正交多项式是在19世纪末由Chebyshev对连分式的研究发展起来的. 随后有越来越多的数学家, Szegö, Bernstein, Akhiezer, Ismail等等, 对它进行研究. 正交多项式有很多有用的性质, 并且广泛应用于数学及数学物理中, 例如近似理论, 随机过程, 随机矩阵. 自20世纪60年代, Géza Freud开始研究正交于实数域$ {{\mathbb R}} $上的指数型权函数$ {\rm e}^{-| x|^m} $, 其中$ m $属于自然数集$ \mathbb{N} $, 的正交多项式$ P_n(x) $以及相关的循环系数$ \beta_n $的渐近行为. 此类最高次项系数为1的(亦称首一的)正交多项式符合递推关系

$ \begin{eqnarray*} xP_n(x) = P_{n+1}(x)+\beta_nP_{n-1}(x). \end{eqnarray*} $

上式中的循环系数$ \beta_n $和Painlevé方程的解有关, 这也是研究正交多项式的一个重要方面. 在应用数学, 计算数学以及物理中, 经常遇到的问题是将上述递推关系中的循环系数$ \beta_n $与权函数的性质联系起来. 这些关系可以转化为特殊的循环系数性质. 相当多的理论研究^[1-3]已经涉及了正交性质和循环系数的关系.

Freud^[4]给出了关于权函数$ |x|^{\rho}{\rm e}^{-|x|^m} $, $ \rho>-1 $, $ m>0 $的循环系数$ \beta_n $的一个猜想

(1.1) $ \begin{equation} \lim\limits_{n\rightarrow \infty}\frac{\beta_n}{n^{\frac{2}{m}}} = \bigg(\frac{\Gamma\left(\frac{m}{2}\right)\Gamma\left(1+\frac{m}{2}\right)}{\Gamma(m+1)}\bigg)^{\frac{2}{m}}, \end{equation} $

并证明了当$ m = 2, \; 4, \; 6 $时猜想成立, 于是大胆猜测对于任意的正偶数$ m $, (1.1) 式都成立. 这个猜想在1985年被Magnus^[5]证明. 随后, Lubinsky^[6]又给出了一个更广义的证明.

关于Freud - 型权函数$ {w}(x;t) = {\rm e}^{-x^4+tx^2}, \; \; t, x\in{{\mathbb R}} , $ Magnus^[7]指出循环系数$ \beta_n $所满足的非线性离散方程$ \beta_n(\beta_{n-1}+\beta_n+\beta_{n+1})+2t\beta_n = n, $正是一个离散的二阶Painlevé方程. 后来, Clarkson^[8-9]又研究了一个更为广义的Freud - 型权函数

$ {w}(x;t, \lambda) = |x|^{2\lambda+1}{\rm e}^{-x^4+tx^2}, \; \; t, \; x\in{{\mathbb R}} , \; \; \lambda>-1, $

它所对应的循环系数$ \beta_n $还可以用含有抛物柱面函数的Wronskian行列式表示. 最近, 王丹等^[19]又讨论了一类六次的Freud - 型权函数

$ {w}(x;t, \lambda) = |x|^{\alpha}{\rm e}^{-x^6+tx^2}, \; \; x, \; t\in{{\mathbb R}} , \; \; \alpha>-1. $

朱孟坤等在文献[27]中构造了一类介于高斯权函数和四次Freud - 型权函数之间的权函数

$ {w}(x;s, \alpha, N) = |x|^{\alpha}{\rm e}^{-N\left[x^2+s\left(x^4-x^2\right)\right]}, \; x\in{{\mathbb R}} , \; \; \alpha>-1, \; \; s\in[0, 1], \; \; N>0, $

得到相关的循环系数满足第一类离散的Painlevé方程, 并讨论了该方程的解. 另外, 这篇文章中建立了正交于该权函数的多项式与一类双合流的Heun方程的关系. 这对正交多项式与Heun方程的交叉研究提供了一个思路. 更多关于Freud - 型权函数的研究, 可参考文献[10-13, 15].

设$ \{P_n(x)\} $为最高次项系数和次数分别为1 (亦称首一的)和$ n $的多项式, 并且正交于一个十次的Freud - 型权函数

(1.2) $ \begin{equation} {w}(x;t, \alpha) = |x|^{\alpha}{\rm e}^{-x^{10}+tx^2}, \; \; t, \; x\in{{\mathbb R}} , \; \alpha>-1. \end{equation} $

根据正交条件, 可知

(1.3) $ \begin{equation} \int_{{{\mathbb R}} }P_m(x)P_n(x){w}(x;t, \alpha){\rm d}x = h_n(t, \alpha)\delta_{(m, n)}, \end{equation} $

这里, $ h_n(t, \alpha)>0 $, $ \delta_{(m, n)} $是指克罗内克$ \delta $函数, 即

$ \delta_{(m, n)} = \left\{ \begin{array}{ll} 0&, \; \; m\neq n, \\ 1&, \; \; m = n. \end{array} \right. $

对正交于Freud - 型权函数(1.2)的多项式而言, 它们的三项循环递推关系式是

(1.4) $ \begin{equation} xP_n(x) = P_{n+1}(x)+\beta_nP_{n-1}(x), \; \; n = 0, \; 1, \; 2, \cdots, \end{equation} $

其中

(1.5) $ \begin{equation} \beta_n(t, \alpha) = \frac{1}{h_{n-1}(t, \alpha)}\int_{{{\mathbb R}} }xP_{n-1}(x)P_n(x){w}(x;t, \alpha){\rm d}x; \end{equation} $

且$ h_n $表示$ P_n $的$ L^2 $ -范数的平方. 当$ n = 0 $或$ 1 $时, 规定

$ \beta_0(t, \alpha)P_{-1}(x) = 0, {\quad} P_0(x) = 1. $

另外, $ P_n(x) $的展开式^[10]为$ P_{2k}(z) = z^{2k}+{\cal P}(2k; t)z^{2k-2}+\dots+P_{2k}(0) $和

$ \begin{eqnarray*} P_{2k+1}(z)& = &z^{2k+1}+{\cal P}(2k+1;t)z^{2k-1}+\dots+Cz\\ & = &z\Big(z^{2k}+{\cal P}(2k+1;t)z^{2k-2}+\dots+C\Big). \end{eqnarray*} $

将(1.4) 式带入(1.5) 式, 可以得到$ \beta_n(t, \alpha) $的另外一种表述

(1.6) $ \begin{equation} \beta_n(t, \alpha) = \frac{h_n(t, \alpha)}{h_{n-1}(t, \alpha)}. \end{equation} $

关于十次Freud - 型权函数(1.2) 的矩定义为

(1.7) $ \begin{equation} \mu_k(t, \alpha) = \int_{{{\mathbb R}} }x^k|x|^{\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x, \; \; k = 0, \; 1, 2, \cdots, \end{equation} $

它所决定的Hankel行列式记为$ D_n(t, \alpha) $. 根据正交多项式的广义定理^{[14, 16-18]}, $ D_n (t, \alpha) $还有两种表现形式

(1.8) $ \begin{equation} D_n(t, \alpha) = \det\Big(\mu_{j+k}(t, \alpha)\Big)_{j, k = 0}^{n-1} = \prod\limits_{j = 0}^{n-1}h_j(t, \alpha), \; \; n\geq 1, \end{equation} $

这里$ D_0 = 1 $, $ D_{-1} = 0 $. 此外, $ \beta_n(t, \alpha) $也可以用$ D_n(t, \alpha) $表示^[12]

(1.9) $ \begin{equation} \beta_n(t, \alpha) = \frac{D_{n+1}(t, \alpha)D_{n-1}(t, \alpha)}{D_n^2(t, \alpha)}. \end{equation} $

近年来, 较大病态Hankel矩阵的最小特征值的渐近行为成为一个研究热点, 具体内容可参考文献[28-38], 当$ \alpha = t = 0 $时, (1.7) 式所决定的Hankel矩阵的最小特征值在本文中也有讨论.

