数学物理学报, 2025, 45(3): 960-971

关于高斯乘积不等式的新结果 (I)

马丽,1,2, 陈蓬颖,1, 韩新方,1,2,*

1海南师范大学数学与统计学院 海口 571158

2海南师范大学数据科学与智慧教育教育部重点实验室 海口 571158

New Results On Gauss Product Inequalities (I)

Ma Li,1,2, Chen Pengying,1, Han Xinfang,1,2,*

1Department of Mathematics and Statistics, Hainan Normal University, Haikou 571158

2Key Laboratory of Data Science and Intelligence Education (Hainan Normal University), Ministry of Education, Haikou 571158

通讯作者: 韩新方，Email: xfanghan@163.com

收稿日期: 2024-05-7   修回日期: 2025-01-13  

基金资助: 海南省自然科学基金(122MS056)
海南省自然科学基金(124MS056)

Received: 2024-05-7   Revised: 2025-01-13  

Fund supported: Hainan Provincial Natural Science Foundation(122MS056)
Hainan Provincial Natural Science Foundation(124MS056)

作者简介 About authors

马丽，Email:malihnsd@163.com;

陈蓬颖，Email:1353525952@qq.com

摘要

设 $(X_1,X_2,X_3)$ 为中心化的高斯随机变量, 其协方差矩阵的对角线元素均为 $1$, 该文借助于超几何函数的性质及因式分解得到了

$E[|X_1^4X_2^3X_3^3|]\geq$E$|X_1^4|$E$|X_2^3|$E$|X_3^3|$,

等号成立当且仅当 $X_1,X_2,X_3$ 相互独立. 从而补充了现有文献中三维高斯乘积不等式的结果.

关键词： 高斯乘积不等式; 正态分布; 超几何函数; 因式分解

Abstract

Let ($X_1$,$X_2$,$X_3$) be a centered Gaussian random vector with $D(X_i)=1$, $i=1,2,3$. By means of the properties of hypergeometric function and factorization, we prove that

$E\big[|X_1^4X_2^3X_3^3|\big]\geq$E$|X_1^4|$E$|X_2^3|$E$|X_3^3|$,

and the equal sign holds if and only if $X_1$,$X_2$,$X_3$ are independent. This complements the results of the three dimensional Gauss product inequality in the existing literature.

Keywords： Gauss Product Inequality; normal Distribution; hypergeometric Function; factorization

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本文引用格式

马丽, 陈蓬颖, 韩新方. 关于高斯乘积不等式的新结果 (I)[J]. 数学物理学报, 2025, 45(3): 960-971

Ma Li, Chen Pengying, Han Xinfang. New Results On Gauss Product Inequalities (I)[J]. Acta Mathematica Scientia, 2025, 45(3): 960-971

1 引言

期望为零的高斯随机向量称为中心高斯随机向量. 关于中心高斯随机向量, 文献 [a] 提出了如下猜想: 对任意 $R^n$ 值中心高斯随机向量 $(X_1,\cdots,X_n)$ 和正实数 $\alpha_1,\cdots\alpha_n$, 有

$E\big[\prod_{j=1}^n|X_j|^{\alpha_j}\big]\geq\prod_{j=1}^nE|X_j|^{\alpha_j}.$

该猜想被称为高斯不等式猜想. 高斯不等式 (1.1) 蕴含着泛函分析中的实极化猜想, 且与统计学中的U猜想有关. 近年来许多学者关注于此猜想的证明.

对复数值中心高斯随机向量, 高斯乘积不等式已经被完全证明了对实数值中心高斯随机向量, 二维 (即 $n=2$) 高斯乘积不等式也已经被解决, 见文献[2,3,4]. 三维高斯乘积不等式只是部分地被证明了. 本文中 $(X_1,\cdots,X_n)$ 均为实数值的中心高斯随机向量, $N$ 表示自然数, $\rho_{ij}$ 为 $X_i$ 和 $X_j$ 的相关系数. 对三维高斯随机向量, 文献[5,6,7] 得到了$\alpha_1=2m_1,\alpha_2=2m_2,\alpha_3=2m_3,m_1,m_2,m_3\in{N}$时,

(1.1) 式成立. 关于更高维的高斯乘积不等式, 文献 [8] 证明了对任意的 $\alpha_1,\cdots,\alpha_n\in(-1,0)$, 任意的 $1\leq{k}\leq{n-1}$, 有

$E[\prod_{j=1}^n|X_j|^{\alpha_j}]\geq E[\prod_{j=1}^k|X_j|^{\alpha_j}]E[\prod_{j=k+1}^n|X_j|^{\alpha_j}].$

文献 [9] 得到了对任意的 $\alpha_1\in(-1,0)$, 有

$E[|X_1|^{-\alpha_1}{\prod_{j=2}^n}|X_j|^2]\geq\prod_{i=2}^n(1-\rho_{1i}^2)E[|X_1|^{-\alpha_1}]{\prod_{j=2}^n}E[|X_j|^2].$

对任意的 $\alpha_1\in(-1,0)$ 和正偶数 $\alpha_2,\alpha_3,\alpha_4$, 有

$E[|X_1|^{-\alpha_1}|X_2|^{\alpha_2}|X_3|^{\alpha_3}|X_4|^{\alpha_4}]\geq \prod_{i=2}^4(1-\rho_{1i}^2)^\frac{\alpha_i}{2}E[|X_1|^{-\alpha_1}]\prod_{j=2}^4E[|X_j|^{\alpha_i}].$

