数学分析英文版-chapter4

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书名:数学分析原理(英文版,第 版)

书名:数学分析原理(英文版,第 版)

书评书名:数学分析原理(英文版,第3版)Principles of Mathematical Analysis (Third Edition)作者:(美)Walter Rudin出版商:机械工业出版社 2004作者介绍Walter Rudin,1921年出生于奥地利维也纳的一个富裕的犹太人家庭,1938年因祖国被纳粹德国占领而逃离奥地利,二次大战期间曾经服役于英国海军,二次大战结束后于1945年移民美国。

1953年Walter Rudin于杜克大学获得数学博士学位,然后在麻省理工学院、罗切斯特大学、耶鲁大学等学校任教。

从1959年起在威斯康星大学麦迪逊分校数学系任教。

他的主要研究领域为调和分析、算子理论和复变函数,是这些研究领域的国际著名学者。

Walter Rudin在麻省理工学院执教期间,写了这本著名的教科书“数学分析原理”作为大学生分析课程的教材,第一版于1953年出版,第二版与第三版分别于1964年与1976年出版。

除“数学分析原理”外,他还著有另外两本名著:“实复分析”(Real and Complex Analysis,1966)和“泛函分析”(Functional Analysis,1973),这些教材已被翻译成13种语言,在世界各地广泛使用。

以“数学分析原理”这本书作为教材的名校有加利福尼亚大学伯克利分校、哈佛大学、麻省理工学院等。

Walter Rudin在威斯康星大学麦迪逊分校数学系任教了32年,于1991年退休。

退休后他写了一部自传小说“我的回忆”(The way I remember it),在书中他描述了他的早年生活、骚乱的战争年代、以及他的数学生涯。

但是Walter Rudin作为数学家而闻名于世的还是这本著名的教科书“数学分析原理”,它被数学界亲切地称为“小鲁丁”(Baby Rudin),而另一本名著“实复分析”则被称为“大鲁丁”(Big Rudin)。

正因为写了这两本数学名著,Walter Rudin 于1993年荣获美国数学会颁发的Leroy P. Steele奖。

数字电路英文版第四单元PPT课件

数字电路英文版第四单元PPT课件
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Rules of Boolean Algebra
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Logical Multiplication

数值分析英文课件

数值分析英文课件
ˆ ∆y = y − y = 1.4 − 1.41421L ≈ 0.0142
or relative forward error of about 1 percent. Since 1.96 = 1.4 , the absolute backward error is
ˆ ∆x = x − x = 1.96 − 2 = 0.04
Computational error = Truncation error + rounding error
• Propagated (传播) vs. computational error 传播)
– x = exact value, – f = exact function,
ˆ x = approx. value ˆ f = its approximation
Backward vs. forward errors
Suppose we want to compute y = f ( x ) , where f : ℜ → ℜ ˆ but obtain approximate value y
Forward Error:
ˆ ˆ ∆y = y − y = f ( x ) − f ( x )
Example of Ill-Posed Problem
x 1 x 1 x 11 1 + 2 + 3 = 2 3 6 1 1 1 13 x1 + x2 + x3 = 3 4 12 2 1 x1 + 1 x2 + 1 x3 = 47 3 4 5 60
2 significant digits rounding
• Problems that are not well-posed are ill-posed.

