数学与应用数学专业论文英文文献翻译

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数学专业外文文献翻译

第3章最小均方算法3．1 引言最小均方(LMS ,least-mean-square)算法是一种搜索算法，它通过对目标函数进行适当的调整[1]—[2],简化了对梯度向量的计算。

由于其计算简单性，LMS 算法和其他与之相关的算法已经广泛应用于白适应滤波的各种应用中[3]-[7]。

为了确定保证稳定性的收敛因子范围，本章考察了LMS 算法的收敛特征。

研究表明，LMS 算法的收敛速度依赖于输入信号相关矩阵的特征值扩展[2]—[6]。

在本章中，讨论了LMS 算法的几个特性，包括在乎稳和非平稳环境下的失调[2]—[9]和跟踪性能[10]-[12]。

本章通过大量仿真举例对分析结果进行了证实。

在附录B 的B ．1节中，通过对LMS 算法中的有限字长效应进行分析，对本章内容做了补充。

LMS 算法是自适应滤波理论中应用最广泛的算法，这有多方面的原因。

LMS 算法的主要特征包括低计算复杂度、在乎稳环境中的收敛性、其均值无俯地收敛到维纳解以及利用有限精度算法实现时的稳定特性等。

3．2 LMS 算法在第2章中，我们利用线性组合器实现自适应滤波器，并导出了其参数的最优解，这对应于多个输入信号的情形。

该解导致在估计参考信号以d()k 时的最小均方误差。

最优(维纳)解由下式给出：10w R p-= (3.1)其中，R=E[()x ()]Tx k k 且p=E[d()x()] k k ，假设d()k 和x()k 联合广义平稳过程。

如果可以得到矩阵R 和向量p 的较好估计，分别记为()R k ∧和()p k ∧，则可以利用如下最陡下降算法搜索式(3．1)的维纳解：w(+1)=w()-g ()w k k k μ∧w()(()()w())k p k R k k μ∧∧=-＋２ (3.2) 其中，k ＝0，1，2，…,g ()w k ∧表示目标函数相对于滤波器系数的梯度向量估计值。

一种可能的解是通过利用R 和p 的瞬时估计值来估计梯度向量，即 ()x()x ()TR k k k ∧=()()x()p k d k k ∧= (3.3) 得到的梯度估计值为()2()x()2x()x ()()T w g k d k k k k w k ∧=-+2x()(()x ()())Tk d k k w k =-+ 2()x()e k k =- (3.4)注意，如果目标函数用瞬时平方误差2()e k 而不是MSE 代替，则上面的梯度估计值代表了真实梯度向量，因为2010()()()()2()2()2()()()()Te k e k e k e k e k e k e k w w k w k w k ⎡⎤∂∂∂∂=⎢⎥∂∂∂∂⎣⎦2()x()e k k =-()w g k ∧= (3.5)由于得到的梯度算法使平方误差的均值最小化．因此它被称为LMS 算法，其更新方程为 (1)()2()x()w k w k e k k μ+=+ (3.6) 其中，收敛因子μ应该在一个范围内取值，以保证收敛性。

数学与应用数学专业论文英文文献翻译

a . , “”.3.1a . , , ),(11y x ),(11y x , )(21x x ≠ , ax. , .. n , ),(k k y x ，n k ,,2,1Λ=, k x ’s, a xn. n , , , ,1-n . , , ..n k y x P k k ,,2,1,)(Λ==，, , , . ..∑∏⎪⎪⎭⎫⎝⎛--=≠k k k j jk jy x x x x x P )( n 1-n , a 1-n . )(x P k x x = , k . ,k , k y ., :0:3; [-5 -6 -1 16];([])0 1 2 3 -5 -6 -1 16)16()6()2)(1()1()2()3)(1()6()2()3)(2()5()6()3)(2)(1()(--+----+---+-----=x x x x x x x x x x x x x P, . , . , 2,1,0=x 3, .. ,523--x xx a .,n n n n c x c x c x c x P ++++=---12211)(Λ, , a⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡------n n n n n n nn n n n y y y c c c x x x x x x x x x M M ΛΛΛΛΛΛΛ21212122212121111111 V a .jn kj k x v -=, a , .. ,V = (x)V = 0 0 0 1 1 1 1 1 8 4 2 1 27 9 3 1 c = V\y’c =1.0000 0.0000 -2.0000 -5.0000, 523--x x .k x . a . , a a . a , ., .v = (), x y , . , u , a . , v, u))(,,()(k u y x terp in k v =, , . u .v = () n = (x); v = ((u)); k = 1w = ((u)); j = [11 1]w = ((j))(x(k)(j)).*w;v = v + w*y(k);, a .u = -.25:.01:3.25;v = ();(,’o’,’-’)3.1.3.1.. ,= (’x’)P = ()(P)-5 (-1/3 x + 1)(-1/2 x + 1)( + 1) - 6 (-1/2 x + 3/2)( + 2)x -1/2 ( + 3)(x - 1)x + 16/3 (x - 2)(1/2 x - 1/2)xa .P = (P)PP =x^3-2*5, a .x = 1:6;y = [16 18 21 17 15 12]; ([x; y])u = .75:.05:6.25;v = ();(,’o’,’-’);1 2 3 4 5 616 18 21 17 15 12 3.2.3.2., , . , , . . a , . .第三章插值多项式插值就是定义一个在特定点取给定值得函数的过程。

数学专业毕业设计文献翻译

附件1：外文资料翻译译文第1章预备知识双曲守恒律系统是应用在出现在交通流，弹性理论，气体动力学，流体动力学等等的各种各样的物理现象的非常重要的数学模型。

一般来说，古典解非线性双曲方程柯西问题解的守恒定律仅仅适时局部存在于初始数据是微小和平滑的.这意味着震波在解决方案里相配的大量时间里出现。

既然解是间断的而且不满足给定的传统偏微分方程式，我们不得不去研究广义的解决方法，或者是满足分布意义的方程式的函数.我们考虑到如下形式的拟线性系统, (1.0.1)这里是代表物理量密度的未知矢量向量，是给定表示保守项的适量函数，这些方程式通常被叫做守恒律.让我们假设一下，是(1.0.1)在初始数据. (1.0.2)下的传统解。

使成为消失在紧凑子集外的函数的一类。

我们用乘以(1.0.1)并且使的部分，得到. (1.0.3)定义1.0.1 有，,有界函数叫做在以原始数据为边界条件下，(1.0.1)初值问题的一个弱解，在(1.0.3)适用于所有.非线性系统守恒理论的一个重要方面是这些方程解的存在疑问性.它正确的帮助解答在手边的已经建立的自然现象的模型的问题，而且如果在问题是适定的.为了得到一个总体的弱解或者一个考虑到双曲守恒律的普遍的解，一个为了在(1.0.1)右手边增加一个微小抛物摄动限：(1.0.4)在这是恒定的.我们首先应该得到一个关于柯西问题(1.0.4),(1.0.2)对于任何一个依据下列抛物方程的一般理论存在的的解的序列：定理1.0.2 (1)对于任意存在的, (1.0.4)的柯西问题在有界可测原始数据(1.0.2)对于无限小的总有一个局部光滑解，仅依赖于以原始数据的.(2)如果解有一个推理的估量对于任意的,于是解在上存在.(3)解满足：如果.( 4)特别的，如果在(1.0.4)系统中的一个解以(1.0.5)形式存在，这里是在上连续函数，,如果(1.0.6) 这里是一个正的恒量，而且当变量趋向无穷大或者趋向于0时，趋向于0.证明.在(1)中的局部存在的结果能简单的通过把收缩映射原则应用到解的积分表现得到，根据半线性抛物系统标准理论.每当我们有一个先验的局部解的评估，明显的本地变量一步一步扩展到，因为逐步变量依据基准.取得局部解的过程清晰地表现在(3)中的解的行为.定理1.0.2的(1)-(3)证明的细节在[LSU,Sm]看到.接下来是Bereux和Sainsaulieu未发表的证明(cf. [Lu9, Pe])我们改写方程式(1.0.5)如下：(1.0.7)当.然后. (1.0.8) 以初值(1.0.8)的解能被格林函数描写：. (1.0.9)由于，,(1.0.9)转化为.(1.0.10)因此对于任意一个，有一个正的下界.在定理1.0.2中获得的解叫做粘性解.然后我们有了粘性解的序列,,如果我们再假如是在关于参数的空间上一致连续，即存在子序列（仍被标记）如下, 在上弱对应(1.0.11) 而且有子序列如下，弱对应(1.0.12) 在习惯于成长适当成长性.如果，a.e.，(1.0.13)然后明显的是(1.01)使在(1.0.4)的趋近于0的一个初始值(1.0.2)的一个弱解.我们如何得到弱连续(1.0.13)的关于粘度解的序列的非线性通量函数？补偿密实度原理就回答了这个问题.为什么这个理论叫补偿密实度？粗略的讲，这个术语源自于下列结果：如果一个函数序列满足(1.0.14)与下列之一或者(1.0.15) 当趋近于0时弱相关，总之，不紧密.然而，明显的，任何一个在(1.0.15)中的弱紧密度能补偿使其成为的紧密度.事实上，如果我们将其相加，得到(1.0.16)当趋近于0时弱相关,与(1.0.14)结合意味着的紧密度.在这本书里，我们的目标是介绍一些补偿紧密度方法对标量守恒律的应用，和一些特殊的两到三个方程式系统.此外，一些具有松弛扰动参量的物理系统也被考虑进来。

数学与应用数学专业概率论的发展大学毕业论文英文文献翻译及原文

毕业设计（论文）外文文献翻译文献、资料中文题目：概率论的发展文献、资料英文题目：The development of probabilitytheory文献、资料来源：文献、资料发表（出版）日期：院（部）：专业：数学与应用数学班级：姓名：学号：指导教师：翻译日期： 2017.02.14毕业论文（设计）英文文献翻译外文文献The development of probability theorySummaryThis paper consist therefore of two parts: The first is concerned with the development of the calyculus of chance before Bernoulli in order to provide a background for the achievement of Ja kob Bernoulli and will emphasize especially the role of Leibniz. The second part deals with the relationship between Leibniz add Bernoulli an d with Bernoulli himself, particularly with the question how it came about that he introduced probability into mathematics.First some preliminary remarks:Ja kob Bernoulli is of special interest to me, because he is the founder of a mathematical theory of probability. That is to say that it is mainly due to him that a concept of probability was introduced into a field of mathematics.TextMathematics could call the calculus of games of chance before Bernoulli. This has another consequence that makes up for a whole programme: The mathematical tools of this calculus should be applied in the whole realm of areas which used a concept of probability. In other words，the Bernoullian probability theory should be applied not only togames of chance and mortality questions but also to fields like jurisprudence, medicine, etc.My paper consists therefore of two parts: The first is concerned with the development of the calculus of chance before Bernoulli in order to provide a background or the achievements of Ja kob Bernoulli and will emphasize especially the role of Leibniz. The second part deals with the relationship between Leibniz and Bernoulli and Bernoulli himself, particularly with the question how it came about that he introduced probability into mathematics.Whenever one asks why something like a calculus of probabilities arose in the 17th century, one already assumes several things: for instance that before the 17th century it did not exist, and that only then and not later did such a calculus emerge. If one examines the quite impressive literature on the history of probability, one finds that it is by no means a foregone conclusion that there was no calculus of probabilities before the 17th century. Even if one disregards numerous references to qualitative and quantitative inquiries in antiquity and among the Arabs and the Jews, which, rather freely interpreted, seem to suggest the application of a kind of probability-concept or the use of statistical methods, it is nevertheless certain that by the end of the 15th century an attempt was being interpreted.People made in some arithmetic works to solve problems of games of chance by computation. But since similar problems form the major part of the early writings on probability in the 17th century, one may be induced to ask why then a calculus of probabilities did not emerge in the late 15th century. One could say many things: For example, that these early game calculations in fact represent one branch of a development which ultimately resulted in a calculus of probabilities. Then why shouldn't one place the origin of the calculus of probabilities before the 17th after all? Quite simply because a suitable concept of probability was missing from the earlier computations. Once the calculus of probabilities had beendeveloped, it became obvious that the older studies of games of chance formed a part of the new discipline.We need not consider the argument that practically all the solutions of problems of games of chance proposed in the 15th and 16th centuries could have been viewed as inexact, and thus at best as approximate, by Fermat in the middle of the 17th century, that is, before the emergence of a calculus of probabilities.The assertion that no concept of probability was applied to games of chance up to the middle of the 17th century can mean either that there existed no concept of probability (or none suitable), or that though such a concept existed it was not applied to games of chance. I consider the latter to be correct, and in this I differ from Hacking, who argues that an appropriate concept of probability was first devised in the 17th century.I should like to mention that Hacking(Mathematician)and I agree ona number of points. For instance, on the significance of the legal tradition and of the practical ("-low") sciences: Hacking makes such factors responsible for the emergence of a new concept of probability, suited to a game calculus, while perceive them as bringing about the transfer and quantification of a pre-existent probability-concept.译文概率论的发展作者；龙腾施耐德摘要本文由两部分构成：首先是提供了一个为有关与发展雅各布 - 前伯努利相关背景，雅各布对数学做出了不可磨灭的贡献。

