数学专业外文文献翻译

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课文2—AB数学专业英语翻译(第二版)吴炯圻

课文2—AB数学专业英语翻译(第二版)吴炯圻

2-A Why study geometry?Why do we study geometry? The student beginning the study of this text may well ask, "What is geometry? What can I expect to gain from this study?2-A为什么研究几何学?为什么我们研究几何学?刚开始学习这篇文章的学生会疑问,“几何是什么?研究几何我们能学到什么呢?Many leading institutions of higher learning have recognized that positive benefits can be gained by all who study this branch of mathematics. This is evident from the fact that they require study of geometry as a prerequisite to matriculation in those schools.许多居领导地位的学术机构承认,所有学习这个数学分支的人都将得到很好的收益。

事实是,他们需要学习几何作为学校入学考试的先决条件。

Geometry had its origin long ago in the measurements by the Babylonians and Egyptians of their lands inundated by the floods of the Nile River. The greek word geometry is derived from geo, meaning "earth," and metron, meaning "measure." As early as 2000 B. C. we find the land surveyors of these people reestablishing vanishing landmarks and boundaries by utilizing the truths of geometry.很早以前,几何学源于测量被尼罗河的洪水淹没了的巴比伦人和埃及人的土地。

数学专业英语中英文对照翻译

数学专业英语中英文对照翻译

2.5笛卡尔几何学的基本概念(basic concepts of Cartesian geometry)课文5-A the coordinate system of Cartesian geometryAs mentioned earlier, one of the applications of the integral is the calculation of area. Ordinarily , we do not talk about area by itself ,instead, we talk about the area of something . This means that we have certain objects (polygonal regions, circular regions, parabolic segments etc.) whose areas we wish to measure. If we hope to arrive at a treatment of area that will enable us to deal with many different kinds of objects, we must first find an effective way to describe these objects.The most primitive way of doing this is by drawing figures, as was done by the ancient Greeks. A much better way was suggested by Rene Descartes, who introduced the subject of analytic geometry (also known as Cartesian geometry). Descartes’ idea was to represent geometric points by numbers. The procedure for points in a plane is this :Two perpendicular reference lines (called coordinate axes) are chosen, one horizont al (called the “x-axis”), the other vertical (the “y-axis”). Their point of intersection denoted by O, is called theorigin. On the x-axis a convenient point is chosen to the right of O and its distance from O is called the unit distance. Vertical distances along the Y-axis are usually measured with the same unit distance ,although sometimes it is convenient to use a different scale on the y-axis. Now each point in the plane (sometimes called the xy-plane) is assigned a pair of numbers, called its coordinates. These numbers tell us how to locate the points.Figure 2-5-1 illustrates some examples.The point with coordinates (3,2) lies three units to the right of the y-axis and two units above the x-axis.The number 3 is called the x-coordinate of the point,2 its y-coordinate. Points to the left of the y-axis have a negative x-coordinate; those below the x-axis have a negtive y-coordinate. The x-coordinateof a point is sometimes called its abscissa and the y-coordinateis called its ordinate.When we write a pair of numberssuch as (a,b) to represent a point, we agree that the abscissa or x-coordinate,a is written first. For this reason, the pair(a,b) is often referred to as an ordered pair. It is clear that two ordered pairs (a,b) and (c,d) represent the same point if and only if we have a=c and b=d. Points (a,b) with both a and b positiveare said to lie in the first quadrant ,those with a<0 and b>0 are in the second quadrant ; and those with a<0 and b<0 are in the third quadrant ; and those with a>0 and b<0 are in the fourthquadrant. Figure 2-5-1 shows one point in each quadrant.The procedure for points in space is similar. We take three mutually perpendicular lines in space intersecting at a point (the origin) . These lines determine three mutually perpendicular planes ,and each point in space can be completely described by specifying , with appropriate regard for signs ,its distances from these planes. We shall discuee three-dimensional Cartesian geometry in more detail later on ; for the present we confine our attention to plane analytic geometry.课文5-A:笛卡尔几何坐标系正如前面所提到的,积分应用的一种是计算面积。

