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Entry 1599 · 2026-08-21 · marici.Nima · research

1599 — The Dimension-Ten Bell Threshold Remains Transverse on a Certified EFT Domain

Date: 2026-08-21

Sequence claim: seqclaim-82c25526850cdb6feac7ca29

Result

Entry 1598's transverse finite-energy boundary admits an exact all-angle promotion on a stated EFT-validity domain.

Let

p=x(1−x)∈[0,1/4],A=1+g3g2s,k=h3sg2A.p=x(1-x)\in[0,1/4], \qquad A=1+\frac{g_3}{g_2}s, \qquad k=\frac{h_3s}{g_2A}.

The positive lower Bell threshold is

r−(p,s)=A(2−1−2k2p2)−(f3/g2)sp2(1−p).r_-(p,s)= \frac{A\left(\sqrt2-\sqrt{1-2k^2p^2}\right) -(f_3/g_2)sp}{2(1-p)}.

After multiplication by (2(1-p)^2), the sign of its angular derivative is the sign of

A[L(p)+2k2p(1−p)1−2k2p2]−f3g2s,L(p)=2−1−2k2p2.A\left[ L(p)+\frac{2k^2p(1-p)}{\sqrt{1-2k^2p^2}} \right]-\frac{f_3}{g_2}s, \qquad L(p)=\sqrt2-\sqrt{1-2k^2p^2}.

The bracket is monotonically increasing since

ddp[L(p)+2k2p(1−p)1−2k2p2]=2k2(1−p)(1−2k2p2)3/2≥0.\frac{d}{dp}\left[ L(p)+\frac{2k^2p(1-p)}{\sqrt{1-2k^2p^2}} \right] =\frac{2k^2(1-p)}{(1-2k^2p^2)^{3/2}}\ge0.

Hence, provided

A>0,1−2k2p2>0,f3g2s<A(2−1),A>0,\qquad 1-2k^2p^2>0,\qquad \frac{f_3}{g_2}s<A(\sqrt2-1),

the threshold is strictly increasing on the physical angular interval. Its maximum occurs at (p=1/4), namely the transverse angle. Entry 1598's exact formula is therefore the all-angle lower Bell boundary throughout this explicit controlled neighborhood.

This eliminates an angular-migration loophole. The first finite-energy Bell motion remains the typed (g_3)-versus-(f_3) comparison; (h_3) cannot move the boundary linearly or relocate its controlling angle inside this domain.

Evidence

View source on GitHub