综上, $n=3,\alpha_1,\alpha_2,\alpha_3$ 均为正偶数时, (1.1) 式已经被证明了. $\alpha_1,\alpha_2,\alpha_3$ 不全为偶数时,(1.1)式是否成立还尚未解决. 本文借鉴了文献 [10] 第六章降阶的方法, 运用超几何函数的性质得到了 $\alpha_1=4,\alpha_2=3,\alpha_3=3$ 时, (1.1) 式是成立的.不同于文献 [10], 本文得到了含有四次方和三次方的不等式, 从而不能直接用文献[10] 的方法考虑判别式与零的关系. 本文通过换元把含有高次项的不等式转化为含有二次项不等式, 再通过基本不等式得到新的变元为负的.通过证明二次项的系数大于 $0$, 一次项的系数小于 $0$, 常数项小于 $0$, 可得(1.1) 式成立.

2 主要结果及其证明

引理 2.1 设 $n\geq3,m_1,\cdots,m_n$ 为自然数, 若对任意的中心高斯随机向量 $(Y_1,\cdots,Y_n)$ 且 $Y_n=\alpha_1Y_1+\cdots+\alpha_{n-1}Y_{n-1}$,$\alpha_1,\cdots,\alpha_{n-1}$ 为常数, 有

$E\Big[\prod_{j=1}^{n-1}|Y_j^{m_j}|Y_n^{2k}\Big]\geq\Big[\prod_{j=1}^{n-1}E|Y_j^{m_j}|\Big]E[Y_n^{2k}],0\leq{k}\leq{m_n},$

则对任意中心高斯随机向量 $(X_1,\cdots,X_n)$,

$E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|X_n^{2m_n}\Big]\geq\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big]E[X_n^{2m_n}].$

证 设 $Z_0=E\big[X_n|X_1\cdots X_{n-1}\big],Z_1=X_n-Z_0$, 那么

$X_n^{2m_n}=(Z_0+Z_1)^{2m_n}=\sum_{i=0}^{2m_n}C_{2m_n}^iZ_0^{2m_n-i}Z_1^i,$

且 $Z_1$ 独立于 $X_1{\cdots}X_{n-1}$. 当 $i$ 为奇数时,

$E[Z_0^{2m_n-i}Z_1^i|X_1{\cdots}X_{n-1}]=Z_0^{2m_n-i}E[Z_1^i]=0,$
$E[X_n^{2m_n}|X_1{\cdots}X_{n-1}]=\sum_{i=0}^{m_n}C_{2m_n}^{2i}Z_0^{2m_n-2i}E(Z_1^{2i}),$

注意到 $Z_0=\alpha_1X_1+\cdots+\alpha_{n-1}X_{n-1}$, 因此,

$E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|Z_0^{2m_n-2i}\Big]\geq\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big]E(Z_0^{2m_n-2i}),$
$\begin{align*} E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|X_n^{2m_n}\Big]&=E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|E(X_n^{2m_n}|X_1\cdots X_{n-1})\Big]\\ &=E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|\Big(\sum_{i=0}^{m_n}C_{2m_n}^{2i}Z_0^{2m_n-2i}E\big(Z_1^{2i}\big)\Big)\Big]\\ &=\sum_{i=0}^{m_n}C_{2m_n}^{2i}E(Z_1^{2i})E\Big[\prod_{j=1}^{n-1}|X_j^{m_j}|Z_0^{2m_n-2i}\Big]\\ &\geq\sum_{i=0}^{m_n}C_{2m_n}^{2i}E(Z_1^{2i})E(Z_0^{2m_n-2i})\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big]\\ &=E\Big[\sum_{i=0}^{m_n}C_{2m_n}^{2i}Z_1^{2i}Z_0^{2m_n-2i}\Big]\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big]\\ &=E\Big[(Z_0+Z_1)^{2m_n}\Big]\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big] =\Big[\prod_{j=1}^{n-1}E|X_j^{m_j}|\Big]E(X_n^{2m_n}). \end{align*}$

推论 2.1 下面两个断言等价

(1) 设 $n\geq3,m_1,\cdots,m_n$ 为自然数, 对任意中心高斯随机向量$(Y_1,\cdots,Y_n)$ 且$Y_n=\alpha_1Y_1+\cdots+\alpha_{n-1}Y_{n-1}$, 其中 $\alpha_1,\cdots,\alpha_{n-1}$ 为常数, 有

$E\Big[\prod_{j=1}^{n-1}|Y_j^{m_j}|Y_n^{2k}\Big]\geq\Big[\prod_{j=1}^{n-1}E|Y_j^{m_j}|\Big]E(Y_n^{2k}),0\leq{k}\leq{m_n};$

(2) 设 $n\geq3,m_1,\cdots,m_n$ 为自然数, 对任意的中心高斯随机向量$(Y_1,\cdots,Y_n)$, 有

$E\Big[\prod_{j=1}^{n-1}|Y_j^{m_j}|Y_n^{2m_n}\Big]\geq\Big[\prod_{j=1}^{n-1}E|Y_j^{m_j}|\Big]E(Y_n^{2m_n}),$

等号成立当且仅当 $Y_1,\cdots,Y_n$ 独立.

