数学专业英语课后答案

数学专业英语

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。

数学专业英语第2章课后答案

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）。

数学专业英语课后答案

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)。

高等数学双语教材答案

高等数学双语教材答案Chapter 1: Limits and ContinuitySection 1: Introduction to Limits and ContinuityThe concept of limits and continuity is fundamental in higher mathematics. In this section, we will introduce the basic definitions and properties associated with limits and continuity.1.1 Definitions of LimitsIn order to understand limits, we need to define what it means for a function to approach a particular value. Let f(x) be a function defined on an open interval containing c, except possibly at c. We say that the limit of f(x) as x approaches c is L, denoted bylim (x→c) f(x) = L, if for every ε > 0, there exists a δ > 0 such that |f(x) - L| < ε whenever 0 < |x - c| < δ.1.2 Basic Limit LawsOnce we have a clear understanding of limits, we can explore some basic laws that govern their behavior. These laws include the sum law, constant multiple law, product law, quotient law, and the power law.1.3 ContinuityA function f(x) is said to be continuous at a point c if three conditions are met: (1) f(c) is defined, (2) the limit of f(x) as x approaches c exists, and (3) the limit of f(x) as x approaches c is equal to f(c). We can also discuss continuity on an interval or at infinity.Chapter 2: DifferentiationSection 1: Introduction to DifferentiationDifferentiation is an important concept in calculus that allows us to find the rate at which a function is changing at any given point. In this section, we will introduce the concept of differentiation and its applications.2.1 Derivative DefinitionThe derivative of a function f(x) at a point c is defined as the limit of the difference quotient as h approaches 0. Mathematically, this can be written as f'(c) = lim (h→0) [(f(c + h) - f(c))/h].2.2 Differentiation RulesThere are several rules that allow us to find the derivative of a function quickly. These rules include the constant rule, power rule, sum rule, difference rule, product rule, quotient rule, and chain rule.2.3 Applications of DifferentiationDifferentiation has many applications in various fields, such as physics, economics, and engineering. It can be used to find maximum and minimum values, determine rates of change, and solve optimization problems.Chapter 3: IntegrationSection 1: Introduction to IntegrationIntegration is the reverse process of differentiation. It enables us to find the area under a curve and solve various mathematical problems. In this section, we will introduce the concept of integration and its applications.3.1 Indefinite IntegralsThe indefinite integral of a function f(x) is the collection of all antiderivatives of f(x). It is denoted by ∫ f(x) dx and represents a family of functions rather than a single value.3.2 Integration TechniquesThere are various techniques for finding antiderivatives and evaluating definite integrals. These techniques include basic integration rules, substitution, integration by parts, and trigonometric substitution.3.3 Applications of IntegrationIntegration has numerous applications, such as finding the area between two curves, calculating the length of curves, determining volumes of solids, and solving differential equations.ConclusionIn conclusion, the study of high-level mathematics, particularly limits, continuity, differentiation, and integration, is crucial for a comprehensive understanding of advanced mathematical concepts. This article has provided a brief overview of these topics, highlighting their definitions, properties, and applications. By mastering these concepts, students can develop strong problem-solving skills and apply them in various academic and real-world scenarios.。

