一道与ODE有关的数列极限求解 - Eufisky - The lost book
涉及到星形线的一个运动轨迹问题
逆神的数学分析题答案总算补全了

一道与ODE有关的数列极限求解

Eufisky posted @ 2014年9月08日 18:07 in 数学分析 with tags ODE 数列极限 , 1047 阅读

一边参考葛神的日志,一边把这题做出来了,在今年中秋节这算是给我的不错礼物.

(准大一问题)已知${U_{n + 2}} = {U_{n + 1}} + \frac{{{U_n}}}{{2n}},{U_1} = 0,{U_2} = 1$,求$\mathop {\lim }\limits_{n \to \infty } \frac{{{U_n}}}{{\sqrt n }}.$
解.设$S\left( x \right) = {U_1} + {U_2}x +  \cdots  + {U_n}{x^{n - 1}} +  \cdots $,则
\begin{align*}S\left( x \right) &= {U_1} + {U_2}x + \sum\limits_{n = 1}^\infty  {{U_{n + 2}}{x^{n + 1}}} \\&= {U_1} + {U_2}x + \sum\limits_{n = 1}^\infty  {\left( {{U_{n + 1}} + \frac{{{U_n}}}{{2n}}} \right){x^{n + 1}}} \\&= {U_1} + {U_2}x + x\left( {S\left( x \right) - {U_1}} \right) + \frac{1}{2}x\int_0^x {S\left( t \right)dt} \\&= x\left( {S\left( x \right) + 1} \right) + \frac{1}{2}x\int_0^x {S\left( t \right)dt}.\end{align*}
\[\frac{{S\left( x \right)}}{x} = S\left( x \right) + 1 + \frac{1}{2}\int_0^x {S\left( t \right)dt} .\]
两边同时求导
\[\frac{{xS'\left( x \right) - S\left( x \right)}}{{{x^2}}} = S'\left( x \right) + \frac12 S\left( x \right) \Rightarrow S = \frac{x}{{{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}{e^{ - \frac{x}{2}}}.\]
注意到
\[\frac{1}{{{{\left( {1 - x} \right)}^{\frac{3}{2}}}}} = \sum\limits_{n = 0}^\infty  {\frac{{\left( {2n + 1} \right)!!}}{{{2^n}n!}}{x^n}} ,{e^{ - \frac{x}{2}}} = \sum\limits_{n = 0}^\infty  {{{\left( { - 1} \right)}^n}\frac{1}{{{2^n}n!}}{x^n}} .\]
\[\frac{x}{{{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}{e^{ - \frac{x}{2}}} = \sum\limits_{n = 0}^\infty  {{c_n}{x^n}} ,\]
其中
\[{c_n} = \sum\limits_{k = 0}^{n - 2} {\frac{{{{\left( { - 1} \right)}^k}\left( {2n - 2k - 3} \right)!!}}{{{2^{n - 2}}k!\left( {n - k - 2} \right)!}}} .\]
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{U_n}}}{{\sqrt n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 0}^{n - 2} {\frac{{{{\left( { - 1} \right)}^k}\left( {2n - 2k - 3} \right)!!}}{{{2^{n - 2}}k!\left( {n - k - 2} \right)!}}} }}{{\sqrt n }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 3} \right)!!}}{{\left( {n - 2} \right)!}} \cdot \sum\limits_{k = 0}^{n - 2} {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}} .\]
注意到
\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 3} \right)!!}}{{\left( {n - 2} \right)!}} = \frac{2}{{\sqrt \pi  }},\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^{n - 2} {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}}  = \frac{1}{e}.\]
事实上
\begin{align*}&\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 3} \right)!!}}{{\left( {n - 2} \right)!}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 2} \right)!\left( {2n - 2} \right)!!}}\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 2} \right)!{2^{n - 1}}\left( {n - 1} \right)!}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{2n - 3}}\sqrt n }}\frac{{\left( {2n - 2} \right)!}}{{\left( {n - 2} \right)!\left( {n - 1} \right)!}}\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{2n - 3}}\sqrt n }} \cdot \frac{{{{\left( {2n - 2} \right)}^{2n - 2}}\sqrt {2\pi \left( {2n - 2} \right)} }}{{{e^{2n - 2}}}} \cdot \frac{{{e^{n - 2}}}}{{{{\left( {n - 2} \right)}^{n - 2}}\sqrt {2\pi \left( {n - 2} \right)} }} \cdot \frac{{{e^{n - 1}}}}{{{{\left( {n - 1} \right)}^{n - 1}}\sqrt {2\pi \left( {n - 1} \right)} }}\\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{2n - 3}}}} \cdot \frac{{{{\left( {2n - 2} \right)}^{2n - 2}}}}{{{{\left( {n - 2} \right)}^{n - 2}}{{\left( {n - 1} \right)}^{n - 1}}}} \cdot \frac{{\sqrt {2\pi \left( {2n - 2} \right)} }}{{\sqrt {2\pi \left( {n - 2} \right)} \sqrt {2\pi \left( {n - 1} \right)} \sqrt n }} \cdot \frac{1}{e}\\&= \frac{1}{{e\sqrt \pi  }}\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{2n - 2}}}}{{{2^{2n - 3}}}} \cdot \frac{{{{\left( {n - 1} \right)}^{n - 1}}}}{{{{\left( {n - 2} \right)}^{n - 2}}}} \cdot \frac{1}{{\sqrt {n\left( {n - 2} \right)} }} = \frac{2}{{e\sqrt \pi  }}\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n - 1} \right)}^{n - 2}}}}{{{{\left( {n - 2} \right)}^{n - 2}}}} \cdot \frac{{\left( {n - 1} \right)}}{{\sqrt {n\left( {n - 2} \right)} }}\\&= \frac{2}{{e\sqrt \pi  }}\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{{n - 2}}} \right)^{n - 2}} = \frac{2}{{\sqrt \pi  }}.\end{align*}
因此
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{U_n}}}{{\sqrt n }} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{2^{n - 2}}\sqrt n }}\frac{{\left( {2n - 3} \right)!!}}{{\left( {n - 2} \right)!}} \cdot \sum\limits_{k = 0}^{n - 2} {\frac{{{{\left( { - 1} \right)}^k}}}{{k!}}}  = \frac{2}{{e\sqrt \pi  }}.\]

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