Hankel矩阵是随机矩阵中最基本的研究对象之一. 其行列式可以代表一个特定随机矩阵的配分函数, 或者可以代表一个随机变量与系综相关的母函数, 亦或者与最大特征值的分布有关. 在给定一个权函数的前提下, 最大或最小特征值的研究又可以提供有关Hankel矩阵非常有用的信息.

在本文第2、3节, 我们将分别给出权函数(1.2)所对应的循环系数$ \beta_n $的行列式表示, 和它所符合的一个非线性离散方程. 基于第3节结果, 我们在第四部分给出了$ \beta_n $的一个近似值. 作为第2–4节的应用, 我们在第5、6节分别得到了其所对应的Hankel行列式满足的一个方程, 和当$ t = \alpha = 0 $时, 得到对应Hankel矩阵的最小特征值的渐近行为.

定义两个新的Hankel行列式

(2.1) $ \begin{eqnarray} \widehat{D}_n(t, \alpha) = \det\Big(\mu_{2j+2k}(t, \alpha)\Big)_{j, k = 0}^{n-1} = \left|\begin{array}{ccccc} \mu_0(t, \alpha) &\mu_2(t, \alpha)& \cdots&\mu_{2n-2}(t, \alpha) \\ \mu_2(t, \alpha) & \mu_4(t, \alpha) & \cdots &\mu_{2n}(t, \alpha)\\ \vdots & \vdots &\ddots&\vdots\\ \mu_{2n-2}(t, \alpha)&\mu_{2n}(t, \alpha)&\cdots &\mu_{4n-4}(t, \alpha) \end{array}\right| \end{eqnarray} $

和

(2.2) $ \begin{eqnarray} \widetilde{D}_n(t, \alpha) = \det\Big(\mu_{2j+2k+2}(t, \alpha)\Big)_{j, k = 0}^{n-1} = \left|\begin{array}{ccccc} \mu_2(t, \alpha) &\mu_4(t, \alpha)& \cdots&\mu_{2n}(t, \alpha) \\ \mu_4(t, \alpha) & \mu_6(t, \alpha) & \cdots &\mu_{2n+2}(t, \alpha)\\ \vdots & \vdots &\ddots&\vdots\\ \mu_{2n}(t, \alpha)&\mu_{2n+2}(t, \alpha)&\cdots &\mu_{4n-2}(t, \alpha) \end{array}\right|, \end{eqnarray} $

它们和(1.8)式可以建立起这样的联系

(2.3) $ \begin{equation} D_{2n}(t, \alpha) = \widehat{D}_n(t, \alpha)\widetilde{D}_n(t, \alpha), \; \; \; D_{2n+1}(t, \alpha) = \widehat{D}_{n+1}(t, \alpha)\widetilde{D}_n(t, \alpha). \end{equation} $

与此同时, Freud -型权函数(1.2)作为定义在实数范围内恒正的对称函数, 所以$ \widehat{D}_n(t, \alpha) $和$ \widetilde{D}_n(t, \alpha) $还可用Wronskian行列式表示.

引理2.1  设$ W(\psi_1, \psi_2, \cdots, \psi_n) $是Wronskian行列式, 对于Freud - 型权函数(1.2),

(2.4) $ \begin{equation} \widehat{D}_n(t, \alpha) = W\bigg(\mu_0(t, \alpha), \frac{{\rm d}\mu_0(t, \alpha)}{{\rm d}t}, \cdots, \frac{{\rm d}^{n-1}\mu_0(t, \alpha)}{{\rm d}t^{n-1}}\bigg) \end{equation} $

和

(2.5) $ \begin{equation} \widetilde{D}_n(t, \alpha) = W\bigg(\frac{{\rm d}\mu_0(t, \alpha)}{{\rm d}t}, \frac{{\rm d}^2 \mu_0(t, \alpha)}{{\rm d}t^2}, \cdots, \frac{{\rm d}^{n}\mu_0(t, \alpha)}{{\rm d}t^{n}}\bigg) \end{equation} $

成立.

证  对于任一个非负整数$ k $, 有

$ \frac{{\rm d}^k\mu_0(t, \alpha)}{{\rm d}t^k} = \frac{{\rm d}^k}{{\rm d}t^k}\int_{{{\mathbb R}} } {w}(x;t, \alpha){\rm d}x = \int_{{{\mathbb R}} }x^{2k}|x|^{\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x = \mu_{2k}(t, \alpha), $

再结合$ \widehat{D}_n(t, \alpha) $和$ \widetilde{D}_n(t, \alpha) $的定义(2.1) 和(2.2), 即可得证(2.4) 和(2.5)式.

注2.1

$ \mu_{2k+1}(t, \alpha) = 0. $

借助Mathematica软件, 我们可以用超几何函数$ {}_1F_4(a_1; b_1, b_2, b_3, b_4; z) $表示0 -阶矩

(2.6) $ \begin{eqnarray} \mu_0(t, \alpha)& = &\int_{{{\mathbb R}} }|x|^{\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x = 2\int_0^{\infty}|x|^{\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x\\ & = &\frac{1}{5}\Gamma\Big(\frac{1+\alpha}{10}\Big)\; _1\textbf{F}_4\Big(\frac{1+\alpha}{10};\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5};\frac{t^5}{5^5}\Big) \\ && +\frac{t}{5}\Gamma\Big(\frac{3+\alpha}{10}\Big)\; _1\textbf{F}_4\Big(\frac{3+\alpha}{10};\frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{6}{5};\frac{t^5}{5^5}\Big)\\ &&+\frac{t^2}{10}\Gamma\Big(\frac{5+\alpha}{10}\Big)\; _1\textbf{F}_4\Big(\frac{5+\alpha}{10};\frac{3}{5}, \frac{4}{5}, \frac{6}{5}, \frac{7}{5};\frac{t^5}{5^5}\Big) \\ && +\frac{t^3}{30}\Gamma\Big(\frac{7+\alpha}{10}\Big)\; _1\textbf{F}_4\Big(\frac{7+\alpha}{10};\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5};\frac{t^5}{5^5}\Big)\\ & &+\frac{t^4}{120}\Gamma\Big(\frac{9+\alpha}{10}\Big)\; _1\textbf{F}_4\Big(\frac{9+\alpha}{10};\frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5};\frac{t^5}{5^5}\Big). \end{eqnarray} $

引理2.2  关于Freud - 型权函数(1.2)的Hankel行列式$ \widetilde{D}_n(t, \alpha) $和$ \widehat{D}_n(t, \alpha) $具有如下关系

(2.7) $ \begin{equation} \widetilde{D}_n(t, \alpha) = \widehat{D}_n(t, \alpha+2). \end{equation} $

证  $ \forall m\in [0, 2n-2] $, 有

$ \begin{eqnarray*} \mu_{2m+2}(t, \alpha)& = &\int_{{{\mathbb R}} }x^{2m+2}|x|^{\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x = 2\int_0^{\infty}x^{2m+2+\alpha}{\rm e}^{-x^{10}+tx^2}{\rm d}x\\ & = &2\int_0^{\infty}x^{2m}|x|^{\alpha+2}{\rm e}^{-x^{10}+tx^2}{\rm d}x = \int_{{{\mathbb R}} }x^{2m}|x|^{\alpha+2}{\rm e}^{-x^{10}+tx^2}{\rm d}x\\ & = &\mu_{2m}(t, \alpha+2), \end{eqnarray*} $

对照(2.1) 和(2.2) 式, 即可得证(2.7)式.

