Timeline for area of triangle from coefficients of its cubic?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2016 at 12:23 | history | edited | Robert Bryant | CC BY-SA 3.0 |
fixed some bad formatting problems that weren't there before; they must have arisen because of some change in MathJax
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| Nov 27, 2011 at 13:19 | history | edited | Robert Bryant | CC BY-SA 3.0 |
corrected number of terms and total degree of P
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| Nov 26, 2011 at 14:15 | comment | added | Robert Bryant | @Dima: If you don't like restricting to $a_1=0$, you could just rewrite $R$, $Q$, and $P$ in terms of the variables $V$, $T = S - \tfrac13a_1{\bar a}_1 = \bar T$ and $b_2 = a_2 - \tfrac13 {a_1}^2$ and $b_3 = a_3 - \tfrac13a_1a_2 + \tfrac2{27}{a_1}^3$ and their conjugates ${\bar b}_2$ and ${\bar b}_3$. That will have the same effect as translating the triangle to get $a_1=0$. I don't think you are going to get simpler formulae than that in the general case. | |
| Nov 26, 2011 at 13:51 | history | edited | Robert Bryant | CC BY-SA 3.0 |
corrected small error and added missing word
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| Nov 26, 2011 at 13:46 | comment | added | Robert Bryant | @Dima: OK. with(Groebner): Q0:= (S-z[1]*w[1]-z[2]*w[2]-z[3]*w[3])*(S-z[1]*w[1]-z[2]*w[3]-z[3]*w[2]) *(S-z[1]*w[2]-z[2]*w[3]-z[3]*w[1])*(S-z[1]*w[3]-z[2]*w[2]-z[3]*w[1]) *(S-z[1]*w[3]-z[2]*w[1]-z[3]*w[2])*(S-z[1]*w[2]-z[2]*w[1]-z[3]*w[3]): Q0:=expand(Q0): B1:=Basis([Q0,s[1]-z[1]-z[2]-z[3],s[2]-z[2]*z[3]-z[3]*z[1]-z[1]*z[2],s[3]-z[1]*z[2]*z[3]], plex(z[1],z[2],z[3],s[1],s[2],s[3],S)): Q1:=expand(op(1,B1)): B2:=Basis([Q1,u[1]-w[1]-w[2]-w[3],u[2]-w[2]*w[3]-w[3]*w[1]-w[1]*w[2],u[3]-w[1]*w[2]*w[3]], plex(w[1],w[2],w[3],u[1],u[2],u[3],S)): Q:=expand(op(1,B2)): | |
| Nov 26, 2011 at 7:31 | comment | added | Dima Pasechnik | @Robert, perhaps you can just post the Maple(?) code you wrote to produce $Q$. | |
| Nov 25, 2011 at 13:56 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added information about the real case
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| Nov 25, 2011 at 13:22 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added some information about the reduced equations
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| Nov 25, 2011 at 12:53 | comment | added | Robert Bryant | @Dima: $Q$ itself has $66$ terms, but the coefficients are quite small, so it's not completely crazy to think that it might contain some useful information. However, I'm a little reluctant to put the whole thing in the answer because it's still quite long, and I'd have to struggle with the typesetting or else leave it as a 'long line'. Let me think about whether I can do something better. One thing that I noticed is that the formulae simplify quite a lot if you assume that the centroid of the triangle is at $z=0$. Maybe I'll post those formulae first, so that you can see the rough shape. | |
| Nov 25, 2011 at 6:42 | comment | added | Dima Pasechnik | It would be interesting to know more about $Q$. After all, if $S$ can be found as a function of $a_k$ and $\overline{a}_k$ then the determinant expression for $VV^\top$ might be more useful than the 1175-term expansion. Could you perhaps post the expansion for $Q$? Thanks in advance. | |
| Nov 25, 2011 at 4:28 | comment | added | Robert Bryant | @Dima: Oh, you are right; thanks for pointing this out. I have edited the answer to correct this. It doesn't really change anything, except that P now has $1175$ terms instead of merely $646$. | |
| Nov 25, 2011 at 4:26 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Corrected a typo and followed through the consequences
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| Nov 25, 2011 at 3:26 | comment | added | Dima Pasechnik | There is a typo in the product of $V$ and $V^\top$. The top left entry should be 3, not 1. I wonder if it affects the outcome... | |
| Nov 25, 2011 at 3:02 | vote | accept | Dima Pasechnik | ||
| Nov 24, 2011 at 14:03 | history | edited | Robert Bryant | CC BY-SA 3.0 |
explained notation more clearly
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| Nov 24, 2011 at 13:33 | history | answered | Robert Bryant | CC BY-SA 3.0 |