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Oct 27, 2021 at 11:44 comment added JoshuaZ @J.J. Not for a general choice of $\theta$. The same construction will go through if one takes a number which has better rational approximations which will get a similar bound.
Oct 27, 2021 at 4:43 comment added J. J. Thank you for your nice remark. This means that one cannot have a lower bound with arbitrary $\sigma\in (0,1)$. Still can we expect a lower bound with $\sigma\in(\frac{1}{2},1)$?
Oct 27, 2021 at 1:42 history answered Jeremy Rouse CC BY-SA 4.0