数学专业英语第二版 课后答案

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。

数学专业英语课后答案2.1数学、方程与比例词组翻译1.数学分支branches of mathematics，算数arithmetics，几何学geometry，代数学algebra，三角学trigonometry，高等数学higher mathematics，初等数学elementary mathematics，高等代数higher algebra，数学分析mathematical analysis，函数论function theory，微分方程differential equation2.命题proposition，公理axiom，公设postulate，定义definition，定理theorem，引理lemma，推论deduction3.形form，数number，数字numeral，数值numerical value，图形figure，公式formula，符号notation(symbol)，记法/记号sign，图表chart4.概念conception，相等equality，成立/真true，不成立/不真untrue，等式equation，恒等式identity，条件等式equation of condition，项/术语term，集set，函数function，常数constant，方程equation，线性方程linear equation，二次方程quadratic equation5.运算operation，加法addition，减法subtraction，乘法multiplication，除法division，证明proof，推理deduction，逻辑推理logical deduction6.测量土地to measure land，推导定理to deduce theorems，指定的运算indicated operation，获得结论to obtain the conclusions，占据中心地位to occupy the centric place汉译英（1）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。

2.12.比:ratio 比例:proportion 利率:interest rate 速率:speed 除:divide 除法:division 商:quotient 同类量:like quantity 项:term 线段:line segment 角:angle 长度:length 宽:width高度:height 维数:dimension 单位:unit 分数:fraction 百分数:percentage3.(1)一条线段和一个角的比没有意义,他们不是相同类型的量.(2)比较式通过说明一个量是另一个量的多少倍做出的,并且这两个量必须依据相同的单位.(5)为了解一个方程,我们必须移项,直到未知项独自处在方程的一边,这样就可以使它等于另一边的某量.4.(1)Measuring the length of a desk, is actually comparing the length of the desk to that of a ruler.(3)Ratio is different from the measurement, it has no units. The ratio of the length and the width of the same book does not vary when the measurement unit changes.(5)60 percent of students in a school are female students, which mean that 60 students out of every 100 students are female students.2.22.初等几何:elementary geometry 三角学:trigonometry 余弦定理:Law of cosines 勾股定理/毕达哥拉斯定理:Gou-Gu theorem/Pythagoras theorem 角:angle 锐角:acute angle 直角:right angle 同终边的角:conterminal angles 仰角:angle of elevation 俯角:angle of depression 全等:congruence 夹角:included angle 三角形:triangle 三角函数:trigonometric function直角边:leg 斜边:hypotenuse 对边:opposite side 临边:adjacent side 始边:initial side 解三角形:solve a triangle 互相依赖:mutually dependent 表示成:be denoted as 定义为:be defined as3.(1)Trigonometric function of the acute angle shows the mutually dependent relations between each sides and acute angle of the right triangle.(3)If two sides and the included angle of an oblique triangle areknown, then the unknown sides and angles can be found by using the law of cosines.(5)Knowing the length of two sides and the measure of the included angle can determine the shape and size of the triangle. In other words, the two triangles made by these data are congruent.4.(1)如果一个角的顶点在一个笛卡尔坐标系的原点并且它的始边沿着x轴正方向，这个角被称为处于标准位置.(3)仰角和俯角是以一条以水平线为参考位置来测量的,如果正被观测的物体在观测者的上方,那么由水平线和视线所形成的角叫做仰角.如果正被观测的物体在观测者的下方,那么由水平线和视线所形成的的角叫做俯角.(5)如果我们知道一个三角形的两条边的长度和对着其中一条边的角度,我们如何解这个三角形呢?这个问题有一点困难来回答,因为所给的信息可能确定两个三角形,一个三角形或者一个也确定不了.2.32.素数:prime 合数:composite 质因数:prime factor/prime divisor 公倍数:common multiple 正素因子: positive prime divisor 除法算式:division equation 最大公因数:greatest common divisor(G.C.D) 最小公倍数: lowest common multiple(L.C.M) 整除:divide by 整除性:divisibility 过程:process 证明:proof 分类:classification 剩余:remainder辗转相除法:Euclidean algorithm 有限集:finite set 无限的:infinitely 可数的countable 终止:terminate 与矛盾:contrary to3.