数学与应用数学英文文献及翻译

（外文翻译从原文第一段开始翻译，翻译了约2000字）勾股定理是已知最早的古代文明定理之一。

这个著名的定理被命名为希腊的数学家和哲学家毕达哥拉斯。

毕达哥拉斯在意大利南部的科托纳创立了毕达哥拉斯学派。

他在数学上有许多贡献，虽然其中一些可能实际上一直是他学生的工作。

毕达哥拉斯定理是毕达哥拉斯最著名的数学贡献。

据传说，毕达哥拉斯在得出此定理很高兴，曾宰杀了牛来祭神，以酬谢神灵的启示。

后来又发现2的平方根是不合理的，因为它不能表示为两个整数比，极大地困扰毕达哥拉斯和他的追随者。

他们在自己的认知中，二是一些单位长度整数倍的长度。

因此2的平方根被认为是不合理的，他们就尝试了知识压制。

它甚至说，谁泄露了这个秘密在海上被淹死。

毕达哥拉斯定理是关于包含一个直角三角形的发言。

毕达哥拉斯定理指出，对一个直角三角形斜边为边长的正方形面积，等于剩余两直角为边长正方形面积的总和图1根据勾股定理，在两个红色正方形的面积之和A和B，等于蓝色的正方形面积，正方形三区因此，毕达哥拉斯定理指出的代数式是：对于一个直角三角形的边长a，b和c，其中c是斜边长度。

虽然记入史册的是著名的毕达哥拉斯定理，但是巴比伦人知道某些特定三角形的结果比毕达哥拉斯早一千年。

现在还不知道希腊人最初如何体现了勾股定理的证明。

如果用欧几里德的算法使用，很可能这是一个证明解剖类型类似于以下内容：六^维-论~文.网“一个大广场边a+ b是分成两个较小的正方形的边a和b分别与两个矩形A和B，这两个矩形各可分为两个相等的直角三角形，有相同的矩形对角线c。

四个三角形可安排在另一侧广场a+b中的数字显示。

在广场的地方就可以表现在两个不同的方式：1。

由于两个长方形和正方形面积的总和：2。

作为一个正方形的面积之和四个三角形：现在，建立上面2个方程，求解得因此，对c的平方等于a和b的平方和（伯顿1991）有许多的勾股定理其他证明方法。

一位来自当代中国人在中国现存最古老的含正式数学理论能找到对Gnoman和天坛圆路径算法的经典文本。

数学与应用数学专业英语翻译

3-c (P38/P33)集合论在讨论任何数学的分支，如分析、代数、几何，最好使用符号和集合理论的专业术语。

在19世纪后期被布勒和康托尔发展的这个学科，已对20世纪数学发展有了深远的影响。

它具有统一的许多看似已断开连接的想法，并有助用优雅有系统的方式减少许多逻辑基础上的数学概念。

集合论彻底处理需要漫长的讨论，我们认为超出了这本书的范围。

幸运的是，基础的概念很少涉及数字，通过非正式的讨论他可能已经发展成为集合论的方法和想法的可操作的知识。

实际上，我们将讨论不是像一个有关精确术语的协议一样的一个新的理论，我们将或多或少的应用到熟悉的想法。

在数学中，'"集'"一词用于表示作为单个实体的对象的聚集，集合也被称为'"群'"部落，'一伙人'"'"，团队'，和‘选举团’这是所有集合的示例。

集合中的单个对象称为元素或集合的成员，他们都称为属于或包含于集合中。

反过来，集合包含或由其元素构成。

我们应该主要对数学对象的集合感兴趣：数集、曲线集、几何图形集等等。

在许多应用程序，通过对集合中的各个对象的性质进行没什么特别性质的假定去处理集合是方便快捷的。

这些称为抽象集。

抽象集理论已经发展到处理任意对象的集合，这种集合，并从这种普遍性理论派生其权力。

9-c（P80/P66）积分的基本属性由积分的定义，它可以推导出以下属性，证明有在第1.27 部分中。

定理1，线性性质对被积函数，如果f 和g 都在[a、b] 上可积，那么对于每一对常量c1 和c2有---。

而且有----------（积分等式成立）注解：利用数学归纳法，线性属性可推广，如下所示：如果--在【a，b】上可积那么对于所有实数-----有-----且---------（积分等式成立）定理2.区间可加性，如果以下三个积分的两个存在，那么第三个也存在，并且我们有...注意：在特别是，如果 f 是在[a、c]上是单调的和在[c、b]上也是单调的，且这两个积分存在，则【a，b】区间上的积分也存在且与其他两个积分的和相等定理3.平移不变性，如果f 在区间[a、b]上可积，那么对于任意实数c，都有… …。

数学与应用数学专业论文英文文献翻译

数学与应用数学专业论文英文文献翻译Chapter 3InterpolationInterpolation is the process of defining a function that takes on specified values at specified points. This chapter concentrates on two closely related interpolants, the piecewise cubic spline and the shape-preserving piecewise cubic named “pchip”.3.1 The Interpolating PolynomialWe all know that two points determine a straight line. More precisely, any two points in the plane, ),(11y x and ),(11y x , with )(21x x ≠ , determine a unique first degree polynomial in x whose graph passes through the two points. There are many different formulas for the polynomial, but they all lead to the same straight line graph.This generalizes to more than two points. Given n points in the plane, ),(k k y x ，n k ,,2,1 =, with distinct k x ’s, there is aunique polynomial in x of degree less than n whose graph passes through the points. It is easiest to remember that n , the number of data points, is also the number of coefficients, although some of the leading coefficients might be zero, so the degree might actually be less than 1-n . Again, there are many different formulas for the polynomial, but they all define the same function.This polynomial is called the interpolating polynomial because it exactly re- produces the given data.n k y x P k k ,,2,1,)( ==，Later, we examine other polynomials, of lower degree, that only approximate the data. They are not interpolating polynomials.The most compact representation of the interpolating polynomial is the La- grange form.∑∏⎪⎪⎭⎫ ⎝⎛--=≠k k k j j k j y x x x x x P )( There are n terms in the sum and 1-n terms in each product, so this expression defines a polynomial of degree at most 1-n . If )(x P is evaluated at k x x = , all the products except the k th are zero. Furthermore, the k th product is equal to one, so the sum is equal to k y and theinterpolation conditions are satisfied.For example, consider the following data set:x=0:3;y=[-5 -6 -1 16];The commanddisp([x;y])displays0 1 2 3-5 -6 -1 16 The Lagrangian form of the polynomial interpolating this data is)16()6()2)(1()1()2()3)(1()6()2()3)(2()5()6()3)(2)(1()(--+----+---+-----=x x x x x x x x x x x x x P We can see that each term is of degree three, so the entire sum has degree at most three. Because the leading term does not vanish, the degree is actually three. Moreover, if we plug in 2,1,0=x or 3, three of the terms vanish and the fourth produces the corresponding value from the data set.Polynomials are usually not represented in their Lagrangian form. More fre- quently, they are written as something like523--x xThe simple powers of x are called monomials and this form of a polynomial is said to be using the power form.The coefficients of an interpolating polynomial using its power form,n n n n c x c x c x c x P ++++=---12211)(can, in principle, be computed by solving a system of simultaneous linear equations⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡------n n n n n n n n n n n y y y c c c x x x x x x x x x 21212122212121111111 The matrix V of this linear system is known as a Vandermonde matrix. Its elements arej n kj k x v -=, The columns of a Vandermonde matrix are sometimes written in the opposite order, but polynomial coefficient vectors in Matlab always have the highest power first.The Matlab function vander generates Vandermonde matrices. For our ex- ample data set,V = vander(x)generatesV =0 0 0 11 1 1 18 4 2 127 9 3 1Thenc = V\y’computes the coefficientsc =1.00000.0000-2.0000-5.0000In fact, the example data was generated from the polynomial 523--x x .One of the exercises asks you to show that Vandermonde matrices are nonsin- gular if the points k x are distinct. But another one of theexercises asks you to show that a Vandermonde matrix can be very badly conditioned. Consequently, using the power form and the Vandermonde matrix is a satisfactory technique for problems involving a few well-spaced and well-scaled data points. But as a general-purpose approach, it is dangerous.In this chapter, we describe several Matlab functions that implement various interpolation algorithms. All of them have the calling sequencev = interp(x,y,u)The first two input arguments, x and y, are vectors of the same length that define the interpolating points. The third input argument, u, is a vector of points where the function is to be evaluated. The output, v, is the same length as u and has elements ))xterpyvuink(k(,,)(Our first such interpolation function, polyinterp, is based on the Lagrange form. The code uses Matlab array operations to evaluate the polynomial at all the components of u simultaneously.function v = polyinterp(x,y,u)n = length(x);v = zeros(size(u));for k = 1:nw = ones(size(u));for j = [1:k-1 k+1:n]w = (u-x(j))./(x(k)-x(j)).*w;endendv = v + w*y(k);To illustrate polyinterp, create a vector of densely spaced evaluation points.u = -.25:.01:3.25;Thenv = polyinterp(x,y,u);plot(x,y,’o’,u,v,’-’)creates figure 3.1.Figure 3.1. polyinterpThe polyinterp function also works correctly with symbolic variables. For example, createsymx = sym(’x’)Then evaluate and display the symbolic form of the interpolating polynomial withP = polyinterp(x,y,symx)pretty(P)produces-5 (-1/3 x + 1)(-1/2 x + 1)(-x + 1) - 6 (-1/2 x + 3/2)(-x + 2)x-1/2 (-x + 3)(x - 1)x + 16/3 (x - 2)(1/2 x - 1/2)xThis expression is a rearrangement of the Lagrange form of the interpolating poly- nomial. Simplifying the Lagrange form withP = simplify(P)changes P to the power formP =x^3-2*x-5Here is another example, with a data set that is used by the other methods in this chapter.x = 1:6;y = [16 18 21 17 15 12];disp([x; y])u = .75:.05:6.25;v = polyinterp(x,y,u);plot(x,y,’o’,u,v,’-’);produces1 2 3 4 5 616 18 21 17 15 12creates figure 3.2.Figure 3.2. Full degree polynomial interpolation Already in this example, with only six nicely spaced points, we canbegin to see the primary difficulty with full-degree polynomial interpolation. In between the data points, especially in the first and last subintervals, the function shows excessive variation. It overshoots the changes in the data values. As a result, full- degree polynomial interpolation is hardly ever used for data and curve fitting. Its primary application is in the derivation of other numerical methods.第三章插值多项式插值就是定义一个在特定点取给定值得函数的过程。

数学专业英语论文(含中文版)