《数值分析》黄仿伦改编英文版课后习题答案

Answers for Exercises —Numerical methods using MatlabChapter 1P10 2. Solution (a) )(x g x = produces an equation 0862=+-x x . Solving it gives the roots 2=x and 4=x .Since 2)2(=g and 4)4(=g , thus, both 2=P and 4=P are fixed points of )(x g . (b) –(d) The iterative rule using )(x g is 22144n n n p p p ---=. The results for part (b)-(d) with starting value 9.10=p and 8.30=p are listed in Table 1.(e) Calculate values of x x g -='4)( at 2=x and 4=x .12)2(>='g , and 10)4(<='g .Since )(x g ' is continuous, there exists a number 0>δ such that1)(<'x g for all ]4,4[δ+δ-∈x .There also exists a number 0>λ such that1)(>'x g for all ]2,2[λ+λ-∈x .Therefore, 4=p is an attractive fixed point. The sequence generated by22144n n n p p p ---=with starting value 8.30=p converges to 4=p . 2=p is a repelling fixed point. The sequence generated by 22144n n n p p p ---=with starting value 9.10=p does not converge to 2=p .P11 4. Find the fixed point for )(x g : )(x g x = gives 2±=p . Find the derivative: 12)(+='x x g .Evaluate )2(-'g and )2(g ': 3)2(-=-'g , 5)2(='g .Both 2-=p and 2=p gives 1)(>'p g . There is no reason to find the solution(s)using the fixed-point iteration.P11 6. Proof ))(()()(010112p p g p g p g p p -ξ'=-=-)()()( 0101p p K p p g -<-ξ'≤P214. False position method: Assume that ],[n n b a contains the root. The equation of the secand line through ))(,(n n a f a and ))(,(n n b f b is )()()()(n nn n n n b x a b a f b f b f y ---=-. Itintersects x -axise at)()())((n n n n n n n a f b f a b b f b c ---= (Eq. 1.36, p18)1981.0)6.1()(,4907.0)4.2()(00-=-==-=f b f f a f ,8301.1)()())((0000000-=---=a f b f a b b f b c ;Since 0095.0)(0-=c f , then ]8301.1,4.2[],[11--=b a . Similarly, we have1.84093- 1=c , ]1.84093- ,4.2[],[22-=b a 1.84139- 2=c , ]1.84139- ,4.2[],[33-=b a -1.841403=c10. Bisection method: Assume that ],[n n b a contains the root. Then 2nn n b a c +=. (a) 1587.1)4(,4;1425.0)3(,300==-==f b f a , then 5.30=c .Since 03746.0)5.3()(0>==f c f , then ]5.3,3[],[11=b a .Similarly, we can obtain ,,,321c c c . The results are listed in Table 3.The values of tan(x) at midpoints are going to zero while the sequence converges(b) Since 0)3tan(<=, there exist a root in )3,1(..0-=, 055741425.1tan(>)1The results using Bisection method are listed in Table 4.Although the sequence converges, the values of tan (x) at midpoints are not going to zero.P36 2. 3)(2--=x x x f has two zeros 2131±=x . (3028.2,3028.121≈-≈x x ) The first derivative of 3)(2--=x x x f is 12)(-='x x f .The Newton-Raphson iterative function is 123)()()(2-+='-=x x x f x f x x g . The Newton-Raphson formula is 12321-+=+n nn p p p , ,2,1,0=n . The results are listed in Table 5 with starting value p 0=1.6 and p 0=0.0 respectively.Obviously, the sequence generated by the starting value p 0=0.0 does not converge.11. Use Newton-raphson method to solve 0)(3=-=A x x f .The derivative of )(x f is 23)(x x f ='.3232)()()(223x Ax x A x x f x f x x g +=+='-=.Newton-Raphoson formula is 32211--+=n n n p Ap p , ,2,1=n .Since 3A p = is a zero of A x x f -=3)( and 10332)(33<=⎥⎦⎤⎢⎣⎡-='=Ap x A p g ,The sequence generated by the recursive formula 32211--+=n n n p Ap p will converge to3A p = for any starting value ],[330δδ+-∈A A p , where 0>δ.·Answers for Exercises —Numerical methods using MatlabChapter 2P44 2. Solution The 4th equation yields 24=x .Substituting 24=x to the 3rd equation gives 53=x .Substituting both 24=x and 53=x to the 2nd equation produces 32-=x . 21=x is obtained by sustituting all 32-=x , 53=x and 24=x to the 1st equation. The value of the determinant of the coefficient matrix is 115573115=⨯⨯⨯=D .4. Proof (a) Calculating the product of the two given upper-triangular matrices gives⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++++=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=33333323232222223313231213112212121111113323221312113323221312110000b a b a b a b a b a b a b a b a b a b a b b b b b b a a a a a a B A . It is also an upper-triangular matrix.(b) Let N N ij a A ⨯=)( and N N ij b B ⨯=)( where 0=ij a and 0=ij b when j i >.Let N N ij c B A C ⨯==)(. According to the definition of product of the two matrices, we have ∑==Nk kjik ij b ac 1for all N j i ,,2,1, =.0=ij c when j i > because 0=ij a and 0=ij b when j i >.That means that the product of the two upper-triangular matrices is also upper triangular.5. Solution From the first equation we have 31=x .Substituting 31=x to the second equation gives 22=x .13=x is obtained from the third equation and 14-=x is attained from the last equation.The value of the determinant of the coefficient is 243)1(42)det(-=⨯-⨯⨯=A7. Proof The formula of the back substitution for an N N ⨯upper-triangular system is N NN a b x =and kkNk j jkj k k a x a b x ∑+=-=1 for 1,,2,1 --=N N k .The process requiresN N=+++111 divisions, 22)1()1(212NN N N N -=-=-+++ multiplications, and2)1(212NN N -=-+++ additions or subtractions.P53 1. Solution Using elementary transformations for the augmented matrix gives330012630464275101263046425232103514642],[3231213121⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=++-+-r r r r r r B AThat means that ⎪⎩⎪⎨⎧=++=++-=-+523 1035 4642321321321x x x x x x x x x is equivalent to⎪⎩⎪⎨⎧==+-=-+33 1263 4642332321x x x x x x The set of solutions is .3,2,1123-===x x x11. Solution Using the algorithm of Gaussian Elimination gives12420010324050110700211242001032409013270021],[212⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡----−−−→−⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=+-r r B A ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+1242001032005011070021324r r ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+21000103200501107002143r r The set of solutions of the system is obtained by the back substitutions,3,2,2234==-=x x x and .11=x(Chasing method for solving tridiagonal linear systems)14. (a) (i) Solution Applying Gaussian elimination with partial pivoting to the augment matrix results in⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---=↔1100320001.0101001.01003001.010030001.010*******],[31r r B A ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−↔+-+-00043.03333.43019933.996667.630001.0100319933.996667.63000043.03333.430001.01003 3231213231r r r r r r ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−−→−+-6806.00625.680019933.996667.630001.01003326667.633333.43r rThe set of solutions is,101.0524,0100.0-623⨯==x x and .105.2400 -61⨯=x15. Solution The N N ⨯Hilbert matrix is defined byN N ij H H ⨯=)( where 11-+=j i H ij for N j i ≤≤,1.(a) The inverse of the 44⨯ Hilbert matrix is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--------=-280042001680140420064802700240168027001200120140240120161H The exact solution is T X )140,240,120,16(--=.(b) The solution is T X )0881.185,0628.310,6053.149,7308.18(--=.>>1 H is ill-conditioned. A miss is as good as a mile. (失之毫厘，谬以千里)P62 5 (a) Solving B LY = gives TY )2,12,6,8(-=. From Y UX = we have TX )2,1,1,3(-=. The product of A and X is TAX )4,10,4,8(--=.That means B AX =(b) Similarly to the part (a), we haveTY )1,12,6,28(=, TX )1,2,1,3(=, and B AX T==)4,23,13,28(.6. ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---=175.113011*********L , ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-----=5.70001040085304011UP72 7. (a) Jacobi Iterative formula is ()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=+++)()()1()()()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults for ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.1 with starting value )0,0,0(0=P .The numerical results show that Jacobi iteration does not converge.(b) Gauss-Seidel Iterative formula is()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=++++++)1()1()1()()1()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.2 with starting value )0,0,0(0=P ’Reasons:Conside the eigenvalues of iterative matricesSplit the coefficient matrix ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----=612114151A into three matrices⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=--=000100150012004000600010001U L D A .The iterative matrix of Jacobi iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=+=-061311041500121041506100010001)(1U L D T JThe spectral raduis of J T is 16800.5)(>=ρJ T . )1176.0,4546405880(i . .-±=λ’ So Jacobi method doesnot converge.Similarly, the iterative matrix of Gauss-Seidel iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=-=-65503200150)(1U L D T G .The spectral radius of G T is 2532.19)(=ρG T >1. )0866.0,2532.19,0(-=λ’ So Gauss-Seidel method does not converge.8. (a) Jacobi Iterative formula is()⎪⎩⎪⎨⎧-+=-+=+-=+++6/225/)8(4/)13()()()1()()()1()()()1(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.3 with starting value )0,0,0(0=P .Jacobi iteration converges to the solution (3, 2, 1)’(b) Gauss-Seidel iterative formula is()⎪⎪⎩⎪⎪⎨⎧+---=+---=+-=++++++)1()1()1()()1()1()()()1(22614/)8(4/)13(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.4 with starting value )0,0,0(0=PGauss-Seidel iteration converges to the solution (3, 2, 1)’Answers for Exercises —Numerical methods using MatlabChapter 3P99 1. Solution (a) The nth order derivative of )sin()(x x f = is )2sin()()(π+=n x x f n .Therefore, !5!3)(535x x x x P +-=, !7!5!3)(7537x x x x x P -+-= and !9!7!5!3)(97539x x x x x x P +-+-=.(b) Estimating the remainder term gives71091075574.2!101!10)5sin()(-⨯≤≤π+=x c x E for 1≤x .(c) Substituting 4π=x to )2sin()()(π+=n x x f n gives ,22)4()4(,22)4()4()3(-=π=π''=π'=πf f f f and 22)4()4()5()4(-=π=πf f .By using Taylor polynomial we have!5)4(22!4)4(22!3)4(22!2)4(22)4(2222)(54325π-+π-+π--π--π-+=x x x x x x P P108 1. (a) Using th e Horner ’s method to find )4(P givesSo )4(P =1.18.(b) From part (a) we have 12.002.002.0)(2-+-=x x x Q . )4()4(Q P =' can be also obtained byusing Horner ’s method.So )4(P '=-0.36 Another method:Hence, P(4)=-0.36.(c) Find )4(I and )1(I firstly.Then=-=⎰)1()4()(41I I dx x P 4.3029.(d) Use Horner ’s method to evaluate P (5.5)Hence, P (5.5)=0.2575.(d) Let 012233)(a x a x a x a x P +++=. There are 4 coefficients needed to found.Substituting four known point ),(i i y x , i =1, 2, 3, 4, into )(x P gives four linear equations with unknowni a , i =1, 2, 3, 4.54.10123=+++a a a a 5.12480123=+++a a a a 42.139270123=+++a a a a 66.05251250123=+++a a a aThe coefficients can be found by solving this linear system: .66.1,2.0,1.0,02.00123=-==-=a a a aP120 1. The values of f (x ) at the given points are listed in Table 3.1:(a) Find the Lagrange coefficient polynomials and 010)(0,1x x x L -=---=.1101)(1,1+=++=x x x LThe interpolating polynomial is x x L f x L f x P =+-=)()0()()1()(1,10,11. (b) ),(21)11()1()(20,2x x x x x L -=----=,110)1)(1()(21,2x x x x L -=--+=),(212)1()(22,2x x x x x L +=+=x x L f x L f x L f x P =++-=)()1()()0()()1()(2,21,20,22. (c) ),2)(1(61)21)(11()2)(1()(0,3---=-------=x x x x x x x L),2)(1)(1(21)20)(10)(10()2)(1)(1()(1,3--+=--+--+=x x x x x x x L),2)(1(21)21(1)11()2()1()(2,3-+-=-+-+=x x x x x x x L),1)(1(61)12(2)12()1()1()(3,3-+=-+-+=x x x x x x x L33,32,31,30,33)()2()()1()()0()()1()(x x L f x L f x L f x L f x P =+++-=(d) ,2212)(0,1x x x L -=--=,1121)(0,1-=--=x x x L 67)()2()()1()(1,10,11-=+=x x L f x L f x P . (e) ),23(21)20)(10()2)(1()(20,2+-=----=x x x x x L ),2()21(1)2()(21,2x x x x x L --=--=),(21)12(2)1()(20,2x x x x x L -=--=.23)()2()()1()()0()(22,21,20,22x x x L f x L f x L f x P -=++=7. (a) Note that each Lagrange polynomial )(,2x L k is of degree at most 2 and )(x g is a combination of)(,2x L k . Hence )(x g is also a polynomial of degree at most 2.(b) For each k x , 2,1,0=k , the Lagrange coefficient polynomial 1)(,2=k k x L , and 0)(,2=k j x L for k j ≠, 2,1,0=j . Therefore, 01)()()()(2,21,20,2=-++=k k k k x L x L x L x g .(c) )(x g is a polynomial of degree 2≤n and has n ≥ 3 zeroes. According to the fundamental theorem of algebra, 0)(=x g for all x .9. Let )()()(x P x f x E N N -=. )(x E N is a polynomial of degree N ≤.)(x f is degree with )(x P N at N +1 points N x x x ,,,10 implies that )(x E N has N +1 zeroes. Therefore, 0)(=x E N for all x , that is, )()(x P x f N = for all x .P131 6. (a) Find the divided-difference table:(b) Find the Newton polynomials with order 1, 2, 3 and 4.)0.1(80.16.3)(1--=x x P , )0.2)(0.1(6.0)0.1(80.160.3)(2--+--=x x x x P ,)0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(3------+--=x x x x x x x P , )0.4)(0.3)(0.2)(0.1(03.0 )0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(4----+------+--=x x x x x x x x x x x P .(c)–(d) The results are listed in Table 3.2P143 6. x x x T 32)(323-=, ]1,1[-∈x .The derivative of )(3x T is 323)(223-⋅='x x T . 