For some fixed point iterations, we get the third order convergence given by
the following error relation:
$|\alpha − x_{n+1}| = \frac14(|\alpha − x_n|)^3$
For what initial values of $x_0$ will these iterations converge? Show this range of possible values of $x_0$ in terms of interval around $\alpha$.
My initial thought is that $|\alpha − x_0|$ should be less than $1$ to make the iteration converge. Therefore, $-1+\alpha<x_0<1+\alpha$
Is my thought is correct solution to this problem?
$|\alpha - x_{n+1}|=\frac14(|\alpha-x_{n}|)^3$
$\frac12|\alpha - x_{n+1}|=\frac18(|\alpha-x_{n}|)^3$
$\frac12|\alpha - x_{n+1}|=(\frac12|\alpha-x_{n}|)^3$ In order to have the iteration converge, $\frac12|\alpha-x_{n}|<1$
Therefore,
$\frac12|\alpha-x_{0}|<1$
$|\alpha-x_{0}|<2$
$\alpha+2>x_{0}>\alpha-2$