Some Properties of Solutions of Periodic Second OrderLinear Differential Equations1. Introduction and main resultsIn this paper, we shall assume that the reader is familiar with the fundamental results and the stardard notations of the Nevanlinna's value distribution theory of meromorphic functions [12, 14, 16]. In addition, we will use the notation )(f σ，)(f μand )(f λto denote respectively the order of growth, the lower order of growth and the exponent of convergence of the zeros of a meromorphic function f ，)(f e σ（[see 8]），the e-type order of f(z), is defined to berf r T f r e ),(log lim)(+∞→=σSimilarly, )(f e λ，the e-type exponent of convergence of the zeros of meromorphic function f , is defined to berf r N f r e )/1,(loglim)(++∞→=λWe say that )(z f has regular order of growth if a meromorphic function )(z f satisfiesrf r T f r log ),(log lim)(+∞→=σWe consider the second order linear differential equation0=+''Af fWhere )()(z e B z A α=is a periodic entire function with period απω/2i =. The complex oscillation theory of (1.1) was first investigated by Bank and Laine [6]. Studies concerning (1.1) have een carried on and various oscillation theorems have been obtained [2{11, 13, 17{19].When )(z A is rational in ze α，Bank and Laine [6] proved the following theoremTheorem A Let )()(z e B z A α=be a periodic entire function with period απω/2i = and rational in zeα.If )(ζB has poles of odd order at both ∞=ζ and 0=ζ, then for everysolution )0)((≠z f of (1.1), +∞=)(f λBank [5] generalized this result: The above conclusion still holds if we just suppose that both ∞=ζ and 0=ζare poles of )(ζB , and at least one is of odd order. In addition, the stronger conclusion)()/1,(l o g r o f r N ≠+ (1.2) holds. When )(z A is transcendental in ze α, Gao [10] proved the following theoremTheorem B Let ∑=+=p j jj b g B 1)/1()(ζζζ,where )(t g is a transcendental entire functionwith 1)(<g σ, p is an odd positive integer and 0≠p b ，Let )()(ze B z A =.Then anynon-trivia solution f of (1.1) must have +∞=)(f λ. In fact, the stronger conclusion (1.2) holds.An example was given in [10] showing that Theorem B does not hold when )(g σis any positive integer. If the order 1)(>g σ , but is not a positive integer, what can we say? Chiang and Gao [8] obtained the following theoremsTheorem 1 Let )()(ze B z A α=,where )()/1()(21ζζζg g B +=,1g and 2g are entire functions with 2g transcendental and )(2g μnot equal to a positive integer or infinity, and 1g arbitrary. If Some properties of solutions of periodic second order linear differential equations )(z f and )2(i z f π+are two linearly independent solutions of (1.1), then+∞=)(f e λOr2)()(121≤+--g f e μλWe remark that the conclusion of Theorem 1 remains valid if we assume )(1g μ isnotequaltoapositiveintegerorinfinity,and2g arbitraryand stillassume )()/1()(21ζζζg g B +=,In the case when 1g is transcendental with its lower order not equal to an integer or infinity and 2g is arbitrary, we need only to consider )/1()()/1()(*21ηηηηg g B B +==in +∞<<η0,ζη/1<.Corollary 1 Let )()(z e B z A α=,where )()/1()(21ζζζg g B +=,1g and 2g are entire functions with 2g transcendental and )(2g μno more than 1/2, and 1g arbitrary.(a) If f is a non-trivial solution of (1.1) with +∞<)(f e λ,then )(z f and )2(i z f π+are linearly dependent.(b)If 1f and 2f are any two linearly independent solutions of (1.1), then +∞=)(21f f e λ.Theorem 2 Let )(ζg be a transcendental entire function and its lower order be no more than 1/2. Let )()(z e B z A =,where ∑=+=p j jj b g B 1)/1()(ζζζand p is an odd positive integer,then +∞=)(f λ for each non-trivial solution f to (1.1). In fact, the stronger conclusion (1.2) holds.We remark that the above conclusion remains valid if∑=--+=pj jjbg B 1)()(ζζζWe note that Theorem 2 generalizes Theorem D when )(g σis a positive integer or infinity but2/1)(≤g μ. Combining Theorem D with Theorem 2, we haveCorollary 2 Let )(ζg be a transcendental entire function. Let )()(z e B z A = where ∑=+=p j jj b g B 1)/1()(ζζζand p is an odd positive integer. Suppose that either (i) or (ii)below holds:(i) )(g σ is not a positive integer or infinity; (ii) 2/1)(≤g μ;then +∞=)(f λfor each non-trivial solution f to (1.1). In fact, the stronger conclusion (1.2) holds.2. Lemmas for the proofs of TheoremsLemma 1 ([7]) Suppose that 2≥k and that 20,.....-k A A are entire functions of period i π2,and that f is a non-trivial solution of0)()()(2)(=+∑-=k i j j z yz A k ySuppose further that f satisfies )()/1,(logr o f r N =+; that 0A is non-constant and rationalin ze ，and that if 3≥k ，then 21,.....-k A A are constants. Then there exists an integer qwith k q ≤≤1 such that )(z f and )2(i q z f π+are linearly dependent. The same conclusionholds if 0A is transcendental in ze ，andf satisfies )()/1,(logr o f r N =+，and if 3≥k ，thenas ∞→r through a set1L of infinite measure, wehave )),((),(j j A r T o A r T =for 2,.....1-=k j .Lemma 2 ([10]) Let )()(z e B z A α=be a periodic entire function with period 12-=απωi and betranscendental in z e α, )(ζB is transcendental and analytic on +∞<<ζ0.If )(ζB has a pole of odd order at ∞=ζ or 0=ζ(including those which can be changed into this case by varying the period of )(z A and Eq . (1.1) has a solution 0)(≠z f which satisfies )()/1,(logr o f r N =+,then )(z f and )(ω+z f are linearly independent. 3. Proofs of main resultsThe proof of main results are based on [8] and [15].Proof of Theorem 1 Let us assume +∞<)(f e λ.Since )(z f and )2(i z f π+are linearly independent, Lemma 1 implies that )(z f and )4(i z f π+must be linearly dependent. Let )2()()(i z f z f z E π+=,Then )(z E satisfies the differential equation222)()()(2))()(()(4z E cz E z E z E z E z A -''-'=, (2.1)Where 0≠c is the Wronskian of 1f and 2f (see [12, p. 5] or [1, p. 354]), and )()2(1z E c i z E =+πor some non-zero constant 1c .Clearly, E E /'and E E /''are both periodic functions with period i π2,while )(z A is periodic by definition. Hence (2.1) shows that 2)(z E is also periodic with period i π2.Thus we can find an analytic function )(ζΦin +∞<<ζ0,so that )()(2z e z E Φ=Substituting this expression into (2.1) yieldsΦΦ''+ΦΦ'-ΦΦ'+Φ=-2222)(43)(4ζζζζcB (2.2)Since both )(ζB and )(ζΦare analytic in }{+∞<<=ζζ1:*C ,the V aliron theory [21, p. 15] gives their representations as)()()(ζζζζb R B n =，)()()(11ζφζζζR n =Φ，（2.3）where n ，1n are some integers, )(ζR and )(1ζR are functions that are analytic and non-vanishing on }{*∞⋃C ，)(ζb and )(ζφ are entire functions. Following the same arguments as used in [8], we have),(),()/1,(),(φρρφρφρS b T N T ++=，（2.4） where )),((),(φρφρT o S =.Furthermore, the following properties hold [8])}(),(max{)()()(222E E E E f eL eR e e e λλλλλ===,)()()(12φλλλ=Φ=E eR ,Where )(2E eR λ(resp, )(2E eL λ) is defined to berE r N R r )/1,(loglim2++∞→(resp, rE r N R r )/1,(loglim2++∞→),Some properties of solutions of periodic second order linear differential equationswhere )/1,(2E r N R (resp. )/1,(2E r N L denotes a counting function that only counts the zeros of 2)(z E in the right-half plane (resp. in the left-half plane), )(1Φλis the exponent of convergence of the zeros of Φ in *C , which is defined to beρρλρlog )/1,(loglim)(1Φ=Φ++∞→NRecall the condition +∞<)(f e λ,we obtain +∞<)(φλ.Now substituting (2.3) into (2.2) yields+'+'+-'+'++=-21112111112)(43)()()()()(4φφζζφφζζζφζζζζζR R n R R n R cb R n n)222)1((1111111112112φφφφζφφζφφζζζ''+''+'''+''+'+'+-R R R R R n R R n n n （2.5）Proof of Corollary 1 We can easily deduce Corollary 1 (a) from Theorem 1 .Proof of Corollary 1 (b). Suppose 1f and 2f are linearlyindependentand +∞<)(21f f e λ,then +∞<)(1f e λ,and +∞<)(2f e λ.We deduce from the conclusion of Corollary 1 (a) that )(z f j and )2(i z f j π+are linearly dependent, j = 1; 2.Let)()()(21z f z f z E =.Then we can find a non-zero constant2c suchthat )()2(2z E c i z E =+π.Repeating the same arguments as used in Theorem 1 by using the fact that 2)(z E is also periodic, we obtain2)()(121≤+--g E e μλ,a contradiction since 2/1)(2≤g μ.Hence +∞=)(21f f e λ.Proof of Theorem 2 Suppose there exists a non-trivial solution f of (1.1) that satisfies)()/1,(logr o f r N =+. We deduce 0)(=f e λ, so )(z f and )2(i z f π+ are linearlydependent by Corollary 1 (a). However, Lemma 2 implies that )(z f and )2(i z f π+are linearly independent. This is a contradiction. Hence )()/1,(log r o f r N ≠+holds for each non-trivial solution f of (1.1). This completes the proof of Theorem 2.Acknowledgments The authors would like to thank the referees for helpful suggestions to improve this paper. References[1] ARSCOTT F M. Periodic Di®erential Equations [M]. The Macmillan Co., New Y ork, 1964. [2] BAESCH A. On the explicit determination of certain solutions of periodic differentialequations of higher order [J]. Results Math., 1996, 29(1-2): 42{55.[3] BAESCH A, STEINMETZ N. Exceptional solutions of nth order periodic linear differentialequations [J].Complex V ariables Theory Appl., 1997, 34(1-2): 7{17.[4] BANK S B. On the explicit determination of certain solutions of periodic differential equations[J]. Complex V ariables Theory Appl., 1993, 23(1-2): 101{121.[5] BANK S B. Three results in the value-distribution theory of solutions of linear differentialequations [J].Kodai Math. J., 1986, 9(2): 225{240.[6] BANK S B, LAINE I. Representations of solutions of periodic second order linear differentialequations [J]. J. Reine Angew. Math., 1983, 344: 1{21.[7] BANK S B, LANGLEY J K. Oscillation theorems for higher order linear differential equationswith entire periodic coe±cients [J]. Comment. Math. Univ. St. Paul., 1992, 41(1): 65{85.[8] CHIANG Y M, GAO Shi'an. On a problem in complex oscillation theory of periodic secondorder lineardifferential equations and some related perturbation results [J]. Ann. Acad. Sci. Fenn. Math., 2002, 27(2):273{290.一些周期性的二阶线性微分方程解的方法1．简介和主要成果在本文中，我们假设读者熟悉的函数的数值分布理论[12，14，16]的基本成果和数学符号。

