结点
- 与 结点 相关的网络例句 [注:此内容来源于网络,仅供参考]
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By computing the system reliability, the performance reliability of the multi-state-node acyclic network system can be evaluated.
通过系统可靠性的计算,评价多态结点非循环网络系统的性能可靠性。
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First, we introduce and discuss the various methods of multivariate polynomial interpolation in the literature. Based on this study, we state multivariate Lagrange interpolation over again from algebraic geometry viewpoint:Given different interpolation nodes A1,A2 .....,An in the affine n-dimensional space Kn, and accordingly function values fi(i = 1,..., m), the question is how to find a polynomial p K[x1, x2,...,xn] satisfying the interpolation conditions:where X=(x1,X2,....,xn). Similarly with univariate problem, we have provedTheorem If the monomial ordering is given, a minimal ordering polynomial satisfying conditions (1) is uniquely exsisted.Such a polynomial can be computed by the Lagrange-Hermite interpolation algorithm introduced in chapter 2. Another statement for Lagrange interpolation problem is:Given monomials 1 ,2 ,.....,m from low degree to high one with respect to the ordering, some arbitrary values fi(i= 1,..., m), find a polynomial p, such thatIf there uniquely exists such an interpolation polynomial p{X, the interpolation problem is called properly posed.
文中首先对现有的多元多项式插值方法作了一个介绍和评述,在此基础上我们从代数几何观点重新讨论了多元Lagrange插值问题:给定n维仿射空间K~n中两两互异的点A_1,A_2,…,A_m,在结点A_i处给定函数值f_i(i=1,…,m),构造多项式p∈K[X_1,X_2,…,X_n],满足Lagrange插值条件:p=f_i,i=1,…,m (1)其中X=(X_1,X_2,…,X_n),与一元情形相似地,本文证明了定理满足插值条件(1)的多项式存在,并且按"序"最低的多项式是唯一的,上述多项式可利用第二章介绍的Lagrange-Hermite插值算法求出,Lagrange插值另一种描述是:按序从低到高给定单项式ω_1,ω_2,…,ω_m,对任意给定的f_1,f_2,…,f_m,构造多项式p,满足插值条件:p=sum from i=1 to m=Ai=f_i,i=1,…,m (2)如果插值多项式p存在且唯一,则称插值问题适定。
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By complete discrimination system of polynomial algebraic equation and resultant theory, a symmetry class of quartic systems with stellar node is studied.
利用多项式代数方程的判别系统理论和结式理论,讨论了具有星形结点的一类对称四次系统的代数分类,并对系统进行全局分析。
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An algebraic classification of the systems is given, and the global phase portraits of systems are studied.
值得指出的是,这一方法适用于具有星形结点的一般多项式微分系统。
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Therefore, the new algorithm can be applied in analysing network with a large number of nodes.
实践证明,该算法对于结点数比较大的网络具有较好的适用性。
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For Figure 5-13 the structure consists of three basic unknowns ---- two angular displacement, a separate node line ...
对于图5-13 的结构共有三个基本未知量----两个角位移、一个独立结点线。。。
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E. On the basis of drawing up the discrete rod element, the preprocessor procedure forming the element formation of arbitrary orientation rock bolt is worked out.
e。在编制离散式杆单元的基础上,编制了生成任意岩体锚杆单元信息的前处理程序,从而克服了常规的有限元算法中必须将锚杆布置于单元结点的常规做法。
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A self-learning algorithm is provided for getting optima in a distributed system without a central task assigner.
本文提出了分布式系统中各独立结点根据自身状态和系统反馈进行自适应,以使系统达到最优状态的一种机制。
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By sampling from the traffic flow and averaging non-overlapping blocks of size m from the original series, the variance asymptotic characteristic of the averaged series is obtained, from which this method shows that the splitting, assembling and merging process of the convergence mechanism will not change the Hurst parmeter of the traffic if the queue length of assemble buffer has a finite second-order moment.
在分析光突发交换边缘结点汇聚机制的基础上,提出了一种基于方差时间图的分析方法,利用"时间离散"和"尺度聚集"的思想,通过构造不同聚集级别m的聚集序列,分析得出业务流时间聚集序列的方差渐近性,从理论上证明了在组装器队长具有有限二阶矩的条件下,汇聚机制的分解、组装和合并过程不会改变业务流的Hurst参数。
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In addition, the smaller granule is obtained by selecting those objects of the larger granule that satisfy the atomic formula.
在一个粒度层次中,每个结点都是数据对象的子集,连接大粒度到小粒度的弧用一个原子公式定义。
- 推荐网络例句
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On the other hand, the more important thing is because the urban housing is a kind of heterogeneity products.
另一方面,更重要的是由于城市住房是一种异质性产品。
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Climate histogram is the fall that collects place measure calm value, cent serves as cross axle for a few equal interval, the area that the frequency that the value appears according to place is accumulated and becomes will be determined inside each interval, discharge the graph that rise with post, also be called histogram.
气候直方图是将所收集的降水量测定值,分为几个相等的区间作为横轴,并将各区间内所测定值依所出现的次数累积而成的面积,用柱子排起来的图形,也叫做柱状图。
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You rap, you know we are not so good at rapping, huh?
你唱吧,你也知道我们并不那么擅长说唱,对吧?