定理 2.1 设 $(X_1,X_2,X_3)$ 为中心化的高斯随机变量, 其协方差矩阵的对角线元素均为 1, 则有

$E\big[|X_1^4X_2^3X_3^3|\big]\geq {E|X_1^4|E|X_2^3|E|X_3^3|}.$

等号成立当且仅当 $X_1,X_2,X_3$ 相互独立.

证 当 {$X_1$} 与 {$X_2$、$X_3$} 独立时, 由二维高斯乘积不等式可得

$E\big[|X_1^4X_2^3X_3^3|\big]\geq {E|X_1^4|E|X_2^3X_3^3|}\geq {E|X_1^4|E|X_2^3|E|X_3^3|}.$

当 {$X_1$} 与 {$X_2,X_3$} 不独立时, 由推论 2.1 只需证明$X_1=a_2X_2+a_3X_3$ 时,

$E[|X_1^{2k}X_2^3X_3^3|]> {E|X_1^{2k}|E|X_2^3|E|X_3^3|},k=0,1,2.$

即可. 由二维高斯乘积不等式可知 $k=0$ 时, (2.2) 式成立. 由文献 [10] 第六章可知 $k=1$ 时, (2.2) 式成立. 下面证明 $k=2$ 时,(2.2) 式成立.$a_2=0$ 时, 由 $X_1=a_3X_3$ 及二维高斯乘积不等式可得 (2.2) 式成立. 当 $a_2\neq0$ 时, 令 $Z_1=\frac{X_1}{a_2},Z_2=X_2,Z_3=\frac{a_3}{a_2}X_3$,$\alpha=\frac{a_3}{a_2}$,则 $Z_1=Z_2+\alpha Z_3$, 且 (2.2) 式等价于

$E[|Z_1^{4}Z_2^3Z_3^3|]> {E|Z_1^{4}|E|Z_2^3|E|Z_3^3|},$

因此, 为了简化符号, 不妨设 $X_1=X_2+{\alpha}X_3$, 只需证明

$E[|X_1^{4}X_2^3X_3^3|]> {E|X_1^{4}|E|X_2^3|E|X_3^3|}$

即可. 设

$F(\alpha,b;c;z)=\sum_{n=0}^{\infty}\frac{(\alpha)_n(b)_n}{(c)_n}\frac{z^n}{n!}$

为超几何函数, $(\alpha)_0=1$, 当 $n\geq1$ 时, $(\alpha)_n=\alpha(\alpha+1)\cdots(\alpha+n-1)$. 设 $x=E[X_2X_3]$, 由于 $X_2$ 与 $-X_2$ 同分布, $X_3$ 与 $-X_3$ 同分布, 不妨设 $\alpha<0$, $0<x<1$.由文献 [10,(2.1),(2.2) 式] 即

$\begin{eqnarray*} &&E\big[X_2^{2m_2}X_3^{2m_3}\big]=(2m_2-1)!!(2m_3-1)!!\big[{var}(X_2)\big]^{m_2}\big[{var}(X_3)\big]^{m_3}F(-m_2,-m_3;\frac{1}{2};x^2),\\ &&E\big[X_2^{2m_2+1}X_3^{2m_3+1}\big]\\ &=&(2m_2+1)!!(2m_3+1)!!\big[{var}(X_2)\big]^\frac{2m_2+1}{2} \big[{var}(X_3)\big]^\frac{2m_3+1}{2}F(-m_2,-m_3;\frac{2}{3};x^2), \end{eqnarray*}$

可得

$\begin{eqnarray*} E|X_1^4|E|X_2^3|E|X_3^3|&=&\frac{8}{\pi}E\big[(X_2+{\alpha}X_3)^4\big]\\ &=&\frac{8}{\pi}\Big[(\alpha^4+1)E(X_2^4)+6\alpha^2E(X_2^2X_3^2)+(4\alpha+4\alpha^3)E(X_2X_3^3)\Big]\\ &=&\frac{8}{\pi}\Big[3(\alpha^4+1)+6\alpha^2F(-1,-1;\frac{1}{2};x^2)+(4\alpha+4\alpha^3)3!!xF(0,-1;\frac{2}{3};x^2)\Big]\\ &=&\frac{8}{\pi}\Big[3(\alpha^4+1)+6\alpha^2(1+2x^2)+3x(4\alpha+4\alpha^3)\Big], \end{eqnarray*}$

由文献 [10,(5.2) 式] 即

$\begin{eqnarray*} E\big[|X_2|^{y_2}|X_3|^{y_3+2}\big]&=&(y_3+1)\frac{2^{\frac{y_2+y_3}{2}}\Gamma(\frac{y_2+1}{2})\Gamma(\frac{y_3+1}{2})}{\pi} F(-\frac{y_3}{2}-1,-\frac{y_2}{2};\frac{1}{2};x^2),\\ E\big[|X_2|^{y_2+2}|X_3|^{y_3}\big]&=&(y_2+1)\frac{2^{\frac{y_2+y_3}{2}}\Gamma(\frac{y_2+1}{2})\Gamma(\frac{y_3+1}{2})}{\pi} F(-\frac{y_2}{2}-1,-\frac{y_3}{2};\frac{1}{2};x^2),\\ E\big[|X_2|^{y_2}X_2|X_3|^{y_3}X_3\big]&=&x(y_2+1)(y_3+1)\frac{2^{\frac{y_2+y_3}{2}}\Gamma(\frac{y_2+1}{2})\Gamma(\frac{y_3+1}{2})}{\pi} F(-\frac{y_2}{2},-\frac{y_3}{2};\frac{3}{2};x^2), \end{eqnarray*}$