最新数学分析第4章PPT课件

最新数学分析第4章PPT课件

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美国algebra1 Chapter04

美国algebra1 Chapter04

CHAPTER 4116C HAPTERT ABLE OF C ONTENTS4-1Solving Equations Using More Than One Operation4-2Simplifying Each Side of anEquation4-3Solving Equations That Have the Variable in Both Sides4-4Using Formulas to SolveProblems4-5Solving for a Variable in T erms of Another Variable4-6T ransforming Formulas4-7Properties of Inequalities4-8Finding and Graphing theSolution Set of an Inequality 4-9Using Inequalities to SolveProblemsChapter SummaryVocabularyReview ExercisesCumulative Review FIRST DEGREE EQUA TIONS AND INEQU ALITIES IN ONE V ARIABLE An equation is an important problem-solving tool.A successful business person must make many deci-sions about business practices.Some of these deci-sions involve known facts,but others require the use of information obtained from equations based on expected trends.For example,an equation can be used to represent the following situation.Helga sews hand-made quilts for sale at a local craft shop.She knows that the mate-rials for the last quilt that she made cost $76 and that it required 44 hours of work to complete the quilt.If Helga received $450 for the quilt,how much did she earn for each hour of work,taking into account the cost of the materials?Most of the problem-solving equations for business are complex.Before you can cope with complex equa-tions,you must learn the basic principles involved in solving any equation.Some Terms and DefinitionsAn equation is a sentence that states that two algebraic expressions are equal.For example,x ϩ3 ϭ9 is an equation in which x ϩ3 is called the left side ,or left member ,and 9 is the right side ,or right member .An equation may be a true sentence such as 5 ϩ2 ϭ7,a false sentence such as 6 Ϫ 3 ϭ4,or an open sentence such as x ϩ3 ϭ9.The number that can replace the variable in an open sentence to make the sentence true is called a root ,or a solution ,of the equation.For example,6 is a root of x + 3 ϭ9.As discussed in Chapter 3,the replacement set or domain is the set of pos-sible values that can be used in place of the variable in an open sentence.If no replacement set is given,the replacement set is the set of real numbers.The set consisting of all elements of the replacement set that are solutions of the open sentence is called the soluti on set of the open sentence.For example,if the replacement set is the set of real numbers,the solution set of x ϩ3 ϭ9 is {6}.If no element of the replacement set makes the open sentence true,the solution set is the empty or null set,or {}.If every element of the domain satisfies an equation,the equation is called an identity .Thus,5 ϩx ϭx Ϫ (Ϫ5) is an iden-tity when the domain is the set of real numbers because every element of the domain makes the sentence true.Two equations that have the same solution set are equivalent equations .To solve an equation is to find its solution set.This is usually done by writing sim-pler equivalent equations.If not every element of the domain makes the sentence true,the equation is called a conditional equation ,or simply an equation.Therefore,x ϩ3 ϭ9 is a conditional equation.Properties of EqualityWhen two numerical or algebraic expressions are equal,it is reasonable to assume that if we change each in the same way,the resulting expressions will be equal.For example:5 ϩ7 ϭ12(5 ϩ7) ϩ3 ϭ12 ϩ3(5 ϩ7) Ϫ8 ϭ12 Ϫ8Ϫ2(5 ϩ7) ϭϪ2(12)These examples suggest the following properties of equality:5 1 735123лSolving Equations Using More Than One Operation 117Properties of Equality1.The addition property of equality.If equals are added to equals,the sums are equal.2.The subtraction property of equality.If equals are subtracted from equals,the differences are equal.3.The multiplication property of equality.If equals are multiplied by equals,the products are equal.4.The division property of equality.If equals are divided by nonzero equals,the quotients are equal.5.The substitution principle.In a statement of equality,a quantity may be substituted for its equal.To solve an equation,you need to work backward or “undo”what has been done by using inverse operations.To undo the addition of a number,add its opposite.For example,to solve the equation x ϩ7 ϭ19,use the addition prop-erty of equality.Add the opposite of 7 to both sides.The variable x is now alone on one side and it is easy to read the solution,x ϭ12.To solve an equation in which the variable has been multiplied by a num-ber,either divide by that number or multiply by its reciprocal.(Remember multiplying by the reciprocal is the same as dividing by the number.) To solve 6x ϭ24,divide both sides by 6 or multiply both sides by .6x ϭ246x ϭ24orx ϭ4x ϭ4To solve ,multiply each side by the reciprocal of which is 3.x ϭ15In the equation 2x ϩ3 ϭ15,there are two operations in the left side:mul-tiplication and addition.In forming the left side of the equation,x was first mul-tiplied by 2,and then 3 was added to the product.To solve this equation,we must undo these operations by using the inverse elements in the reverse order.Since the last operation was to add 3,the first step in solving the equation is to add its opposite,Ϫ3,to both sides of the equation or subtract 3 from both sides(3)x 35(3)5x35513x 35516(6x )516(24)6x6524616x 1 7 5 1927 27x512118First Degree Equations and Inequalities in One Variableof the equation.Here we are using either the addition or the subtraction prop-erty of equality.orNow we have a simpler equation that has the same solution set as the original and includes only multiplication by 2.To solve this simpler equation,we multi-ply both sides of the equation by ,the reciprocal of 2,or divide both sides of the equation by 2.Here we can use either the multiplication or the division property of equality.orAfter an equation has been solved,we check the equation,that is,we verify that the solution does in fact make the given equation true by replacing the vari-able with the solution and performing any computations.Check:2x ϩ3 ϭ152(6) ϩ3 ϭ1512 ϩ3 ϭ1515 ϭ 15 ✔To find the solution of the equation 2x ϩ3 ϭ15,we used several properties of the four basic operations and of equality.The solution below shows the math-ematical principle that we used in each step.2x ϩ3 ϭ15Given(2x ϩ3) ϩ(Ϫ3)ϭ15 ϩ(Ϫ3)Addition property of equality 2x ϩ[3 ϩ(Ϫ3)]ϭ15 ϩ(Ϫ3)Associative property of addition 2x ϩ0ϭ12Additive inverse property 2x ϭ12Additive identity property ϭMultiplication property of equality ϭAssociative property of multiplication 1x ϭ6Multiplicative inverse property x ϭ6Multiplicative identity propertyThese steps and properties are necessary to justify the solution of an equationof this form.However,when solving an equation,we do not need to write each of the steps,as shown in the examples that follow.12(12)C 12(2)D x12(12)12(2x )2x 5122x 25122x 562x 51212(2x )512(12)x 56122x 1 323 2x551523122x 1 35152x 1 3 1 (23)515 1 (23)2x 512Solving Equations Using More Than One Operation 119EXAMPLE 1Solve and check:7x ϩ15 ϭ71Solution How to Proceed(1)Write the equation:(2)Add Ϫ15,the opposite of ϩ15 toeach side:(3)Since multiplication and division areϭinverse operations,divide each x ϭ8side by 7:(4)Check the solution.Write the solution7x ϩ15 ϭ71in place of x and perform the computations:71 ϭ71 ✔Answer x ϭ8Note:The check is based on the substitution principle.EXAMPLE 2Find the solution set and check:ϭ Ϫ18SolutionAnswer The solution set is {Ϫ20}.EXAMPLE 3Solve and check:7 Ϫx ϭ9Solution METHOD 1.Think of 7 Ϫ x as 7 ϩ(Ϫ1x ).35x 2 656 1 15 5717(8) 1 15 5715677x 7120First Degree Equations and Inequalities in One VariableϭϪ12ϭx ϭϪ2053(212)53A 35x B 35x 16 1635x 2 65 218Addition property of equality Multiplication property of equalityCheckϪ6ϭϪ18Ϫ18 ϭ Ϫ18 ✔212 2 6 521835(220) 2 6 5 21835x7x ϩ15ϭ71Ϫ15Ϫ157x ϭ56CheckAddition property of equality Division property of equalityx ϭϪ2METHOD 2.Add x to both sides of the equation so that the variable has apositive coefficient.How to Proceed(1)Write the equation:7 Ϫ x ϭ9(2)Add x to each side of the equation:7 Ϫ x ϩx ϭ9 ϩx7 ϭ9 ϩx(3) Add Ϫ9 to each side of the equation:Ϫ9 ϩ7 ϭϪ9 ϩ9 ϩxϪ2 ϭxThe check is the same as for Method 1.Answer {Ϫ2} or x ϭϪ2Writing About Mathematics1.Is it possible for the equation 2x ϩ5 ϭ0 to have a solution in the set of positive real num-bers? Explain your answer.2.Max wants to solve the equation 7x ϩ15 ϭ71.He begins by multiplying both sides of the equation by ,the reciprocal of the coefficient of x .a.Is it possible for Max to solve the equation if he begins in this way? If so,what would be the result of multiplying by and what would be his next step?b.In this section you learned to solve the equation 7x ϩ15 ϭ71 by first adding the opposite of 15,Ϫ15,to both sides of the equation.Which method do you think is better? Explain your answer.Developing SkillsIn 3 and 4,write a complete solution for each equation,listing the property used in each step.3.3x ϩ5 ϭ354.ϭ1512x 211717EXERCISES21x 21522171(2x ) 5 927272x 52Solving Equations Using More Than One Operation 1217 Ϫ x ϭ99 ϭ9✔7 1 2 597 2 (؊2) 59In 5–32,solve and check each equation.5.55 ϭ6a ϩ76.17 ϭ8c Ϫ 77.9 Ϫ 1x ϭ78.11 ϭ15t ϩ169.15 Ϫ a ϭ310.11 ϭϪ6d Ϫ 111.8 Ϫ y = 112.ϭ1213.ϭϪ814.12 ϭ15.16.ϭ3017.7.2 ϭ18.ϭ519.