0)(3='x T yields 21±=x . Evaluating )(3x T at 21±=x and 1±=x gives 1)1(3-=-T , 1)21(3=-T , 1)21(3-=T and 1)1(3=T .Therefore, 1))(max(3=x T , 1))(min(3-=x T .10. When 2=N , the Chebyshev nodes are ,23)6/5cos(0-=π=x ,01=x and 23)6/cos(2=π=x .Calculating the Lagrange coefficient polynomials based on 210,,x x x can produce the following results:,323)232(23)23()(20,2x x x x x L +-=⨯-⨯--=,341)23(23)23()23()(21,2x x x x L -=-⨯-+=.32323232)23()(22,2x x x x x L +=⨯⨯+=The proof is finished.Answers for Exercises —Numerical methods using MatlabChapter 4P157 1(a). Solution The sums for obtaining Normal equations are listed in Table 4.1The normal equations are ,710=A 135=B . Then ,7.0=A 6.2=B .The least-squares line is 6.27.0+=x y .2449.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f EP158 4. Proof Suppose the linear-squares line is B Ax y += where A and B satisfiesthe Normal equations ∑∑===+N k k Nk ky xAB N 11and ∑∑∑====+Nk k k N k k N k k y x x A x B 1121.y y N x A B N N B x N A B x A N k k Nk k N k k ==⎪⎪⎭⎫ ⎝⎛+=+⎪⎪⎭⎫ ⎝⎛=+∑∑∑===111111 meas thatthe point ),(y x lies on the linear-squares line B Ax y +=.5. First eliminating B on the Normal equations∑∑===+Nk k Nk k y x A B N 11and ∑∑∑====+Nk k k Nk k Nk k y x x A x B 1121gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑===Nk k N k k N k k k y x y x N D A 1111 where 2112⎪⎪⎭⎫ ⎝⎛-=∑∑==N k k N k k x x N D . Substituting A into the first equation gets⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛+-=∑∑∑∑∑=====Nk k Nk k N k k k N k k N k k y x N y x x y N D D B 12111111. Note that ∑∑∑∑∑∑∑∑========⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=N k k N k k N k k N k k N k k N k k N k k Nk k y x N y x y x x N N y N D 12111212112111. Simplifying B gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑∑====Nk k k N k k Nk k N k k y x x y x D B 111121.8(b). The sums needed in the Normal equations are listed in Table 4.26177.142==∑∑k kk x y x A )2(=M5606.063==∑∑kkkxy x B )3(=MHence, 26177.1x y = and 35606.0x y =.0.3594 )(51)(21512222=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Ax y Ax E , 1.1649 )(51)(21512332=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Bx y Bx E .26175.1x y =fits the given data better.P171 2(c). The sums for normal equations are listed in Table 4.3.Using the formula∑∑∑∑∑∑∑========++=++52545352515135125151512515k kk k k k k k k k kk k k ky x x A x B x C y x A xB Cproduces the system with unkowns A , B , and CSolving the obove system gives .6.0,1.0,5.2-=-==C B A The fitting curve is .6.01.05.22--=x x yP172 4. (a) Translate points in x-y plane into X-Y plane using y Y x X ln ,==. The results arelisted in Table 4.4.The Normal equationsgive the system .793410,110,22105=+-==+A C B A C ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A B .8648.0155,2196.455-=+=+A B A BThen -0.50844=A , 1.3524=B . Thus 866731.3524.e e C B ===.The fitting curve is xe .y 50844.086673-=, and 1190.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(b) Translate points in x-y plane into X-Y plane using yY x X 1,==. The results are listed in Table 4.5.The Normal equationsgive the system Then 2432.0=A , 30280.0=B .The fitting curve is 30280.02432.01+=x y and 5548.4)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(c) It is easy to see that the exponential function is better comparing with errors in part (a) and part (b)..1620.5155,7300.255=+=+A B A B ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A BP188 1. (a) Derivativing )(x S gives 232132)(x a x a a x S ++='. Substituting the conditions intothe derivative pruduces the system of equations . 012428420321 32132103213210⎪⎪⎩⎪⎪⎨⎧=++=+++=++=+++a a a a a a a a a a a a a a (b) Solving the linear system of equations in (a) gives 29,,12,63210-==-==a a a a . The cubic polynomial is 3229126)(x x x x S -+-=.Figure: Graph of the cubic polynomial4. Step 1 Find the quantities: 3,1210===h h h , 21/)20(/)(0010-=-=-=h y y d13/)03(/)(1121=-=-=h y y d , 6667.03/)31(/)(2232-=-=-=h y y d18)(6011=-=d d u , 10)(6122-=-=d d uStep 2 Use ⎪⎪⎩⎪⎪⎨⎧-'-=⎪⎭⎫ ⎝⎛++'--=+⎪⎭⎫⎝⎛+))((3232))((32232322211100121110d x S u m h h m h x S d u m h m h h to obtain the linear system⎩⎨⎧-=+=+0001.155.1032135.72121m m m m .The solutions are 5161.2,8065.321-==m m . Step 3 Compute 0m and 3m using clamaped boundary.4.90322))((310000-=-'-=m x S d h m , 2.92482))((322323=--'=md x S h m Step 4 Find the spline coefficients16)2(,210001,000,0-=+-===m m h d s y s , 1.45166,-2.45162013,002,0=-===h m m s m s ;-1.54856)2(,021111,110,1=+-===m m h d s y s , -0.35136,1.903321123,112,1=-===h m m s m s ;0.38716)2(,332221,220,2=+-===m m h d s y s , 0.30236,-1.258122233,222,2=-===h m m s m s ;Step 5 The cubic spline is320)3(4516.1)3(4516.2)3(2)()(+++-+-==x x x x S x S for 23-≤≤-x ,321)2(3513.0)2(903.1)2(5484.1)()(+-+++-==x x x x S x S for 12≤≤-x , and322)1(3023.0)1(2581.1)1(3871.03)()(-+---+==x x x x S x S for 41≤≤x .5. Calculate the quantities: 3,1210===h h h , 20-=d ,11=d , 6667.02-=d ,181=u , 102-=u . ( Same values as Ex. 4)Substituting }{j h , }{j d and }{j u into ()()⎩⎨⎧=++=++22211112111022u m h h m h u m h m h h gives ⎩⎨⎧-=+=+1012318382121m m m mSolve the linear equation to obtain .5402.1,8276.221-==m m In addition, .030==m m Use formula (4. 65) to find the spline coefficients:4713.26)2(,210001,000,0-=+-===m m h d s y s , 4713.06,020013,002,0=-===h m m s m s ;-1.05756)2(,021111,110,1=+-===m m h d s y s , -0.24276,4138.121123,112,1=-===h m m s m s ;8735.06)2(,332221,220,2=+-===m m h d s y s , 0856.06,7701.0-22233,222,2=-===h m m s m s .Therefore, 30)3(4713.0)3(4713.22)(+++-=x x x S , for 23-≤≤-x ;321)2(2427.0)2(4138.1)2(0575.1)(+-+++-=x x x x S , for 12≤≤-x322)1(0856.0)1(7701.0)1(8735.03)(-+---+=x x x x S for 41≤≤x .Answers for Exercises —Numerical methods using MatlabChapter 5P209 1(b). Solution LetThe result of using the trapezoidal rule with h =1 isUsing Simpson’s rule with h=1/2, we haveFor Simpson’s 3/8 rule with h=1/3, we obtainThe result of using the Boole’s rule with h=1/4 is4. Proof Integrate )(1x P over ],[10x x .110102012101)(2)(2)(xx x x x x x x h f x x h f dx x P -+--=⎰=)(210f f h +. The Quadrature formula )(2)(1010f f hdx x f x x +≈⎰is called the trapezoidal rule.6. Solution The Simpson ’s rule is)4(3)(21010f f f hdx x f x x ++≈⎰. It will suffice to apply Simpson ’s rule over the interval [0, 2] with the test functions32,,,1)(x x x x f = and 4,x . For the first four functions, since)1141(31212+⨯+==⎰dx , )2140(31220+⨯+==⎰xdx , )4140(3138202+⨯+==⎰dx x , )8140(314203+⨯+==⎰dx x , the Simpson ’s rule is exact. But for 4)(x x f =,)16140(3153224+⨯+≠=⎰dx x . .Therefore, the degree of precision of Simpson ’s rule is n =3.T he Simpson’s rule and the Simpson ’s 3/8 rule have the same degree of precision n =3.).4cos(1)(x e x f x -+=..f f f f h dx x f 3797691))1()0((21)(2)(1010=+=+≈⎰.9583190))1()5.0(4)0((61)4(3)(21010. f f f f f f h dx x f =++=++≈⎰.9869270 ))1()3/2(3)3/1(3)0(( 8/1 )33(83)(321010.f f f f f f f f hdx x f =+++=+++≈⎰.008761 ))1(7)4/3(32)2/1(12)4/1(32)0(7( 90/1 )73212327(452)(432101.f f f f f f f f f f hdx x f =++++=++++≈⎰P220 3(a) Solution When 3)(x x f =for 10≤≤x , ⎰+π=123912dx x x area .The values of 2391)(x x x g +=at 11 sample points (M =10) are listed in the Table 5.1:(i) Using the composite Trapezoidal rule ∑-=++=110)()()((2),(M k k M x g h x g x g hh g T , the computation is)9156.11084.16098.03719.01563.00710.00280.00081.00010.0(101)1623.30(201)101,(++++++++++=g T=)2160.4(101)1623.3(201+=0.1576+0.4216=0.5792.(ii) Using the composite Simposon ’s rule ∑∑-=--=+++=11121120)(34)(32)()((3),(M k k M k k M x g h x g h x g x g h h g S , the computation is)9156.16098.01563.00280.00010.0(304)1084.13719.00710.00081.0(302 )1623.30(301)101,(++++++++++=g S=)7106.2(304)5054.1(302 )1623.3(301++=0.5672.7. (a) Because the formula)2()1()0()(2102g w g w g w dt t g ++=⎰is exact for the three functions 1)(=t g ,x t g =)(, and 2)(x t g =, we obtain three equations with unkowns 0w , 1w , and 2w :2210=++w w w , 2221=+w w ,38421=+w w . Solving this linear system gives 310=w , 341=w and 312=w .Thus, ())2()1(4)0(31)(20g g g dt t g ++=⎰(b) Let ht x x +=0 and denote ,01h x x +=.202h x x +=Then the change of variable ht x x +=0 translates ],[20x x into [0, 2] and converts the integral expresion dx x f )( into dt ht x hf )(0+. Hence,()())()(4)(3)2()1(4)0(3)()()(21022002x f x f x f h g g g hdt t g h dt ht x f h dx x f x x ++=++==+=⎰⎰⎰. The formula ())()(4)(3)(21020x f x f x f hdx x f x x ++=⎰ is known as the Simpson ’s rule over ],[20x x .8(a).9(a).P234 1(a) Let 212sin )(x xx f +=. The Romberg table with three rows for ⎰+3212sin dx x xis given as follows:Where04191.0)02794.0(23)106sin 0(23))3()0((23)0()0,0(-=-=+=+==f f T R , 04418.0)5.113sin (5.1204191.0)5.1(5.12)0()1()0,1(2=++-=+==f T T R ,3800.0)25.215.4sin 75.015.1sin (75.0204418.0))25.2()75.0((75.02)1()2()0,2(22=++++=++==f f T T R , 07288.03)04191.0(04418.043)0,0()0,1(4)1()1,1(=--⨯=-==R R S R ,].6/,6/[],[ and cos )(Let ππ-==b a x x f have we , and cos )( ,sin )( Since Mab h x x f x x f -=-=''-='.10513/123/ )(12),(922-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤''--=M c f h a b h f E T .1039.2/)( and )4375( 9.4374 So,4-⨯≈-==>M a b h M M ].6/,6/[],[ andcos )(Let ππ-==b a x x f ,cos )( ,sin )( , cos )( ,sin )( Since )4(x x f x x f x x f x x f =='''-=''-='.105123/1803/ )(180),(92)4(4-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤--=M c f h a b h f E S havewe ,2 and M a b h -=.1027.92/)( and )565( 8.564 So,4-⨯≈-==>M a b h M M4919.0304418.03800.043)0,1()0,2(4)2()1,2(=-⨯=-==R R S R ,5198.0307288.04919.01615)1,1()1,2(16)2()2,2(=-⨯=-==R R B R ,2. Proof If L J T J =∞→)(lim , thenL LL J T J T J S J J =-=--=∞→∞→343)1()(4lim)(lim andL LL J S J S J B J J =-=--=∞→∞→151615)1()(16lim )(lim .9. (a) Let 78)(x x f =. 0)()8(=x f implies 4=K . Thus 256)4,4(=R .(b) Let 1011)(x x f =.0)()11(=x f implies 5=K . Thus 2048)5,5(=R .10. (a) Do variable translation t x =. Thendt t dt t t dx x tx ⎰⎰⎰=⋅===121122.That means the two integrals dx x ⎰1anddt t ⎰122have the same numerical value.(b) Let 22)(t t f = and x x g =)(.Use dt t R ⎰≈122)1,1( means that the truncation error is )()4(n f k ξ approximately.Note that 0)()3(=t f . It means )1,1(212R dt t =⎰.But for x x g =)(, 0)()(=x g n is not true for all ]1,0[∈x and any integer 0>n .Thus the Romberg sequence is faster for dt t ⎰122 than fordx x ⎰1even though they have the samenumerical value.P242 1 (a) Applying the change of variable 22ab x a b t ++-=to dt t ⎰256 givesdx x dt t x t ⎰⎰-+=+⋅==115125)1(66.Thus the two integrals are dt t ⎰256 anddx x ⎰-+⋅115)1(6equivalent.(b)315315311525)1(6)1(6)()1(66=-=-+++=≈+⋅=⎰⎰x x x x f G dx x dt t =0.0809 +58.5857=58.6667If using )(3f G to approximate the integral, The result is535055353115205)1(695)1(698)1(695)()1(66==-=-+++++=≈+⋅=⎰⎰x x x x x x f G dx x dt t64105.5965956.0000 98 0.0035 95=⨯+⨯+⨯=6. Analysis: The fact that the degree of precision of N -point Gauss-Legendra integration is 2N -1 impliesthat the error term can be represented in the form )()()2(c kf f E N N =.(a) Since dt t dx x tx ⎰⎰-+=+==117127)1(88, and ()0)1()8(7=+t implies 82=K . Thus =256)(4=f G .(b),)1(111111101210dt t dx x tx ⎰⎰-+=+==and ()0)1()11(10=+t implies 122=K .Thusdx x⎰21011=2048)(6=f G .7. The n th Legendre polynomial is defined by The first five polynomials areThe roots of them are same as ones in Table 5.8.11. The conditions that the relation is exact for the functions means the three equations:326.0 6.0 0)6.0( )6.0(2 3132111321=+=+-=++w w w w w w w Sloving the system gives 98 ,95 231===w w w . ))6.0((95)0(98))6.0((95)(212111f f f dx x f ++-≈⎰- is called three-point Gauss-Legendre rule.Answers for Exercises —Numerical methods using MatlabChapter 6P249 1. (a) Proof Differentiate 22)(2+-+=-t t Ce t y t .22)(-+-='-t Ce t y tSubstitute )(t y and )(t y ' into the right-hand side of the equation y t y -='2.side left )(22)22(side right 222='=-+-=+-+-=-=--t y t Ce t t Ce t y t t t(b) Solution Let y t y t f -=2),(. Then 1),(-=y t f y for any ),(y t .So, the Lipschitz constant is 1=L .()[],2,11!21)(1)(20=-⋅==n x dx d n x P x P nnn n n ()()()3303581)(3521)(1321)()(1)(244332210+-=-=-===x x x P x x P x x P x x P x P 2,,1)(x x x f =12 . Integrate both side of )(t f y =' over [a , b ]: ⎰⎰='=-babadt t f dt y a y b y )()()(.Then,)()()(a y b y dt t f ba-=⎰, where )(t y is the solution of the I. V . P)(t f y =', for b t a ≤≤ with 0)(=a y . That means that the definite integral⎰badt t f )( can becomputed using the two values )(a y and )(b y of the solution )(t y of the given I. V . P.. 14. Solution Separate the two variables of the equation 211t y +=' into the form dt tdy 211+=. Integrate dt tdy 211+=and yeild the general solution C t y +=arctan . The initial-value condition 0)0(=y means that 0=C . The solution for the I. V . P. is t y arctan =.P257 3. (a)-(c) The formula using Euler ’s method to solve the I. V . P. ty y -=', 1)0(=y canbe represented in the form k k k y ht y )1(1-=+. When 2.0=h and 1.0=h , the results are listed in Table 6.1.(d) The F. G . E. does decrease half approximatelly as expacted when h is halved.6. When 02.0=a , 00004.0=b and 10=h , the Euler ’s formula for 2bP aP P -=' is in the form210004.02.1k k k P P P -=+. With 1.760=P , the missing entries can be filled in the table.。