应用数学专业外文翻译

本科毕业论文外文翻译外文译文题目（中文）：具体数学：汉诺塔问题学院:专业:学号:学生姓名:指导教师:日期: 二○一二年六月1 Recurrent ProblemsTHIS CHAPTER EXPLORES three sample problems that give a feel for what’s to c ome. They have two traits in common: They’ve all been investigated repeatedly by mathe maticians; and their solutions all use the idea of recurrence, in which the solution to eac h problem depends on the solutions to smaller instances of the same problem.1.1 THE TOWER OF HANOILet’s look first at a neat little puzzle called the Tower of Hanoi,invented by the Fr ench mathematician Edouard Lucas in 1883. We are given a tower of eight disks, initiall y stacked in decreasing size on one of three pegs:The objective is to transfer the entire tower to one of the other pegs, movingonly one disk at a time and never moving a larger one onto a smaller.Lucas furnished his toy with a romantic legend about a much larger Tower of Brah ma, which supposedly has 64 disks of pure gold resting on three diamond needles. At th e beginning of time, he said, God placed these golden disks on the first needle and orda ined that a group of priests should transfer them to the third, according to the rules abov e. The priests reportedly work day and night at their task. When they finish, the Tower will crumble and the world will end.It's not immediately obvious that the puzzle has a solution, but a little thought (or h aving seen the problem before) convinces us that it does. Now the question arises：What's the best we can do？That is，how many moves are necessary and suff i cient to perfor m the task?The best way to tackle a question like this is to generalize it a bit. The Tower of Brahma has 64 disks and the Tower of Hanoi has 8；let's consider what happens if ther e are TL disks.One advantage of this generalization is that we can scale the problem down even m ore. In fact, we'll see repeatedly in this book that it's advantageous to LOOK AT SMAL L CASES first. It's easy to see how to transfer a tower that contains only one or two di sks. And a small amount of experimentation shows how to transfer a tower of three.The next step in solving the problem is to introduce appropriate notation：NAME ANO CONQUER. Let's say that T n is the minimum number of moves that will t ransfer n disks from one peg to another under Lucas's rules. Then T1is obviously 1 , an d T2= 3.We can also get another piece of data for free, by considering the smallest case of all：Clearly T0= 0，because no moves at all are needed to transfer a tower of n = 0 disks! Smart mathematicians are not ashamed to think small,because general patterns are easier to perceive when the extreme cases are well understood(even when they are trivial).But now let's change our perspective and try to think big；how can we transfer a la rge tower? Experiments with three disks show that the winning idea is to transfer the top two disks to the middle peg, then move the third, then bring the other two onto it. Thi s gives us a clue for transferring n disks in general：We first transfer the n−1 smallest t o a different peg (requiring T n-1moves), then move the largest (requiring one move), and finally transfer the n−1 smallest back onto the largest (req uiring another T n-1moves). Th us we can transfer n disks (for n > 0)in at most 2T n-1+1 moves：T n≤2T n—1+1，for n > 0.This formula uses '≤' instead of '=' because our construction proves only that 2T n—1+1 mo ves suffice; we haven't shown that 2T n—1+1 moves are necessary. A clever person might be able to think of a shortcut.But is there a better way? Actually no. At some point we must move the largest d isk. When we do, the n−1 smallest must be on a single peg, and it has taken at least Tmoves to put them there. We might move the largest disk more than once, if we're n n−1ot too alert. But after moving the largest disk for the last time, we must trans fr the n−1 smallest disks (which must again be on a single peg)back onto the largest；this too re quires T n−1moves. HenceT n≥ 2T n—1+1，for n > 0.These two inequalities, together with the trivial solution for n = 0, yieldT0=0；T n=2T n—1+1 , for n > 0. (1.1)(Notice that these formulas are consistent with the known values T1= 1 and T2= 3. Our experience with small cases has not only helped us to discover a general formula, it has also provided a convenient way to check that we haven't made a foolish error. Such che cks will be especially valuable when we get into more complicated maneuvers in later ch apters.)A set of equalities like (1.1) is called a recurrence (a. k. a. recurrence relation or r ecursion relation). It gives a boundary value and an equation for the general value in ter ms of earlier ones. Sometimes we refer to the general equation alone as a recurrence, alt hough technically it needs a boundary value to be complete.The recurrence allows us to compute T n for any n we like. But nobody really like to co m pute fro m a recurrence，when n is large；it takes too long. The recurrence only gives indirect, "local" information. A solution to the recurrence would make us much h appier. That is, we'd like a nice, neat, "closed form" for Tn that lets us compute it quic kly，even for large n. With a closed form, we can understand what T n really is.So how do we solve a recurrence? One way is to guess the correct solution,then to prove that our guess is correct. And our best hope for guessing the solution is t o look (again) at small cases. So we compute, successively,T3= 2×3+1= 7; T4= 2×7+1= 15; T5= 2×15+1= 31; T6= 2×31+1= 63.Aha! It certainly looks as ifTn = 2n−1，for n≥0. (1.2)At least this works for n≤6.Mathematical induction is a general way to prove that some statement aboutthe integer n is true for all n≥n0. First we prove the statement when n has its smallest v alue,no; this is called the basis. Then we prove the statement for n > n0,assuming that it has already been proved for all values between n0and n−1, inclusive; this is called th e induction. Such a proof gives infinitely many results with only a finite amount of wo rk.Recurrences are ideally set up for mathematical induction. In our case, for exampl e，(1.2) follows easily from (1.1)：The basis is trivial，since T0 = 20−1= 0.And the indu ction follows for n > 0 if we assume that (1.2) holds when n is replaced by n−1：T n= 2T n+1= 2(2n−1−1)+1=2n−1.Hence (1.2) holds for n as well. Good! Our quest for T n has ended successfully.Of course the priests' task hasn't ended；they're still dutifully moving disks,and wil l be for a while, because for n = 64 there are 264−1 moves (about 18 quintillion). Even at the impossible rate of one move per microsecond, they will need more than 5000 cent uries to transfer the Tower of Brahma. Lucas's original puzzle is a bit more practical, It requires 28−1 = 255 moves, which takes about four minutes for the quick of hand.The Tower of Hanoi recurrence is typical of many that arise in applications of all kinds. In finding a closed-form expression for some quantity of interest like T n we go t hrough three stages：1 Look at small cases. This gives us insight into the problem and helps us in stages2 and 3.2 Find and prove a mathematical expression for the quantity of interest.For the Tower of Hanoi, this is the recurrence (1.1) that allows us, given the inc lination，to compute T n for any n.3 Find and prove a closed form for our mathematical expression.For the Tower of Hanoi, this is the recurrence solution (1.2).The third stage is the one we will concentrate on throughout this book. In fact, we'll fre quently skip stages I and 2 entirely, because a mathematical expression will be given to us as a starting point. But even then, we'll be getting into subproblems whose solutions will take us through all three stages.Our analysis of the Tower of Hanoi led to the correct answer, but it r equired an“i nductive leap”；we relied on a lucky guess about the answer. One of the main objectives of this book is to explain how a person can solve recurrences without being clairvoyant. For example, we'll see that recurrence (1.1) can be simplified by adding 1 to both sides of the equations：T0+ 1= 1;T n + 1= 2T n-1+ 2, for n >0.Now if we let U n= T n+1，we haveU0 =1；U n= 2U n-1，for n > 0. (1.3)It doesn't take genius to discover that the solution to this recurrence is just U n= 2n；he nce T n= 2n −1. Even a computer could discover this.Concrete MathematicsR. L. Graham, D. E. Knuth, O. Patashnik《Concrete Mathematics》，1.1 ，The Tower Of HanoiR. L. Graham, D. E. Knuth, O. PatashnikSixth printing, Printed in the United States of America1989 by Addison-Wesley Publishing Company，Reference 1-4 pages具体数学R.L.格雷厄姆，D.E.克努特，O.帕塔希尼克《具体数学》，1.1，汉诺塔R.L.格雷厄姆，D.E.克努特，O.帕塔希尼克第一版第六次印刷于美国，韦斯利出版公司，1989年，引用1-4页1 递归问题本章将通过对三个样本问题的分析来探讨递归的思想。

数学外文翻译外文文献英文文献具体数学

Concrete MathematicsR. L. Graham, D. E. Knuth, O. Patashnik《Concrete Mathematics》，1.3 THE JOSEPHUS PROBLEM R. L. Graham, D. E. Knuth, O. Patashnik Sixth printing, Printed in the United States of America 1989 by Addison-Wesley Publishing Company，Reference 1-4pages具体数学R.L.格雷厄姆，D.E.克努特，O.帕塔希尼克《具体数学》，1.3，约瑟夫环问题R.L.格雷厄姆，D.E.克努特，O.帕塔希尼克第一版第六次印刷于美国，韦斯利出版公司，1989年，引用8-16页1.递归问题本章研究三个样本问题。

这三个样本问题给出了递归问题的感性知识。

它们有两个共同的特点：它们都是数学家们一直反复地研究的问题；它们的解都用了递归的概念，按递归概念，每个问题的解都依赖于相同问题的若干较小场合的解。

2.约瑟夫环问题我们最后一个例子是一个以Flavius Josephus命名的古老的问题的变形，他是第一世纪一个著名的历史学家。

据传说，如果没有Josephus的数学天赋，他就不可能活下来而成为著名的学者。

在犹太|罗马战争中，他是被罗马人困在一个山洞中的41个犹太叛军之一，这些叛军宁死不屈，决定在罗马人俘虏他们之前自杀，他们站成一个圈，从一开始，依次杀掉编号是三的倍数的人，直到一个人也不剩。

但是在这些叛军中的Josephus和他没有被告发的同伴觉得这么做毫无意义，所以他快速的计算出他和他的朋友应该站在这个恶毒的圆圈的哪个位置。

在我们的变形了的问题中，我们以n个人开始，从1到n编号围成一个圈，我们每次消灭第二个人直到只剩下一个人。

例如，这里我们以设n= 10做开始。

数学专业英语论文(含中文版)