可得

$\begin{eqnarray*} E\big[|X_1^4X_2^3X_3^3|\big]&=&E\big[(X_2^4+4{\alpha}X_2^3X_3+6\alpha^2X_2^2X_3^2+4\alpha^3X_2X_3^3+\alpha^4X_3^4)|X_2^3X_3^3|\big]\\ &=&E\big[|X_2^7X_3^3|+4{\alpha}X_2^3X_3|X_2^3X_3^3|+6\alpha^2|X_2^5X_3^5|+4\alpha^3X_2X_3^3|X_2^3X_3^3|+\alpha^4|X_2^3X_3^7|\big]\\ &=&12\frac{8{\Gamma(3)}{\Gamma(2)}}{\pi}F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)+192x\alpha\frac{8{\Gamma(3)}{\Gamma(2)}}{\pi}F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)\\ &&+48\alpha^2\frac{8{\Gamma(3)}{\Gamma(2)}}{\pi}F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)+192x\alpha^3\frac{8{\Gamma(3)}{\Gamma(2)}}{\pi}F(-\frac{3}{2},-\frac{5}{2};\frac{3}{2};x^2)\\ &&+12\alpha^4\frac{8{\Gamma(3)}{\Gamma(2)}}{\pi}F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2). \end{eqnarray*}$

注意到 $F(-\frac{y_2}{2},-\frac{y_3+2}{2};\frac{3}{2};x^2)=F(-\frac{y_3+2}{2},-\frac{y_2}{2};\frac{3}{2};x^2)$, 从而 (2.1) 式等价于

$\begin{eqnarray*} &&8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)+128x{\alpha}F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+32\alpha^2F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)\\ &&+128x{\alpha^3}F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+8{\alpha^4}F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)\\ &\geq&(\alpha^4+1)+2\alpha^2(1+2x^2)+x(4\alpha+4\alpha^3), \end{eqnarray*}$

上式两边同除以 ${\alpha^2}$, 可得

$\begin{matrix} &&(\alpha^2+\frac{1}{\alpha^2})\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\Big]+(\alpha+\frac{1}{\alpha})\Big[128xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-4x\Big]\nonumber\\ &&+32F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)-2(1+2x^2)>0, \end{matrix}$

令 $\alpha+\frac{1}{\alpha}+2=b$, 则 $\alpha^2+\frac{1}{\alpha^2}=b^2-4b+2,b\leq0$, (2.4) 式等价于

$\begin{eqnarray*} &&b^2\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\Big]-4b\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1-32xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+x\Big]\\ &&+16F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-256xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+32F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)-4+8x-4x^2>0. \end{eqnarray*}$

由

$\frac{\rm d}{{\rm d}x}F(a,b;c;x)=\frac{ab}{c}F(a+1,b+1;c+1;x),$
$F(a,b;c;x)=(1-x)^{c-a-b}F(c-a,c-b;c;x),$

可得当 $0\leq{x}\leq1$ 时,

$\frac{\rm d}{{\rm d}x}\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\Big]=168xF(-\frac{5}{2},-\frac{1}{2};\frac{3}{2};x^2)=168x(1-x^2)^{\frac{9}{2}}F(3,2;\frac{3}{2};x^2)\geq0.$

因此, 二次项系数

$8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\geq8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};0)-1=7\geq0.$

由

$(1-z)F(\alpha,b;c;z)=F(\alpha-1,b;c;z)-c^{-1}(c-b)zF(\alpha,b;c+1;z),$

可得

$\begin{eqnarray*} &&(1-x^2)F(-\frac{y_2}{2}-1,-\frac{y_2}{2};\frac{1}{2};x^2)+(y_2+1)x^2F(-\frac{y_2}{2}-1,-\frac{y_2}{2};\frac{3}{2};x^2)\\ &=&F(-\frac{y_2}{2}-2,-\frac{y_2}{2};\frac{1}{2};x^2),\\ &&(1-x^2)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)+4x^2F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)=F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2),\\ &&(1-x^2)F(-\frac{3}{2},-\frac{5}{2};\frac{1}{2};x^2)=F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)-6x^2F(-\frac{3}{2},-\frac{5}{2};\frac{3}{2};x^2). \end{eqnarray*}$

令

$\begin{eqnarray*} F(b,x)&=&b^2\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\Big]\\ &&-4b\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1-32xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+x\Big ]\\ &&+16F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-256xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)+32F(-\frac{5}{2},-\frac{5}{2};\frac{1}{2};x^2)-4(1-x)^2\\ &=&b^2\Big[8F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-1\Big]\\ &&-4b\Big[8(1-x^2)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)+32x(x-1)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-1+x\Big]\\ &&+48(1-x^2)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)+256(x^2-x)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-4(1-x)^2\\ &=&8b^2\Big[F(-\frac{7}{2},-\frac{3}{2};\frac{1}{2};x^2)-\frac{1}{8}\Big]\\ &&+32b(1-x)\Big[4xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-(1+x)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)+\frac{1}{8}\Big]\\ &&+48(1-x^2)F(-\frac{3}{2},-\frac{5}{2};\frac{1}{2};x^2)-256x(1-x)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-4(x-1)^2, \end{eqnarray*}$

