翻译论文001:Wallis不等式的最佳界 - Eufisky - The lost book
来自天书的不等式证明

翻译论文001:Wallis不等式的最佳界

Eufisky posted @ 2014年7月15日 03:30 in 不等式 with tags 不等式 论文 , 1258 阅读

Wallis不等式的最佳界

CHAO-PING   CHEN   AND   FENG   QI

(Communicated by Carmen C. Chicone)

摘要:对于所有自然数$n$,将$n!!$记为双阶乘.则有:\[\frac{1}{{\sqrt {\pi \left( {n + \frac{4}{\pi } - 1} \right)} }} \le \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} < \frac{1}{{\sqrt {\pi \left( {n + \frac{1}{4}} \right)} }}.\]其中的常数$\frac{4}{\pi}-1$和$\frac{1}{4}$是最佳可能值。从这篇论文可知,著名的Wallis不等式被加强了。

1.介绍

对任一给定的正整数$m$,双阶乘可以记作\[\left( {2m} \right)!! = \prod\limits_{i = 1}^m {\left( {2i} \right)}  \text{and}   \left( {2m - 1} \right)!! = \prod\limits_{i = 1}^m {\left( {2i - 1} \right)}. \]令\begin{align}{P_n} = \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}}.\tag{1}\end{align}则我们对$n>1$有\[\frac{1}{{2\sqrt n }} < \frac{{\sqrt 2 }}{{\sqrt {\left( {2n + 1} \right)\pi } }} < {P_n} < \frac{2}{{\sqrt {\left( {4n + 1} \right)\pi } }} < \frac{1}{{\sqrt {3n + 1} }} < \frac{1}{{\sqrt {2n + 1} }} < \frac{1}{{\sqrt {2n} }}.\]此不等式在【18,p.103】中被称为Wallis不等式。

$\text{(2)}$中$P_n$的上下界频繁被数学家引用和应用。$\text{(2)}$中最小上界$\frac{2}{\sqrt{(4n+1)\pi}}$和最大下界$\frac{\sqrt{2}}{\sqrt{(2n+1)\pi}}$,即是,不等式\[\frac{{\sqrt 2 }}{{\sqrt {\left( {2n + 1} \right)\pi } }} < {P_n} < \frac{2}{{\sqrt {\left( {4n + 1} \right)\pi } }}\tag{3}\]是由N. D. Kazarino得到的 。见【16,pp.47-48和pp.65-67】。我们可以把不等式$\text{(3)}$改写成\[\frac{1}{{\sqrt {\pi \left( {n + \frac{1}{2}} \right)} }} < {P_n} < \frac{2}{{\sqrt {\pi \left( {n + \frac{1}{4}} \right)} }}\tag{4}\],对$n\in \mathbb{N}$均成立。

公式$\text{(4)}$使用的重要性是为了通过在$\text{(6)}$中取$x=\frac{\pi}{2}$给出一个特殊情形下的Wallis公式(见4,p.259):\[\frac{\pi }{2} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left[ {\left( {2n} \right)!!} \right]}^2}}}{{{{\left[ {\left( {2n - 1} \right)!!} \right]}^2}\left( {2n + 1} \right)}} = \prod\limits_{n = 1}^\infty  {\left[ {\frac{{{{\left( {2n} \right)}^2}}}{{\left( {2n - 1} \right)\left( {2n + 1} \right)}}} \right]} .\]Wallis公式最初是伴随着正弦函数的无穷乘积展开式而来的(见【12,23】):\[\sin x = x\prod\limits_{n = 1}^\infty  {\left( {1 - \frac{{{x^2}}}{{{\pi ^2}{n^2}}}} \right)} .\tag{6}\]Wallis公式也可以表示成\[\frac{\pi }{2} = {\left[ {{4^{\zeta \left( 0 \right)}}{e^{ - \zeta '\left( 0 \right)}}} \right]^2};\]见【12】,其中$\zeta$是Riemann zeta 函数【11】。