Differential CalculusNewton and Leibniz,quite independently of one another,were largely responsible for developing the ideas of integral calculus to the point where hitherto insurmountable problems could be solved by more or less routine methods.The successful accomplishments of these men were primarily due to the fact that they were able to fuse together the integral calculus with the second main branch of calculus,differential calculus.In this article, we give su ﬃcient conditions for controllability of some partial neutral functional di ﬀerential equations with inﬁnite delay. We suppose that the linear part is not necessarily densely deﬁned but satisﬁes the resolvent estimates of the Hille -Yosida theorem. The results are obtained using the integrated semigroups theory. An application is given to illustrate our abstract result. Key words Controllability; integrated semigroup; integral solution; inﬁnity delay1 IntroductionIn this article, we establish a result about controllability to the following class of partial neutral functional di ﬀerential equations with inﬁnite delay:0,),()(0≥⎪⎩⎪⎨⎧∈=++=∂∂t x xt t F t Cu ADxt Dxt tβφ (1) where the state variable (.)x takes values in a Banach space ).,(E and the control (.)u is given in []0),,,0(2>T U T L ,the Banach space of admissible control functions with U a Banach space. Cis a bounded linear operator from U into E, A : D(A) ⊆ E → E is a linear operator on E, B is the phase space of functions mapping (−∞, 0] into E, which will be speciﬁed later, D is a bounded linear operator from B into E deﬁned byB D D ∈-=ϕϕϕϕ,)0(00D is a bounded linear operator from B into E and for each x : (−∞, T ] → E, T > 0, and t ∈ [0,T ], xt represents, as usual, the mapping from (−∞, 0] into E deﬁned by]0,(),()(-∞∈+=θθθt x xtF is an E-valued nonlinear continuous mapping on B ⨯ℜ+.The problem of controllability of linear and nonlinear systems repr esented by ODE in ﬁnit dimensional space was extensively studied. Many authors extended the controllability concept to inﬁnite dimensional systems in Banach space with unbounded operators. Up to now, there are a lot of works on this topic, see, for example, [4, 7, 10, 21]. There are many systems that can be written as abstract neutral evolution equations with inﬁnite delay to study [23]. In recent years, the theory of neutral functional di ﬀerential equations with inﬁnite delay in inﬁnitedimension was deve loped and it is still a ﬁeld of research (see, for instance, [2, 9, 14, 15] and the references therein). Meanwhile, the controllability problem of such systems was also discussed by many mathematicians, see, for example, [5, 8]. The objective of this article is to discuss the controllability for Eq. (1), where the linear part is supposed to be non-densely deﬁned but satisﬁes the resolvent estimates of the Hille-Yosida theorem. We shall assume conditions that assure global existence and give the su ﬃcient conditions for controllability of some partial neutral functional di ﬀerential equations with inﬁnite delay. The results are obtained using the integrated semigroups theory and Banach ﬁxed point theorem. Besides, we make use of the notion of integral solution and we do not use the analytic semigroups theory.Treating equations with inﬁnite delay such as Eq. (1), we need to introduce the phase space B. To avoid repetitions and understand the interesting properties of the phase space, suppose that ).,(B B is a (semi)normed abstract linear space of functions mapping (−∞, 0] into E, and satisﬁes the following fundamental axioms that were ﬁrst introduced in [13] and widely discussedin [16].(A)There exist a positive constant H and functions K(.), M(.):++ℜ→ℜ,with K continuous and M locally bounded, such that, for any ℜ∈σand 0>a ,if x : (−∞, σ + a] → E, B x ∈σ and (.)x is continuous on [σ, σ+a], then, for every t in [σ, σ+a], the following conditions hold:(i) B xt ∈, (ii) Bt x H t x ≤)(,which is equivalent toB H ϕϕ≤)0(or every B ∈ϕ(iii) Bσσσσx t M s x t K xtts B)()(sup )(-+-≤≤≤(A) For the function (.)x in (A), t → xt is a B -valued continuous function for t in [σ, σ + a]. (B) The space B is complete.Throughout this article, we also assume that the operator A satisﬁes the Hille -Yosida condition :(H1) There exist and ℜ∈ω，such that )(),(A ρω⊂+∞ and {}M N n A I n n ≤≥∈---ωλλωλ,:)()(sup （2） Let A0 be the part of operator A in )(A D deﬁned by{}⎩⎨⎧∈=∈∈=)(,,)(:)()(000A D x for Ax x A A D Ax A D x A D It is well known that )()(0A D A D =and the operator 0A generates a strongly continuoussemigroup ))((00≥t t T on )(A D .Recall that [19] for all )(A D x ∈ and 0≥t ,one has )()(000A D xds s T f t∈ andx t T x sds s T A t )(0)(00=+⎪⎭⎫ ⎝⎛⎰. We also recall that 00))((≥t t T coincides on )(A D with the derivative of the locally Lipschitz integrated semigroup 0))((≥t t S generated by A on E, which is, according to [3, 17, 18],a family of bounded linear operators on E, that satisﬁes(i) S(0) = 0, (ii) for any y ∈ E, t → S(t)y is strongly continuous with values in E,(iii)⎰-+=sdr r s r t S t S s S 0))()(()()(for all t, s ≥ 0, and for any τ > 0 there exists aconstant l(τ) > 0, such thats t l s S t S -≤-)()()(τ or all t, s ∈ [0, τ] .The C0-semigroup 0))((≥'t t S is exponentially bounded, that is, there exist two constantsM and ω,such that t e M t S ω≤')( for all t ≥ 0.Notice that the controllability of a class of non-de nsely deﬁned functional di ﬀerential equations was studied in [12] in the ﬁnite delay case.、2 Main ResultsWe start with introducing the following deﬁnition.Deﬁnition 1 Let T > 0 and ϕ ∈ B. We consider the following deﬁnition.We say that a function x := x(., ϕ) : (−∞, T ) → E, 0 < T ≤ +∞, is an integral solution of Eq. (1) if(i) x is continuous on [0, T ) ,(ii) ⎰∈tA D Dxsds 0)( for t ∈ [0, T ) ,(iii) ⎰⎰+++=ts tt ds x s F s Cu Dxsds A D Dx 0),()(ϕfor t ∈ [0, T ) ,(iv))()(t t x ϕ= for all t ∈ (−∞, 0].We deduce from [1] and [22] that integral solutions of Eq. (1) are given for ϕ ∈ B, such that )(A D D ∈ϕ by the following system⎪⎩⎪⎨⎧-∞∈=∈+-'+'=⎰+∞→],0,(),()(),,0[,)),()(()(lim )(0t t t x t t ds x s F s Cu B s t S D t S Dxt ts ϕλϕλ 、 (3)Where 1)(--=A I B λλλ.To obtain global existence and uniqueness, we supposed as in [1] that (H2) 1)0(0<D K .(H3) E F →B ⨯+∞],0[:is continuous and there exists 0β> 0, such that B -≤-21021),(),(ϕϕβϕϕt F t F for ϕ1, ϕ2 ∈ B and t ≥ 0. (4)Using Theorem 7 in [1], we obtain the following result.Theorem 1 Assume that (H1), (H2), and (H3) hold. Let ϕ ∈ B such that D ϕ ∈ D(A). Then, there exists a unique integral solution x(., ϕ) of Eq. (1), deﬁned on (−∞,+∞) .Deﬁnition 2 Under the above conditions, Eq. (1) is said to be controllable on the interval J = [0, δ], δ > 0, if for every initial function ϕ ∈ B with D ϕ ∈ D(A) and for any e1 ∈ D(A), there exists a control u ∈ L2(J,U), such that the solution x(.) of Eq. (1) satisﬁes 1)(e x =δ.Theorem 2 Suppose that(H1), (H2), and (H3) hold. Let x(.) be the integral solution of Eq. (1) on (−∞, δ) , δ > 0, and assume that (see [20]) the linear operator W from U into D(A) deﬁned byds s Cu B s S Wu )()(limλδλδ⎰-'=+∞→，（5）nduces an invertible operator W ~on KerW U J L /),(2，such that there exist positive constants1N and 2N satisfying 1N C ≤and 21~N W ≤-，then, Eq. (1) is controllable on J providedthat1))(2221000<++δδωδωδβδβK e M N N e M D ，（6）Where)(max :0t K K t δδ≤≤=.Proof Following [1], when the integral solution x(.) of Eq. (1) exists on (−∞, δ) , δ > 0, it is given for all t ∈ [0, δ] byds s Cu s t S dt d ds x s F s t S dt d D t S x D t x tt s t ⎰⎰-+-+'+=000)()(),()()()(ϕOrdsx s B s t S D t S x D t x tst ⎰-'+'+=+∞→00),()(lim)()(λλϕds s Cu B s t S t⎰-'++∞→0)()(limλλThen, an arbitrary integral solution x(.) of Eq. (1) on (−∞, δ) , δ > 0, satisﬁes x(δ) = e1 if and only ifdss Cu B s t S ds x s F s S d d D S x D e ts ⎰⎰-'+-+'+=+∞→001)()(lim),()()(λλδδδδϕδThis implies that, by use of (5), it su ﬃces to take, for all t ∈ J,{})()()(lim~)(01t ds s Cu B s t S W t u t⎰-'=+∞→-λλ{})(),()(lim )(~011t ds x s B s t S D S x D e W ts⎰-'-'--=+∞→-λλϕδδin order to have x(δ) = e1. Hence, we must take the control as above, and consequently, the proof is reduced to the existence of the integral solution given for all t ∈ [0, δ] by⎰-+'+=ts t ds z s F s t S dtd D t S z D t Pz 00),()()(:))((ϕ {ϕδδδD S z D z W C s t S dt d t )()(~)(001'---=⎰- ds s d z F B S )}(),()(limτττδτλδλ⎰-'-+∞→Without loss of generality, suppose thatω ≥ 0. Using similar arguments as in [1], we can seehat, for every1z ,)(2ϕδZ z ∈and t ∈ [0, δ] ,∞-+≤-210021)())(())((z z K e M D t Pz t Pz δδωβAs K is continuous and1)0(0<K D ,we can choose δ > 0 small enough, such that1)2221000<++δδωδωδββK e M N N e M D .Then, P is a strict contraction in )(ϕδZ ,and the ﬁxed point of P gives the unique integral olution x(., ϕ) on (−∞, δ] that veriﬁes x(δ) = e1.Remark 1 Suppose that all linear operators W from U into D(A) deﬁned byds s Cu B s b S Wu )()(limλδλ⎰-'=+∞→0 ≤ a < b ≤ T, T > 0, induce invertible operators W ~ on KerW U b a L /)],,([2,such that thereexist positive constants N1 and N2 satisfying 1N C ≤ and21~N W ≤-,taking NT =δ,N large enough and following [1]. A similar argument as the above proof can be used inductively in 11],)1(,[-≤≤+N n n n δδ,to see that Eq. (1) is controllable on [0, T ] for all T > 0.Acknowledgements The authors would like to thank Prof. Khalil Ezzinbi and Prof. Pierre Magal for the fruitful discussions.References[1] Adimy M, Bouzahir H, Ezzinbi K. Existence and stability for some partial neutral functional di ﬀerenti al equations with inﬁnite delay. J Math Anal Appl, 2004, 294: 438–461[2] Adimy M, Ezzinbi K. A class of linear partial neutral functional diﬀerential equations withnondense domain. J Dif Eq, 1998, 147: 285–332[3] Arendt W. Resolvent positive operators and integrated semigroups. Proc London Math Soc,1987, 54(3):321–349[4] Atmania R, Mazouzi S. Controllability of semilinear integrodiﬀerential equations withnonlocal conditions. Electronic J of Diﬀ Eq, 2005, 2005: 1–9[5] Balachandran K, Anandhi E R. Controllability of neutral integrodiﬀerential inﬁnite delaysystems in Banach spaces. Taiwanese J Math, 2004, 8: 689–702[6] Balasubramaniam P, Ntouyas S K. Controllability for neutral stochastic functional diﬀerentialinclusionswith inﬁnite delay in abst ract space. J Math Anal Appl, 2006, 324(1): 161–176、[7] Balachandran K, Balasubramaniam P, Dauer J P. Local null controllability of nonlinearfunctional diﬀer-ential systems in Banach space. J Optim Theory Appl, 1996, 88: 61–75 [8] Balasubramaniam P, Loganathan C. Controllability of functional diﬀerential equations withunboundeddelay in Banach space. J Indian Math Soc, 2001, 68: 191–203[9] Bouzahir H. On neutral functional diﬀerential equations. Fixed Point Theory, 2005, 5: 11–21 The study of differential equations is one part of mathematics that, perhaps more than any other, has been directly inspired by mechanics, astronomy, and mathematical physics. Its history began in the 17th century when Newton, Leibniz, and the Bernoullis solved some simple differential equation arising from problems in geometry and mechanics. There early discoveries, beginning about 1690, gradually led to the development of a lot of “special tricks” for solving certain special kinds of differential equations. Although these special tricks are applicable in mechanics and geometry, so their study is of practical importance.微分方程牛顿和莱布尼茨，完全相互独立，主要负责开发积分学思想的地步，迄今无法解决的问题可以解决更多或更少的常规方法。

数学专业英语论文(含中文版)

