Timeline for Diophantine approximation

7 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:57	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 12, 2011 at 0:48	comment	added	Oleg Eroshkin		@Felipe Good point. Than even this conjecture (wide open) is too weak for vanvu.
May 12, 2011 at 0:41	comment	added	Felipe Voloch		@Oleg But he seems to want $N$ to be the same as $n$, which is related to the degree of $\alpha$ and so hidden in the ineffective constants in Roth's theorem and conjectural generalizations.
May 12, 2011 at 0:28	comment	added	Oleg Eroshkin		Roth's theorem gives bound for $\|q\alpha-p\|$. This is the linear polynomial. Vanvu need such bounds for polynomials of degree $d$. Conjecturally for algebraic number $\alpha$ and $\epsilon>0$ there are only finitely many polynomials $P$ of degree $d$ with integer coefficients of absolute values at most $N$ with $0<\|P(\alpha)\|<N^{-d-1-\epsilon}$.
May 12, 2011 at 0:16	comment	added	Fedor Petrov		@Oleg: I am afraid that Roth's theorem wors for any specific polynomial, while $n$ does change...
May 12, 2011 at 0:09	comment	added	Oleg Eroshkin		This is Liouville's inequality. For $d=1$, Roth's theorem gives required bounds. The generalization of Roth theorem for polynomials of higher degree is open. I doubt that the special case of $\alpha$ is significantly easier.
May 11, 2011 at 23:52	history	answered	Felipe Voloch	CC BY-SA 3.0