## Appendix B Mathematical Theorems and Proofs
Then the falling factorial satisfies_  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]7  
Proof. : Express the falling factorial as  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]8  
Take logarithms to convert the product into a sum:  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]9  
Since \(m\) is large, we can approximate the sum by using the midpoint rule of integrals:  
\[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]0 \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]1  
For \(k=o(m)\) and large \(m\), Taylor series expansions produce  
\[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]2 \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]3 \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]4  
and  
\[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]5 \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]6  
Substituting these approximations back yields  
\[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]7Exponentiating gives  
\[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]8 \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]9 (101) \[-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda}=0.\]0  
The next lemma gives the asymptotic upper bound on the growth rate of the parameteric scalar \(\hat{\chi}^{\lambda}_{n,m}\). **Lemma A3**. : _Let \(\lambda>-1/2\) and \(m\in\mathbb{Z}^{+}:m\gg\lambda\). Then_  
\[\hat{\chi}^{\lambda}_{n,m}=\left\{ \begin{array}{ll}O\Bigl{(}\left(\frac{4}{ \varepsilon}\right)^{m}m^{m+\frac{1}{2}-\lambda}\Bigr{)}&\text{if $n\sim m$,}\\ O\bigl{(}n^{2m}\bigr{)}&\forall_{t}\,n\gg m.\end{array} \right. \tag{103}\]  
Proof. : Let \(n\sim m\). Then \(n=m+k:k=o(m)\), and we can write \((n)_{m}\approx m^{m+1/2}e^{-m}\) by Lemma A2. Substituting this approximation and the sharp inequalities of the Gamma function [36] (InEquation (96)) into Equation (35) give  
\[\chi(n,m,\lambda)\gtrsim\hat{\ell}_{\lambda}\,m^{m+1/2}e^{-m}\cdot m^{-m- \lambda}n^{-n-2\lambda+\frac{1}{2}}(m+n)^{m+n+2\lambda-\frac{1}{2}}\sim\ell_{ \lambda}\left(\frac{4}{e}\right)^{m}m^{m+\frac{1}{2}-\lambda}, \tag{104}\]  
where \(\ell_{\lambda}=\frac{4^{\lambda}}{\sqrt{2}}\hat{\ell}_{\lambda}\), and \(\hat{\ell}_{\lambda}\in\mathbb{R}^{+}\) is a \(\lambda\)-dependent constant. Now, consider the case when \(n\gg m\). Applying [PERSON]'s approximation of factorials and the same sharp inequalities of the Gamma function on the parameteric scalar \(\hat{\chi}^{\lambda}_{n,m}\) give  
\[\hat{\chi}^{\lambda}_{n,m}\gtrsim h^{\lambda}_{m}n^{\frac{1}{2}+n}(n-m)^{\frac{1}{2}(-1+3m-3n)}(n+2 \lambda-1)^{\frac{1}{2}-n-2\lambda}\times \tag{105}\] \[(n+m+2\lambda-1)^{n+m+2\lambda-\frac{1}{2}}\sinh^{\frac{n-m}{2}} \left(\frac{1}{n-m}\right)\times\] (106) \[\left[(n+2\lambda-1)\sinh\biggl{(}\frac{1}{n+2\lambda-1}\biggr{)} \right]^{\frac{1}{2}(1-n-2\lambda)}\times\] (107) \[\left[(n+m+2\lambda-1)\sinh\biggl{(}\frac{1}{n+m+2\lambda-1} \biggr{)}\right]^{\frac{1}{2}(n+m-1)+\lambda}\] (108) \[\sim h^{\lambda}_{m}n^{\frac{1}{2}(5m-n)}\sinh^{\frac{n-m}{2}} \left(\frac{1}{n-m}\right), \tag{109}\]  
where \(h^{\lambda}_{m}=e^{5m/2}h^{\lambda}_{m}\), and \(h^{\lambda}_{m}\) is a constant dependent on the parameters \(m\) and \(\lambda\). Since \(1/(n-m)\) is small \(\forall_{t}\,n\), we have  
\[\sinh^{\frac{n-n}{2}}\left(\frac{1}{n-m}\right)\approx\left(\frac{1}{n-m} \right)^{\frac{n-m}{2}}\sim n^{\frac{n-m}{2}}\quad\forall n\gg m. \tag{110}\]  
Substituting (110) into (109) yields  
\[\hat{\chi}^{\lambda}_{n,m}\lesssim h^{\lambda}_{m}n^{2m},\]  
from which the proof is accomplished.