Five-vertex vertex risk

The additive-kernel transform from Tetrahedron vertex risk also resolves the vertex coefficient on a simplex with five vertices. The result is one sum with four harmonic states, and its leading constant involves \(\zeta(2)-\zeta(3)\). A positivity argument for the transformed diagonals applies to every fixed vertex count.

All statements use the uniform-simplex assumptions of Theory of the skeleton risk. The sum has \(D\) terms. It is not a bounded-length classical-special-function expression. Five-vertex vertex degree recurrence gives an explicit certified order-seventeen degree recurrence with initial values and no exceptional positive integer indices. The broader construction and fixed-dimension leading constants are in Vertex coefficients at every fixed vertex count. The five-vertex leading term has an explicit remainder bound in Quantitative five-vertex vertex asymptotics. Together with the nonvertex edge upper bound, this also completes the all-degree risk comparison in Five-vertex risk comparison at every degree.

An exact harmonic weight

Write \(H_j^{(r)}=\sum_{k=1}^j k^{-r}\), \(H_j=H_j^{(1)}\), and define the depth-two finite harmonic sum

\[A_j=\sum_{k=1}^j\frac{H_{k-1}}{k^2}.\]

All four states vanish at zero. They update by \(H_j^{(r)}=H_{j-1}^{(r)}+j^{-r}\) and \(A_j=A_{j-1}+H_{j-1}/j^2\); no inner summation is needed during evaluation. For \(n=j+1\), abbreviate \(H=H_j\), \(H_2=H_j^{(2)}\), \(H_3=H_j^{(3)}\), and \(A=A_j\). Define

(90)\[\begin{split}\Psi_j={}&\frac{24H^3}{n(n+1)(n+2)(n+3)}\\ &-\frac{12(5n^3+7n^2-6n-12)H^2} {n^2(n+1)^2(n+2)^2(n+3)}\\ &+\frac{24(2n+3)HH_2}{n(n+1)(n+2)(n+3)}\\ &-\frac{24(2n^4+15n^3+44n^2+47n+24)H} {n^2(n+1)^2(n+2)^2(n+3)}\\ &+\frac{12(4n^4+22n^3+41n^2+26n+12)H_2} {n^2(n+1)(n+2)^2(n+3)}\\ &-\frac{48H_3}{n(n+1)(n+3)} -\frac{48A}{n(n+2)} +\frac{144}{n^2(n+1)^2(n+2)^2(n+3)}.\end{split}\]

Theorem 45 (Five-vertex vertex coefficient at every degree)

For every integer \(D\geq1\), let \(W_{D,j}=\binom{D-1}j\binom{D+1}j\) and \(Z_D=\binom{2D}{D-1}\). Then

(91)\[a_0(5,D)=\frac{600} {Z_D(2D+1)(2D+2)(2D+3)(2D+4)} \sum_{j=0}^{D-1}W_{D,j}\Psi_j.\]

Every \(\Psi_j\) is strictly positive, with \(\Psi_0=1\). All denominators in the formula are nonzero on the stated domain.

Proof. Put \(g(t)=-(1-t)\log(1-t)/t\) and \(\phi_r(j)=[t^j]g(t)^r/(1-t)\). Besides the three formulas in Tetrahedron vertex risk, differentiating \(\binom\alpha{j+4}\) four times at \(\alpha=3\) gives

\[\phi_4(j)=\frac{24\left( -H^3+\tfrac{11}2H^2-6H+1 +(3H-\tfrac{11}2)H_2-2H_3\right)} {n(n+1)(n+2)(n+3)}.\]

For sequences, let \(*\) denote ordinary convolution and let \(g_k=[t^k]g(t)\). Expand the additive kernel to the fourth power. Terms in which one of \(g(x),g(y)\) is absent combine by convolution. The result is

(92)\[[x^jy^j]\frac{(g(x)+g(y)-g(xy))^4}{(1-x)(1-y)} =-\phi_4(j)+8\phi_1(j)\phi_3(j)+6\phi_2(j)^2 -24\bigl(g*(\phi_1\phi_2)\bigr)_j +12\bigl(g^2*\phi_1^2\bigr)_j.\]

Here \(g^2\) denotes the coefficient sequence of the square of the series, while the products of \(\phi\) sequences are pointwise. In particular, the coefficient \(-24\) includes both the \(g(x)^2g(y)g(xy)\) term and its symmetric counterpart.

