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Jul 3, 2018 at 22:21 vote accept kodlu
Jul 3, 2018 at 14:07 comment added Cherng-tiao Perng By considering $x_0=\frac \pi 2+\epsilon$ with $\epsilon$ sufficiently small, it still follows that $f_{a_1,\cdots,a_m}(x_0)<1$.
Jul 3, 2018 at 9:27 comment added kodlu Please see edit
Jul 3, 2018 at 9:07 history answered Cherng-tiao Perng CC BY-SA 4.0