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What is the order of convergence of modified Newton's? $$x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)},$$ where $m$ is the multiplicity. I know it is $2$, but can someone explain it briefly?

Ѕᴀᴀᴅ
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7237372728
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Proof of Quadratic Convergence

  • Taylor expension at $x$:

$$f(x_n) = f(x) + f'(x)(x_n-x)+\frac{1}{2}f''(\xi_n)(x_n-x)^2 = f(x)+ \frac{1}{2}f''(\xi_n)(x_n-x)^2,$$where $\xi_n$ between $x$ and $x_k$.

  • Therefore,

$$\begin{split}\lim_{n\to \infty}\frac{|x_{n+1} - x|}{|x_n - x|^2} &= \lim_{n\to \infty} \frac{|f(x_n) - x|}{|x_n-x|^2} = \frac{1}{2}\lim_{k\to\infty} \frac{|f''(x_n)(x_n-x)^2|}{|x_n-x|^2}\\ &= \frac{1}{2}|f''(x)| \end{split}$$

Bo Lan
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    That's only the case $m=1$. – J.G. Apr 10 '20 at 12:49
  • This problem have been solved. See https://math.stackexchange.com/questions/2362853/quadratic-rate-of-convergence-of-modified-nr-method – Bo Lan Apr 10 '20 at 12:55
  • That comment should be on the OP, not your answer. In fact, it should be given as an is-a-duplicate reason for closure. – J.G. Apr 10 '20 at 12:58
  • @J.G. am stuck!! Once I see: If m is the multiplicity then the order is m.And other times I see : the order is 2..Which one is correct? – 7237372728 Apr 10 '20 at 13:13
  • @user754312 $m$ is the multiplicity of the root $f$ to which we seek convergence; $2$ is the order of convergence, i.e. if $a$ is the root $|x_{n+1}-a|\sim A|x_n-a|^2$ for some $A\ne0$. – J.G. Apr 10 '20 at 14:07