关于代数体函数的增长与其例外函数个数的关系

ON THE RELATIONS BETWEEN THE NUMBER OF EXCEPTIONAL FUNCTIONS AND THE GROWTH OF ALGEBROID FUNCTIONS

设ｆ（ｚ）为ｎ值的超越代数体函数，其级为λ（λ＞０）．证明了：如果ｆ（ｚ）具有ｎ＋１个Ｂｏｒｅｌ例外函数，则ｆ（ｚ）是正规增长的，级λ为正整数或无穷．如０＜λ＜∞且不为整数，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例外函数个数。

Let f(z)be a n-valued transcendental algebroid function and p be the number of Borel exceptional functions of f(z),q be the number of Nevanlinna exceptional functions of deficiency one,if the order of f(z) is not an integer or infinite,then,p+q≤n.