定理2.1   正交于Freud - 型权函数(1.2)的首一的正交多项式$ P_n(x) $满足

(2.8) $ \begin{equation} P_{n+1}(x) = xP_n(x)-\beta_n(t, \alpha)P_{n-1}(x), \end{equation} $

这里

(2.9) $ \begin{equation} \beta_{2n}(t, \alpha) = \frac{\widehat{D}_{n+1}(t, \alpha)\widehat{D}_{n-1}(t, \alpha+2)}{\widehat{D}_n(t, \alpha)\widehat{D}_n(t, \alpha+2)} = \frac{\rm d}{{\rm d}t}\ln\frac{\widehat{D}_n(t, \alpha+2)}{\widehat{D}_n(t, \alpha)}, \end{equation} $

(2.10) $ \begin{equation} \beta_{2n+1}(t, \alpha) = \frac{\widehat{D}_{n}(t, \alpha)\widehat{D}_{n+1}(t, \alpha+2)}{\widehat{D}_{n+1}(t, \alpha)\widehat{D}_n(t, \alpha+2)} = \frac{\rm d}{{\rm d}t}\ln\frac{\widehat{D}_{n+1}(t, \alpha)}{\widehat{D}_n(t, \alpha+2)}, \end{equation} $

并且$ P_0(x) = 1 $, $ \beta_0(t, \alpha)P_{-1}(x) = 0 $.

证  根据(1.9)和(2.3)式, 有

$ \beta_{2n}(t, \alpha) = \frac{\widehat{D}_{n+1}(t, \alpha)\widetilde{D}_{n-1}(t, \alpha)}{\widetilde{D}_{n}(t, \alpha)\widehat{D}_{n}(t, \alpha)}{, } $

用引理2.2, 代换上式的$ \widetilde{D}_{n-1}(t, \alpha) $和$ \widetilde{D}_n(t, \alpha) $, 可得

$ \beta_{2n}(t, \alpha) = \frac{\widehat{D}_{n+1}(t, \alpha)\widehat{D}_{n-1}(t, \alpha+2)}{\widehat{D}_{n}(t, \alpha+2)\widehat{D}_{n}(t, \alpha)}{.} $

另一方面, Vein和Dale^[20]已经证明

$ \widehat{D}_{n+1}(t, \alpha)\widetilde{D}_{n-1}(t, \alpha) = \widehat{D}_{n}(t, \alpha)\frac{{\rm d}\widetilde{D}_{n+1}(t, \alpha)}{{\rm d}t}-\widetilde{D}_{n}(t, \alpha)\frac{{\rm d}\widehat{D}_{n}(t, \alpha)}{{\rm d}t}, $

那么$ \beta_{2n}(t, \alpha) $也可写成

$ \beta_{2n}(t, \alpha) = \frac{1}{\widetilde{D}_{n}(t, \alpha)}\frac{{\rm d}\widetilde{D}_{n} (t, \alpha)}{{\rm d}t}-\frac{1}{\widehat{D}_{n}(t, \alpha)}\frac{{\rm d}\widehat{D}_{n}(t, \alpha)}{{\rm d}t} = \frac{\rm d}{{\rm d}t}\ln\frac{\widetilde{D}_{n}(t, \alpha)}{\widehat{D}_{n}(t, \alpha)}, $

再结合引理2.2, 可得

$ \beta_{2n}(t, \alpha) = \frac{\rm d}{{\rm d}t}\ln\frac{\widehat{D}_{n}(t, \alpha+2)}{\widehat{D}_{n}(t, \alpha)}. $

至此, 得证(2.9)式. 同理可得(2.10)式.

注2.2  $ \beta_0(t, \alpha) = 0 $, $ \beta_1(t, \alpha) = \frac{\rm d}{{\rm d}t}\ln\mu_0(t, \alpha) $.

定理3.1   对于(1.4) 式中的循环系数$ \beta_n(t, \alpha) $符合如下的非线性离散方程

(3.1) $ \begin{eqnarray} & &10\beta_n\Big(\beta_{n-4}\beta_{n-3}\beta_{n-2}\beta_{n-1}+\beta_{n-3}^2\beta_{n-2}\beta_{n-1}+2\beta_{n-3}\beta_{n-2}^2\beta_{n-1}+\beta_{n-2}^3\beta_{n-1} \\ &&+2\beta_{n-3}\beta_{n-2}\beta_{n-1}^2+3\beta_{n-2}^2\beta_{n-1}^2+3\beta_{n-2}\beta_{n-1}^3+\beta_{n-1}^4 +2\beta_{n-3}\beta_{n-2}\beta_{n-1}\beta_n\\ &&+2\beta_{n-2}^2\beta_{n-1}\beta_{n}+6\beta_{n-2}\beta_{n-1}^2\beta_{n} +4\beta_{n-1}^3\beta_{n}+3\beta_{n-2}\beta_{n-1}\beta_{n}^2+6\beta_{n-1}^2\beta_{n}^2 \\ &&+4\beta_{n-1}\beta_n^3+\beta_n^4+\beta_{n-3}\beta_{n-2}\beta_{n-1}\beta_{n+1} +\beta_{n-2}^2\beta_{n-1}\beta_{n+1}+2\beta_{n-2}\beta_{n-1}^2\beta_{n+1}\\ & &+\beta_{n-1}^3\beta_{n+1}+4\beta_{n-2}\beta_{n-1}\beta_{n}\beta_{n+1} +6\beta_{n-1}^2\beta_{n}\beta_{n+1}+9\beta_{n-1}\beta_{n}^2\beta_{n+1}+4\beta_{n}^3\beta_{n+1}\\ &&+\beta_{n-2}\beta_{n-1}\beta_{n+1}^2+\beta_{n-1}^2\beta_{n+1}^2+6\beta_{n-1} \beta_{n}\beta_{n+1}^2+6\beta_{n}^2\beta_{n+1}^2+\beta_{n-1}\beta_{n+1}^3 +4\beta_{n}\beta_{n+1}^3\\ & &+\beta_{n+1}^4+\beta_{n-2}\beta_{n-1}\beta_{n+1}\beta_{n+2}+\beta_{n-1}^2\beta_{n+1}\beta_{n+2} +3\beta_{n-1}\beta_n\beta_{n+1}\beta_{n+2}+2\beta_{n}^2\beta_{n+1}\beta_{n+2} \\ & &+\beta_{n-1}\beta_{n+1}^2\beta_{n+2}+4\beta_{n}\beta_{n+1}^2\beta_{n+2}+2\beta_{n+1}^3\beta_{n+2} +\beta_{n+1}^2\beta_{n+2}^2+\beta_{n-1}\beta_{n}\beta_{n+1}\beta_{n+3} \\ & & +\beta_{n}^2\beta_{n+1}\beta_{n+3}+\beta_{n-1}\beta_{n+1}^2\beta_{n+3} +2\beta_{n}\beta_{n+1}^2\beta_{n+3}+\beta_{n+1}^3\beta_{n+3} +\beta_{n-1}\beta_{n+1}\beta_{n+2}\beta_{n+3}\\ & &+3\beta_{n}\beta_{n+1}\beta_{n+2}\beta_{n+3} +3\beta_{n+1}^2\beta_{n+2}\beta_{n+3}+2\beta_{n+1}\beta_{n+2}^2\beta_{n+3}+\beta_{n-1}\beta_{n+1}\beta_{n+3}^2 \\ && +\beta_{n}\beta_{n+1}\beta_{n+3}^2+\beta_{n+1}^2\beta_{n+3}^2+2\beta_{n+1}\beta_{n+2}\beta_{n+3}^2 +\beta_{n+1}\beta_{n+2}\beta_{n+3}\beta_{n+4}\Big)-2t\beta_{n}\\ & = &n+\frac{\alpha}{2}-\frac{(-1)^n\alpha}{2}. \end{eqnarray} $

