数列的单调性与极限毕业论文.doc

数学分析中 一类递推数列的单调性与极限 张荣权（100210111002）的单调性与收敛性问题,同时也推广与包含了近期一些文献中的结果。 关键词： 递推数列；单调性；不动点；收敛 。 The Limits and Monotonicity of a Recursive Sequence in Mathematical Analysis Zhang Rong-quan (Department of Mathematics and Statistics,Anhui Normal University,Wuhui,Anhui ,241000,China) Abstract: The sequence is the focus of the high school algebra is one of the college entrance examination examination of key, in recent ten years in high have a large proportion of the examination. These questions not only examines the sequence, each digit and compares the sequence, the foundation of the sequence limit knowledge, basic skills, basic thoughts and methods, as well as the basic mathematical induction method, and effectively test logical reasoning ability, operation ability, and the use of relevant knowledge and methods, the ability to analyze and solve problems. The same sequence in the higher mathematics is very important in the position, it is the follow-up study a series of related theorem foundation。This article from some recursion sequence of recursive formula, the analysis and study, some recursion sequence monotonicity and convergence problem, also contain some recent promotion and documents of the results。 Key words: recursive sequence ; monotonicity; fixed point; convergence. 1 引言 在近期的一些文献中,讨论了形如 （） 的递推数列的极限问题[1-7],这类数列的极限问题经常出现在研究生入学试题与大学数学考试试题中,在高等数学中占有重要的地位. 研究结果表明,这类递推数列极限的存在性与求法往往与它的迭代函数的不动点相关联,该递推数列的迭代函数为 , 注意到 不变号,它启发我们从迭代函数的不动点与导函数的不变号两方面考虑这类问题. 本文将给出联系迭代函数的不动点与导函数的几个实用命题,把现行文献[1-7]中的相关结论进行拓广,通过这些命题使我们可以统一处理有关例子,揭示这类试题的背景与思想方法. 命题与证明 定义1[1]　对于函数,若数列满足,,,则数列称为递推数列,称为数列的迭代函数,称为初始值. 命题1[1]设函数在上连续,在上可导,且,, .设,则递推数列（）收敛. 证明 只需证明数列单调有界,可用归纳法证之. 1ο当时,由于,,因此 , 又,所以 , 而,故有 , 从而结论成立. 2ο假设当时,结论成立,即 . 当时,由于,则有 , 即得 ,也即, 从而当时,结论成立. 故命题1得证. 命题 2[1]设函数在上连续,在上可导,且,, .设,则递推数列（,）收敛. 证明 类似于命题1,可以证明数列单调递减并且有界,即 ， 从而数列收敛. 定义 2[1]　对于函数,若存在实数,使,则称为的不动点. 注:在命题1,命题2的条件下,若还满足“在上有唯一的不动点”条件,易知数列必收敛于该不动点. 事实上,在满足所给条件的情况下,由数学分析[8]中的确界原理及上确界的定义,对于命题1中的数列,必为其上确界.任给,按上确界的定义,存在数列中的某一项,