The two remaining convolutions reduce to finite harmonic sums. Since

\[g_0=1,\quad g_k=-\frac1{k(k+1)},\qquad [t^0]g^2=1,\quad [t^k]g^2=\frac{4H_{k-1}-6}{k(k+1)(k+2)}\quad(k\geq1),\]

their contribution, with \(n=j+1\), is

\[-24\bigl(g*(\phi_1\phi_2)\bigr)_j +12\bigl(g^2*\phi_1^2\bigr)_j =-\frac{48(1-H)}{n^2(n+1)}+\frac{12}{n^2} +\sum_{k=1}^{n-1}\bigl(R_H(k,n)H_{k-1}+R_0(k,n)\bigr),\]

where

\[\begin{split}p(k,n)&=\frac1{k^2(k+1)(n-k)(n-k+1)},\\ q(k,n)&=\frac1{k(k+1)(k+2)(n-k)^2},\\ R_H(k,n)&=48(q(k,n)-p(k,n)),\\ R_0(k,n)&=48p(k,n)-72q(k,n).\end{split}\]

Partial fractions use only the seven denominators \(k,k^2,k+1,k+2,n-k,(n-k)^2,n-k+1\). The sums without \(H_{k-1}\) are immediate harmonic numbers. For the remaining sums, put \(T=(H^2-H_2)/2\); their respective values are

\[\begin{split}\sum_{k=1}^{n-1}\frac{H_{k-1}}k&=T,\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{k^2}&=A,\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{k+1} &=T+H/n-1+1/n,\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{k+2} &=T+H\left(1/n+1/(n+1)\right)+1/(n(n+1)) -7/4+1/(2n)+3/(2(n+1)),\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{n-k}&=H^2-H_2,\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{(n-k)^2}&=A+HH_2-H_3,\\ \sum_{k=1}^{n-1}\frac{H_{k-1}}{n-k+1}&=H^2-H_2+2H/n-H.\end{split}\]

These identities follow by changing the order of the finite sums; for example, the penultimate sum is \(\sum_{i+r\leq n-1}1/(ir^2)\), and partial fractions at \(i+r=s\) give \(2A+\sum_{s\leq n-1}H_{s-1}^{(2)}/s\) \({}=A+HH_2-H_3\). The formulas also hold when \(n=1\) and the sums are empty. Substitution and collection give exactly (90).

The binomial-transform diagonalization now gives

\[a_0(5,D)=\frac{25(D-1)!(D+1)!}{(5)_{2D}} \sum_j W_{D,j}\Psi_j.\]

Use \((D-1)!(D+1)!Z_D=(2D)!\) to obtain (91). Positivity follows from the general argument below.

Positive diagonals for every vertex count

Proposition 6 (Outside-kernel powers remain positive after transformation)

For every integer \(q\geq1\) and \(j\geq0\),

\[[x^jy^j]\frac{(g(x)+g(y)-g(xy))^q}{(1-x)(1-y)}>0.\]

Proof. Let \(T_1,\ldots,T_q\) be independent exponential random variables of mean one, and define the homogeneous polynomial

\[Q_h(T)=\sum_{\alpha_1+\cdots+\alpha_q=h} \prod_{r=1}^q\frac{T_r^{\alpha_r}}{(\alpha_r+1)!}.\]

Since \(\mathbb E[T_r^k]=k!\), multiplication of the outside kernel \(F\) gives

\[[x^hy^l]F(x,y)^q=\mathbb E[Q_h(T)Q_l(T)].\]

The signed binomial transform in each index therefore identifies the diagonal in the proposition with

\[\mathbb E\!\left[ \left(\sum_{h=0}^j(-1)^h\binom jh Q_h(T)\right)^2\right].\]

The polynomial inside the square is nonzero: its highest homogeneous part is \((-1)^jQ_j\), including the case \(j=0\). Exponential density is strictly positive on the positive orthant, so the expectation is strictly positive. This proves positivity without requiring each term of a collected harmonic expression to be positive.

