The vertex leading constant at large vertex count

Keep the uniform-simplex assumptions and the positive leading constant \(C_M\) of Vertex coefficients at every fixed vertex count. This chapter concerns that constant as \(M\) grows. The degree limit used to define \(C_M\) remains a separate fixed-\(M\) theorem; no simultaneous \(M,D\) limit follows by substitution.

A quantified large-M theorem

Write \(q=M-1\), \(n=M-2\), \(L=\log2\) and \(a=L-1/2>0\). Define

\[B_M=\frac{M^2(M-1)!\,(2\log2)^{M-1}\sqrt{M-1}} {4\sqrt{\pi\log2\,(\log2-1/2)}}.\]

Theorem 34 (Explicit large-vertex asymptotic constant)

For every integer \(M\geq1026\),

(68)\[\left|\frac{C_M}{B_M}-1\right|\leq\frac{25}{M-2}.\]

Consequently \(C_M\sim B_M\) as \(M\to\infty\). In particular, the relative error is at most one percent for every integer \(M\geq2502\). These constants and the starting vertex count are conservative.

Proof. The unique maximum and the exact normalization. Put \(h(\theta)=2\operatorname{Re}g(e^{i\theta})\) with the existing \(g(z)=-(1-z)\log(1-z)/z\). For \(0\leq x<\pi\), its explicit real form is

\[h(\pi-x)=4\cos^2(x/2)\log(2\cos(x/2))+x\sin x.\]

It extends continuously to zero at \(x=\pi\), is positive before that point, and strictly decreases. To check monotonicity, set \(c=\cos\theta\) in the already established positive integral:

\[\operatorname{Re}g(e^{i\theta}) =\int_0^1\frac{(1+u)(1-c)}{1-2uc+u^2}\,du.\]

The derivative of its integrand with respect to \(c\) is \(-(1+u)(1-u)^2/(1-2uc+u^2)^2<0\) for \(u<1\). This proves the unique maximum \(H=h(\pi)=4L\). Set \(f(x)=h(\pi-x)/H\), \(A(x)=\sec^2(x/2)\) and

\[I_n=\int_0^\pi f(x)^n A(x)\,dx,\qquad \alpha=\frac{a}{4L}.\]

Symmetry and \(|1-e^{i(\pi-x)}|^2=4\cos^2(x/2)\) give exactly

(69)\[\frac{C_M}{M^2q!\,(2L)^q}=\frac{q}{4\pi L}I_n.\]

Local estimates with explicit constants. The elementary bounds \(2/3<L<7/10\) imply \(1/16<\alpha<1/12\). They follow, for example, by bounding the positive series \(L=2\sum_{j\geq0}3^{-2j-1}/(2j+1)\). For \(0\leq x\leq1/2\), direct differentiation gives

\[\frac{d^4}{dx^4}h(\pi-x) =x\sin x+2\cos x\log(2\cos(x/2)) +\frac{\cos x}{1+\cos x}-2.\]

Its absolute value is at most one on this interval. Indeed, \(\cos x\geq7/8\), \(\cos(x/2)\geq31/32\), and \(\log(2\cos(x/2))\geq1/2\); the last inequality follows from \(e<3<(31/16)^2\). A lower bound is \(7/8+7/15-2>-1\). An upper bound is \(1/4+2(7/10)+1/2-2<1\). Taylor’s theorem, including the vanishing odd derivatives, yields

\[|f(x)-1+\alpha x^2|\leq x^4/64.\]

Monotonicity and this bound imply \(0\leq1-f(x)\leq(33/256)x^2\leq33/1024\). The inequality \(|\log(1-u)+u|\leq u^2/(2(1-u))\) therefore gives

\[|\log f(x)+\alpha x^2|\leq x^4/32,\qquad \log f(x)\leq-x^2/20.\]

For the second assertion one may use \(f(x)\leq1-(\alpha-1/256)x^2\) and \(\alpha-1/256>15/256>1/20\). Also \(1\leq A(x)\leq9/8\) and \(A(x)-1=\tan^2(x/2)\leq x^2/2\).

With \(G_n=\int_0^\infty e^{-\alpha nx^2}\,dx =\sqrt\pi/(2\sqrt{\alpha n})\), the mean-value inequality for exponentials gives

\[\begin{split}\left|\int_0^{1/2}f(x)^nA(x)\,dx -\int_0^{1/2}e^{-\alpha nx^2}\,dx\right| &\leq\int_0^\infty \left(\frac{9n}{256}x^4+\frac12x^2\right)e^{-nx^2/20}\,dx\\ &\leq \frac{21}{n}G_n.\end{split}\]

The Gaussian moments evaluate the ratio in the last line to at most

\[\frac{2\sqrt\alpha}{n} \left(\frac{27\,20^{5/2}}{2048} +\frac{20^{3/2}}8\right)<\frac{21}{n}.\]

For an entirely rational check, use \(2\sqrt\alpha<3/5\), \(\sqrt{20}<9/2\); the resulting numerator is \(5373/256<21\).