高等数学英文教材答案

高等数学英文教材答案Title: Answer Key for Advanced Mathematics English TextbookI. IntroductionAdvanced Mathematics is a fundamental subject for students pursuing a higher education in STEM fields. To aid in their learning, an English textbook for Advanced Mathematics has been developed. This article serves as an answer key for the textbook, providing students with accurate solutions to exercises and guiding them towards a better understanding of the subject.II. Chapter 1: Functions and Their Graphs1.1 Linear FunctionsExercise 1:Solution: The equation of the line is y = 2x - 3.Exercise 2:Solution: The slope of the line passing through the points (-1, 4) and (3, -2) is -1. The equation of the line is y = -x + 3.1.2 Quadratic FunctionsExercise 1:Solution: The vertex of the parabola y = x^2 - 2x + 1 is (1, 0).Exercise 2:Solution: The equation of the parabola passing through the points (-1, 3), (0, 1), and (2, -1) is y = -x^2 + x + 1.III. Chapter 2: Differentiation2.1 Derivative of FunctionsExercise 1:Solution: The derivative of f(x) = 3x^2 - 2x + 1 is f'(x) = 6x - 2.Exercise 2:Solution: The derivative of g(x) = 5x^3 + 2x^2 - 4x is g'(x) = 15x^2 + 4x - 4.2.2 Rules of DifferentiationExercise 1:Solution: The derivative of h(x) = sin(x) + cos(x) is h'(x) = cos(x) -sin(x).Exercise 2:Solution: The derivative of i(x) = ln(x^2) is i'(x) = 2/x.IV. Chapter 3: Integration3.1 Indefinite IntegralsExercise 1:Solution: The integral of f(x) = 3x^2 - 2x + 1 is F(x) = x^3 - x^2 + x + C, where C is the constant of integration.Exercise 2:Solution: The integral of g(x) = 5x^3 + 2x^2 - 4x is G(x) = x^4/4 +2x^3/3 - 2x^2/2 + C.3.2 Definite IntegralsExercise 1:Solution: The definite integral of h(x) = sin(x) + cos(x) from 0 to π/2 is ∫[0, π/2] (sin(x) + cos(x)) dx = 2.Exercise 2:Solution: The definite integral of i(x) = ln(x) from 1 to e is ∫[1, e] ln(x) dx = 1.V. ConclusionThis answer key provides students using the Advanced Mathematics English textbook with accurate solutions to exercises, enabling them to verify their work and further enhance their understanding of the subject. By following the provided solutions, students can gain confidence in their problem-solving abilities and excel in the field of mathematics.。

大学高数英语无机答案

第11章固体结构1、指出下列物质哪些是金属晶体？哪些是离子晶体？哪些是共价键晶体(又称原子晶体)？哪些是分子晶体？Au (s) AlF3 (s) Ag (s) B2O3 (s) BCl3 (s) CaCl2 (s)H2O (s) BN (s) C (石墨)H2C2O4 (s) Fe (s) SiC (s)CuC2O4 (s) KNO3 (s) Al (s) Si (s)解：金属晶体：Au(s) Ag(s) Fe(s) Al(s)离子晶体：AlF3(s) CaCl 2(s) CuC2O4(s) KNO3(s)共价键晶体：BN(s) C(石墨) SiC(s) Si(s)分子晶体：B2O3(s) BCl3(s) H2O(s) H2C2O4(s)2、大多数晶态物质都存在同质多晶现象。

即在不同的热力学条件(温度、压力等)下，由于晶体内部粒子(原子、离子或分子)的热运动，它们在三维空间的排列方式将会发生一些变化。

例如：α-Fe (体心立方) 906℃γ-Fe (面心立方)α-CsCl (简单立方) 445℃β-CsCl (面心立方， NaCl型结构)α-NH4Cl (简单立方) 184β-NH4Cl (面心立方，NaCl型结构) 试问，同一种物质的不同类型的晶体，它们的晶面角是否相同或者守恒？晶面角守恒的本质原因是什么？解：根据晶面角守恒定律，同一种晶体晶面大小和形状会随外界的条件不同而变化，但同一种晶体的相应晶面（或晶棱）间的夹角却不受外界条件的影响，它们保持恒定不变的值。

晶面角守恒决定于晶体内部的周期性结构。

解：I2，正交晶系；H2C2O4，单斜晶系；NaCl，立方晶系；β-TiCl3，正交晶系；α-As，三方晶系；Sn（白锡），四方晶系；CuSO4.5H2O，三斜晶系。

4、试画出金属Na和Mg单质的分子轨道能级图，并据此解释其导电性。

解：根据金属能带理论,金属Na和Mg基态时的电子填充情况如下图所示:Na的3s能带半充满，在电场的作用下其电子获得能量可借助空轨道发生定向移动，所以能够导电。

数学专业英语第二版2.5-c

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)。

微积分高等数学英文练习题以及答案(2)

Exercise 2-1 Concept of Derivative1. The motion of an object along the s -axis follows the law2s t t =+(m). Find:(1) the average speed of the object during the time interval from the 1st second to the 2nd second; 解：2(2)(1)(22)(11)421s s v -==+-+=- (m/sec)(2) the Instantaneous velocity of the object at the 2nd second.解：()()12v t s t t '==+；所以(2)(2)5v s '== (m/sec)2. Compare the following limits with the definition of the derivative and then point out the relation between A and 0()f x '. (Assume that ()0x f ' exists) (1) ()()A xx f x x f x =∆-∆-→∆000lim ；解：()()()0000lim x fx x fx A f x x∆→-∆-'=-=--∆(2) ()A x f n x f n n =⎥⎦⎤⎢⎣⎡-⎪⎭⎫ ⎝⎛+∞→001lim解：()()0001lim1n f x f x n A f x n→∞⎡⎤⎛⎫+- ⎪⎢⎥⎝⎭⎣⎦'==(3) ()()00limh fx h fx h A h→+--=.(Hint:()()()()()()000000f x h f x h f x h f x f x h f x hhh+--+---=-)解：()()()()00000lim →+---⎛⎫=- ⎪⎝⎭h f x h f x f x h f x A h h ()()()()()000000lim 2→+---⎛⎫'=+= ⎪-⎝⎭h f x hf x f x h f x f x h h3. Find 0()lim sin 2→x f x x, if (0)0f = and (0)2f '=.解：00()()(0)lim lim sin 20sin 2→→-=⋅-xx f x f x f xx x x00()(0)lim lim 0sin 21(0)12→→-=⋅-'=⋅=x x f x fxx x f4. Find derivatives of the following functions by using the derivative formula of power functions:(1)y =解：32==y x ，所以311223322-'==y x x(2)y =；解：13-==y x，所以141331133---'=-=-y xx(3)3y x =解：1163355+===y x x x，所以1125161655'==y x x5. Take two points with the abscissa 11=x and 33=x on the parabola 2x y =and draw a secant line through the two points. Find the point on the parabola at which the tangent line to the curve is parallel to this secant line.解：当11=x 时，11y =；当23x =时，29y =。

高等数学教材答案下册英语

高等数学教材答案下册英语Unit 1: Functions and Their GraphsChapter 1: Linear Functions1.1 Functions and Their Representations1.2 Domain and Range1.3 Linear Functions and EquationsChapter 2: Quadratic Functions2.1 Graphs of Quadratic Functions2.2 Solving Quadratic Equations2.3 Quadratic Functions and Their Transformations Chapter 3: Exponential and Logarithmic Functions3.1 Exponential Functions and Their Graphs3.2 Logarithmic Functions and Their Graphs3.3 Exponential and Logarithmic EquationsUnit 2: Limits and ContinuityChapter 4: Limits and Continuity4.1 Limits and Their Properties4.2 Continuity and Its Properties4.3 Computing LimitsChapter 5: Derivatives5.1 The Derivative and its Interpretation5.2 Differentiation Techniques5.3 Applications of DerivativesChapter 6: Higher-Order Derivatives6.1 Higher-Order Derivatives and Their Interpretations 6.2 The Chain Rule6.3 Implicit DifferentiationUnit 3: IntegrationChapter 7: Antiderivatives and Indefinite Integrals 7.1 Antiderivatives and Their Properties7.2 Indefinite Integrals7.3 Substitution MethodChapter 8: Definite Integrals and Their Applications 8.1 Definite Integrals and Their Properties8.2 Applications of Definite Integrals8.3 Numerical IntegrationChapter 9: Techniques of Integration9.1 Integration by Parts9.2 Trigonometric Integrals9.3 Trigonometric SubstitutionUnit 4: Differential Equations and Applications Chapter 10: First-Order Differential Equations 10.1 Separable Differential Equations10.2 Linear Differential Equations10.3 Applications of Differential Equations Chapter 11: Applications of Differential Calculus 11.1 Optimization11.2 Related Rates11.3 Newton's MethodChapter 12: Sequences and Series12.1 Sequences and Their Limits12.2 Infinite Series12.3 Convergence TestsUnit 5: Multivariable CalculusChapter 13: Functions of Several Variables 13.1 Functions of Two or More Variables13.2 Partial Derivatives13.3 Optimization of Functions of Two VariablesChapter 14: Multiple Integrals14.1 Double Integrals14.2 Triple Integrals14.3 Applications of Multiple IntegralsChapter 15: Vector Calculus15.1 Vector Fields15.2 Line Integrals15.3 Green's TheoremChapter 16: Differential Calculus of Vector Fields16.1 Gradient Fields and Potential Functions16.2 Divergence and Curl16.3 Stokes' TheoremI hope the above chapters and sections provide a comprehensive overview of the answers to the exercises and problems in the textbook. Remember to utilize this answer key as a useful tool to check your understanding and progress in studying advanced mathematics.。

