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
Best Answer
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 $$G (x)=x-\frac {x-g (x)}{1-g'(x)} $$