1. ## Newton's method

We assume that f(x), f'(x) and f''(x) are continuous in [a,b], and that for some $\alpha \in (a,b)$ , we have $f( \alpha )= 0$ and $f'( \alpha ) \neq 0$ . We show that if $x_{0}$ is chosen close enough to $\alpha$ , the iterates
$x_{n+1} = x_{n}- \frac{f(x_{n})}{f'(x_{n})}$
converge to $\alpha$ .

I tried to use Taylor's expansion for $f( \alpha )$ (centered at $x_{n}$ ), and I got to this expression

$lim_{n \rightarrow \infty} (\alpha -x_{n+1})= lim_{n \rightarrow \infty} - \frac{1}{2} f''(c) \frac{( \alpha - x_{n} )^{2}}{f'(x_{n})}$

where $c \in ( \alpha , x_{n} )$
and I guess I want the right hand side to be 0 to get to the answer. But I am not sure how to prove this.

2. Unfortunately what You are trying [I suppose...] to demonstrate not always is true... for example $f(x)= x^{\frac{1}{3}}$ is continous in $(-\infty,+ \infty)$ and it is clearly $f(0)=0$ but if You set $x_{0} = a$ no matter how $a$ is 'near' to $0$ the Newton's iterations poroduce a divergent sequence... a very unpleasant situation! ...

Kind regards

$\chi$ $\sigma$

3. The hypotheses state that f '(x) and f "(x) must be continuous which is not the case for $x^{1/3}$ in $( - \infty , \infty)$

Would that be a reason why it's not working for this specific function?