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- From: israel@unixg.ubc.ca (Robert B. Israel)
- Subject: Re: f: Q -> Q , f analytic
- Message-ID: <israel.713952031@unixg.ubc.ca>
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- Organization: University of British Columbia, Vancouver, B.C., Canada
- References: <1992Aug14.142252.8804@unibi.uni-bielefeld.de>
- Date: Sun, 16 Aug 1992 08:00:31 GMT
- Lines: 34
-
- In <1992Aug14.142252.8804@unibi.uni-bielefeld.de> achim@unibi.uni-bielefeld.de (Achim Flammenkamp) writes:
-
- >A analytic number theoretic question (or ?):
-
- >Given a function with domain and target C excepted a finite number of points.
- >It maps Q into Q (Q denotes the set of rational numbers) and
- >it is meromorph on C (this implies uniquely extendable from Q to C).
-
- >Must this function be representable as the quotient of two polynomies with
- >integer coefficients or are there other functions left (present a counter-
- >example, please) ?
-
-
-
- Let {q_n} be an enumeration of Q. Let {P_n} be an enumeration of
- polynomials with rational coefficients. Let f_n be an entire
- function (a polynomial will do) satisfying:
- (i) f_n(q_j) = 0 for j < n
- (ii) f_n(z) is real when z is real.
- (iii) sum_{j=1}^n f_j(q_n) is rational, but is not P_n(q_n).
- (iv) |f_n(z)| < 2^(-n) for |z| <= n
- Let F(z) = sum_{n=1}^infinity f_n(z). The series converges
- uniformly on compact sets, so F is an entire function. For each
- rational q_n, F(q_n) = sum_{j=1}^n f_j(q_n) is rational, so F
- maps Q into Q. Moreover, F(q_n) <> P_n(q_n), so F is not a
- polynomial with rational coefficients. Since it's entire, it's
- also not a quotient of polynomials with rational coefficients.
-
- >achim
- --
- Robert Israel israel@math.ubc.ca
- Department of Mathematics or israel@unixg.ubc.ca
- University of British Columbia
- Vancouver, BC, Canada V6T 1Y4
-