The iteration $x_{k + 1} = \frac{1}{2}(x_{k}^{2} + c)$ where $0 < c < 1$ has two fixed points $\zeta_{1}$, $\zeta_{2}$, where $0 < \zeta_{1} < 1 < \zeta_{2}$. Show that
$$x_{k + 1} - \zeta_{1} = \frac{1}{2}(x_{k} + \zeta_{1})(x_{k} - \zeta_{1}) $$
and show $\lim_{k\to\infty} x_{k} = \zeta_1$ whenever $0 \leq x_{0} < \zeta_{2}$
I've done the first part, but I can't prove the limit. I have no idea how to show the limit. To do the first part, I just subtracted the iteration equation with the fixed point equation and used difference of squares.
Does anyone have any ideas?
Let $f(x)=(x^2+c)/2$. Then $x<f(x)<\zeta_1$ if $x\in[0,\zeta_1)$ and $\zeta_1<f(x)<x$ if $x\in(\zeta_1,\zeta_2)$.
If $x_0=\zeta_1$, the $x_n=\zeta_1$ for all $n$.
Take now $x_0\in[0,\zeta_1)$. Then, using the first of the above inequalities, it is easy to see that the sequence of iterates $\{x_n\}$ is increasing and bounded by $\zeta_1$. Therefore it converges, and the limit is precisely \zeta_1$.
Similarly, if $x\in(\zeta_1,\zeta_2)$, then $\{x_n\}$ is decreasing, bounded below by $\zeta_1$ and converges to $\zeta_1$.