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How quickly does Newton's Method for the approximation of the Lambert W function converge?

$$w_{j+1}=w_j-\frac{w_je^{w_j}-x}{e^{w_j}+w_je^{w_j}}$$

or simplified:

$$w_{j+1}=\frac{xe^{-w_j}+w_j^2}{w_j+1}$$

where $x$ replaces $z$ in the equation, because I'm only concerned with real numbers.

I understand that given $\infty$ iterations, it will converge to be absolutely accurate.

For example, after $n$ iteration, it will be accurate to $m$ decimal places.

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  • $\begingroup$ In connection with your previous question (math.stackexchange.com/q/1950644). I think that instead of deleting this question, you could have asked this second question in the same framework. Moreover, I advise you to try to have an idea by numerical computations and ask us in a second step how you can improve the accuracy. $\endgroup$ – Jean Marie Oct 3 '16 at 22:00
  • $\begingroup$ @JeanMarie I deleted that other question because I found the solution on my own, and didn't think it merited me "answering" the question there. (If anything, the first part of my question was a duplicate, and the second part belongs on Stack Overflow more than here) $\endgroup$ – esote Oct 3 '16 at 22:12
  • $\begingroup$ Technically speaking, have you considered simplifying your recurrence expression as $w_{j+1}=\frac{w_j^2+xe^{-w_j}}{1+w_j}$ in order to improve its analysis and moreover its convergence speed ? $\endgroup$ – Jean Marie Oct 3 '16 at 22:28
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    $\begingroup$ Most numerical approximation methods increase the number of digits accurate by doubling or tripling, or in some way along those lines. For example, I do think the number of digits accurate when solving $x=\sqrt{c}$ doubles per iteration, once convergence sets in. $\endgroup$ – Simply Beautiful Art Oct 4 '16 at 20:24
  • $\begingroup$ @Simple Art Not most numericalapproximation methods ; quadratic convergence is almost specific to Newton's method... $\endgroup$ – Jean Marie Oct 4 '16 at 21:59

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