Let

$$ F(z)=\sum_{n\ge0}f_nz^n $$

be entire, nonpolynomial, and coefficientwise positive. Extend \(f_n=0\) for \(n<0\), and form the one-sided Toeplitz matrix

$$ T=[f_{c-r}]_{r,c\ge0}. $$

Here is the theorem:

If \(T\) is totally nonnegative through order \(p+1\), then every order-\(p\) minor that is not structurally forced to vanish is strictly positive.

The loss of one order is real. \(PF_p\) alone is not being upgraded by optimism. But \(PF_{p+1}\) buys strictness everywhere at order \(p\), regardless of how far the minor is shifted.

Suppose one minor vanishes#

Take ordered row and column sets

$$ R=\{r_1<\cdots

with \(r_j\le c_j\), exactly the condition that the minor is not forced to be zero by the hard edge.

Assume \(\det T[R,C]=0\). By induction, the nested \((p-1)\)-minor \(B\) is already strictly positive. Eliminate the final selected column against the first \(p-1\) columns and call the residual at row \(s\) by \(e_s\).

Every lower order-\(p\) enclosure is

$$ B e_s, $$

so total nonnegativity gives \(e_s\ge0\). Every rightward order-\(p\) enclosure gives another nonnegative determinant \(D_N\).

Now place both extensions into one order-\((p+1)\) minor. Exact expansion gives

$$ \det T[R\cup\{s\},C\cup\{N\}] =-e_sD_N. $$

The left side is nonnegative. The right side is nonpositive. Therefore

$$ e_sD_N=0 $$

for every lower row \(s\) and every future column \(N\).

That minus sign is the entire machine.

The two exits are both impossible#

If every \(D_N\) vanishes, then all future columns restricted to \(R\) lie in one fixed \((p-1)\)-dimensional span. The coefficients \(f_n\) satisfy an eventual constant-coefficient recurrence. Therefore \(F\) is rational.

But a rational entire function is a polynomial, contradicting the hypotheses.

So some \(D_N>0\). Then every lower residual \(e_s\) vanishes. Propagate this relation to row \(c_p\), where the one-sided hard edge deletes all earlier selected columns but leaves

$$ T_{c_p,c_p}=f_0>0. $$

The residual is simultaneously zero and \(f_0\). Contradiction.

Thus the original minor was strict.

The Xi payoff is absurdly large and still finite#

The verified zeta-zero height gives a finite Pólya-frequency order

$$ M=9{,}425{,}328{,}785{,}007 $$

for the normalized Xi coefficient sequence and every quotient obtained by deleting one certified critical-line factor.

Applying the theorem with \(p=r\) proves strict endpoint-cofactor positivity and rectangular-minor positivity for

$$ 1\le r\le M-1 $$

and every shift \(k\ge1\). The old finite shift ceiling disappears entirely.

The rank ceiling does not disappear. Unknown ranks begin at \(M\), because rank \(r\) consumes \(PF_{r+1}\). This is a spectacular finite wedge, not an all-order theorem and not an RH proof. The construction tells us exactly what one extra PF order buys, and exactly where the purchased strictness runs out.

Notebook references: K-0073, C-0215, C-0026, C-0207