Cheuksing

2020-07-21

文章字数: 2.1k

导读

定义1

对函数\(f(x)\),若存在实数\(x_0\),满足\(f(x_0)=x_0\),则称\(x_0\)为\(f(x)\)的不动点。

对此定义有两方面的理解∶

(1)代数意义:若方程\(f(x)=x\)有实数根\(x_0\),则\(y=f(x)\)有不动点\(x_0\).

(2)几何意义:若函数\(y=f(x)\)与\(y=x\)有交点\((x_0,y_0)\), 则\(x_0\)为\(y=f(x)\)的不动点.

利用递推数列\(f(n)\)的不动点,可以将某些由递推关系\(a_n=f(a_n-1)\)所确定的数列转化为较易求通项的数列(如等差数列或等比数列)，这种方法称为不动点法.下面举例说明两种常见的递推数列如何用不动点法求其通项公式.

定义2

若数列{\(a_n\)}满足\(a_n=f(a_n-1)\),则称\(f(x)\)为数列\({a_n}\)的特征函数.

定义3

方程\(f(x)=x\)称为函数\(f(x)\)的不动点方程(特征方程),其根称为函数\(f(x)\)的不动点.

定理1

设\(f(x)=ax+b\)\((a≠0且a≠1)\),\(\left\{a_n\right\}\)满足递推关系\(a_n=f(a_n-1)(n≥2)\),\(p\)为\(f(x)\)的不动点,则\(a_n-p=a(a_n-1-p)\).具体证明步骤参见例题解法.

已知数列\({a_n}\)中\(a_{1}=1,a_{n+1}=-\dfrac{1}{3}a_{n}+\dfrac{2}{3}\)，求\(a_{n}\).
解析
特征方程为\(x=-\dfrac{1}{3}x+\dfrac{2}{3}\)，解得不动点\(x=\dfrac{1}{2}\)，
由\(a_{n+1}=-\dfrac{1}{3}a_{n}+\dfrac{2}{3}\)，得
\(a_{n+1}-\dfrac{1}{2}=-\dfrac{1}{3}a_{n}+\dfrac{1}{6}=-\dfrac{1}{3}(a_{n}-\dfrac{1}{2}),\)
所以\(\left\{a_{n}-\dfrac{1}{2}\right\}\)是以\(\dfrac{1}{2}\)为首项，\(-\dfrac{1}{3}\)为公比的等比数列，
则有\(a_{n}-\dfrac{1}{2}=\dfrac{1}{2}\times(-\dfrac{1}{3})^{n-1},\)
即\(a_{n}=\dfrac{1}{2}+\dfrac{1}{2}\times(-\dfrac{1}{3})^{n-1}.\)

定理2

数列\({a_{n}}\)满足\(a_{n+1}=\dfrac{aa_{n}+b}{ca_{n}+d}(c\neq0,ad-bc\neq0)\)，特征函数为\(f(x)=\dfrac{ax+b}{cx+d}\)，且首项\(a_{1}\neq f(a_{1}).\)

(1)若\(f(x)\)有两个相异不动点\(p,q\)，则\(\dfrac{a_{n+1}-p}{a_{n+1}-q}=\) \(\dfrac{a-cp}{a-cq}\cdot\dfrac{a_{n}-p}{a_{n}-q}\);

(2)若\(f(x)\)只有一个不动点\(p\)，且\(p\neq -d\)，则\(\dfrac{1}{a_{n+1}-p}=\) \(\dfrac{2c}{a+d}+\dfrac{1}{a_{n}-p}.\)

