[{"type":"text","text":"Now inserting the new §3.1 subsection after the proof, before the remark."},{"type":"tool_use","id":"Edit_310","name":"Edit","input":{"replace_all":false,"file_path":"D:\\Python\\Quantum\\research\\paper\\paperA.tex","old_string":"\\emph{Step 5 (finite verification).} $\\lambda_n > 0$ for $n \\le 3000$\nby exact high-precision computation (this project, two independent\nmethods, agreement to $4.8\\times10^{-25}$ relative), and for\n$n \\le 5\\times10^5$ by Misguich--Voros \\cite{voros2017li}. The ranges\noverlap the $n = 10^5$ split by a factor $5$ (and the crossover\n$N_0 = 8\\times 10^4$ by $6.25$).\n\\end{proof}","new_string":"\\emph{Step 5 (finite verification; the declared certificate\n\\texttt{LiVerifiedUpTo($10^5$)}).} The base range $n < 10^5$ is covered\nby rigorous high-precision computation of the ordinary Keiper--Li\ncoefficients: for all $n \\le 10^5$ by Johansson \\cite{johansson2015}\n(interval-arithmetic error bounds throughout; the full table\n$\\lambda_0, \\dots, \\lambda_{100000}$ all positive, matching the\nRH-true asymptotic, with $\\lambda_{100000} = 4.6258\\ldots$ certified\nto $\\sim 2900$ digits), and independently for all $n \\le 3000$ by this\nproject (two methods --- the arithmetic side via Bombieri--Lagarias\nThm~2 and the zero side --- agreeing to $4.8\\times10^{-25}$ relative,\nand $\\lambda_1$ agreeing with its closed form to better than $10^{-28}$\nrelative). The crossover $N_0 = 8\\times 10^4$ is covered by a factor\n$1.25$.\\footnote{An earlier draft cited Voros's discretized Keiper--Li\ncomputations \\cite{voros2017li} (with Misguich) for this range; those\nconcern a \\emph{deformed} sequence $\\Lambda_n$, not the ordinary\n$\\lambda_n$, and the citation has been replaced by Johansson's rigorous\nordinary-$\\lambda_n$ table.}\n\\end{proof}\n\n\\subsection{Repair of the right-horizontal estimate}\n\\label{subsec:dogleg}\n\nThe geometric-decrease conclusion following\n\\cite[(6.12)]{lagarias2007li} is not implied by the displayed bounds\nwhen $T \\asymp \\sqrt n$: the per-step majorant ratio there is $e/2 > 1$\n(the zeta factor halves per unit step while the kernel ratio is $\\le e$),\nso the right-tail envelope grows like $e^{0.066\\sqrt n}$ rather than\ndecreasing ($\\sim 6\\times 10^{10}$ at $n = 10^5$ measured; E-031). In\nthe zeta specialization we replace this step by the following zero-free\ncontour deformation (D-073/D-074).\n\nKeep the cut height $T = \\sqrt n$ (so the bridge and the endpoint\n$N_1 = (T_0-1)^2$ are untouched) and deform only the right horizontal\nsegment $[3, 2\\sqrt n]$ at height $T$ into a \\emph{dog-leg}: lift\nvertically from $3 + iT$ to $3 + iU$ with $U = \\tfrac32 \\sqrt n$ (the\nleft vertical), traverse horizontally at height $U$ (the high\nsegment), and drop back at $2\\sqrt n$ (the right vertical). The\nrectangle this traverses lies entirely in $\\Re w \\ge 3$, where $\\zeta$\nis zero-free and pole-free \\emph{unconditionally}, so Cauchy's theorem\ngives the exact identity\n\\[\nH_{\\mathrm{low}} \\;=\\; H_{\\mathrm{high}} - V_{\\mathrm{right}} +\n  V_{\\mathrm{left}}\n\\]\n(\\texttt{upper\\_right\\_horizontal\\_eq\\_dogleg} in the formal\ndevelopment). Each dog-leg piece has an honest bound: the left vertical\ncontributes $\\le \\tfrac{\\sqrt n}{2}(e^{5/2}+1)\\cdot\\tfrac12$; the high\nsegment contracts geometrically (the repaired stepping ratio\n$q = e^{4/9}/2 < 1$, certified) and contributes $\\le 8.38$; and the\nright vertical is exponentially small, $\\le \\tfrac14 I_{\\mathrm{Maj}}(n)$.\nThe charge replacing the broken ``$3.3$'' is therefore\n$\\mathrm{DL}(n) = \\tfrac{\\sqrt n + 1}{4}(e^{5/2}+1) + 8.38 +\n\\tfrac14 I_{\\mathrm{Maj}}(n)$ per side, with the margin essentially\nunchanged (margins $13\\,160$ / $88\\,911$ / $2.458\\times10^{26}$ at\n$8\\times10^4$ / $10^5$ / $N_1$). We record for completeness that the\nper-side bound at $T = \\sqrt n$ appears to hold empirically by a strong\nhidden cancellation (E-032: true-integral to envelope ratio down to\n$5.6\\times10^{-10}$); whether that cancellation admits a proof is an\nopen question we do not rely on."},"extra_content":null}]