摘要
本文主要研究了全纯函数的差分算子分担一个值的唯一性问题,并且得到了:若f与g为超级ρ2<1的两个非常数的超越全纯函数,n,k,m为满足n≥5k+4m+13的整数,c是满足f(z+c)-f(z)≠0且g(z+c)-g(z)■0的非零常数,则若f(z)~n(f(z)~m[-1)(f(z+c)-f(z))]^(k)与g(z)~n(g(z)~m[-1)(g(z+c)-g(z))]^(k)IM分担1,则f=tg,其中t为满足t^(n+1)=1与t^m=1的常数.
In this paper, we deal with the value distribution of difference operators of entire functions and obtain that. Let f and g be transcendental entire functions of p2 〈 1 , n,k,m are three integers satisfying n ≥ 5k + 4m + 13 , c is a nonzero complex constant such that f(z + c) - f(z) ≠ 0 and g(z + c) - g(z)≡/0, and if [f(z)n(f(z)m - 1)(f(z +c) -f(z))](k) and [g(z)n(g(z)m - 1)(g(z +c) -g(z))](k) share 1 IM, thenf=tg foraconstanttwith tn+1 = land tm= 1.
出处
《四川大学学报(自然科学版)》
CAS
CSCD
北大核心
2015年第6期1199-1207,共9页
Journal of Sichuan University(Natural Science Edition)
基金
国家自然科学基金(11501068)
关键词
唯一性
全纯函数
差分算子.
Uniqueness
Entire functions
Difference operators.