Modified Newton's method - denominator 0

Asked Mar 24 '24 at 06:47

Active Mar 24 '24 at 06:54

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While I find information on Modified Newton's method, I have read this post from mathstackexchange.

In short, a modified Newton's method, $x_{n+1} = x_n - k\frac{f(x_n)}{f'(x_n)} ,$ such that ,in the case of $k>1$ , and $f$ has a root of multiplicity $k$ at $r$ and $h(r) \neq 0$. it can be written in the form $f(x)=(x−r)^kh(x)$ where $h(r)≠0$.

Also, we have $x_{k+1} = g(x_k)$ and $g(x) := x - k f(x)/f'(x)$

While I show that it is quadratically convergent, we encounter a equation like:

$e_{n+1} = x_{n+1} - r = g(x_n) - r \approx g(r) + g'(r) (x_n - r) - r = g'(r) e_n$

By Taylor's expansions.

Here, I am not able to understand that $g'(r) $ has a denominator 0, i.e,

$g'(r) = 1-t(r)/f'(r)^2$ , for some $t(r)$ but we know that $f'(r)=0. $

So is it not valid answer?

edited Mar 24 '24 at 06:54

asked Mar 24 '24 at 06:47

JAEMTO

At a root with higher multiplicity, the curve is tangent to the ordinate axis to a higher order. Which means that the derivatives below that order=multiplicity are zero. – Lutz Lehmann Mar 24 '24 at 07:39
@LutzLehmann Could be more specific please? – JAEMTO Mar 24 '24 at 22:23
1

What is the derivative of $f$ in terms of $h$? – Lutz Lehmann Mar 24 '24 at 22:31
@LutzLehmann Say, $k=2$ then $f' = (x-r)h(x) + (x-r)^2h'(x)$ – JAEMTO Mar 25 '24 at 06:07
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Yes, so the factors $(x-r)$ cancel and the limit is finite. – Lutz Lehmann Mar 25 '24 at 06:58

Modified Newton's method - denominator 0

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