Fixed point Iteration method with parameters

Question

We want to approach the number $\alpha =\sqrt[3]{2}$. The function $f(x) = x^{3} - 2$ has $\alpha$ as a root. Now take a function $g$ so that $\alpha$ is a fixed point, $g(\alpha) = \alpha$. Use $ g(x)=\frac{x^{3}-2+kx}{k} $ and find $k$ so we can approach $\alpha$ from Fixed point Iteration Method in less that $10$ steps. In the exercise there is no initial point or approximation so I used mine.

I want to know if there is a method to find the parameter $k$ depending on the exercise. For instance, I wish to know how to find $k$ in this case. I wrote and algorithm and found $k$, not with a method but while trying a couple of numbers.

My python program:

def g(x):
    return (x**3 - 2 +k*x)/k
k=-5
x0=1.5
tol=1.e-7
nmax=100
x=g(x0)
e=abs(x-x0)
n=1
while n< nmax and e > tol:
    x0=x
    x=g(x0)
    e=abs(x-x0)
    n+=1
print ('Steps: %d Approximate solution: %.10f' %(n,x))

I get: Steps: $7$ Approximate solution: $1.2599210492$

Answer 1

Close to the fixed point the linearization is $$x_{n+1}=g(α)+g'(α)(x_n-α)\implies x_{n+1}-α=g'(α)(x_n-α).$$ For fast convergence you want to have $|g'(α)|$ as small as possible (and smaller than $1$ for any convergence at all). Now $$ g'(x)=\frac{3x^2}k+1, $$ so ideally you need $k=-3α^2=-\sqrt[3]{108}$, but any value close to it will do, for instance $k=-5$ (as $5^3=125$).

The interval on which $g$ is contracting is given by $\frac35x^2< 2\implies |x|<\sqrt{\frac{10}3}$, which is true for $|x|\le\frac53$. For smaller contraction factors the interval will be correspondingly smaller, for $q=\frac12$ this gives $\frac56\le x^2\le \frac52$, etc.