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The Newton's method that I know is defined as follows:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

However, I've recently encountered a paper that talks about a one-parameter family of Newton's method (page 4, equation 2.8), defined as follows:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)-pf(x_n)}$$

What is this $p$ parameter in the equation above? Why is it not present in the first equation? What is this parameter useful for? What is that one-parameter Newton's method?

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"One-parameter family" simply means that one real parameter, $p$, appears in the definition. That method is simply a generalization of Newton's method exposed in the first equation; you can't derive it from the first.

The classical Newton's method has slower convergence when $f'(\alpha)=0$, $\alpha$ being the sought solution: the purpose of this generalization is obtaining a Newton-like method that does not encounter trouble when $f'(\alpha)=0$. I hope this solves your issues!

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  • $\begingroup$ Thank you for the response. How does introducing the $p$ makes the method faster than the classical one? What value should $p$ have to make it work better? $\endgroup$ – Marc Andreson Jan 24 '15 at 8:51
  • $\begingroup$ This is explained in your paper. Check around Equation (2.9), there is an asymptotic expression for the error which involves $p$. $\endgroup$ – Federico Poloni Jan 24 '15 at 9:04
  • $\begingroup$ So as to speed up the root finding and get the second-order (quadratic) convergence we need to find a $p$ that will make the denominator the largest one? Could you please include in your response a brief explanation how this is calculated from the error equation (I haven't seen before the asymptotic expressions)? $\endgroup$ – Marc Andreson Jan 24 '15 at 9:14
  • $\begingroup$ It's the first time I see that generalization, too, so I can't give you more information than what can be read in the paper. Have you checked reference [2]? $\endgroup$ – Federico Poloni Jan 24 '15 at 9:33
  • $\begingroup$ I don't have the access to this [2] paper. But basically, is my assumption correct that $p$ is supposed to make the denominator the largest one? $\endgroup$ – Marc Andreson Jan 24 '15 at 9:35

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