Ϫ2 ϭ20.21.4a ϩ0.2 ϭ522.4 ϭ3t Ϫ 0.223.ϭ524.13 ϭ25.ϭ4726.0.04c ϩ1.6 ϭ027.15x ϩ14 ϭ1928.8 ϭ18c Ϫ 129.30.0.8r ϩ19 ϭ2031.32.842 Ϫ162m ϭϪ616Applying Skills33.The formula F ϭgives the relationship between the Fahrenheit temperature F andthe Celsius temperature C .Solve the equation 59 ϭto find the temperature in degrees Celsius when the Fahrenheit temperature is 59°.34.When Kurt orders from a catalog,he pays $3.50 for shipping and handling in addition to thecost of the goods that he purchases.Kurt paid $33.20 when he ordered six pairs of socks.Solve the equation 6x ϩ3.50 ϭ33.20 to find x ,the price of one pair of socks.35.When Mattie rents a car for one day,the cost is $29.00 plus $0.20 a mile.On her last trip,Mattie paid $66.40 for the car for one day.Find the number of miles,m ,that Mattie drove by solving the equation 29 ϩ0.20x ϭ66.40.36.On his last trip to the post office,Hal paid $4.30 to mail a package and bought some 39-centstamps.He paid a total of $13.66.Find s ,the number of stamps that he bought,by solving the equation 0.39s ϩ4.30 ϭ13.66.An equation is often written in such a way that one or both sides are not in sim-plest form.Before starting to solve the equation by using additive and multi-plicative inverses,you should simplify each side by removing parentheses if necessary and adding like terms.Recall that an algebraic expression that is a number,a variable,or a prod-uct or quotient of numbers and variables is called a term .First-degree equations in one variable contain two kinds of terms,terms that are constants and terms that contain the variable to the first power only.95C 13295C 13213w 1 652217514 2 x45t 1 7 5 2 23y14x 1 119d 2 1251712y5 1 3a 4 1 94m 5235m 5t 4545234y 23x 3a 8122First Degree Equations and Inequalities in One VariableLike and Unlike TermsTwo or more terms that contain the same variable or variables,with corre-sponding variables having the same exponents,are called like terms or similar terms .For example,the following pairs are like terms.6k and k5x 2and Ϫ7x 29ab and 0.4aband Two terms are unli ke terms when they contain different variables,or thesame variable or variables with different exponents.For example,the following pairs are unlike terms.3x and 4y5x 2and 5x 39ab and 0.4aand To add like terms,we use the distributive property of multiplication overaddition.9x ϩ2x ϭ(9 ϩ2)x ϭ11x Ϫ16d ϩ3d ϭ(–16 ϩ3)d ϭϪ13dNote that in the above examples,when like terms are added:1.The sum has the same variable factor as the original terms.2.The numerical coefficient of the sum is the sum of the numerical coeffi-cients of the terms that were added.The sum of like terms can be expressed as a single term.The sum of unlike terms cannot be expressed as a single term.For example,the sum of 2x and 3cannot be written as a single term but is written 2x ϩ3.EXAMPLE 1Solve and check:2x ϩ3x ϩ4 ϭϪ6Solution How to Proceed Check(1)Write the equation:2x ϩ3x ϩ4 ϭϪ6(2)Simplify the left side bycombining like terms:(3)Add Ϫ4,the additiveinverse of ϩ4,to Ϫ6 ϭϪ6✔each side:(4)Multiply by ,themultiplicative inverse of 5:(5)Simplify each side.x ϭϪ2Answer Ϫ215(5x )515(210)1524 2 6 1 4 5262(22) 1 3(22) 1 4 52647x 2y383x 3y22113x 2y392x 2y3Simplifying Each Side of an Equation 1232x ϩ3x ϩ4ϭϪ65x ϩ4ϭϪ6Ϫ4Ϫ45xϭϪ10Note:When solving equations,remember to check the answer in the original equation and not in the simplified one.The algebraic expression that is on one side of an equation may contain e the distributive property to remove the parentheses solving the equation.The following examples illustrate how the distributive and asso-ciative properties are used to do this.EXAMPLE 2Solve and check:27x Ϫ3(x Ϫ6) ϭ6Solution Since Ϫ3(x Ϫ6) means that (x Ϫ6) is to be multiplied by Ϫ3,we will use thedistributive property to remove parentheses and then combine like terms.Note that for this solution,in the first three steps the left side is being simplified.These steps apply only to the left side and only change the form but not the numerical value.The next two steps undo the operations of addition and multi-plication that make up the expression 24x ϩ18.Since adding Ϫ18 and dividing by 24 will change the value of the left side,the right side must be changed in the same way to retain the equality.How to Proceed(1)Write the equation:27x Ϫ3(x Ϫ6) ϭ6(2)Use the distributive property:27x Ϫ3x ϩ18ϭ6(3)Combine like terms:(4)Use the addition property of equality.Add Ϫ18,the additive inverse of ϩ18,to each side:(5)Use the division property ofequality.Divide each side by 24:(6)Simplify each side:x ϭCheck(1)Write the equation:27x Ϫ3(x Ϫ6) ϭ6(2)Replace x by (3)Perform the indicated computation:6 ϭ6 ✔Answer x ϭ212122 562272 1 392 562272 1 1832 5627A 212B 2 3A 2612B 5 627A 212B 2 3A 212 2 6B 5 621221224x24521224124First Degree Equations and Inequalities in One Variable24x ϩ18ϭ6Ϫ18Ϫ1824x ϭϪ12Simplifying Each Side of an Equation125 Representing Two Numbers withthe Same VariableProblems often involve finding two or more different numbers.It is useful toexpress these numbers in terms of the same variable.For example,if you knowthe sum of two numbers,you can express the second in terms of the sum and thefirst number.•If the sum of two numbers is 12 and one of the numbers is 5,then theother number is 12 Ϫ5 or 7.•If the sum of two numbers is 12 and one of the numbers is 9,then theother number is 12 Ϫ9 or 3.•If the sum of two numbers is 12 and one of the numbers is x,then theother number is 12 Ϫx.A problem can often be solved algebraically in more than one way by writ-ing and solving different equations,as shown in the example that follows.Themethods used to obtain the solution are different,but both use the facts statedin the problem and arrive at the same solution.EXAMPLE 3The sum of two numbers is 43.The larger number minus the smaller number is5.Find the numbers.Solution This problem states two facts:The sum of the numbers is 43.The larger number minus the smaller number is 5.In other words,the larger number is 5 more than the smaller.(1)Represent each number in terms of the same variable using Fact 1:the sum of the numbers is 43.Let xϭthe larger number.Then,43 Ϫxϭthe smaller number.(2)Write an equation using Fact 2:The larger number minus the smaller number is 5.___________________________________↓↓↓↓↓xϪ(43 Ϫx)ϭ5(3)Solve the equation.(a) Write the equation:x Ϫ(43 Ϫx ) ϭ5(b) To subtract (43 Ϫx ),add its opposite:x ϩ(Ϫ43 ϩx )ϭ5(c)Combine like terms:(d) Add the opposite of Ϫ43 to each side:(e) Divide each side by 2:x ϭ24(4)Find the numbers.The larger number ϭx ϭ24.The smaller number ϭ43 Ϫx ϭ43 Ϫ24 ϭ19.Check A word problem is checked by comparing the proposed solution with the factsstated in the original wording of the problem.Substituting numbers in theequation is not sufficient since the equation formed may not be correct.The sum of the numbers is 43:24 ϩ19 ϭ43.The larger number minus the smaller number is 5:24 Ϫ19 ϭ5.Reverse the way in which the facts are used.(1)Represent each number in terms of the same variable using Fact 2:the larger number is 5 more than the smaller.Let x ϭthe smaller number.Then,x ϩ5 ϭthe larger number.(2)Write an equation using the first fact.(3)Solve the equation.(a) Write the equation:x ϩ(x ϩ5) ϭ43(b) Combine like terms:(c) Add the opposite of 5 to each side:(d) Divide each side by 2:x ϭ19(4)Find the numbers.The smaller number ϭx ϭ19.The larger number ϭx ϩ5 ϭ19 ϩ5 ϭ24.(5)Check.(See the first solution.)Answer The numbers are 24 and 19.2x 25382The sum of the numbers is 43.______________________↓↓↓x ϩ(x ϩ5)ϭ43Alternate Solution 2x 25482126First Degree Equations and Inequalities in One Variable2x Ϫ43ϭ5ϩ43ϩ432x ϭ482x ϩ5ϭ43Ϫ5Ϫ52x ϭ38Writing About Mathematics1.Two students are each solving a problem that states that the difference between two num-bers is 12.Irene represents one number by x and the other number by x ϩ12.Henry repre-sents one number by x and the other number by x Ϫ12.Explain why both students arecorrect.2.A problem states that the sum of two numbers is 27.The numbers can be represented by x and 27 Ϫx .Is it possible to determine which is the larger number and which is the smaller number? Explain your answer.Developing SkillsIn 3–28,solve and check each equation.3.x ϩ(x Ϫ6) ϭ204.x Ϫ(12 Ϫx ) ϭ385.(15x ϩ7) Ϫ12 ϭ46.