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Math Help - Newton's method

  1. #1
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    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.
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    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?
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