Logarithmic singularities and the zeros of the second derivative
The author previously proved that whenever \(f\) is meromorphic of finite order and \(f^{(k)}\) has finitely many zeros for some \(k\geq 2\), then \(f\) has finitely many poles only [see Bull. Lond. Math. Soc. 35, 97--108 (2003; Zbl 1022.30035)]. In this paper, the following improvement is proved: Suppose \(f\) is of finite lower order, and \(f^{(k)}\) has finitely many zeros for some \(k\geq 2\). Assume further that there exists \(M\in (0,\infty)\) so that if \(\zeta \) is a pole of \(f\) of multiplicity \(m_{\zeta }\), then \(m_{\zeta}\leq M+|\zeta |^{M}\). Then \(f\) must be of finite order, having finitely many poles only. The case when \(f^{(k)}\) is replaced by a linear differential polynomial \(L(f)=f^{(k)}+a_{k-1}f^{(k-1)}+\cdots +a_{0}f\) with constant coefficients will be treated as well. As an example of the obtained results, suppose that \(f\) is meromorphic of finite order, and \(L(f)\) has finitely many zeros only. If then the equation \(x{k}+a_{k-1}x^{k-1}+\cdots +a_{0}=0\) has a multiple root, then \(f\) has finitely many poles.
scientific article; zbMATH DE number 5648474