## Appendix B Mathematical Theorems and Proofs  
**Theorem A1**. : _The function \(T_{1}(\lambda)=(1+2\lambda)^{-1/2-2\lambda}\) is strictly concave over the interval \(-0.5<\lambda<0\), and attains its maximum value at_  
\[\lambda^{*}=\frac{-e+e^{W(e/2)}}{2e}\approx-0.1351,\]  
_rounded to 4 significant digits._  
Proof. : Notice that \(\ln T_{1}(\lambda)=(-1/2-2\lambda)\ln(1+2\lambda)\). Thus,  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\] \[\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda) =\frac{-6-8\lambda}{(1+2\lambda)^{2}}.\]  
For \(\lambda\in\mathbb{R}^{-}_{-1/2}\), we find that \((1+2\lambda)^{2}>0\), and \(-6-8\lambda\) is linear and decreasing in \(\lambda\). Thus, \(\frac{d^{2}}{d\lambda^{2}}\ln T_{1}(\lambda)<0\) for all \(\lambda\in\mathbb{R}^{-}_{-1/2}\). Since the logarithmic second derivative is negative, \(\ln T_{1}(\lambda)\) is strictly concave on \(\mathbb{R}^{-}_{-1/2}\). Since the logarithm is a strictly increasing function and \(\ln T_{1}(\lambda)\) is strictly concave, then \(T_{1}(\lambda)\) itself is also strictly concave on the interval \(\mathbb{R}^{-}_{-1/2}\). This proves the first part of the theorem. Now, to prove that \(T_{1}\) attains its maximum at \(\lambda^{*}\), we set the logarithmic derivative equal to zero:  
\[-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda}=0.\]  
Let \(x=1+2\lambda\):  
\[-2\ln x-\frac{2x-1}{x}=0\implies 2x\ln x+2x-1=0.\]  
Let \(x=e^{\mu}\): \(2 ue^{\mu}+2e^{\mu}-1=0\implies e^{\mu+1}(u+1)=\frac{e}{2}\). The solution of this the transcendental equation is \(u=W(\frac{e}{2})-1\). Thus, \(x=e^{W(e/2)}/e=1+2\lambda\), from which the maximum point \(\lambda=\lambda^{*}\) is determined. This completes the proof of the theorem. Figure A1 shows a sketch of \(T_{1}\) on the interval \(\mathbb{R}^{-}_{-1/2}\). The following lemma establishes the asymptotic equivalence of \((n_{q}+1)!\hat{K}_{n_{q}+1}^{\lambda_{q}}\ \forall_{t}\,n_{q}\). **Lemma A1**. : _Let \(\lambda_{q}>-\frac{1}{2}\) and consider the leading coefficient of the \((n_{q}+1)\)th-degree, \(\lambda_{q}\)-indexed SG polynomial defined by_  
\[\hat{K}_{n_{q}}^{\lambda_{q}}=\frac{2^{n_{q}-1}\Gamma(2\lambda_{q}+1)\Gamma(n_{ q}+\lambda_{q})}{\Gamma(\lambda_{q}+1)\Gamma(n_{q}+2\lambda_{q})}.\]  
_Then,_  
\[(n_{q}+1)!\,\hat{K}_{n_{q}+1}^{\lambda_{q}}\sim\Theta_{\lambda_{q}}n_{q}^{3/2- \lambda_{q}}\left(\frac{2n_{q}}{e}\right)^{n_{q}}\quad\forall_{t}\,n_{q},\]  
_where_  
\[\Theta_{\lambda_{q}}=\frac{\sqrt{2\pi}\,\Gamma(2\lambda_{q}+1)}{\Gamma( \lambda_{q}+1)}.\] (A1)  
Proof. : We begin by expressing \(\hat{K}_{n_{q}+1}^{\lambda_{q}}\) explicitly:  
\[\hat{K}_{n_{q}+1}^{\lambda_{q}}=\frac{2^{n_{q}}\Gamma(2\lambda_{q}+1)\Gamma(n_ {q}+\lambda_{q}+1)}{\Gamma(\lambda_{q}+1)\Gamma(n_{q}+2\lambda_{q}+1)}.\]  
Since  
\[(n+1)!\hat{K}_{n_{q}+1}^{\lambda_{q}}=O\bigg{(}n_{q}^{3/2-\lambda_{q}}\bigg{(} \frac{2n_{q}}{e}\bigg{)}^{n_{q}}\bigg{)},\quad\text{as $n_{q}\to\infty$},\] (A2)  
\(\forall_{t}\,\lambda_{q}>-1/2\) by [43] (Lemma 2.2) and [34] (Lemma 4.2). Then,  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]0 (A3)  
[PERSON]'s approximations to the factorial and gamma functions \(\ \forall_{t}\,n_{q}\) give  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]1 \[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]2 \[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]3  
Therefore,  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]4 (A4)Multiplying by the remaining terms:  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]5  
This shows that  
\[\frac{d}{d\lambda}\ln T_{1}(\lambda) =-2\ln(1+2\lambda)-\frac{1+4\lambda}{1+2\lambda},\]6  
which completes the proof. The next lemma is needed for the proof of the following lemma. **Lemma A2** (Falling Factorial Approximation). : _Let \(n=m+k:k=o(m)\)\(\forall_{t}\,m\in\mathbb{Z}^{+}\).