(14 Ϫ3c )ϩ7c ϭ947.x ϩ(4x ϩ32) ϭ128.7x Ϫ(4x Ϫ39) ϭ09.5(x ϩ2) ϭ2010.3(y Ϫ9) ϭ3011.8(2c Ϫ1) ϭ5612.6(3c Ϫ1) ϭϪ4213.30 ϭ2(10 Ϫy )14.4(c ϩ1) ϭ3215.25 Ϫ 2(t Ϫ5) ϭ1916.18 ϭϪ6x ϩ4(2x ϩ3)17.55 ϭ4 ϩ3(m ϩ2)18.5(x Ϫ3) Ϫ30 ϭ1019.3(2b ϩ1) Ϫ7 ϭ5020.5(3c Ϫ2) ϩ8 ϭ4321.7r Ϫ(6r Ϫ5) ϭ722.8b Ϫ4(b Ϫ2) ϭ2423.5m Ϫ2(m Ϫ5) ϭ1724.28y Ϫ6(3y Ϫ5) ϭ4025.3(a Ϫ5) Ϫ2(2a ϩ1) ϭ026.0.04(2r ϩ1) Ϫ0.03(2r Ϫ5) ϭ0.2927.0.3a ϩ(0.2a Ϫ0.5) ϩ0.2(a ϩ2) ϭ1.328.Applying SkillsIn 29–33,write and solve an equation for each problem.Follow these steps:a.List two facts in the problem.b.Choose a variable to represent one of the numbers to be determined.e one of the facts to write any other unknown numbers in terms of the chosen variable.e the second fact to write an equation.e.Solve the equation.34(8 1 4x ) 2 13(6x 1 3) 5 9EXERCISESSimplifying Each Side of an Equation 127f.Answer the question.g.Check your answer using the words of the problem.29.Sandi bought 6 yards of material.She wants to cut it into two pieces so that the differencebetween the lengths of the two pieces will be 1.5 yards.What should be the length of each piece?30.The Tigers won eight games more than they lost,and there were no ties.If the Tigers played78 games,how many games did they lose?31.This month Erica saved $20 more than last month.For the two months,she saved a total of$70.How much did she save each month?32.On a bus tour,there are 100 passengers on three buses.Two of the buses each carry fourfewer passengers than the third bus.How many passengers are on each bus?33.For a football game,of the seats in the stadium were filled.There were 31,000 empty seatsat the game.What is the stadium’s seating capacity?A variable represents a number.As you know,any number may be added toboth sides of an equation without changing the solution set.Therefore,the samevariable (or the same multiple of the same variable) may be added to or sub-tracted from both sides of an equation without changing the solution set.For instance,to solve 8x ϭ30 ϩ5x ,write an equivalent equation that hasonly a constant in the right side.To do this,eliminate 5x from the right side byadding its opposite,Ϫ5x ,to each side of the equation.METHOD 1METHOD 2Check8x ϭ30 ϩ5x 8x ϭ30 ϩ5x 8x ϩ(Ϫ5x ) ϭ30 ϩ5x ϩ(Ϫ5x )8(10) 3x ϭ3080 x ϭ1080 ϭ80✔Answer:x ϭ10To solve an equation that has the variable in both sides,transform it into anequivalent equation in which the variable appears in only one side.Then,solvethe equation.5 30 1 505 30 1 5(10)45128First Degree Equations and Inequalities in One Variable8x ϭ30 ϩ5x Ϫ5x Ϫ5x 3x ϭ30x ϭ10EXAMPLE 1Solve and check:7x ϭ63 Ϫ2xSolution How to ProceedCheck (1)Write the equation:7x ϭ63 Ϫ2x (2)Add 2x to each side of7(7) the equation:49 49 ϭ49✔(3)Divide each side of theequation by 9:(4)Simplify each side:x ϭ7Answer x ϭ7To solve an equation that has both a variable and a constant in both sides,first write an equivalent equation with only a variable term on one side.Thensolve the simplified equation.The following example shows how this can bedone.EXAMPLE 2Solve and check:3y ϩ7 ϭ5y Ϫ3Solution METHOD 1METHOD 2Check3y ϩ7 ϭ5y Ϫ33(5) ϩ7 15 ϩ7 22 ϭ22 ✔ϭy ϭ5y ϭ5Answer y ϭ5A graphing calculator can be used to check an equation.The calculator candetermine whether a given statement of equality or inequality is true or false.Ifthe statement is true,the calculator will display 1;if the statement is false,the calculator will display 0.The symbols for equality and inequality are found in the menu.TEST 10252y22102222y225 25 2 35 5(5) 2 39x 956395 63 2 145 63 2 2(7)Solving Equations That Have the Variable in Both Sides 1297x ϭ63 Ϫ2x ϩ2x ϩ 2x 9x ϭ633y ϩ7ϭ5y Ϫ3Ϫ5y Ϫ5y Ϫ2y ϩ7ϭϪ3Ϫ7Ϫ7Ϫ2y ϭϪ103y ϩ7ϭ5y Ϫ3Ϫ3y Ϫ3y 7ϭ2y Ϫ3ϩ3ϩ310ϭ2yTo check that y ϭ5 is the solution to the equation 3y ϩ7 ϭ5y Ϫ3,firststore 5 as the value of y .then enter the equation to be checked.ENTER :5 3 75 3 DISPLAY :EXAMPLE 3The larger of two numbers is 4 times the smaller.If the larger number exceedsthe smaller number by 15,find the numbers.Note:When s represents the smaller number and 4s represents the largernumber,“the larger number exceeds the smaller by 15”has the followinge any one of them.1.The larger equals 15 more than the smaller,written as 4s = 15 ϩs .2.The larger decreased by 15 equals the smaller,written as 4s Ϫ15 ϭs .3.The larger decreased by the smaller is 15,written as 4s Ϫs ϭ15.Solution Let s = the smaller number.Then 4s = the larger number.Check The larger number,20,is 4 times the smaller number,5.The larger number,20,exceeds the smaller number,5,by 15.Answer The larger number is 20;the smaller number is 5.4s ϭ15 ϩs Ϫs Ϫs 3s ϭ15s ϭ54s ϭ4(5) ϭ20The larger is 15 more than the smaller._____________________________↓↓↓↓↓4s ϭ15ϩs4s ϭ15 ϩsENTER ؊ALPHA ENTER TEST 2nd ؉ALPHA ENTERALPHA STO Ł130First Degree Equations and Inequalities in One VariableThe calculator displays 1 which indi-cates that the statement of equality is true for the value that has been stored for y .EXAMPLE 4In his will,Uncle Clarence left $5,000 to his two nieces.Emma’s share is to be$500 more than Clara’s.How much should each niece receive?Solution (1)Use the fact that the sum of the two shares is $5,000 to express each sharein terms of a variable.Let x ϭClara’s share.Then 5,000 Ϫx ϭEmma’s share.(2)Use the fact that Emma’s share is $500 more than Clara’s share to writean equation.(3)Solve the equation to find Clara’s share.2,250 ϭxClara’s share is x ϭ$2,250.(4)Find Emma’s share:5,000 Ϫx ϭ5,000 Ϫ2,250 ϭ$2,750.(1) Use the fact that Emma’s share is $500 more than Clara’s share to expresseach share in terms of a variable.Let x ϭClara’s share.Then x ϩ500 ϭEmma’s share.(2)Use the fact that the sum of the two shares is $5,000 to write an equation.(3)Solve the equation to find Clara’s sharex ϩ(x ϩ500) ϭ5,000Clara’s share is x ϭ$2,250.2x ϩ500ϭ5,000Ϫ500Ϫ5002x ϭ4,500x ϭ2,250Clara’s share plus Emma’s share is $5,000.__________________________↓↓↓↓↓x ϩ(x ϩ 500)ϭ5,000Alternate Solution 5,000Ϫx ϭ500 ϩx ϩx ϩx 5,000ϭ500 ϩ2x Ϫ500Ϫ5004,500ϭ2xEmma’s share is $500 more than Clara’s share.___________________________________↓↓↓↓↓5,000 Ϫx ϭ500ϩxSolving Equations That Have the Variable in Both Sides 131(4)Find Emma’s share:x ϩ500 ϭ2250 ϩ500 ϭ$2,750.Check $2,750 is $500 more than $2,250,and $2,750 ϩ$2,250 ϭ$5,000.Answer Clara’s share is $2,250,and Emma’s share is $2,750.Writing About Mathematicsus said that he finds it easier to work with integers than with fractions.Therefore,in order to solve the equation ,he began by multiplying both sides of the equation by 4.3a Ϫ28 ϭ2a ϩ12Do you agree with Milus that this is a correct way of obtaining the solution? If so,what mathematical principle is Milus using?2.Katie said that Example 3 could be solved by letting equal the smaller number and x equal the larger number.Is Katie correct? If so,what equation would she write to solve the problem?Developing SkillsIn 3–36,solve and check each equation.3.7x ϭ10 ϩ2x4.9x ϭ44 Ϫ2x5.5c ϭ28 ϩc6.y ϭ4y ϩ307.2d ϭ36 ϩ5d 8.9.0.8m ϭ0.2m ϩ2410.8y ϭ90 Ϫ2y 11.2.3x ϩ36 ϭ0.3x 12.13.5a Ϫ40 ϭ3a 14.5c ϭ2c Ϫ8115.x ϭ9x Ϫ7216.0.5m Ϫ30 ϭ1.1m 17.18.7r ϩ10 ϭ3r ϩ5019.4y ϩ20 ϭ5y ϩ920.7x ϩ8 ϭ6x ϩ121.x ϩ4 ϭ9x ϩ422.9x Ϫ3 ϭ2x ϩ4623.y ϩ30 ϭ12y Ϫ1424.c ϩ20 ϭ55 Ϫ4c25.2d ϩ36 ϭϪ3d Ϫ5426.7y Ϫ5 ϭ9y ϩ2927.3m Ϫ(m ϩ1) ϭ6m ϩ128.x Ϫ3(1 Ϫx ) ϭ47 Ϫx 29.3b Ϫ8 ϭ10 ϩ(4 Ϫ8b )30.31.18 Ϫ4n ϭ8 Ϫ2(1 ϩ8n )32.8c ϩ1 ϭ7c Ϫ2(7 ϩc )33.8a Ϫ3(5 ϩ2a ) ϭ85 Ϫ3a34.4(3x Ϫ5) ϭ5x ϩ2( x ϩ15)35.3m Ϫ5m Ϫ12 ϭ7m Ϫ88 Ϫ536.5 Ϫ3(a ϩ6) ϭa Ϫ1 ϩ8a 23t 2 1154(16 2 t ) 2 13t 414c 5934c 1 44234x 1 2453x214y 5114y 2 8x 44A 34a 2 7B 54A 12a 1 3B 34a 2 7512a 1 3EXERCISES132First Degree Equations and Inequalities in One VariableSolving Equations That Have the Variable in Both Sides133In 37–42,a.write an equation to represent each problem,and b.solve the equation to find each number.37.Eight times a number equals 35 more than the number.Find the number.38.Six times a number equals 3 times the number,increased by 24.Find the number.39.If 3 times a number is increased by 22,the result is 14 less than 7 times the number.Find thenumber.40.The greater of two numbers is 1 more than twice the smaller.Three times the greaterexceeds 5 times the smaller by 10.Find the numbers.41.The second of three numbers is 6 more than the first.The third number is twice the first.The sum of the three numbers is 26.Find the three numbers.42.The second of three numbers is 1 less than the first.The third number is 5 less than the sec-ond.If the first number is twice as large as the third,find the three numbers.Applying SkillsIn 43–50,use an algebraic solution to solve each problem.43.It took the Gibbons family 2 days to travel 925 miles to their vacation home.They traveled75 miles more on the first day than on the second.How many miles did they travel eachday?44.During the first 6 month of last year,the interest on an investment was $130 less than dur-ing the second 6 months.The total interest for the year was $1,450.What was the interest for each 6-month period?45.Gemma has 7 more five-dollar bills than ten-dollar bills.The value of the five-dollar billsequals the value of the ten-dollar bills.How many five-dollar bills and ten-dollar bills does she have?46.Leonard wants to save $100 in the next 2 months.He knows that in the second month he willbe able to save $20 more than during the first month.How much should he save each month?47.The ABC Company charges $75 a day plus $0.05 a mile to rent a car.How many miles didMrs.Kiley drive if she paid $92.40 to rent a car for one day?48.Kesha drove from Buffalo to Syracuse at an average rate of 48 miles per hour.On thereturn trip along the same road she was able to travel at an average rate of 60 miles perhour.The trip from Buffalo to Syracuse took one-half hour longer than the return trip.How long did the return trip take?49.Carrie and Crystal live at equal distances from school.Carries walks to school at an averagerate of 3 miles per hour and Crystal rides her bicycle at an average rate of 9 miles per hour.It takes Carrie 20 minutes longer than Crystal to get to school.How far from school doCrystal and Carrie live?50.Emmanuel and Anthony contributed equal amounts to the purchase of a gift for a friend.Emmanuel contributed his share in five-dollar bills and Anthony gave his share in one-dollar bills.Anthony needed 12 more bills than Emmanuel.How much did each contribute toward the gift?。