具体证明步骤参见例题解法。

已知数列{\(a_{n}\)}中，\(a_{1}=3,a_{n}=\dfrac{4a_{n-1}-2}{a_{n-1}+1}\)，求\({a_{n}}\)的通项.
解析
因为{\(a_{n}\)}的特征函数为.\(f(x)=\dfrac{4x-2}{x+1},\)
则特征方程为\(\dfrac{4x-2}{x+1}=x\)，即\(x^{2}-3x+2=0,\)
解得\(x_{1}=1,x_{2}=2,\)，
则\(a_{n}-1=\dfrac{4a_{n-1}-2}{a_{n-1}+1}-1=\dfrac{3(a_{n-1}-1)}{a_{n-1}+1} \tag{1}\)
\(a_{n}-2=\dfrac{4a_{n-1}-2}{a_{n-1}+1}-2=\dfrac{2(a_{n-1}-2)}{a_{n-1}+1}\tag{2}\)
则\((1) \div (2)\)得：
\(\dfrac{a_{n}-1}{a_{n}-2}=\dfrac{3}{2}\cdot\dfrac{a_{n-1}-1}{a_{n-1}-2},\)
∴数列\(\left\{\dfrac{a_{n}-1}{a_{n}-2}\right\}\)是公比为\(\dfrac{3}{2}\)的等比数列，
∴\(\dfrac{a_{n}-1}{a_{n}-2}=\dfrac{a_{1}-1}{a_{1}-2}\cdot(\dfrac{3}{2})^{n-1}.\)
∵\(a_{1}=3\)，∴\(\dfrac{a_{n}-1}{a_{n}-2}=2\cdot(\dfrac{3}{2})^{n-1}\)，
即\(a_{n}=\dfrac{2^{n-2}-2\cdot3^{n-1}}{2^{n-2}-3^{n-1}}.\)
总结
这是典型的用不动点法求数列通项公式的题目，其一般解题方法为:
(1)由特征方程求出不动点\(x_{1}\)，\(x_{2}\);
(2)列出数列\(b_{n}=a_{n}-x_{1},c_{n}=a_{n}-x_{2}\)
(3)由\(\dfrac{b_{n}}{c_{n}}\)得出关系，从而解出\(a_{n}\).
在数列\(\left\{a_{n}\right\}\)中，\(a_{1}=2\)，且\(a_{n+1}=\dfrac{2a_{n}-4}{a_{n}+6}\)，求其通项公式\(a_{n}\).
解析
特征方程为\(x=\dfrac{2x-4}{x+6}\)，解得\(x_{1}=x_{2}=-2.\)
令\(a_{n}-(-2)=b_{n}\)，代入原递推式得\(b_{n+1}=\dfrac{4b_{n}}{b_{n}+4}.\)
因为\(b_{1}=a_{1}+2=4\neq0\)，所以\(b_{k}\neq0(k\in N_{+}),\)
故\(\dfrac{1}{b_{n+1}}=\dfrac{1}{b_{n}}+\dfrac{1}{4}.\)
因此，\(\dfrac{1}{b_{n}}=\dfrac{1}{b_{1}}+(n-1)\cdot\dfrac{1}{4}=\dfrac{n}{4}\)，
从而\(b_{n}=\dfrac{4}{n}\)
又因为\(a_{n}=b_{n}-2\)，
所以\(a_{n}=\dfrac{4}{n}-2=\dfrac{4-2n}{n}\)
(2009江西理22)各项均为正数的数列{\(a_{n}\)},\(a_{1}=a,a_{2}=b\)，且对满足\(m+n=p+q\)的正整数\(m，n,p,q\)都有\(\dfrac{a_{m}+a_{n}}{(1+a_{m})(1+a_{n})}=\dfrac{a_{p}+a_{q}}{(1+a_{p})(1+a_{q})}.\)
(1)当\(a=\dfrac{1}{2},b=\dfrac{4}{5}\)时，求通项\(a_{n}\);
(2)证明:对任意a，存在与a有关的常数λ，使得对于每个正整数n，都有\(\dfrac{1}{\lambda}\leq a_{n}\leq\lambda.\).
解析
(1)由\(\dfrac{a_{m}+a_{n}}{(1+a_{m})(1+a_{n})}=\dfrac{a_{p}+a_{q}}{(1+a_{p})(1+a_{q})}\)
得\(\dfrac{a_{1}+a_{n}}{(1+a_{1})(1+a_{n})}=\dfrac{a_{2}+a_{n-1}}{(1+a_{2})(1+a_{n-1})}.\)
将\(a_{1}=\dfrac{1}{2},a_{2}=\dfrac{4}{5}\)代入化简得\(a_{n}=\dfrac{2a_{n-1}+1}{a_{n-1}+2}.\)
所以\(\dfrac{1-a_{n}}{1+a_{n}}=\dfrac{1}{3}\cdot\dfrac{1-a_{n-1}}{1+a_{n-1}}\)，
故数列{\(\dfrac{1-a_{n}}{1+a_{n}}\)}为等比数列，
从而\(\dfrac{1-a_{n}}{1+a_{n}}=\dfrac{1}{3^{n}}\)，
即\(a_{n}=\dfrac{3^{n}-1}{3^{n}+1}.\)
(2)略.

定理3

设函数\(f(x)=\dfrac{ax^{2}+bx+c}{ex+f}(a\neq 0,e\neq0)\)有两个不同的不动点\(p,q\)，且由\(a_{n}=f(a_{n-1})\)确定数列{\(a_{n}\)},那么当且仅当\(b=0,e=2a\)时，\(\dfrac{a_{n}-p}{a_{n}-q}=(\dfrac{a_{n-1}-p}{a_{n-1}-q})^{2}\).此时\(f(x)=\)\(\dfrac{ax^{2}+c}{2ax+f}(a\neq 0)\).

具体证明步骤参见例题解法.

