首页 > 代码库 > [再寄小读者之数学篇](2014-05-25 非线性递归数列的敛散性)

[再寄小读者之数学篇](2014-05-25 非线性递归数列的敛散性)

2024-07-06 00:47:30   224人阅读

数列{xn}<script id="MathJax-Element-1" type="math/tex">\begin{Bmatrix} {x}_{n} \end{Bmatrix}</script>满足如下定义:

a>0,b>0;x1=a,x2=b;xn+2=2+1x2n+1+1x2n,n≥1.
<script id="MathJax-Element-2" type="math/tex; mode=display">a>0,\quad b>0; \qquad {x}_{1}=a,\quad{x}_{2}=b ;\qquad {x}_{n+2}=2+\cfrac{1}{{x}_{n+1}^{2}}+\cfrac{1}{{x}_{n}^{2}},\qquad n\geq 1.</script> 讨论该数列 {xn}<script id="MathJax-Element-3" type="math/tex">\begin{Bmatrix} {x}_{n} \end{Bmatrix}</script> 的敛散性.

证明: (来自 magic9901) 设

limn→+∞ˉˉˉˉˉˉˉˉxn=A,limn→+∞ˉˉˉˉˉˉˉˉxn=B
<script id="MathJax-Element-4" type="math/tex; mode=display">\lim\limits_{\overline{n\to+\infty}}x_n=A,\quad\overline{\lim\limits_{n\to+\infty}}x_n=B</script> 于是就有 2<A?B<52<script id="MathJax-Element-5" type="math/tex">2，则由递推关系式:
xn+2=2+1x2n+1+1x2n
<script id="MathJax-Element-6" type="math/tex; mode=display">x_{n+2}=2+\cfrac{1}{x_{n+1}^2}+\cfrac{1}{x_n^2}</script> 就得到:
A==?==limn→+∞ˉˉˉˉˉˉˉˉxn+2limn→+∞ˉˉˉˉˉˉˉˉ(2+1x2n+1+1x2n)2+limn→+∞ˉˉˉˉˉˉˉˉ1x2n+1+limn→+∞ˉˉˉˉˉˉˉˉ1x2n2+1limn→+∞ˉˉˉˉˉˉˉˉx2n+1+1limn→+∞ˉˉˉˉˉˉˉˉx2n2+2B2.
<script id="MathJax-Element-7" type="math/tex; mode=display">\begin{eqnarray*}A&=&\lim\limits_{\overline{n\to+\infty}}x_{n+2}\\ &=&\lim\limits_{\overline{n\to+\infty}}\left(2+\cfrac{1}{x_{n+1}^2}+\cfrac{1}{x_n^2}\right)\\ &\geqslant&2+\lim\limits_{\overline{n\to+\infty}}\cfrac{1}{x_{n+1}^2}+\lim\limits_{\overline{n\to+\infty}}\cfrac{1}{x_n^2}\\ &=&2+\cfrac{1}{\overline{\lim\limits_{n\to+\infty}}x_{n+1}^2}+\cfrac{1}{\overline{\lim\limits_{n\to+\infty}}x_n^2}\\ &=&2+\cfrac{2}{B^2}.\end{eqnarray*}</script> 同理可以得到:
B?2+2A2
<script id="MathJax-Element-8" type="math/tex; mode=display">B\leqslant 2+\cfrac{2}{A^2}</script> 因此就有:
0?B?A?2A2?2B2=2(A+B)A2B2(B?A)
<script id="MathJax-Element-9" type="math/tex; mode=display">0\leqslant B-A\leqslant \cfrac{2}{A^2}-\cfrac{2}{B^2}=\cfrac{2(A+B)}{A^2B^2}(B-A)</script> 注意到:
128625<2(A+B)A2B2<58
<script id="MathJax-Element-10" type="math/tex; mode=display">\cfrac{128}{625}<\cfrac{2(A+B)}{A^2B^2}<\cfrac{5}{8}</script> 则得到: A=B<script id="MathJax-Element-11" type="math/tex">A=B</script>, 所以就有:
limn→+∞xn=A=B=x∈(2,52).
<script id="MathJax-Element-12" type="math/tex; mode=display">\lim\limits_{n\to+\infty}x_n=A=B=x\in\sex{2,\cfrac{5}{2}}.</script> 


联系
我们