Timeline for Can Davenport's estimate be extended to cubic polynomials with non-zero discriminant?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2017 at 13:43 | comment | added | Stanley Yao Xiao | The answer is surely yes, as I suspect Davenport's argument will hold verbatim by replacing the assumption "irreducible" with "having non-zero discriminant". | |
| Dec 5, 2017 at 22:15 | history | edited | Alessandro Pezzoni | CC BY-SA 3.0 |
added 63 characters in body
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| Dec 5, 2017 at 22:15 | comment | added | Alessandro Pezzoni | Oh, right. $D(P)$ stands for the discriminant of the polynomial $P$. | |
| Dec 5, 2017 at 22:14 | comment | added | Gerry Myerson | You forgot to tell us what $D$ stands for. | |
| Dec 5, 2017 at 21:48 | review | First posts | |||
| Dec 5, 2017 at 22:10 | |||||
| Dec 5, 2017 at 21:43 | history | asked | Alessandro Pezzoni | CC BY-SA 3.0 |