重新给定

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。

Its encryption method is to express the letter in accordance with the order given in a matrix, and then use the key provided in accordance with the order re-arrangement of the letters in the matrix to form a ciphertext.

它的加密方法是将明文中的字母按照给定的顺序安排在一个矩阵中，然后用根据密钥提供的顺序重新组合矩阵中的字母，形成密文。

The parametric variable of above process was re-expressed by using linear interpolation method in the corresponding time period, and then, the recurrence subscript and the model value in the previous control cycle was inputted to gain the crystal model weight of the corresponding length increment in the end.

先确定晶体重量、时间和长度初始值，再以给定初始时间确定晶体生长阶段，在相应时间段用线性内插法重新表示上述过程参变量，然后引入循环下标及前一控制周期模型量，得到相应长度增量的晶体模型重量。

That calculated the result of raising a given number to a given power to use appropriate using declarations rather than accessing library names through a std:: prefix.

用适当的 using声明，而不用 std：：，访问标准库中名字的方法，重新编写第 2.3节的程序，计算一给定数的给定次幂的结果。

After regular to faintness library matrix ω undertakes bizarre value is decomposed, the basis is given threshold value abandons bizarre value appropriately, express u σ what get matrix ω approximatelyv(kt, undertake to u and v the standard is mixed processing, use protruding to wrap a technology to undertake is not enjoying a sex and normal state sex processing, use the u after processing,σ and v generate new regular library before after subject function is mixed matrix, library of new and tectonic regulation, the function of new regular library and former regular library are approximate, but regular number decreases significantly.

本篇论文剩余部分，请联系九旭论文代写www.991lw.com网站客服qq：论文写作：1210196822 论文发表： 739878127 论文咨询：1210196822②qq.com，联系电话：013155803009。关键词：——模糊规则库；奇异值分解；凸包；约简；梗概：对模糊规则库后件矩阵ω进行奇异值分解，根据给定的阈值适当舍弃奇异值，得到矩阵ω的近似表示uσv(kt，对u和v进行标准和处理，并利用凸包技术进行非负性和正态性处理，利用处理后的u，σ和v生成新规则库的前件隶属函数和后件矩阵，重新构造规则库，新规则库的功能和原规则库近似，但规则总数显著减少。

You must also be willing to reexamine your beliefs about what is the "right approach" to clinical care in a gien situation.

你还应该乐于重新审视你的观点：即在给定的情况下，什么才是临床护理的正确方法。

Given after a lot of data points to find these kind of cluster center, and then reclassified.

详细说明：给定很多数据点之后找这些类的聚类中心，然后重新分类。

A parametric surface was reconstructed at a local area of each sample point, and a believing neighbor area was generated based on a given error. A new parametric surface was recomputed, and a larger believing area was generated. The surfel with largest believing area was reconstructed, and the original model was approximated by the surfel in an as large as possible local area.

该算法在每个采样点附近重建一个函数曲面，根据给定误差得到置信邻域，重新计算函数曲面，得到更大的置信邻域，如此反复迭代，产生一个具有最大置信邻域，并在更大范围内逼近原模型的面元。

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。

The present invention solves these problems by providing the possibility to configure or reconfigure a rate matching attribute for a given transport channel without having to perform unnecessary transport format configurations/reconfigurations.

本发明通过提供配置或重新配置对于给定的输送信道的速率匹配属性的可能性而不必执行不必要的输送格式配置/重新配置来解决这些问题。

With Death guitarist Schuldiner adopting vocal duties, the band made a major impact on the scene.

随着死亡的吉他手Schuldiner接受主唱的职务，乐队在现实中树立了重要的影响。

But he could still end up breakfasting on Swiss-government issue muesli because all six are accused of nicking around 45 million pounds they should have paid to FIFA.

不过他最后仍有可能沦为瑞士政府&议事餐桌&上的一道早餐，因为这所有六个人都被指控把本应支付给国际足联的大约4500万英镑骗了个精光。