Rearranging to get an iterative function (fixed point)

Question

I don't quite get why things are rearranged the way they are when trying to get an equation to be used in fixed point iteration. For example,

$x^3+2x+5=0$

could be rearranged to give

$\:x=-\frac{5}{x^2+2}\:or\:x=-\sqrt[3]{2x+5}$

But apparently not

$x=-\frac{x^3+5}{2}$.

Why so?

Answer 1

The problem is that you generally want the fixed point iteration mapping to be contractive, at least in a neighborhood of the fixed point. Otherwise you start with a small error and end up with a larger error.

The situation is easier to see in a case where the exact solution of the equation is easily constructed. Look at something like $x^2-x-12=0$, and you're trying to find the root $x=4$. If you use $x=x^2-12$, the problem is that although indeed $4=4^2-12$ (i.e. the desired solution is a fixed point of the mapping), if you have an $x$ close to $4$ instead, $x^2-12$ is generally further away from $4$ than $x$ was. For example $3.9^2-12=3.21$.

The standard way to check this is to compute the absolute value of the derivative of the fixed point function at the fixed point, which in this example is $8$. If it is larger than $1$, then the mapping is not contractive near the fixed point, so the iteration (usually) does not converge.

Answer 2

The key idea is to rewrite $f(x)=0$ in the form $x=\phi(x)$ such that $\vert \phi'(x)\vert <1$ for some $x$ in the vicinity of the root.

The picture above depicts the iteration process $x_{n+1}=\phi(x_n)$ for $n=0,1,2,...$ which guarantees the convergence as $\vert \phi'(x)\vert <1$ as the movement along the cobweb cycle indeed takes you towards the point of intersection of the curves.

You can easily see by drawing the graph that the iteration may diverge (this time the cobweb cycle will take you away from the point of intersection of the curves) rather if we relax the condition $\vert \phi'(x)\vert <1$.