Timeline for Decomposition of symmetric homogeneous polynomials
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2014 at 21:03 | vote | accept | AatG | ||
| Sep 18, 2014 at 8:44 | answer | added | Ilya Bogdanov | timeline score: 4 | |
| Sep 17, 2014 at 7:43 | comment | added | Pietro Majer | Even assuming that the polynomial $P$ is homogeneous, at least for even $r$ the statement can't be true. If it were true ,one could iterate , and eventually write $P$ as a sum of r-powers without remainder (thus getting $P\ge0$) | |
| Sep 16, 2014 at 23:18 | comment | added | darij grinberg | Maybe you just want the original polynomial to be homogeneous? | |
| Sep 16, 2014 at 22:49 | comment | added | Ilya Bogdanov | Still, for $x^2+y^2+x+y$ it is impossible: $x$ and $y$ cannot appear in neither of your summands since all of them are homogeneous of degree $2$. | |
| Sep 16, 2014 at 21:50 | comment | added | AatG | Thanks for asking for the clarification. I have done so: the linear polynomials do not have constant terms, and the field is R. | |
| Sep 16, 2014 at 21:50 | history | edited | AatG | CC BY-SA 3.0 |
explained what I mean by linear polynomials, and what field the problem is set in, and gave an example
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| S Sep 16, 2014 at 5:00 | history | suggested | Mostafa Mirabi | CC BY-SA 3.0 |
corrected spelling
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| Sep 16, 2014 at 4:28 | review | Suggested edits | |||
| S Sep 16, 2014 at 5:00 | |||||
| Sep 16, 2014 at 1:31 | comment | added | darij grinberg | Please elaborate more, as it seems that your question is ambiguous or misstated. For $r=2$ and base field $\mathbb{R}$, an $r$-th power of a linear polynomial always has nonnegative constant term (provided that your notion of a "linear polynomial" allows constant terms to begin with), so a polynomial like $x^2+x+y^2+y$ cannot be obtained in this way. | |
| Sep 15, 2014 at 23:25 | history | asked | AatG | CC BY-SA 3.0 |