Show fixed point iteration does not converge to a given point

Question

I was given this question. No idea how to show this. Can this be done by substituting initial guesses or in more theoretical way ?

Consider the fixed point iteration $x_{n+1} = g(x_n)$ where $g(x) = \tan^{−1} (2x)$. Clearly, $x = 0$ is a fixed point of $g(x)$. Show that fixed point iteration will not converge to this fixed point

Answer 1

hint

$$g (x)=\arctan (2x) $$ $$g'(x)=\frac {2}{1+4x^2} $$

check that near $0$,

$$|g'(x)|>1$$

To use Newton's method, replace $g (x) $ by $$G (x)=x-\frac {x-g (x)}{1-g'(x)} $$

Answer 2

The fixed point iteration scheme will not converge to the fixed point,as

$g(x) = \tan^{-1}(2x) \implies |g'(x)| = \frac{2}{1 + 4x^2} > 1$,when $x \in (-0.5,0.5)$ .

the fixed point iteration scheme is convergent when $|g'(x)| < 1$.

For reference you can refere this note - here which has details through computational examples and also includes Newton's method!

Answer 3

Recall that one of the necessary conditions under Banach Fixed-Point theorem for the fixed point iteration to converge is that $f$ must be a contraction on a closed interval containing zero. Also recall that $f$ is a contraction on $[a,b]$ if $|f'(x)| = L < 1 \ \forall x \in [a,b]$. But $$f'(x) = \frac{2}{1+4x^2} \implies |f'(0)| = 2 > 1.$$ Thus, there cannot exist a closed interval containing 0 on which $f$ is a contraction. Therefore, by Banach fixed-point theorem, the fixed point iteration does not converge to 0.