By Melvyn B. Nathanson (auth.), David Chudnovsky, Gregory Chudnovsky (eds.)

This notable quantity is devoted to Mel Nathanson, a number one authoritative professional for a number of a long time within the region of combinatorial and additive quantity idea. Nathanson's various effects were largely released in first class journals and in a couple of first-class graduate textbooks (GTM Springer) and reference works. For a number of a long time, Mel Nathanson's seminal principles and ends up in combinatorial and additive quantity thought have inspired graduate scholars and researchers alike. The invited survey articles during this quantity replicate the paintings of amazing mathematicians in quantity concept, and symbolize quite a lot of vital subject matters in present research.

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Lemma 3. V; ˛I N 1/: Proof. We work by induction on d . For d D 1, the result obviously holds since ˛1 62 Z. 50 R. Graham and K. d 1/-tuples . Assume that W < V . v2 vd ; hŒ˛1 w1 ˛2 ; ˛3 ; : : : ; ˛d iI 1/ Thus, we may assume that w1 < v1 , and so w1 wd 1 < v1 vd 1 . V; ˛I N 1/: Lemma 4. V; ˛I N 1/. 0d ; ˇI then ff˛i gW 1 Ä i Ä d g D ffˇi gW 1 Ä i Ä d g. Proof. ˇ/W Can You Hear the Shape of a Beatty Sequence? ˛/ N C Az is determined by the sequence. Therefore, the set of its roots 1 f˛f˛i gi g is also determined by the sequence.

Graham and K. ˛/W N Œ˛i 2 fb˛i c; d˛i eg : @xi We have assumed that ˛N is irrational to guarantee that all of these expressions arise as limit points, and we assumed that ˛i are algebraically independent to guarantee that all of these expressions correspond to distinct real numbers. ˛/ N ˇˇ : @x i From here, we apply ad hoc arguments that depend on the special structure of S . d 1/-th root, learning the value of Q ˛i by each element of the set L yields the set Q ˛i . x/ N D i;j D1 xi xj , then we have learned 8 9 d < = X L D ˛i C ˛j W 1 Ä i Ä d : : ; j D1 Can You Hear the Shape of a Beatty Sequence?

Theorems 2 and 3 rely on an extension of the sum-product theorem to Cartesian products. 0 Theorem 4. Fix " > 0. ı/ ! x; ax/jx 2 Fp g such that jA \ Lj > p 1 ı0 : 9 The Sum-Product in Zq D Z=qZ Because of the presence of subrings when q is composite, additional restrictions on A Zq are needed. The following gives a uniform statement in the modulus q. Theorem 1 ([B3]). ı1 ; ı2 / > 0 such that the following holds. A/j > q1ı2 for all q1 jq with q1 > q " where map. q1 W Zq ! Zq1 is the quotient Then jA C Aj C jA:Aj > q jAj: Remark.