Timeline for The solutions of a system of polynomials
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2013 at 17:59 | comment | added | Emil Jeřábek | I don’t know if it’s of any use, but if $f(z)$ denotes the polynomial $(1-x_1z)^{m_1}\cdots(1-x_nz)^{m_n}$, then the first $n-1$ equations are equivalent to the condition that the logarithmic derivative $f'(z)/f(z)$ has a root of order at least $n-1$ at the origin (this follows easily enough from the Cauchy integral formula). | |
| Nov 27, 2013 at 17:01 | answer | added | Jason Starr | timeline score: 11 | |
| Nov 27, 2013 at 16:40 | comment | added | Lev Borisov | Do you by any chance want $x_1^{m_1}...x_n^{m_n}=1$? | |
| Nov 27, 2013 at 15:45 | comment | added | Julian Rosen | For $m_1=\ldots=m_n=1$, $(\zeta^{\sigma(1)},\ldots,\zeta^{\sigma(n)})$ is a solution only when $n$ is odd. For $n$ even, $(\zeta)(\zeta^2)\ldots(\zeta^n)=-1$. | |
| Nov 27, 2013 at 14:43 | answer | added | Francesco Polizzi | timeline score: 5 | |
| Nov 27, 2013 at 12:57 | review | First posts | |||
| Nov 27, 2013 at 13:01 | |||||
| Nov 27, 2013 at 12:51 | history | edited | Zhihua Chang | CC BY-SA 3.0 |
added 39 characters in body
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| Nov 27, 2013 at 12:41 | history | asked | Zhihua Chang | CC BY-SA 3.0 |