高等数学的英文版教材答案

高等数学的英文版教材答案I. Introduction to Higher MathematicsHigher Mathematics is a fundamental subject that plays a crucial role in various fields of study, such as engineering, physics, economics, and computer science. To fully comprehend the concepts and principles of Higher Mathematics, it is essential to have access to accurate and comprehensive textbooks. This article aims to provide an overview of the English version of a Higher Mathematics textbook, focusing on its content, structure, and the importance of having reliable answer keys.II. Content of the English Version of Higher Mathematics TextbookThe English version of the Higher Mathematics textbook covers a wide range of mathematical topics, including calculus, linear algebra, differential equations, probability theory, and mathematical modeling. Each chapter delves into specific concepts and provides detailed explanations, accompanied by illustrative examples to enhance understanding.1. Calculus:The section on calculus is divided into differential calculus and integral calculus. It introduces key principles such as limits, derivatives, and integrals. Various calculus techniques, such as the chain rule, product rule, and fundamental theorem of calculus, are thoroughly explained.2. Linear Algebra:The linear algebra section encompasses topics like vector spaces, matrices, determinants, and eigenvalues. It illustrates the fundamentaltheories of linear algebra and introduces essential concepts, including linear transformations, spanning sets, and eigenvectors.3. Differential Equations:In the differential equations chapter, students learn about different types of differential equations, such as first-order, second-order, and higher-order equations. The solution methods for both homogeneous and non-homogeneous equations are outlined, along with applications in various fields.4. Probability Theory:The section on probability theory introduces students to basic concepts, such as random variables, probability distributions, and expected values. It covers topics like binomial, Poisson, and normal distributions, enabling students to apply probability theory in real-life situations.5. Mathematical Modeling:The mathematical modeling section explores the process of formulating a mathematical model to describe real-world phenomena. It highlights the importance of applying mathematical techniques to analyze and solve practical problems.III. Structure of the English Version of Higher Mathematics TextbookThe English version of the Higher Mathematics textbook follows a well-organized structure, with each chapter building upon the previous one. It presents complex mathematical content in a systematic manner, ensuring a progressive learning experience for students.1. Chapter Organization:Each chapter begins with an introduction that outlines the key concepts and learning objectives. It is followed by concise explanations and illustrative examples that aid in conceptual understanding. The chapters conclude with a summary of the main points covered, allowing for a quick review.2. Exercises and Problems:Throughout the textbook, numerous exercises and problems are provided to reinforce the learned concepts. Students can practice solving mathematical problems to enhance their problem-solving skills and apply the newly acquired knowledge. The textbook also offers answers to the exercises for self-assessment.IV. Importance of Reliable Answer KeysHaving reliable answer keys is crucial for students studying Higher Mathematics. These answer keys serve as a valuable resource to verify the accuracy of their solutions and improve their problem-solving abilities. Furthermore, they provide immediate feedback, allowing students to identify and rectify any mistakes made during the learning process.V. ConclusionThe English version of the Higher Mathematics textbook serves as a comprehensive guide for students studying Higher Mathematics. With its well-structured content, illustrative examples, and reliable answer keys, the textbook enhances students' understanding of mathematical concepts and facilitates their learning experience. It equips them with essential skillsrequired for various fields, enabling them to excel in their academic and professional pursuits.。

数学专业英语答案之2.2 几何与三角

2.2 几何与三角词组翻译1.学会institution，建筑师 architect, 机械师 machinist, 制图员draftsman, 测量者surveyor, 木匠carpenter2.点point, 端点endpoint, 线line, 直线straight line, 线段 line segment, 曲线curved line, 折线 broken line, 射线ray , 平面 plane,曲面 curved surface3.立体solid, 柱体cylinder, 立方体cube,球 sphere, 棱锥pyramid,圆锥 cone ,4.圆circle,圆心 center, 直径diameter, 半径radius, 半圆semicircle, 弦chord, 弧arc, 优弧major arc, 劣弧minor arc5.角angle, 边side, 三角形triangle, 直角三角形right triangle,斜边 hypotenuse, 直角边right-angle side6.长度length,宽度 breadth/width,厚度 thickness, 位置position7.几何的geometrical,立体的 three-dimensional , 弯曲的curved,等距离的equidistant ,无限的 infinite8.培养创造力train originality,必须的毅力 necessary perseverance , 提高鉴赏力raise/improve the appreciation ability9.消失了的边界vanishing boundaries/landmarks,有序性和优美感 orderliness and sense of beauty, 几何图形大量存在geometric forms abound in , 定理成立的先决条件a prerequisite to a theorem 汉译英1.许多专家都认为数学是学习其他科学技术的必备基础和先决条件。

英文高等数学教材答案

英文高等数学教材答案Chapter 1: Functions and their Graphs1.1 Introduction to Functions1.1.1 Definition of a FunctionA function is a relation that assigns a unique output value to each input value. It can be represented symbolically as f(x) or y = f(x), where x is the input variable and y is the output variable.1.1.2 Notation and TerminologyIn function notation, f(x) represents the output value corresponding to the input value x. The domain of a function is the set of all possible input values, while the range is the set of all possible output values.1.2 Graphs of Functions1.2.1 Cartesian Coordinate SystemThe Cartesian coordinate system consists of two perpendicular number lines, the x-axis and the y-axis. The point of intersection is called the origin and is labeled (0, 0). The x-coordinate represents the horizontal distance from the origin, while the y-coordinate represents the vertical distance.1.2.2 Graphical Representation of FunctionsThe graph of a function is a visual representation that shows the relationship between the input and output values. It consists of all points (x, f(x)) where x is in the domain of the function. The shape of the graph depends on the nature of the function.1.3 Properties of Functions1.3.1 Even and Odd FunctionsAn even function is symmetric with respect to the y-axis, meaning f(-x) = f(x) for all x in the domain. An odd function is symmetric with respect to the origin, meaning f(-x) = -f(x) for all x in the domain.1.3.2 Increasing and Decreasing FunctionsA function is increasing if the output values increase as the input values increase. It is decreasing if the output values decrease as the input values increase. A function can also be constant if the output values remain the same for all inputs.Chapter 2: Limits and Continuity2.1 Introduction to Limits2.1.1 Limit of a FunctionThe limit of a function f(x) as x approaches a particular value c, denoted as lim[x→c] f(x), describes the behavior of the function near that point. It represents the value that the function approaches as x gets arbitrarily close to c.2.1.2 One-Sided LimitsOne-sided limits consider the behavior of the function from only one side of the point. The limit from the left, lim[x→c-] f(x), looks at the behavior as x approaches c from values less than c. The limit from the right, lim[x→c+] f(x), considers the behavior as x approaches c from values greater than c.2.2 Techniques for Evaluating Limits2.2.1 Direct SubstitutionDirect substitution involves substituting the value of the input variable directly into the function to find the limit. This method works when the function is continuous at that point and doesn't result in an indeterminate form (e.g., 0/0 or ∞/∞).2.2.2 Factoring and CancellationFactoring and cancellation can be used to simplify expressions and eliminate common factors before applying direct substitution. This technique is particularly useful when dealing with polynomial functions.2.3 ContinuityA function is continuous at a point if the limit of the function exists at that point, the function is defined at that point, and the left and right limits are equal. A function is called continuous on an interval if it is continuous at every point within that interval.Chapter 3: Derivatives3.1 Introduction to Derivatives3.1.1 Definition of the DerivativeThe derivative of a function f(x) represents the rate at which the function changes with respect to its input variable. It is denoted as f'(x) or dy/dx and is defined as the limit of the difference quotient Δy/Δx as Δx approaches zero.3.1.2 Interpretation of the DerivativeThe derivative represents the slope of the tangent line to the graph of the function at a given point. It provides information about the rate of change, instantaneous velocity, and concavity of the function.3.2 Techniques for Finding Derivatives3.2.1 Power RuleThe power rule states that if f(x) = x^n, where n is a constant, then f'(x) = nx^(n-1). This rule allows us to find the derivative of polynomial functions.3.2.2 Chain RuleThe chain rule is used to find the derivative of composite functions. If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). This rule is particularly useful when dealing with functions that are composed of multiple functions.3.3 Applications of Derivatives3.3.1 Rate of ChangeThe derivative represents the rate of change of a function. It can be used to determine the instantaneous rate of change at a specific point or the average rate of change over a given interval.3.3.2 OptimizationDerivatives can be used to optimize functions by finding the maximum or minimum values. This involves finding the critical points where the derivative is zero or undefined and analyzing the behavior of the function around those points.Note: This is just a sample outline for the article on the answer key for an English advanced mathematics textbook. The actual content will depend on the specific exercises and problems in the textbook, which cannot be provided without access to the textbook itself.。

2023大学_数学专业英语试题及参考答案

2023数学专业英语试题及参考答案数学专业英语试题一、词汇及短语1. For a long period of the history of mathematics, the centric place of mathematical methods was occupied by the logical deductions “在数学史的很长的时期内，是逻辑推理一直占据数学方法的中心地位”2. An equation is a statement of the equality between two equal numbers or number symbols.equation ：“方程”“等式”等式是关于两个数或数的符号相等的'一种陈述3. In such an equation either the two members are alike, or become alike on performance of the indicated operation. 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。

注“two members”表等号的两端alike 相同的一样的On the performance of …中的“on”引导一个介词短语做状语Either…or…4. is true “成立”5. to more and change the terms移次和变形without making the equation untrue 保持方程同解数学专业英语试题二、句型及典型翻译1. change the terms about 变形2. full of ：有许多的充满的例 The streets are full of people as on a holiday(像假日一样，街上行人川流不息)3. in groups of ten…4. match something against sb. “匹配”例 Long ago ,when people had to count many things ,they matched them against their fingers. 古时候，当人们必须数东西时，在那些东西和自己的手指之间配对。

数学专业英语答案2.3集合论的基本概念

2.3集合论的基本概念单词、词组1.1集set，子集subset，真子集proper subset，全集universal subset，空集void/ empty set，基地集the underlying set 1.2正数positive number，偶数even integer，图形diagram，文氏图V enn diagram，哑标dummy index，大括号brace 1.3可以被整除的be divisible by，两两不同的distinct from each other，确定的definite，无关紧要的irrelevant/inessential 1.4一样的结论the same conclusion，等同的效果equivalent effect，用大括号表示集sets are designated by braces，把这个图形记作A：this diagram is designated by letter A，区别对象to distinguish between objects，证明定理to prove theorems，把结论可视化to visualize conclusions/consequences汉译英2.1由于小于10且能被3整除的正整数组成的集是整数集的子集。

