Timeline for Does there exist a homogeneous polynomial $F$ whose partial derivatives satisfy the following inequalities?
Current License: CC BY-SA 3.0
12 events
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Dec 3, 2017 at 18:42 | comment | added | Rodrigo de Azevedo | Are $n$ and $d$ inputs? Or can one choose them? | |
S Dec 3, 2017 at 18:37 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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Dec 3, 2017 at 18:05 | review | Suggested edits | |||
S Dec 3, 2017 at 18:37 | |||||
S Dec 3, 2017 at 18:02 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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Dec 3, 2017 at 18:00 | review | Suggested edits | |||
S Dec 3, 2017 at 18:02 | |||||
Dec 3, 2017 at 17:17 | comment | added | Johnny T. | @JasonStarr True. Thanks very much! | |
Dec 3, 2017 at 16:34 | comment | added | Jason Starr | There are still plenty of examples. For instance, fix an integer $d$ with $2\leq d \leq n$, and consider $F(x_1,x_2,\dots,x_n) = \sum_{1\leq i_1<i_2<\dots<i_d \leq n} a_{i_1,i_2,\dots,i_d}x_{i_1}x_{i_2}\cdots x_{i_d},$ i.e., a general linear combination of square-free monomials of total degree $d$. The partial derivatives $\partial^2 F/\partial x_i^2$ are all zero, but the singular set does not equal all of $W$. | |
Dec 3, 2017 at 16:20 | history | edited | Johnny T. | CC BY-SA 3.0 |
added 33 characters in body
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Dec 3, 2017 at 16:19 | comment | added | Johnny T. | @JasonStarr Let me fixed this... | |
Dec 3, 2017 at 15:59 | comment | added | Jason Starr | @gcousin. More generally, if $W$ is contained in the singular locus of $F$, e.g., for $F=G^2$. Perhaps the OP would like to specify that $W$ is not contained in the singular locus of $F$. | |
Dec 3, 2017 at 15:52 | comment | added | gcousin | F=0 does the job ;-) | |
Dec 3, 2017 at 15:36 | history | asked | Johnny T. | CC BY-SA 3.0 |