Timeline for SOS polynomials with rational coefficients
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2020 at 8:53 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
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| Sep 28, 2020 at 21:52 | vote | accept | Gautam | ||
| Sep 28, 2020 at 10:05 | history | edited | Glorfindel | CC BY-SA 4.0 |
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| Sep 28, 2020 at 4:32 | history | became hot network question | |||
| Sep 28, 2020 at 0:27 | answer | added | Noam D. Elkies | timeline score: 20 | |
| Sep 27, 2020 at 22:17 | comment | added | user347489 | I guess the idea is that $p(x)=(f(x)-ig(x))(f(x)+ig(x))$? But how is finding this factorization easier than the proposed problem? | |
| Sep 27, 2020 at 21:54 | comment | added | Gautam | Fedor, I didn't quite understand your comment. Are you suggesting we first factor $p(x)$ over the rationals and then use this factorization to obtain the desired decomposition? Can you please elaborate? | |
| Sep 27, 2020 at 21:22 | comment | added | Fedor Petrov | This is about factorization in $\mathbb{Q}[i]$, which may be done efficiently. | |
| Sep 27, 2020 at 20:53 | history | edited | Gautam | CC BY-SA 4.0 |
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| Sep 27, 2020 at 20:53 | comment | added | Gautam | Yes, you may assume that $p(x)$ has a representation as a sum of squares of rational polynomials (though this may involve more than two squares!). The tricky part is finding a representation as a sum of two squares, and also doing it efficiently. I edited the question for clarity. | |
| Sep 27, 2020 at 20:41 | comment | added | Olivier Bégassat | Is this always possible? How would you express the constant polynomial 3 as a sum of two squares of rational polynomials? EDIT: I guess your assumption is that $p$ is the SOS of rational polynomials? | |
| Sep 27, 2020 at 20:31 | history | asked | Gautam | CC BY-SA 4.0 |