$ f$ is an homographic function.

Consider the equation $ f(x)=x$ .

  1. If this equation has a double root $ \alpha$ , then define a sequence $ (b_n)$ by $ b_n = a_n - \alpha$ . The sequence $ (b_n)$ is arithmetic.
  2. If this equation has a two distinct real solutions $ \alpha$ and $ \beta$ , then define a sequence $ (b_n)$ by $ b_n = \frac {a_n - \alpha}{a_n - \beta}$ . The sequence $ (b_n)$ is geometric. The choice of the notations has no importance.

Example 3.9.2   Let $ (a_n)$ be the sequence defined by its first term $ a_0=1$ and the recurrence relation $ a_{n+1} = \frac {2}{a_n+1}$ .

Let $ f(x)= \frac {2}{x+1}$ . We have:

$\displaystyle f(x)=x \Longleftrightarrow \frac {2}{x+1} =x \Longleftrightarrow x^2 +x -2=0.$    

This equation has two real solutions: 1 and $ -2$ . Define:

$\displaystyle b_n=\frac {a_n - 1}{a_n +2}$    

We have:

$\displaystyle b_{n+1}$ $\displaystyle =\frac {a_{n+1} - 1}{a_{n+1} + 2} =\frac {\frac {2}{a_n+1} -1}{\frac {2}{a_n+1} +2} =\frac {2 - a_n -1}{2 + 2a_n+2} =\frac {-a_n+1}{2a_n+4}$    
$\displaystyle \quad$ $\displaystyle = \frac {\frac {2b_n+1}{1-b_n} +1}{ 2 \frac {2b_n+1}{1-b_n} + 4} =-\frac 12 b_n.$    

The sequence $ (b_n)$ is geometric with ratio equal to $ -\frac 12$ , therefore it is converhent and has limit equal to 0. As $ a_n=\frac {1+2b_n}{1-b_n}$ , by Thm 7.1 the sequence $ (a_n)$ is convergent and its limit is equal to 1.

Noah Dana-Picard 2007-12-28