6
$\begingroup$

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*}

$\endgroup$
5
  • $\begingroup$ 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
  • 1
    $\begingroup$ Just a textbook problem for a class homework. $\endgroup$
    – clay
    Oct 2, 2017 at 22:20

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.