下面证明

$\begin{eqnarray*} &&8(1-x^2)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)+32x(x-1)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-1+x\\ &=&(1-x)\Big[8(1+x)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)-32xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-1\Big]\geq0. \end{eqnarray*}$

从而$F(b,x)$的第二项大于等于零.由 (2.6) 式和附录中 (3.2) 式可得

$\begin{eqnarray*} &&8(1+x)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)-32xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-1\\ &=&8(1+x)(1-x^2)^{\frac{9}{2}}F(2,3;\frac{1}{2};x^2)-32x(1-x^2)^{\frac{11}{2}}F(4,3;\frac{3}{2};x^2)-1\\ &=&(1+x)(1-x^2)^{\frac{3}{2}}(22.5+30x^2)F(1,1;\frac{1}{2};x^2)-(1+x)(1-x^2)^{\frac{3}{2}}(14.5+3x^2)\\ &&-(1-x^2)^{\frac{3}{2}}(\frac{7.5+90x^2+60x^4}{2x})F(1,1;\frac{1}{2};x^2)+(1-x^2)^{\frac{3}{2}}(\frac{7.5+41x^2+4x^4}{2x})-1\\ &=&(1-x^2)^{\frac{3}{2}}F(1,1;\frac{1}{2};x^2)(22.5-\frac{7.5}{2x}-\frac{135x}{6}+30x^2)\\ &&-1+(1-x^2)^{\frac{3}{2}}(-14.5+\frac{22.5}{6x}+6x-3x^2-x^3)\\ &=&\sin^{-1}x(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3)+\sqrt{1-x^2}(22.5-\frac{7.5}{2x}-\frac{135x}{6}+30x^2)-1\\ &&+\sqrt{1-x^2}(-14.5+\frac{22.5}{6x}+\frac{13.5x}{6}+11.5x^2-7x^3+3x^4+x^5)\\ &=&\sin^{-1}x(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3)\\ &&+\sqrt{1-x^2}(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1, \end{eqnarray*}$

由于$0<x<1$, $x<{\sin^{-1}x}<\frac{x}{\sqrt{1-x^2}}$, 所以只需证

$\begin{matrix} &&x(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3) \\ &&+\sqrt{1-x^2}(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1 >0 \end{matrix}$

和

$\begin{matrix} &&\frac{x}{\sqrt{1-x^2}}(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3) \\ &&+\sqrt{1-x^2}(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1 >0. \end{matrix}$

即可.

先证 (2.8) 式. 由于

$\begin{eqnarray*} &&8-20.25x+41.5x^2-7x^3+3x^4+x^5\\ &=&8-20.25x+34.5x^2+7x^2(1-x)+3x^4+x^5\\ &=&34.5\Big[x^2-\frac{20.25x}{34.5}+(\frac{20.25}{34.5\times2})^2\Big]-\frac{(20.25)^2}{34.5\times4}+8+7x^2(1-x)+3x^4+x^5\\ &=&34.5(x-\frac{20.25}{69})^2+5.028533+7x^2(1-x)+3x^4+x^5>0 \end{eqnarray*}$

和 $\sqrt{1-x^2}> 1-x$, 所以只需证

$x(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3)+(1-x)(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1>0,$

即可.

$\begin{eqnarray*} &&x(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3)+(1-x)(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1\\ &=&7-32x+84.25x^2-71x^3+40x^4-2x^5-x^6\\ &=&50(x-0.32)^2+1.88+37x^2(x-\frac{71}{74})^2+\frac{7}{37}x^2+x^4(1-x)(3+x)>0. \end{eqnarray*}$

因此, (2.8) 式成立. 下面证 (2.9) 式.

$\begin{eqnarray*} &&\frac{x}{\sqrt{1-x^2}}(22.5x-\frac{7.5}{2}-\frac{135x^2}{6}+30x^3)\\ &&+\sqrt{1-x^2}(8-20.25x+41.5x^2-7x^3+3x^4+x^5)-1\\ &=&\frac{1}{\sqrt{1-x^2}}(8-24x+56x^2-9.25x^3-8.5x^4+8x^5-3x^6-x^7)-1\\ &=&\frac{1}{\sqrt{1-x^2}}\Big[3.2+30(x-0.4)^2+8.25x^2+9.25x^2(1-x)+8.5x^2(1-x^2)+4x^5+3x^5(1-x)\\ &&+x^5(1-x^2)\Big]-1\\ &\geq&\frac{1}{\sqrt{1-x^2}}-1 >0. \end{eqnarray*}$

综上所述: $8(1+x)F(-\frac{5}{2},-\frac{3}{2};\frac{1}{2};x^2)-32xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-1>0$.

