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指数函数项级数与Hadamard乘积 被引量:1

The series of exponential terms and Hadamard product
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摘要 在本文中,证明了下述定理,设sum sum k=1 to ∞ c_k,sum from j=1 to ∞ d_j均为绝对收敛级数且其各项均不为零.设(λ_k),(μ_j)为两个有界复数序列,每一序列的任惠两项均不相等,并设μ=|μ_1|>|μ_2|≥…。于是,F(z)=e·f(z)·g(z)均为正指型函数,这里f(z),g(z)分别表示f(z)g(z)的相伴函数·表示Hadamard乘积算子.设f(z)的指示图为Ⅰ:令μ_j=ρ_je_j^(iφ)(ρ_j>0,0≤φ_j<2π)(j=1,2,…)。把Ⅰ旋转角—φ_j,并进行比ρj的相似变换得到凸集Ⅰ_j。如果二重序列(λ_kμ_j)(k=1,2,…,j=1,2,…)的任何两项均不相等,则F(z)的指标图Ⅰ 为点集Ⅰ_1,Ⅰ_2,…的凸壳 从此定理出发,在适当条件下,证明F(z)与f(z)有公共Julia线以及二者同为多零点整函数。 In this paper we prove the following.Theorem.Suppose that ∑(c_k)k from 1 to ∞ and ∑(d_j) j from 1 to ∞ are both absolutely convergent with non-zero terms Suppose that (λ_k),(μ_j) are both bounded sequecnes and thatany two terms in each sequence are not equal, ane that μ=|μ_1|>|μ_2|≥|μ_3|≥…。Then f(z) =∑(c_ke~λk^z) k from 1 to ∞ g(z) = ∑(d_je~μj^k) j from 1μ to ∞, F(z) = e~?·f(z)·e(z) all are functions ofpositive exponential type, ehere f(z), g(z) are functions associated with f(z)g(z) respectively and '·' is a Hadamard product operator .Let the indicatordiagram of f(z) be I and μ_j= ρ_je^(iφj)(ρ_j>0,0≤φ_j<2π)(j=1, 2, …), we denote the set obtained from I by rotation an angle-φ_j and under a similar transformation of ratio ρ_j by Ij if any two terms of the double sequence (λ_kμ_j)(k=1,2,…;j=1, 2,…)are not eqnal, then the iudicator diagram I, of F(z) is theconvex hull of the set I_1,I_2,…。 From this theorem we prove that unber suitable couditious, F(z) and f(z)have common lincs of Julia aud they are both integral functions with manyzeros.
作者 莫叶
机构地区 山东大学
出处 《工程数学学报》 CSCD 1989年第3期9-15,共7页 Chinese Journal of Engineering Mathematics
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同被引文献6

  • 1杨定恭.某些P叶函数的Hadamard乘积[J].数学杂志,1983,8(3):217-225.
  • 2吴卓人.关于Sakaguchi函数与Hadamard乘积[J].中国科学,1985,2:104-110.
  • 3爱尔台里.高级超越函数(第二册)[Z].,..
  • 4王安斌.超球级数所定义的整函数[J].牡丹江大学学报,2003,.
  • 5王安斌.超球级数所定义整函数的极大项[J].湖南数学年刊,1998,18(1):32-34.
  • 6王安斌.关于超球级数的增长性质[J].Journal of Mathematical Research and Exposition,2003,23(3):510-514. 被引量:6

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