The set consisting of those positive integers less than 10 which are divisible by 3 is a subset of the set of all integers.2.2如果方便，我们通过在括号中列举元素的办法来表示集。

When convenient，we shall designate sets by displaying the elements in braces.2.3用符号￠表示集的包含关系，也就是说，式子A￠B表示A包含于B。

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。

《数值分析》黄仿伦改编英文版课后习题答案

Answers for Exercises —Numerical methods using MatlabChapter 1P10 2. Solution (a) )(x g x = produces an equation 0862=+-x x . Solving it gives the roots 2=x and 4=x .Since 2)2(=g and 4)4(=g , thus, both 2=P and 4=P are fixed points of )(x g . (b) –(d) The iterative rule using )(x g is 22144n n n p p p ---=. The results for part (b)-(d) with starting value 9.10=p and 8.30=p are listed in Table 1.(e) Calculate values of x x g -='4)( at 2=x and 4=x .12)2(>='g , and 10)4(<='g .Since )(x g ' is continuous, there exists a number 0>δ such that1)(<'x g for all ]4,4[δ+δ-∈x .There also exists a number 0>λ such that1)(>'x g for all ]2,2[λ+λ-∈x .Therefore, 4=p is an attractive fixed point. The sequence generated by22144n n n p p p ---=with starting value 8.30=p converges to 4=p . 2=p is a repelling fixed point. The sequence generated by 22144n n n p p p ---=with starting value 9.10=p does not converge to 2=p .P11 4. Find the fixed point for )(x g : )(x g x = gives 2±=p . Find the derivative: 12)(+='x x g .Evaluate )2(-'g and )2(g ': 3)2(-=-'g , 5)2(='g .Both 2-=p and 2=p gives 1)(>'p g . There is no reason to find the solution(s)using the fixed-point iteration.P11 6. Proof ))(()()(010112p p g p g p g p p -ξ'=-=-)()()( 0101p p K p p g -<-ξ'≤P214. False position method: Assume that ],[n n b a contains the root. The equation of the secand line through ))(,(n n a f a and ))(,(n n b f b is )()()()(n nn n n n b x a b a f b f b f y ---=-. Itintersects x -axise at)()())((n n n n n n n a f b f a b b f b c ---= (Eq. 1.36, p18)1981.0)6.1()(,4907.0)4.2()(00-=-==-=f b f f a f ,8301.1)()())((0000000-=---=a f b f a b b f b c ;Since 0095.0)(0-=c f , then ]8301.1,4.2[],[11--=b a . Similarly, we have1.84093- 1=c , ]1.84093- ,4.2[],[22-=b a 1.84139- 2=c , ]1.84139- ,4.2[],[33-=b a -1.841403=c10. Bisection method: Assume that ],[n n b a contains the root. Then 2nn n b a c +=. (a) 1587.1)4(,4;1425.0)3(,300==-==f b f a , then 5.30=c .Since 03746.0)5.3()(0>==f c f , then ]5.3,3[],[11=b a .Similarly, we can obtain ,,,321c c c . The results are listed in Table 3.The values of tan(x) at midpoints are going to zero while the sequence converges(b) Since 0)3tan(<=, there exist a root in )3,1(..0-=, 055741425.1tan(>)1The results using Bisection method are listed in Table 4.Although the sequence converges, the values of tan (x) at midpoints are not going to zero.P36 2. 3)(2--=x x x f has two zeros 2131±=x . (3028.2,3028.121≈-≈x x ) The first derivative of 3)(2--=x x x f is 12)(-='x x f .The Newton-Raphson iterative function is 123)()()(2-+='-=x x x f x f x x g . The Newton-Raphson formula is 12321-+=+n nn p p p , ,2,1,0=n . The results are listed in Table 5 with starting value p 0=1.6 and p 0=0.0 respectively.Obviously, the sequence generated by the starting value p 0=0.0 does not converge.11. Use Newton-raphson method to solve 0)(3=-=A x x f .The derivative of )(x f is 23)(x x f ='.3232)()()(223x Ax x A x x f x f x x g +=+='-=.Newton-Raphoson formula is 32211--+=n n n p Ap p , ,2,1=n .Since 3A p = is a zero of A x x f -=3)( and 10332)(33<=⎥⎦⎤⎢⎣⎡-='=Ap x A p g ,The sequence generated by the recursive formula 32211--+=n n n p Ap p will converge to3A p = for any starting value ],[330δδ+-∈A A p , where 0>δ.·Answers for Exercises —Numerical methods using MatlabChapter 2P44 2. Solution The 4th equation yields 24=x .Substituting 24=x to the 3rd equation gives 53=x .Substituting both 24=x and 53=x to the 2nd equation produces 32-=x . 21=x is obtained by sustituting all 32-=x , 53=x and 24=x to the 1st equation. The value of the determinant of the coefficient matrix is 115573115=⨯⨯⨯=D .4. Proof (a) Calculating the product of the two given upper-triangular matrices gives⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++++=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=33333323232222223313231213112212121111113323221312113323221312110000b a b a b a b a b a b a b a b a b a b a b b b b b b a a a a a a B A . It is also an upper-triangular matrix.(b) Let N N ij a A ⨯=)( and N N ij b B ⨯=)( where 0=ij a and 0=ij b when j i >.Let N N ij c B A C ⨯==)(. According to the definition of product of the two matrices, we have ∑==Nk kjik ij b ac 1for all N j i ,,2,1, =.0=ij c when j i > because 0=ij a and 0=ij b when j i >.That means that the product of the two upper-triangular matrices is also upper triangular.5. Solution From the first equation we have 31=x .Substituting 31=x to the second equation gives 22=x .13=x is obtained from the third equation and 14-=x is attained from the last equation.The value of the determinant of the coefficient is 243)1(42)det(-=⨯-⨯⨯=A7. Proof The formula of the back substitution for an N N ⨯upper-triangular system is N NN a b x =and kkNk j jkj k k a x a b x ∑+=-=1 for 1,,2,1 --=N N k .The process requiresN N=+++111 divisions, 22)1()1(212NN N N N -=-=-+++ multiplications, and2)1(212NN N -=-+++ additions or subtractions.P53 1. Solution Using elementary transformations for the augmented matrix gives330012630464275101263046425232103514642],[3231213121⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=++-+-r r r r r r B AThat means that ⎪⎩⎪⎨⎧=++=++-=-+523 1035 4642321321321x x x x x x x x x is equivalent to⎪⎩⎪⎨⎧==+-=-+33 1263 4642332321x x x x x x The set of solutions is .3,2,1123-===x x x11. Solution Using the algorithm of Gaussian Elimination gives12420010324050110700211242001032409013270021],[212⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡----−−−→−⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=+-r r B A ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+1242001032005011070021324r r ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡------−−→−+21000103200501107002143r r The set of solutions of the system is obtained by the back substitutions,3,2,2234==-=x x x and .11=x(Chasing method for solving tridiagonal linear systems)14. (a) (i) Solution Applying Gaussian elimination with partial pivoting to the augment matrix results in⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---=↔1100320001.0101001.01003001.010030001.010*******],[31r r B A ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--−−→−↔+-+-00043.03333.43019933.996667.630001.0100319933.996667.63000043.03333.430001.01003 3231213231r r r r r r ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---−−−−→−+-6806.00625.680019933.996667.630001.01003326667.633333.43r rThe set of solutions is,101.0524,0100.0-623⨯==x x and .105.2400 -61⨯=x15. Solution The N N ⨯Hilbert matrix is defined byN N ij H H ⨯=)( where 11-+=j i H ij for N j i ≤≤,1.(a) The inverse of the 44⨯ Hilbert matrix is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--------=-280042001680140420064802700240168027001200120140240120161H The exact solution is T X )140,240,120,16(--=.(b) The solution is T X )0881.185,0628.310,6053.149,7308.18(--=.>>1 H is ill-conditioned. A miss is as good as a mile. (失之毫厘，谬以千里)P62 5 (a) Solving B LY = gives TY )2,12,6,8(-=. From Y UX = we have TX )2,1,1,3(-=. The product of A and X is TAX )4,10,4,8(--=.That means B AX =(b) Similarly to the part (a), we haveTY )1,12,6,28(=, TX )1,2,1,3(=, and B AX T==)4,23,13,28(.6. ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---=175.113011*********L , ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-----=5.70001040085304011UP72 7. (a) Jacobi Iterative formula is ()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=+++)()()1()()()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults for ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.1 with starting value )0,0,0(0=P .The numerical results show that Jacobi iteration does not converge.(b) Gauss-Seidel Iterative formula is()⎪⎪⎩⎪⎪⎨⎧-+=+-=++-=++++++)1()1()1()()1()1()()()1(226141358k k k k k k k k k y x z z x y z y x for ,2,1,0=kResults ),,()()()(k k k k z y x P =’, ,3,2,1=k are listed in Table 2.2 with starting value )0,0,0(0=P ’Reasons:Conside the eigenvalues of iterative matricesSplit the coefficient matrix ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----=612114151A into three matrices⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=--=000100150012004000600010001U L D A .The iterative matrix of Jacobi iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=+=-061311041500121041506100010001)(1U L D T JThe spectral raduis of J T is 16800.5)(>=ρJ T . )1176.0,4546405880(i . .