(1)We need to study by which integers an integer is divisible, that is , what factor it has. Specially, it is sometime required that an integer is expressed as the product of its prime factors.(3)The number 1 is neither a prime nor a composite number;A composite number in addition to being divisible by 1 and itself, can also be divisible by some prime number.(5)The number of the primes bounded above by any given finite integer N can be found by using the method of the sieve Eratosthenes.4.(1)数论中一个重要的问题是哥德巴赫猜想,它是关于偶数作为两个奇素数和的表示.(3)一个数,形如2p-1的素数被称为梅森素数.求出5个这样的数.(5)任意给定的整数m和素数p,p的仅有的正因子是p和1,因此仅有的可能的p和m的正公因子是p和1.因此,我们有结论:如果p是一个素数,m是任意整数,那么p整除m,要么(p,m)=1.2.42.集:set 子集:subset 真子集:proper subset 全集:universe 补集:complement 抽象集:abstract set 并集:union 交集:intersection 元素:element/member 组成:comprise/constitute包含:contain 术语:terminology 概念:concept 上有界:bounded above 上界:upper bound 最小的上界:least upper bound 完备性公理:completeness axiom3.(1)Set theory has become one of the common theoretical foundation and the important tools in many branches of mathematics.(3)Set S itself is the improper subset of S; if set T is a subset of S but not S, then T is called a proper subset of S.(5)The subset T of set S can often be denoted by {x}, that is, T consists of those elements x for which P(x) holds.(7)This example makes the following question become clear, that is, why may two straight lines in the space neither intersect nor parallel.4.(1)设N是所有自然数的集合,如果S是所有偶数的集合,那么它在N中的补集是所有奇数的集合.(3)一个非空集合S称为由上界的,如果存在一个数c具有属性:x<=c对于所有S中的x.这样一个数字c被称为S的上界.(5)从任意两个对象x和y，我们可以形成序列（x，y），它被称为一个有序对，除非x=y，否则它当然不同于（y，x）.如果S和T是任意集合，我们用S*T表示所有有序对（x,y),其中x术语S，y属于T.在R.笛卡尔展示了如何通过实轴和它自己的笛卡尔积来描述平面的点之后，集合S*T被称为S和T的笛卡尔积.2.52.竖直线:vertical line 水平线:horizontal line 数对:pairs of numbers 有序对:ordered pairs 纵坐标:ordinate 横坐标:abscissas 一一对应:one-to-one 对应点:corresponding points圆锥曲线:conic sections 非空图形:non vacuous graph 直立圆锥:right circular cone 定值角:constant angle 母线:generating line 双曲线:hyperbola 抛物线:parabola 椭圆:ellipse退化的:degenerate 非退化的:nondegenerate任意的:arbitrarily 相容的:consistent 在几何上:geometrically 二次方程:quadratic equation 判别式:discriminant 行列式:determinant3.(1)In the planar rectangular coordinate system, one can set up aone-to-one correspondence between points and ordered pairs of numbers and also a one-to-one correspondence between conic sections and quadratic equation.(3)The symbol can be used to denote the set of ordered pairs（x，y）such that the ordinate is equal to the cube of the abscissa.（5）According to the values of the discriminate，the non-degenerate graph of Equation （iii） maybe known to be a parabola， a hyperbolaor an ellipse.4.（1）在例1，我们既用了图形，也用了代数的代入法解一个方程组（其中一个方程式二次的，另一个是线性的）。

+...+an a4 +...+an SL = n + ( a2 +a3a ) + ( a1 +a3 ++ ) + ... + ( a2 1 1)a1 1)an SL = n + (n− + ... + (n− a1 an a1 +...+ak−1 +ak +...+an ...+an−1 ) + · · · + ( a1 +a2 + ) ak an
1 3 SL = (∑3 j=1 a j )(∑k=1 ak ) 1 1 1 SL = (∑3 j=1 a j )( a1 + a2 + a3 ) 1 1 1 +a +a ) SL = (a1 + a2 + a3 )( a 1 2 3 1 1 1 1 1 1 1 1 1 SL = a1 ( a +a +a ) + a2 ( a +a +a ) + a3 ( a +a +a ) 1 2 3 1 2 3 1 2 3 a3 a3 a2 a2 a1 1 +a SL = 1 + a a3 + a1 + 1 + a3 + a1 + a2 + 1 2 +a3 +a3 +a2 SL = 3 + a2a + a1a + a1a 1 2 3 ak 3 SR = ∑3 j=1 ∑k=1 ak a3 a2 a1 SR = ∑3 j=1 ( a1 + a2 + a3 )
SL = n + (n − 1)n = n2 = SR