下面证明 $F(b,x)$ 的第三项大于等于零. 令

$\begin{matrix} G(x)&=&48(1-x^2)F(-\frac{3}{2},-\frac{5}{2};\frac{1}{2};x^2)-256x(1-x)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-4(x-1)^2\nonumber\\ &=&4(1-x)\Big[12(1+x)F(-\frac{3}{2},-\frac{5}{2};\frac{1}{2};x^2)-(1-x)-64xF(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)\Big]. \end{matrix}$

注意到 $G(1)=0$, 只需证明 $0<x<1$ 时, $G'(x)<0$. 由 (2.5) 式可得

$\begin{eqnarray*} \frac{G'(x)}{8}&=&-12xF(-\frac{3}{2},-\frac{5}{2};\frac{1}{2};x^2)+90x(1-x^2)F(-\frac{1}{2},-\frac{3}{2};\frac{3}{2};x^2)-(x-1)\\ &&-32(1-2x)F(-\frac{5}{2},-\frac{3}{2};\frac{3}{2};x^2)-160x^2(1-x)F(-\frac{3}{2},-\frac{1}{2};\frac{5}{2};x^2)\\ &=&-12x(1-x^2)^{\frac{9}{2}}F(2,3;\frac{1}{2};x^2)+90x(1-x^2)^{\frac{9}{2}}F(2,3;\frac{3}{2};x^2)-(x-1)\\ &&-32(1-2x)(1-x^2)^{\frac{11}{2}}F(4,3;\frac{3}{2};x^2)-160x^2(1-x)(1-x^2)^{\frac{9}{2}}F(4,3;\frac{5}{2};x^2), \end{eqnarray*}$

由附录 (3.3) 式和

$\begin{eqnarray*} &&F(1,1;\frac{3}{2};x^2)=(1-x^2)^{-\frac{1}{2}}F(\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2)=(1-x^2)^{-\frac{1}{2}}\frac{\sin^{-1}x}{x},\\ &&(1-x^2)F(1,1;\frac{1}{2};x^2)=1+x^2F(1,1;\frac{3}{2};x^2)=1+x^2(1-x^2)^{-\frac{1}{2}}\frac{\sin^{-1}x}{x},\\ &&(1-x^2)^{\frac{3}{2}}F(1,1;\frac{1}{2};x^2)=(1-x^2)^{\frac{1}{2}}+x\sin^{-1}{x}, \end{eqnarray*}$

可得

$\begin{eqnarray*} \frac{G'(x)}{8}&=&-12x(1-x^2)^{\frac{9}{2}}F(2,3;\frac{1}{2};x^2)+90x(1-x^2)^{\frac{9}{2}}F(2,3;\frac{3}{2};x^2)-(x-1)\\ &&-32(1-2x)(1-x^2)^{\frac{11}{2}}F(4,3;\frac{3}{2};x^2)-160x^2(1-x)(1-x^2)^{\frac{9}{2}}F(4,3;\frac{5}{2};x^2)\\ &=&-3x(1-x^2)^{\frac{3}{2}}\Big[\frac{22.5+30x^2}{2}F(1,1;\frac{1}{2};x^2)-\frac{14.5+3x^2}{2}\Big]\\ &&+22.5(1-x^2)^{\frac{5}{2}}\Big[\frac{1.5+6x^2}{2x}F(1,1;\frac{1}{2};x^2)-\frac{1.5+x^2}{2x}]-(x-1)\\ &&-(1-2x)(1-x^2)^{\frac{3}{2}}\Big[\frac{7.5+90x^2+60x^4}{2x^2}F(1,1;\frac{1}{2},x^2)-\frac{7.5+41x^2+4x^4}{2x^2}\Big]\\ &&-5(1-x)(1-x^2)^{\frac{3}{2}}\Big[\frac{-0.75+9x^2+18x^4}{x^2}F(1,1;\frac{1}{2};x^2)+\frac{0.75-7.5x^2-2x^4}{x^2}\Big]\\ &=&\frac{41.25x-180x^2+303.75x^3-240x^4+75x^5}{2x^2}(1-x^2)^{\frac{3}{2}}F(1,1;\frac{1}{2};x^2)\\ &&-(x-1)+(1-x^2)^{\frac{3}{2}}\frac{-41.25x+116x^2-102.25x^3+24x^4+3.5x^5}{2x^2}\\ &=&\frac{41.25x-180x^2+303.75x^3-240x^4+75x^5}{2x}\sin^{-1}x\\ &&+\frac{41.25x-180x^2+303.75x^3-240x^4+75x^5}{2x^2}(1-x^2)^{\frac{1}{2}}\\ &&+(1-x^2)^{\frac{3}{2}}\frac{-41.25x+116x^2-102.25x^3+24x^4+3.5x^5}{2x^2}-(x-1)\\ &=&\frac{41.25-180x+303.75x^2-240x^3+75x^4}{2}\sin^{-1}x\\ &&+(1-x^2)^{\frac{1}{2}}\frac{-64+242.75x-332x^2+180.75x^3-24x^4-3.5x^5}{2}-(x-1)\\ &=&\frac{(x-1)(75x^3-165x^2+138.75x-41.25)}{2}\sin^{-1}x\\ &&+(1-x^2)^{\frac{1}{2}}\frac{(x-1)(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-(x-1), \end{eqnarray*}$

因此,

$\begin{eqnarray*} \frac{G'(x)}{8(1-x)}&=&\frac{-64+178.75x-153.25x^2+27.5x^3+3.5x^4}{2}(1-x^2)^{\frac{1}{2}}+1\\ &&+\frac{41.25-138.75x+165x^2-75x^3}{2}\sin^{-1}x. \end{eqnarray*}$

由于 $0<x<1$, $x\leq{\sin^{-1}x}\leq\frac{x}{\sqrt{1-x^2}}$, 因此, 只需要证

$\begin{matrix} &&\frac{(75x^3-165x^2+138.75x-41.25)x}{2}\nonumber\\ &&+(1-x^2)^{\frac{1}{2}}\frac{(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-1>0 \end{matrix}$

