Let [tex]x_{m}[/tex] and [tex]x_{m+1}[/tex] be two successive iterates when Newton's method is applied to a polynomial [tex]p(z)[/tex] of degree n. We prove that there is a zero of [tex]p(z)[/tex] in the disk(adsbygoogle = window.adsbygoogle || []).push({});

[tex] {z \in \textbf{C}: |z - x_{m}| \leq n|x_{m+1} - x_{m}| } [/tex].

I suppose we may use [tex]p'(z)/ p(z) = \sum ^{n}_{j=1} 1/ \left( z-r_{j} \right) [/tex] where [tex]r_{1}, r_{2}, \cdots , r_{n}[/tex] are roots of p.

We need to show there is a root [tex]\alpha[/tex] of [tex]p(z)[/tex] for which [tex] |\alpha - x_{m}| \leq n|x_{m+1} - x_{m}|} [/tex] i.e. for which [tex] \frac{n}{|\alpha - x_{m}|} \geq \left| \frac{p'(x_{m})}{p(x_{m})} \right| = \sum ^{n}_{j=1} 1/ |x_{m}-r_{j} |[/tex]

But I am not sure how to proceed from there.

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# Homework Help: Question about Newton's Method

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