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Absorb the telescoping factor (10 p^2)^{g-1} p^6 (no new κ) - #4485
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Mister-Meeseeks merged 1 commit intoOct 22, 2026
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Co-authored-by: Cursor <cursoragent@cursor.com>
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Question
#4484 absorbs
g (10 p^2)^g exp(-p^2)whenp ≥ max(64, 4K + 2C)and16 g ceil(log2 p) ≤ p. Ag-fold triangle inequality carriesg - 1inverse ceilings and one section error. Doesg (10 p^2)^{g-1} p^6 exp(-p^2)stay below2^{-C p}under that same gate?Result
Let
p ≥ 2,g ≥ 1,K ≥ 0andC ≥ 0be integers, and letL = ceil(log2 p). Ifp ≥ max(64, 4K + 2C)and16 g L ≤ p, thenThe case
K = 0is the monomial in the question. Relative to #4484 the extra factor isp^4 / 10, and it fits the same power-of-two budget: the exponent is at most-C p - 4. Wheneverg log p = o(p), every fixed pairK, Cmeets the gate for largep. Atg = p^2the telescoping error is greater than 1 for every integerp ≥ 2.Claim type: one absorption bound. No new κ. The ceiling
10 p^2and the section errorp^6 exp(-p^2)stay hypotheses. Replacing a ceiling by the approximant norm10 p^2 + p^6 exp(-p^2)is not proved here, and the #4100 notes are unchanged.This branch starts from canonical main. It does not replay #4484.
Test plan
PASS_GAUSSIAN_TENSOR_FOLDandoutcome=completedon this machine. Maintainer replay has not been run.Base
2a13eb1827784206202db90e5dc8327cf8a250db, priority revision 7, selection refresh κ =23/10000.Prepared by James Reed with substantial Cursor Grok 4.7 assistance.