A simultaneous vertex-count and degree regime ============================================== Retain the uniform-simplex assumptions, the positive constant :math:`C_M` of :doc:`skeleton_risk_vertex_general` and the explicit finite-degree remainder from :doc:`skeleton_risk_vertex_general_asymptotics`. Here both :math:`M` and :math:`D` may grow. The argument controls the entire dependence on :math:`M`; it does not substitute a growing value into a fixed-:math:`M` limit. A constructive uniform regime ------------------------------ For integer :math:`D\geq2^{1024}`, put .. math:: N=\lfloor\log_2D\rfloor,\qquad \ell=\lceil\log_2(N+2)\rceil,\qquad h(D)=1+\left\lfloor\frac{N}{16\ell}\right\rfloor. .. prf:theorem:: A growing-vertex vertex asymptotic :label: thm-risk-joint-vertex-growth For every pair of integers :math:`D\geq2^{1024}` and :math:`4\leq M\leq h(D)`, .. math:: :label: risk-joint-vertex-error \left|\frac{D^{M+1}a_0(M,D)}{C_M}-1\right| \leq D^{-3/4}. In particular, :math:`h(D)\to\infty`, with :math:`h(D)\sim\log D/(16\log\log D)`, and the relative error tends to zero uniformly over all the indicated vertex counts. The start :math:`2^{1024}` is astronomical. This is a conservative theoretical regime, not a claim of useful finite-degree precision near that scale. The integer ceiling can decrease when :math:`\ell` jumps; no monotonicity of :math:`h` is needed or asserted. .. prf:proof:: **A uniform lower bound for the positive constant.** Put :math:`q=M-1\geq3` and :math:`L=\log2`. The positive integral for :math:`c_q` has kernel :math:`H(\theta)=2\operatorname{Re}g(e^{i\theta})`. Its monotonicity on :math:`[0,\pi]` follows by differentiating the positive integral with respect to :math:`c=\cos\theta`: .. math:: \frac{\partial}{\partial c} \frac{(1+u)(1-c)}{1-2uc+u^2} =-\frac{(1+u)(1-u)^2}{(1-2uc+u^2)^2}<0\quad(0\leq u<1). At :math:`\theta=\pi/2`, the explicit logarithm gives :math:`H(\pi/2)=L+\pi/2>2`. On the two arcs :math:`|\theta|\in[\pi/2,\pi]`, of total length :math:`\pi`, also :math:`|1-e^{i\theta}|^2\leq4`. Therefore .. math:: c_q>q\,2^{q-4},\qquad C_M=M^2q!2^{2-q}c_q>\frac{M^2q!q}{4}. This bound holds for every :math:`M\geq4`. **Control of every error-budget constant.** Use the notation of the finite-degree error proof, made explicit here. For :math:`1\leq r\leq q`, let .. math:: P_r&=r!r^{r-1},\\ B_r&=2^{r+1}r\,r!(r+1)^{r-1},\\ A_r&=\sum_{a=1}^{r-1}\binom ra P_aP_{r-a},\\ \Delta_r&=2^{r-2}\sum_{a=1}^{r-1}\binom ra (B_aP_{r-a}+P_aB_{r-a}). The last two quantities are needed only for :math:`r\geq2`. The elementary estimates :math:`r!\leq q^r`, :math:`r\leq q` and :math:`r+1\leq2q` give .. math:: P_r\leq q^{2r},\quad B_r\leq(4q^2)^r,\quad A_r\leq(2q^2)^r,\quad\Delta_r\leq(10q^2)^r. For :math:`A_r`, sum the binomial coefficients. For :math:`\Delta_r`, extend the positive sums to both endpoints and use :math:`2^{r-2}(2\,5^r)q^{2r}=(10q^2)^r/2`. The explicit limiting and diagonal constants are .. math:: J_q={}&2qA_{q-1},\\ K_q={}&P_q+A_q+q\bigl(2^q\Delta_{q-1} +(20+3^{q-2})A_{q-1}\bigr)\\ &+\sum_{k=2}^{q-2}\binom qk B_k \bigl(2^{q-k+2}\Delta_{q-k}+5\,2^kA_{q-k}\bigr) +8(1+q^2+J_q). Empty sums, including the sum at :math:`q=3`, are zero. Directly :math:`J_q\leq(2q^2)^q`. The two terms multiplied by :math:`q` in :math:`K_q` are at most :math:`(20q^2)^q` and :math:`(6q^2)^q`, respectively; use :math:`20+3^{q-2}\leq2\,3^q` for :math:`q\geq3`. The two binomial sums are at most :math:`4(24q^2)^q` and :math:`5(10q^2)^q`. The last term is at most :math:`16(2q^2)^q`, since :math:`1+q^2\leq(2q^2)^q`. It follows that .. math:: K_q\leq29(24q^2)^q\leq(96q^2)^q, because :math:`29\leq4^q` for :math:`q\geq3`. The already proved finite-degree remainder is .. math:: |a_0(M,D)-C_MD^{-M-1}| &\leq R_M(1+\log D)^{q-1}D^{-M-2},\qquad D\geq256,\\ R_M&=\frac{M^2q!}{2^q} \bigl(96K_q+(273+q(q+1))J_q\bigr). Since :math:`273+q(q+1)\leq32q^2`, division by the lower bound for :math:`C_M` gives .. math:: \frac{R_M}{C_M} &\leq96(96q^2)^q+32q^2(2q^2)^q\\ &\leq(480q^2)^q+(32q^2)^q\leq(512q^2)^q. In the first line, :math:`2^{2-q}/q\leq1` was discarded. For the second line, :math:`96\leq5^q` and :math:`32q^2\leq16^q` for :math:`q\geq3`; the latter follows from :math:`q^2\leq4^q` and :math:`32\leq4^q`. This controls every constant in the old error theorem with a single explicit bound, uniformly in :math:`M`. **The growing range.** Here :math:`N\geq1024`, and :math:`q\leq N/(16\ell)\leq N+2`. Also :math:`512\leq N+2`, :math:`1+\log D