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Cse547 Chapter 1, problem 9
(a)
y If we want to get P(n‐1) from P(n), we have to eliminate xn by constant or x1…xn‐1. y If using constant, the left seems good, the right hand side is hard to go on: the constant c is hard to deal withFra bibliotek(b)
y The About method uses P(n), then P(2). y What if we use P(2), then P(n)? y The method is to aggregate x(2i‐1) and x(2i) into one part:
(b)
(c)
y The idea is to use a special version of mathematical induction as this form: y Basis: the statement holds when n=1or2. y Inductive: if the statement holds for n<=k, then the same statement holds for n=k+1
(c)
y Basis: for n=1, x1·(x1/1)1. So we have P(1) for n=2, x1+x2·((x1+x2)/2)2. So we have P(2)
(c)
y Inductive step: Suppose for n<=k, we have P(n) then for n=k+1, we do a case analysis to show P(n) still holds:

1-A What is mathematics1-A 数学是什么？Mathematics comes from man' s social practice, for example, industrial and agricultural production, commercial activities, military operations and scientific and technological researches. And in turn, mathematics serves the practice and plays a great role in all fields. No modern scientific and technological branches could be regularly developed without the application of mathematics.数学来源于人类的社会实践，例如，工农业生产，商业活动，军事行动和科学技术研究。

总之，数学服务于实践且在众多领域扮演重要角色。

没有数学的应用，现代科学技术分支也不会得到广泛地发展。

From the early need of man came the concepts of numbers and forms. Then, geometry developed out of problems of measuring land, and trigonometry came from problems of surveying. To deal with some more complex practical problems, man established and then solved equation with unknown numbers, thus algebra occurred. Before 17th century, man confined himself to the elementary mathematics, i. e. geometry, trigonometry and algebra, in which only the constants were considered.早期数字的概念和形成源于人类的需要。

Discrete Mathematics 10
=
= xm(x + m)n
2009-04-21
Proof continue
Case 4: integers m<0 and n>0
+n – Ifxm (m+n)>=0 = x(x +1)...(x + m+ n −1) (x + m)(x + m+1)...x(x +1)...(x + m+ n +1) = (x + m)(x + m+1)...(x −1)
• x m and x m +1 are similar, only differ in one factor.
2009-04-21 Discrete Mathematics 3
• Conjecture, for integers m,n>=0 xm+1 = xm(x + m) • More general version, for integers m,n>=0
−n
= xm(x + m)n
2009-04-21 Discrete Mathematics 9
• If (m+n)<0
x
m+n
Case 3 continue
1
=
(x + m+ n)−(m+n) 1 = (x + m+ n)(x + m+ n +1)...(x −1) x(x +1)...(x + m−1) = (x + m+ n)(x + m+ n +1)...(x −1)x...(x + m−1) xm (x + m+ n)−n

