美国数学月刊无穷乘积与西西新年祝福题
求
若$a_i$是超越方程\[\left( {\cos x} \right)\left( {\cosh x} \right) + 1 = 0\]的正实数根从小到大排成的数列, 求证:
$$\sum_{i=1}^{\infty}{a_{i}^{-6}\left(\frac{\sin a_i-\sinh a_i}{\cos a_i+\cosh a_i}\right)^2}=\frac{1}{80}.$$
首先可以肯定,这个方程里$a_i$肯定是解不出的.其次我们有\[{\left( {\sinh {a_i}} \right)^2} = {\left( {\cosh {a_i}} \right)^2} - 1 = \frac{1}{{{{\cos }^2}{a_i}}} - 1 = {\tan ^2}{a_i} \Rightarrow \sinh {a_i} = \left| {\tan {a_i}} \right|.\]
当$\sinh {a_i} = \tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} - \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\tan ^2}\frac{{{a_i}}}{2}.\]
当$\sinh {a_i} = -\tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} + \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\cot ^2}\frac{{{a_i}}}{2}.\]事实上,两种情况都会出现.接下来大家一起来思考下哈!
以下几个也是不同寻常的题,正是因为莫名其妙、不明觉厉才想一探究竟,希望大家一起来玩!
1、求无穷积分$$\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\frac{dxdydz}{\prod\limits_{cyc}{\left[242x^5-\left(y-1\right)^5-\left(z+1\right)^5\right]}}}}}.$$
2、设\[{\left( {1 + \frac{1}{x}} \right)^x} = e\left( {1 - \sum\limits_{k = 1}^\infty {\frac{{{d_k}}}{{{{\left( {\frac{{11}}{{12}} + x} \right)}^k}}}} } \right),\]求
\[{\sum\limits_{k = 1}^\infty {\frac{1}{{1 + d_k^2}}} }.\]
(2017年10月AMM征解题)求证
记
Problems of the Miklós Schweitzer Memorial Competition
1、AOPS论坛
2、匈牙利语版本
3、英文版本
4、其它试题
也就是求\[\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{\prod\limits_{k = 0}^n {\left( {a + kd} \right)} }}} .\]
解.首先有$$\prod_{k=0}^n{\frac{1}{a+kd}}=\frac{\Gamma \left( \frac{a}{d} \right)}{d^{n+1}\Gamma \left( \frac{a}{d}+n+1 \right)},$$
又因为$$\gamma \left( s,x \right) =\sum_{k=0}^{\infty}{\frac{x^se^{-x}x^k}{s\left( s+1 \right) ...\left( s+k \right)}}=x^s\,\Gamma \left( s \right) \,e^{-x}\sum_{k=0}^{\infty}{\frac{x^k}{\Gamma \left( s+k+1 \right)}},$$我们有
$g(x) = x^a f(x^d)$ satifies $g'(x) = x^{a-1} + x^{d-1} g(x)$. Solve the associated differential equation and conclude.
令$a\in (0,\pi)$,设$n$为正整数.证明$$\int_0^{\pi}{\frac{\cos \left( nx \right) -\cos \left( na \right)}{\cos x-\cos a}dx}=\pi \frac{\sin \left( na \right)}{\sin a}.$$
Since
Thus $$ f_{2n}\left(\frac {\pi}{2n + 1}\right) < \frac {\pi}{2n + 1}\cdot\frac {2\pi}{2n + 1} - \frac {\pi}{2(2n + 1)} < 0.$$
Euler-Maclaurin求和公式估计梯形积分公式的误差
西西在大学群里的一道题,也是2014年第六届非数竞赛预赛最后一题的推广:
设${A_n} = \frac{n}{{{n^2} + 1}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}$,求极限
\[\mathop {\lim }\limits_{n \to \infty } {n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right).\]
这里提供个一般的方法.
Euler-Maclaurin求和公式
设函数$f\in C^{(2m+2)}[a,b],h=\frac{b-a}{n},x_i=a+ih,i=0,1,\cdots,n$,则
\begin{align*}\frac{{b - a}}{n}\sum\limits_{i = 1}^n {\frac{1}{2}\left[ {f\left( {{x_{i - 1}}} \right) + f\left( {{x_i}} \right)} \right]} - \int_a^b {f\left( x \right)dx} = &\sum\limits_{k = 1}^m {\frac{{{B_{2k}}}}{{\left( {2k} \right)!}}{h^{2k}}\left[ {{f^{\left( {2k - 1} \right)}}\left( b \right) - {f^{\left( {2k - 1} \right)}}\left( a \right)} \right]} \\&+ \frac{{{B_{2m + 2}}}}{{\left( {2m + 2} \right)!}}{h^{2m + 2}}{f^{\left( {2m + 2} \right)}}\left( \xi \right)\left( {b - a} \right),\end{align*}
其中$\xi\in [a,b]$, $B_{2k}(k=1,2,\cdots,m+1)$是Bernoulli数且$B_2=\frac16,B_4=-\frac{1}{30},B_6=\frac{1}{42}$.
解:取$a=0,b=1,f(x)=\frac{1}{1+x^2}$,则$h=\frac1n,x_i=\frac{i}{n},A_n=\frac{1}{n}\sum\limits_{i = 1}^n {f\left( {{x_i}} \right)}$,则
\begin{align*}&{A_n} + \frac{1}{{4n}} - \frac{\pi }{4} = \frac{1}{2}\left[ {\left( {{A_n} - \frac{1}{{2n}} + \frac{1}{n}} \right) + {A_n}} \right] - \frac{\pi }{4} = \frac{{{B_2}}}{{2!}} \cdot \frac{1}{{{n^2}}}\left[ {f'\left( 1 \right) - f'\left( 0 \right)} \right]\\+ &\frac{{{B_4}}}{{4!}} \cdot \frac{1}{{{n^4}}}\left[ {f'''\left( 1 \right) - f'''\left( 0 \right)} \right] + \frac{{{B_6}}}{{6!}} \cdot \frac{1}{{{n^6}}}\left[ {{f^{\left( 5 \right)}}\left( 1 \right) - {f^{\left( 5 \right)}}\left( 0 \right)} \right] + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^8}}}{f^{\left( 8 \right)}}\left( \xi \right),\end{align*}
其中$\xi\in [0,1]$,也即
\[{n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right) = \frac{1}{{2016}} + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^2}}}{f^{\left( 8 \right)}}\left( \xi \right),\]
注意到${f^{\left( 8 \right)}}\left( \xi \right)$有界,因此$n\to\infty$时,所求极限为$\frac{1}{{2016}}$.
西西爆难积分题求解
在这里,主要展示西西12年7月在百度贴吧数学吧中贴出的30个积分题的求解,本文中主要参考的是自己以前的摘录,不知何故,与西哥的版本有些出入,但基本包含了西哥的所有问题,来源于网友的解答均会注明出处。
参阅:[1]http://tieba.baidu.com/p/2114477017
[2]http://tieba.baidu.com/p/3148596990
[3]http://tieba.baidu.com/p/2121721064