Wallis公式的使用Hadamard乘积【10】从Riemann zeta 函数$\zeta(s)$,由$\zeta '\left( 0 \right)$得到的一个归功于Y. L. Yung的推导可以在【12】中被发现。Wallis公式也可以倒过来从没有使用Hadamard乘积而来的Wallis公式中得到$\zeta '\left( 0 \right)$的值【22】。

注意到Wallis正弦(余弦)公式【13,14】可表达为如下式子:

\begin{align*}\int_0^{\frac{\pi }{2}} {{{\sin }^n}xdx}  = \int_0^{\frac{\pi }{2}} {{{\cos }^n}xdx}  = \frac{{\sqrt \pi  \Gamma \left( {\frac{{n + 1}}{2}} \right)}}{{n\Gamma \left( {\frac{n}{2}} \right)}} = \left\{ \begin{array}{l}\frac{\pi }{2} \cdot \frac{{\left( {n - 1} \right)!!}}{{n!!}}&&n\text{为偶数时}\\\frac{{\left( {n - 1} \right)!!}}{{n!!}}&&n\text{为奇数时},\end{array} \right.\end{align*}

其中$\Gamma$是gamma函数。

涉及到$P_n$的不等式由第二作者他的合作者们在【21】中通过使用Tchebyshe 积分不等式得到。

阶乘和他们的连续性延拓充当了重要角色,比如,在组合数学,图论和特殊函数领域里。

为了了解到Wallis公式的更多信息,请参看【1,p.258】,【5,6,7】,【8,pp.17-28】,【15,p.468】,【17,pp.63-64】,和其中的参考文献部分。

在本论文中,我们将改善不等式$\text{(4)}$。更确切地说,我们将寻求两个最佳的可能常数值A和B使得双向不等式\[\frac{1}{{\sqrt {\pi \left( {n + A} \right)} }} \le {P_n} \le \frac{1}{{\sqrt {\pi \left( {n + B} \right)} }}\tag{9}\]对所有自然数$n$成立。换句话说,在$\text{(9)}$中的常数$A=\frac{4}{\pi}-1$和$B=\frac{1}{4}$不能分别被更小和更大的数取代。

2.引理

引理1.对$x>0$,我们有\[\frac{{2x + 1}}{{x\left( {4x + 1} \right)}} < \frac{{\Gamma '\left( {x + \frac{1}{2}} \right)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}} - \frac{{\Gamma '\left( x \right)}}{{\Gamma \left( x \right)}},\tag{10}\]

\[{x^{b - a}}\frac{{\Gamma \left( {x + a} \right)}}{{\Gamma \left( {x + b} \right)}} = 1 + \frac{{\left( {a - b} \right)\left( {a + b - 1} \right)}}{{2x}} + O\left( {\frac{1}{{{x^2}}}} \right),x \to \infty .\tag{11}\]

不等式$\text{(10)}$的证明在【2,3,19】中已给出,渐近展开式$\text{(11)}$的证明可以从【9】和【20,p.378】中找到。也可以见【1,p.257】

备注1.在$\text{(10)}$中用$x+\frac12$取代$x$得到\[\frac{{4x + 4}}{{\left( {2x + 1} \right)\left( {4x + 3} \right)}} < \frac{{\Gamma '\left( {x + 1} \right)}}{{\Gamma \left( {x + 1} \right)}} - \frac{{\Gamma '\left( {x + \frac{1}{2}} \right)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}}.\tag{12}\]

引理1.对$x>0$,我们有\[\frac{{\Gamma \left( {x + 1} \right)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}} < \frac{{2x + 1}}{{\sqrt {4x + 3} }}.\]