数学专业英语课后答案1、He kept walking up and down, which was a sure（）that he was very worried. [单选题] *A. sign(正确答案)B. characterC. natureD. end2、20．Jerry is hard-working. It’s not ______ that he can pass the exam easily. [单选题] * A．surpriseB．surprising (正确答案)C．surprisedD．surprises3、--What’s the weather like today?--It’s _______. [单选题] *A. rainB. windy(正确答案)C. sunD. wind4、Many of my classmates are working _______volunteers. [单选题] *A. as(正确答案)B. toC. atD. like5、Mary is interested ______ hiking. [单选题] *A. onB. byC. in(正确答案)D. at6、We need a _______ when we travel around a new place. [单选题] *A. guide(正确答案)B. touristC. painterD. teacher7、While they were in discussion, their manager came in by chance. [单选题] *A. 抓住时机C. 碰巧(正确答案)D. 及时8、Our campus is _____ big that we need a bike to make it. [单选题] *A. veryB. so(正确答案)C. suchD. much9、57．Next week will be Lisa's birthday. I will send her a birthday present ________ post. [单选题] *A．withB．forC．by(正确答案)D．in10、The manager demanded that all employees _____ on time. [单选题] *A. be(正确答案)B. areC. to be11、He has two sisters but I have not _____. [单选题] *A. noneB. someC. onesD. any(正确答案)12、______! It’s not the end of the world. Let’s try it again.（）[单选题] *A. Put upB. Set upC. Cheer up(正确答案)D. Pick up13、My brother usually _______ his room after school. But now he _______ soccer. [单选题] *A. cleans; playsB. cleaning; playingC. cleans; is playing(正确答案)D. cleans; is playing the14、The museum is _______ in the northeast of Changsha. [单选题] *A. sitB. located(正确答案)C. liesD. stand15、______ my great joy, I met an old friend I haven' t seen for years ______ my way ______ town. [单选题] *A. To, in, forB. To, on, to(正确答案)C. With, in, toD. For, in, for16、In the closet（）a pair of trousers his parents bought for his birthday. [单选题] *A. lyingB. lies(正确答案)c. lieD. is lain17、44．—Hi, Lucy. You ________ very beautiful in the new dress today.—Thank you very much. [单选题] *A．look(正确答案)B．watchC．look atD．see18、They all choose me ______ our class monitor.（）[单选题] *A. as(正确答案)B. inC. withD. on19、My mother’s birthday is coming. I want to buy a new shirt ______ her.（）[单选题] *A. atB. for(正确答案)C. toD. with20、--Could you please tell me _______ to get to the nearest supermarket?--Sorry, I am a stranger here. [单选题] *A. whatB. how(正确答案)C. whenD. why21、?I am good at schoolwork. I often help my classmates _______ English. [单选题] *A. atB. toC. inD. with(正确答案)22、He was very excited to read the news _____ Mo Yan had won the Nobel Prize for literature [单选题] *A. whichB. whatC. howD. that(正确答案)23、Becky is having a great time ______ her aunt in Shanghai. （）[单选题] *A. to visitB. visitedC. visitsD. visiting(正确答案)24、In the future, people ______ a new kind of clothes that will be warm when they are cold, and cool when they’re hot.（）[单选题] *A. wearB. woreC. are wearingD. will wear(正确答案)25、This message is _______. We are all _______ at it. [单选题] *A. surprising; surprisingB. surprised; surprisedC. surprising; surprised(正确答案)D. surprised; surprising26、Your homework must_______ tomorrow. [单选题] *A. hand inB. is handed inC. hands inD. be handed in(正确答案)27、Since the war their country has taken many important steps to improve its economic situation. [单选题] *A. 制定B. 提出C. 讨论D. 采取(正确答案)28、How _______ Grace grows! She’s almost as tall as her mother now. [单选题] *A. cuteB. strongC. fast(正确答案)D. clever29、_____ is not known yet. [单选题] *A. Although he is serious about itB. No matter how we will do the taskC. Whether we will go outing or not(正确答案)D. Unless they come to see us30、You cannot see the doctor _____ you have made an appointment with him. [单选题] *A. exceptB.evenC. howeverD.unless(正确答案)。

我们马上就会发现(9.1)的每一个解都一定是f(x)=Cex 这种形式，这里C可以是任何常数。
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.
近年来，在数学和许多各种不同的领域中，矩阵的应 用一直以惊人的速度不断增加。在研究量子力学时， 矩阵理论在现代物理学上起着主要的作用。
Matrix methods are used to solve problems in applied differential equations, specifically, in the area of aerodynamics, stress and structure analysis. One of the most powerful mathematical methods for psychological studies is factor analysis, a subject that makes wide use of matrix methods.
1. 理解微分方程的分类。 2. 理解矩阵学习的重要性。
Or a radioactive substance may be disintegrating at a known rate and we may be required to determine the amount of material present after a given time.

数学专业英语【第二版】1- A 什么是数学数学来自于人的社会实践，例如，工业和农业生产、商业活动、军事行动和科研工作。

与数学反过来，为实践服务和所有字段中的伟大作用。

没有现代的科学和技术分支机构可以定期制定中的数学，应用无早有需要的人来了数字和形式的概念。

然后，开发出的几何土地和三角测量的问题来自测量的问题。

若要对付一些更复杂的实际问题，男子成立，然后解决方程未知号码，因此代数发生。

17 世纪前, 男子向自己限于小学数学，即几何、三角和代数，只有常量被认为在其中。

17 世纪产业的快速发展促进了经济和技术的进展和所需变量的数量、处理从常量到带来两个分支的数学-解析几何和微积分，属于高等数学，现在有很多分支机构，其中有数学分析、高等代数、微分方程的高等数学中的可变数量的飞跃函数理论等。

