## A Generalised Decic Freud-Type Weight

Wang Dan,1, Zhu Mengkun,2, Yang Chen,1, Wang Xiaoli,2

 基金资助: 澳门科学技术发展基金.  FDCT023/2017/A1澳门大学基金.  MYRG2018-00125FST国家自然科学基金.  11801292广东省自然科学基金.  2021A1515010361

 Fund supported: the Macau Science and Technology Development Fund.  FDCT023/2017/A1the Universidade de Macau.  MYRG2018-00125FSTthe NSFC.  11801292the NSF of Guangdong Province.  2021A1515010361

YangChen,E-mail:yayangchen@um.edu.mo , E-mail：yayangchen@um.edu.mo

Abstract

In this paper, the authors focus on a generalised decic Freud-type weight functionand study the properties of the orthogonal polynomials with respect to this weight. The difference-differential equations of their associated recurrence coefficients are derived; meanwhile, the authors also find the asymptotic behavior of recurrence coefficients via above mentioned equations. What's more, the authors discuss the Hankel determinant in regard to this weight as $n\rightarrow\infty$, and calculate the smallest eigenvalues of large Hankel matrices generated by this weight when $\alpha=t=0$.

Keywords： Orthogonal Polynomials ; Recurrence coefficients ; Hankel determinant ; Smallest eigenvalue

Wang Dan, Zhu Mengkun, Yang Chen, Wang Xiaoli. A Generalised Decic Freud-Type Weight. Acta Mathematica Scientia[J], 2021, 41(4): 921-935 doi:

## 1 引言

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

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

$\{P_n(x)\}$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

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

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

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

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

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

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

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

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

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

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.

## 2 循环系数的行列式表达

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

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

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

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

## 3 循环系数满足的差分方程

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

为了方便起见, 记

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

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

$\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}$

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

## 4 循环系数的近似值

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

### 图 4

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

$\bullet $$n 是奇数时 \begin{eqnarray} \beta_n(t, \alpha)& = &\frac{n^{\frac{1}{5}}}{\phi}+\frac{\phi^3t}{3150n^{\frac{3}{5}}}-\frac{608-167\phi^5\alpha}{119700\phi n^{\frac{4}{5}}}-\frac{\phi^7 t^2}{9922500n^{\frac{7}{5}}}+\frac{(23104+37101\phi^5\alpha)\phi^3 t}{7164045000n^{\frac{8}{5}}}\\ & &+\frac{1056355312-21820256\phi^5\alpha-653\phi^{10}\alpha^2}{14328090000\phi n^{\frac{9}{5}}}+{\cal O}(n^{-2}), \end{eqnarray} 这里, \phi = \sqrt[5]{1260} . 通过观察(3.1) 式, (-1)^n 取决于 n 是奇数或偶数. 于是, 不妨令 $$\beta_n = \left\{ \begin{array}{ll} u_n&, \; \; {{ n }是偶数{, }} \\ v_n&, \; \; {{ n }是奇数{, }} \end{array} \right.$$ 这里通过(4.1)式, 可知 这就意味着, u_n$$ v_n$也满足(3.1) 式, 分别是

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

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

## 5 Hankel行列式满足的方程

$$$D_{2n+1} = {\rm e}^{\Theta_n{+C}}D_{2n},$$$

在(2.9) 和(2.10) 式的两边同时进行积分, 可得

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

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

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

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

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

$C$为常数, 定理5.1得证.

## 6 Hankel矩阵的最小特征值

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

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