# 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.

A kickstart would be helpful.

• 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

## 1 Answer

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.

Scanning your comment, it is clear that you arrived at the correct result, but doubted yourself. Doubt can be reduced by experiments and experiments will often reveal mistakes which saves you the pain of trying to prove a fallacy. Confidence is established by writing the analysis out in full as I have done above.