Results 1 to 3 of 3

Thread: Newton's method

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    98

    Newton's method

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

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

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

    where $\displaystyle c \in ( \alpha , x_{n} )$
    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Unfortunately what You are trying [I suppose...] to demonstrate not always is true... for example $\displaystyle f(x)= x^{\frac{1}{3}}$ is continous in $\displaystyle (-\infty,+ \infty)$ and it is clearly $\displaystyle f(0)=0$ but if You set $\displaystyle x_{0} = a$ no matter how $\displaystyle a$ is 'near' to $\displaystyle 0$ the Newton's iterations poroduce a divergent sequence... a very unpleasant situation! ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2009
    Posts
    98
    The hypotheses state that f '(x) and f "(x) must be continuous which is not the case for $\displaystyle x^{1/3}$ in $\displaystyle ( - \infty , \infty)$

    Would that be a reason why it's not working for this specific function?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Newton's Method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 6th 2009, 03:10 PM
  2. Newton's method
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Oct 17th 2009, 12:46 PM
  3. Newton's Method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 15th 2009, 08:03 PM
  4. Newton's Method help!!
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Oct 29th 2008, 05:27 PM
  5. Newton's Method
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Dec 12th 2007, 08:53 AM

Search Tags


/mathhelpforum @mathhelpforum