整函数及其微分多项式分担一个多项式

张国威,陈昂

(1.安阳师范学院数学与统计学院,河南 安阳 455000;2.教育部考试中心,北京 100084)

整函数及其微分多项式分担一个多项式

张国威1,陈昂2

(1.安阳师范学院数学与统计学院,河南 安阳 455000;2.教育部考试中心,北京 100084)

将Br¨uck猜想目前得到的几个结论进行了推广,研究了整函数及其微分多项式分担的一个多项式时的问题,并且得到了一个与之相关的复微分方程的解的性质.另外,还得到了一个定理,这个定理改进了一些已知的结果.

整函数;Nevanlinna理论;唯一性;分担值

1 引言及主要结果

本文中使用了Nevanlinna经典理论的一些基本符号和基本定理,关于这部分详细内容可见文献[1-4].令f(z)和g(z)为复平面ℂ上的两个非常数亚纯函数,并且令P(z)为一个多项式或者是一个有限数.用deg P(z)来定义多项式P(z)的级.用f(z)=P(z)⇒g(z)=P(z)来表示:当f(z)-P(z)=0时可推得g(z)-P(z)=0.若有f(z)=P(z)⇒g(z)=P(z)且g(z)=P(z)⇒f(z)=P(z),则将其表示为f(z)=P(z)⇔g(z)=P(z),并且称f(z)和g(z)分担P(z)IM(不计零点重数).如果f(z)-P(z)和g(z)-P(z)有相同的零点并且这些零点的重数相同,则称f(z)和g(z)分担P(z)CM(计零点重数)[1].更进一步,用记号σ(f), ν(f)来定义f(z)的级和超级.下面给出定义:

2 几个引理

3 定理1.3的证明

证明将分两种情况讨论.

情况1如果P(z)是个多项式.如果f不是个超越的整函数,由于方程(1)的解均为多项式,因此由方程(1),可知eP=c是个常数,则ν(f)=σ(eP)=0,易知定理1.3的结论成立.

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Entire functions that share one polynomial with their linear differential polynomials

Zhang Guowei1,Chen Ang2

(1. School of Mathematics and Statistics, Anyang Normal University, Anyang 455000, China; 2. National Education Examinations Authority, Beijing 100084, China)

In this paper,we improve some known results about Bruck's conjecture. We study the problem that entire function and its linear differential polynomial share a polynomial and obtain some properties of solution of the related complex differential equation. Moreover, we get a theorem which improves some known results.

entire functions,Nevan linna theory,uniqueness,share value

O174.5

A

1008-5513(2012)02-0196-05

2010-12-10.

河南省教育厅重点项目(12A 110002).

张国威(1981-),博士,讲师,研究方向：值分布论,复微分方程,复动力系统等.

2010 MSC:30D 35