已知数列{\(a_{n}\)}满足\(a_{n}=\dfrac{3a_{n-1}^{2}+2}{6a_{n-1}-5}(n\geq2)\)，首项\(a_{1}=\dfrac{19}{6}\)，求其通项公式\(a_{n}\).
解析
特征方程为\(x=\dfrac{3x^{2}+2}{6x-5}\)，
得\(3x^{2}-5x-2=0,\)则\((3x+1)(x-2)=0\)，
故\(x_{1}=-\dfrac{1}{3},x_{2}=2\)是函数\(f(x)=\dfrac{3x^{2}+2}{6x-5}\)的两个不动点.
则\(a_{n}-(-\dfrac{1}{3})=a_{n}+\dfrac{1}{3}=\dfrac{3a_{n-1}^{2}+2}{6a_{n-1}-5}-(-\dfrac{1}{3})=\)
\(\dfrac{3(a_{n-1}+\dfrac{1}{3})^{2}}{6a_{n-1}-5}\tag{1}\)
\(a_{n}-2=\dfrac{3a_{n-1}^{2}+2}{6a_{n-1}-5}-2=\dfrac{3(a_{n-1}-2)^{2}}{6a_{n-1}-5}\tag{2}\)
则\((1)\div (2)\)得\(\dfrac{a_{n}+\dfrac{1}{3}}{a_{n}-2}=\left(\dfrac{a_{n-1}+\dfrac{1}{3}}{a_{n-1}-2}\right)^{2}\),
所以由选代法得：
\(\dfrac{a_{n}+\dfrac{1}{3}}{a_{n}-2}=\left(\dfrac{a_{n-1}+\dfrac{1}{3}}{a_{n-1}-2}\right)^{2}=\left (\dfrac{a_{n-2}+\dfrac{1}{3}}{a_{n-2}-2}\right)^{2}=\)\(\cdots=\left(\dfrac{a_{1}+\dfrac{1}{3}}{a_{1}-2}\right)^{2^{n-1}}\)\(=3^{2^{n-1}}\),
则\(a_{n}=\dfrac{6\cdot3^{2^{n-1}}+1}{3\cdot3^{2^{n-1}}-3}.\)
(2007广东理21)已知函数\(f(x)=x^{2}+x-1\)，\(α,β\)是方程\(f(x)=0\)的两个根\((\alpha>\beta\),\(f^{\prime}(x)\)是\(f(x)\)的导数，设 \(a_{1}=1,a_{n+1}=a_{n}-\dfrac{f(a_{n})}{f^{\prime}(a_{n})}(n=1,2,\cdots).\)
(1)求\(α,β\)的值;
(2)证明:对任意的正整数\(n\)，都有\(a_{n}>\alpha\);
(3)记\(b_{n}=\ln\dfrac{a_{n}-\beta}{a_{n}-\alpha}(n=1,2,\cdots))\),求数列{\(b_{n}\)}的前n项和 \(S_{n}\).
解析
(1)∵\(f(x)=x^{2}+x-1\)，α,β是方程\(f(x)=0\)的两个根
\((\alpha>\beta),\)
∴\(\alpha=\dfrac{-1+\sqrt{5}}{2},\beta=\dfrac{-1-\sqrt{5}}{2}.\)
(2)\(f^{\prime}(x)=2x+1\)，
\(a_{n+1}=a_{n}-\dfrac{a_{n}^{2}+a_{n}-1}{2a_{n}+1}=a_{n}-\dfrac{1/2a_{n}(2a_{n}+1)+1/4(2a_{n}+1)-5/4}{2a_{n}+1}\)
\(=\dfrac{1}{4}(2a_{n}+1)+\dfrac{5/4}{2a_{n}+1}-\dfrac{1}{2},\)，
∵\(a_{1}=1\)，∴由基本不等式可知\(a_{2}\geq\dfrac{\sqrt{5}-1}{2}>0\)(当且仅当\(a_{1}=\dfrac{\sqrt{5}-1}{2}\)时取等号)，
∴\(a_{2}>\dfrac{\sqrt{5}-1}{2}>0\)，同样\(a_{3}>\dfrac{\sqrt{5}-1}{2},\cdots,a_{n}>\dfrac{\sqrt{5}-1}{2}=\alpha\)
\((n=1,2,\cdots).\)
(3)\(a_{n+1}-\beta=a_{n}-\beta-\dfrac{(a_{n}-\alpha)(a_{n}-\beta)}{2a_{n}+1}=\dfrac{a_{n}-\beta}{2a_{n}+1}(a_{n}+1+\alpha\)
而\(\alpha+\beta=-1\)，即\(\alpha+1=-\beta a_{n+1}-\beta=\dfrac{(a_{n}-\beta)^{2}}{2a_{n}+1}\)
同理\(a_{n+1}-\alpha=\dfrac{(a_{n}-\alpha)^{2}}{2a_{n}+1}b_{n+1}=2b_{n}\)
又\(b_{1}=\ln\dfrac{1-\beta}{1-\alpha}=\ln\dfrac{3+\sqrt{5}}{3-\sqrt{5}}=2\ln\dfrac{3+\sqrt{5}}{2}\)
\(S_{n}=2(2^{n}-1)\ln\dfrac{3+\sqrt{5}}{2}.\)