Eufisky - The lost book

Euler Sum的若干研究

本文主要展示并分享有关Euler Sum的若干求解问题。

问题一：求\[\sum_{n=1}^\infty{\frac{H_n}{n^q}}\]

Solution.

\begin{align}&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\&=(k+1)\zeta(k+4)+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\&=(k+1)\zeta(k+4)\\&+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\&-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\&=(k+1)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\&=(k+5)\zeta(k+4)+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}-4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\&=(k+5)\zeta(k+4)-2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12}\end{align}
 
Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields
$$\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)=(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13}$$
and finally
$$\sum_{m=1}^\infty\frac{H_m}{m^q}=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14}$$
 
Explanation
 
$\hphantom{0}(1)$ expand $\zeta$  
$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest  
$\hphantom{0}(3)$ simplify terms  
$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$  
$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation  
$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$  
$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$  
$\hphantom{0}(8)$ $m\mapsto m-n$  
$\hphantom{0}(9)$ subtract and add the terms for $m=n$  
$(10)$ combine $\zeta(k+4)$ and change the order of summation  
$(11)$ $H_m=\sum_{n=1}^m\frac1n$  
$(12)$ combine sums  
 

 

 

 

 

参考文献

[1]

[2]

数学分析 级数 Comments(0) 2014年7月23日 01:50

积分不等式的

这个模块，我将展现并分享对积分不等式的一些研究成果。

 

 

 

 

参考：

[1] 积分不等式的研究;

[2] 积分不等式的证明方法;

数学分析 积分不等式 Comments(0) 2014年7月23日 00:42

解析几何竞赛题

第四届全国大学生数学竞赛决赛（数学组）解析几何试题

设$A$为正整数，直线$L$与双曲线$x^2-y^2=2(x>0)$所围成的面积为$A$，证明：

(1)上述$L$被双曲线$x^2-y^2=2(x>0)$所截线段的中点的轨迹为双曲线；

(2)$L$总是(1)中轨迹曲线的切线.

证明. (1)不妨设直线$L$的方程为$x=my+l(m^2<1)$，直线$L$与双曲线$x^2-y^2=2(x>0)$的交点$P,Q$分别为$(x_1,y_1),(x_2,y_2)$$(\text{其中}y_1<y_2)$,

联立方程，有\[\left\{ \begin{array}{l}x = my + l\\{x^2} - {y^2} = 2\end{array} \right. \Rightarrow \left( {{m^2} - 1} \right){y^2} + 2mly + {l^2} - 2 = 0.\]

由韦达定理，我们有：\[{y_1} + {y_2} = \frac{{2ml}}{{1 - {m^2}}},{y_1}{y_2} = \frac{{{l^2} - 2}}{{{m^2} - 1}},{y_2} - {y_1} = \sqrt {{{\left( {{y_2} + {y_1}} \right)}^2} - 4{y_2}{y_1}}  = \frac{{2\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.\]

