Hi,

**problem:**

The iteration defined by $\displaystyle x_{k+1} = \frac{1}{2}(x^2_k+c)$, where $\displaystyle 0<c<1$,

has two fixed points $\displaystyle \xi_1, \xi_2$ where $\displaystyle 0<\xi_1<1<\xi_2$. Show that

$\displaystyle x_{k+1} - \xi_1 = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1), k=0,1,2,\cdots,$

and deduce that $\displaystyle \lim_{k\rightarrow \infty}x_k=\xi_1 \;\textrm{if}\; 0\leq x_0<\xi_2$. How does the iteration behave for other values of $\displaystyle x_0$?

**attempt at solution:**

$\displaystyle x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k+c) - \xi_1

$

and since,

$\displaystyle \frac{1}{2}(\xi^2_1+c)=\xi_1 \Rightarrow c = 2\xi_1-\xi^2_1$

we have that,

$\displaystyle x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k-\xi^2_1) = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1)$

I am a bit confused regarding the limit-question. If $\displaystyle 0\leq x_0 < \xi_2$ and $\displaystyle 1<\xi_2$ it means that $\displaystyle x_0$ could be any number larger than 1 but smaller than $\displaystyle \xi_2$. I figure that I could then choose some $\displaystyle x_0$ such that $\displaystyle x_1 > x_0$, in other words we would be moving away from $\displaystyle \xi_1$.

Now if $\displaystyle 0 \leq x_0 < \xi_1$ instead of $\displaystyle 0\leq x_0 <\xi_2$, I could see how $\displaystyle \lim_{x\rightarrow \infty}x_k= \xi_1$.

As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

Thanks.