数学专业英语第二版的课文翻译

数学专业英语第二版的课文翻译

2-A Why study geometry Many leading institutions of higher learning have recognized that positive benefits can be gained by all who study this branch of mathematics. This is evident from the fact that they require study of geometry as a prerequisite to matriculation in those schools. 许多居于领导地位的学术机构承认,所有学习这个数学分支的人都将得到确实的受益,许多学校把几何的学习作为入学考试的先决条件,从这一点上可以证明。

Geometry had its origin long ago in the measurement by the Babylonians and Egyptians of their lands inundated by the floods of the Nile River. The greek word geometry is derived from geo, meaning “earth” and metron, meaning “measure” . As early as 2000 . we find the land surveyors of these people re-establishing vanishing landmarks and boundaries by utilizing the truths of geometry . 几何学起源于很久以前巴比伦人和埃及人测量他们被尼罗河洪水淹没的土地,希腊语几何来源于geo ,意思是”土地“,和metron 意思是”测量“。

数学专业英语课文翻译

数学专业英语课文翻译

1-A:什么是数学数学来源于人类的社会实践,包括工农业的劳动,商业、军事和科学技术研究等活动。

反过来,数学服务于实践并在所有领域扮演一个重要的角色。

没有数学的应用,现代化科学和技术的分支都不能有规律的发展。

从早期人类的需求引出了数和形状。

然后,几何学因测量陆续的发展出来,三角学来自于勘探问题。

为了处理一些更复杂的实践问题,人们建立了方程,通过求解方程的未知数,从而代数学出现了。

17世纪之前,人们局限于初等数学,例如几何、三角和代数,那些只考虑常数。

17世纪工业的迅速发展促进了经济学和科技的发展,并且我们需要处理变量。

从常数到变量的跳跃带来了两个属于高等数学的新的数学分支,解析几何和微积分学。

现在,高等数学中有了许多分支,数学分析、高等代数、微分方程、函数论等。

数学家们研究概念和命题。

公理、公社、定义和定理都是命题。

符号是一种特别并且很重要的数学工具,它常用于表示概念和命题。

公式、图形和表格充满着不同的符号。

阿拉伯数字1,2,3,4,5,6,7,8,9,0和加”+”减”-”乘”*”除”/”等号”=”使我们最熟悉的数学符号。

主要通过逻辑推导和计算来获得数学结论。

在数学史的很长的时期内,逻辑推论一直占据数学方法的中心地位。

现在,自从电子计算机迅速发展和广泛应用,计算的角色越来越重要。

现在,计算不仅用来处理信息与数据,而且用来完成一些在以前只能靠逻辑推理来做的工作,例如证明大多数的几何定理。

1-B:等式等式是关于两个数或数的符号相等的一种陈述。

因此a(a-5)=a^2-5a和x-3=5是等式。

等式有两种,恒等式和条件等式。

算术和代数恒等式是等式。

这种等式的两端要么一样,要么经过执行指定的运算后变成一样。

因此12-2=2+8,(m-n)(m+n)=m^2-n^2是恒等式。

含有字母的恒等式对其中字母的任何一组数值都成立。

因此恒等式x(a+2)=ax+2x变成3(7+2)=21+6或27=27,比如当x=3和a=7。

数学外文翻译

数学外文翻译

Power Series Expansion and Its ApplicationsIn the previous section, we discuss the convergence of power series, in its convergence region, the power series always converges to a function. For the simple power series, but also with itemized derivative, or quadrature methods, find this and function. This section will discuss another issue, for an arbitrary function ()f x , can be expanded in a power series, and launched into.Whether the power series ()f x as and function? The following discussion will address this issue. 1 Maclaurin (Maclaurin) formulaPolynomial power series can be seen as an extension of reality, so consider the function ()f x can expand into power series, you can from the function ()f x and polynomials start to solve this problem. To this end, to give here without proof the following formula.Taylor (Taylor) formula, if the function ()f x at 0x x = in a neighborhood that until the derivative of order 1n +, then in the neighborhood of the following formula :20000()()()()()()n n f x f x x x x x x x r x =+-+-++-+… (9-5-1)Among10()()n n r x x x +=-That ()n r x for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.If so 00x =, get2()(0)()n n f x f x x x r x=+++++…, (9-5-2) At this point,(1)(1)111()()()(1)!(1)!n n n n n f f x r x x x n n ξθ+++++==++ (01θ<<).That (9-5-2) type formula for the Maclaurin.Formula shows that any function ()f x as long as until the 1n +derivative, n can be equal to a polynomial and a remainder.We call the following power series()2(0)(0)()(0)(0)2!!n nf f f x f f x x x n '''=+++++…… (9-5-3) For the Maclaurin series.So, is it to ()f x for the Sum functions? If the order Maclaurin series (9-5-3) the first 1n + itemsand for 1()n S x +, which()21(0)(0)()(0)(0)2!!n nn f f S x f f x x x n +'''=++++…Then, the series (9-5-3) converges to the function ()f x the conditions1lim ()()n n s x f x +→∞=.Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between theknown1()()()n n f x S x r x +=+Thus, when()0n r x =There,1()()n f x S x +=Vice versa. That if1lim ()()n n s x f x +→∞=,Units must()0n r x =.This indicates that the Maclaurin series (9-5-3) to ()f x and function as the Maclaurin formula (9-5-2) of the remainder term ()0n r x → (when n →∞).In this way, we get a function ()f x the power series expansion:()()0(0)(0)()(0)(0)!!n n n nn f f f x x f f x x n n ∞='==++++∑……. (9-5-4) It is the function ()f x the power series expression, if, the function of the power series expansion is unique. In fact, assuming the function f (x ) can be expressed as power series20120()n n n n n f x a x a a x a x a x ∞===+++++∑……, (9-5-5)Well, according to the convergence of power series can be itemized within the nature of derivation,and then make 0x = (power series apparently converges in the 0x = point), it is easy to get()2012(0)(0)(0),(0),,,,,2!!n nn f f a f a f x a x a x n '''====…….Substituting them into (9-5-5) type, income and ()f x the Maclaurin expansion of (9-5-4) identical. In summary, if the function f (x ) contains zero in a range of arbitrary order derivative, and in this range of Maclaurin formula in the remain der to zero as the limit (when n → ∞,), then , the function f (x ) can start forming as (9-5-4) type of power series.Power Series()20000000()()()()()()()()1!2!!n n f x f x f x f x f x x x x x x x n '''=+-+-++-……,Known as the Taylor series.Second, primary function of power series expansionMaclaurin formula using the function ()f x expanded in power series method, called the direct expansion method.Example 1Test the function ()x f x e =expanded in power series of x . Solution because()()n x f x e =,(1,2,3,)n =…Therefore()(0)(0)(0)(0)1n f f f f '''====…,So we get the power series21112!!n x x x n +++++……, (9-5-6) Obviously, (9-5-6)type convergence interval (,)-∞+∞, As (9-5-6)whether type ()x f x e = is Sum function, that is, whether it converges to ()xf x e = , but also examine remainder ()n r x . Because1e ()(1)!xn n r x x n θ+=+ (01θ<<),且x x x θθ≤≤,Therefore11e e ()(1)!(1)!xx n n n r x x x n n θ++=<++,Noting the value of any set x ,xe is a fixed constant, while the series (9-5-6) is absolutely convergent, sothe general when the item when n →∞, 10(1)!n xn +→+ , so when n → ∞,there10(1)!n xx e n +→+, From thislim ()0n n r x →∞=This indicates that the series (9-5-6) does converge to ()x f x e =, therefore21112!!x n e x x x n =+++++…… (x -∞<<+∞). Such use of Maclaurin formula are expanded in power series method, although the procedure is clear,but operators are often too Cumbersome, so it is generally more convenient to use the following power series expansion method.Prior to this, we have been a functionx-11, xe and sin x power series expansion, the use of these known expansion by power series of operations, we can achieve many functions of power series expansion. This demand function of power series expansion method is called indirect expansion .Example 2Find the function ()cos f x x =,0x =,Department in the power series expansion. Solution because(sin )cos x x '=,And3521111sin (1)3!5!(21)!n n x x x x x n +=-+-+-++……,(x -∞<<+∞)Therefore, the power series can be itemized according to the rules of derivation can be342111cos 1(1)2!4!