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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。

In this paper, multivariate Lagrange interpolation problems are systemically discussed,and through amending Cramer Strange Proposition and proving interpolation theory, the geometry construction of the properly posed sets of nodes along the plane algebraic curve is studied, at the same time, a number of practical constructing methods are given.

本文系统阐述了多元Lagrange插值问题，通过修正Cramer奇论并用插值法加以证明，进一步对沿平面代数曲线插值唯一可解点组的几何结构进行深入的研究，同时得到了一些实用性较强的构造方法。

We posed the concept of sufficient intersection about s(1≤s≤n) algebraic hypersurfaces in n-dimensional space and proved the dimension of polynomial space Pm(which denotes the space of all multivariate polynomials of total degree≤m) on the algebraic manifold S=s(f1,…, fs) where f1(X=0,…, f s=0denote s algebraic hypersurfaces of sufficient intersection, then gave a convenient expression for dimension calculation by using the backw ard difference operator.

给出了n维空间中s(1≤s≤n)个代数超曲面充分相交的概念，证明了n元m次多项式空间Pm在充分相交的代数流形S=s(f1，…， fs)（f1=0，…， fs=0表示s个代数超曲面）上的维数，并利用倒差分算子给出一个方便计算的表达式；构造了沿代数流形上插值适定结点组的叠加插值法；证明了在充分相交的代数流形上任意次插值适定结点组的存在性；给出代数流形上插值适定结点组的性质和判定条件。

Answerer: The answerer assures that the group will be able to elaborate their ideas and answer questions posed by other groups or the instructor during the sharing period of the class.

回答者：回答者的保证，该集团将能够阐述自己的想法，并回答问题，所造成的其他团体或导师在分享一段上课。

In order to find a stable approximate solution of inverse problem, the general theory about regularization of ill-posed problems is introduced.

为了求出反问题的稳定近似解，采用了不适定问题正则化的一般理论。

However, it is 28 by the longitudinal arch posed side by side.

但它却是由 28道拱纵向并列构成的。

Arithmetic progression: starting from the second, an increase each by the former posed a constant sequence, such as the odd-numbered 1,3,5,7 ... geometric progression: from the second onwards, each of which is th power of the number of previous ...

算术级数：从第二项起，每一项均由前一项加一个常数所构成的序列，如奇数1，3，5，7…几何级数：从第二项起，每一项是前一项的多少次方。。。

Proudmoore's armada posed a serious threat to the stability of the region.

海军上将普罗德摩尔的舰队对该地区的稳定造成了严重的威胁。

At this point, the armature is not posed, but I'm still hoping to get a sense of the final product even at this early stage.

在这一点上，电枢是不是构成的，但我还是希望得到的感觉最终产品，甚至在此早期阶段。

Having little sympathy for Arminianism, Ryle was equally aware of the threat posed by high Calvinism.

很少同情亚米纽斯主义，赖尔也同样意识到所构成的威胁高加尔文主义。

The shaping method of noncircular part and the tool holder's radial motion characters in noncircular turning process are discussed in detail in the thesis.

论文详细研究了非圆零件的成型方法和加工过程中刀架的径向运动规律。

I have not really liked him,I do not like his this kind of disposition.

我没有真的喜欢他，我不喜欢他的这种性格。