Differential CalculusNewton and Leibniz,quite independently of one another,were largely responsible for developing the ideas of integral calculus to the point where hitherto insurmountable problems could be solved by more or less routine methods.The successfulaccomplishments of these men were primarily due to the fact that they were able to fuse together the integral calculus with the second main branch ofcalculus,differential calculus.In this article, we give su ﬃcient conditions for controllability of some partial neutral functional di ﬀerential equations with in ﬁnite delay. We suppose that the linear part is not necessarily densely de ﬁned but satis ﬁes the resolvent estimates of the Hille-Yosida theorem. The results are obtained using the integrated semigroups theory. An application is given to illustrate our abstract result. Key words Controllability; integrated semigroup; integral solution; in ﬁnity delay1 IntroductionIn this article, we establish a result about controllability to the following class of partial neutral functional di ﬀerential equations with in ﬁnite delay:0,),()(0≥⎪⎩⎪⎨⎧∈=++=∂∂t x xt t F t Cu ADxt Dxt t βφ (1) where the state variable (.)x takes values in a Banach space ).,(E and the control(.)u isgiven in []0),,,0(2>T U T L ,the Banach space of admissible control functions with U a Banach space. C is a bounded linear operator from U into E, A : D(A) ⊆ E → E is a linear operator on E, B is the phase space of functions mapping (−∞, 0] into E, which will be speci ﬁed later, D is a bounded linear operator from B into E de ﬁned byB D D ∈-=ϕϕϕϕ,)0(00D is a bounded linear operator from B into E and for each x : (−∞, T ] → E, T > 0, and t ∈ [0, T ], xt represents, as usual, the mapping from (−∞, 0] into E de ﬁned by]0,(),()(-∞∈+=θθθt x xtF is an E-valued nonlinear continuous mapping on B ⨯ℜ+.The problem of controllability of linear and nonlinear systems represented by ODE in ﬁnit dimensional space was extensively studied. Many authors extended the controllability concept to in ﬁnite dimensional systems in Banach space with unbounded operators. Up to now, there are a lot of works on this topic, see, for example, [4, 7, 10, 21]. There are many systems that can be written as abstract neutral evolution equations with in ﬁnite delay to study [23]. In recent years, the theory of neutral functional di ﬀerential equations with in ﬁnite delay in in ﬁnite dimension was developed and it is still a ﬁeld of research (see, for instance, [2, 9, 14, 15] and the references therein). Meanwhile, the controllability problem of such systems was also discussed by many mathematicians, see, for example, [5, 8]. The objective of this article is to discuss the controllability for Eq. (1), where the linear part is supposed to be non-densely de ﬁned but satis ﬁes the resolventestimates of the Hille-Yosida theorem. We shall assume conditions that assure global existence and give the suﬃcient conditions for controllability of some partial neutral functional diﬀerential equations with inﬁnite delay. The results are obtained using the integrated semigroups theory and Banach ﬁxed point theorem. Besides, we make use of the notion of integral solution and we do not use the analytic semigroups theory.Treating equations with in ﬁnite delay such as Eq. (1), we need to introduce the phase space B. To avoid repetitions and understand the interesting properties of the phase space, suppose that ).,(B B is a (semi)normed abstract linear space offunctions mapping (−∞, 0] into E, and satis ﬁes the following fundamental axioms that were ﬁrst introduced in [13] and widely discussed in [16].(A) There exist a positive constant H and functions K(.),M(.):++ℜ→ℜ,with K continuous and M locally bounded, such that, forany ℜ∈σand 0>a ,if x : (−∞, σ + a] → E, B x ∈σ and (.)x iscontinuous on [σ, σ+a], then, for every t in [σ, σ+a], the followingconditions hold:(i) B xt ∈, (ii) B tx H t x ≤)(,which is equivalent to B H ϕϕ≤)0(or every B ∈ϕ (iii) B σσσσx t M s x t K xt t s B )()(sup )(-+-≤≤≤(A) For the function (.)x in (A), t → xt is a B -valued continuous function for t in [σ, σ + a].(B) The space B is complete.Throughout this article, we also assume that the operator A satis ﬁes the Hille-Yosida condition : (H1) There exist and ℜ∈ω，such that )(),(A ρω⊂+∞ and {}M N n A I n n ≤≥∈---ωλλωλ,:)()(sup （2） Let A0 be the part of operator A in )(A D de ﬁned by {}⎩⎨⎧∈=∈∈=)(,,)(:)()(000A D x for Ax x A A D Ax A D x A D It is well known that )()(0A D A D =and the operator 0A generates a strongly continuous semigroup ))((00≥t t T on )(A D . Recall that [19] for all )(A D x ∈ and 0≥t ,one has )()(000A D xds s T f t ∈ and x t T x sds s T A t )(0)(00=+⎪⎭⎫ ⎝⎛⎰. We also recall that 00))((≥t t T coincides on )(A D with the derivative of the locally Lipschitz integrated semigroup 0))((≥t t S generated by A on E, which is, according to [3, 17, 18], a family of bounded linear operators on E, that satis ﬁes(i) S(0) = 0,(ii) for any y ∈ E, t → S(t)y is strong ly continuous with values in E, (iii) ⎰-+=sdr r s r t S t S s S 0))()(()()(for all t, s ≥ 0, and for any τ > 0 there exists a constant l(τ) > 0, such thats t l s S t S -≤-)()()(τ or all t, s ∈ [0, τ] .The C0-semigroup 0))((≥'t t S is exponentially bounded, that is, there exist two constants M and ω,such that t e M t S ω≤')( for all t ≥ 0.Notice that the controllability of a class of non-densely de ﬁned functional di ﬀerential equations was studied in [12] in the ﬁnite delay case.、2 Main ResultsWe start with introducing the following de ﬁnition.De ﬁnition 1 Let T > 0 and ϕ ∈ B. We consider the following de ﬁnition.We say that a function x := x(., ϕ) : (−∞, T ) → E, 0 < T ≤ +∞, is an integral solution of Eq. (1) if(i) x is continuous on [0, T ) ,(ii)⎰∈t A D Dxsds 0)( for t ∈ [0, T ) , (iii)⎰⎰+++=t s t t ds x s F s Cu Dxsds A D Dx 00),()(ϕfor t ∈ [0, T ) ,(iv) )()(t t x ϕ= for all t ∈ (−∞, 0].We deduce from [1] and [22] that integral solutions of Eq. (1) are given for ϕ ∈ B, such that )(A D D ∈ϕ by the following system⎪⎩⎪⎨⎧-∞∈=∈+-'+'=⎰+∞→],0,(),()(),,0[,)),()(()(lim )(0t t t x t t ds x s F s Cu B s t S D t S Dxt t s ϕλϕλ 、 (3)Where 1)(--=A I B λλλ.To obtain global existence and uniqueness, we supposed as in [1] that (H2) 1)0(0<D K .(H3) E F →B ⨯+∞],0[:is continuous and there exists 0β> 0, such that B -≤-21021),(),(ϕϕβϕϕt F t F for ϕ1, ϕ2 ∈ B and t ≥ 0. (4) Using Theorem 7 in [1], we obtain the following result.Theorem 1 Assume that (H1), (H2), and (H3) hold. Let ϕ ∈ B such that D ϕ ∈ D(A). Then, there exists a unique integral solution x(., ϕ) of Eq. (1), de ﬁned on (−∞,+∞) .De ﬁnition 2 Under the above conditions, Eq. (1) is said to be controllable on the interval J = [0, δ], δ > 0, if for every initial function ϕ ∈ B with D ϕ ∈ D(A) and for any e1 ∈ D(A), there exists a control u ∈ L2(J,U), such that the solution x(.) of Eq. (1) satis ﬁes 1)(e x =δ.Theorem 2 Suppose that(H1), (H2), and (H3) hold. Let x(.) be the integral solution of Eq. (1) on (−∞, δ) , δ > 0, and assume tha t (see [20]) the linear operator W from U into D(A) de ﬁned byds s Cu B s S Wu )()(limλδλδ⎰-'=+∞→，（5）nduces an invertible operator W ~on KerW U J L /),(2，such that there exist positive constants 1N and 2N satisfying 1N C ≤and 21~N W≤-，then, Eq. (1) is controllable on J provided that 1))(2221000<++δδωδωδβδβK e M N N e M D ，（6） Where )(max :0t K K t δδ≤≤=.Proof Following [1], when the integral solution x(.) of Eq. (1) exists on (−∞, δ) , δ > 0, it is given for all t ∈ [0, δ] byds s Cu s t S dt d ds x s F s t S dt d D t S x D t x t t s t ⎰⎰-+-+'+=000)()(),()()()(ϕ Ords x s B s t S D t S x D t x t s t ⎰-'+'+=+∞→00),()(lim)()(λλϕ ds s Cu B s t S t⎰-'++∞→0)()(lim λλ Then, an arbitrary integral solution x(.) of Eq. (1) on (−∞, δ) , δ > 0, satis ﬁes x(δ) = e1 if and only ifds s Cu B s t S ds x s F s S d d D S x D e t s ⎰⎰-'+-+'+=+∞→0001)()(lim ),()()(λλδδδδϕδThis implies that, by use of (5), it su ﬃces to take, for all t ∈ J, {})()()(lim~)(01t ds s Cu B s t S W t u t ⎰-'=+∞→-λλ {})(),()(lim )(~0011t ds x s B s t S D S x D e W t s ⎰-'-'--=+∞→-λλϕδδin order to have x(δ) = e1. Hence, we must take the control as above, and consequently, the proof is reduced to the existence of the integral solution given for all t ∈[0, δ] by⎰-+'+=t s t ds z s F s t S dt d D t S z D t Pz 00),()()(:))((ϕ {ϕδδδD S z D z W C s t S dt d t )()(~)(001'---=⎰- ds s d z F B S )}(),()(lim 0τττδτλδλ⎰-'-+∞→ Without loss of generality, suppose that ω ≥ 0. Using simila r arguments as in [1], we can see hat, for every 1z ,)(2ϕδZ z ∈and t ∈ [0, δ] ,∞-+≤-210021)())(())((z z K e M D t Pz t Pz δδωβAs K is continuous and 1)0(0<K D ,we can choose δ > 0 small enough, such that1)2221000<++δδωδωδββK e M N N e M D .Then, P is a strict contraction in )(ϕδZ ,and the ﬁxed point of P gives the unique integral olution x(., ϕ) on (−∞, δ] that veri ﬁes x(δ) = e1.Remark 1 Suppose that all linear operators W from U into D(A) de ﬁned byds s Cu B s b S Wu )()(lim 0λδλ⎰-'=+∞→0 ≤ a < b ≤ T, T > 0, induce invertible operators W ~ on KerW U b a L /)],,([2,such that there exist positive constants N1 and N2 satisfying 1N C ≤ and21~N W ≤-,taking N T =δ,N large enough and following [1]. A similar argumentas the above proof can be used inductively in 11],)1(,[-≤≤+N n n n δδ,to see that Eq. (1) is controllable on [0, T ] for all T > 0.Acknowledgements The authors would like to thank Prof. Khalil Ezzinbi and Prof. Pierre Magal for the fruitful discussions.References[1] Adimy M, Bouzahir H, Ezzinbi K. Existence and stability for some partial neutralfunctional di ﬀerential equations with in ﬁnite delay. J Math Anal Appl, 2004, 294: 438–461[2] Adimy M, Ezzinbi K. A class of linear partial neutral functional di ﬀerentialequations with nondense domain. J Dif Eq, 1998, 147: 285–332[3] Arendt W. Resolvent positive operators and integrated semigroups. Proc LondonMath Soc, 1987, 54(3):321–349[4] Atmania R, Mazouzi S. Controllability of semilinear integrodi ﬀerentialequations with nonlocal conditions. Electronic J of Di ﬀ Eq, 2005, 2005: 1–9[5] Balachandran K, Anandhi E R. Controllability of neutral integrodi ﬀerentialin ﬁnite delay systems in Banach spaces. Taiwanese J Math, 2004, 8: 689–702[6] Balasubramaniam P, Ntouyas S K. Controllability for neutral stochasticfunctional di ﬀerential inclusionswith in ﬁnite delay in abstract space. J Math Anal Appl, 2006, 324(1): 161–176、[7] Balachandran K, Balasubramaniam P, Dauer J P. Local null controllability ofnonlinear functional di ﬀer-ential systems in Banach space. J Optim Theory Appl, 1996, 88: 61–75[8] Balasubramaniam P, Loganathan C. Controllability of functional di ﬀerentialequations with unboundeddelay in Banach space. J Indian Math Soc, 2001, 68: 191–203[9] Bouzahir H. On neutral functional diﬀerential equations. Fixed Point Theory, 2005, 5: 11–21The study of differential equations is one part of mathematics that, perhaps more than any other, has been directly inspired by mechanics, astronomy, and mathematical physics. Its history began in the 17th century when Newton, Leibniz, and the Bernoullis solved some simple differential equation arising from problems in geometry and mechanics. There early discoveries, beginning about 1690, gradually led to the development of a lot of “special tricks” for solving certain special kinds of differential equations. Although these special tricks are applicable in mechanics and geometry, so their study is of practical importance.微分方程牛顿和莱布尼茨，完全相互独立，主要负责开发积分学思想的地步，迄今无法解决的问题可以解决更多或更少的常规方法。

数学与应用数学专业级数第及向量代数和空间解析几何大学毕业论文外文文献翻译

毕业设计（论文）外文文献翻译文献、资料中文题目：第八章级数第九章向量代数和空间解析几何文献、资料英文题目：文献、资料来源：文献、资料发表（出版）日期：院（部）：专业：数学与应用数学班级：姓名：学号：指导教师：翻译日期： 2017.02.14外文翻译译文第八章级数级数是微积分研究的一个重要组成部分，代表功能作为“无限资金”。

要做到这一点就要求用熟悉的加法运算扩展一组有限的数字“添加无限多的数字”。

想要解决这一点我们需要通过考虑序列来解决限制的过程。

8.1 数项级数当我们有表达式 ++++n u u u 21，我们也许认为我们得到序列,,,21 u u 并且得到所有元素的无限和，通过普通加法我们不能获得这个和，当n 足够大时，我们可认为这和近似于n u u u +++ 21，当n 越大时，近似值越精确。

表达式 ++++n u u u 21可以看做序列 ),(,),(),(,21321211n u u u u u u u u u ++++++，我们把它叫做级数8.1.1 基本概念定义8.1 如果{}n u 是一个序列并且n n u u u s +++= 21，那么序列{}n s 叫做无穷级数。

这个无穷级数表示为++++=∑+∞=n n n u u u u211其中 ,,,,21n u u u 叫做级数的一般项。

,,,,21n s s s 叫做级数的部分和。

定义8.2 ∑+∞=1n n u是一级数并且{}n s 是这个级数的部分和序列。

如果n n s ∞→lim 存在并且等于s ，则所给级数收敛并且s 是级数的和，记作s u n n =∑+∞=1，若n n s ∞→lim 不存在，则称级数发散并且级数没有和。

如果一个无穷级数有和s ，我们称级数收敛于s 。

对于一个收敛级数∑+∞=1n n u 是用来表示级数和级数的和。

使用相同符号不应该混淆，因为从上下文的解释中可以得到正确的解释。

例1考虑收敛的几何级数 +++++n ar ar ar a 2， )0(≠a 其中a 是第一项且不等于0，其中r 为公比。

数学与应用数学英文文献及翻译

数学与应用数学英文文献及翻译-勾股定理(外文翻译从原文第一段开始翻译,翻译了约2000字)勾股定理是已知最早的古代文明定理之一。

这个著名的定理被命名为希腊的数学家和哲学家毕达哥拉斯。

毕达哥拉斯在意大利南部的科托纳创立了毕达哥拉斯学派。

他在数学上有许多贡献,虽然其中一些可能实际上一直是他学生的工作。

毕达哥拉斯定理是毕达哥拉斯最著名的数学贡献。

据传说,毕达哥拉斯在得出此定理很高兴,曾宰杀了牛来祭神,以酬谢神灵的启示。

后来又发现2的平方根是不合理的,因为它不能表示为两个整数比,极大地困扰毕达哥拉斯和他的追随者。

他们在自己的认知中,二是一些单位长度整数倍的长度。

因此2的平方根被认为是不合理的,他们就尝试了知识压制。

它甚至说,谁泄露了这个秘密在海上被淹死。

毕达哥拉斯定理是关于包含一个直角三角形的发言。

毕达哥拉斯定理指出,对一个直角三角形斜边为边长的正方形面积,等于剩余两直角为边长正方形面积的总和图1根据勾股定理,在两个红色正方形的面积之和A和B,等于蓝色的正方形面积,正方形三区因此,毕达哥拉斯定理指出的代数式是:对于一个直角三角形的边长a,b和c,其中c是斜边长度。

