recurrence relation: $x_{n+1} = x^2_n - 2x_n + 2$

Question

$$x_0 = \frac32; x_{n+1} = x^2_n - 2x_n + 2$$

$$\Rightarrow x = x^2 - 2x +2 \Rightarrow x^2 - 3x +2 = 0 \Rightarrow x = {1;2} $$

How to determine which one is the limit, i.e. $\lim_{n\rightarrow \infty} x_n = 1$ or $\lim_{n\rightarrow \infty} x_n = 2$ ?

Answer 1

HINT: The function $f(x)=x^2-3x+2$ is negative between $1$ and $2$, so if $1<x_n<2$, then $x_{n+1}-x_n=f(x_n)<0$, and $x_{n+1}<x_n$. This should tell you which of the two roots in the only possible limit if $x_0=\frac32$.

Note that you also have to show that the sequence actually does converge; showing that it’s monotone and bounded would do this. The fact that $x_n^2-2x_n+2=(x_n-1)^2+1$ should help you here.

Answer 2

$x_{n+1}-1 = (x_n - 1)^2$ whence $x_n-1=(x_0-1)^{2^n}$.