整函數及其微分多項式分擔一個多項式

張國威,陳昂

(1.安陽師范學院數學與統計學院,河南 安陽 455000;2.教育部考試中心,北京 100084)

整函數及其微分多項式分擔一個多項式

張國威1,陳昂2

(1.安陽師范學院數學與統計學院,河南 安陽 455000;2.教育部考試中心,北京 100084)

將Br¨uck猜想目前得到的幾個結論進行了推廣,研究了整函數及其微分多項式分擔的一個多項式時的問題,并且得到了一個與之相關的復微分方程的解的性質.另外,還得到了一個定理,這個定理改進了一些已知的結果.

整函數;Nevanlinna理論;唯一性;分擔值

1 引言及主要結果

本文中使用了Nevanlinna經典理論的一些基本符號和基本定理,關于這部分詳細內容可見文獻[1-4].令f(z)和g(z)為復平面?上的兩個非常數亞純函數,并且令P(z)為一個多項式或者是一個有限數.用deg P(z)來定義多項式P(z)的級.用f(z)=P(z)?g(z)=P(z)來表示:當f(z)-P(z)=0時可推得g(z)-P(z)=0.若有f(z)=P(z)?g(z)=P(z)且g(z)=P(z)?f(z)=P(z),則將其表示為f(z)=P(z)?g(z)=P(z),并且稱f(z)和g(z)分擔P(z)IM(不計零點重數).如果f(z)-P(z)和g(z)-P(z)有相同的零點并且這些零點的重數相同,則稱f(z)和g(z)分擔P(z)CM(計零點重數)[1].更進一步,用記號σ(f), ν(f)來定義f(z)的級和超級.下面給出定義:

2 幾個引理

3 定理1.3的證明

證明將分兩種情況討論.

情況1如果P(z)是個多項式.如果f不是個超越的整函數,由于方程(1)的解均為多項式,因此由方程(1),可知eP=c是個常數,則ν(f)=σ(eP)=0,易知定理1.3的結論成立.

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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.

河南省教育廳重點項目(12A 110002).

張國威(1981-),博士,講師,研究方向：值分布論,復微分方程,復動力系統等.

2010 MSC:30D 35