How to prove sequence converges quadratically to a root of multiplicity

[i_a_n]

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)$ )

 

[MarkFL]

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?

 

[I like Serena]

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 http://www.mathhelpboards.com/f16/optimization-problem-Newtons-method-3447/#post15282.
Heck, you can even copy and paste it, and tweak it a little to be more generalized.
How far can you get?

 

[I like Serena]

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.