数学家研究理念和主张。

所有命题公理、假设、定义和定理都。

符号是一种特殊和功能强大的数学工具，用于表示很多时候的理念和主张。

公式、数字和图表是阿拉伯数字1,2,3,4,5,6,7,8,9,0 与另外的符号"+"、减法"-"，乘"*"，除"\"和平等"="。

数学中的结论得到主要由逻辑推理和计算。

长期的数学史上，以中心地点的数学方法被占领逻辑扣除。

现在，由于电子计算机是迅速发展和广泛应用，计算的作用变得越来越多重要。

在我们这个时代计算不只用于处理大量的信息和数据，而且还进行一些只是可以做的工作较早前的逻辑推理，例如，大部分的几何定理的证明。

1--B 方程方程是平等的语句的两个相等的数字或数字符号之间。

因此(a-5）= 一5a 和x 3 = 5 是方程。

方程的两种——身份和方程的条件。

方程的算术或代数的身份。

这种方程中两名成员是相似的或成为相似的指示操作的性能。

因此12-2=2+8,(m+n)(m-n) = m n 是身份。

1—c 比与测量今天的思想沟通往往根据编号和数量的比较。

英语（第二版）习题册答案Unit 1Lesson 11．（1）s （2）i （3）o （4）e （5）u （6）e （7）a （8）i（9）c （10）a,o2．（1）—（b）；（2）—（d）；（3）—（a）；（4）—（e）；（5）—（c）；（6）—（g）；（7）—（f）；（8）—（i）；（9）—（h）；（10）—（j）3．（1）car （2）bus （3）truck （4）tractor （5）ambulance （6）camper （7）garbage truck （8）sprinkler （9）mail car （10）concrete mixer4．A—（2）； B—（1）； C—（4）； D—（5）； E—（3）5．（1）sprayed （2）abilities （3）drive （4）decorate （5）equip6．（1）medium bus （2）mini car （3）dump truck （4）inspecting car（5）garbage truck （6）light truck （7）racing car （8）concrete mixer（9）fire truck （10）training school7．（1）汽车是一种由发动机驱动的陆上交通工具。

汽车的主要用途是载运乘客或货物。

（2）汽车的类型很多。

各类汽车的总体构造有所不同，但它们的基本组成是一致的。

（3）发动机将燃料的能量转换成机械能。

底盘接收发动机的动力，使汽车正常行驶。

（4）车身用于安置驾驶员、旅客或装载货物。

汽车电气系统由电源和用电设备组成。

8．BCDAELesson 21．（1）u （2）a （3）r （4）e,i （5）d （6）x,i （7）o,a （8）o （9）u （10）i2．（1）—（c）； （2）—（a）； （3）—（b）； （4）—（g）； （5）—（d）（6）—（e）； （7）—（f）； （8）—（i）； （9）—（j）； （10）—（h）3．（1）no smoking （2）caution （3）wear a safety helmet （4）wear protective shoes （5）wear a gas mask （6）wear protective gloves （7）wear work clothes（8）wear protective glasses （9）wear protective clothing4．A—（2）； B—（3）； C—（1）； D—（5）； E—（4）5．（1）slipped （2）tunnels （3）abolished （4）alarm （5）pedestrians· 1 ·6．（1）non-motorized vehicle （2）non-motorized lane （3）driving license（4）protective gloves （5）work clothes （6）gas mask （7）protective clothing （8）cut off （9）good ventilation （10）security problem7．（1）在汽车喷涂作业过程中，操作现场必须通风良好。

Last update: December 9, 2002
1.2 − 2 Insertion sort beats merge sort when 8n2 < 64n lg n, ⇒ n < 8 lg n, ⇒ 2n/8 < n. This is true for 2 n 43 (found by using a calculator). Rewrite merge sort to use insertion sort for input of size 43 or less in order to improve the running time. 1−1 We assume that all months are 30 days and all years are 365.
n
Θ
i=1
i
= Θ(n2 )
This holds for both the best- and worst-case running time. 2.2 − 3 Given that each element is equally likely to be the one searched for and the element searched for is present in the array, a linear search will on the average have to search through half the elements. This is because half the time the wanted element will be in the first half and half the time it will be in the second half. Both the worst-case and average-case of L INEAR -S EARCH is Θ(n). 3

