← Session 14 — Phase 5 kickoff: paper draft + honesty audit
thinking (232 chars)
The true definition (D-004, line 69):
R(ψ) := [∫ψ² + ∬|u−v|ψ(u)ψ(v)dudv] / (∫ψ)² (even ψ ∈ C([−½,½]), ψ>0).
Fix the formula in the Background section. Also the chain: N^s_0 ≥ (2 − R(ψ) − o(1))N. Good. Let me fix the R(ψ) display.
Found it — D-004 line 69: `R(ψ) := [∫ψ² + ∬|u−v|ψ(u)ψ(v)dudv] / (∫ψ)²`. Fixing the formula.
tool: Edit
input
{
"file_path": "D:\\Python\\Quantum\\research\\paper\\draft.tex",
"old_string": "\\emph{The rank--trace inequality.} Writing $\\tilde G$ in inertia form,\na rank--trace comparison between the positive eigenspace of $\\tilde G$\nand the positive part of $P$ converts positivity of the prime side\ninto a lower bound on the number of simple on-line zeros in the\nwindow. The strength of the certificate is governed by the\n\\emph{window ratio}\n\\[\nR(\\psi) \\;=\\; \\frac{\\|\\psi\\|_2^4}{2\\int |\\psi'|^2 \\cdot\n(\\text{second-moment kernel})},\n\\]\nwith $R(\\psi_0) = 4/3$ for the triangle window and\n$R(\\psi_{MT}) = \\tfrac12 + \\tfrac{1}{\\sqrt2}\\cot\\tfrac{1}{\\sqrt2}\n= 1.32749\\ldots$ for the Montgomery--Taylor window, giving the\nfractions $2 - R(\\psi)$: $0.6725$ of zeros simple and on the line,\n$0.83625$ distinct.",
"new_string": "\\emph{The rank--trace inequality.} Writing $\\tilde G$ in inertia form,\na rank--trace comparison between the positive eigenspace of $\\tilde G$\nand the positive part of $P$ converts the second-moment input\n$\\|\\tilde G\\|_{HS}^2 = (R(\\psi) + o(1))\\,N$ into a lower bound on the\nnumber of simple on-line zeros in the window. The strength of the\ncertificate is governed by the \\emph{window ratio}\n\\[\nR(\\psi) \\;=\\; \\frac{\\int \\psi^2 \\;+\\;\n\\iint |u-v|\\,\\psi(u)\\psi(v)\\,du\\,dv}{\\bigl(\\int \\psi\\bigr)^2}\n\\qquad (\\psi \\in C([-\\tfrac12,\\tfrac12]) \\text{ even}, \\ \\psi > 0),\n\\]\nwith $R(\\psi_0) = 4/3$ for the triangle window and\n$R(\\psi_{MT}) = \\tfrac12 + \\tfrac{1}{\\sqrt2}\\cot\\tfrac{1}{\\sqrt2}\n= 1.32749\\ldots$ for the Montgomery--Taylor window (both verified\nsymbolically to $60$ digits in this project), giving the certified\nfractions $2 - R(\\psi)$: $0.6725$ of zeros simple and on the line,\n$0.83625$ distinct."
}
raw content_json (database row)
[{"type":"thinking","thinking":"The true definition (D-004, line 69):\n\nR(ψ) := [∫ψ² + ∬|u−v|ψ(u)ψ(v)dudv] / (∫ψ)² (even ψ ∈ C([−½,½]), ψ>0).\n\nFix the formula in the Background section. Also the chain: N^s_0 ≥ (2 − R(ψ) − o(1))N. Good. Let me fix the R(ψ) display.","signature":""},{"type":"text","text":"Found it — D-004 line 69: `R(ψ) := [∫ψ² + ∬|u−v|ψ(u)ψ(v)dudv] / (∫ψ)²`. Fixing the formula."},{"type":"tool_use","id":"Edit_13","name":"Edit","input":{"file_path":"D:\\Python\\Quantum\\research\\paper\\draft.tex","old_string":"\\emph{The rank--trace inequality.} Writing $\\tilde G$ in inertia form,\na rank--trace comparison between the positive eigenspace of $\\tilde G$\nand the positive part of $P$ converts positivity of the prime side\ninto a lower bound on the number of simple on-line zeros in the\nwindow. The strength of the certificate is governed by the\n\\emph{window ratio}\n\\[\nR(\\psi) \\;=\\; \\frac{\\|\\psi\\|_2^4}{2\\int |\\psi'|^2 \\cdot\n(\\text{second-moment kernel})},\n\\]\nwith $R(\\psi_0) = 4/3$ for the triangle window and\n$R(\\psi_{MT}) = \\tfrac12 + \\tfrac{1}{\\sqrt2}\\cot\\tfrac{1}{\\sqrt2}\n= 1.32749\\ldots$ for the Montgomery--Taylor window, giving the\nfractions $2 - R(\\psi)$: $0.6725$ of zeros simple and on the line,\n$0.83625$ distinct.","new_string":"\\emph{The rank--trace inequality.} Writing $\\tilde G$ in inertia form,\na rank--trace comparison between the positive eigenspace of $\\tilde G$\nand the positive part of $P$ converts the second-moment input\n$\\|\\tilde G\\|_{HS}^2 = (R(\\psi) + o(1))\\,N$ into a lower bound on the\nnumber of simple on-line zeros in the window. The strength of the\ncertificate is governed by the \\emph{window ratio}\n\\[\nR(\\psi) \\;=\\; \\frac{\\int \\psi^2 \\;+\\;\n\\iint |u-v|\\,\\psi(u)\\psi(v)\\,du\\,dv}{\\bigl(\\int \\psi\\bigr)^2}\n\\qquad (\\psi \\in C([-\\tfrac12,\\tfrac12]) \\text{ even}, \\ \\psi > 0),\n\\]\nwith $R(\\psi_0) = 4/3$ for the triangle window and\n$R(\\psi_{MT}) = \\tfrac12 + \\tfrac{1}{\\sqrt2}\\cot\\tfrac{1}{\\sqrt2}\n= 1.32749\\ldots$ for the Montgomery--Taylor window (both verified\nsymbolically to $60$ digits in this project), giving the certified\nfractions $2 - R(\\psi)$: $0.6725$ of zeros simple and on the line,\n$0.83625$ distinct."},"extra_content":null}]