由题意得：

\begin{align}A &= \int_{{y_1}}^{{y_2}} {\left( {my + l - \sqrt {{y^2} + 2} } \right)dy}  = \left[ {\frac{{m{y^2}}}{2} + ly - \left( {\frac{{y\sqrt {{y^2} + 2} }}{2} + \ln \left( {y + \sqrt {{y^2} + 2} } \right)} \right)} \right]_{{y_1}}^{{y_2}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{y_2}\sqrt {y_2^2 + 2} }}{2} - \frac{{{y_1}\sqrt {y_1^2 + 2} }}{2}} \right) - \ln \frac{{{y_2} + \sqrt {y_2^2 + 2} }}{{{y_1} + \sqrt {y_1^2 + 2} }}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left( {\frac{{{x_2}{y_2}}}{2} - \frac{{{x_1}{y_1}}}{2}} \right) - \ln \frac{{{x_2} + {y_2}}}{{{x_1} + {y_1}}}\\&= \frac{m}{2}\left( {y_2^2 - y_1^2} \right) + l\left( {{y_2} - {y_1}} \right) - \left[ {\frac{m}{2}\left( {y_2^2 - y_1^2} \right) + \frac{l}{2}\left( {{y_2} - {y_1}} \right)} \right] - \ln \frac{{{x_1}{x_2} - {y_1}{y_2} + {x_1}{y_2} - {x_2}{y_1}}}{2}\\&= \frac{l}{2}\left( {{y_2} - {y_1}} \right) - \ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2}\end{align}
又
\begin{align}&\ln \frac{{\left( {{m^2} - 1} \right){y_1}{y_2} + ml\left( {{y_1} + {y_2}} \right) + l\left( {{y_2} - {y_1}} \right) + {l^2}}}{2} = \ln \frac{{{l^2} - 2 + \frac{{2{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{2l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} + {l^2}}}{2}\\&= \ln \left( {{l^2} - 1 + \frac{{{m^2}{l^2}}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right) = \ln \left( {\frac{{{m^2} + {l^2} - 1}}{{1 - {m^2}}} + \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}} \right).\end{align}
故我们有
\[A = \frac{{l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}} - \ln \frac{{{m^2} + {l^2} - 1 + l\sqrt {2{m^2} + {l^2} - 2} }}{{1 - {m^2}}}.\]
设$P,Q$中点为$M(x,y)$，则有
\[\left\{ \begin{array}{l}x = \frac{{{x_1} + {x_2}}}{2} = \frac{{{y_1} + {y_2}}}{2}m + l = \frac{l}{{1 - {m^2}}}\\y = \frac{{{y_1} + {y_2}}}{2} = \frac{{ml}}{{1 - {m^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}m = \frac{y}{x}\\l = \frac{{{x^2} - {y^2}}}{x}\end{array} \right..\]
由此得
\[A = \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)}  - \ln \left[ {\left( {{x^2} - {y^2}} \right) + \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)}  - 1} \right].\]
令\[t = \sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)} ,\]
则有
\[A = t - \ln \left( {t + \sqrt {{t^2} + 1} } \right).\tag{1}\]
记$f\left( t \right) = t - \ln \left( {t + \sqrt {{t^2} + 1} } \right)$，注意到$f'\left( x \right) = 1 - \frac{1}{{\sqrt {{t^2} + 1} }} > 0$，其值域显然为$(0,+\infty)$，故方程$\text{(1)}$的解唯一，记作$t_0$.因此我们有\[\sqrt {\left( {{x^2} - {y^2}} \right)\left( {{x^2} - {y^2} - 2} \right)}  = t_0 \Rightarrow {x^2} - {y^2} = 1 + \sqrt {1 + {t_0^2}}.\]

即\[x^2-y^2=C(C>2).\]为双曲线轨迹.

(2)再之，由\[2x - 2yy' = 0 \Rightarrow y' = \frac{x}{y} = \frac{1}{m} = {k_L}\]及直线$L$经过点$M$可知，直线$L$总为$M$的轨迹曲线的切线.

解析几何 解析几何 竞赛题 Comments(1) 2014年7月20日 20:08

很好的封面

代码：

 

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 Rhön-Gymnasium-Gymnasium \\
 in \\
 Bad Neustadt

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Hier steht noch immer der gleiche Text, aber jetzt auf der zweiten Seite.

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LaTeX latex 封面 Comments(0) 2014年7月18日 03:40

