Timeline for Finding a solution to a simple geometric set of equalities
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2016 at 11:32 | comment | added | Pietro Majer | good point; indeed $u_1,\dots u_{n}$ are $n$ points on the given ellipse, corresponding to the $n$ degree of freedom needed to fix a solution. | |
| Dec 21, 2016 at 1:28 | comment | added | Robert Israel | It was always going to be underdetermined: you have $n$ equations and $2n$ degrees of freedom, so you should expect the solution space to be $n$-dimensional (if nonempty). Yes, I know that's not always the case, but generically... | |
| Dec 20, 2016 at 23:21 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 20, 2016 at 23:19 | comment | added | Pietro Majer | It seems the most general solution is determined if we fix freely the similarity class of each triangle $(p_k,x_k,x_{k+1} )$, that is any $(0,1,u_k)$, with $|1-u_k|=a(1+|u_k|)$ provided the non-degeneracy condition $u_0u_1\dots u_{n-1}\neq1$ holds: then there exists a unique solution. | |
| Dec 20, 2016 at 23:13 | vote | accept | Tom Solberg | ||
| Dec 20, 2016 at 23:13 | comment | added | Tom Solberg | Ah, so my problem is underdetermined, and admits many solutions. Thank you! | |
| Dec 20, 2016 at 23:04 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 20, 2016 at 22:25 | history | answered | Pietro Majer | CC BY-SA 3.0 |