-±=λ’ So Jacobi method doesnot converge.Similarly, the iterative matrix of Gauss-Seidel iteration is⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=-=-65503200150)(1U L D T G .The spectral radius of G T is 2532.19)(=ρG T >1. )0866.0,2532.19,0(-=λ’ So Gauss-Seidel method does not converge.8. (a) Jacobi Iterative formula is()⎪⎩⎪⎨⎧-+=-+=+-=+++6/225/)8(4/)13()()()1()()()1()()()1(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.3 with starting value )0,0,0(0=P .Jacobi iteration converges to the solution (3, 2, 1)’(b) Gauss-Seidel iterative formula is()⎪⎪⎩⎪⎪⎨⎧+---=+---=+-=++++++)1()1()1()()1()1()()()1(22614/)8(4/)13(k k k k k k k k k y x z z x y z y x for ,2,1,0=k ),,()()()(k k k k z y x P =’ for 10,,2,1 =k are listed in Table 2.4 with starting value )0,0,0(0=PGauss-Seidel iteration converges to the solution (3, 2, 1)’Answers for Exercises —Numerical methods using MatlabChapter 3P99 1. Solution (a) The nth order derivative of )sin()(x x f = is )2sin()()(π+=n x x f n .Therefore, !5!3)(535x x x x P +-=, !7!5!3)(7537x x x x x P -+-= and !9!7!5!3)(97539x x x x x x P +-+-=.(b) Estimating the remainder term gives71091075574.2!101!10)5sin()(-⨯≤≤π+=x c x E for 1≤x .(c) Substituting 4π=x to )2sin()()(π+=n x x f n gives ,22)4()4(,22)4()4()3(-=π=π''=π'=πf f f f and 22)4()4()5()4(-=π=πf f .By using Taylor polynomial we have!5)4(22!4)4(22!3)4(22!2)4(22)4(2222)(54325π-+π-+π--π--π-+=x x x x x x P P108 1. (a) Using th e Horner ’s method to find )4(P givesSo )4(P =1.18.(b) From part (a) we have 12.002.002.0)(2-+-=x x x Q . )4()4(Q P =' can be also obtained byusing Horner ’s method.So )4(P '=-0.36 Another method:Hence, P(4)=-0.36.(c) Find )4(I and )1(I firstly.Then=-=⎰)1()4()(41I I dx x P 4.3029.(d) Use Horner ’s method to evaluate P (5.5)Hence, P (5.5)=0.2575.(d) Let 012233)(a x a x a x a x P +++=. There are 4 coefficients needed to found.Substituting four known point ),(i i y x , i =1, 2, 3, 4, into )(x P gives four linear equations with unknowni a , i =1, 2, 3, 4.54.10123=+++a a a a 5.12480123=+++a a a a 42.139270123=+++a a a a 66.05251250123=+++a a a aThe coefficients can be found by solving this linear system: .66.1,2.0,1.0,02.00123=-==-=a a a aP120 1. The values of f (x ) at the given points are listed in Table 3.1:(a) Find the Lagrange coefficient polynomials and 010)(0,1x x x L -=---=.1101)(1,1+=++=x x x LThe interpolating polynomial is x x L f x L f x P =+-=)()0()()1()(1,10,11. (b) ),(21)11()1()(20,2x x x x x L -=----=,110)1)(1()(21,2x x x x L -=--+=),(212)1()(22,2x x x x x L +=+=x x L f x L f x L f x P =++-=)()1()()0()()1()(2,21,20,22. (c) ),2)(1(61)21)(11()2)(1()(0,3---=-------=x x x x x x x L),2)(1)(1(21)20)(10)(10()2)(1)(1()(1,3--+=--+--+=x x x x x x x L),2)(1(21)21(1)11()2()1()(2,3-+-=-+-+=x x x x x x x L),1)(1(61)12(2)12()1()1()(3,3-+=-+-+=x x x x x x x L33,32,31,30,33)()2()()1()()0()()1()(x x L f x L f x L f x L f x P =+++-=(d) ,2212)(0,1x x x L -=--=,1121)(0,1-=--=x x x L 67)()2()()1()(1,10,11-=+=x x L f x L f x P . (e) ),23(21)20)(10()2)(1()(20,2+-=----=x x x x x L ),2()21(1)2()(21,2x x x x x L --=--=),(21)12(2)1()(20,2x x x x x L -=--=.23)()2()()1()()0()(22,21,20,22x x x L f x L f x L f x P -=++=7. (a) Note that each Lagrange polynomial )(,2x L k is of degree at most 2 and )(x g is a combination of)(,2x L k . Hence )(x g is also a polynomial of degree at most 2.(b) For each k x , 2,1,0=k , the Lagrange coefficient polynomial 1)(,2=k k x L , and 0)(,2=k j x L for k j ≠, 2,1,0=j . Therefore, 01)()()()(2,21,20,2=-++=k k k k x L x L x L x g .(c) )(x g is a polynomial of degree 2≤n and has n ≥ 3 zeroes. According to the fundamental theorem of algebra, 0)(=x g for all x .9. Let )()()(x P x f x E N N -=. )(x E N is a polynomial of degree N ≤.)(x f is degree with )(x P N at N +1 points N x x x ,,,10 implies that )(x E N has N +1 zeroes. Therefore, 0)(=x E N for all x , that is, )()(x P x f N = for all x .P131 6. (a) Find the divided-difference table:(b) Find the Newton polynomials with order 1, 2, 3 and 4.)0.1(80.16.3)(1--=x x P , )0.2)(0.1(6.0)0.1(80.160.3)(2--+--=x x x x P ,)0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(3------+--=x x x x x x x P , )0.4)(0.3)(0.2)(0.1(03.0 )0.3)(0.2)(0.1(15.0)0.2)(0.1(6.0)0.1(80.16.3)(4----+------+--=x x x x x x x x x x x P .(c)–(d) The results are listed in Table 3.2P143 6. x x x T 32)(323-=, ]1,1[-∈x .The derivative of )(3x T is 323)(223-⋅='x x T . 0)(3='x T yields 21±=x . Evaluating )(3x T at 21±=x and 1±=x gives 1)1(3-=-T , 1)21(3=-T , 1)21(3-=T and 1)1(3=T .Therefore, 1))(max(3=x T , 1))(min(3-=x T .10. When 2=N , the Chebyshev nodes are ,23)6/5cos(0-=π=x ,01=x and 23)6/cos(2=π=x .Calculating the Lagrange coefficient polynomials based on 210,,x x x can produce the following results:,323)232(23)23()(20,2x x x x x L +-=⨯-⨯--=,341)23(23)23()23()(21,2x x x x L -=-⨯-+=.32323232)23()(22,2x x x x x L +=⨯⨯+=The proof is finished.Answers for Exercises —Numerical methods using MatlabChapter 4P157 1(a). Solution The sums for obtaining Normal equations are listed in Table 4.1The normal equations are ,710=A 135=B . Then ,7.0=A 6.2=B .The least-squares line is 6.27.0+=x y .2449.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f EP158 4. Proof Suppose the linear-squares line is B Ax y += where A and B satisfiesthe Normal equations ∑∑===+N k k Nk ky xAB N 11and ∑∑∑====+Nk k k N k k N k k y x x A x B 1121.y y N x A B N N B x N A B x A N k k Nk k N k k ==⎪⎪⎭⎫ ⎝⎛+=+⎪⎪⎭⎫ ⎝⎛=+∑∑∑===111111 meas thatthe point ),(y x lies on the linear-squares line B Ax y +=.5. First eliminating B on the Normal equations∑∑===+Nk k Nk k y x A B N 11and ∑∑∑====+Nk k k Nk k Nk k y x x A x B 1121gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑===Nk k N k k N k k k y x y x N D A 1111 where 2112⎪⎪⎭⎫ ⎝⎛-=∑∑==N k k N k k x x N D . Substituting A into the first equation gets⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛+-=∑∑∑∑∑=====Nk k Nk k N k k k N k k N k k y x N y x x y N D D B 12111111. Note that ∑∑∑∑∑∑∑∑========⎪⎪⎭⎫ ⎝⎛-=⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛-=N k k N k k N k k N k k N k k N k k N k k Nk k y x N y x y x x N N y N D 12111212112111. Simplifying B gives⎪⎪⎭⎫ ⎝⎛-=∑∑∑∑====Nk k k N k k Nk k N k k y x x y x D B 111121.8(b). The sums needed in the Normal equations are listed in Table 4.26177.142==∑∑k kk x y x A )2(=M5606.063==∑∑kkkxy x B )3(=MHence, 26177.1x y = and 35606.0x y =.0.3594 )(51)(21512222=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Ax y Ax E , 1.1649 )(51)(21512332=⎪⎪⎭⎫ ⎝⎛-=∑=k k k Bx y Bx E .26175.1x y =fits the given data better.P171 2(c). The sums for normal equations are listed in Table 4.3.Using the formula∑∑∑∑∑∑∑========++=++52545352515135125151512515k kk k k k k k k k kk k k ky x x A x B x C y x A xB Cproduces the system with unkowns A , B , and CSolving the obove system gives .6.0,1.0,5.2-=-==C B A The fitting curve is .6.01.05.22--=x x yP172 4. (a) Translate points in x-y plane into X-Y plane using y Y x X ln ,==. The results arelisted in Table 4.4.The Normal equationsgive the system .793410,110,22105=+-==+A C B A C ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A B .8648.0155,2196.455-=+=+A B A BThen -0.50844=A , 1.3524=B . Thus 866731.3524.e e C B ===.The fitting curve is xe .y 50844.086673-=, and 1190.0)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(b) Translate points in x-y plane into X-Y plane using yY x X 1,==. The results are listed in Table 4.5.The Normal equationsgive the system Then 2432.0=A , 30280.0=B .The fitting curve is 30280.02432.01+=x y and 5548.4)((51)(215122=⎪⎪⎭⎫ ⎝⎛-=∑=k k k x f y f E .(c) It is easy to see that the exponential function is better comparing with errors in part (a) and part (b)..1620.5155,7300.255=+=+A B A B ∑∑∑∑∑======+=+515125151515k kk k k k k k kk k Y X X A X B Y X A BP188 1. (a) Derivativing )(x S gives 232132)(x a x a a x S ++='. Substituting the conditions intothe derivative pruduces the system of equations . 012428420321 32132103213210⎪⎪⎩⎪⎪⎨⎧=++=+++=++=+++a a a a a a a a a a a a a a (b) Solving the linear system of equations in (a) gives 29,,12,63210-==-==a a a a . The cubic polynomial is 3229126)(x x x x S -+-=.Figure: Graph of the cubic polynomial4. Step 1 Find the quantities: 3,1210===h h h , 21/)20(/)(0010-=-=-=h y y d13/)03(/)(1121=-=-=h y y d , 6667.03/)31(/)(2232-=-=-=h y y d18)(6011=-=d d u , 10)(6122-=-=d d uStep 2 Use ⎪⎪⎩⎪⎪⎨⎧-'-=⎪⎭⎫ ⎝⎛++'--=+⎪⎭⎫⎝⎛+))((3232))((32232322211100121110d x S u m h h m h x S d u m h m h h to obtain the linear system⎩⎨⎧-=+=+0001.155.1032135.72121m m m m .The solutions are 5161.2,8065.321-==m m . Step 3 Compute 0m and 3m using clamaped boundary.4.90322))((310000-=-'-=m x S d h m , 2.92482))((322323=--'=md x S h m Step 4 Find the spline coefficients16)2(,210001,000,0-=+-===m m h d s y s , 1.45166,-2.45162013,002,0=-===h m m s m s ;-1.54856)2(,021111,110,1=+-===m m h d s y s , -0.35136,1.903321123,112,1=-===h m m s m s ;0.38716)2(,332221,220,2=+-===m m h d s y s , 0.30236,-1.258122233,222,2=-===h m m s m s ;Step 5 The cubic spline is320)3(4516.1)3(4516.2)3(2)()(+++-+-==x x x x S x S for 23-≤≤-x ,321)2(3513.0)2(903.1)2(5484.1)()(+-+++-==x x x x S x S for 12≤≤-x , and322)1(3023.0)1(2581.1)1(3871.03)()(-+---+==x x x x S x S for 41≤≤x .5. Calculate the quantities: 3,1210===h h h , 20-=d ,11=d , 6667.02-=d ,181=u , 102-=u . ( Same values as Ex. 4)Substituting }{j h , }{j d and }{j u into ()()⎩⎨⎧=++=++22211112111022u m h h m h u m h m h h gives ⎩⎨⎧-=+=+1012318382121m m m mSolve the linear equation to obtain .5402.1,8276.221-==m m In addition, .030==m m Use formula (4. 