Lascu_Chapter4

Lascu_Chapter4

Copyright Atomic Dog Publishing, 2 2008
Types of Research

Exploratory Research:
- Research conducted early in the research process that helps further define a problem or identify additional problems that need to be investigated.
• STEP 2 Develop the international research plan
International Promotion Research
• Allows the firm to evaluate the extent to which the promotional mix effectively targets the market. Examples:





Studies of premiums, coupons, and deals: - Identify practices in each country and determine the legality of using promotions and consumers’ response. Advertising effectiveness research: - Examine the effectiveness of ads aimed at individual markets. Media research: - Determines media availability and which communication channels are the most effective. Sales force compensation, quota, and territory studies: - Help determine the appropriate strategies for different markets.

4 Fourier Series

4 Fourier Series

An alternative form for the FOURIER SERIES
x(t ) = c0 cn cosnw1t n
n =1

c0 = a0
an = cn cosn
cn = a b
2 n
2 n
bn n = arctan a n
1 jw0t 1 jw0t 1 j (p 4) j 2w0t 1 j (p 4) j 2w0t x(t ) = 1 1 e 1 e e e e e 2 2 2j 2j
Thus, the Fourier series coefficients for this example are :
0 = 0
1 = 0.15 π
1 = 0.15 π
2 = 0.25 π
2 = 0.25 π
Spectrum in terms of Complex Exponential
F nw 1
n
0.5
2w 1 w
0.5
1.12
1
1.12
0.15 π
2w 1
w1
0.25 π
2w π
2w 1 w
0.25 π
Comparison
Spectrum in terms of Trigonometric Functions
cn c 1 c0
1
2.24
n
0.25 π
c2
1
O
w1
2w 1 w
w1
O
2w 1 w
Spectrum in terms of Complex Exponential

Principles of Mathematical Analysis(数学分析原理),THIRD

Principles of Mathematical Analysis(数学分析原理),THIRD

Principles of Mathematical Analysis(数学分析原理),THIRD EDITION,WALTER RUDIN著这是一部现代数学名著,一直受到数学界的推崇。

