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Thread Subject:
nonlinear equations

Subject: nonlinear equations

From: yunzhi cheng

Date: 17 Jan, 2008 14:52:02

Message: 1 of 3

I use Newton-Raphson method to solve F(x)=0.
x is n-dimension vector
F is n nonlinear equations.
It runs well.
But you know, Newton-Raphson needs a good initiate guess of x.

I just found a new method especially for my F(x) which is
proven to be less insensitive to initiate guess than
Newton-Raphson. I wonder if I can conclude that the new
method is robuster than Newton-Raphson?

Subject: nonlinear equations

From: John D'Errico

Date: 17 Jan, 2008 15:54:02

Message: 2 of 3

"yunzhi cheng" <sjtu_yh@yahoo.com> wrote in message
<fmnq2i$eqj$1@fred.mathworks.com>...
> I use Newton-Raphson method to solve F(x)=0.
> x is n-dimension vector
> F is n nonlinear equations.
> It runs well.
> But you know, Newton-Raphson needs a good initiate guess of x.
>
> I just found a new method especially for my F(x) which is
> proven to be less insensitive to initiate guess than
> Newton-Raphson. I wonder if I can conclude that the new
> method is robuster than Newton-Raphson?

How can one single example prove robustness?

John

Subject: nonlinear equations

From: yunzhi cheng

Date: 17 Jan, 2008 23:38:02

Message: 3 of 3

Thanks for your reply.
F(x) are some special functions in my major.
I just want to conclude that the new method is robuster than
Newton-Raphson for such F(x).
I wonder how to prove it?
Can I prove it by generating some cases which the new method
converge but N-R diverge?

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