Every endpoint cofactor is positive for fourteen Xi shifts

For the normalized Xi series

$$ A(z)=\frac{\xi(1/2+\sqrt z)}{\xi(1/2)}, $$

let \(\Delta_m(r,k)\) be the endpoint-completed cofactor in the rank-\(r\), shift-\(k\) Radau system. The new theorem says that

$$ \Delta_m(r,k)>0 $$

for every

$$ 1\le k\le14,\qquad r\ge1,\qquad0\le m\le r. $$

There is also strict adjacent dominance:

$$ \frac14\Delta_{m+1}(r,k)\lt\Delta_m(r,k) \qquad(m\lt r). $$

These inequalities make the alternating endpoint formula behave, so every corresponding rectangular Toeplitz minor is strictly positive at every rank. The same conclusion survives deletion of any one certified critical-line zero through shift \(13\), uniformly in which zero is removed.

The result is close enough to the earlier fifteen-shift theorem to be confusing, but it is not the same theorem. The earlier result signs the final rectangular minors. This one signs every cofactor from which the endpoint formula is assembled, and also proves the inequality between neighboring cofactors. It is a substantially more rigid statement.

The two estimates point in opposite directions#

Write the reciprocal Xi source as a finite head of positive poles plus a contour tail. Expanding an endpoint cofactor by pole packets gives one distinguished all-head packet and many mixed packets in which one or more head poles have been replaced by tail points.

For a mixed packet missing head pole \(x_j\), the one-hole action contains

$$ \left(\frac{y}{x_j}\right)^{m-k}, $$

where \(y\lt x_j\) is the contour radius. This decays geometrically as the cofactor index \(m\) grows. After all holes are collected, a product bound

$$ \prod_j(1+B_j^{(m)})\lt2 $$

forces positivity. This settles the high-\(m\) region uniformly in rank.

And then it stops helping. If \(m\) is fixed near zero while \(r\) grows, that exponent does not move. An estimate that decays only in \(m\) cannot prove an all-rank theorem for the low cofactors.

The repair is to group packets before estimating them.

Take the complete \(k\)-pole head \(X=(x_1,\ldots,x_k)\) and sum every packet that contains this head plus exactly one tail point \(z\). Schur branching gives the exact identity

$$ s_{(r^k,m)}(X,z) = \left(\prod_i x_i\right)^r z^m \sum_{\ell=0}^{r-m}h_\ell(X^{-1})z^\ell. $$

The cross-boundary factors supply the head annihilator

$$ P_X(z)=\prod_i\left(1-\frac z{x_i}\right). $$

Multiplying by \(P_X\) cancels the infinite complete-homogeneous series. What remains is a truncated kernel \(K_{r-m,X}(z)\) that converges explicitly to one. The grouped packet therefore becomes

$$ \text{positive head constant}\times [t^m]\frac1{U(-t)} +\text{error}, $$

where \(U\) is the factor-deleted reciprocal tail. Now the error decays in \(r-m\), exactly the direction missing from the one-hole estimate.

So the proof has two geometric wedges:

$$ \text{high }m:\quad (y/x_j)^m, $$

$$ \text{low }m:\quad (y/x_j)^r \ \text{after normalization}. $$

One estimate comes down from the top of the cofactor range. The other comes up from the bottom as rank grows. They meet.

The absurd finite bridge#

The coefficient of the complete factor-deleted tail still needs a positive baseline. This is where the certified finite Pólya-frequency order enters:

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

Below rank \(M\), the finite theorem already signs every required cofactor. Above it, the reciprocal tail remains Pólya-frequency through every coefficient index the grouped estimate needs. Dual Jacobi--Trudi gives a positive lower bound, while rectangle log-concavity controls the adjacent ratio by

$$ \frac{A'(0)}{A(0)} =0.0231049931154\ldots . $$

The endpoint scalar is \(\alpha=1/4\), so the adjacent-dominance margin is larger than

$$ 1-\frac14\frac{A'(0)}{A(0)}>0.994. $$

The rigorous omitted-packet errors are not merely smaller than this margin. Their logarithms are below \(-2.3\times10^{11}\) in the worst certified low-index case. The finite bridge is so enormous that, by the time the all-rank contour argument takes over, the error has become comic.

This proves every endpoint cofactor and its adjacent inequality for shifts \(1\) through \(14\), at every rank. For a one-zero deflation, either the deleted factor lies inside the chosen head or outside it; choosing the next untouched certified factor gives the same argument through shift \(13\), uniformly over the deleted zero.

The quantifier that still refuses to move is the shift. Nothing here controls arbitrarily long positive heads or supplies factor-deleted tail baselines uniformly in \(k\). Fourteen fixed shifts are complete. Growing shift remains the wall, which is exactly where an RH-scale obstruction ought to be hiding.

Notebook references: C-0223, C-0222, K-0089, K-0090, C-0207, C-0187