Timeline for Constructing a system of two cubic polynomial equations with exactly 9 real solutions in Maple [closed]
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 26 at 12:58 | history | closed |
Daniele Tampieri user44191 Jonathan Beardsley Daniel Weber Ben McKay |
Not suitable for this site | |
| Jul 29 at 8:19 | review | Close votes | |||
| Aug 26 at 12:58 | |||||
| Jul 29 at 8:08 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Jul 29 at 8:01 | comment | added | Jules Lamers | In general it is considered good etiquette to wait at least a couple of days after one posts on math.SE, and then mention the original question if one does crosspost. This avoids unnecessary duplicate work. | |
| Jul 29 at 3:11 | comment | added | A. Brik | If so, I apologize I'm new here. | |
| Jul 29 at 3:09 | history | edited | A. Brik | CC BY-SA 4.0 |
added 174 characters in body
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| Jul 29 at 3:05 | comment | added | A. Brik | Is it bad to crosspost? | |
| Jul 29 at 3:00 | comment | added | Will Jagy | math.stackexchange.com/questions/4951785/… | |
| Jul 29 at 3:00 | comment | added | A. Brik | okay thanks, I'll try. I've also noticed that $(0,0)$ is already one of the solutions. Someone suggested taking a cubic in $x$ with distinct real roots and the same for $y$ and then making the coupled equations through the linear transformation $x+y=X, x-y=Y$. | |
| Jul 29 at 2:59 | comment | added | Will Jagy | Crossposted from MSE, where he received enough hints in comments | |
| Jul 29 at 2:21 | comment | added | Noam D. Elkies | @GeraldEdgar I see. But now I also see that neither equation has a constant term, so $(0,0)$ is automatically a solution. OK, use $x(x-1)(x-2) = y^3 - y = 0$, and optionally deform without making the solutions $(0,\pm 1)$ go to the left of the $y$-axis. | |
| Jul 29 at 2:16 | comment | added | Gerald Edgar | @NoamD.Elkies: I started with $x^3-x = y^3-y = 0$, then noticed the condition $x$ non-negative. Also, it seems the first one has no $y^3$ term. | |
| Jul 29 at 2:13 | comment | added | Noam D. Elkies | Gerald Edgar suggested, but deleted, using $(x-1)(x-2)(x-3) = (y-1)(y-2)(y-3) = 0$. One could more generally use any independent linear combinations of two such polynomials; or more interesting, start from such a system (or translate to the simpler-looking $x^3-x = y^3-y = 0$) and deform a bit so the solutions are real but not obvious. | |
| S Jul 29 at 2:00 | review | First questions | |||
| Jul 29 at 8:02 | |||||
| S Jul 29 at 2:00 | history | asked | A. Brik | CC BY-SA 4.0 |