和

$\begin{matrix} &&\frac{x}{\sqrt{1-x^2}}\frac{(75x^3-165x^2+138.75x-41.25)}{2}\nonumber\\ &&+(1-x^2)^{\frac{1}{2}}\frac{(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-1>0. \end{matrix}$

先证 (2.11) 式. 用 Mathematica 软件对$64-178.75x+153.25x^2-27.5x^3-3.5x^4$ 进行因式分解, 可得

$\begin{eqnarray*} && 64-178.75x+153.25x^2-27.5x^3-3.5x^4\\ &=&-3.5(-2.55432+x)(11.9059+x)[(x-0.747225)^2+0.04293179]>0, \end{eqnarray*}$

所以 (2.11) 式左侧为

$\begin{eqnarray*} &&\frac{75x^4-165x^3+138.75x^2-41.25x}{2}\\ &&+(1-x^2)^{\frac{1}{2}}\frac{(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-1\\ &\geq&\frac{75x^4-165x^3+138.75x^2-41.25x}{2}\\ &&+(1-x^2)\frac{(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-1\\ &=&\frac{62-220x+228x^2-13.75x^3-81.75x^4+27.5x^5+3.5x^6}{2}\\ &=&\frac{3.5(1.81006+x)(10.0682+x)(2.12587-2.67622x+x^2)(0.457239-1.34367x+x^2)}{2}\\ &=&\frac{3.5(1.81006+x)(10.0682+x)\big[(x-1.33811)^2+0.3353316279\big]\big[(x-0.67244)^2+0.0050634464\big]}{2}\\ &>&0, \end{eqnarray*}$

其中倒数第二个等号用到了 Mathematica 因式分解. 下面证明 (2.12) 式.

$\begin{eqnarray*} &&\frac{x}{\sqrt{1-x^2}}\frac{(75x^3-165x^2+138.75x-41.25)}{2}\\ &&+(1-x^2)^{\frac{1}{2}}\frac{(64-178.75x+153.25x^2-27.5x^3-3.5x^4)}{2}-1\\ &=&\frac{64-220x+228x^2-13.75x^3-81.75x^4+27.5x^5+3.5x^6}{2}(1-x^2)^{-\frac{1}{2}}-1\\ &=&\frac{64-220x+228x^2-13.75x^3-81.75x^4+27.5x^5+3.5x^6-2(1-x^2)^{\frac{1}{2}}}{2(1-x^2)^{\frac{1}{2}}}\\ &\geq&\frac{62-220x+228x^2-13.75x^3-81.75x^4+27.5x^5+3.5x^6}{2(1-x^2)^{\frac{1}{2}}}\\ &=&\frac{3.5(1.81006+x)(10.0682+x)(2.12587-2.67622x+x^2)(0.457239-1.34367x+x^2)}{2(1-x^2)^{\frac{1}{2}}}\\ &=&\frac{3.5(1.81006\!+\!x)(10.0682\!+\!x)[(x\!-\!1.33811)^2\!+\!0.3353316279][(x\!-\!0.67244)^2\!+\!0.0050634464]}{2(1-x^2)^{\frac{1}{2}}}\\ &>&0. \end{eqnarray*}$

因此, $G'(x)<0$. 定理证明完毕.

3 附录

本节给出第二节中用到的超几何函数的详细计算过程.

由相邻函数 (2.7) 式和

$\big[2\alpha-c+(b-\alpha)z\big]F(\alpha,b;c;z)=\alpha(1-z)F(\alpha+1,b;c;z)-(c-\alpha)F(\alpha-1,b;c;z),$

可得

$\begin{matrix} F(3,2;\frac{1}{2};x^2)&=&\frac{1.5+2x^2}{1-x^2}F(1,3;\frac{1}{2};x^2)-\frac{0.5}{1-x^2}F(0,3;\frac{1}{2};x^2)\nonumber\\ &=&\frac{1.5+2x^2}{1-x^2}\frac{15}{8(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{1.5+2x^2}{1-x^2}\frac{7-2x^2}{8(1-x^2)^2}-\frac{0.5}{1-x^2}\nonumber\\ &=&\frac{22.5+30x^2}{8(1-x^2)^3}F(1,1;\frac{1}{2};x^2)-\frac{14.5+3x^2}{8(1-x^2)^3},\nonumber\\ F(4,3;\frac{3}{2},x^2)&=&\frac{(6-\frac{3}{2})F(3,3;\frac{3}{2};x^2)-\frac{3}{2}F(2,3;\frac{3}{2};x^2)}{3(1-x^2)}\nonumber\\ &=&\frac{3F(3,3;\frac{3}{2};x^2)-F(2,3;\frac{3}{2};x^2)}{2(1-x^2)}\nonumber\\ &=&\frac{13.5+108x^2+36x^4}{64x^2(1-x^2)^4}F(1,1;\frac{1}{2};x^2)-\frac{13.5-39x^2}{64x^2(1-x^2)^4}\nonumber\\ &-&\frac{1.5+6x^2}{16x^2(1-x^2)^3}F(1,1;\frac{1}{2};x^2)+\frac{1.5+x^2}{16x^2(1-x^2)^3}\nonumber\\ &=&\frac{7.5+90x^2+60x^4}{64x^2(1-x^2)^4}F(1,1;\frac{1}{2};x^2)-\frac{7.5+41x^2+4x^4}{64x^2(1-x^2)^4}. \end{matrix}$