翻译论文001：Wallis不等式的最佳界

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

[1] M. Abramowitz and I. A. Stegun (Eds.), Handbook of Mathematical Functions with Formulas,Graphs, and Mathematical Tables, National Bureau of Standards, Applied Mathematics Series 55, 9th printing, Dover, New York, 1972. MR 34:8607
[2] H. Alzer, On some inequalities for the gamma and psi functions, Math. Comp. 66 (1997),373{389. MR 97e:33004
[3] G. D. Anderson, R. W. Barnard, K. C. Richards, M. K. Vamanamurthy, and M. Vuorinen,
Inequalities for zero-balanced hypergeometric functions, Trans. Amer. Math. Soc. 347 (1995),1713{1723. MR 95m:33002
[4] P. S. Bullen, A Dictionary of Inequalities, Pitman Monographs and Surveys in Pure and Applied Mathematics 97, Addison-Wesley, Longman Limited, 1998. MR 2000e:26001
[5] Ch.-P. Chen and F. Qi, Improvement of lower bound in Wallis' inequality, RGMIA Res. Rep.Coll. 5 (2002), suppl., Art. 23. Available online at http://rgmia.vu.edu.au/v5(E).html.
[6] Ch.-P. Chen and F. Qi, The best bounds in Wallis' inequality, RGMIA Res. Rep. Coll. 5(2002), no. 4, Art. 13. Available online at http://rgmia.vu.edu.au/v5n4.html.
[7] Ch.-P. Chen and F. Qi, A new proof of the best bounds in Wallis' inequality, RGMIA Res.Rep. Coll. 6 (2003), no. 2, Art. 2. Available online at http://rgmia.vu.edu.au/v6n2.html.
[8] S. R. Finch, Archimedes' Constant, $\S$ 1.4 in Mathematical Constants, Cambridge Univ. Press,Cambridge, England, 2003. Available online at http://pauillac.inria.fr/algo/bsolve/.
[9] C. L. Frenzer, Error bounds for asymptotic expansions of the ratio of two gamma functions,SIAM J. Math. Anal. 18 (1987), 890{896. MR 88d:33001
[10] http://mathworld.wolfram.com/HadamardProduct.html.
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[12] http://mathworld.wolfram.com/WallisFormula.html.
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[15] H. Je reys and B. S. Je reys, Wallis's Formula for $\pi$, $\S$ 15.07 in Methods of Mathematical Physics, 3rd ed. Cambridge Univ. Press, Cambridge, England, 1988.
[16] N. D. Kazarino , Analytic Inequalities, Holt, Rhinehart and Winston, New York, 1961. MR 41:5577
[17] J. F. Kenney and E. S. Keeping, Mathematics of Statistics, Part 2, 2nd ed., Van Nostrand,Princeton, New Jersey, 1951.
[18] J.-Ch. Kuang, Changyong Budengsh (Applied Inequalities), 2nd edition, Hunan Education Press, Changsha, China, 1993. (Chinese) MR 95j:26001
[19] Y. L. Luke, Inequalities for the gamma function and its logarithmic derivative, Math. Balkanica(N. S.) 2 (1972), 118{123. MR 50:10338
[20] A. F. Nikiforov and V. B. Uvarov, Special Functions of Mathematical Physics, Birkhauser,Basel, 1988. MR 89h:33001
[21] F. Qi, L.-H. Cui, and S.-L. Xu, Some inequalities constructed by Tchebyshe 's integral inequality,Math. Inequal. Appl. 2 (1999), no. 4, 517{528. MR 2000m:26027
[22] J. Sondow, Analytic continuation of Riemann's zeta function and values at negative integers via Euler's transformation of series, Proc. Amer. Math. Soc. 120 (1994),421{424. MR 94d:11066
[23] E. W.Weisstein, Concise Encyclopedia of Mathematics CD-ROM, CD-ROM edition 1.0, May 20, 1999. Available online at http://www.math.pku.edu.cn/stu/eresource/wsxy/sxrjjc/wk/Encyclopedia/math/w/w009.htm

不等式 不等式 论文 Comments(1) 2014年7月15日 03:30

双阶乘的估计

双阶乘除式的一个估计
\[\boxed{\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} \sim \sqrt {n\pi } .}\]
 
