Theorem 1
If $g \in [a,b]$ and $g(x) \in [a,b] \forall x \in [a,b]$, then $g$ has a fixed point in $[a,b].$
If in addition, $g'(x)$ exists on $(a,b)$ and a positive constant $k < 1$ exists with
$$|g'(x)| \leq k, \text{ for all } \in (a, b)$$
then the fixed point in $[a,b] is unique.Fixed-point Theorem
Let $g \in C[a,b]$ be such that $g(x) \in [a,b]$, for all $x$ in $[a,b]$. Suppose, in addition, that $g'$ exists on $(a,b)$ and that a constant $0 < k < 1$ exists with
$$|g'(x)| \leq k, \text{ for all } x \in (a, b)$$
Then, for any number $p_0$ in $[a,b]$, the sequence defined by
$$p_n = g(p_{n-1}), n \geq 1$$
converges to the unique fixed-point in $[a,b]$
These are two theorems that I have learned, and I'm having a hard time with this problem:
Given a function $f(x)$, how can we find the interval $[a,b]$ on which fixed-point iteration will converge?
Besides guess and check, I couldn't find any other way to solve this problem. I tried to link the above theorems, but it involves two variables, so I have a feeling it can't be solved algebraically. I wonder is there a general way to find the interval of convergence rather trial and error? Thank you.
Best Answer
The iteration converges for any starting point in an interval stable by $f$ and such that $|f'|\leqslant k$ on this interval, with $k<1$ (this is a mere rephrasing of the theorem). If $f(x)=(x+3/x)/2$ for example, $f'(x)=(1-3/x^2)/2$ hence any closed interval included in $(1,+\infty)$ will do.