← CPL / 10 · Refined Exact B70 Certificate
The previous version, to make exact positivity easier to establish, lowered the numerical dual's $c_0$ by $5\times10^{-5}$ and obtained $B_{70}^{cert}=3.667777612662112\ldots$; this round keeps shrinking the rationalization safety margin down to $\delta=25024291/312500000000000=8.00777312\times10^{-8}$, and all three multiplicity patterns can still be proved non-negative by exact-rational Bernstein subdivision, giving the sharper $\boxed{B_{70}^{cert}=35186790600709/9589237500000=3.669404433950979\ldots}$ — only about $5.6\times10^{-6}$ away from the previous round's numerical crossing at $\approx3.66941$. The document honestly records a genuine QCI case: why the safety margin cannot simply be taken to zero — taking the numerical decimals directly as the exact dual, the $(2,2)$ pattern's nominal exact-rational quartic has an extremely small negative trough (about $-8.0\times10^{-8}$), and the $(2,1,1)$ pattern's nominal polynomial has a minimum value of about $-6.72\times10^{-8}$; this is not a case of the Bernstein method “failing to prove it,” but rather that taking the numbers as exact really does produce tiny violations, so a sufficient margin must be kept. The document concludes by recommending that the next step should not be to keep chasing the seventh and eighth decimal digits of $B$, but instead to replace the single boundary row $S(4)$ with a genuine continuous support strip $\alpha\in[1,1+\delta]$ — only then does the problem start to enter the same quantitative regime as Claude's actual $\sigma_{70}\approx1.04$.
Relationship to other documents, stated as far as possible in the document's own words, not my interpretation.
“A certificate that looks valid numerically must still retain a sufficient margin once you move into exact arithmetic — this is precisely a small-scale example in the QCI sense.” — from this document's Section 2, “Why Can't the Safety Margin Simply Be Taken to Zero?”
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