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I was asked to calculate the relative and absolute precision of my implementation of Newton's method and I wanted to make sure I understand it correctly.

According to the method:

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

So if I understand correctly:

  1. Absolute error is $\frac{|f(x_n)|}{|f'(x_n)|}$
  2. Relative error is $\frac{|x_{n+1}|}{|x_n|}$

Is that correct?

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  • $\begingroup$ Which is the same as mine, isn't it? $\endgroup$ Dec 17, 2016 at 18:31

1 Answer 1

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If:

  • $x_{\infty}$ is precise value you want to find

  • $x_{n+1}$ your best approximation

then absolute error $\epsilon$ will be :

$$\epsilon = | x_{\infty} - x_{n+1}|$$

and relative error $\eta$ will be:

$$ \eta = \frac{\epsilon}{|x_{\infty}|}$$

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