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Improve this relatively simple FindRoot code?

Todor Latev
Posted 8 years ago
POSTED BY: Todor Latev
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Perhaps you've found these tutorials, but in case you haven't:
POSTED BY: Michael Rogers
Todor Latev
Todor Latev
Posted 8 years ago

I think I may have found a possible answer to my question: My code is a pure Newton Rapshon method whereas Mathematica actually uses a dampened version of this method with some builtin intelligence as to how to pick a damping factor to prevent the method from diverging in certain cases Now I need to figure out what that extra intelligence is
POSTED BY: Todor Latev
Todor Latev
Todor Latev
Posted 8 years ago

Thanks Jim, but the question here is not about how to find the roots in those particular examples, but rather - if FindRoot is using the Newton Raphson method to search for roots, what kind of tweaks is it doing that result in a different behavior in that particular case. If FindRoot manages to work properly with the initial {10,10} /I know it's far and that's on purpose/ I'd like to be able to do the same manually to replicate the exact intermediate states
POSTED BY: Todor Latev
Jim Baldwin
Jim Baldwin, Retired
Posted 8 years ago

You can't beat good starting values. Using {10.,10.} is a bit too far away. Try {2.,6.}. (And your first example has 3 roots.) This doesn't match with each iteration but both do end up with the same result (when more iterations are added to the NestList command).
POSTED BY: Jim Baldwin
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