作为Rudin的分析学经典著作之一,本书在西方各国乃至我国均有着广泛而深远的影响,被许多高校用做数学分析课的必选教材。

本书涵盖了高等微积分学的丰富内容,最精彩的部分集中在基础拓扑结构、函数项序列与级数、多变量函数以及微分形式的积分等章节。

第3版经过增删与修订,更加符合学生的阅读习惯与思考方式。

本书内容相当精练,结构简单明了,这也是Rudin著作的一大特色。

与其说这是一部教科书,不如说这是一部字典。

英文版的PREFACEThis book is intended to serve as a text for the course in analysis that is usually taken by adva nced undergraduates or by first-year students who study mathe-matics.The present edition covers essentially the same topics as the second one, with some additions, a few minor omissions, and considerable rearrangement. I hope that these changes will make the material more accessible amd more attractive to the students who take such a course.Experience has convinced me that it is pedagogically unsound (though logically correct) to start o ff with the construction of the real numbers from the rational ones. At the beginning, most studen ts simply fail to appreciate the need for doing this. Accordingly, the real number system is introd uced as an ordered field with the least-upper-bound property, and a few interesting applications o f this property are quickly made. However, Dedekind's construction is not omitted. It is now in a n Appendix to Chapter 1, where it may be studied and enjoyed whenever the time seems ripe.The material on functions of several variables is almost completely rewritten, with many details fill ed in, and with more examples and more motivation. The proof of the inverse function theorem--the key item in Chapter 9--is X PREFACEsimplified by means of the fixed point theorem about contraction mappings. Differential forms are discussed in much greater detail. Several applications of Stokes' theorem are included. As regard s other changes, the chapter on the Riemann-Stieltjes integral has been trimmed a bit, a short d o-it-yourself section on the gamma function has been added to Chapter 8, and there is a large n umber of new exercises, most of them with fairly detailed hints.I have also included several references to articles appearing in the American Mathematical Month ly and in Mathematics Magazine, in the hope that students will develop the habit of looking into t he journal literature. Most of these references were kindly supplied by R. B. Burckel.Over the years, many people, students as well as teachers, have sent me corrections, criticisms, and other comments concerning the previous editions of this book. I have appreciated these, an d I take this opportunity to express my sincere thanks to all who have written me.WALTER RUDIN。

