Timeline for How many solutions do you get from simultaneous polynomial equations?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2016 at 6:32 | vote | accept | Hugh | ||
| Mar 15, 2016 at 15:49 | answer | added | Daniel Lichtblau | timeline score: 4 | |
| Mar 15, 2016 at 12:04 | comment | added | Hugh | Thanks for all the comments. I thought this was a simple question but it appears to be complex. @DanielLichtblau and others I have 24 solutions with the values I am using for the known parameters. Can I assume that there will always be 24 solutions to these equations whatever parameters I use? Thanks | |
| Mar 14, 2016 at 22:41 | comment | added | Daniel Lichtblau |
An important question is "expect from what"? You have the number of solutions (assuming NSolve is correct). Are there parameters that you intend to vary? If so, what are the specifics?
|
|
| Mar 14, 2016 at 22:39 | comment | added | Daniel Lichtblau | @J.M. That looks like the Bezout bound. For many common families of examples it is too high. | |
| Mar 14, 2016 at 19:22 | comment | added | george2079 | @J.M. can you give the formula from that paper here? | |
| Mar 14, 2016 at 19:21 | comment | added | george2079 |
For this particular example, two of your equations are linear in one unknown, so you can Eliminate[eqns , {c,k}]. The result is a twelfth order polynomial in only e ( 12 solutions ), and an equation quadratic in a , so 2x12->24 . (I ran eliminate on a numerical example to get that.. )
|
|
| Mar 14, 2016 at 18:47 | comment | added | J. M.'s missing motivation♦ | Have you seen this? | |
| Mar 14, 2016 at 18:26 | history | edited | Dr. belisarius | CC BY-SA 3.0 |
deleted 146 characters in body
|
| Mar 14, 2016 at 16:03 | history | asked | Hugh | CC BY-SA 3.0 |