(2)!n nx x x x n =-+-+-+……,(x -∞<<+∞) Third, the function power series expansion of the application exampleThe application of power series expansion is extensive, for example, can use it to set some numerical or other approximate calculation of integral value.Example 3 Using the expansion to estimate arctan x the value of π.Solution because πarctan14= Because of357arctan 357x x x x x =-+-+…, (11x -≤≤),So there1114arctan14(1)357π==-+-+…Available right end of the first n items of the series and as an approximation of π. However, the convergence is very slow progression to get enough items to get more accurate estimates of πvalue.此外文文献选自于:Walter.Rudin.数学分析原理(英文版)[M].北京:机械工业出版社.幂级数的展开及其应用在上一节中,我们讨论了幂级数的收敛性,在其收敛域内,幂级数总是收敛于一个和函数.对于一些简单的幂级数,还可以借助逐项求导或求积分的方法,求出这个和函数.本节将要讨论另外一个问题,对于任意一个函数()f x ,能否将其展开成一个幂级数,以及展开成的幂级数是否以()f x 为和函数?下面的讨论将解决这一问题.一、 马克劳林(Maclaurin)公式幂级数实际上可以视为多项式的延伸,因此在考虑函数()f x 能否展开成幂级数时,可以从函数()f x 与多项式的关系入手来解决这个问题.为此,这里不加证明地给出如下的公式.泰勒(Taylor)公式 如果函数()f x 在0x x =的某一邻域内,有直到1n +阶的导数,则在这个邻域内有如下公式:()20000000()()()()()()()()()2!!n n n f x f x f x f x f x x x x x x x r x n '''=+-+-++-+…,(9-5-1)其中(1)10()()()(1)!n n n f r x x x n ξ++=-+.称()n r x 为拉格朗日型余项.称(9-5-1)式为泰勒公式. 如果令00x =,就得到2()(0)()n n f x f x x x r x =+++++…, (9-5-2)此时,(1)(1)111()()()(1)!(1)!n n n n n f f x r x x x n n ξθ+++++==++, (01θ<<).称(9-5-2)式为马克劳林公式.公式说明,任一函数()f x 只要有直到1n +阶导数,就可等于某个n 次多项式与一个余项的和. 我们称下列幂级数()2(0)(0)()(0)(0)2!!n nf f f x f f x x x n '''=+++++…… (9-5-3)为马克劳林级数.那么,它是否以()f x 为和函数呢?若令马克劳林级数(9-5-3)的前1n +项和为1()n S x +,即()21(0)(0)()(0)(0)2!!n nn f f S x f f x x x n +'''=++++…,那么,级数(9-5-3)收敛于函数()f x 的条件为1lim ()()n n s x f x +→∞=.注意到马克劳林公式(9-5-2)与马克劳林级数(9-5-3)的关系,可知1()()()n n f x S x r x +=+.于是,当()0n r x =时,有1()()n f x S x +=.反之亦然.即若1lim ()()n n s x f x +→∞=则必有()0n r x =.这表明,马克劳林级数(9-5-3)以()f x 为和函数⇔马克劳林公式(9-5-2)中的余项()0n r x → (当n →∞时).这样,我们就得到了函数()f x 的幂级数展开式:()()20(0)(0)(0)()(0)(0)!2!!n n n nn f f f f x x f f x x x n n ∞='''==+++++∑……(9-5-4) 它就是函数()f x 的幂级数表达式,也就是说,函数的幂级数展开式是唯一的.事实上,假设函数()f x 可以表示为幂级数20120()n n n n n f x a x a a x a x a x ∞===+++++∑……, (9-5-5)那么,根据幂级数在收敛域内可逐项求导的性质,再令0x =(幂级数显然在0x =点收敛),就容易得到()2012(0)(0)(0),(0),,,,,2!!n nn f f a f a f x a x a x n '''====…….将它们代入(9-5-5)式,所得与()f x 的马克劳林展开式(9-5-4)完全相同.综上所述,如果函数()f x 在包含零的某区间内有任意阶导数,且在此区间内的马克劳林公式中的余项以零为极限(当n →∞时),那么,函数()f x 就可展开成形如(9-5-4)式的幂级数.幂级数()00000()()()()()()1!!n n f x f x f x f x x x x x n '=+-++-……,称为泰勒级数.二、 初等函数的幂级数展开式利用马克劳林公式将函数()f x 展开成幂级数的方法,称为直接展开法. 例1 试将函数()x f x e =展开成x 的幂级数. 解 因为()()n x f x e =, (1,2,3,)n =…所以()(0)(0)(0)(0)1n f f f f '''====…,于是我们得到幂级数21112!!n x x x n +++++……, (9-5-6) 显然,(9-5-6)式的收敛区间为(,)-∞+∞,至于(9-5-6)式是否以()x f x e =为和函数,即它是否收敛于()xf x e =,还要考察余项()n r x .因为1e ()(1)!xn n r x x n θ+=+ (01θ<<), 且x x x θθ≤≤,所以11e e ()(1)!(1)!xx n n n r x x x n n θ++=<++.注意到对任一确定的x 值,xe 是一个确定的常数,而级数(9-5-6)是绝对收敛的,因此其一般项当n →∞时,10(1)!n xn +→+,所以当n →∞时,有10(1)!n xx e n +→+, 由此可知lim ()0n n r x →∞=.这表明级数(9-5-6)确实收敛于()x f x e =,因此有21112!!x n e x x x n =+++++…… (x -∞<<+∞). 这种运用马克劳林公式将函数展开成幂级数的方法,虽然程序明确,但是运算往往过于繁琐,因此人们普遍采用下面的比较简便的幂级数展开法.在此之前,我们已经得到了函数x-11,xe 及sin x 的幂级数展开式,运用这几个已知的展开式,通过幂级数的运算,可以求得许多函数的幂级数展开式.这种求函数的幂级数展开式的方法称为间接展开法.例2 试求函数()cos f x x =在0x =处的幂级数展开式. 解 因为(sin )cos x x '=,而3521111sin (1)3!5!(21)!n n x x x x x n +=-+-+-++……,(x -∞<<+∞), 所以根据幂级数可逐项求导的法则,可得342111cos 1(1)2!4!(2)!n nx x x x n =-+-+-+……,(x -∞<<+∞). 三、 函数幂级数展开的应用举例幂级数展开式的应用很广泛,例如可利用它来对某些数值或定积分值等进行近似计算. 例3 利用arctan x 的展开式估计π的值. 解 由于πarctan14=,又因357arctan 357x x x x x =-+-+…, (11x -≤≤),所以有1114arctan14(1)357π==-+-+….可用右端级数的前n 项之和作为π的近似值.但由于级数收敛的速度非常慢,要取足够多的项才能得到π的较精确的估计值.此外文文献选自于:Walter.Rudin.数学分析原理(英文版)[M].北京:机械工业出版社.。

