Timeline for answer to NSolve gives additional solutions that don't satisfy the equations! by Daniel Lichtblau
Current License: CC BY-SA 3.0
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| Nov 1, 2012 at 16:40 | comment | added | dbm | @Licthblau, Excellent! Thanks a lot for your detailed explanation. In short, we need to be extra careful while using NSolve for systems with approximate coefficients. I already knew discriminant variety and homotopy continuation method. But I didn't know about the near discriminant variety issue, and specially that it could affect solving systems so drastically. Thanks a lot again! dbm368 | |
| Nov 1, 2012 at 14:19 | comment | added | Daniel Lichtblau | Edited to add explanation. | |
| Nov 1, 2012 at 14:19 | history | edited | Daniel Lichtblau | CC BY-SA 3.0 |
Explained arisal of extraneous large solutions
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| Nov 1, 2012 at 0:17 | history | edited | J. M.'s missing motivation | CC BY-SA 3.0 |
deleted 1 characters in body
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| Oct 31, 2012 at 23:57 | comment | added | dbm | Thanks for your comment and your request to others to stop down-voting. The problem comes from particle theory where I have a nonlinear parametric system of equations and I want to find all real solutions of it. In particular, I want to see if the no. of real solutions is constant for the parametric system. The present system is at a particular parameter-point (a generic point). I started having doubt about the infinite solutions was from my cross-checking the solutions with homotopy continuation method which gave only two solutions always instead of 8 by NSolve. Thanks again. | |
| Oct 31, 2012 at 23:39 | comment | added | Daniel Lichtblau | I'll edit my response tomorrow to address that issue of "infinite" solutions. Meanwhile you might mention where the system came from or the nature of the problem. That might help when I add an explanation. | |
| Oct 31, 2012 at 23:04 | comment | added | dbm | That worked! However, if I still think there is more fundamental problem here: NSolve gives solutions at infinity. If you see the actual solutions in sol, some variables take values of the order 10^16. These are solutions at infinity for a reasonable numerical precision say 10^13 or so. So there is a catch22 here. If we choose WorkingPrecision->50, the solutions at infinity are considered as affine solutions, if we choose the default WorkingPrecision then these solutions still appear but don't satisfy the equations! May be I am explaining it not properly?! | |
| Oct 31, 2012 at 22:51 | history | answered | Daniel Lichtblau | CC BY-SA 3.0 |