亚纯函数及其n阶导数权分担两个值

徐洪焱,易才凤

(1.景德镇陶瓷学院信息学院,江西景德镇 333403;2.江西师范大学数信学院,江西南昌 330027)

亚纯函数及其n阶导数权分担两个值

徐洪焱1,易才凤2

(1.景德镇陶瓷学院信息学院,江西景德镇 333403;2.江西师范大学数信学院,江西南昌 330027)

研究亚纯函数及其n阶导数权分担两个值的唯一性问题.得到了:如果两个非常数亚纯函数f,g分担(∞,∞),f(n)与g(n)分担(1,0),n(≥0)为一整数,且满足△C0:= (4n+6)λ+δn+1(0,f)+δn+1(0,g)+δn+2(0,f)+δn+2(0,g)+δn(0,f)＞4n+10,其中λ=max{min{Θ(∞,f),Θ(0,f)},min{Θ(∞,g),Θ(0,g)}},那么f(n)·g(n)≡1,或者f≡g.该结果改进了前人的有关定理.

亚纯函数;权分担;唯一性

1 引言及主要结果

2 引理

3 定理1.1的证明

这样很容易知道H的极点只可能发生以下几种情况:(1)F,G的重级零点;(2)F,G的重级极点;(3)F,G的1-值点且其重数不相等;(4)F'的零点但非F(F−1)的零点;(5)G'的零点但非G(G−1)的零点.

(i)由(i)的条件,知F,G分担(1,0),(∞,∞),再根据H的表达式很容易知道F,G的极点不是H的极点.

如果H/≡0,由引理2.7,有

类似于情形1也可得到矛盾.

如果a−1=0,则f(n)≡g(n).由此方程可得到f=g+p(z),这里p(z)为多项式,显然T(r,f)=T(r,g)+S(r,f).如果p(z)/≡0,根据引理2.2,有

由(1.1)式很容易得到T(r,f)≤S(r,f),r∈I矛盾.因此p(z)≡0,即f≡g.

综合三种情形,定理1.1(i)得证.下证定理1.1(ii).

(ii)因为f(n)与g(n)分担(1,0),所以F,G分担(1,0).

如果H/≡0,则(3.1)式变为

4 定理1.2的证明

如果H≡0,类似于定理1.1可以证得f(n)·g(n)≡1,或者f≡g,即定理1.2(i)得证.

类似于定理1.1(ii)证明方法并结合(4.1)-(4.4)式及f,g分担(∞,0),也很容易证明.这里略之.

综上所述,则完成了定理1.2的证明.

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Mermorphic functions concerning their n-th derivative sharing two values with weight

XU Hong-yan1,YI Cai-feng2

(1.Department of Informatics and Engineering,Jingdezhen Ceramic Institute,Jingdezhen333403,China; 2.Institute of Mathematics and Informatics,Jiangxi Normal University,Nanchang330027,China)

In this paper,we deal with the uniqueness problem of meromorphic functions concerning their n-th derivative sharing two values with weight and obtain the following theorem:if two nonconstant meromorphic functions f,g share(∞,∞),fn,gnshare(1,0),and satisfy△C0:=(4n+6)λ+δn+1(0,f)+δn+1(0,g)+δn+2(0,f)+ δn+2(0,g)＞4n+10,where λ=max{min{Θ(∞,f),Θ(0,f)},min{Θ(∞,g),Θ(0,g)}},then either f(n)·g(n)≡1 or f≡g,where n(＞0)is an integer.These results extend the former theorems.

meromorphic function,weighted sharing,uniqueness

O174.52

A

1008-5513(2009)04-0777-09

2008-04-21.

国家自然科学基金(10871108),景德镇陶瓷学院科研项目(景陶院发[2009]86号).

徐洪焱(1980-),硕士,研究方向：复分析.

2000MSC:30D30,30D35