1. Fixed point iteration doubt

Hi! I was asked to find an approximation for a root of $\displaystyle f(x)=2sin(x\pi)+x=0$ on the interval [1, 2] using the fixed point iteration method, with p0=1.

So, I got $\displaystyle g(x)=x=(\arcsin(-x/2))/\pi$

However, that way I just get numbers very close to zero (which is a solution for the equation, but it's not on the interval I want).

Anyway, I checked the answer on the back of the book and they used $\displaystyle g(x)=x=(\arcsin(-x/2))/\pi+2$ and got $\displaystyle x \simeq 1.68$

I just don't know from where does that +2 come? Could anyone explain me, please?

2. Re: Fixed point iteration doubt

It's because the inverse sine is multivalued. When you attempt to evaluate $\displaystyle \arcsin (-1/2),$ your calculator will return a negative value whereas what you want is the value that is $\displaystyle 2\pi$ further on (if that makes sense).

The iteration that you need is

$\displaystyle \pi x = \arcsin(-x/2) + 2\pi,$

from which

$\displaystyle x = \frac{1}{\pi} \arcsin(-x/2) + 2.$

3. Re: Fixed point iteration doubt

I get it now. Thanks!