# Algebraic numbers and Fourier analysis (Heath mathematical by Raphael Salem Posted by By Raphael Salem

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A Property of the Set of Numbers of the Class S 20 We can now give the new proof of the theorem stating the closure of S. Let w be a limit point of the set S, and suppose first w > 1. Let {Os} be an infinite sequence of numbers of S, tending to w as s ~ 00. Denote by Ps(z) the irreducible polynomial with rational integral coefficients and having the root Os and let Ks be its degree (the coefficient of ZK. being 1). Let PROOF. Qs(z) = ZK·p s (~) be the reciprocal polynomial. The rational function Ps/ Qs is regular for I z I < 1 except for a single pole at z = Os-I, and its expansion around the origin P _s Qs L

It follows that the function k-I 'f z) fi\ J,L - 1 - rz k-I + 1 - J,Lr-1z +" 'Yi +" ~ 1 - ajZ ~ 'Yi 1 - a/z has, in the neighborhood of the origin, an expansion ClO L: anzn o with rational integral coefficients. The only singularity of fez) for the pole l/r. We have ClO k-l ClO L: (a - J,Lrn)zn = L: a zn ono n J,L = J,L 1 - rz 1 - r-Iz m( l-az 1 »1 - 2 and ( 1 - r-1z >--. - r +1 r 1 IS k-l + ~I 1 -'YiajZ + ~I 1 - 'Yia/z . By well-known properties of linear functions we have for I a CR - -1- - ) Iz I < I= I and I z I < 1 Applications to Power Series,.

A particular case of the theorem can be stated in the following simple way. Let be a power series with rational integral coefficients, converging for I z I < 1. Let S be the set of values taken by f(z) when I z I < 1. If the derived set S' is not the whole plane, f(z) is a rational function. In other words iff(z) is not a rational function, it takes in the unit circle values arbitrarily close to any given number a. It is interesting to observe that the result would become false if we replace the whole unit circle by a circular sector.