Timeline for what part of using vieta's formulas violates quintic non-solvability?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2011 at 14:08 | comment | added | Jacques Carette | If you have k integer roots, k! will pop out, I would think (i.e. all permutations). I got to n! first from working out n=2,3,4 explicitly (using a CAS), and extrapolating. Then I remembered Bezout's theorem. | |
| Nov 5, 2011 at 13:55 | comment | added | Cris Stringfellow | Sorry I was thinking over Z. Over Z how many solutions pop out? Is it still n!. I don't think so. By the way, how do you work out that there are n! solutions? | |
| Nov 5, 2011 at 13:12 | comment | added | Jacques Carette | No, plugging in a particular order of coefficients does not fix it. The system of equations you get inherently has n! solutions. | |
| Nov 4, 2011 at 14:42 | comment | added | Cris Stringfellow | Yes the technical detail that all the roots (in the Vieta formula) are cyclically permutable. It is obvious. But I don't see how that returns n! answers, if you plug in a particular order of coefficients that should fix the answer permutation....Any more detail here that is worth discussing? Will solving Jacobian in Newton's method produce n! with equal probability? Appreciate the pdfs. | |
| Nov 3, 2011 at 20:40 | history | answered | Jacques Carette | CC BY-SA 3.0 |