Sets of points in the planeWe have already shown that there is a one-to-one correspondence be tween points in a plane and pairs of numbers (x,y) . Certain sets of points in the plane may be of special interest. For example , we may wish to exa mine the set of point comprising the circumference of a certain circle , or the set of points constituting the interior of a certain triangle . One may w onder if such sets of points may be succinctly described in compact mathe matical notation.We may write{(x,y)|y=2x} （1）to describe the set of ordered pairs (x,y) , or corresponding points , such that the ordinate is equal to twice the abscissas. In effect ,then, the vertical line in (1) is read “such that” . By “the graph of the set of ordered pair s” is meant the set of all points of the plane corresponding to the set of ordered pairs. The student will readily infer that the set of points constituting the graph lies on a straight line.Consistent the set{(x,y)|y=x^2}Consistent with our previous interpretation , this symbol represents the se t of ordered pairs (x,y) such that the ordinate is equal to the square of the abscissa. Here ,the total graph comprises a simple recognizable geometric al figure , a curve known as a parabola.on the basis of these two example ,one may be tempted to believe that an y ar-bitrarily drawn curve , which of course determines a set of points ord ered pairs, could be described succinctly by a simple equation. Unfortunat ely ,this is not the case. For example , the broken line in figure 2-2-3 is on e of such curves.Consider now the set{(x,y)|y>2x} （2）to describe the set of points (x,y) whose ordinate is greater than twice its abscissa. In this case ,our set of point constitutes not a curve , but a region of the coordinate plane.建立点在平面上我们已经表明,在平面之间存在一一对应点和坐标(x,y)。

2.4 整数、有理数与实数4－A Integers and rational numbersThere exist certain subsets of R which are distinguished because they have special properties not shared by all real numbers. In this section we shall discuss such subsets, the integers and the rational numbers.有一些R的子集很著名，因为他们具有实数所不具备的特殊性质。

在本节我们将讨论这样的子集，整数集和有理数集。

To introduce the positive integers we begin with the number 1, whose existence is guaranteed by Axiom 4. The number 1+1 is denoted by 2, the number 2+1 by 3, and so on. The numbers 1,2,3,…, obtained in this way by repeated addition of 1 are all positive, and they are called the positive integers.我们从数字1开始介绍正整数，公理4保证了1的存在性。

1+1用2表示，2+1用3表示，以此类推，由1重复累加的方式得到的数字1,2,3，…都是正的，它们被叫做正整数。

Strictly speaking, this description of the positive integers is not entirely complete because we have not explained in detail what we mean by the expressions “and so on”, or “repeated addition of 1”.严格地说，这种关于正整数的描述是不完整的，因为我们没有详细解释“等等”或者“1的重复累加”的含义。

