I'm solving some numerical analysis exercises and I still can not find a way to solve the following
Consider the function $f (x) = x^3 - x - 1$. For the equation $f (x) = 0$ answer the following:
.
If we rearrange $f(x)=x(x^2-1)-1$, we arrive at the iteration:
$x_{n+1} = \frac{1}{x_n^2-1}$
When reorganizing $f (x) = (x-1)(x^2+x+1)-1$ we obtain:
$x_{n+1} = 1 + \frac{x}{x^2+x+1}$
demonstrate the equivalence of these two expressions.
How do these iterations behave? (Take start point $x_0 = 1.5$ and calculate some $x_n$ values).
Analyze the convergence or divergence of the previous interactions near zero $x$*.
How should I do it to get to the right way?
Regarding 1, taking into account that the first expression is not defined for $x=1$, the expressions are only equivalent for $x \ne 1$.
Regarding 3, just recall that the fixed point theorem establishes local convergence when $|g'(x^*)|<1$, and almost sure divergence when $|g'(x^*)|>1$. In the first case, $$ |g'(x^*)| = \left|\dfrac{-2x^*}{((x^*)^2-1)^2}\right|>1 $$ and so you conclude that the fixed point iterations will diverge unless, by pure luck, one of the iterations hit $x^*$.
In the second case, this iteration function is continuous, invariant and contractive in $[1,2]$ and therefore the iterations will converge to $x^*$ for any $x_0\in [1,2]$. Moreover, considering that $|g'|\leq 3/49$, we can say that $$ |x_n-x^*| \leq (3/49)^n |x_0-x^*| \leq (3/49)^n. $$