不动点方法探隅 1

不动点方法探隅 1

若 \(a_{n+1}=\frac{aa_{n}+b}{ca_n+d}\) ，求通项

\[\begin{equation} \begin{aligned} &设& a_{n+1}-\lambda&=\frac{(a-\lambda c)a_{n}+(b-\lambda d)}{ca_n+d}\\ &要满足&a_{n+1}-\lambda&=\frac{(a-\lambda c)(a_n-\lambda)}{ca_n+d}\\ && &=\frac{(a-\lambda c)a_n-\lambda(a-\lambda c)}{ca_n+d}\\ &即& (b-\lambda d)&=-\lambda(a-\lambda c)\\ &即& (b+\lambda a)&=\lambda(\lambda c+d)\\ &&\lambda&=\frac{\lambda a+b}{\lambda c+d}\\ &解得\lambda\ 是\ x_1,x_2，代入. &&\begin{cases} a_{n+1}-x_1&=\frac{(a-cx_1)(a_n-x_1)}{ca_n+d}\\ a_{n+1}-x_2&=\frac{(a-cx_2)(a_n-x_2)}{ca_n+d}\\ \end{cases}\\ &if\ x_1!=x_2& \frac{a_{n+1}-x_1}{a_{n+1}-x_2}&=\frac{\frac{(a-cx_1)(a_n-x_1)}{ca_n+d}}{\frac{(a-cx_2)(a_n-x_2)}{ca_n+d}}\\ &&&=k\frac{a_{n}-x_1}{a_{n}-x_2}\\ &if\ x_1==x_2& \frac{1}{a_{n+1}-\lambda}&=\frac{ca_n+d}{(a-\lambda c)(a_n-\lambda)}\\ && \frac{1}{a_{n+1}-\lambda}&=\frac{ca_n+d-(a-\lambda c)+(a-\lambda c)}{(a-\lambda c)(a_n-\lambda)}\\ && &=\frac{ca_n+d-(a-\lambda c)}{(a-\lambda c)(a_n-\lambda)}+\frac{1}{a_n-\lambda}\\ &&&=\frac{c(a_n-\lambda)+d-a+2\lambda c}{(a-\lambda c)(a_n-\lambda)}+\frac{1}{a_n-\lambda}\\ &考察c\lambda^2+(d-a)\lambda-b=0& \lambda=\frac{a-d}{2c}\\ &so&\frac{1}{a_{n+1}-\lambda}&=\frac{c(a_n-\lambda)}{(a-\lambda c)(a_n-\lambda)}+\frac{1}{a_n-\lambda}\\ &&&=\frac{c}{a-\lambda c}+\frac{1}{a_n-\lambda}\\ &&&=k+\frac{1}{a_n-\lambda} \end{aligned} \end{equation} \]