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Sep 14, 2021 at 22:21 comment added No One No problem, any help or clarification is appreciated!
Sep 14, 2021 at 21:29 comment added Ronnie Pavlov Ah, thanks. I don't know how I read "all large enough" as "infinitely many." I guess the same proof should give you a version of your main inequality for all large enough $T$ with a multiplicative factor of $(1 + \epsilon)$ (i.e. the second quantity is less than $(1+\epsilon)T$), but according to the paper you sent, these multiplicative factors change the problem quite drastically. So I don't have any knowledge of how to get this stronger result. Sorry for the unhelpful answer!
Sep 14, 2021 at 17:25 comment added No One The point is that we need to show "for all large enough $T$'s" not just for infinitely many $T$'s (which was the Dirichlet's theorem).
Sep 14, 2021 at 17:04 comment added No One please see people.brandeis.edu/~kleinboc/Pub/kwadcompjournal.pdf#page=2# at the bottom (last paragraph). The authors claim that for $m=n=1$, this is the whole space. But otherwise it is a set of full measure
Sep 14, 2021 at 14:51 history answered Ronnie Pavlov CC BY-SA 4.0