用mittagleffler函数系统对级数的相对增长作了初步说明

O. Mulyava
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引用次数: 0

摘要

对于一个正则收敛的${\Bbb C}$级数$F_{\varrho}(z)=\sum\limits_{n=1}^{\infty} a_n E_{\varrho}(\lambda_nz)$,其中$0<\varrho<+\infty$和$E_{\varrho}(z)=\sum\limits_{k=0}^{\infty}\frac{z^k}{\Gamma(1+k/\varrho)}$是Mittag-Leffler函数,研究了函数$E_{\varrho}^{-1} (M_{F_{\varrho}}(r))$的渐近行为,其中$M_f(r)=\max\{|f(z)|:\,|z|=r\}$。例如,证明了对于所有的$n\ge 1$,如果$\varlimsup\limits_{n\to \infty}\frac{\ln\,\ln\,n}{\ln\,\lambda_n}\le \varrho$和$a_n\ge 0$,则$\varlimsup\limits_{r\to+\infty}\frac{\ln\,E^{-1}_{\varrho}(M_{F_{\varrho}}(r))}{\ln\,r}=\frac{1}{1-\overline{\gamma}\varrho}$,其中$\overline{\gamma}=\varlimsup\limits_{n\to\infty}\frac{\ln\,\lambda_n}{\ln\,\ln\,(1/a_n)}$。对于Laplace-Stiltjes型积分$I_{\varrho}(r) = \int\limits_{0}^{\infty}a(x)E_{\varrho}(r x) d F(x)$,也得到了类似的结果。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
ELEMENTARY REMARKS TO THE RELATIVE GROWTH OF SERIES BY THE SYSTEM OF MITTAG-LEFFLER FUNCTIONS
For a regularly converging in ${\Bbb C}$ series $F_{\varrho}(z)=\sum\limits_{n=1}^{\infty} a_n E_{\varrho}(\lambda_nz)$, where $0<\varrho<+\infty$ and $E_{\varrho}(z)=\sum\limits_{k=0}^{\infty}\frac{z^k}{\Gamma(1+k/\varrho)}$ is the Mittag-Leffler function, it is investigated the asymptotic behavior of the function $E_{\varrho}^{-1} (M_{F_{\varrho}}(r))$, where $M_f(r)=\max\{|f(z)|:\,|z|=r\}$. For example, it is proved that if $\varlimsup\limits_{n\to \infty}\frac{\ln\,\ln\,n}{\ln\,\lambda_n}\le \varrho$ and $a_n\ge 0$ for all $n\ge 1$, then $\varlimsup\limits_{r\to+\infty}\frac{\ln\,E^{-1}_{\varrho}(M_{F_{\varrho}}(r))}{\ln\,r}=\frac{1}{1-\overline{\gamma}\varrho}$, where $\overline{\gamma}=\varlimsup\limits_{n\to\infty}\frac{\ln\,\lambda_n}{\ln\,\ln\,(1/a_n)}$. A similar result is obtained for the Laplace-Stiltjes type integral $I_{\varrho}(r) = \int\limits_{0}^{\infty}a(x)E_{\varrho}(r x) d F(x)$.
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