) − 1/(w+1)? Let me redo: r(w) := ψ(w) − log w + 1/(2w). ψ(w) = ψ(w+2) − 1/w − 1/(w+1). r(w) = ψ(w+2) − 1/w − 1/(w+1) − log w + 1/(2w). ψ(w+2) = log(w+2) − 1/(2(w+2)) + r(w+2). So r(w) = log(w+2) − log w − 1/(2(w+2)) + r(w+2) − 1/w − 1/(w+1) + 1/(2w) = r(w+2) + log(1+2/w) − 1/(2(w+2)) − 1/(2w) − 1/(w+1) ✓ matches.\n\nVerify E's leading term numerically quickly as a sanity check, then also verify the full claim |r(w)| ≤ 0.6/|w|² numerically at sample points (mpmath digamma). Quick check then write D-039, update D-037/R-004 H2 status, commit. Then the valley theorem attack (the big one) — that's a deeper analysis project; start it after D-039 is banked.","signature":""},{"type":"text","text":"Starting D-039. The proof route: Binet's second formula for Re w > 0 + recurrence shift, with explicit remainder bounds. Quick numerical sanity check of the derived bound first."},{"type":"tool_use","id":"Bash_264","name":"Bash","input":{"command":"cd /d/Python/Quantum && ./.venv/Scripts/python.exe - <<'EOF'\nimport mpmath as mp\nmp.mp.dps = 40\ndef r(w):\n    return mp.digamma(w) - mp.log(w) + 1/(2*w)\n# claim: |r(w)| <= 0.6/|w|^2 for |w|>=5, Re w >= -1.5\nworst = 0.0; worstw = None\nfor sig in (-1.25, -0.75, -0.25, 0.25, 1.0, 2.0):\n    for t in (5.0, 5.5, 7.0, 10.0, 20.0, 50.0, 100.0):\n        w = mp.mpc(sig, t)\n        if abs(w) < 5: continue\n        val = abs(r(w)) * abs(w)**2\n        if val > worst: worst, worstw = val, (sig, t)\nprint(\"max |r(w)|*|w|^2 over sample:\", mp.nstr(worst, 6), \"at\", worstw, \" (bound 0.6)\")\n# also check leading asymptotic: r(w) ~ -1/(12 w^2)\nw = mp.mpc(-1.0, 50.0)\nprint(\"r(w)*w^2 at w=-1+50i:\", mp.nstr(r(w)*w**2, 8), \" (expect ~ -1/12 = -0.0833)\")\nEOF","description":"Numerically sanity-check the digamma bound"},"extra_content":null}]