关于代数体函数的增长与其例外函数个数的关系

THE GROWTH AND EXCEPTIONAL FUNCTION OF AN ALGEBROID FUNCTION

设f（z）为n值代数体函数，如果f（z）具有n＋1个Borel例外函数，则f（z）是正规增长的，其级为正整数或无穷。如果f（z）的级ρ（0＜ρ＜∞）不为整数，记P为f（z）的Borel例外函数个数，q为f（z）的亏量等于1的Nevanlinna例外值个数，则P＋q≤n．

Let f(y) be an n-values algebroid function suppose it has n + 1-Borel exceptional functions then it is regular groathful and its order is integer or infinite,let p be the number of Borel exceptional functions of f(y) q be the number of Nevanlinna exceptional values of f(z)whose defcective number is 1 if the order p of f(z) is not an integer then ρ + q ≤ n.