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Newton's method for finding zeroes of a function uses an iterative step like this:

\begin{align*} x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \\ \end{align*}

The following variation will instead find poles of a function. why? \begin{align*} x_{k+1} = x_k + \frac{f(x_k)}{f'(x_k)} \\ \end{align*}

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  • $\begingroup$ Why do you say that it will find the poles of a function? Try $f(x) = {1 \over x}$ for example, this gives $x_{n+1} = 2 x_n$, and $f(x_n) \to 0$. $\endgroup$
    – copper.hat
    Oct 2, 2017 at 19:32
  • $\begingroup$ With $f(x) = 1/x$, $f(x)/f'(x) = -x$, so normal Newton is $x_{k+1} = 2x_k$, while the pole variation of Newton's Method yields $x_{k+1} = 0$ for $f(x) = 1/x$. $\endgroup$
    – clay
    Oct 2, 2017 at 20:11
  • $\begingroup$ Oops, my mistake. $\endgroup$
    – copper.hat
    Oct 2, 2017 at 20:13
  • $\begingroup$ I am curious how/where you came across this fact? $\endgroup$
    – copper.hat
    Oct 2, 2017 at 21:16
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    $\begingroup$ Just a textbook problem for a class homework. $\endgroup$
    – clay
    Oct 2, 2017 at 22:20

1 Answer 1

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Taking the limit as $k\to\infty$ in both expressions gives $f(x_k)/f'(x_k)\to 0$. When searching for zeros, it is natural that $f(x_k)\to 0$. However, when searching for poles, it is not, but $f'(x_k)\to \pm\infty$ is. Since searching the poles of $f$ is equivalent to searching the zeros of $1/f$, we have \begin{aligned} x_{k+1} &= x_k - \frac{1/f(x_k)}{-f'(x_k)/f(x_k)^2} \, ,\\ &= x_k + \frac{f(x_k)}{f'(x_k)} \, . \end{aligned} The simplification by $f(x_k)^2$ is responsible for the confusion.

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