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Problem 987

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  1. vf_285881bba6f1b7fd

    theoretical

    Erdős Problem #987: declared status 'proved'. Formalized: no. Let x1,x2,(0,1)x_1,x_2,\ldots \in (0,1) be an infinite sequence and letAk=lim supnjne(kxj),A_k=\limsup_{n\to \infty}\left\lvert \sum_{j\leq n} e(kx_j)\right\rvert,where e(x)=e2πixe(x)=e^{2\pi ix}. Is it true thatlim supkAk=?\limsup_{k\to \infty} A_k=\infty?Is it possible for Ak=o(k)A_k=o(k)? Current best: Erd\H{o}s [Er64b] remarks it is 'easy to see' thatlim supk(supnjne(kxj))=.\limsup_{k\to \infty}\left(\sup_n\left\lvert \sum_{j\leq n} e(kx_j)\right\rvert\right)=\infty.Erd\H{o}s [Er65b] later found a 'very easy' proof that AklogkA_k\gg \log k for infinitely many kk. Clunie [Cl67] proved that Akk1/2A_k\gg k^{1/2} infinitely often, and that there exist sequences with AkkA_k\leq k for all kk. Tao has independently found a proof that Akk1/2A_k\gg k^{1/2} infinitely often (see the comment section). Liu [Li69] showed that, for any ϵ>0\epsilon>0, Akk1ϵA_k\gg k^{1-\epsilon} infinitely often, under the additional assumption that there are only a finite number of distinct points. Prize: no. Tags: analysis, discrepancy.

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