证  为了方便起见, 记

$ {w}(x) = |x|^{\alpha}{\rm e}^{-x^{10}+tx^2} = |x|^{\alpha}{w}_0(x) = |x|^{\alpha}{\rm e}^{-v(x)}, \; \; x, t\in{{\mathbb R}} , \; \; \alpha>-1, $

这里$ {w}_0(x) = {\rm e}^{-x^{10}+tx^2}, \; v(x) = x^{10}-tx^2. $反复应用(1.4)式, 可以得到

(3.2) $ \begin{eqnarray} x^5P_n(x)& = &P_{n+5}(x)+\beta_{n+4}P_{n+3}(x)+(P_{n+3}(x)+\beta_{n+2}P_{n+1}(x))(\beta_n+\beta_{n+1}+\beta_{n+2}+\beta_{n+3})\\ & &+(P_{n+1}(x)+\beta_nP_{n-1}(x))(\beta_n\beta_{n-1}+\beta_n^2+2\beta_n\beta_{n+1}+\beta_{n+1}^2+\beta_{n+1}\beta_{n+2})\\ &&+(P_{n-1}(x)+\beta_{n-2}P_{n-3}(x))\beta_{n-1}\beta_n(\beta_{n-2}+\beta_{n-1}+\beta_n+\beta_{n+1})\\ & &+(P_{n-3}(x)+\beta_{n-4}P_{n-5}(x))\beta_{n-3}\beta_{n-2}\beta_{n-1}\beta_n, \end{eqnarray} $

(3.3) $ \begin{eqnarray} x^4P_{n-1}(x) & = &P_{n+3}(x)+\beta_{n+2}P_{n+1}(x)+(P_{n+1}(x)+\beta_{n}P_{n-1}(x))(\beta_n+\beta_{n+1}+\beta_{n-1})\\ &&+(P_{n-1}(x)+\beta_{n-2}P_{n-3}(x))\beta_{n-1}(\beta_{n-2}+\beta_{n-1}+\beta_{n})\\ &&+(P_{n-3}(x)+\beta_{n-4}P_{n-5}(x))\beta_{n-3}\beta_{n-2}\beta_{n-1}. \end{eqnarray} $

将(3.2) 和(3.3) 式带入到如下的积分计算中, 并结合(1.3) 式, 我们得到

(3.4) $ \begin{eqnarray} &&\int_{{{\mathbb R}} }\Big(P_n(x)P_{n-1}(x)|x|^{\alpha}\Big)^{\prime}{w}_0(x){\rm d}x \\ & = &P_n(x)P_{n-1}(x){w}(x)\mid_{-\infty}^{+\infty} -\int_{{{\mathbb R}} }P_n(x)P_{n-1}(x)|x|^{\alpha}{\rm d}{w}_0(x)\\ & = &\int_{{{\mathbb R}} }P_{n-1}(x)P_n(x){w}(x)(10x^9-2tx){\rm d}x\\ & = &10\int_{{{\mathbb R}} }(x^5P_n(x))(x^4P_{n-1}(x)){w}(x){\rm d}x-2t\int_{{{\mathbb R}} }xP_{n-1}(x)P_n(x){w}(x){\rm d}x\\ & = &10\beta_n\Big[\beta_{n-4}\beta_{n-3}\beta_{n-2}\beta_{n-1}+\beta_{n-3}^2\beta_{n-2}\beta_{n-1} +2\beta_{n-3}\beta_{n-2}^2\beta_{n-1}+\beta_{n-2}^3\beta_{n-1}\\ & &+2\beta_{n-3}\beta_{n-2}\beta_{n-1}^2+3\beta_{n-2}^2\beta_{n-1}^2+3\beta_{n-2}\beta_{n-1}^3 +\beta_{n-1}^4+2\beta_{n-3}\beta_{n-2}\beta_{n-1}\beta_n\\ &&+2\beta_{n-2}^2\beta_{n-1}\beta_{n}+6\beta_{n-2}\beta_{n-1}^2\beta_{n}+4\beta_{n-1}^3\beta_{n} +3\beta_{n-2}\beta_{n-1}\beta_{n}^2+6\beta_{n-1}^2\beta_{n}^2+4\beta_{n-1}\beta_n^3\\ &&+\beta_n^4+\beta_{n-3}\beta_{n-2}\beta_{n-1}\beta_{n+1}+\beta_{n-2}^2\beta_{n-1}\beta_{n+1}+2\beta_{n-2} \beta_{n-1}^2\beta_{n+1}+\beta_{n-1}^3\beta_{n+1}\\ &&+4\beta_{n-2}\beta_{n-1}\beta_{n}\beta_{n+1}+6\beta_{n-1}^2\beta_{n}\beta_{n+1}+9\beta_{n-1} \beta_{n}^2\beta_{n+1}+4\beta_{n}^3\beta_{n+1}+\beta_{n-2}\beta_{n-1}\beta_{n+1}^2\\ &&+\beta_{n-1}^2\beta_{n+1}^2+6\beta_{n-1}\beta_{n}\beta_{n+1}^2+6\beta_{n}^2\beta_{n+1}^2 +\beta_{n-1}\beta_{n+1}^3+4\beta_{n}\beta_{n+1}^3+\beta_{n+1}^4\\ & &+\beta_{n-2}\beta_{n-1}\beta_{n+1}\beta_{n+2}+\beta_{n-1}^2\beta_{n+1}\beta_{n+2}+3 \beta_{n-1}\beta_n\beta_{n+1}\beta_{n+2}+2\beta_{n}^2\beta_{n+1}\beta_{n+2}\\ &&+\beta_{n-1}\beta_{n+1}^2\beta_{n+2}+4\beta_{n}\beta_{n+1}^2\beta_{n+2}+2\beta_{n+1}^3 \beta_{n+2}+\beta_{n+1}^2\beta_{n+2}^2+\beta_{n-1}\beta_{n}\beta_{n+1}\beta_{n+3}\\ &&+\beta_{n}^2\beta_{n+1}\beta_{n+3}+\beta_{n-1}\beta_{n+1}^2\beta_{n+3}+2\beta_{n}\beta_{n+1}^2 \beta_{n+3}+\beta_{n+1}^3\beta_{n+3}+\beta_{n-1}\beta_{n+1}\beta_{n+2}\beta_{n+3}\\ &&+3\beta_{n}\beta_{n+1}\beta_{n+2}\beta_{n+3}+3\beta_{n+1}^2\beta_{n+2}\beta_{n+3} +2\beta_{n+1}\beta_{n+2}^2\beta_{n+3}+\beta_{n-1}\beta_{n+1}\beta_{n+3}^2\\ &&+\beta_{n}\beta_{n+1}\beta_{n+3}^2+\beta_{n+1}^2\beta_{n+3}^2+2\beta_{n+1}\beta_{n+2} \beta_{n+3}^2+\beta_{n+1}\beta_{n+2}\beta_{n+3}\beta_{n+4}\Big]h_{n-1}\\ &&-2t\beta_{n}h_{n-1}. \end{eqnarray} $

另一方面

(3.5) $ \begin{eqnarray} & &\int_{{{\mathbb R}} }\Big(P_n(x)P_{n-1}(x)|x|^{\alpha}\Big)^{\prime}{w}_0(x){\rm d}x\\ & = &\int_{{{\mathbb R}} }P_n^{\prime}(x)P_{n-1}(x){w}(x){\rm d}x+\int_{{{\mathbb R}} }P_n(x)P_{n-1}^{\prime}(x){w}(x){\rm d}x+\alpha\int_{{{\mathbb R}} }\frac{P_n(x)P_{n-1}(x)}{x}{w}(x){\rm d}x\\ & = &n h_{n-1}+\alpha h_{n-1}\Delta_n, \end{eqnarray} $

这里$ \Delta_n = \frac{1-(-1)^n}{2} $, 理由是: 当$ n $是奇数时, $ \frac{P_n(x)}{x} $是$ n-1 $次多项式; 当$ n $是偶数时, $ \frac{P_{n-1}(x)}{x} $是$ n-2 $次多项式^[21]. 合并(3.4)和(3.5)式, 同时约去$ h_{n-1} $, 结果得证.

