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与导数分担有理函数的整函数

, PP. 44-47

Keywords: 整函数,有理函数,惟一性

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Abstract:

主要证明了以下定理:设f是超越整函数,r是非常数有理函数,k、m是两个不同的正整数,d=(k,m)是k、m的最大公约数.若f,f(k),(fm)cm分担r,那么f=f(d).

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