跳转至

数列不动点浅探

数列不动点可以辅助我们快速解决与数列通项公式有关的问题。

基础知识

定义

  1. 已知函数 \(f(x)\),若存在 \(x_0\) 使得 \(f(x_0) = x_0\),则称 \(x_0\)\(f(x)\) 的一个不动点。

  2. 若数列 \(\{a_n\}\) 满足 \(a_{n + 1} = g(a_n) \quad(n \in \mathbb{N}^{*})\),则称 \(g(x)\) 为数列 \(\{a_n\}\) 的递推函数。

定理一

\(f(x) = ax + b\quad(a \notin \{0, 1\})\) 为数列 \(\{a_n\}\) 的递推函数,且 \(p\)\(f(x)\) 的不动点,则数列 \(\{a_n - p\}\) 为公比为 \(a\) 的等比数列。

证明

\(\because\) \(p\)\(f(x)\) 的不动点

\(\therefore p = f(p) = ap + b\),即 \(b - p = -ap\)

\(a_{n + 1} = f(a_n) = a\cdot a_n + b\)

\(\begin{aligned}\therefore a_{n + 1} - p & = a\cdot a_n + b - p \\ & = a\cdot a_n - ap \\ & = a(a_n - p)\end{aligned}\)

\(\therefore\) \(\{a_n - p\}\) 为公比为 \(a\) 的等比数列

定理二

  1. \(f(x) = \dfrac{ax + b}{cx + d}\quad(c \neq 0, ad - bc \neq 0)\) 为数列 \(\{a_n\}\) 的递推函数,且 \(p, q\)\(f(x)\) 的两 相异 不动点,则数列 \(\{\dfrac{a_n - p}{a_n - q}\}\) 为公比为 \(\dfrac{a - pc}{a - qc}\) 的等比数列。

    证明

    \(x = f(x) = \dfrac{ax + b}{cx + d}\)\(cx^2 + (d - a)x - b = 0\)

    \(x_1 = p, x_2 = q\) 代入得 \(\begin{cases}cp^2 + (d - a)p - b = 0 \\ cq^2 + (d - a)q - b = 0\end{cases}\)

    整理得 \(\begin{cases}p = \dfrac{pd - b}{a - pc} \\ q = \dfrac{qd - b}{a - qc}\end{cases}\)

    \(\begin{aligned}\therefore \dfrac{a_{n + 1} - p}{a_{n + 1} - q} & = \dfrac{\dfrac{a\cdot a_n + b}{c\cdot a_n + d} - p}{\dfrac{a\cdot a_n + b}{c\cdot a_n + d} - q} \\ & = \dfrac{(a - pc)a_n + b - pd}{(a - qc)a_n + b - qd} \\ & = \dfrac{a - pc}{a - qc} \cdot \dfrac{a_n - \dfrac{pd - b}{a - pc}}{a_n - \dfrac{qd - b}{a - qc}} \\ & = \dfrac{a_n - p}{a_n - q}\cdot\dfrac{a - pc}{a - qc}\end{aligned}\)

    \(\therefore\) \(\{\dfrac{a_n - p}{a_n - q}\}\) 为公比为 \(\dfrac{a - pc}{a - qc}\) 的等比数列

  2. \(f(x) = \dfrac{ax + b}{cx + d}\quad(c \neq 0, ad - bc \neq 0)\) 为数列 \(\{a_n\}\) 的递推函数,且 \(p\)\(f(x)\)唯一 不动点,则数列 \(\{\dfrac{1}{a_n - p}\}\) 为公差为 \(\dfrac{2c}{a + d}\) 的等差数列。

    证明

    \(x = f(x) = \dfrac{ax + b}{cx + d}\)\(cx^2 + (d - a)x - b = 0\)

    \(\therefore cp^2 + (d - a)p - b = 0\),即 \(b - pd = cp^2 - ap\)

    \(\begin{aligned}\therefore \dfrac{1}{a_{n + 1} - p} & = \dfrac{1}{\dfrac{a\cdot a_n + b}{c\cdot a_n + d} - p} \\ & = \dfrac{c\cdot a_n + d}{(a - cp)\cdot a_n + b - pd} \\ & = \dfrac{c\cdot a_n + d}{(a - cp)\cdot a_n + cp^2 - ap} \\ & = \dfrac{c\cdot a_n + d}{(a - cp)(a_n - p)} \\ & = \dfrac{1}{a - cp}\cdot\dfrac{c(a_n - p) + cp + d}{a_n - p} \\ & = \dfrac{c}{a - cp} + \dfrac{cp + d}{a - cp}\cdot\dfrac{1}{a_n - p} \\ & = \dfrac{1}{a_n - p} + \dfrac{2c}{a + d}\end{aligned}\)

