Timeline for Is it possile for all real algebraic numbers to have continued fractions with bounded partial quotients ?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 23, 2010 at 12:43 | vote | accept | Chua KS | ||
| Jun 16, 2010 at 7:22 | comment | added | Wadim Zudilin | @Gerry, hmmmm. Let call them rationalities! :-) | |
| Jun 16, 2010 at 7:02 | comment | added | Gerry Myerson | @Wadim, you'll also need a name for algebraic numbers of degree 1. | |
| Jun 16, 2010 at 5:46 | comment | added | Wadim Zudilin | @Gerry: I call algebraic numbers of degree 2 quadratic irrationalities. :-) Sorry, for confusion. | |
| Jun 16, 2010 at 5:34 | comment | added | Gerry Myerson | @Wadim, you are of course referring to algebraic numbers of degree at least three. @OP, "Khinchin Thm 23" isn't a useful way to give a reference. I suspect you'll find that it's an "if and only if" statement, so that if algebraic numbers of degree at least three all have unbounded partial quotients (which, as others have mentioned, is where the smart money is), then there is no such $\alpha$ for which such a $c>0$ exists. | |
| Jun 16, 2010 at 5:22 | comment | added | Wadim Zudilin | The question of boundedness of partial quotients of algebraic numbers is addressed in several article by Enrico Bombieri and Alf van der Poorten, and there are too computational articles by Serge Lang (included in the last edition of his Diophantine Approximations, as far as I remember). It is likely that the continued fractions of algebraic numbers are "random" (behave as generic continued fractions following the Gauss-Kuzmin statistics), so there partial quotients are unbounded. See the answer of Kevin to mathoverflow.net/questions/23111. | |
| Jun 16, 2010 at 4:29 | answer | added | Boyarsky | timeline score: 11 | |
| Jun 16, 2010 at 4:11 | history | asked | Chua KS | CC BY-SA 2.5 |