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As you know convergence order of newton raphson method is:

$ E_{t,i+1}=-\frac {f''(x_r)}{2f'(x_r)}E_{t,i}^2 $

What's newton raphson method convergence order if $f'(x_r)=0$.

H.H
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    Well, check $f(x) = x^4$ – user251257 Nov 17 '16 at 18:56
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    The convergence becomes very very slow. Does one speak of convergence order in that case ? I don't know. – Jean Marie Nov 17 '16 at 18:57
  • How can I do that??Is it correct to just check a single function? – H.H Nov 17 '16 at 18:59
  • You get linear convergence with convergence factor $1-\frac1m$, where $m$ is the multiplicity of the root. Practically related is that Newton's method for polynomials, when you start far outside the root region, only sees the leading term and thus converges slowly, linearly with rate $1-\frac1{\deg p}$ until it sees the root cluster closely as single roots. – Lutz Lehmann Nov 17 '16 at 19:09
  • Could you prove that! – H.H Nov 17 '16 at 19:35

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In the hypothesis of the Newton's method there is one about $f'$ calculated on the sequence $\{x_r\}$, i.e. $$ f'(x_r) \neq 0 \quad \forall \, x_r \in \{x_r\} $$

From a geometrical point of view if for an $r$ happen $f'(x_r) = 0$ you find a point such that has got the derivate parallel to $x$ axis, so you can't intersect it.

Instead if your case is $$ f(\alpha) = 0 \quad f'(\alpha) = 0 $$ i.e $\alpha$ is the root (I change the notation because $x_r$ is simular to a sequence) and the derivate calculate in $\alpha$ is zero ($\alpha$ is a root with multiciplity $>1$) see this answer. Dominique's answer explains the convergence argument in general for multiple roots.

Community
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Mauro Vanzetto
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