Timeline for Constructing M-curves à la Hilbert
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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Nov 16, 2020 at 21:22 | vote | accept | Jose Capco | ||
Nov 15, 2020 at 13:16 | history | edited | Jose Capco | CC BY-SA 4.0 |
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Nov 15, 2020 at 12:39 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
title fix
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Nov 15, 2020 at 11:23 | history | edited | Jose Capco | CC BY-SA 4.0 |
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Nov 15, 2020 at 11:04 | answer | added | Jose Capco | timeline score: 2 | |
Nov 15, 2020 at 8:48 | comment | added | Jose Capco | You could write this as an answer so that I can mark the post as answerd once I checked this computationally. Thanks for the tip! | |
Nov 15, 2020 at 8:46 | comment | added | Jose Capco | Ah! That is a good idea. We should be able to assume that the intersection with the original ellipse in the construction is 0-dimensional and it is then easy to check if the points are real or not using almost any computer algebra system. Yeah, the vertical line construction should be wrong (it does not follow Hilbert's algorithm precisely). I think, 8 real intersection with one component 0 with the others is intended (otherwise you might have less component in the next M-curve). But I think I can modify the lines to make this happen (cont'd) | |
Nov 15, 2020 at 8:08 | comment | added | Zach Teitler | Hmm, those "prongs" in the drawing are wildly exaggerated. They cross a line 8 times. Going beyond just visually looking for "prongs", have you tried computationally finding the number of intersections of your perturbed curves with the original ones? With Macaulay2 it appears that in your example with horizontal lines, the perturbed curve has 8 real intersection points with each ellipse (but in the example with vertical lines, the intersection points are nonreal). Is it that you want 8 real intersections with one, and 0 with the other? Anyway try computational checking instead of visual. | |
Nov 14, 2020 at 17:55 | history | asked | Jose Capco | CC BY-SA 4.0 |