证明：
\begin{align*}\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}} =\frac{{{{\left[ {\left( {2n} \right)!!} \right]}^2}}}{{\left( {2n} \right)!}} =\frac{{{2^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}} \sim \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \frac{{{2^{2n}}{{\left[ {{{\left( {\frac{n}{e}} \right)}^n}\sqrt {2\pi n} } \right]}^2}}}{{{{\left( {\frac{{2n}}{e}} \right)}^{2n}}\sqrt {4\pi n} }} = \sqrt {n\pi }. \end{align*}
由此可对如下一个关于$\sin x$的整数次幂在$[-\frac{\pi}{2},\frac{\pi}{2}]$上的积分进行数值计算：
\[\int_0^{\frac{\pi }{2}} {{{\sin }^n}xdx}  = \left\{ \begin{array}{l}\frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \sim \frac{1}{2}\sqrt{\frac{\pi }{m}} ,n = 2m\\\frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \sim \frac{{\sqrt {m\pi } }}{{2m + 1}},n = 2m + 1\end{array} \right.,m \in {N_ + }.\]
特殊地，我们有：
\[0.250037 \approx I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\sin }^{100}}xdx}  = 2\int_0^{\frac{\pi }{2}} {{{\sin }^{100}}xdx}  = \frac{{99!!}}{{100!!}}\pi  \sim \frac{1}{{\sqrt {50\pi } }}\pi  = \frac{1}{5}\sqrt {\frac{\pi }{2}}  = 0.25003696.\]
由上可知，此误差是很小的。

 

数学分析 数值计算 双阶乘 Comments(0) 2014年7月12日 04:29

一个级数求解

\[\sum\limits_{n = 1}^\infty  {\frac{{\ln n}}{{{n^2}}}}  =  - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma  + \ln 2 + \ln \pi } \right).\]
证明：(Glaisher–Kinkelin constant)\[\boxed{\ln A = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^2}}}{4} - \left( {\frac{{{n^2}}}{2} + \frac{n}{2} + \frac{1}{{12}}} \right)\ln n + \sum\limits_{k = 1}^n {k\ln k} } \right].}\]
(Riemann zeta函数的导函数)\[\boxed{\zeta '\left( s \right) =  - \sum\limits_{k = 1}^\infty  {\frac{{\ln k}}{{{k^s}}}} .}\]
(Gamma 函数)
\[\Gamma \left( s \right) = \int_0^\infty  {{x^{s - 1}}{e^{ - x}}dx}. \]
首先，证明$\zeta'(-1)=\frac{1}{12}-\ln A$.
 
再证明$\Gamma'(2)=1-\gamma.$
\[\Gamma '\left( 2 \right) = \int_0^\infty  {x{e^{ - x}}\ln xdx}  = \int_0^\infty  {\left( {{e^{ - x}} + {e^{ - x}}\ln x} \right)dx}  = 1 - \gamma .\]
 
利用
\[\zeta \left( s \right) = {2^s}{\pi ^{s - 1}}\sin \frac{{\pi s}}{2}\Gamma \left( {1 - s} \right)\zeta \left( {1 - s} \right).\]
令$s=-1$，我们有$\zeta{-1}=-\frac{1}{12}.$
两边同取对数得
\[\ln \zeta \left( s \right) = s\ln 2 + \left( {s - 1} \right)\ln \pi  + \ln \sin \frac{{\pi s}}{2} + \ln \Gamma \left( {1 - s} \right) + \ln \zeta \left( {1 - s} \right).\]
求导，得到\[\frac{{\zeta '\left( s \right)}}{{\zeta \left( s \right)}} = \ln \left( {2\pi } \right) + \frac{\pi }{{2\tan \frac{{\pi s}}{2}}} - \frac{{\Gamma '\left( {1 - s} \right)}}{{\Gamma \left( {1 - s} \right)}} - \frac{{\zeta '\left( {1 - s} \right)}}{{\zeta \left( {1 - s} \right)}}.\]
令$s=-1$，我们有
\[\frac{{\zeta '\left( { - 1} \right)}}{{\zeta \left( { - 1} \right)}} = 12\ln A - 1 = \ln \left( {2\pi } \right) - 1 + \gamma  - \frac{{\zeta '\left( 2 \right)}}{{\zeta \left( 2 \right)}}.\]
 
\[ \Rightarrow \zeta '\left( 2 \right) = \frac{{{\pi ^2}}}{6}\left( {\ln \left( {2\pi } \right) - 12\ln A + \gamma } \right).\]
 