注3.1  方程(3.1)是一个八阶差分方程, 属于离散的Painlevé I方程族, 可参考文献[22]. 类似地, 我们可称该方程为$ d_8{ P}_{I} $.

利用Freud^[4]中的结论, 关于权函数$ {w}(x;0, \alpha) = |x|^{\alpha}{\rm e}^{-x^{10}} $的循环系数$ \beta_n $的渐近行为可表示为

(4.1) $ \begin{equation} \lim\limits_{n\rightarrow \infty}\frac{\beta_n}{\sqrt[5]{n}} = \bigg(\frac{\Gamma(5)\Gamma(6)}{\Gamma(11)}\bigg)^{\frac{1}{5}} = \frac{1}{\sqrt[5]{1260}}, \end{equation} $

于是, 可以得到关于权函数(1.2)的循环系数$ \beta_n(t, \alpha) $的近似值.

定理4.1   对于(1.4) 式中的循环系数$ \beta_n(t, \alpha) $, 有

$ \bullet $ 当$ n $是偶数时

(4.2) $ \begin{eqnarray} \beta_n(t, \alpha)& = &\frac{n^{\frac{1}{5}}}{\phi}+\frac{\phi^3t}{3150n^{\frac{3}{5}}}-\frac{152+37\phi^5\alpha}{29925\phi n^{\frac{4}{5}}}-\frac{\phi^7 t^2}{9922500n^{\frac{7}{5}}}+\frac{(2888-4773\phi^5\alpha)\phi^3 t}{895505625n^{\frac{8}{5}}}\\ & &+\frac{1056355312+21866464\phi^5\alpha+11317\phi^{10}\alpha^2}{14328090000\phi n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

$ \bullet $ 当$ n $是奇数时

(4.3) $ \begin{eqnarray} \beta_n(t, \alpha)& = &\frac{n^{\frac{1}{5}}}{\phi}+\frac{\phi^3t}{3150n^{\frac{3}{5}}}-\frac{608-167\phi^5\alpha}{119700\phi n^{\frac{4}{5}}}-\frac{\phi^7 t^2}{9922500n^{\frac{7}{5}}}+\frac{(23104+37101\phi^5\alpha)\phi^3 t}{7164045000n^{\frac{8}{5}}}\\ & &+\frac{1056355312-21820256\phi^5\alpha-653\phi^{10}\alpha^2}{14328090000\phi n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

这里, $ \phi = \sqrt[5]{1260} $.

证  通过观察(3.1) 式, $ (-1)^n $取决于$ n $是奇数或偶数. 于是, 不妨令

(4.4) $ \begin{equation} \beta_n = \left\{ \begin{array}{ll} u_n&, \; \; {{ n }是偶数{, }} \\ v_n&, \; \; {{ n }是奇数{, }} \end{array} \right. \end{equation} $

这里通过(4.1)式, 可知

$ \lim\limits_{n\rightarrow \infty}\frac{u_n}{\sqrt[5]{n}} = \frac{1}{\sqrt[5]{1260}}, \; \; \lim\limits_{n\rightarrow \infty}\frac{v_n}{\sqrt[5]{n}} = \frac{1}{\sqrt[5]{1260}}. $

这就意味着, $ u_n $和$ v_n $也满足(3.1) 式, 分别是

(4.5) $ \begin{eqnarray} & &10u_n[u_{n-4}v_{n-3}u_{n-2}v_{n-1}+v_{n-3}^2u_{n-2}v_{n-1}+2v_{n-3}u_{n-2}^2v_{n-1}+u_{n-2}^3v_{n-1} +2v_{n-3}u_{n-2}v_{n-1}^2\\ & &+3u_{n-2}^2v_{n-1}^2+3u_{n-2}v_{n-1}^3+v_{n-1}^4+2v_{n-3}u_{n-2}v_{n-1}u_n+2u_{n-2}^2v_{n-1}u_{n}+6u_{n-2}v_{n-1}^2u_{n}\\ & &+4v_{n-1}^3u_{n}+3u_{n-2}v_{n-1}u_{n}^2+6v_{n-1}^2u_{n}^2+4v_{n-1}u_n^3+u_n^4+v_{n-3}u_{n-2}v_{n-1}v_{n+1}\\ & &+u_{n-2}^2v_{n-1}v_{n+1}+2u_{n-2}v_{n-1}^2v_{n+1}+v_{n-1}^3v_{n+1}+4u_{n-2}v_{n-1}u_{n}v_{n+1}+6v_{n-1}^2u_{n}v_{n+1}\\ &&+9v_{n-1}u_{n}^2v_{n+1}+4u_{n}^3v_{n+1}+u_{n-2}v_{n-1}v_{n+1}^2+v_{n-1}^2v_{n+1}^2+6v_{n-1}u_{n}v_{n+1}^2+6u_{n}^2v_{n+1}^2\\ &&+v_{n-1}v_{n+1}^3+4u_{n}v_{n+1}^3+v_{n+1}^4+u_{n-2}v_{n-1}v_{n+1}u_{n+2}+v_{n-1}^2v_{n+1}u_{n+2}+3v_{n-1}u_nv_{n+1}u_{n+2}\\ &&+2u_{n}^2v_{n+1}u_{n+2}+v_{n-1}v_{n+1}^2u_{n+2}+4u_{n}v_{n+1}^2u_{n+2}+2v_{n+1}^3u_{n+2}+v_{n+1}^2u_{n+2}^2\\ &&+v_{n-1}u_{n}v_{n+1}v_{n+3}+u_{n}^2v_{n+1}v_{n+3}+v_{n-1}v_{n+1}^2v_{n+3}+2u_{n}v_{n+1}^2v_{n+3}+v_{n+1}^3v_{n+3}\\ &&+v_{n-1}v_{n+1}u_{n+2}v_{n+3}+3u_{n}v_{n+1}u_{n+2}v_{n+3}+3v_{n+1}^2u_{n+2}v_{n+3}+2v_{n+1}u_{n+2}^2v_{n+3}\\ &&+v_{n-1}v_{n+1}v_{n+3}^2+u_{n}v_{n+1}v_{n+3}^2+v_{n+1}^2v_{n+3}^2+2v_{n+1}u_{n+2}v_{n+3}^2+v_{n+1}u_{n+2}v_{n+3}u_{n+4}]\\ &&-2tu_{n} = n \end{eqnarray} $

