Timeline for Does anyone know anything about the 2-valuation of the discriminant of a polynomial?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2016 at 10:02 | comment | added | Lior Bary-Soroker | I made a mistake; the $n(n-1)$ is the degree in the roots, it should be replaced by $2n-2$, the degree in the coefficients | |
| Sep 26, 2016 at 8:00 | comment | added | Lior Bary-Soroker | The degree of the discriminant is $n(n-1)$ so the typical size is $B^{n(n-1)}$ and we have only $2$ options modulo $4$ so the 2-valuation is $\geq k$ in about $B^{n(n-1)}/2^{k}$. Does this make sense? | |
| Sep 25, 2016 at 12:11 | comment | added | Igor Rivin | What would be the "naive expectation" for numbers of the relevant size? | |
| Sep 25, 2016 at 11:44 | comment | added | Ofir Gorodetsky | These properties seem to hold already in the toy example $n=2$. Namely, let $\Delta = b^2-4ac$. If $v_2(b^2) > v_2(4ac)$, then necessarily the 2-valuation is even. If $v_2(b^2) < v_2(4ac)$, the 2-valuation is even with probability $\frac{5}{9}$. | |
| Sep 25, 2016 at 11:28 | history | asked | Lior Bary-Soroker | CC BY-SA 3.0 |