因此我们得到\[\sum\limits_{n = 1}^\infty  {\frac{{\ln n}}{{{n^2}}}}  =  - \frac{{{\pi ^2}}}{6}\left( { - 12\ln A + \gamma  + \ln 2 + \ln \pi } \right).\]

数学分析 级数 Comments(1) 2014年6月24日 23:28

想看《笑傲江湖》的请戳进来

    利用周末时间，更新了TeXlive2014，把金庸先生的巨作《笑傲江湖》重新排版了下，最大的特色是采用超强大的淡黄色纸张颜色，高清无码，拥有防近视的技能。里面内容和图片均源于网络，本人仅仅是收集整理，一切权限归于网络和金庸老先生所有。想看的可以在下面到下面的百度云链接里下载，限全文完整转载，仅供学习交流之用，勿用作商业用途！！！  

百度云链接：http://pan.baidu.com/s/17y8MU

自创 重排 自创 武侠 Comments(0) 2014年6月21日 02:27

用tikz设计一个灰色页眉

页眉页脚的设计前文涉及不多，这里分享一个稍微可变的页眉设计，把tikz嵌入进来进行设计制作，我们可以制作更多变的样式效果出来。

效果图如下：

headdesign20140123010456

代码如下：

 

\documentclass[12pt,twoside]{scrbook}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[ngerman]{babel}

\usepackage[table,dvipsnames,svgnames]{xcolor}
\usepackage{calc}
%Schrift
\usepackage{mathptmx,charter,courier}
\usepackage[scaled]{helvet}
\usepackage{microtype}

\usepackage{blindtext}

% TikZ-Pakete
\usepackage{tikz}
\usetikzlibrary{positioning,backgrounds,shapes}

% Kopf- und Fußzeile
\usepackage[automark]{scrpage2}
\clearscrheadings
\clearscrplain

\tikzset{pagenumber/.style={rectangle,rounded corners=5pt,inner sep=6pt, fill=gray!20,draw=gray!30,line width=0.1em}}

\lohead{\begin{tikzpicture}[remember picture,overlay]
\begin{pgfonlayer}{background}
\draw[fill=gray!10, fill opacity=1.0, draw=none] (-.65\paperwidth,0cm) rectangle (\paperwidth+2cm,2cm);
\end{pgfonlayer}
\path (current page.north west) node[anchor=west,xshift=1.58cm, yshift=-0.9cm,pagenumber] {\parbox{\widthof{\Large 1111}}{\centering \color{black} \bfseries\Large \thepage}};
\end{tikzpicture}}
\rohead{\begin{tikzpicture}[remember picture,overlay]
\path (current page.north east) node[anchor=east,xshift=-3.4cm, yshift=-0.9cm] {\color{black}\upshape Beispieldokument};
\end{tikzpicture}}

\lehead{\begin{tikzpicture}[remember picture,overlay]
\begin{pgfonlayer}{background}
\draw[fill=gray!10, fill opacity=1.0, draw=none] (-.65\paperwidth,0cm) rectangle (\paperwidth+2cm,2cm);
\end{pgfonlayer}
\path (current page.north west) node[anchor=west,xshift=3.4cm, yshift=-1.0cm] {\color{black}\upshape \Large \headmark};
\end{tikzpicture}}
\rehead{\begin{tikzpicture}[remember picture,overlay]
\path (current page.north east) node[anchor=east,xshift=-1.58cm, yshift=-0.9cm,pagenumber] {\parbox{\widthof{\Large 1111}}{\centering \color{black}\bfseries\Large \thepage}};
\end{tikzpicture}}
\setlength{\headsep}{1.1cm}

\begin{document}
\pagestyle{scrheadings}
\chapter{Einleitung}
\Blindtext

\Blindtext

\chapter{Hauptteil}
\blindtext

\end{document}

选自：http://web.slzm.de/blog/latex/latex-schone-kopfzeilen/#more-345

LaTeX LaTeX 页眉设计 Comments(1) 2014年6月21日 00:38

Tikz设计页眉设计样例分享

效果图：

enter image description here

代码如下：

\documentclass{book}
\usepackage[a6paper]{geometry}
\usepackage{fancyhdr}
\usepackage{tikzpagenodes}
\usetikzlibrary{shapes.geometric}
\usepackage{lipsum}

