Computational Methods and Function Theory 3 (2003), No. 1, 55--78
Copyright Heldermann Verlag 2003
Walter Bergweiler
bergweiler@math.uni-kiel.de, Mathematisches Seminar, Christian-Albrechts-Universität zu Kiel, Ludewig-Meyn-Str.~4, D-24098 Kiel, Germany.
Walter K. Hayman
Department of Mathematics, Imperial College London, London, SW7 2AZ, U.K.
We consider the zeros of transcendental entire solutions $f$ of the functional equation $$ \sum_{j=0}^m a_j(z)f(c^jz)= Q(z), $$ where $c\in\C$, $0<|c|<1$, 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 $\{z_{n,\mu}:\,\mu\in\{1,2,\dots,M\},n\in\N\}$ where for each~$\mu$ in $\{1,2,\dots,M\}$ there exists $A_\mu$ in $\C\setminus\{0\}$ with $z_{n,\mu}\sim A_\mu\;c^{-N n}$ as $n\to\infty$. The proof is achieved by showing that $f$ behaves asymptotically like a product of $\theta$-functions. The hypothesis on the Newton-Puiseux diagram is satisfied, e.g., if for each positive $\sigma$ and each real $\tau$ the line $\{(x,y)\in \R^2:\, y=\sigma x +\tau\}$ 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 $Q\equiv 0$, 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.
Keywords: q-difference equation, q-series, theta function, Newton-Puiseux diagram.
MSC 2000: 39A13, 39B32, 33D99, 30D05.
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