由 (2.7) 式和 (3.1) 式可得

$\begin{matrix} F(3,3;\frac{1}{2};x^2)&=&\frac{3.5+x^2}{2(1-x^2)}F(2,3;\frac{1}{2};x^2)-\frac{1.5}{2(1-x^2)}F(1,3;\frac{1}{2};x^2)\nonumber\\ &=&\frac{3.5+x^2}{2(1-x^2)}\Big[\frac{1.5+2x^2}{1-x^2}F(1,3;\frac{1}{2};x^2)-\frac{0.5}{1-x^2}F(0,3;\frac{1}{2};x^2)\Big]\nonumber\\ &&-\frac{1.5}{2(1-x^2)} F(1,3;\frac{1}{2};x^2)\nonumber\\ &=&\frac{7.5+20x^2+4x^4}{4(1-x^2)^2}F(1,3;\frac{1}{2};x^2)-\frac{3.5+x^2}{4(1-x^2)^2}\nonumber\\ &=&\frac{7.5+20x^2+4x^4}{4(1-x^2)^2}\Big[\frac{15}{8(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{7-2x^2}{8(1-x^2)^2}\Big]-\frac{3.5+x^2}{4(1-x^2)^2}\nonumber\\ &=&\frac{112.5+300x^2+60x^4}{32(1-x^2)^4}F(1,1;\frac{1}{2};x^2)-\frac{80.5+77x^2}{32(1-x^2)^4},\nonumber\\ F(3,1;\frac{1}{2};x^2)&=&\frac{15}{8(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{7-2x^2}{8(1-x^2)^2},\nonumber\\ F(3,3;\frac{3}{2};x^2)&=&\frac{1}{5x^2}(1-x^2)F(3,3;\frac{1}{2};x^2)-\frac{1}{5x^2}F(2,3;\frac{1}{2};x^2)\nonumber\\ &=&\frac{22.5+60x^2+12x^4}{32x^2(1-x^2)^3}F(1,1;\frac{1}{2};x^2)-\frac{16.1+15.4x^2}{32x^2(1-x^2)^3}\nonumber\\ &&-\frac{18+24x^2}{32x^2(1-x^2)^3}F(1,1;\frac{1}{2};x^2)+\frac{11.6+2.4x^2}{32x^2(1-x^2)^3}\nonumber\\ &=&\frac{4.5+36x^2+12x^4}{32x^2(1-x^2)^3}F(1,1;\frac{1}{2};x^2)-\frac{4.5+13x^2}{32x^2(1-x^2)^3},\nonumber\\ F(2,2;\frac{1}{2};x^2) &=&\frac{1.5+x^2}{1-x^2}F(1,2;\frac{1}{2};x^2)-\frac{0.5}{1-x^2}F(0,2;\frac{1}{2};x^2)\nonumber\\ &=&\frac{1.5+x^2}{1-x^2}\Big[\frac{1.5}{1-x^2}F(1,1;\frac{1}{2};x^2)-\frac{0.5}{1-x^2}F(0,1;\frac{1}{2};x^2)\Big]-\frac{0.5}{1-x^2}\nonumber\\ &=&\frac{9+6x^2}{4(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{5}{4(1-x^2)^2},\nonumber\\ F(3,2;\frac{3}{2};x^2) &=&\frac{1-x^2}{3x^2}F(3,2;\frac{1}{2};x^2)-\frac{1}{3x^2}F(2,2;\frac{1}{2};x^2)\nonumber\\ &=&\frac{1-x^2}{3x^2}\Big[\frac{22.5+30x^2}{8(1-x^2)^3}F(1,1;\frac{1}{2};x^2)-\frac{14.5+3x^2}{8(1-x^2)^3}\Big]\nonumber\\ &&-\frac{1}{3x^2}\Big[\frac{9+6x^2}{4(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{5}{4(1-x^2)^2}\Big]\nonumber\\ &=&\frac{1.5+6x^2}{8x^2(1-x^2)^2}F(1,1;\frac{1}{2};x^2)-\frac{1.5+x^2}{8x^2(1-x^2)^2},\nonumber\\ F(4,3;\frac{5}{2};x^2)&=&(\frac{1}{x^2}-1)F(4,3;\frac{3}{2};x^2)-\frac{1}{x^2}F(3,3;\frac{3}{2};x^2)\nonumber\\ &=&(\frac{1}{x^2}-1)\Big[\frac{7.5+90x^2+60x^4}{64x^2(1-x^2)^4}F(1,1;\frac{1}{2},x^2)-\frac{7.5+41x^2+4x^4}{64x^2(1-x^2)^4}\Big]\nonumber\\ &&-\frac{1}{x^2}\Big[\frac{4.5+36x^2+12x^4}{32x^2(1-x^2)^3}F(1,1;\frac{1}{2};x^2)-\frac{4.5+13x^2}{32x^2(1-x^2)^3}\Big]\nonumber\\ &=&\frac{-0.75+9x^2+18x^4}{32x^4(1-x^2)^3}F(1,1;\frac{1}{2};x^2)+\frac{0.75-7.5x^2-2x^4}{32x^4(1-x^2)^3}. \end{matrix}$

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