\pagestyle{fancy}
\fancyhf{}
\renewcommand\headrulewidth{0pt}
\fancyhead[OC]{\begin{tikzpicture}[remember picture,overlay]
\node[diamond,draw,font=\small\itshape] at (current page header area.south west) (dia) {\thepage};
\draw[double=white] (dia.east) -- (current page header area.south east);
\end{tikzpicture}}
\fancyhead[EC]{\begin{tikzpicture}[remember picture,overlay]
\node[diamond,draw,font=\small\itshape] at (current page header area.south east) (dia) {\thepage};
\draw[double=white] (dia.west) -- (current page header area.south west);
\end{tikzpicture}}
\fancyhead[OR]{\small\nouppercase\leftmark}
\fancyhead[EL]{\small\nouppercase\rightmark}

\begin{document}

\chapter{Test chapter}
\lipsum[2]
\section{Test section}
\lipsum[2]\lipsum[2]\lipsum[2]\lipsum[2]\lipsum[2]

\end{document}

效果图：

enter image description here

代码如下：

 

\documentclass{book}
\usepackage[a6paper]{geometry}
\usepackage{fancyhdr}
\usepackage{tikzpagenodes}
\usetikzlibrary{shapes.geometric}
\usepackage{lipsum}

\pagestyle{fancy}
\fancyhf{}
\renewcommand\headrulewidth{0pt}
\fancyhead[OC]{\begin{tikzpicture}[remember picture,overlay]
\node[diamond,draw,font=\small\itshape] at (current page header area.south west) (dia) {\thepage};
\draw (dia.3) -- (current page header area.south east|-dia.3);
\draw (dia.357) -- ([xshift=-7pt]current page header area.south east|-dia.357);
\end{tikzpicture}}
\fancyhead[EC]{\begin{tikzpicture}[remember picture,overlay]
\node[diamond,draw,font=\small\itshape] at (current page header area.south east) (dia) {\thepage};
\draw (dia.177) -- (current page header area.south west|-dia.177);
\draw (dia.183) -- ([xshift=7pt]current page header area.south west|-dia.183);
\end{tikzpicture}}
\fancyhead[OR]{\small\nouppercase\leftmark}
\fancyhead[EL]{\small\nouppercase\rightmark}

\begin{document}

\chapter{Test chapter}
\lipsum[2]
\section{Test section}
\lipsum[2]\lipsum[2]\lipsum[2]\lipsum[2]\lipsum[2]

\end{document}

效果图：

enter image description here

enter image description here

代码如下：

 

\documentclass{book}
\usepackage[a6paper]{geometry}% just for the example
\usepackage{fancyhdr}
\usepackage{stackengine}
\usepackage{graphicx}
\usepackage{lipsum}% just to generate text for the example

\pagestyle{fancy}
\fancyhf{}
\renewcommand\headrulewidth{0pt}
\fancyhead[OC]{%
  \def\stackalignment{c}%
  \topinset{\itshape\thepage}{\scalebox{5}{\(\diamond\)}}{1.5ex}{}%
  \rule{-.2ex}{0ex}%
  \def\stackalignment{l}%
  \stackon[1pt]{\rule[2.8ex]{2.5in}{.1ex}}{\rule{2.55in}{.1ex}}}
\fancyhead[EC]{%
  \def\stackalignment{r}%
  \stackon[1pt]{\rule[2.8ex]{2.5in}{.1ex}}{\rule{2.55in}{.1ex}}%
  \rule{-.2ex}{0ex}%
  \def\stackalignment{c}%
  \topinset{\itshape\thepage}{\scalebox{5}{\(\diamond\)}}{1.5ex}{}}
\begin{document}
\lipsum[2]\lipsum[2]\lipsum[2]\lipsum[2]
\end{document}

LaTeX latex 页眉设计 Comments(63) 2014年6月20日 23:28