虽然记入史册的是著名的毕达哥拉斯定理,但是巴比伦人知道某些特定三角形的结果比毕达哥拉斯早一千年。

现在还不知道希腊人最初如何体现了勾股定理的证明。

如果用欧几里德的算法使用,很可能这是一个证明解剖类型类似于以下内容:六^维-论~文.网“一个大广场边a+ b是分成两个较小的正方形的边a和b分别与两个矩形A 和B,这两个矩形各可分为两个相等的直角三角形,有相同的矩形对角线c。

四个三角形可安排在另一侧广场a+b中的数字显示。

在广场的地方就可以表现在两个不同的方式:1。

由于两个长方形和正方形面积的总和:2。

作为一个正方形的面积之和四个三角形:现在,建立上面2个方程,求解得因此,对c的平方等于a和b的平方和(伯顿1991)有许多的勾股定理其他证明方法。

一位来自当代中国人在中国现存最古老的含正式数学理论能找到对Gnoman和天坛圆路径算法的经典文本。

数学与应用数学专业英语

数学与应用数学专业英语English:Mathematics and Applied Mathematics is a field of study that focuses on the mathematical principles and their practical applications. Students in this major will study a wide range of mathematical topics including calculus, algebra, differential equations, probability, and statistics. They will also learn how to apply these mathematical concepts to solve real-world problems in fields such as engineering, economics, physics, and computer science. This major requires strong analytical and problem-solving skills, as well as proficiency in computer programming and data analysis. Graduates with a degree in Mathematics and Applied Mathematics have a wide range of career opportunities, including roles in finance, research, teaching, and data analysis.中文翻译:数学与应用数学是一门致力于数学原理及其实际应用的研究领域。

在这个专业中，学生将学习包括微积分、代数、微分方程、概率和统计在内的各种数学课题。

数学专业英语论文(英文翻译中文)

