Show that there is no $\lambda \in (0,1]$ so that $$ x_{k+1}=x_k-\lambda\frac{f(x_k)}{f'(x_k)}$$
is global convergent.
I know that Newton ($\lambda =1)$ isn't global convergent for $f(x)= arctan(x)$, my idea was to construct a function $f(x,\lambda)$ so that the above method isn't global convergent. But I dont find any.
Greetings.
A simple way of constructing a non-convergent sequence is to try to construct a cycle where the iteration just goes back and forth without converging. Take for example $f(x) = ax^3 + bx + c$ with $a,b,c\not=0$ (simplest polynomial where this can happen) and start with $x_0 = 0$ and compute $x_1$ and $x_2$ and enforce $x_2 = 0$. This will give you an (over-determined) equation system for $a,b,c$ so just pick some values that satisfy it and double-check that $x_1$ is not a root and you have your counter-example. For example if $\lambda = \frac{1}{2}$ then $f(x) = 3x^3 - 5x - 10$ works as $x_0 = 0$ gives $x_1 = -1$ and $x_2 = 0$.