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How to prove sequence converges quadratically to a root of multiplicity

ianchenmu

Member
Feb 3, 2013
74
A function f has a root of multiplicity $m>1$ at the point $ x_*$ if $f(x_*)=f'(x_*)=...=f^{(m-1)}(x_*)=0$. Assume that the iteration$ x_{k+1}=x_k-mf(x_k)/f'(x_k)$ converges to $x_*$. If$ f^{(m)}(x_*)≠0$, prove that this sequence converges quadratically.


(We may use the Taylor's series, but I cannot get the result we need to prove.
Expand
expand $f(x_k)$ around $x_*$ until $m$-th order derivative term, which has the form
$(x_k - x_*)^m f^{(m)} (x_k) / m!$, and similarly for $ f ' (x_k)$ )


 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: how to prove sequence converges quadratically to a root of multiplicity

In an effort to let our helpers know where you are stuck, can you post your working so far and/or your thoughts on what you should try?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,876
Re: how to prove sequence converges quadratically to a root of multiplicity

A function f has a root of multiplicity $m>1$ at the point $ x_*$ if $f(x_*)=f'(x_*)=...=f^{(m-1)}(x_*)=0$. Assume that the iteration$ x_{k+1}=x_k-mf(x_k)/f'(x_k)$ converges to $x_*$. If$ f^{(m)}(x_*)≠0$, prove that this sequence converges quadratically.


(We may use the Taylor's series, but I cannot get the result we need to prove.
Expand
expand $f(x_k)$ around $x_*$ until $m$-th order derivative term, which has the form
$(x_k - x_*)^m f^{(m)} (x_k) / m!$, and similarly for $ f ' (x_k)$ )


Hey!
This looks a lot like the other thread, where I posted this.
Heck, you can even copy and paste it, and tweak it a little to be more generalized.
How far can you get?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,876
Re: how to prove sequence converges quadratically to a root of multiplicity

Hmm, I thought you would be able to do this.
I guess I was wrong.
Ah well, it wouldn't be the first time I was wrong.