证明.定义对正实数$x$,有:\[f(x)=\ln(2x+1)-\frac12\ln(4x+3)-\ln\Gamma(x+1)+\frac12\ln\Gamma(x+\frac12).\]对$f(x)$求导给出了\[f'\left( x \right) = \frac{2}{{2x + 1}} - \frac{2}{{4x + 3}} - \left[ {\frac{{\Gamma '\left( {x + 1} \right)}}{{\Gamma \left( {x + 1} \right)}} - \frac{{\Gamma '\left( {x + \frac{1}{2}} \right)}}{{\Gamma \left( {x + \frac{1}{2}} \right)}}} \right].\]利用$\text{(12)}$,我们得到\[f'\left( x \right) > \frac{2}{{2x + 1}} - \frac{2}{{4x + 3}} - \frac{{4x + 4}}{{\left( {2x + 1} \right)\left( {4x + 3} \right)}} = 0.\]因此,$f(x)$在$(0,\infty)$上是严格递增的且\[f(x)>f(0)=\frac12\ln\frac{\pi}{3},\],由此引出不等式$\text{(13)}$。

推论1.对于所有自然数$n$,我们有\[\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}} < \frac{{2n + 1}}{{\sqrt {4n + 3} }}.\tag{14}\]

推论2.数列\[\left\{ {{Q_n}} \right\}_{n = 1}^\infty \underline{\underline \triangle} \left\{ {{{\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]}^2} - n} \right\}_{n = 1}^\infty \]是严格递增的。

证明.不等式$Q_{n+1}<Q_n$等价于\[{\left[ {\frac{{\Gamma \left( {n + 2} \right)}}{{\Gamma \left( {n + \frac{3}{2}} \right)}}} \right]^2} - \left( {n + 1} \right) < {\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2} - n,\]利用$\Gamma(x+1)=x\Gamma(x)$可以改写成\[{\left[ {\frac{{n + 1}}{{n + \frac{1}{2}}} \cdot \frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2} - 1 < {\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2},\]\[\left[ {\frac{{{{\left( {n + 1} \right)}^2}}}{{{{\left( {n + \frac{1}{2}} \right)}^2}}} - 1} \right]{\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2} < 1,\]\[{\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2} < \frac{{{{\left( {n + \frac{1}{2}} \right)}^2}}}{{{{\left( {n + 1} \right)}^2} - {{\left( {n + \frac{1}{2}} \right)}^2}}},\]\[\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}} < \frac{{2n + 1}}{{\sqrt {4n + 3} }},\]此为不等式$\text{(14)}$中取$x=n$的特殊情形。因此,单调性的证明由此得到。

3.主要结果

现在我们给出本篇论文的主要结果。

定理1.对于所有自然数$n$,我们有\[\frac{1}{{\sqrt {\pi \left( {n + \frac{4}{\pi } - 1} \right)} }} \le \frac{{\left( {2n - 1} \right)!!}}{{\left( {2n} \right)!!}} < \frac{1}{{\sqrt {\pi \left( {n + \frac{1}{4}} \right)} }}.\tag{16}\]其中的常数$\frac{4}{\pi}-1$和$\frac14$是最佳可能值。

证明.因为\[\Gamma \left( {n + 1} \right) = n!,\Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n - 1} \right)!!}}{{{2^n}}}\sqrt \pi  ,{2^n}n! = \left( {2n} \right)!!,\]双向不等式$\text{(16)}$等价于\[\frac{1}{4} < {Q_n} = {\left[ {\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}}} \right]^2} - n \le \frac{4}{\pi } - 1.\tag{17}\]从推论2中数列$Q_n$的单调性可以看出\[\mathop {\lim }\limits_{n \to \infty } {Q_n} < {Q_n} \le {Q_1} = \frac{4}{\pi } - 1.\]利用渐近公式$\text{(11)}$,我们可以推断出\[{Q_n} = n\left[ {{n^{ - \frac{1}{2}}}\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}} - 1} \right]\left[ {{n^{ - \frac{1}{2}}}\frac{{\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {n + \frac{1}{2}} \right)}} + 1} \right]\]即是\[\mathop {\lim }\limits_{n \to \infty } {Q_n} = \frac{1}{4}.\]因此,不等式$\text{(17)}$由此得到.证毕.

致谢

作者对匿名的审阅人和编辑,Carmen Chicone教授的许多有价值的评论和语言表达上的纠正表示感谢。

参考文献

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