和

(4.6) $ \begin{eqnarray} & &10v_n[v_{n-4}u_{n-3}v_{n-2}u_{n-1}+u_{n-3}^2v_{n-2}u_{n-1}+2u_{n-3}v_{n-2}^2u_{n-1}+v_{n-2}^3u_{n-1} +2u_{n-3}v_{n-2}u_{n-1}^2\\ &&+3v_{n-2}^2u_{n-1}^2+3v_{n-2}u_{n-1}^3+u_{n-1}^4+2u_{n-3}v_{n-2}u_{n-1}v_n+2v_{n-2}^2u_{n-1}v_{n}+6v_{n-2}u_{n-1}^2v_{n}\\ &&+4u_{n-1}^3v_{n}+3v_{n-2}u_{n-1}v_{n}^2+6u_{n-1}^2v_{n}^2+4u_{n-1}v_n^3+v_n^4+u_{n-3}v_{n-2}u_{n-1}u_{n+1}\\ & &+v_{n-2}^2u_{n-1}u_{n+1}+2v_{n-2}u_{n-1}^2u_{n+1}+u_{n-1}^3u_{n+1}+4v_{n-2}u_{n-1}v_{n}u_{n+1}+6u_{n-1}^2v_{n}u_{n+1}\\ & &+9u_{n-1}v_{n}^2u_{n+1}+4v_{n}^3u_{n+1}+v_{n-2}u_{n-1}u_{n+1}^2+u_{n-1}^2u_{n+1}^2+6u_{n-1}v_{n}u_{n+1}^2+6v_{n}^2u_{n+1}^2\\ & &+u_{n-1}u_{n+1}^3+4v_{n}u_{n+1}^3+u_{n+1}^4+v_{n-2}u_{n-1}u_{n+1}v_{n+2}+u_{n-1}^2u_{n+1}v_{n+2}+3u_{n-1}v_nu_{n+1}v_{n+2}\\ & &+2v_{n}^2u_{n+1}v_{n+2}+u_{n-1}u_{n+1}^2v_{n+2}+4v_{n}u_{n+1}^2v_{n+2}+2u_{n+1}^3v_{n+2}+u_{n+1}^2v_{n+2}^2\\ & &+u_{n-1}v_{n}u_{n+1}u_{n+3}+v_{n}^2u_{n+1}u_{n+3}+u_{n-1}u_{n+1}^2u_{n+3}+2v_{n}u_{n+1}^2u_{n+3}+u_{n+1}^3u_{n+3}\\ & &+u_{n-1}u_{n+1}v_{n+2}u_{n+3}+3v_{n}u_{n+1}v_{n+2}u_{n+3}+3u_{n+1}^2v_{n+2}u_{n+3}+2u_{n+1}v_{n+2}^2u_{n+3}\\ & &+u_{n-1}u_{n+1}u_{n+3}^2+v_{n}u_{n+1}u_{n+3}^2+u_{n+1}^2u_{n+3}^2+2u_{n+1}v_{n+2}u_{n+3}^2+u_{n+1}v_{n+2}u_{n+3}v_{n+4}]\\ & &-2tv_{n} = n+\alpha. \end{eqnarray} $

设存在常数$ a_j $和$ b_j $, $ j = 0, 1, 2, \cdots, 9 $, 使得^{[8-9, 27]}

(4.7) $ \begin{equation} u_n = \frac{\sqrt[5]{n}}{\sqrt[5]{1260}}+\sum\limits_{j = 0}^{9}\frac{a_j}{n^{\frac{j}{5}}}+{\cal O}(n^{-2}), \; \; \; v_n = \frac{\sqrt[5]{n}}{\sqrt[5]{1260}} +\sum\limits_{j = 0}^{9}\frac{b_j}{n^{\frac{j}{5}}}+{\cal O}(n^{-2}). \end{equation} $

进而

(4.8) $ \begin{eqnarray} u_{n\pm 4}& = &\frac{\sqrt[5]{n}}{\phi}+a_0+\frac{a_1}{\sqrt[5]{n}}+\frac{a_2}{n^{\frac{2}{5}}}+\frac{a_3}{n^{\frac{3}{5}}}+\frac{a_4\pm\frac{4}{5\phi}}{n^{\frac{4}{5}}}+\frac{a_5}{n} +\frac{a_6\mp\frac{4a_1}{5}}{n^{\frac{6}{5}}}+\frac{a_7\mp\frac{8a_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{a_8\mp\frac{12a_3}{5}}{n^{\frac{8}{5}}}+\frac{a_9\mp\frac{16a_4}{5}-\frac{32}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.9) $ \begin{eqnarray} v_{n\pm 4}& = &\frac{\sqrt[5]{n}}{\phi}+b_0+\frac{b_1}{\sqrt[5]{n}}+\frac{b_2}{n^{\frac{2}{5}}}+\frac{b_3}{n^{\frac{3}{5}}}+\frac{b_4\pm\frac{4}{5\phi}}{n^{\frac{4}{5}}}+\frac{b_5}{n} +\frac{b_6\mp\frac{4b_1}{5}}{n^{\frac{6}{5}}}+\frac{b_7\mp\frac{8b_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{b_8\mp\frac{12b_3}{5}}{n^{\frac{8}{5}}}+\frac{b_9\mp\frac{16b_4}{5}-\frac{32}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.10) $ \begin{eqnarray} u_{n\pm 3}& = &\frac{\sqrt[5]{n}}{\phi}+a_0+\frac{a_1}{\sqrt[5]{n}}+\frac{a_2}{n^{\frac{2}{5}}}+\frac{a_3}{n^{\frac{3}{5}}}+\frac{a_4\pm\frac{3}{5\phi}}{n^{\frac{4}{5}}}+\frac{a_5}{n} +\frac{a_6\mp\frac{3a_1}{5}}{n^{\frac{6}{5}}}+\frac{a_7\mp\frac{6a_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{a_8\mp\frac{9a_3}{5}}{n^{\frac{8}{5}}}+\frac{a_9\mp\frac{12a_4}{5}-\frac{8}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.11) $ \begin{eqnarray} v_{n\pm 3}& = &\frac{\sqrt[5]{n}}{\phi}+b_0+\frac{b_1}{\sqrt[5]{n}}+\frac{b_2}{n^{\frac{2}{5}}}+\frac{b_3}{n^{\frac{3}{5}}}+\frac{b_4\pm\frac{3}{5\phi}}{n^{\frac{4}{5}}}+\frac{b_5}{n} +\frac{b_6\mp\frac{3b_1}{5}}{n^{\frac{6}{5}}}+\frac{b_7\mp\frac{6b_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{b_8\mp\frac{9b_3}{5}}{n^{\frac{8}{5}}}+\frac{b_9\mp\frac{12b_4}{5}-\frac{8}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.12) $ \begin{eqnarray} u_{n\pm 2}& = &\frac{\sqrt[5]{n}}{\phi}+a_0+\frac{a_1}{\sqrt[5]{n}}+\frac{a_2}{n^{\frac{2}{5}}}+\frac{a_3}{n^{\frac{3}{5}}}+\frac{a_4\pm\frac{2}{5\phi}}{n^{\frac{4}{5}}}+\frac{a_5}{n} +\frac{a_6\mp\frac{2a_1}{5}}{n^{\frac{6}{5}}}+\frac{a_7\mp\frac{4a_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{a_8\mp\frac{6a_3}{5}}{n^{\frac{8}{5}}}+\frac{a_9\mp\frac{8a_4}{5}-\frac{8}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.13) $ \begin{eqnarray} v_{n\pm 2}& = &\frac{\sqrt[5]{n}}{\phi}+b_0+\frac{b_1}{\sqrt[5]{n}}+\frac{b_2}{n^{\frac{2}{5}}}+\frac{b_3}{n^{\frac{3}{5}}}+\frac{b_4\pm\frac{2}{5\phi}}{n^{\frac{4}{5}}}+\frac{b_5}{n} +\frac{b_6\mp\frac{2b_1}{5}}{n^{\frac{6}{5}}}+\frac{b_7\mp\frac{4b_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{b_8\mp\frac{6b_3}{5}}{n^{\frac{8}{5}}}+\frac{b_9\mp\frac{8b_4}{5}-\frac{8}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.14) $ \begin{eqnarray} u_{n\pm 1}& = &\frac{\sqrt[5]{n}}{\phi}+a_0+\frac{a_1}{\sqrt[5]{n}}+\frac{a_2}{n^{\frac{2}{5}}}+\frac{a_3}{n^{\frac{3}{5}}}+\frac{a_4\pm\frac{1}{5\phi}}{n^{\frac{4}{5}}}+\frac{a_5}{n} +\frac{a_6\mp\frac{a_1}{5}}{n^{\frac{6}{5}}}+\frac{a_7\mp\frac{2a_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{a_8\mp\frac{3a_3}{5}}{n^{\frac{8}{5}}}+\frac{a_9\mp\frac{4a_4}{5}-\frac{2}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

