查询词典 hypergeometric
- 与 hypergeometric 相关的网络例句 [注:此内容来源于网络,仅供参考]
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In part II,we consider the recurrence formula of double hypergeometric terms.
在文章的第二部分,我们考虑了双超几何项的递推公式。
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In the frame of quantum mechanics, Schrodinger equation has been deduced to the hypergeometric equati...
在量子力学框架内,利用这一相互作用势成功地将系统的Schrodinger方程化为超几何方程,从而简化了系统本征值和本征态问题的计算和讨论。
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Using the two basic recurrence formulae that double hypergeometric terms satisfy, we get the general form of such representation.
利用双超几何项的两个基本递推关系,我们导出了这种表示的一般形式。
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In 1993, Yen in her doctoral dissertation [45] gave the first a priori estimate of the number for hypergeometric identities, it is extremely large.
1993年,Yen在她的博士论文中第一次对超几何恒等式给出了这个项数的一个先验估计,但是非常大。
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Multivariate hypergeometric functions, multivariate Jacobi polynomials and h-harmonic polynomials connected with root systems and Coxeter groups are introduced.
多元超几何函数,多元Jacobi多项式和h与根系考克斯特集团有关的调和多项式介绍。
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With the help of a transformation formula of basic hypergeometric functions, known identities, and Jacobi triple, we have established some new Rogers-Ramanujan type identities.
2我们对Agarwal与Singh(见文献[1])推导出的一个新的Rogers--Ramanujan型恒等式〓给出了一个新的、更为简单的证法,并得出一系列新的Rogers-Ramanujan型恒等式。
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While, Zeilberger and Wilf [39] showed that proper hypergeometric terms are holonomic and conjectured that the inverse assertion also holds.
所幸的是Zeilberger和Wilf在中证明了正则的超几何项是完整的并猜想其逆命题也成立。
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Peoplr usually think that for hypergeometric experiment, the traials are dependent and the probability of success therefore is not constant.
超几何实验中,每次试行间并不是互相独立,因此有些人以为成功的机率并不像二项实验般是固定不变的。
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From this theorem, we prove the existence of the recurrence for the admissible q-proper hypergeometric sums, and obtain an estimate of the order J of the recurrence.
在此基础上证明admissible q-proper超几何和的迭代关系的存在性,并且得到迭代关系阶数J的一个估计。
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Based on the Collins integral formula, the propagation of generalized hypergeometric beams in paraxial ABCD optical systems is studied and the analytical expressions for complex amplitude at output plane are obtained.
利用柯林积分公式研究了广义超几何光束在ABCD近轴光学系统中的传播行为,得到了光束在输出平面复振幅的解析表达式。
- 推荐网络例句
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We have no common name for a mime of Sophron or Xenarchus and a Socratic Conversation; and we should still be without one even if the imitation in the two instances were in trimeters or elegiacs or some other kind of verse--though it is the way with people to tack on 'poet' to the name of a metre, and talk of elegiac-poets and epic-poets, thinking that they call them poets not by reason of the imitative nature of their work, but indiscriminately by reason of the metre they write in.
索夫农 、森那库斯和苏格拉底式的对话采用的模仿没有一个公共的名称;三音步诗、挽歌体或其他类型的诗的模仿也没有——人们把&诗人&这一名词和格律名称结合到一起,称之为挽歌体诗人或者史诗诗人,他们被称为诗人,似乎只是因为遵守格律写作,而非他们作品的模仿本质。
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The relationship between communicative competence and grammar teaching should be that of the ends and the means.
交际能力和语法的关系应该是目标与途径的关系。
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This is not paper type of business,it's people business,with such huge money involved.
这不是纸上谈兵式的交易,这是人与人的业务,而且涉及金额巨大。