The singular part is controlled, not discarded. Although \(A(x)\) is singular at \(x=\pi\), the exact second moment is finite. The known \(c_3=\pi^2\) gives

\[I_2=\frac{\pi^3}{12L^2}<12.\]

Thus \(I_n\) exists for all \(n\geq2\). Monotonicity and the local estimate give \(f(x)\leq f(1/2)\leq79/80\) for \(x\geq1/2\), so

\[\frac1{G_n}\int_{1/2}^\pi f(x)^nA(x)\,dx \leq4\sqrt n\,(79/80)^{n-2}.\]

Here \(G_n>3/\sqrt n\) follows from \(\alpha<1/12\) and \(\pi>3\). Separately, the elementary Gaussian-tail bound gives

\[\frac1{G_n}\int_{1/2}^\infty e^{-\alpha nx^2}\,dx \leq\frac{16}{3\sqrt n}e^{-n/64}.\]

For every real \(n\geq1024\), each of these two displayed right sides is at most \(1/n\). For the first, dominate \((79/80)^{n-2}\) by \(e^{-(n-2)/80}\) and differentiate the logarithm of \(n^{3/2}e^{-(n-2)/80}\); it decreases after \(n=120\). At \(n=1024\), \(e^{1022/80}>2^{18}>4\,1024^{3/2}\), using \(18\log2<18(7/10)<1022/80\). The second follows in the same way: \(n^{1/2}e^{-n/64}\) decreases after \(n=32\), and at 1024 its required bound follows from \(e^{16}>2^{16}>512/3\). Combining the three errors proves

\[\left|I_n/G_n-1\right|\leq24/n\qquad(n\geq1024).\]

The exact normalization makes this a relative bound with leading term \(\widetilde B_M=B_M\sqrt{q/n}\). Finally, \(1\leq\sqrt{q/n}=\sqrt{1+1/n}\leq1+1/(2n)\), and hence

\[|C_M/B_M-1|\leq(24/n)(1+1/(2n))+1/(2n)\leq25/n.\]

This proves the theorem, including the prefactor and remainder.

Certified evaluation and independent integration

The standard-library evaluator tools.risk_research.vertex_large_dimension saves exact rational intervals for the normalized constant in (69). It encloses \(\log2\) by its positive series, \(\pi\) by Machin’s alternating arctangent formula, and square roots by integer square roots. Removing the factorial and exponential scale keeps reports short even at large \(M\).

The simple relative radius is \(25/(M-2)\) for \(M\geq1026\). For \(4\leq M<1026\), the same proof gives a weaker valid radius. Put

\[E_n=\frac{21}{n}+4\sqrt n(79/80)^{n-2} +\frac{16}{3\sqrt n}(64/65)^n.\]

Then \((1+1/(2n))E_n+1/(2n)\) bounds \(|C_M/B_M-1|\); use \(e^{-1/64}<64/65\). The tool rounds \(\sqrt n\) outward and clips a negative lower constant endpoint to zero. The small-\(M\) fallback can be wide; the existing exact-series enclosure remains useful there.

Optional python-flint restores the factorial/exponential scale with Arb. A separate integration path directly evaluates the analytic interval of \(I_n\); it does not use the large-\(M\) error bound. At a cutoff \(0<b<\pi\), its omitted endpoint interval satisfies

\[0\leq\int_b^\pi f(x)^nA(x)\,dx\leq12f(b)^{n-2}.\]

The callback forwards the analytic flag to the logarithm, and both the cutoff kernel and finiteness are checked. This integration path covers the singular neighborhood by a proved tail, including \(n=2\). Exact rational-series constant intervals provide another independent comparison at moderate \(M\).

For example, from a source checkout:

$ .venv/bin/python -m tools.risk_research.vertex_large_dimension --vertices 2502 --relative-target 1/100 --output .research/skeleton-risk/constant2502.json
$ .venv/bin/python -m tools.risk_research.vertex_large_dimension --vertices 1026 --with-arb --integral-check --output .research/skeleton-risk/constant1026.json

The second command requires optional python-flint. Precision settings are restored after ball calculations. Saved ball strings are rounded outward; source changes or computation failures preserve an existing output file. No public sampling API or mandatory package dependency is added.

The theorem concerns \(C_M\) alone. A simultaneous growing-\(M,D\) vertex asymptotic requires separate control of the \(M\) dependence in the finite-degree remainder from Explicit vertex errors at every fixed vertex count.