Given $c>0$ and the iteration $x_{n+1}=\frac{{x_n}^2+c}{2}$, show that $\lim_{n\to ∞} x_n=a$, if $0\le x_0\le b$.
What is happening with {${x_n}$} if $x_0<0$ or if $x_0 \ge b$ ?
Note: $a$ and $b$ are two fixed points of $f(x)$. I was able to find them, with the requested condition $0<a<1<b$ ( it was a part of a previous exercise).
To prove the limit, I thought at $|x_n-a|=|f(x_{n-1})-f(a)|=|f'(d)||x_{n-1}-a| \le ... \le k^n|x_0-a|$. This would go to $0$ if $0<k<1$, but I don't know that. I know there are some theorems about that limit, but I can't use them since they were not part of my course. So, how can I approach this?
As $f'(x)=x$, the interval where you have contraction is $|x|<1$. With $|c|<1$ one also easily confirms that $f$ maps this interval into itself, thus satisfying the conditions of the Banach fixed-point theorem.
In general, the condition of $f'$ at a root/fixed-point also holds, by continuity, in a small neighborhood, so if the root/fixed-point is attracting, there is some interval around it where the fixed-point iteration is contracting, if the root is repelling,...