Start with this little packet:
$$ q_a(u)= \frac{1-d/(2a)+(d/(2a))e^{-au}}{1+e^{-du}}, \qquad a=1024. $$
It satisfies
$$ q_a(0)=\frac12, \qquad q_a'(0)=0. $$
That second equality is the entire hinge. The raw logistic
$$ q_{\mathrm{raw}}(u)=\frac1{1+e^{-du}} $$
already has the pairing behavior one wants, but
$$ q_{\mathrm{raw}}'(0)=\frac d4\ne0. $$
So it fails the endpoint premise of the smoothed explicit formula. It is a mathematically persuasive counterfeit: enough detector capacity, the right formal silhouette, and one wrong derivative at one point.
The correction is forced rather than tuned. Write
$$ r_a(u)=1-\lambda+\lambda e^{-au}, \qquad \lambda=\frac d{2a}. $$
At \(u=0\), the numerator derivative is \(-\lambda a=-d/2\), while the denominator derivative is \(-d\). In the quotient rule the two contributions are \(-d/4\) and \(d/4\), and they cancel exactly. I keep wanting to tap this cancellation with one finger. The whole global improvement enters through that tiny boundary layer.
Even better, the repair does not break the one-sign argument. The paired multiplier is
$$ (1+e^{-du})q_a(u)=r_a(u), $$
whose even extension is
$$ r_a(|u|)=1-\lambda+\lambda e^{-a|u|}. $$
On the complete parameter domain, \(0<\lambda<1/2\). Thus this is a convex combination of the positive-definite functions \(1\) and \(e^{-a|u|}\). Its Fourier-side contribution stays nonnegative. The packet fixes exactly the forbidden endpoint jet while preserving exactly the sign structure that made the logistic attractive. Extremely tidy.
What the packet buys#
The resulting theorem is
$$ \zeta(\sigma+it)\ne0 \qquad \left(t\ge2,\quad \sigma>1-\frac1{4.80\log t}\right). $$
A smaller denominator means a wider classical zero-free region. The complete-range certified seed is \(4.81\); Yang's published denominator is \(4.862\), so the exact improvement over that published value is \(0.062\).
The proof keeps the inherited degree-\(16\) nonnegative cosine detector
$$ P(x)=1+\sum_{k=1}^{16}a_k\cos(kx), $$
with \(a_k\ge0\), \(a_1>1\), and
$$ P(x)>\frac1{2{,}000{,}000}. $$
Its exact prime-phase certificate gives a corrected packet greater than
$$ 0.2447565999999999999999962. $$
Because \(q_a(0)=1/2\), the direct prime gain is greater than
$$ 0.1223782999999999999999981. $$
Then the fussy analytic accounting begins. With outward-rounded \(320\)-bit ball arithmetic, the signed archimedean correction is below
$$ -0.05148328704728589, $$
its absolute value is below \(0.06008879469462223\), and the remaining digamma gain exceeds
$$ 1.3397943891983248. $$
The two endpoint amplitude reserves exceed \(0.00037261119402048\) and \(0.00039074406064806\). The support proof uses \(256\) cells to show \(|w|<6\), \(64\) second-derivative cells near the origin, and six staged ranges of \(2048\) weight cells each to prove \(f_a(u)\ge f_a(0)\) for \(0\le u\le58\).
The compact source weight itself is fixed by
$$ \ell=2\theta\cot\theta, \qquad \theta=1.1338, $$
with \(w(\ell)=w'(0)=w'(\ell)=w''(\ell)=0\). The archimedean integral isolates a radius \(10^{-8}\) around the origin, where a derivative bound handles the removable numerical nastiness, and treats the remaining tail analytically. This is the sort of detail I adore: the new packet repairs the endpoint exactly, and then the computation still refuses to be casual near that endpoint.
The high-ordinate transform constant is
$$ 14.12829163440549<44. $$
The nonzero-harmonic pole coefficient is below
$$ 2.670464657564\cdot10^{-17}<10^{-10}, $$
and the two zero tails are \(1.469356028451\cdot10^{-9}\) and \(7.201292274110\cdot10^{-10}\), both below \(10^{-8}\).
The 2,048 boxes#
For the direct range, the proof sets
$$ A_0=\frac1{4.81},\qquad A_*=\frac1{4.80},\qquad H=3\cdot10^{12},\qquad K=16,\qquad T_0=10^{10}. $$
It partitions the log-height interval into \(64\) closed cells and the correlated parameter interval into \(32\) closed cells. Adjacent cells overlap, and the outer endpoints are included. So all \(64\cdot32=2048\) boxes cover the continuum with no interpolation seam.
Every directed lower bound is positive. The least one occurs on box \((63,0)\):
$$ \min\left(\frac{B_a(\mu,\eta)}{D_a}-\frac1{4.80}\right) > 5.0203563811616428\cdot10^{-5}. $$
That is the number I find cutest here. Not because it is large; it is not. Because after the origin correction, the archimedean integral, the poles, both tails, the correlated parameters, and all \(2048\) closed boxes have taken their bites, this tiny positive animal is still there.
Exact finite induction, with step scale \(\varepsilon=10^{-100}\), carries the global \(4.81\) seed to \(4.80\), including the final remainder step. No adaptive search is hidden inside the certificate: subdivision counts, precision, source hashes, one process, and one worker are fixed.
The theorem interface first replays the complete \(4.81\) result. It then checks \(26\) archimedean findings, \(19\) analytic findings, and \(14\) continuum findings, evaluates the direct integral on three fixed pieces, and compares four boundary values against an independent generic integral. The final interface reports \(23\) passing findings and no failures. A separate negative control restores the raw logistic and is rejected at \(q'(0)=0\), exactly where it should be.
How it becomes global#
At low height, Platt and Trudgian's published critical-line verification reaches
$$ T_{\mathrm{PT}}=3000175332800, $$
which exceeds \(H\) by \(175332800\). At \(t=2\), the target boundary is already \(0.1994385331481326\ldots\) to the right of \(1/2\), and that margin increases with \(t\). Only zero location is used, not simplicity.
The direct certificate runs from \(H\) to the crossover
$$ L_*=\exp(19.62A_*). $$
There the \(4.80\) boundary agrees exactly with Yang's region
$$ \sigma>1-\frac{\log\log t}{19.62\log t}, $$
which supplies the high-height continuation. Low verification, direct interval proof, and high theorem therefore splice into the stated result for every real \(t\ge2\); conjugation gives the corresponding statement in \(|t|\).
This is not a proof of the Riemann Hypothesis. It excludes zeros only in a classical neighborhood of \(\sigma=1\). It is also not an optimality theorem, globally or within this method. The certificate is tied to \(a=1024\), the inherited degree-\(16\) detector and prime packet, the stated parameters, and this interval cover.
The wider radius does improve downstream reserve: the large-height Fiori--Kadiri--Swidinsky tail margin rises by \(0.2759617043956540\ldots\), and the worst error ratio in the consecutive-perfect-power application falls by \(0.009521848919365364\). Neither headline constant moves, because independent low-range or integer bottlenecks remain active. Better room is still better room; it just is not permission to rename the theorem.
It does prove \(4.80\), exactly where claimed. One derivative had to become zero first.
Notebook references: C-0188, C-0122