数学分析英文版 chapter4

数学分析英文版 chapter4

CHAPTER4LIMITS AND CONTINUITY4.1 INTRODUCTIONThe reader is already familiar with the limit concept as introduced in elementarycalculus where, in fact, several kinds of limits are usually presented. For example, thelimit of sequence of real numbers {x n }, denoted symbolically by writingn n xlim ∞→=A ,means that for every number ε >0 there is an integer N such thatA x n -<ε wheneve r n ≥N.This limit process conveys the intuitive idea that x n can be made arbitrarily close toA provided that n is sufficiently large. There is also the limit of a function, indicatedby notation such as)(lim x f px →=A ,which means that for every ε >0 there is another number δ>0 such thatA x f -)(<ε whenever 0<p x -<δ.This conveys the idea that f(x) can be made arbitrarily close to A by taking xsufficiently close to p.Applications of calculus to geometrical and physical problems in 3-space and tofunctions of several variables make it necessary to extend these concepts to R n . It isjust as easy to go one step further and introduce limits in the more general setting ofmetric spaces. This achieves a simplification in the theory by stripping it ofunnecessary restrictions and at the same time covers nearly all the important aspectsneeded in analysis.First we discuss limits of sequences of points in a metric space, then we discusslimits of functions and the concept of continuity.4.2 CONVERGENT SEQUENCES IN A METRIC SPACEDefinition4.1. A sequence {x n } of points in a metric space(S,d) is said to converge ifthere is a point p in S with the following property:For every ε >0 there is an integer N such thatd(x n ,p) <ε whenever n ≥NWe also say that {x n } converges to p and we write x n →p as n →∞, or simply x n→p. If there is no such p in S, the sequence {x n} is said to diverge. NOTE. The definition of convergence implies thatxn →p if and only if d(xn,p)→0.The convergence of the sequence {d(xn,p)} to 0 takes place in the Euclidean metric space R1.Examples1.In Euclidean space R1,a sequence {xn } is called increasing if xn≤x1+nfor all n.If an increasing sequence is bounded above (that is, if xn≤M for some M>0 and alln),then {xn } converges to the supremum of its range, sup{x1,x2,...}. Similarly,{xn } is called decreasing if x1+n≤xnfor all n. Every decreasing sequence whichis bounded below converges to the infimum of its range, For example, {1/n} converges to 0.2.If {an } and {bn} are real sequences converging to 0, then { an+ bn} alsoconverges to 0. If 0≤cn ≤anfor all n and if { an} converges to 0,then { cn} alsoconverges to 0. these elementary properties of sequences in R1can be used to simplify some of the proofs concerning limits in a general metric space.3.In the complex plane C, let zn =1+n2-+(2-1/n)i. Then { zn} converges to 1 + 2ibecaused(zn ,1+2i)2=2)21(izn+-=01124→+nnas n→∞,so d(zn,1+2i)→0.Theorem4.2. A sequence {xn} in a metric space (S,d) can converge to at most one point in S.Proof. Assume that xn →p and xn→q.we will prove that p=q. By the triangleinequality we have0≤d(p,q)≤d(p, xn )+d(xn,q).Since d(p, xn )→0 and d(xn,q) →0 this implies that d(p,q)=0,so p=q.If a sequence {xn} converges, the unique point to which it converges is called thelimit of the sequence and is denoted by lim xn or by lim∞→nxn.Example. In Euclidean space R1we have lim∞→n1/n=0. The same sequence in the metric subspace T=(0,1] dose not converge because the only candidate for the limit is 0 and 0∉T. This example shows that the convergence or divergence of a sequence depends on the underlying space as well as on the metric.Theorem 4.3. In a metric space (S, d), assume xn →p and let T={x1,x2,...} be therange of {xn}. Then:a)T is bounded.b)p is an adherent point of T.Proof. a) Let N be the integer corresponding to ε=1 in the definition ofconvergence. Then every xnwith n ≥N lies in the ball B (p; 1 ), so every point in T lies in the ball B(p;1 ), wherer=1+max {d(p, x1),…,d(p, x1-N)}.Therefore T is bounded.b) Since every ball B(p; ε) contains a point of T, p is an adherent point of T. NOTE. If T is infinite, every ball B (p;ε) contains infinitely many points of T, so p is an accumulation point of T.The next theorem provides a converse to part (b).Theorem 4.4.Given a metric space (S, d) and a subset T ⊆S. If a point p in S is an adherent point of T, then there is a sequence {xn} of point in T which converges to p.Proof. For every integer n≥ 1 there is a point xn in T with d(p, xn) ≤1/n. Henced(p, xn )→0, so xn→p.Theorem 4.5.In a metric space (S , d ) a sequence {xn} converges to p if, and only if,every subsequence {xk(n)} converges to p.Proof. Assume xn →p and consider any subsequence{xk(n)}. For every ε>0 there isan N such that n ≥N implies d(xn ,p) < ε. Since {xk(n)} is a subsequence , thereis an integer M such that k(n)≥N for n ≥M.H ence n ≥M implies d(x)(nk,p)< ε,which proves that x)(nk →p. The converse statement holds trivially since {xn} is itselfa subsequence.4.3 CAUCHY SEQUENCESIf a sequence{xn} converges to a limit p, its terms must ultimately become close to p and hence close to each other. This property is stated more formally in the next theorem.Theorem 4.6.Assume that {xn} converges in a metric space (S, d). Then for every ε> 0 there is an integer N such thatd(xn ,xm) <εwhenever n ≥N and m ≥N.Proof. Let p=lim xn . Given ε>0, let N be such that d(xn,p)< ε/2 whenever n ≥N.Then d(xm,p) < ε/2 if m≥N. If both n ≥N and m≥N the triangle inequality gives usd(xn ,xm) ≤d(xn,p)+ d(p, xm) ≤2ε+2ε=ε.4.7 Definition of a Cauchy Sequence. A sequence {xn} in a metric space (S, d) is called a Cauchy sequence if it satisfies the following condition (called the Cauchy condition):For every ε>0 there is an integer N such thatd(xn ,xm)< εwhenever n ≥N and m ≥N.Theorem 4.6 states that every convergent sequence is a Cauchy sequence. The converse is not true in a general metric space. For example, the sequence {1/n}is a Cauchy Sequence in the Euclidean subspace T=(0,1] of R1,but sequence dose not converge in T. However, the converse of Theorem 4.6 is true in every Euclidean space R k.Theorem 4.8. In Euclidean space R k every Cauchy sequence is convergent.Proof. Let{xn } be a Cauchy sequence in R k and let T={x1,x2,...} be the range ofthe sequence. If T is finite, then all except a finite number of the terms {xn} are equaland hence {xn} converges to this common value.Now suppose T is infinite. We use the Bolzano-Weierstrass theorem to show thatT has an accumulation point p , and then we show that {x n} converges to p. First we need to know that T is bounded. This follows from the Cauchy condition. Infact, when ε =1 there is an N such that n ≥ N implies N n x x - < 1. This meansthat all points x n with n ≥ N lie inside a ball of radius 1 about x N as center, so Tlies inside a ball of radius 1+M about 0, where M is the largest of the numbers1x ,…, N x . Therefore, since T is a bounded infinite set it has an accumulationpoint p in R k (by the Bolzano-Weierstrass theorem).We show next that {x n }converges to p .Given ε>0 there is an N such that m n x x -<ε/2 whenever n ≥ N and m ≥ N .The ball B (p; ε/2) contains a point x m with m ≥ N . Hence if n ≥ N we have≤-p x n m n x x - + p x m -<2ε+2ε=ε,so lim x n =p . This completes the proof.Examples1. Theorem 4.8 is often used for proving the convergence of a sequence when thelimit is not known in advance. For example, consider the sequence in R 1defined byx n =1-nn 1)1(...413121--++-+. If m> n ≥ N, we find(by taking successive terms in pairs) thatn m x x -=m n n 1...2111±++-+<Nn 11≤, so n m x x -<ε as soon as N >1/ε. Therefore {x n } is a Cauchy sequence andhence it converges to some limit. It can be shown(see Exercise8.18) that this limit islog2, a fact which is not immediately obvious.2.Given a real sequence {a n } such that 12++-n n a a ≤21n n a a -+1 for all n ≥ 1. We can prove that {a n } converges without knowing its limit. Let b n =n n a a -+1.Then 0≤b 1+n ≤ b n /2 so, by induction, b 1+n ≤b 1/2n . Hence b n →0. Also,if m>n we have a m -a n =∑-=+-11)(m n k k k a a ; hence n m a a -≤∑-=1m n k k b ≤ b n (1+n m --++121...21)<2 b n . This implies that {a n } is a Cauchy sequence, so {a n } converges.4.4 COMPLETE METRIC SPACESDefinition 4.9. A metric space (S, d) is called complete if every Cauchy sequence inS converges in S. A subset T of S is called complete if the metric subspace (T, d) iscomplete.Example 1. Every Euclidean space R k is complete (Theorem 4.8). In particular, R 1is complete, but the subspace T= (0, 1] is not complete.Example 2. The space R n with the metric d(x , y ) = max n i ≤≤1i i y x - is complete.The next theorem relates completeness with compactness.Theorem 4.10. In any metric space (S, d) every compact subset T is complete.Proof. Let {x n} be a Cauchy sequence in T and let A ={x 1, x 2, ...} denote the range of {x n }. If A is finite, then {x n } converges to one of the elements of A , hence {x n} converges in T .If A is infinite, Theorem3.38 tells us that A has an accumulation point p in T sinceT is compact. We show next that x n →p. Given ε>0, choose N so that n ≥ N andm ≥ N implies d(x n ,x m ) <ε/2. The ball B (p ; ε/2) contains a point x m with m ≥N. Therefore if n ≥ N the triangle inequality gives us),(),(),(p x d x x d p x d m m n n +≤< 2ε+2ε=ε, so x n→p. Therefore every Cauchy sequence in T has a limit in T , so T is complete.4.5 LIMIT OF A FUNCTIONIn this section we consider two metric spaces (S,s d )and (T, T d ), where s d andT d denote the respective metrics. Let A be a subset of S and let T A f →: be afunction from A to T.Definition 4.11 If p is an accumulation point of A and if T b ∈, the notationb x f px =→)(lim , (1)is defined to mean the following:For every ε>0 there is a δ>0 such thatε<)),((b x f d T whenever and p x A x ,,≠∈ δ<),(p x d S .The symbol in (1) is read “the limit of f(x), as x tends to p, is b,” or “f(x)approaches b as x approaches p.” We sometimes indicate this by writing b x f →)(as p x →.The definition conveys the intuitive idea that f(x) can be made arbitrarily close tob by taking x sufficiently close to p. (See Fig.4.1.) We require that p be anaccumulation point of A to make certain that there will be points x in A sufficientlyclose to p, with p x ≠. However, p need not be in the domain of f, and b need not bein the range of f.NOTE. The definition can also be formulated in terms of balls. Thus, (1) holds if,and only if, for every ball )(b B T , there is a ball )(p B S such that )(p B S ∩A is notempty and such that)(x f ∈)(b B T whenever x ∈)(p B S ∩A, p x ≠.When formulated this way, the definition is meaningful when p or b (or both) are in the extended real number system R *or in the extended complex number system C *. However, in what follows, it is to be understood that p and b are finite unless it is explicitly stated that they can be infinite.The next theorem relates limits of functions to limits of convergent sequences. Theorem 4.12. Assume p is an accumulation point of A and assume b ∈T. Then b x f px =→)(lim , (2)if, and only if,,)(lim b xf n n =∞→ (3)for every sequence }{n x of points in A – {p} which converges to p.Proof. If (2) holds, then for every ε> 0 there is a δ such thatε<)),((b x f d T whenever Aand x ∈ 0<δ<),(p x d S . (4) Now take any sequence }{n x in A – {p} which converges to p. For the δ in (4), there is an integer N such that n ≥ N implies ),(p x d n S < δ. Therefore (4) implies ε<)),((b x f d n T for n ≥ N . and hence )}({n x f converges to b. Therefore (2) implies(3).To prove the converse we assume that (3) holds and that (2) is false and arrive at a contradiction. If (2) is false, then for someε>0 and every δ>0 there is a point xin A (where x may depend on δ) such that0<δ<),(p x d S but ε≥)),((b x f d T . (5)Taking δ=1/n, n=1,2,…, this means there is a corresponding sequence of points }{n x in A-{p} such that 0<),(p x d n S <1/n but ε≥)),((b x f d n T .Clearly, this sequence }{n x converges to p but the sequence )}({n x f does not converge to b, contradicting (3).NOTE Theorems 4.12 and 4.2 together show that a function cannot have two different limits as p x → .。

数值分析Chapter 4

数值分析Chapter 4
Finally,
n k 1
2
Av0 v1 u1 max(v1 ) max( Av0 ) A v0 v2 u2 max(v2 ) max( A2 v0 ) vk Ak v0 uk max(vk ) max( Ak v0 )
2
i k [a1 x1 ai ( ) xi ] 1 x1 i2 lim uk lim n k k i k max( x1 ) k max{1 [a1 x1 ai ( ) xi ]} 1 i2
1
1
10
2013-12-9
1.3 Inverse Iteration
We can use the inverse iteration to solve the eigenvalue with smallest absolute value and its eigenvector. Let
| 1 || 2 | | n 1 || n |
2013-12-9 5
Note 1. The speed of convergence is determined by the 2 value | | . When the value is small, the speed is fast.
Note 2.Even the dominant eigenvalue is multiple root, the result still holds.
QT AQ diag (1 , 2 ,, n )
2013-12-9
13
2.1 Diagonalization of a real symmetric matrix
The idea of Jacobi’s method is finding s series of orthogonal matrices {S k } , such that

