# Fixed Point Iteration $x^3 - 3 = 0$

I am having trouble solving $$x^3 - 3 = 0$$ using the fixed point iteration method.

It is advised in the problem to put $$g(x)$$ in a form similar to $$g(x) = x + c(x^2 - 5)$$ for $$x^2 - 5 = 0$$ but I am not sure what value for c I should employ in this question.

• In fixed-point iteration method for $g$, the method converges if $|g'(x)|\le 1$. Now using this fact choose your $c$. – Sujit Bhattacharyya Mar 9 at 14:22
• That's what I am unsure of. Approaching it like that I arrive at $\frac{-2}{6.24}\ < c < 0$ this doesn't seem right? – number8 Mar 9 at 15:48
Your goal is to solve the non-linear equation using the functional iteration $$x_{n+1} = g(x_n)$$ for some suitable value of $$x_0$$. To that end you require a function $$g$$ for which $$g(z) = z$$ if and only if $$z^3 = 3$$. Moreover, local convergence is ensured provided $$|g'(z)| < 1$$. Your textbook suggest functions of the type $$g(x) = x + c(x^3-3),$$ for some $$c \not =0$$. It is clear that $$g(z)=z$$ if and only if $$z^3-3 = 0$$, so there is at least a glimmer of hope. In general, the derivative of $$g$$ is $$g'(x) = 1 + 3 c x^2,$$ but we care mainly for the specific value $$g'(z)$$, where $$z = 3^{\frac{1}{3}}$$. The inequalities $$-1 < g'(z) = 1 + c \cdot 3^{\frac{5}{3}} < 1$$ will secure local convergence. Equivalently, $$-2\cdot3^{-\frac{5}{3}} < c <0 .$$ Not all values of $$c$$ are equally good and a bad choice will reduce your choices for $$x_0$$, but this is perhaps a subject for another question.