数学专业英语课文翻译(吴炯圻)第二章2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

数学专业英语课文翻译(吴炯圻)第二章2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

数学专业英语3—A符号指示集一组的概念如此广泛利用整个现代数学的认识是所需的所有大学生。

集是通过集合中一种抽象方式的东西的数学家谈的一种手段。

集,通常用大写字母:A、B、C、进程运行·、X、Y、Z ;由小写字母指定元素:a、b 的c、进程运行·,若x、y z.我们用特殊符号x∈S 意味着x 是S 的一个元素或属于美国的x如果x 不属于S,我们写xS.≠当方便时,我们应指定集的元素显示在括号内;例如,由符号表示的积极甚至整数小于10 集{2,468} {2,4.6,进程运行·} 作为显示的所有积极甚至整数集,而三个点等的发生。

点的和等等的意思是清楚时,才使用。

上市的大括号内的一组成员方法有时称为名册符号。

涉及到另一组的第一次基本概念是平等的集。

DEFINITIONOFSETEQUALITY。

两组A 和B,据说是平等的(或相同的)如果它们包含完全相同的元素,在这种情况下,我们写A = B。

如果其中一套包含在另一个元素,我们说这些集是不平等,我们写A = B。

EXAMPLE1。

根据对这一定义,由于他们都是由构成的这四个整数2,4.6 和8 两套{2,468} 和{2,864} 一律平等。

因此,当我们用来描述一组的名册符号,元素的显示的顺序无关。

动作。

集{2,468} 和{2,2,4,4,6,8} 是平等的即使在第二组,每个元素2 和4 两次列出。

这两组包含的四个要素2,468 和无他人;因此,定义要求我们称之为这些集平等。

此示例显示了我们也不坚持名册符号中列出的对象是不同。

类似的例子是一组在密西西比州,其值等于{M、我、s、p} 一组单词中的字母,组成四个不同字母M、我、s 和体育3 —B 子集S.从给定的集S,我们可能会形成新集,称为.的子集例如,组成的那些正整数小于10 整除4 (集合{8 毫米})的一组一般是的所有甚至小于10.整数集的一个子集,我们有以下的定义。

