Zeros of Solutions of a Functional Equation



We consider the zeros of transcendental entire solutions f of the functional equation
$$\sum_{j=o}^{m}a_{j}(z)f(c^{j}z) = Q(z)$$
where \(c\in {\rm C},0<|c|<1,\) and Q and the a j are polynomials. Under a suitable hypothesis concerning the associated Newton-Puiseux diagram we show that the zeros of f are asymptotic to certain geometric progressions. More precisely, with this hypothesis there exist positive integers M and N such that the zero set can written in the form \(\lbrace z_{n,u}:\mu \in \lbrace 1,2,\dots, M\rbrace, n\in {\rm N}\rbrace\) where for each μ in \(\lbrace 1,2,\dots, M\rbrace \) there exists A μ in ℂ {0} with \(z_{n,\mu}\sim A_{\mu}\ c^{-Nn}\ {\rm as}\ n\rightarrow \infty\). The proof is achieved by showing that f behaves asymptotically like a product of θ-functions. The hypothesis on the Newton-Puiseux diagram is satisfied, e.g., if for each positive σ and each real τ the line \(\lbrace (x,y)\in {\rm R}^2:y=\sigma x+\tau \rbrace\) contains at most two points of the form (j, deg(a j)). In particular, this is the case if all a j are linear, in which case the above conclusion follows with M = 1 which means that the zeros are asymptotic to only one geometric progression. The hypothesis on the Newton-Puiseux diagram is also satisfied if m = 1. If m = 1 and Q ≡ 0, however, we have a much simpler and more precise result. We illustrate our results by a number of examples. In particular, we show that if the hypothesis on the Newton-Puiseux diagram is not satisfied, then the zeros of the solutions need not be asymptotic to a finite number of geometric progessions.


q-difference equation q-series theta function Newton-Puiseux diagram 

2000 MSC

39A13 39B32 33D99 30D05 


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© Heldermann  Verlag 2003

Authors and Affiliations

  1. 1.Mathematisches SeminarChristian-Albrechts-Universität zu KielKielGermany
  2. 2.Department of MathematicsImperial College LondonLondonUK

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