The leading asymptotic constant

Corollary 15 (Five-vertex vertex asymptotic)

As \(D\to\infty\),

(93)\[a_0(5,D)=\frac{7200(\zeta(2)-\zeta(3))}{D^6} +O\!\left(\frac{\log D}{D^7}\right).\]

Proof. The Euler sum \(A_j\) tends to \(\zeta(3)\). Indeed, monotone convergence and the power series for the logarithm give

\[\sum_{k=1}^\infty\frac{H_{k-1}}{k^2} =\frac12\int_0^1\frac{\log^2(1-t)}t\,dt =\frac12\int_0^1\frac{\log^2t}{1-t}\,dt =\zeta(3).\]

Moreover, for \(j\geq1\),

\[0\leq\zeta(2)-H_j^{(2)}\leq1/j,\qquad 0\leq\zeta(3)-A_j\leq(2+\log j)/j.\]

The second inequality follows from \(H_{k-1}\leq1+\log k\) and comparison with the decreasing function \((1+\log x)/x^2\). The coefficient of \(H_2\) in (90) differs from \(48/(n(n+2))\) by

\[-\frac{12(2n^3+3n^2-2n-12)} {n^2(n+1)(n+2)^2(n+3)}=O(n^{-3}).\]

All the other terms apart from \(-48A/(n(n+2))\) are \(O(\log n/n^3)\); the cubic harmonic term is \(O(\log^3n/n^4)=O(\log n/n^3)\). Consequently

\[\Psi_j=\frac{48(\zeta(2)-\zeta(3))}{(j+1)(j+3)} +O\!\left(\frac{\log(j+2)}{(j+1)^3}\right).\]

Let \(J\) have probabilities \(W_{D,j}/Z_D\). The mean, variance and small-index bound proved in Tetrahedron vertex risk apply unchanged:

\[\mathbb E[J]=\frac{D^2-1}{2D},\qquad \operatorname{Var}(J)=O(D),\qquad \mathbb P(J\leq\lfloor D/4\rfloor)\leq4\,2^{-D/12} \quad(D\geq12).\]

On the complementary event, the remainder is uniformly \(O(\log D/D^3)\). The sequence \(\Psi_j\) is bounded, so the omitted event contributes an exponentially small amount. Taylor expansion of \(1/((x+1)(x+3))\) about the mean, with second derivative \(O(D^{-4})\) and variance \(O(D)\), gives

\[\mathbb E[\Psi_J] =\frac{192(\zeta(2)-\zeta(3))}{D^2} +O(\log D/D^3).\]

Finally, (91) is \(600\mathbb E[\Psi_J]/\prod_{r=1}^4(2D+r)\). Expanding this rational factor proves the assertion.

Reproduction and limits

tools.risk_research.vertex_five.five_vertex_coefficient evaluates (91) using exact rational arithmetic, four harmonic states, a binomial weight and an accumulator. The number of rational states is fixed, while their bit lengths increase with \(D\). Floating-point evaluation of the ungrouped weight can lose precision through cancellation; the exact evaluator avoids that loss.

The tests compare the new expression with the original scalar kernel sum at degrees 1 through 20, independent Gram propagation at degrees 1 through 5, and direct signed binomial transforms of \(F^4\). After the candidate was frozen, a separate integer-polynomial \(F^4\) computation at degrees 193, 227 and 251 agreed exactly. The committed holdout file records the candidate and oracle source hashes. These checks support the implementation; the algebraic and asymptotic arguments above establish the mathematical identities.