参考答案目录第五章 (3)5.3同角三角比的关系和诱导公式 (3)5.3.1同角三角比的关系 (3)5.3.2诱导公式 (4)5.3.5三角式的化简与证明 (6)5.4两角和与差的余弦、正弦与正切 (7)5.4.1两角和与差的余弦 (7)5.4.2两角和与差的正弦 (9)5.4.3两角和与差的正切 (11)5.4.4两角和与差的余弦、正弦和正切的应用 (12)5.5二倍角与半角的正弦、余弦和正切 (14)5.5.1倍角公式 (14)5.5.2半角的正弦、余弦和正切 (15)5.5.3万能公式和降幂公式 (17)5.6正弦定理、余弦定理和解斜三角形 (19)5.6.1正弦定理 (19)5.6.2余弦定理 (20)5.6.3正弦、余弦定理的应用（一） (21)5.6.3正弦、余弦定理的应用（二） (23)第六章三角函数 (25)6.1三角函数的图像与性质 (25)6.1.1正弦函数和余弦函数的图像 (25)6.1.2正弦函数和余弦函数的值域与最值（一） (26)6.1.3正弦函数和余弦函数的值域与最值（二） (28)6.1.4正弦函数和余弦函数的周期 (30)6.1.5正弦函数和余弦函数的奇偶性与单调性 (32)6.2正切函数的图像与性质 (33)6.3函数sin()y A x ωϕ=+的图像与性质 (35)6.3.1函数sin()y A x ωϕ=+的图像变换 (35)6.3.2函数sin()y A x ωϕ=+的图像与性质 (36)6.3.3函数sin()y A x ωϕ=+的图像与性质的应用 (37)6.4反三角函数与最简三角方程 (39)6.4.1反三角函数的图像与性质 (39)6.4.2反三角函数的求值与化简 (41)6.4.3反三角函数的图像与性质的应用 (43)6.5最简三角方程 (45)6.5.1最简三角方程的解集（一） (45)6.5.2最简三角方程的解集（二） (46)第七章平面向量的坐标表示 (48)7.1向量的加减法 (48)7.1.1向量的概念和向量的加减法 (48)7.1.2实数与向量的乘积 (50)第八章 (51)8.2等差数列 (51)8.2.1等差数列及其通项公式（一） (51)8.2.2等差数列及其通项公式（二） (52)8.2.3等差数列的前n项和（一） (53)8.2.4等差数列的前n项和（二） (55)第五章5.3同角三角比的关系和诱导公式5.3.1同角三角比的关系iPreview （i 预习）（1）倒数关系：sin csc 1,cos sec 1,tan cot 1αααααα⋅=⋅=⋅=.（2）商数关系：sin cos tan (cos 0),cot (sin 0)cos sin αααααααα=≠=≠.（3）平方关系：222222sin co s 1,1tan sec (co s 0),1co t csc (sin 0)αααααααα+=+=≠+=≠.iSelftest （i 自测）1.C2.A3.D4.B5.DiEvolve （i 演变）例133455sin ,tan ,cot ,sec ,csc 54343ααααα=-=-=-==-演变若α为第四象限答案同上；若α为第一象限：33455sin ,tan ,cot ,sec ,csc 54343ααααα=====例2当α为第一象限角时，51212sin ,cos ,cot 13135ααα===当α为第三象限角时，51212sin ,cos ,cot 13135ααα=-=-=演变163326(1)-;(2)-;(3)75169169例379125 (1);(2),(3) 512512-iPractice（i练习）1.4 32.13 22 -+3.3 2 -4.{1,3}-5.(2,9)6.07.1 33或8.(1)当A是锐角时，4cos5α=，3tan4α=；(2)当A是钝角时，4cos5α=-，3tan4α=-。

CSE547 Discrete Mathematics
Chapter 1, Problem 7
P4: Problem 7 on Page 17
Let H(n) = J(n+1) – J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n + 1) = J(2n+2) – J(2n + 1) = (2J(n+1) – 1) – (2J(n) + 1) = 2H(n) – 2, for all n ≥ 1. Therefore it seems possible to prove that H(n) = 2 for all n, by induction on n. What’s wrong here?
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Detailed Solution - continue
Let’s check whether H(2n) = 2, for all n ≥ 1 as the problem stated:
H(2n) = J(2n + 1) – J(2n) H(n) = J(n+1) – J(n) = [2J(n) + 1] – [2J(n) - 1] Eq. (1.8) = 2J(n) + 1 – 2J(n) + 1 Take the [ ] out = 2 YES ! Algebra
Eq. (1.8): J(1) = 1 J(2n) = 2J(n) – 1, for n ≥ 1; J(2n + 1) = 2J(n) + 1, for n ≥ 1.
H(n) = J(n+1) – J(n) Algebra Algebra Eq. (1.8) Take the [ ] out Algebra H(n) = J(n+1) – J(n)

Prove by induction: P(n)=(n-1)n(n+1)/6 +n+1
It works for P(0)=1; Assume it holds when n is replaced with n-1 since P(n) = P(n-1) + n(n-1)/2 +1 then P(n) = (n-2)(n-1)(n)/6 + n +n(n-1)/2 +1 And : P(n) = (n-1)n(n+1)/6 + n+1
• After these moves all the n disks will be in order on peg B. Thus we can see that the total moves required to transfer the n disks is P(n) = 3P(n-1) + 2. • We want to guess the close form so we look at small cases: where we know P(1)=2; • P(2)=3*2 +2 =8, • P(3)=3*8+2=26,.. • We suggest the solution to this recurrence as: P(n) =3^n -1.
The problem can be solved recursively as follows:
First consider case, n=1, where we have to move a single disk from A to B. Since direct moves are disallowed this requires 2 moves and hence P(1) = 2.