The growth of entire functions in the terms of generalized orders
Let $\Phi$ be a convex function on $[x_0,+\infty)$ such that$\frac{\Phi(x)}x\to+\infty$, $x\to+\infty$, $f(z)=\sum_{n=0}^\infty a_nz^n$ — a transcendental entire function, let $M(r,f)$ be the maximum modulus of $f$ and let$$\rho_\Phi(f)=\limsup_{r\to +\infty}\frac{\ln\ln M(r,f)}{\ln\Phi(\ln r)},\qua...
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Veröffentlicht in: | Karpats'kì matematinì publìkacìï 2013-01, Vol.4 (1), p.28-35 |
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Hauptverfasser: | , |
Format: | Artikel |
Sprache: | eng |
Online-Zugang: | Volltext |
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Zusammenfassung: | Let $\Phi$ be a convex function on $[x_0,+\infty)$ such that$\frac{\Phi(x)}x\to+\infty$, $x\to+\infty$, $f(z)=\sum_{n=0}^\infty a_nz^n$ — a transcendental entire function, let $M(r,f)$ be the maximum modulus of $f$ and let$$\rho_\Phi(f)=\limsup_{r\to +\infty}\frac{\ln\ln M(r,f)}{\ln\Phi(\ln r)},\quad c_{\Phi}=\limsup_{x\to +\infty}\frac{\ln x}{\ln\Phi(x)},\quad d_{\Phi}=\limsup\limits_{x\to +\infty}\frac{\ln\ln\Phi'_+(x)}{\ln\Phi(x)}.$$It is proved that for every transcendental entire function $f$ the generalized order $\rho_\Phi(f)$ is independent on the arguments of the coefficients $a_n$ (or defined by the sequence $(|a_n|)$) if and only if the inequality $d_{\Phi}\le c_{\Phi}$ holds. |
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ISSN: | 2075-9827 2313-0210 |
DOI: | 10.15330/cmp.4.1.28-35 |