(4.15) $ \begin{eqnarray} v_{n\pm 1}& = &\frac{\sqrt[5]{n}}{\phi}+b_0+\frac{b_1}{\sqrt[5]{n}}+\frac{b_2}{n^{\frac{2}{5}}}+\frac{b_3}{n^{\frac{3}{5}}}+\frac{b_4\pm\frac{1}{5\phi}}{n^{\frac{4}{5}}}+\frac{b_5}{n} +\frac{b_6\mp\frac{b_1}{5}}{n^{\frac{6}{5}}}+\frac{b_7\mp\frac{2b_2}{5}}{n^{\frac{7}{5}}}\\ & &+\frac{b_8\mp\frac{3b_3}{5}}{n^{\frac{8}{5}}}+\frac{b_9\mp\frac{4b_4}{5}-\frac{2}{25\phi}}{n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} $

这里$ \phi = \sqrt[5]{1260} $. 将(4.7)–(4.15) 式代入(4.5) 式和(4.6)式, 可以计算出

$ a_0 = b_0 = 0, \; \; a_1 = b_1 = 0, \; \; a_2 = b_2 = 0, \; \; a_3 = b_3 = \frac{\phi^3 t}{3150}, \; \; a_4 = \frac{-37\phi^5\alpha-152}{29925\phi}, $

$ b_4 = \frac{167\phi^5\alpha-608}{119700\phi}, \; \; \; a_5 = b_5 = a_6 = b_6 = 0, \; \; \; a_7 = b_7 = -\frac{\phi^7 t^2}{9922500}, $

$ a_8 = \frac{(2888-4773\phi^5\alpha)\phi^3 t}{895505625}, \; \; \; b_8 = \frac{(23104+37101\phi^5\alpha)\phi^3 t}{7164045000}, $

$ a_9 = \frac{11317\phi^{10}\alpha^2+21866464\phi^5\alpha+1056355312}{14328090000\phi}, $

$ b_9 = \frac{-653\phi^{10}\alpha^2-21820256\phi^5\alpha+1056355312}{14328090000\phi}. $

定理4.1得证.

定理5.1   关于Freud - 型权函数(1.2), 当$ n\rightarrow \infty $时, 对应的Hankel行列式满足关系式

(5.1) $ \begin{equation} D_{2n+1} = {\rm e}^{\Theta_n{+C}}D_{2n}, \end{equation} $

这里, $ C $为任意常数,

$ \begin{eqnarray*} \Theta_n& = &\frac{2^{\frac{4}{5}}tn^{\frac{1}{5}}}{3^{\frac{2}{5}}\times 35^{\frac{1}{5}}}+\frac{t^2}{5\times 3^{\frac{4}{5}}\times 70^{\frac{2}{5}}n^{\frac{3}{5}}}+\frac{(299+315\alpha)t}{3150\times 3^{\frac{2}{5}}\times 70^{\frac{1}{5}}n^{\frac{4}{5}}}-\frac{2^{\frac{2}{5}}t^3}{225\times 3^{\frac{1}{5}}\times {35}^{\frac{3}{5}}n^{\frac{7}{5}}}\nonumber\\ & &-\frac{(945\alpha+913)t^2}{31500\times 3^{\frac{4}{5}}\times 70^{\frac{2}{5}}n^{\frac{8}{5}}} +\frac{(264533850\alpha^2-313931205\alpha+16010711)t}{895505625\times 3^{\frac{2}{5}}\times 70^{\frac{1}{5}}n^{\frac{9}{5}}} +{\cal O}\left(n^{-2}\right). \end{eqnarray*} $

证  在(2.9) 和(2.10) 式的两边同时进行积分, 可得

(5.2) $ \begin{equation} \int\beta_{2n}(t, \alpha){\rm d}t = \ln\widehat{D}_n(t, \alpha+2)-\ln\widehat{D}(t, \alpha){+C_1} \end{equation} $

和

(5.3) $ \begin{equation} \int\beta_{2n+1}(t, \alpha){\rm d}t = \ln\widehat{D}_{n+1}(t, \alpha+2)-\ln\widehat{D}(t, \alpha+2){+C_2}. \end{equation} $

联立(5.2) 和(5.3) 式, 并且在消去$ \ln\widehat{D}_n(t, \alpha+2) $的同时, 应用(2.3) 式, 可得

(5.4) $ \begin{equation} \int(\beta_{2n}(t, \alpha)+\beta_{2n+1}(t, \alpha)){\rm d}t = \ln\widehat{D}_{n+1}(t, \alpha)-\ln\widehat{D}_n(t, \alpha){+C_3} = \ln\frac{D_{2n+1}(t, \alpha)}{D_{2n}(t, \alpha)}{+C_3}, \end{equation} $

其中$ C_1-C_3 $为任意常数. 又根据(4.2) 和(4.3) 式, 有

(5.5) $ \begin{eqnarray} \beta_{2n}(t, \alpha)& = &\frac{n^{\frac{1}{5}}}{3^{\frac{2}{5}}\times 70^{\frac{1}{5}}}+\frac{t}{5\times 3^{\frac{4}{5}}\times 70^{\frac{2}{5}}n^{\frac{3}{5}}}-\frac{t^2}{75\times 3^{\frac{1}{5}}\times 70^{\frac{3}{5}}n^{\frac{7}{5}}}-\frac{(38+11655\alpha)2^{\frac{4}{5}}}{29925\times 3^{\frac{2}{5}}\times 35^{\frac{1}{5}}n^{\frac{4}{5}}}\\ & &-\frac{2^{\frac{8}{5}}(1503495\alpha-722)\times 3^{\frac{1}{5}}\times 35^{\frac{3}{5}}t}{298501875n^{\frac{8}{5}}}+\frac{1122929325\alpha^2+1721984040\alpha+66022207}{3582022500\times 3^{\frac{2}{5}}\times 70^{\frac{1}{5}}n^{\frac{9}{5}}} \\ && +{\cal O}\left((2n)^{-2}\right), \end{eqnarray} $

(5.6) $ \begin{eqnarray} \beta_{2n+1}(t, \alpha)& = &\frac{(2n+1)^{\frac{1}{5}}}{6^{\frac{2}{5}}\times 35^{\frac{1}{5}}}+\frac{2^{\frac{1}{5}}t}{5\times 3^{\frac{4}{5}}\times 35^{\frac{2}{5}}(2n+1)^{\frac{3}{5}}}-\frac{2^{\frac{4}{5}}t^2}{75\times 3^{\frac{1}{5}}\times 35^{\frac{3}{5}}(2n+1)^{\frac{7}{5}}}\\ & &+\frac{52605\alpha-152}{29925\times 6^{\frac{2}{5}}\times 35^{\frac{1}{5}}(2n+1)^{\frac{4}{5}}} +\frac{2^{\frac{1}{5}}(11686815\alpha+5776)t}{2842875\times 3^{\frac{4}{5}}\times 35^{\frac{2}{5}}(2n+1)^{\frac{8}{5}}}\\ & &-\frac{64793925\alpha^2+1718345160\alpha-66022207}{895505625\times 6^{\frac{2}{5}}\times 35^{\frac{1}{5}}(2n+1)^{\frac{9}{5}}}+{\cal O}\left((2n+1)^{-2}\right), \end{eqnarray} $

