On the fixed-point iteration

Question

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?

Answer 1

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.

Answer 2

Too long for a comment.

If you use Newton method for finding the zero of $$f(x)=x^4-\sin \left(\frac{\pi x}{2}\right)$$ starting with $x_0=1.1$, you will get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.1000000000000000000 \\ 1 & 1.0144640865676508848 \\ 2 & 1.0003650875077377200 \\ 3 & 1.0000002408222162678 \\ 4 & 1.0000000000001048802 \\ 5 & 1.0000000000000000000 \end{array} \right)$$ Nice convergence, no overshoot of the solution.

Now, suppose that you use $x_0=0.9$; the problem will be quite different $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.90000000000000000000 \\ 1 & 1.0241776781772581515 \\ 2 & 1.0009966763457768217 \\ 3 & 1.0000017919396538515 \\ 4 & 1.0000000000058069133 \\ 5 & 1.0000000000000000000 \end{array} \right)$$ This is related to Darboux theorem; you will converge smoothly if yous start at a poinf $x_0$ such that $f(x_0)\times f''(x_0) >0$. This was the case for the first example but not for the second.

As Ross Millikan answered, if you look instead for the zero of function $$g(x)=x-\sqrt[4]{\sin \left(\frac{\pi x}{2}\right)}$$ for each of the above cases, you would have $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.1000000000000000000 \\ 1 & 1.0029268046440802235 \\ 2 & 1.0000026372829963012 \\ 3 & 1.0000000000021451740 \\ 4 & 1.0000000000000000000 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.90000000000000000000 \\ 1 & 1.0033137508866509540 \\ 2 & 1.0000033799192375030 \\ 3 & 1.0000000000035233964 \\ 4 & 1.0000000000000000000 \end{array} \right)$$