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Why the specified alpha equation has one positive root

The specified electromagnetic closure has exactly one positive simple root. The original convexity proof establishes this on the whole positive half line, without a numerical sign scan. It is a theorem about the stated function and its inputs.

Statement and assumptions

Reading edition of EM.FP.POSITIVE_ROOT.CONVEXITY.01, dated 13 October 2026. The contract’s verdict is PARTIAL: the mathematical uniqueness question is settled for the specified function, while physical selection is a separate question. The published ledger classification remains Numerical.

Fix the function and inputs

Use c = 1/(8π), p = 1/(6π), d = 48c⁴ and s = 5/4. The physical budget used in the paper is M = 41. The proof below only needs M > 0, so uniqueness itself does not select the value 41.

q(x)=de−2x,u(x)=q(x)(1−q(x))−s,φ(x)=p+u(x),C=45c6M>0.q(x)=d e^{-2x},\qquad u(x)=q(x)(1-q(x))^{-s},\qquad \varphi(x)=p+u(x),\qquad C=\frac45c^6M>0.

In words: The seam response is a positive baseline plus a decaying correction, and C is positive.

F(x)=x3−2c3x2+Clog⁡φ(x).F(x)=x^3-2c^3x^2+C\log\varphi(x).

In words: The closure is the cubic expression plus the logarithm of the specified seam response.

This is the same equation as F = x³ − 2c³x² − C log(1/φ) = 0. Because φ depends on x, the complete function is not a cubic polynomial.

The function is negative before the cubic turns positive

For x ≥ 0, π > 3 implies c < 1/24, d < 1/6912 < 1/1000 and p < 1/18. Therefore 0 < q ≤ d < 1/1000. Since s = 5/4 < 2, the seam response stays strictly between zero and one.

(1−q)−s<(1000999)2<2,0<φ(x)<118+1500<1.(1-q)^{-s}<\left(\frac{1000}{999}\right)^2<2,\qquad 0<\varphi(x)<\frac1{18}+\frac1{500}<1.

In words: The response is positive and below one, so its logarithm is negative.

The function is smooth. Set t = 2c³. For 0 < x ≤ t, the cubic part is nonpositive and the logarithmic part is strictly negative. There can be no root in this interval.

F(x)=x2(x−t)+Clog⁡φ(x)<0(0<x≤t),F(t)<0.F(x)=x^2(x-t)+C\log\varphi(x)<0\qquad(0<x\le t),\qquad F(t)<0.

In words: The closure stays negative up to and including t.

The logarithmic response is strictly convex

Put ℓ = log u and w = u/(p + u), so 0 < w < 1. Differentiating the specified response gives exact identities.

ℓ′=−2−2sq1−q,ℓ′′=4sq(1−q)2>0.\ell'=-2-\frac{2sq}{1-q},\qquad\ell''=\frac{4sq}{(1-q)^2}>0.

In words: The second derivative of the logarithm of u is positive.

(log⁡φ)′′=wℓ′′+w(1−w)(ℓ′)2>0.(\log\varphi)''=w\ell''+w(1-w)(\ell')^2>0.

In words: Both contributions to the second derivative of log phi are positive.

The second identity follows directly by differentiating log(p + exp ℓ). It gives a strict sign for the whole response rather than relying on an estimate that the correction is small.

Existence, uniqueness and a simple crossing

On x ≥ t, the second derivative of F is strictly positive. As x tends to infinity, φ tends to p > 0, so the logarithm has a finite limit while the cubic tends to infinity. Continuity and F(t) < 0 therefore give a root r > t.

F′′(x)=6x−4c3+C(log⁡φ)′′>8c3>0(x≥t).F''(x)=6x-4c^3+C(\log\varphi)''>8c^3>0\qquad(x\ge t).

In words: The full closure is strictly convex after t.

At any root r > t, strict convexity makes the derivative greater than the secant slope from t to r. That slope is positive.

F′(r)>F(r)−F(t)r−t=−F(t)r−t>0.F'(r)>\frac{F(r)-F(t)}{r-t}=\frac{-F(t)}{r-t}>0.

In words: Every root crosses with a strictly positive derivative.

After the first root, the derivative is positive and increasing. A second crossing is impossible. Together with the negative interval before t, this proves exactly one positive root on the whole positive half line. Its positive derivative also proves that it is simple.

Root location and what the theorem establishes

For M = 41, the original proof reports opposite definite signs from the existing interval implementation F_iv at the following endpoints. Combined with the analytic uniqueness result, this encloses the unique root.

0.0072973525622097<α⋆<0.0072973525622100,α⋆−1≈137.0359992168407.0.0072973525622097<\alpha_\star<0.0072973525622100,\qquad\alpha_\star^{-1}\approx137.0359992168407.

In words: The unique root has reciprocal approximately 137.0359992168407.

The analytic proof, the reported numerical enclosure and a fresh execution are distinct evidence. This page reproduces the argument and cites the existing calculation; it does not report a new run.

The same proof works for every positive M. The original contract also differentiates the root with respect to M, using its positive derivative at the crossing.

drdM=−(4/5)c6log⁡φ(r)F′(r)>0.\frac{dr}{dM}=-\frac{(4/5)c^6\log\varphi(r)}{F'(r)}>0.

In words: Increasing the positive budget increases its unique positive root.

The theorem establishes uniqueness for the supplied equation. Deriving its physical electromagnetic response and identifying the result with the renormalized Thomson coupling require the original source and observable map. Those physical obligations are separate from the existence and uniqueness proof.

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