将(5.5) 和(5.6) 式代入(5.4) 式, 并且进行积分运算, 可得

$ \begin{eqnarray*} \ln\frac{D_{2n+1}(t, \alpha)}{D_{2n}(t, \alpha)} & = &\frac{2^{\frac{4}{5}}tn^{\frac{1}{5}}}{3^{\frac{2}{5}}\times 35^{\frac{1}{5}}}+\frac{t^2}{5\times 3^{\frac{4}{5}}\times 70^{\frac{2}{5}}n^{\frac{3}{5}}}+\frac{(299+315\alpha)t}{3150\times 3^{\frac{2}{5}}\times 70^{\frac{1}{5}}n^{\frac{4}{5}}}-\frac{2^{\frac{2}{5}}t^3}{225\times 3^{\frac{1}{5}}\times {35}^{\frac{3}{5}}n^{\frac{7}{5}}}\nonumber\\ &&-\frac{(945\alpha+913)t^2}{31500\times 3^{\frac{4}{5}}\times 70^{\frac{2}{5}}n^{\frac{8}{5}}} +\frac{(264533850\alpha^2-313931205\alpha+16010711)t}{895505625\times 3^{\frac{2}{5}}\times 70^{\frac{1}{5}}n^{\frac{9}{5}}}\nonumber\\ & &+{\cal O}\left(n^{-2}\right){+C}, \end{eqnarray*} $

$ C $为常数, 定理5.1得证.

在这一节, 我们讨论由权函数

(6.1) $ \begin{equation} {w}_{E}(x) = {\rm e}^{-x^{10}}, \; \; \; \; x\in{{\mathbb R}} \end{equation} $

生成的Hankel矩阵的最小特征值在$ n\rightarrow \infty $时的渐近行为. 显然, 这个权函数是(1.2) 式当$ t = \alpha = 0 $时的特例. 该节要用到的方法, 可参见文献[23, 31]. 它可被用于求任意一个指数型权函数$ {\rm e}^{-2Q(x)} $生成的Hankel矩阵的最小特征值在高维时的渐近行为, 这里$ Q(x) $是偶函数并且所符合的条件请参阅文献[31]. 对于函数(6.1)而言, $ Q(x) = \frac{1}{2}x^{10} $. 根据平衡测度以及Mhaskar-Rakhmanor-Saff数字$ a_n $ (参见文献[23, 31]), 循环系数$ \beta_n $的近似值可以表出, 这里$ a_n $是方程$ n = \frac{2}{\pi}\int_0^1\frac{a_ntQ^{\prime}(a_nt)}{\sqrt{1-t^2}}{\rm d}t $唯一的正根. 除此之外, 当$ n\rightarrow \infty $时, $ \beta_n $和$ a_n $的关系式^[23]是

(6.2) $ \begin{equation} a_n^2 = 4\beta_n. \end{equation} $

接下来, 我们给出本文关于最小特征值的结论.

定理6.1  关于权函数(6.1), 它的最大Hankel矩阵所对应的最小特征值可以表示成

(6.3) $ \begin{equation} \lambda_n\simeq\frac{n^{\frac{9}{20}}}{{\cal K}^{\frac{1}{2}}}\exp\left(-\frac{525n^{\frac{1}{10}}}{864{\cal K}^9}+\frac{25n^{\frac{3}{10}}}{84{\cal K}^7}-\frac{3n^{\frac{1}{2}}}{10{\cal K}^5} +\frac{10n^{\frac{7}{10}}}{21{\cal K}^3}-\frac{20n^{\frac{9}{10}}}{9{\cal K}}\right), \end{equation} $

这里$ {\cal K} = \left(\frac{256}{315}\right)^{\frac{1}{10}} $.

证  不妨令$ v_E(x) = -\ln{w}_E(x) = x^{10} $, 则$ v_E^{\prime}(x) = 10x^9 $. 对于函数(6.1), 根据库仑流体方法^[24-26]及其补充条件$ \int_a^b\frac{xv^{\prime}(x)}{\sqrt{(b-x)(x-a)}}{\rm d}x = 2\pi n $可知

$ 2\pi n = \int_{-b_E}^{b_E}\frac{10x^{10}}{\sqrt{b_E^2-x^2}}{\rm d}x = 20\int_0^{b_E}\frac{x^{10}}{\sqrt{b_E^2-x^2}}{\rm d}x = 20b_E^{10}\int_0^{\frac{\pi}{2}}\sin^{10}\theta {\rm d}\theta = \frac{315\pi}{128}b_E^{10}, $

于是

$ b_E^2\simeq \left(\frac{256n}{315}\right)^{\frac{1}{5}}. $

从而^[24], 有

(6.4) $ \begin{eqnarray} \beta_n\simeq\left(\frac{b_E-a_E}{4}\right)^2 = \frac{b_E^2}{4} = \left(\frac{n}{1260}\right)^{\frac{1}{5}}, \; \; \; n\rightarrow \infty, \end{eqnarray} $

这个结果也对应了(4.1)式. 将(6.4) 式应用于(6.2)式, 有

$ a_n = \left(\frac{256n}{315}\right)^{\frac{1}{10}} = {\cal K}n^{\frac{1}{10}}, $

这里$ {\cal K} = \left(\frac{256}{315}\right)^{\frac{1}{10}} $.

基于Chen和Lubinsky的理论^[31], 同时利用$ \ln(x+\sqrt{1+x^2}) $在$ x = 0 $处的麦克劳林展开式

$ \ln(x+\sqrt{1+x^2}) = \sum\limits_{k = 0}^{\infty}(-1)^k\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}x^{2k+1}, \; \; \; |x|\leq 1, $

可以计算出最小特征值$ \lambda_n $如下

$ \begin{eqnarray*} \lambda_n&\simeq& \sqrt{\frac{n}{a_n}}\exp\Big(-2\int_0^n\ln\Big(\frac{1}{a_s}+\sqrt{1+\frac{1}{a_s^2}}\Big){\rm d}s\Big)\nonumber\\ & = &\sqrt{\frac{n}{a_n}}\exp\Big(-2\int_0^n\sum\limits_{k = 0}^{\infty}(-1)^k\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}a_s^{-2k-1}{\rm d}s\Big)\nonumber\\ & = &\frac{n^{\frac{9}{20}}}{{\cal K}^{\frac{1}{2}}}\exp\Big(-2\sum\limits_{k = 0}^{\infty}(-1)^k\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}\int_0^n{\cal K}^{-2k-1}s^{\frac{-2k-1}{10}}{\rm d}s\Big)\nonumber\\ & = &\frac{n^{\frac{9}{20}}}{{\cal K}^{\frac{1}{2}}}\exp\Big(-2n\sum\limits_{k = 0}^{\infty}(-1)^k\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}\frac{10a_n^{-2k-1}}{9-2k}\Big)\nonumber\\ &\simeq &\frac{n^{\frac{9}{20}}}{{\cal K}^{\frac{1}{2}}}\exp\Big(-2n\Big(\frac{10}{9{\cal K}n^{\frac{1}{10}}}-\frac{5}{21{\cal K}^3n^{\frac{3}{10}}} +\frac{3}{20{\cal K}^5n^{\frac{1}{2}}}-\frac{25}{168{\cal K}^7n^{\frac{7}{10}}}+\frac{175}{576{\cal K}^9n^{\frac{9}{10}}}+{\cal O}(n^{-1})\Big)\Big)\nonumber\\ &\simeq &\frac{n^{\frac{9}{20}}}{{\cal K}^{\frac{1}{2}}}\exp\Big(-\frac{175n^{\frac{1}{10}}}{288{\cal K}^9}+\frac{25n^{\frac{3}{10}}}{84{\cal K}^7}-\frac{3n^{\frac{1}{2}}}{10{\cal K}^5} +\frac{10n^{\frac{7}{10}}}{21{\cal K}^3}-\frac{20n^{\frac{9}{10}}}{9{\cal K}}\Big). \end{eqnarray*} $

定理6.1得证.