fixed-point iteration find convergence condition

Question

We want find root $\alpha$ of the equation be method fixed-point iteration. The function is $f(x)=x^3-2$ the root $\alpha$ so that $f(\alpha)=0$
The interation is: $x_{k+1}=g_w(x_{k})$
$g_w(x)=(1-w)x^3+(1-w/4)x+3(w-1)+\frac{3w}{4x^2}$
I need find $w$ that this interation converge to a root $\alpha$.
So first the root is $\alpha=2^{1/3}$.
Now i substitute in the equation the root:
$g_w(2^{1/3})=2(1-w)+(1-w/4)2^{1/3}+3(w-1)+\frac{3w}{4(2)^{2/3}} {\overset{!}{=}2^{1/3}} $
Is it enought to find $w$ from this last equation?
I get: $w=0.86$
If i try solve it in another way, so I make:
$g_w'(x)=3(1-w)x^2+(1-w/4)+-2\frac{3w}{4x^3}$
$g_w'(2^{1/3})=3(1-w)2^{2/3}+(1-w/4)+-2\frac{3w}{4 \bullet2}=4.76(1-w)+1-0.25w-0.75w=5.76-5.76w$
$|g_w'(2^{1/3})|{\overset{!}{<}}1$
$|5.76-5.76w|{\overset{!}{<}}1$
$-1<5.76-5.76w<1$
$1.17>w>0.82$
Is it right?

Answer 1

A sensible fixed-point iteration to find solutions to $f(x)=0$ should have the form $$ g(x)=x-h(x)f(x). $$ Transforming the given iteration to this form gives $$ g_w(x)=x-\left((w-1)+\frac{w}{4x^2}\right)(x^3-3). $$ This means that this iteration was constructed for $f(x)=x^3-3$, not for the cube root of $2$.

For the convergence, at least locally, one can now explore conditions for $|g_w'(3^{1/3})|<1$, the smaller the better. Alternatively, check when the interval $[1,2]$ is mapped into itself by $g_w$.