Timeline for Simultaneous diophantine approximation with polynomial bound
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2013 at 4:08 | review | Suggested edits | |||
| Dec 12, 2013 at 7:41 | |||||
| Dec 1, 2011 at 18:19 | comment | added | Alan Haynes | @Marcin I think your question has an answer but it is hard to determine exactly what you are asking. For example why can't you recover $q$ if $r$ does not grow with $n$? No matter what $r$ is, for any $q$, if the RHS is less that $1/2q$ then there is always a choice of $p_i$'s that works. That is also why I asked if you really to say mean $1/poly (n)$ on the RHS. If $q$ grows along any sequence which is exponential in $n$ but the bound on the right hand side is only $1/poly (n)$ then for every large enough $q$ there will be many choices for $p_i$'s. | |
| Nov 30, 2011 at 23:06 | comment | added | Marcin Kotowski | @AH: I do mean poly(n) - poly(q) would be exponential in n. r must grow with n, otherwise, you can't even hope to recover q. | |
| Nov 23, 2011 at 22:20 | comment | added | Alan Haynes | @Marcin There are two things that are confusing me. First, do you really mean to allow $r$ to grow with $n$? It seems like the dependence of the problem on $r$ is artificial since you can just multiply it through the whole equation. Secondly, do you really mean $poly(n)$ on the RHS of your inequality, or should it be $poly(q)$? | |
| Oct 15, 2011 at 23:13 | history | edited | Marcin Kotowski | CC BY-SA 3.0 |
added 115 characters in body
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| Oct 15, 2011 at 23:04 | history | asked | Marcin Kotowski | CC BY-SA 3.0 |