If I use starting value x0 = 1 for the following function Newton-Raphson oscillates between -0.113356775 and 0.113356775. If I however use other values, I manage to find the root.
Why is that? Any help is appreciated.
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@lulu, that does indeed seem to answer my question, thank you. Do you maybe know why the "trap" occurs? It seems that it has something do with the absolute sign. – LLScheme May 07 '23 at 11:06
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1@Moo, that helps a lot thank you – LLScheme May 07 '23 at 11:08
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1In general, with numerical methods, the idea is to not get overly concerned with isolated failures. Dither the starting values...most errors involving "friendly" functions evaporate quickly (as you noticed with this case). There are, of course, truly pathological examples where it is hard to get the numerical method to find the root you want. Numerical methods depend on context. – lulu May 07 '23 at 11:09
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The Newton iteration is, from a certain point-of-view, just another fixed-point iteration. And a common feature-of-interest of fixed-point iterations is periodic points or cycles. Usually most of them are unstable, but if a cycle is stable, points from some neighborhood will converge to it, and it is not perturbed by floating-point errors. – Lutz Lehmann May 07 '23 at 13:46