美国数学月刊无穷乘积与西西新年祝福题

Eufisky posted @ 2017年10月29日 04:59 in 数学分析 with tags 西西无穷乘积 AMM , 1191 阅读

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！

$\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\frac{dxdydz}{\prod\limits_{cyc}{\left[242x^5-\left(y-1\right)^5-\left(z+1\right)^5\right]}}}}}.$

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

$\prod\limits_{j \ge 1} {{e^{ - 1/j}}\left( {1 + \frac{1}{j} + \frac{1}{{2{j^2}}}} \right)} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

记

${x_n} = \prod\limits_{k = 1}^n {\left( {1 + \frac{1}{k} + \frac{1}{{2{k^2}}}} \right)} = \prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} ,$

则

$\begin{align*}\frac{{\prod\limits_{k = 1}^{2n} {\left( {1 + \frac{1}{{{k^2}}}} \right)} }}{{{x_n}}} &= \frac{{\prod\limits_{k = 1}^{2n} {\frac{{{k^2} + 1}}{{{k^2}}}} }}{{\prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} }} = \frac{{\left( {{1^2} + 1} \right)\left( {{2^2} + 1} \right)\left( {{4^2} + 1} \right) \cdots \left[ {{{\left( {2n} \right)}^2} + 1} \right]}}{{{1^2}{3^2} \cdots {{\left( {2n - 1} \right)}^2}\left[ {{{\left( {2n + 1} \right)}^2} + 1} \right]}}\\&= 2\prod\limits_{k = 1}^n {\left( {1 + \frac{1}{{4{k^2}}}} \right) \cdot } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{4{n^2} + 4n + 2}},\end{align*}$

由Wallis公式可知

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.$

由

$\mathrm{sinh} x$ 的无穷乘积

$\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} = \prod\limits_{k = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{k^2}}}} \right)}$

可知

$\prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{{k^2}}}} \right)} = \frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }},\quad \prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{4{k^2}}}} \right)} = \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi },$

而调和数列

${H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + o\left( 1 \right),$

故

$\mathop {\lim }\limits_{n \to \infty } n\prod\limits_{k = 1}^n {{e^{ - 1/k}}} = {e^{ - \gamma }}.$

因此所求积分为

$\frac{{\frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }}}}{{{e^\gamma } \times \frac{\pi }{2} \times \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi }}} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

事实上，我们还有

$\begin{align*}\cosh \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{n^2}}}} \right)} ,\\\cos \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sin \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{{x^2}}}{{{n^2}}}} \right)} ,\end{align*}$

另外

$\begin{align*}\sqrt 2 \sin \left( {\frac{{x + 1}}{4}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{{{{\left( { - 1} \right)}^n}x}}{{2n + 1}}} \right)} ,\\\sqrt {x + 1} \sin \left( {\frac{{\sqrt {x + 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} - 1}}} \right)} ,\\- \sqrt {x - 1} \mathrm{csch}\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{{\sqrt {x - 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} + 1}}} \right)} ,\\- \sqrt { - x - 1} \mathrm{csch}\left( {\frac{\pi }{{\sqrt a }}} \right)\sin \left( {\frac{{\sqrt { - x - 1} }}{{\sqrt a }}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{x}{{a{n^2} + 1}}} \right)} ,\\\frac{{{e^{ - \gamma x}}}}{{\Gamma \left( {1 + x} \right)}} &= \prod\limits_{n = 1}^\infty {\frac{{1 + x/n}}{{{e^{x/n}}}}},\end{align*}$

对于求和,我们有

$\begin{align*}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} - {x^2}}}} &= \frac{1}{{2{x^2}}} - \frac{\pi }{{2x}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {{n^2} - {x^2}} \right)}^2}}}} &= - \frac{1}{{2{x^4}}} - \frac{{{\pi ^2}}}{{4{x^2}}}\mathrm{csc}^2\left( {\pi x} \right) + \frac{\pi }{{4{x^3}}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} - {x^2}}}} &= \frac{\pi }{{4x}}\tan \left( {\frac{\pi }{2}x} \right),&& \left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left[ {{{\left( {2n - 1} \right)}^2} - {x^2}} \right]}^2}}}} &= \frac{{{\pi ^2}}}{{16{x^2}}}\sec \left( {\frac{\pi }{2}x} \right) - \frac{\pi }{{8{x^3}}}\tan \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {x^2}}}} &= \frac{\pi }{{2x}}\coth \left( {\pi x} \right) - \frac{1}{{2{x^2}}}, &&\left| x \right| < \infty\\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} + {x^2}}}} &= \frac{\pi }{{4x}}\tanh \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty\end{align*}$

其中

$\mathrm{sinh}x=\frac{e^x-e^{-x}}2,\mathrm{cosh}x=\frac{e^x+e^ {-x}}2,\mathrm{csch}x=\frac2{e^x-e^ {-x}},\mathrm{tanh}x=\frac{e^x-e^ {-x}}{e^x+e^ {-x}},\mathrm{coth}x=\frac{e^x+e^ {-x}}{e^x-e^ {-x}}$ .

The Weierstrass factorization theorem. Sometimes called the Weierstrass product/factor theorem.

Let

$f$ be an entire function, and let

$\{a_n\}$ be the non-zero zeros of

$ƒ$ repeated according to multiplicity; suppose also that

$ƒ''$ has a zero at

$z= 0$ of order

$m\geq 0$ (a zero of order

$m=0$ at

$z=0$ means

$f(0)\neq 0$ .

Then there exists an entire function

$g$ and a sequence of integers

$\{p_n\}$ such that

$f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right).$

====Examples of factorization====

$\begin{align*}\sin \pi z &= \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)\\\cos \pi z &= \prod_{q \in \mathbb{Z}, \, q \; \text{odd} } \left(1-\frac{2z}{q}\right)e^{2z/q} = \prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+\tfrac{1}{2}} \right)^2 \right) \end{align*}$

The cosine identity can be seen as special case of

$\frac{1}{\Gamma(s-z)\Gamma(s+z)} = \frac{1}{\Gamma(s)^2}\prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+s} \right)^2 \right)$

for

$s=\tfrac{1}{2}$ .

Mittag-Leffler's theorem.

== Pole expansions of meromorphic functions ==

Here are some examples of pole expansions of meromorphic functions:

$\begin{align*}\frac{1}{\sin(z)}&= \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{z-n\pi}= \frac{1}{z} + 2z\sum_{n=1}^\infty (-1)^n \frac{1}{z^2 - (n\,\pi)^2},\\\cot(z) &\equiv \frac{\cos (z)}{\sin (z)}= \sum_{n \in \mathbb{Z}} \frac{1}{z-n\pi}= \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2 - (k\,\pi)^2},\\\frac{1}{\sin^2(z)} &= \sum_{n \in \mathbb{Z}} \frac{1}{(z-n\,\pi)^2},\\\frac{1}{z \sin(z)}&= \frac{1}{z^2} + \sum_{n \neq 0} \frac{(-1)^n}{\pi n(z-\pi n)}= \frac{1}{z^2} + \sum_{n=1}^\infty \frac{(-1)^n}{n\,\pi} \frac{2z}{z^2 - (n\,\pi)^2}.\end{align*}$

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2022年8月24日 23:30

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