子集的定义。

数学专业英语论文(含中文版)

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。

数学专业英语课文句子翻译ok

数学专业英语课文句子翻译ok

There are many other less direct benefits the students of geometry may gain……mathematicinans to culture andcivilization学习几何的学生可以获得许多其他不太直接的利益,这些人当中必须包括训练在英语语言的精确使用和分析一个新情况或者问题直达要害的能力,以及利用毅力,创造性和逻辑思维解决是问题。

欣赏大自然的创作将是几何研究的副产品。

学生还应该发展数学和数学家们对我们的文化和文明作出的贡献的认知。

10-A Although dependence and independence are properties of sets of elements, we also apply these terms to the elements themselves. For example, the elements in an independent set are called independent elements. 虽然相关和无关是元素集的属性,我们也适用于这些元素本身。

例如,在一个独立设定的元素被称为独立元素。

If s is finite set, the foregoing definition agrees with that given in Chapter 8 for the space nV. However, the present definition is not restricted to finite sets. 如果S是有限集,同意上述定义与第8章中给出的空间nV,然而,目前的定义不局限于有限集。

If a subset T of a set S is dependent, then S itself is dependent. This is logically equivalent to the statement that every subset of an independent set is independent. 如果集合S的子集T是相关的,然后S本身是相关的,这在逻辑上相当于每一个独立设置的子集是独立的语句。

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1 / 29 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 2 / 29

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: 10wRp(3.1)

其中,R=E[()x()]T

xkk

且p=E[d()x()] kk,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()Rk和()pk,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()wkkk

w()(()()w())kpkRkk+2(3.2)

其中,k=0,1,2,…,g()wk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()TRkkk ()()x()pkdkk(3.3)

得到的梯度估计值为 3 / 29

()2()x()2x()x()()Twgkdkkkkwk 2x()(()x()())Tkdkkwk 2()x()ekk(3.4)

注意,如果目标函数用瞬时平方误差2()ek

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()()()()Tekekekekekekekwwkwkwk





L

2()x()ekk ()wgk(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()wkwkekk(3.6) 其中,收敛因子应该在一个范围内取值,以保证收敛性。

图3.1表示了对延迟线输入x()k的LMS算法实现。典型情况是,LMS算法的每次迭代需要N+2次乘法(用于滤波器系数的更新),而且还需要N+1次乘法(用于产生误差信号)。

LMS算法的详细描述见算法3.1 4 / 29 图3.1LMS自适应RH滤波器

算法3.1 LMS算法 5 / 29

Initialization x(0)(0)[000]TwL Dofor0k ()()x()()Tekdkkwk (1)()2()x()wkwkekk 需要指出的是,初始化并不一定要像在算法3.1小那样将白

适应滤波器的系数被创始化为零:比如,如果知道最优系数的粗略值,则可以利用这些值构成w(0),这样可以减少到达0w的

邻域所需的迭代次数。 3.3 LMS算法的一些特性 在本节中,描述丁在平稳环境下与LMS算法收敛特性相关的主要特性。这里给出的信息对于理解收敛因子对LMS算法的各个收敛方面的影响是很重要的。 3.3.1 梯度特性 正如第2章中所指出的(见式(2.79)),在MSE曲面上完成搜索最优系数向量解的理想梯度方向为 ()2{[x()x()]()[()x()]}TwgkEkkwkEdkk 2[()]Rwkp(3.7) 在LMS算法中,利用R和p的瞬时估计值确定搜索方向,即 6 / 29

()2[x()x()()()x()]Twgkkkwkdkk(3.8)

