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)。
高中数学英文试题及答案
高中数学英文试题及答案High School Mathematics English Exam Questions and AnswersQuestion 1:Solve the following quadratic equation for x:\[ 2x^2 - 7x + 3 = 0 \]Answer 1:To solve the quadratic equation \( 2x^2 - 7x + 3 = 0 \), we can use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 2 \), \( b = -7 \), and \( c = 3 \). Plugging in the values, we get:\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} \]\[ x = \frac{7 \pm \sqrt{49 - 24}}{4} \]\[ x = \frac{7 \pm \sqrt{25}}{4} \]\[ x = \frac{7 \pm 5}{4} \]So, the solutions are \( x = 3 \) and \( x = \frac{1}{2} \).Question 2:Find the derivative of the function \( f(x) = 3x^4 - 2x^3 + x^2 + 5 \).Answer 2:To find the derivative of the function \( f(x) = 3x^4 - 2x^3 + x^2 + 5 \), we apply the power rule for derivatives:\[ f'(x) = 4 \cdot 3x^3 - 3 \cdot 2x^2 + 2 \cdot x + 0 \]\[ f'(x) = 12x^3 - 6x^2 + 2x \]Question 3:Evaluate the definite integral of the function \( g(x) = 4x - 3 \) from \( x = 1 \) to \( x = 4 \).Answer 3:To evaluate the definite integral of \( g(x) = 4x - 3 \) from \( x = 1 \) to \( x = 4 \), we integrate the function and then subtract the value of the integral at the lower limit from the value at the upper limit:\[ \int_{1}^{4} (4x - 3) \, dx = \left[ 2x^2 - 3x\right]_{1}^{4} \]\[ = (2 \cdot 4^2 - 3 \cdot 4) - (2 \cdot 1^2 - 3 \cdot 1) \] \[ = (32 - 12) - (2 - 3) \]\[ = 20 - (-1) \]\[ = 21 \]Question 4:Simplify the expression \( \frac{2x^2 - 4x + 2}{x - 1} \) by factoring.Answer 4:To simplify the expression \( \frac{2x^2 - 4x + 2}{x - 1} \), we factor out the greatest common factor from the numerator: \[ \frac{2(x^2 - 2x + 1)}{x - 1} \]Notice that the numerator is a perfect square trinomial:\[ \frac{2(x - 1)^2}{x - 1} \]Now, we can cancel out the common factor \( (x - 1) \) from the numerator and denominator:\[ 2(x - 1) \]So, the simplified expression is \( 2x - 2 \).Question 5:Determine the equation of the line that passes through the points \( (2, 3) \) and \( (-1, -2) \).Answer 5:To find the equation of the line passing through the points \( (2, 3) \) and \( (-1, -2) \), we first find the slope \( m \) of the line:\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{-1 - 2} = \frac{-5}{-3} = \frac{5}{3} \]Now, using the point-slope form of a line equation \( y - y_1 = m(x - x_1) \), we can plug in one of the points and the。
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）。
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.。
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.。
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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。
Mathematics comes from man’s social practice, for example, industrial and agricultural production, commercial activities, military operations and scientific and technological researches.（2）如果没有运用数学，任何一个科学技术分支都不可能正常地发展。
No modern scientific and technological branches could be regularly developed without the application of mathematics.（3）符号在数学中起着非常重要的作用，它常用于表示概念和命题。
Notations are a special and powerful tool of mathematics and are used to express conceptions and propositions very often.（4）17 世纪之前，人们局限于初等数学，即几何、三角和代数，那时只考虑常数。
Before 17th century, man confined himself to the elementary mathematics, i. e. , geometry, trigonometry and algebra, in which only the constants were considered.（5）方程与算数的等式不同在于它含有可以参加运算的未知量。
Equation is different from arithmetic identity in that it contains unknown quantity which can join operations.（6）方程又称为条件等式，因为其中的未知量通常只允许取某些特定的值。
Equipment is called an equation of condition in that it is true only for certain values of unknown quantities in it.（7）方程很有用，可以用它来解决许多实际应用问题。
Equations are of very great use. We can use equations in many mathematical problems.（8）解方程时要进行一系列移项和同解变形，最后求出它的根，即未知量的值。
To solve the equation means to move and change the terms about without making the equation untrue, until the root of the equation is obtained, which is the value of unknown term.英译汉1.Algebra has evolved from the operations and rules of arithmetic.The study of arithmetic begins withaddition,multiplication,subtraction,and division ofnumbers:4+7,37×682,49-22,40÷8.In algebra we introduce symbols or letters—such as a,b,c,d,x,y,z—to denote arbitrary numbers and，instead of special cases，we often consider general statements：a+b,cd,x-y,x÷a.代数是从算术的运算和规则当中逐渐发展起来的，算术的研究是从数的加减乘除开始的。
例如4+7，37×682，49-22，40÷8。
在代数学里,我们采用符号或字母。
例如a,b,c,d,x,y,z来表示任意的数字，而不考虑那些特殊情况。
我们经常考虑的是一般的表达式，例如a+b,cd,x-y,x÷a。
2.The language of algebra serves a twofold purpose.First,we may use it as a shorthand to abbreviate and simplify long or complicated statements.Second,it proves a convenient means of generalizing many specific statements.代数的语言有两个作用。
第一个是使用它作为一种速记法去缩减和减化那些又长又复杂的表达。
第二，它被证明是一种概括许多具体的表达方式的便捷途径。
3.Many expressions involve two or more operations.Grouping symbols tell us which operation is to be done first.The common grouping symbols are parentheses,(),brackets.[],and the fraction bar,—.For example,in the expression 2(3+4),we do the addition first and then we do the multiplication:2(3+4)=2(7)=14.许多数学表达式包含两个或更多的运算。
分组符号告诉我们哪一个运算先做。
常见的分组符号是圆括号，方括号和分数线。
例如，在数学表达公式2（3+4）里。
我们先做加法再做乘法2（3+4）=2（7）=142.2 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。
Many experts recognize that mathematics is the necessary foundation and prerequisite of studying other science technology.（2）西方国家的专家认为几何起源于巴比伦和埃及人的土地测量技术，其实中国古代的数学家对几何做了许多出色的研究。