1
$\begingroup$

well, could any one tell me how to prove this in the relation with newton raphson method? if $y$ is a root of $f(x)=0$ with multiplicity $p$,then iterative formula becomes $$x_{n+1}=x_n-p[f(x_n)/f'(x_n)]$$

we know in general for simple root the iteration formula is $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

  • 0
    What exactly would you like to see proved? I'm guessing that what you actually want might be an explanation why this is a good method for finding multiple roots?2012-08-05
  • 0
    No, I just need to prove if there is some multiple root, will this be iterative formula?2012-08-05
  • 0
    This is clearly an iterative formula; there's nothing to prove there. You must be after some property of this iterative formula; either informally that it's a useful formula for finding multiple roots (which can't be proved, only made plausible, because it's informal), or some formalization of that claim that you want to see proved; in the latter case you should provide that formalization.2012-08-05
  • 0
    @joriki we know in general for simple root the iteration formula is $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ , I want to know for what if some root of multiplicity $p$?2012-08-05
  • 0
    Yes, I understand that. I was merely pointing out that you haven't told us what you want to see "proved", so all you can expect as an answer is an informal explanation why this formula is useful for finding multiple roots; if you want a proof, you have to first provide a formal statement to be proved.2012-08-05

1 Answers 1

2

Your formula is discussed in William J Gilbert, Newton's method for multiple roots, Comput. & Graphics 18 (1994) 227-229, available here. Gilbert cites equations 8.6-13 in Ralston and Rabinowitz, A First Course in Numerical Analysis, McGraw-Hill 1978.

  • 0
    See that book [here](http://books.google.com/books?hl=en&id=czHV-1bEFl0C&pg=PA354).2012-08-06