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Answer #4371's open question 1: one coefficient's error path through #4100's transforms carries one factor of T (m = 1), so C = 5 suffices for b >= 13 (no new κ) - #4415
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…through CrocSwap#4100's transforms carries one factor of T (m = 1), so C = 5 suffices for b >= 13 (no new κ) Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
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Summary
This adds
research/one-factor-of-t/. It answers open question 1 of unpaidmoon's #4371: how many factors ofTdoes one coefficient pick up through #4100PROOF.md's three source transforms, the pointwise product and the finalMscale? #4371 showed the product reading costs2^{Θ(p²)}, and that one factor givesC = 5(forb ≥ 13) orC = 8.Claim type: a written sup-norm error ledger plus a fixed-point simulation. No new κ.
Result:
m = 1o(p)scale2^γ. These are the permutations, the unimodular chirp, twist and twiddle diagonals, theH0butterflies, the bounded ring products, and the normalized Gaussian factors via their row-sum bounds.×Mstep. It is the one exception. Even it multiplies only rounding made inside a synthetic convolution, because the exact mapw ↦ (a*w)/Tis a cyclic average, so input errors pass through it unamplified.≲ 3·2^{2γ}·M·poly(p)·2^{−P}. So Audit the 2^{-4p} amplification sentence in #4100. #4371's one-factorCtable applies.Simulation
The simulation runs the scalar case
M = TatP = 40with truncation toward zero, forT = 2^4..2^14.4·M·ℓ·2^{−P}.M·ℓ.error/(M·ℓ·2^{−P})falls from 0.68 to 0.34.M².error/(M²·2^{−P})falls from1.7·10⁻¹to2.9·10⁻⁴.Evidence
Head:
ebfa9588f5a1ca4c595b3ff837f24917d5dd7caf. Base: canonical main12fa8c5c, priority revision 7.PROOF.mdis pinned.Not covered
GAUSSIAN.md's circuits, taken as contractions up to2^γ.S²·2^{2b+4}.Task and overlap
uniform-transfer-audit, refined.Credit
Cinna Davis (#4100) and @unpaidmoon (#4371).
Prepared by Ryanson Jonathan with substantial Claude Code (Claude Opus 5.5) assistance; no independent human review. Apache-2.0. This PR does not merge anything or change the frontier.
🤖 Generated with Claude Code