Suppose we are trying to solve $x^4 – \sin( \pi x /2) = 0$ by fixed point iteration $x_{n+1} = g(x_n)$ for suitable $g(x)$ a starting guess $x_0 = 1.1$ Will it succed in finding the root $x=1$ if $g(x) = \dfrac{ \sin (\pi x / 2 ) }{x^3}$?
Well, one way to tell is if the root $r \in [a,b]$ and if $|g(x)| \leq 1$ for all $x \in [a,b]$, then fixed point iteration converges for all starting point $x \in [a,b]$.
So, my idea is to to choose an interval containing the root x=1 and the starting point $x_0 =1.1$. For example, we can take $[1,2]$ and we know
$$ g'(x) = \frac{ \pi x \cos( \pi x /2) – 6 \sin (\pi x /2) }{x^4}$$ we have $|g(x)| \leq |\pi x \cos( \pi x /2) – 6 \sin (\pi x /2) | $ since $\frac{1}{x^4} \leq 1$ on $[1,2]$. Now, Im having some difficulties in bounded the last expression so as to obtain the bound of $1$. Any suggestions?
Best Answer
If we just evaluate the derivative (supplying the missing factor $2$ in the denominator) at $x=1$ we get $g'(x)|_{x=1}=-3$
You can do fixed point iteration but need to find a different formulation. As $\sin \left(\frac {\pi x}2 \right)$ is flat at $x=1$ my thought would be $$x=\left(\sin \left(\frac {\pi x}2 \right)\right)^{1/4}$$
Starting with $1.1$ two iterations got to $0.999997$ and two more got to $1$ within the accuracy of my spreadsheet.