正如所期望的,由式(3.8)所确定的方向与式(3.7)所确定的方向

很不同。因此,当通过利用LMS算法计算更加有效的梯度方向时,收敛特性与最陡下降算法的收敛特性并不相同。 从平均的意义上讲,可以说LMS梯度方向具有接近理想梯度方向的趋势,因为对于固定购系数向量w,有

[()]2{[x()x()][()x()]}TwEgkEkkwEdkk wg(3.9)

因此,向量g()wk可以解释为wg的无偏瞬时估计值。在具有遍历

件的环境中,如果对于一个固定的w,利用大量的输入和参考信号来计算向量g()wk,则平均方向趋近于wg,即

11lim()MwwMigkigM(3.10)

3.3.2 系数向量的收敛特性

假设一个系数向量为w。的未知FIR滤波器,被一个具备相同阶数的白适应FIR滤波器利用LMS算法进行辨识。在未知系统输出令附加了测量白噪声n(k),其均值为零,方差为2

n

在每一次迭代中,自适应滤波器系数相对于理想系数向量0w,

的误差由N+1维向量描述: 7 / 29

0()()wkwkw(3.11) 利用这种定义,LMS算法也可以另外描述为 (1)()2()x()wkwkekk 0()2x()[x()x()()]TTwkkkwkwk

0()2x()[x()()]Twkkekwk

0[2x()x()]()2()x()TIkkwkekk(3.12) 其中,0()ek为最优输出误差.它由下式给出:

00()()x()Tekdkwk 00x()()x()TTwknkwk ()nk(3.13) 于是,系数向量中的期望误差为

0[(1)]{[2x()x()]()2[()x()]}TEwkEIkkwkEekk(3.14) 假设x()k的元素与()wk和0()ek的元素统计独立,则式(314)可以简

化为

[(1)]{2[x()x()]}[()]TEwkIEkkEwk (2)[()]IREwk(3.15) 如果我们假设参数的偏差只依赖于以前的输入信号向量,则第一个假设成立,而在第二个假设中,我们也考虑了最优解对应的误差信号与输入信号向量的元素正交。由上述表达式可得 8 / 29

1[(1)](2)[(0)]kEwkIREw

(3.16)

如果将式(3.15)左乘QT(其中Q为通过一个相似变换使R对角化的酉矩阵),则可以得到 [(1)](2)[()]TTTEQwkIQRQEQwk '[(1)]Ewk

'(2)[()]IEwk

01'

1200012[()]0012NEwk







LMMMOM(3.17)

其中,'(1)(1)TwkQwk

为旋转系数误差向量。应用旋转可以

得到一个产生对角矩阵的方程,从而更加易于分析方程的动态特性。另外.上述关系可以表示为 '1'[(1)](2)[(0)]kEwkIEw

101'11(12)000(12)[(0)]00(12)kkkNEw











LMMMOM(3.18)

该方程说明.为了保证系数在平均意义上收敛,LMS算法的收敛因子必须在如下范围内选取:

max10

(3.19) 9 / 29

其中,max为R的最大持征值。在该范围内的值保证了当

k

时,式(3.18)中对角矩阵的所有元素趋近于零.这是因为对于i

=0,l,…,N,有1(12)1

i

。因此,对于较大的k值,'[(1)]Ewk

趋近于零。 按照上述方法选取的值确保了系数向量的平均值接近于员优系数向量0w比该指出的是,如果矩阵R具有大的特征值扩展,

则建议选择远小于上界值。因此,系数的收敛速度将主要取决于最小特征值,它对应于式(3.18)中的最慢模式。 上述分析中的关键假设是所谓的独立件理论[4],它考虑了当i0,1,…,k时,所有向量()xi均为统计独立的情况。这个假设允许我们考虑在式(3.14)中()wk独立于()x()T

xkk

。尽管在x()k由

延迟线元素组成时,这个假设并不是非常有效,但是由它得到的理论结果与实验结果能够很好地吻合。 3.3.3 系数误差向量协方差矩阵 在本节中,我们将推导得出自适应滤波器系数误差的二阶统计量表达式。由于对于大的k值,()wk的平均值为零,因此系数误差向量的协方差的定义为

00cov[()][()()]{[()][()]}TTwkEwkwkEwkwwkw(3.20) 将式(3.12)代人式(3.20),可以得到

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