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Oct 6, 2016 at 14:24 comment added Kagaratsch @MichaelE2 my intent was to find the slope of the curve close to the spot where it bends back. Having a few numeric points in this subregion allows to do a fit.
Oct 4, 2016 at 15:45 answer added Michael E2 timeline score: 1
Oct 4, 2016 at 15:24 comment added Michael E2 Just how do you want to use the "numerical intersection"? I don't see the purpose in having a few scattered points to very high precision. (1) For plotting, increasing precision beyond a few digits does not improve the image. (2) Given a point close the curve, should one keep x or y fixed in trying to improve its position, or should one move parallel to the gradient or some other direction? Your example fixes x, but that seems arbitrary and inappropriate for all possible curves.
Oct 4, 2016 at 1:01 answer added Bob Hanlon timeline score: 2
Oct 3, 2016 at 21:29 history tweeted twitter.com/StackMma/status/783056402017288192
Oct 3, 2016 at 20:06 answer added george2079 timeline score: 3
Oct 3, 2016 at 19:43 comment added george2079 related . mathematica.stackexchange.com/q/75352/2079
Oct 3, 2016 at 18:58 answer added Kagaratsch timeline score: 3
Oct 3, 2016 at 18:40 comment added Kagaratsch @march Thank you, this helped me alot! I adopted the solution in this link to my case, with a few tweaks. Since I need a lot of precision, I'll post my tweaks below in case if others might be interested.
Oct 3, 2016 at 18:13 comment added march You can extract the points from the plot and refine them using FindRoot. See, for instance, here. This is actually a pretty common method; there should be many examples of this on this very site.
Oct 3, 2016 at 18:13 history edited J. M.'s missing motivation
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Oct 3, 2016 at 18:05 history asked Kagaratsch CC BY-SA 3.0