fixed point iteration algebra problem

Question

I am looking at an example which finds the root of:

$$ f(x) = \cos(3x) \tag 1$$

using the fixed point iteration method. It uses

$$ g(x) = \frac{2x+\cos(3x)}{2} \tag 2$$

However, it was my understanding that to get $g(x)$ I should make (1) equal to zero, and then re-arrange to get $x=$ . The simplest method being to just add $x$ to each side. So,

$$ 0=\cos(3x) \tag 3$$ $$ x=\cos(3x)+x \tag4$$

So how do I get from (4) to (2), or have I misunderstood something.

Thanks.

Answer 1

When we write:

$$x_{n+1} = g(x_n)$$

we need:

$$|g'(r)| \lt 1$$

for convergence, where $r$ is the actual root.

If we find an example root of $\cos 3x = 0$, we have $x = \dfrac{\pi}{6}$ (of course there are many given the periodic function).

If we calculate this for your choice of $g(x) = x + \cos(3x)$, we have:

$$\left|g'\left(\dfrac{\pi}{6}\right)\right| = \left|1 - 3 \sin\left(\dfrac{\pi}{2}\right)\right| = |-2| \implies ~ \mbox{diverges}$$

However, if we use their choice of $g(x) = \dfrac{2x+cos(3x)}{2}$, we have:

$$\left|g'\left(\dfrac{\pi}{6}\right)\right| = \left|1 - \dfrac{3}{2} \sin\left(\dfrac{\pi}{2}\right)\right| = \left|-\dfrac{1}{2}\right| \implies ~ \mbox{converges}$$

Indeed, for a starting point of $x_0 = 1$, this one converges to the root:

$$x = 0.5235987756$$

Of course, this is $\dfrac{\pi}{6}$.

Answer 2

We want the iteration $x_{n+1}=g(x_n)$ to converge, where $g(x)=x+a\cos(3x)$. (When $x$ is near the solution.)

This is guaranteed if $\lvert g'(x)\rvert <1$, near the root, where $g'(x)=1-3a\sin(3x)$.

The idea is to pick $a$, so that $\lvert g'(x)\rvert <1$.

Apparently, at the root $\sin 3x=\pm 1$, and hence $a$ should be picked, so that $$ \lvert 1\pm 3a\rvert <1. $$ If $\sin 3x=1$, then best pick for $a=1/3$, while if $\sin 3x=-1$, then pick $a=-1/3$.