    \(\therefore\) \(\{\dfrac{1}{a_n - p}\}\) 为公差为 \(\dfrac{2c}{a + d}\) 的等差数列

定理三

\(f(x) = \dfrac{ax^2 + bx + c}{ex + f}\) 为数列 \(\{a_n\}\) 的递推函数,且 \(p, q\)\(f(x)\) 的两 相异 不动点,则 当且仅当 \(b = 0, e = 2a\) 时有 \(\dfrac{a_{n + 1} - p}{a_{n + 1} - q} = (\dfrac{a_n - p}{a_n - q})^2\)

证明

\(\because\) \(p, q\)\(f(x)\) 的两相异不动点

\(\therefore \begin{cases}c - pf = (e - a)p^2 - bp \\ c - qf = (e - a)q^2 - bq\end{cases}\)

\(\begin{aligned}\therefore \dfrac{a_{n + 1} - p}{a_{n + 1} - q} & = \dfrac{a\cdot a_n^2 + b\cdot a_n + c - p(e\cdot a_n + f)}{a\cdot a_n^2 + b\cdot a_n + c - q(e\cdot a_n + f)} \\ & = \dfrac{a\cdot a_n^2 + (b - ep)\cdot a_n + c - pf}{a\cdot a_n^2 + (b - eq)\cdot a_n + c - qf} \\ & = \dfrac{a\cdot a_n^2 + (b - ep)\cdot a_n + (e - a)\cdot p^2 - bp}{a\cdot a_n^2 - (b - eq)\cdot a_n + (e - a)\cdot q^2 - bq}\end{aligned}\)

\(\begin{aligned}\therefore \dfrac{a_{n + 1} - p}{a_{n + 1} - q} = (\dfrac{a_n - p}{a_n - q})^2 & \iff \dfrac{a\cdot a_n^2 + (b - ep)\cdot a_n + (e - a)\cdot p^2 - bp}{a\cdot a_n^2 - (b - eq)\cdot a_n + (e - a)\cdot q^2 - bq} = \dfrac{a_n^2 - 2p\cdot a_n + p^2}{a_n^2 - 2q\cdot a_n + q^2} \\ & \iff \begin{cases}b - ep = -2ap \\ b - eq = -2aq\end{cases} \\ & \iff \begin{cases}(2a - e)p + b = 0 \\ (2a - e)q + b = 0\end{cases} \\ & \iff \begin{cases}b = 0 \\ e = 2a\end{cases} \quad(p \neq q)\end{aligned}\)

例题讲解

例1

设数列 \(\{a_n\}\) 满足 \(a_1 = 1, a_{n + 1} = \dfrac{a_n + 2}{a_n}\),求数列 \(\{a_n\}\) 的通项公式。

解析

特征方程为 \(x = \dfrac{x + 2}{x}\),解得 \(x_1 = -1, x_2 = 2\)

\(\therefore\) \(\{\dfrac{a_n + 1}{a_n - 2}\}\) 为首项为 \(-2\),公比为 \(-2\) 的等比数列

\(\therefore \dfrac{a_n +1}{a_n - 2} = (-2)^n\)

\(\therefore a_n = \dfrac{2\cdot (-2)^n +1}{(-2)^n - 1}\)

例2

数列 \(\{a_n\}\) 满足下列关系:\(a_1 = 2a, a_{n + 1} = 2a - \dfrac{a^2}{a_n}\quad(a \neq 0)\),求数列 \(\{a_n\}\) 的通项公式。

解析

特征方程为 \(x = \dfrac{2ax - a^2}{x}\),解得 \(x_1 = x_2 = a\)

\(\therefore\) \(\{\dfrac{1}{a_n - a}\}\) 为首项为 \(\dfrac{1}{a}\),公比为 \(\dfrac{1}{a}\) 的等比数列

\(\therefore \dfrac{1}{a_n - a} = \dfrac{n}{a}\)

\(\therefore a_n = \dfrac{a + na}{n}\)

例3

已知数列 \(\{a_n\}\) 满足 \(a_1 = 2, a_{n + 1} = \dfrac{a_n^2 + 2}{2a_n}\),求数列 \(\{a_n\}\) 的通项公式。

解析

特征方程为 \(x = \dfrac{x^2 + 2}{2x}\),解得 \(x_1 = \sqrt{2}, x_2 = -\sqrt{2}\)

\(\therefore \dfrac{a_{n + 1} - \sqrt{2}}{a_{n + 1} + \sqrt{2}} = (\dfrac{a_n - \sqrt{2}}{a_n + \sqrt{2}})^2\)

\(\therefore \dfrac{a_n - \sqrt{2}}{a_n + \sqrt{2}} = (3 - 2\sqrt{2})^{2^{n - 1}} = (\sqrt{2} - 1)^{2^n}\)

\(\therefore a_n = \sqrt{2}\cdot\dfrac{1 + (\sqrt{2} - 1)^{2^n}}{1 - (\sqrt{2} - 1)^{2^n}}\)

总结

无技巧,就是套公式。