65) to find the spline coefficients:4713.26)2(,210001,000,0-=+-===m m h d s y s , 4713.06,020013,002,0=-===h m m s m s ;-1.05756)2(,021111,110,1=+-===m m h d s y s , -0.24276,4138.121123,112,1=-===h m m s m s ;8735.06)2(,332221,220,2=+-===m m h d s y s , 0856.06,7701.0-22233,222,2=-===h m m s m s .Therefore, 30)3(4713.0)3(4713.22)(+++-=x x x S , for 23-≤≤-x ;321)2(2427.0)2(4138.1)2(0575.1)(+-+++-=x x x x S , for 12≤≤-x322)1(0856.0)1(7701.0)1(8735.03)(-+---+=x x x x S for 41≤≤x .Answers for Exercises —Numerical methods using MatlabChapter 5P209 1(b). Solution LetThe result of using the trapezoidal rule with h =1 isUsing Simpson’s rule with h=1/2, we haveFor Simpson’s 3/8 rule with h=1/3, we obtainThe result of using the Boole’s rule with h=1/4 is4. Proof Integrate )(1x P over ],[10x x .110102012101)(2)(2)(xx x x x x x x h f x x h f dx x P -+--=⎰=)(210f f h +. The Quadrature formula )(2)(1010f f hdx x f x x +≈⎰is called the trapezoidal rule.6. Solution The Simpson ’s rule is)4(3)(21010f f f hdx x f x x ++≈⎰. It will suffice to apply Simpson ’s rule over the interval [0, 2] with the test functions32,,,1)(x x x x f = and 4,x . For the first four functions, since)1141(31212+⨯+==⎰dx , )2140(31220+⨯+==⎰xdx , )4140(3138202+⨯+==⎰dx x , )8140(314203+⨯+==⎰dx x , the Simpson ’s rule is exact. But for 4)(x x f =,)16140(3153224+⨯+≠=⎰dx x . .Therefore, the degree of precision of Simpson ’s rule is n =3.T he Simpson’s rule and the Simpson ’s 3/8 rule have the same degree of precision n =3.).4cos(1)(x e x f x -+=..f f f f h dx x f 3797691))1()0((21)(2)(1010=+=+≈⎰.9583190))1()5.0(4)0((61)4(3)(21010. f f f f f f h dx x f =++=++≈⎰.9869270 ))1()3/2(3)3/1(3)0(( 8/1 )33(83)(321010.f f f f f f f f hdx x f =+++=+++≈⎰.008761 ))1(7)4/3(32)2/1(12)4/1(32)0(7( 90/1 )73212327(452)(432101.f f f f f f f f f f hdx x f =++++=++++≈⎰P220 3(a) Solution When 3)(x x f =for 10≤≤x , ⎰+π=123912dx x x area .The values of 2391)(x x x g +=at 11 sample points (M =10) are listed in the Table 5.1:(i) Using the composite Trapezoidal rule ∑-=++=110)()()((2),(M k k M x g h x g x g hh g T , the computation is)9156.11084.16098.03719.01563.00710.00280.00081.00010.0(101)1623.30(201)101,(++++++++++=g T=)2160.4(101)1623.3(201+=0.1576+0.4216=0.5792.(ii) Using the composite Simposon ’s rule ∑∑-=--=+++=11121120)(34)(32)()((3),(M k k M k k M x g h x g h x g x g h h g S , the computation is)9156.16098.01563.00280.00010.0(304)1084.13719.00710.00081.0(302 )1623.30(301)101,(++++++++++=g S=)7106.2(304)5054.1(302 )1623.3(301++=0.5672.7. (a) Because the formula)2()1()0()(2102g w g w g w dt t g ++=⎰is exact for the three functions 1)(=t g ,x t g =)(, and 2)(x t g =, we obtain three equations with unkowns 0w , 1w , and 2w :2210=++w w w , 2221=+w w ,38421=+w w . Solving this linear system gives 310=w , 341=w and 312=w .Thus, ())2()1(4)0(31)(20g g g dt t g ++=⎰(b) Let ht x x +=0 and denote ,01h x x +=.202h x x +=Then the change of variable ht x x +=0 translates ],[20x x into [0, 2] and converts the integral expresion dx x f )( into dt ht x hf )(0+. Hence,()())()(4)(3)2()1(4)0(3)()()(21022002x f x f x f h g g g hdt t g h dt ht x f h dx x f x x ++=++==+=⎰⎰⎰. The formula ())()(4)(3)(21020x f x f x f hdx x f x x ++=⎰ is known as the Simpson ’s rule over ],[20x x .8(a).9(a).P234 1(a) Let 212sin )(x xx f +=. The Romberg table with three rows for ⎰+3212sin dx x xis given as follows:Where04191.0)02794.0(23)106sin 0(23))3()0((23)0()0,0(-=-=+=+==f f T R , 04418.0)5.113sin (5.1204191.0)5.1(5.12)0()1()0,1(2=++-=+==f T T R ,3800.0)25.215.4sin 75.015.1sin (75.0204418.0))25.2()75.0((75.02)1()2()0,2(22=++++=++==f f T T R , 07288.03)04191.0(04418.043)0,0()0,1(4)1()1,1(=--⨯=-==R R S R ,].6/,6/[],[ and cos )(Let ππ-==b a x x f have we , and cos )( ,sin )( Since Mab h x x f x x f -=-=''-='.10513/123/ )(12),(922-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤''--=M c f h a b h f E T .1039.2/)( and )4375( 9.4374 So,4-⨯≈-==>M a b h M M ].6/,6/[],[ andcos )(Let ππ-==b a x x f ,cos )( ,sin )( , cos )( ,sin )( Since )4(x x f x x f x x f x x f =='''-=''-='.105123/1803/ )(180),(92)4(4-⨯<⨯⎪⎭⎫ ⎝⎛ππ≤--=M c f h a b h f E S havewe ,2 and M a b h -=.1027.92/)( and )565( 8.564 So,4-⨯≈-==>M a b h M M4919.0304418.03800.043)0,1()0,2(4)2()1,2(=-⨯=-==R R S R ,5198.0307288.04919.01615)1,1()1,2(16)2()2,2(=-⨯=-==R R B R ,2. Proof If L J T J =∞→)(lim , thenL LL J T J T J S J J =-=--=∞→∞→343)1()(4lim)(lim andL LL J S J S J B J J =-=--=∞→∞→151615)1()(16lim )(lim .9. (a) Let 78)(x x f =. 0)()8(=x f implies 4=K . Thus 256)4,4(=R .(b) Let 1011)(x x f =.0)()11(=x f implies 5=K . Thus 2048)5,5(=R .10. (a) Do variable translation t x =. Thendt t dt t t dx x tx ⎰⎰⎰=⋅===121122.That means the two integrals dx x ⎰1anddt t ⎰122have the same numerical value.(b) Let 22)(t t f = and x x g =)(.Use dt t R ⎰≈122)1,1( means that the truncation error is )()4(n f k ξ approximately.Note that 0)()3(=t f . It means )1,1(212R dt t =⎰.But for x x g =)(, 0)()(=x g n is not true for all ]1,0[∈x and any integer 0>n .Thus the Romberg sequence is faster for dt t ⎰122 than fordx x ⎰1even though they have the samenumerical value.P242 1 (a) Applying the change of variable 22ab x a b t ++-=to dt t ⎰256 givesdx x dt t x t ⎰⎰-+=+⋅==115125)1(66.Thus the two integrals are dt t ⎰256 anddx x ⎰-+⋅115)1(6equivalent.(b)315315311525)1(6)1(6)()1(66=-=-+++=≈+⋅=⎰⎰x x x x f G dx x dt t =0.0809 +58.5857=58.6667If using )(3f G to approximate the integral, The result is535055353115205)1(695)1(698)1(695)()1(66==-=-+++++=≈+⋅=⎰⎰x x x x x x f G dx x dt t64105.5965956.0000 98 0.0035 95=⨯+⨯+⨯=6. Analysis: The fact that the degree of precision of N -point Gauss-Legendra integration is 2N -1 impliesthat the error term can be represented in the form )()()2(c kf f E N N =.(a) Since dt t dx x tx ⎰⎰-+=+==117127)1(88, and ()0)1()8(7=+t implies 82=K . Thus =256)(4=f G .(b),)1(111111101210dt t dx x tx ⎰⎰-+=+==and ()0)1()11(10=+t implies 122=K .Thusdx x⎰21011=2048)(6=f G .7. The n th Legendre polynomial is defined by The first five polynomials areThe roots of them are same as ones in Table 5.8.11. The conditions that the relation is exact for the functions means the three equations:326.0 6.0 0)6.0( )6.0(2 3132111321=+=+-=++w w w w w w w Sloving the system gives 98 ,95 231===w w w . ))6.0((95)0(98))6.0((95)(212111f f f dx x f ++-≈⎰- is called three-point Gauss-Legendre rule.Answers for Exercises —Numerical methods using MatlabChapter 6P249 1. (a) Proof Differentiate 22)(2+-+=-t t Ce t y t .22)(-+-='-t Ce t y tSubstitute )(t y and )(t y ' into the right-hand side of the equation y t y -='2.side left )(22)22(side right 222='=-+-=+-+-=-=--t y t Ce t t Ce t y t t t(b) Solution Let y t y t f -=2),(. Then 1),(-=y t f y for any ),(y t .So, the Lipschitz constant is 1=L .()[],2,11!21)(1)(20=-⋅==n x dx d n x P x P nnn n n ()()()3303581)(3521)(1321)()(1)(244332210+-=-=-===x x x P x x P x x P x x P x P 2,,1)(x x x f =12 . Integrate both side of )(t f y =' over [a , b ]: ⎰⎰='=-babadt t f dt y a y b y )()()(.Then,)()()(a y b y dt t f ba-=⎰, where )(t y is the solution of the I. V . P)(t f y =', for b t a ≤≤ with 0)(=a y . That means that the definite integral⎰badt t f )( can becomputed using the two values )(a y and )(b y of the solution )(t y of the given I. V . P.. 14. Solution Separate the two variables of the equation 211t y +=' into the form dt tdy 211+=. Integrate dt tdy 211+=and yeild the general solution C t y +=arctan . The initial-value condition 0)0(=y means that 0=C . The solution for the I. V . P. is t y arctan =.P257 3. (a)-(c) The formula using Euler ’s method to solve the I. V . P. ty y -=', 1)0(=y canbe represented in the form k k k y ht y )1(1-=+. When 2.0=h and 1.0=h , the results are listed in Table 6.1.(d) The F. G . E. does decrease half approximatelly as expacted when h is halved.6. When 02.0=a , 00004.0=b and 10=h , the Euler ’s formula for 2bP aP P -=' is in the form210004.02.1k k k P P P -=+. With 1.760=P , the missing entries can be filled in the table.。

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