Math problems about BaseballSummaryBaseball is a popular bat-and-ball game involving both athletics and wisdom. There are strict restrictions on the material, size and manufacture of the bat.It is vital important to transfer the maximum energy to the ball in order to give it the fastest batted speed during the hitting process.Firstly, this paper locates the center-of-percussion (COP) and the viberational node based on the single pendulum theory and the analysis of bat vibration.With the help of the synthesizing optimization approach, a mathematical model is developed to execute the optimized positioning for the “sweet spot”, and the best hitting spot turns out not to be at the end of the bat. Secondly, based on the basic model hypothesis, taking the physical and material attributes of the bat as parameters, the moment of inertia and the highest batted ball speed (BBS) of the “sweet spot” are evaluated using different parameter values, which enables a quantified comparison to be made on the performance of different bats.Thus finally explained why Major League Baseball prohibits “corking” and metal bats.In problem I, taking the COP and the viberational node as two decisive factors of the “sweet zone”, models are developed respectively to study the hitting effect from the angle of energy conversion.Because the different “sweet spots” decided by COP and the viberational node reflect different form of energy conversion, the “space-distance” concept is introduced and the “Technique for Order Preferenceby Similarity to Ideal Solution (TOPSIS) is used to locate the “sweet zone” step by step. And thus, it is proved that t he “sweet spot” is not at the end of the bat from the two angles of specific quantitative relationship of the hitting effects and the inference of energy conversion.In problem II, applying new physical parameters of a corked bat into the model developed in Problem I, the moment of inertia and the BBS of the corked bat and the original wood bat under the same conditions are calculated. The result shows that the corking bat reduces the BBS and the collision performance rather than enhancing the “sweet spot” effect. On the other hand, the corking bat reduces the moment of inertia of the bat, which makes the bat can be controlled easier. By comparing the two Team # 8038 Page 2 of 20 conflicting impacts comprehensively, the conclusion is drawn that the corked bat will be advantageous to the same player in the game, for which Major League Baseball prohibits “corking”.In problem III, adopting the similar method used in Problem II, that is, applying different physical parameters into the model developed in Problem I, calculate the moment of inertia and the BBS of the bats constructed by different material to analyze the impact of the bat material on the hitting effect. The data simulation of metal bats performance and wood bats performance shows that the performance of the metal bat is improved for the moment of inertia is reduced and the BBS is increased. Our model and method successfully explain why Major League Baseball, for the sake of fair competition, prohibits metal bats.In the end, an evaluation of the model developed in this paper is given, listing itsadvantages s and limitations, and providing suggestions on measuring the performance of a bat.Restatement of the ProblemExplain the “sweet spot” on a baseball bat.Every hitter knows that there is a spot on the fat part of a baseball bat where maximum power is transferred to the ball when hit.Why isn’t this spot at the end of the bat? A simple explanation based on torque might seem to identify the end of the bat as the sweet spot, but this is known to be empirically incorrect. Develop a model that helps explain this empirical finding.Some players believe that “corking” a bat (hollowing out a cylinder in the head of the bat and filling it with cork or rubber, then replacing a wood cap) enhances the “sweet spot” effect. Augment your model to confirm or deny this effect. Does this explain why Major League Baseball prohibits “corking”?Does the material out of which the bat is constructed matter? That is, does this model predict different behavior for wood (usually ash) or metal (usually aluminum) bats? Is this why Major League Baseball prohibits metal bats?2.1 Analysis of Problem IFirst explain the “sweet spot” on a baseball bat, and then develop a model that helps ex plain why this spot isn’t at the end of the bat.[1]There are a multitude of definitions of the sweet spot:1) the location which produces least vibrational sensation (sting) in the batter's hands2) the location which produces maximum batted ball speed3) the location where maximum energy is transferred to the ball4) the location where coefficient of restitution is maximum5) the center of percussionFor most bats all of these "sweet spots" are at different locations on the bat, so one is often forced to define the sweet spot as a region.If explained based on torque, this “sweet spot” might be at the end of the bat, which is known to be empirically incorrect.This paper is going to explain this empirical paradox by exploring the location of the sweet spot from a reasonable angle.Based on necessary analysis, it can be known that the sweet zone, which is decided by the center-of-percussion (COP) and the vibrational node, produces the hitting effect abiding by the law of energy conversion.The two different sweet spots respectively decided by the COP and the viberational node reflect different energy conversions, which forms a two-factor influence.2.2 Analysis of Problem IIProblem II is to explain whether “corking” a bat enhances the “sweet spot” effect and why Major League Baseball prohibits “corking”.[4]In order to find out what changes will occur after corking the bat, the changes of the bat’s parameters should be analyzed first:1) The mass of the corked bat reduces slightly than before;2) Less mass (lower moment of inertia) means faster swing speed;3) The mass center of the bat moves towards the handle;4) The coefficient of restitution of the bat becomes smaller than before;5) Less mass means a less effective collision；6) The moment of inertia becomes smaller.[5][6]By analyzing the changes of the above parameters of a corked bat, whether the hitting effect of the sweet spot has been changed could be identified and then the reason for prohibiting “corking” might be cle ar.2.3 Analysis of Problem IIIFirst, explain whether the bat material imposes impacts on the hitting effect; then, develop a model to predict different behavior for wood or metal bats to find out the reason why Major League Baseball prohibits metal bats?The mass (M) and the center of mass (CM) of the bat are different because of the material out of which the bat is constructed. The changes of the location of COP and moment of inertia ( I bat ) could be inferred.[2][3]Above physical attributes influence not only the swing speed of the player (the less the moment of inertia-- I bat is, the faster the swing speed is) but also the sweet spot effect of the ball which can be reflected by the maximum batted ball speed (BBS). The BBS of different material can be got by analyzing the material parameters that affect the moment of inertia.Then, it can be proved that the hitting effects of different bat material are different.3.Model Assumptions and Symbols3.1 Model Assumptions1) The collision discussed in th is paper refers to the vertical collision on the “sweet spot”;2) The process discussed refers to the whole continuous momentary process starting from the moment the bat contacts the ball until the moment the ball departs from the bat;3) Both the bat and the ball discussed are under common conditions.3.2 Instructions Symbolsa kinematic factor kthe rotational inertia of the object about its pivot point I0the mass of the physical pendulum Mthe location of the center-of-mass relative to the pivot point dthe distance between the undetermined COP and the pivot Lthe gravitational field strength gthe moment-of-inertia of the bat as measured about the pivot point on the handleI the swing period of the bat on its axis round the pivotTthe length of the bat sthe distance from the pivot point where the ball hits the bat zvibration frequency fthe mass of the ball m4.Modeling and Solution4.1 Modeling and Solution to Problem I 4.1.1 Model Preparation 1) Analysis of the pushing force or pressure exerted on hands[1] Team # 8038 Page 7 of 20 Fig. 4-1 As showed in Fig. 4-1:If an impact force F were to strike the bat at the center-of-mass (CM) then point P would experience a translational acceleration - the entire bat would attempt to accelerate to the left in the same direction as the applied force, without rotating about the pivot point.If a player was holding the bat in his/her hands, this would result in an impulsive force felt inthe hands.If the impact force F strikes the bat below the center-of-mass, but above the center-of-percussion, point P would experience both a translational acceleration in the direction of the force and a rotational acceleration in the opposite direction as the bat attempts to rotate about its center-of-mass. The translational acceleration to the left would be greater than the rotational acceleration to the right and a player would still feel an impulsive force in the hands.If the impact force strikes the bat below the center-of-percussion, then point P would still experience oppositely directed translational and rotational accelerations, but now the rotational acceleration would be greater.If the impact force strikes the bat precisely at the center-of-percussion, then the translational acceleration and the rotational acceleration in the opposite direction exactly cancel each other.method: Instead of being distributed throughout the entire object, let the mass of the physical pendulum M be concentrated at a single point located at a distance L from the pivot point.This point mass swinging from the end of a string is now a "simple" pendulum, and its period would be the same as that of the original physical pendulum if the distance L wasThis location L is known as the "center-of-oscillation". A solid object which oscillates about a fixed pivot point is called a physical pendulum.When displaced from its equilibrium position the force of gravity will attempt to return the object to its equilibrium position, while its inertia will cause it to overshoot.As a result of this interplay between restoring force and inertia the object will swing back and forth, repeating its cyclic motion in a constant amount of time. This time, called the period, depends on the mass of the object M , the location of the center-of-mass relative to the pivot point d , the rotational inertia of the object about its pivot point I 0 and the gravitational field strength g according to4.1.2 Solutions to the two “sweet spot” regions1) Locating the COP[1][4]Determining the parameters：a. mass of the bat M ;b. length of the bat S (the distance between Block 1 and Block 5 in Fig 4-3);c. distance between the pivot and the center-of-mass d ( the distance between Block2 and Block3 in Fig. 4-3);d .swing period of the bat on its axis round the pivot T (take an adult male as an example: the distance between the pivot and the knob of the bat is 16.8cm (the distance between Block 1 and Block 2 in Fig. 4-3);e.distance between the undetermined COP and the pivot L (the distance between Block 2 and Block 4 in Fig. 4-3, that is the turning radius) .Fig.4-3 Table 4-1Block1 knobBlock2 pivotBlock 3 the center-of-mass（CM）Block 4 t he center of percussion (COP)Block 5the end of the batCalculation method of COP[1][4]:distance between the undetermined COP and the pivot:T 2g L= 4π 2 ( g is the gravity acceleration）（4-3）moment of inertia:I0 = T 2 MgL 4π 2 ( L is the turning radius, M is the mass) （4-4）Results:The reaction force on the pivot is less than 10% of the bat-and-ball collision force. When the ball falls on any point in the “sweet spot” region, the area where the collision force reduction is less than 10% is (0.9 L ,1.1L) cm, which is called “Sweet Zone 1”.2) Determining the vibrational nodeThe contact between bat and ball, we consider it a process of wave ransmission.When the bat excited by a baseball of rapid flight, all of these modes, (as well as some additional higher frequency modes) are excited and the bat vibrates .We depend on the frequency modes ,list the following two modes:The fundamental bending mode has two nodes, or positions of zero displacement). One is about 6-1/2 inches from the barrel end close to the sweet spot of the bat. The other at about 24 inches from the barrel end (6 inches from the handle) at approximately the location of a right-handed hitter's right hand.Fundamental bending mode 1 (215 Hz) The second bending mode has three nodes, about 4.5 inches from the barrel end, a second near the middle of the bat, and the third at about the location of a right-handed hitter's left hand.Second bending mode 2 (670 Hz) The figures show the two bending modes of a freely supported baseball bat.The handle end of the bat is at the right, and the barrel end is at the left. The numbers on the axis represent inches (this data is for a 30 inch Little League wood baseball bat). These figures were obtained from a modal analysis experiment. In this opinion we prefer to follow the convention used by Rod Cross[2] who defines the sweet zone as Team # 8038 Page 11 of 20 the region located between the nodes of the first and second modes of vibration (between about 4-7 inches from the barrel end of a 30-inch Little League bat).The solving time in accordance with the searching times and backtrack times. It isobjective to consider the two indices together.4.1.3 Optimization Modelwood bat (ash)swing period T 0.12sbat mass M 876.0 15gbat length S 86.4 cmCM position d 41.62cmcoefficient of restitution BBCOR 0.4892initialvelocity vin 7.7m /sswing speed vbat 15.3 m/sball mass mball850.5gAdopting the parameters in the above table and based on the quantitative regions in sweet zone 1 and 2 in 4.1.2, the following can be drawn:[2] Sweet zone 1 is (0.9 L ,1.1L) = (50cm , 57.8358cm)Sweet zone 2 is ( L* , L* ) = (48.41cm,55.23cm)define the position of Block 2 which is the pivot as the origin of the number axis, and x as a random point on the number axis.Optimization modeling[2]The TOPSIS method is a technique for order preference by similarity to ideal solution whose basic idea is to transform the integrated optimal region problem into seeking the difference among evaluation objects—“distance”. That is, to determine the most ideal position and the acceptable most unsatisfactory position according to certain principals, and then calculate the distance between each evaluation object and Team # 8038 Page 12 of 20 the most ideal position and the distance between each evaluation object and the acceptable most unsatisfactory position.Finally, the “sweet zone” can be drawn by an integrated c omparison.Step 1 : Standardization of the extent value Standardization is performed via range transformation,x * = a dimensionless quantity，and x * ∈[0,1]Step 2:x min = min{0.9 L, L* }x max = max{1.1L, L* }x ∈( x min , x max ) ;Step 3: Calculating the distance The Euclidean distance of the positive ideal position is:The Euclidean distance of the negative ideal position is:Step 4: Seeking the integrated optimal region The integrated evaluation index of theevaluation object is:……………………………（4-5）Optimization positioningConsidering bat material physical attributes of normal wood, when the period is T = 0.12s and the vibration frequency is f = 520 HZ, the ideal “sweet zone” extent can be drawn as [51.32cm , 55.046cm] .As this consequence showed, the “sweet spot” cannot be at the end the bat. This conclusion can also be verified by the model for problem II.4.1.4 Verifying the “sweet spot” is not at the end of the bat1) Analyzed from the hitting effect According to Formula 4-11 and Table 4-2, the maximum batted-ball-speed of Team # 8038 Page 13 of 20 the “sweet spot” can be calculated as BBS sweet = 27.4 m / s , and the maximum batted-ball-speed of the bat end can be calculated as BBS end = 22.64 m / s . It is obvious that the “sweet spot” is not at the end of the bat.2) Analyzed from the energy According to the definition of “sweet spot” and the method of locating the “sweet spot”, energy loss should be minimized in order to transfer the maximum energy to the ball.When considering the “sweet spot” region from angle of torque, the position for maximum torque is no doubt at the end of the bat. But this position is also the maximum rebounded point according to the theory of force interaction. Rebound wastes the energy which originally could send the ball further.To sum up the above points: it can be proved that the “sweet spot” is not at the end of the bat by studying the quantitative relationship of the hitting effect and the inference of the energy transformation.4.2 Modeling and Solution to Problem II4.2.1 Model Preparation 1) Introduction to corked bat[5][6]: Fig 4-7As shown in Fig 4-7, Corking a bat the traditional way is a relatively easy thing to do. You just drill a hole in the end of the bat, about 1-inch in diameter, and about 10-inches deep. You fill the hole with cork, super balls, or styrofoam - if you leave the hole empty the bat sounds quite different, enough to give you away. Then you glue a wooden plug, like a 1-inch dowel, in to the end. Finally you sand the end to cover the evidence.Some sources suggest smearing a bit of glue on the end of the bat and sprinkling sawdust over it so help camouflage the work you have done.2) Situation studied：Situation of the best hitting effect: vertical collision occurs between the bat and the ball, and the energy loss of the collision is less than 10% and more than 90% of the momentum transfers from the bat to the ball (the hitting point is the “sweet spot”). Team # 8038 Page 14 of 203) Analysis of COR After the collision the ball rebounded backwards and the bat rotated about its pivot. The ratio of ball speeds (outgoing / incoming) is termed the collision efficiency, e A . A kinematic factor k , which is essentially the effective massof the bat, is defined as…………………………………………………………（4-6）I bat where I nat is the moment-of-inertia of the bat as measured about the pivot point on the handle, and z is the distance from the pivot point where the ball hits the bat. Once the kinematic factor k has been determined and the collision efficiency e A has been measured, the BBCOR is calculated from…………………………………………（4-7）Physical parameters vary with the material:The hitting effect of the “sweet spot” varies with the d ifferent bat material.It is related with the mass of the ball M , the center-of-mass ( CM ), the location of the center-of-mass d , the location of COP L , the coefficient of restitution BBCOR and the moment-of-inertia of the bat I bat .4.2.2 Controlling variable method analysisM is the mass of the object；is the location of the center-of-mass relative to the d pivot point；is the gravitational field strength；bat is the moment-of-inertia of the bat g I as measured about the pivot point on the handle; z is the distance from the pivot point where the ball hits the bat；vinl speed just before collision. The following formulas are got by sorting the above variables[1]:…………………………………………… （4-8 ）is the incoming ball speed；vbat is the bat swing………………………………………………（4-9）…………………………………………（4-10）Associating the above three formulas with formula (4-6) and (4-7), the formulas among BBS , the mass M , the center-of-mass ( CM ), the location of COP, the coefficient of restitution BBCOR and the moment-of-inertia of the bat I bat are:………………………（4-11）……………………………………………………………（4-12）………………………………………………………………（4-13）It can be known form formula (4-11), (4-12) and (4-13):1) When the coefficient of restitution BBCOR and mass M of the material changes, BBS will change;2) When mass M and the location of center-of mass CM changes, I bat changes, which is the dominant factor deciding the swing speed.4.2.3 Analysis of corked bat and wood bat [5][6]It makes the game unfair to increase the hitting accuracy by corking the bat.4.2.4 Reason for prohibiting corking[4]If the swing speed is unchanged, the corked bat cannot hit the ball as far as the wood bat, but it grants the player more reaction time and increases the accuracy. Influenced by a multitude of random factors, vertical collision cannot be assured in each hitting.The following figure shows the situation of vertical collision between the bat and the ball:In order to realize the best hitting effect, all of the BBS drawn from the above calculating results are assumed to be vertical collision.But in a professional baseball game, because the hitting accuracy is also one of the decisive factors, increasing the hitting accuracy equals to enhance the hitting effect. After cording the bat, the moment-of-inertia of the bat reduces, which improves the player’s capability of controlling the bat.Thus, the hitting is more accurate, which makes the game unfair.To sum up, in order to avoiding the unfairness of a game, Major League Baseball prohibits “corking”4.3 Modeling and Solution to Problem IIIAccording to the model developed in Problem I, the hitting effect of the “sweet spot” depends on the mass of the ball M , the center-of-mass ( CM ), the location of CM d , the location of COP L , the coefficient of restitution BBCOR and the moment-of-inertia of the bat I bat . An analysis of metal bat and wood bat is made.4.3.1 Analysis of metal bat and wood bat [8][9]I bat (1) is the moment-of-inertia of the metal bat, and I bat (2) is moment-of-inertia of the wood bat.Conclusion: Because the hitting part is hollow for the metal bat, the CM is closer tothe handle of bat for an aluminum bat than a wood bat. I bat of metal bat is less than I bat of the wood bat, which increases the swing speed.It means the professional players are able to watch the ball travel an additional 5-6 feet before having to commit to a swing, which makes the hitting more accurate to damage the fairness of the game.4.3.2 Reason for prohibiting the metal bat [4]Through the studies on the above models: 【4.3.1-(1)】proves the best hitting effect of a metal bat is better than a wood bat.【4.3.1-(2）】proves the hitting accuracy of a metal bat is better than a wood bat.To sum up，the metal bat is better than the wood bat in both the two factors, which makes the game unfair.And that’s why Major League Baseball prohibits metal bat.5.Strengths and Weaknesses of the Model5.1.Strengths1) The model, with the help of the single-pendulum theory and the analysis of the vibration of the bat and the ball, locates the COP and vibrational node of bats respectively, and locates the “sweet spot” influenced by multitudes of factors utilizing the integrated optimization method.The overall optimized solution makes the “sweet spot” more persuasive.2) The Model analyzes the integrated performance of different material bats from the aspects of the maximum initial velocity (BBS) and the hitting accuracy, and explains why corked bats are prohibited successfully.3) Deriving results from the controlling variable method analysis and taking the Law of Energy Conservation and the theories of structural dynamics as foundation enable to avoid the complicated mechanical analysis and derivation.5.2 WeaknessesThe model fails to evaluate the performance of bats exactly from the angle of the game relationship between the maximum initial velocity (BBS) and the accuracy when evaluating the hitting effect.6.References [1]/~drussell/bats-new/sweetspot.html [2] Mathematical Modeling Contest: Selection and Comment on Award-winning Papers [3].au/~cross/baseball.htm[4]Adair, R.K.1994.The Physics of Baseball. New York: Harper Perennial.[5]D.A.Russell,"Hoop frequency as a predictor of performance for softball bats," Engineering of Sport 5 Vol.2, pp.641-647 (International Sports Engineering Association, 2004).Proceedings of the 5th International Conference on the Engineering of Sport, UC Davis, September 11-15, 2004.[6] A.M.Nathan, "Some Remarks on Corked Bats" (June 10, 2003)[7] ESPN Baseball Tonight, on June 3, 2003 aired a nice segment in which Buck Showalter showed how to cork a bat, drilling the hole, filling it with cork, and plugging the end.[8] R.M. Greenwald R.M., L.H.Penna , and J.J.Crisco,"Differences in Batted Ball Speed with Wood and Aluminum Baseball Bats: A Batting Cage Study," J. Appl.Biomech., 17, 241-252 (2001).[9] J.J.Crisco, R.M.Greenwald, J.D.Blume, and L.H.Penna, "Batting performance ofwood and metal baseball bats," Med. Sci. Sports Exerc., 34(10), 1675-1684 (2002) [10] Robert K. Adair, The Physics of Baseball, 3rd Ed., (Harper Collins, 2002)[11] R. Cross, "The sweet spot of a baseball bat," Am. J. Phys., 66(9), 771-779 (1998)[12] A. M. Nathan, "The dynamics of the baseball-bat collision," Am. J. Phys., 68(11), 979-990 (2000)[13] K.Koenig, J.S.Dillard, D.K.Nance, and D.B.Shafer, "The effects of support conditions on baseball bat testing," Engineering of Sport 5 Vol.2, pp.87-93 (International Sports Engineering Association, 2004).Proceedings of the 5th International Conference on the Engineering of Sport, UC Davis, September 11-15, 2004.棒球的数学问题简介：棒球是一个受欢迎的bat-and-ball游戏,既包括体育和智慧。

数学专业英语翻译3.4.5

3.4数学的应用与应用数学3.4.5Construction of mathematical model（建立数学模型）In the preceding discussion we viewed modeling as a process and considered briefly the form of the model.Now let's focus attention on the construction of mathematical models.（在前面的讨论中，我们将建模视为一个过程，并简要地仔细思考了模型的形式。

现在让我们把注意力集中在数学模型的构建上。

）We begin by presenting an outline of a peocedure that is helpful in constructing models.In the next section,we illustrate the various steps in the procedure by discussing several real-world examples.（我们首先介绍一种有助于构建模型的概述。

在下一节中，我们通过讨论几个实际示例来说明过程中的各个步骤。

）STEP1Identify the problem（确定问题）What is the problem you would like to explore?Typically this is a difficult step because in real-life situations no one simply hands you a mathematical problem to solve.（你想探索的问题是什么？通常这是一个困难的步骤，因为在现实生活中，没有人只是给你一个数学问题来解决。

）Usually you have to sort through large amounts of data and identify some particular aspect of the situation to study.Moreover,it is imperative to be sufficiently precise(ultimately)in the formulation of the problem to allow for translation of the verbal statements describing the problem into mathematical symbology.（通常你必须对大量数据进行排序，并确定要研究的某些特定情况。

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