数学分析中的英文单词和短语

数学分析中的英文单词和短语

数学分析中的英文单词和短语第一章实数集与函数第二章数列极限Chapter 2 Limits of Sequences第三章函数极限Chapter 3 Limits of Functions第四章函数的连续性Chapter 4 Continuity of Functions第六章 微分中值定理及其应用Chapter 6 Mean Value Theorems of Differentials and their Applications第七章 实数的完备性Chapter 7 Completeness of Real Numbers第八章 不定积分Chapter 8 Indefinite Integrals第九章 定积分Chapter 9 Definite Integrals第十章定积分的应用Chapter 10 Applications of Definite Integrals第十一章反常积分Chapter 11 Improper Integrals 第十二章数项级数Chapter 12 Series of Number Terms第十三章函数列与函数项级数Chapter 13 Sequences of Functions andSeries of Functions第十四章 幂级数Chapter 14 Power Series第十五章 傅里叶级数Chapter 15 Fourier Series第十六章 多元函数的极限与连续Chapter 16 Limits and Continuity of Functions of Several Variavles第十七章多元函数微分学Chapter 17 Differential Calculus of Functions of Several Variables第十八章隐函数定理及其应用Chapter 18 Implicit Funciton Theorems and their Applications第十九章含参量积分Chapter 19 Integrals with Parameters第二十章重积分Chapter 20 Multiple Integrals第二十一章曲线积分Chapter 21 Curvilinear Integrals 第二十二章曲面积分Chapter 22 Surface Integrals。

[理学]概率论与数理统计英文 第四章-精品文档

[理学]概率论与数理统计英文 第四章-精品文档
(a) Find the probability that a bulb selected at random from the product of company X has life longer than 450 hrs.
(b) Select 5 bulbs randomly from the product of company X, what is the probability that at least 3 of them has life longer than 450 hrs.
, (4.1.4)
Where is the mean ofX, is referred to as thestandard deviation.
We easily get
.(4.1.5)
Example4.1.5
Determining the mean and variance using the probability density function
Note Equation (4.5.1) really gives a density function, since
Theorem4.5.1The mean and variance of a continuous random variableXhaving exponential distribution with parameter is given by
.
ProofSince the probability density function ofXis (4.5.1), we have
Example4.5.1.
Assume that the lifeYof bulbs produced by company X has exponential distribution with mean .
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lim xn
lim 或者 n xn 来表示.
3.柯西数列
如果一个数列
x 收敛于一个极限 p, 这个数列的项最终将会越来越接近 p, 并且
n
之后的项与项之间会十分接近, 以至于充分后面的任何两项之差的绝对值可小于 给定的任意小正数.这一特性在下面的定理中很好的反映出来了. 定理 2.设
n
x 在度量空间 s, d 中收敛,则对任给的 0, 存在正整数 N , 使得当
3 CAUCHY SEQUENCES
If a sequence {x } converges to a limit p, its terms must ultimately become close to p n and hence close to each other. This property is stated more formally in the next theorem. Theorem 2. Assume that {x } converges in a metric space (S, d). Then for every n whenever n N and m N.
s, d 中的数列 xn为柯西数列,如果它满足
n N, m N
m
0,
存在正整数
N,
使得当
n
时有
x x
.
定理 2.表明任意的收敛数列都是柯西数列,但反之,并不是在所有的度量空间中
1 1 在欧几里得度量空间 T 0,1 R 中是柯 的柯西数列都是收敛的.例如, 数列 n
a Cauchy sequence if it satisfies the following condition (called the Cauchy condition): For every >0 there is an integer N such that d(x n ,x m )<

n n n n

0 cn an ,
则数列
c 也收敛于 0.
n
这些欧几里得度量空间里数列的基本特性可以用来简化一些一般空间里的有关 于极限的证明. 定理 1.若数列
p 证明:设 是
x 在度量空间 S , d 中收敛,则它只有一个极限.
n n n
x 的一个极限, q 也是 x 的一个极限.我们证明: p q.
We also say that{x n } converges to p and we write x n →p as n →∞, or simply
3
→p. If there is no such p in S, the sequence {x } is said to diverge.
n
NOTE. The
n
这里需要注意:收敛的规定要求 xn p 当且仅当 d
1
x , p 0.
n
另外还有关于收敛的一些性质需要了解: (1).若实数列
n n
a ,b 均收敛于 0,则数列 a b 也收敛于 0. n (2).设数列 a 收敛于 0,若存在数列 c ,对于所有的正整数 都
lim x
n
n
=A,
means that for every number >0 there is an integer N such that
xn A <
whenever n ≥N.
This limit process conveys the intuitive idea that x n can be made arbitrarily close to A provided that n is sufficiently large. Then we discuss limits of sequences of points in a metric space and the Cauchy Sequences. 2 CONVERGENT SEQUENCES IN A METRIC SPACE Definition1. A sequence {x n } of points in a metric space(S,d) is said to converge if there is a point p in S with the following property: For every >0 there is an integer N such that d(x n ,p) < whenever n ≥N xn
0 d p, q d p, xn d xn , q ,
由三角不等式我们可得
又因为
p, q
均为
x 的极限,由 d p, x 0, d x , q 0, 有
n n n
d p, q 0,
如果一个数列
n
即证
p q.
x 收敛,我们就把它收敛到的这个数叫做这个数列的极限,用
时有
n N, m N
x x
n
m
.
lim p. 0, N 0, n N , m N 证明:设 n an 由数列极限定义,对任给的 存在 当
时有
x
因而 x x
n
p m

2
,
x
p n

2
,
m
x
m
p
x
n
p

2


2
.
2
柯西数列的定义:我们称度量空间 以下条件(也称柯西条件) : 对任给的
西数列,但是这个数列在这个空间中并不收敛.然而,所有的柯西数列在欧几里 得空间 R
k
中都是收敛的.
CHAPTER4 LIMITS OF SEQUENCE 1 INTRODUCTION
The reader is already familiar with the limit concept as introduced in elementary calculus where, in fact, several kinds of limits are usually presented. For example, the limit of sequence of real numbers {x n }, denoted symbolically by writing
4
inequality gives us d(x n ,x m ) d(x n ,p)+ d(p, x m ) Definition of a Cauchy Sequence.
+ = . 2 2
A sequence {x } in a metric space (S, d) is called n
5
姓名:郭庆
学号:20130512013
数列的收敛
1.简介:
关于极限的概念大家应该都不陌生,事实上,在初等微积分中就有很多极限经常 出现 . 例如,实数列
x 的极限用符号表示为:
n
lim xn A,
n
也就是说,设
x 为数列, A 为定数 . 若对任给的正数 ,总存在正整数 N ,使
definition of convergence implies that x n →p if and only if d(x n ,p) →0.
Note:: 1.If {a n } and {b n } are real sequences converging to 0, then { a n + b n } also converges to 0. 2.If 0≤c n ≤a n for all n and if { a n } converges to 0,then { c n } also converges to 0. these elementary properties of sequences in R 1 can be used to simplify some of the proofs concerning limits in a general metric space. Theorem1. point in S. Proof. Assume that x n → p and x n →q. we will prove that p=q. By the triangle inequality we have 0≤d(p,q) ≤d(p, x n )+d(x n ,q). Since d(p, x n )→0 and d(x n ,q) →0 this implies that d(p,q)=0,so p=q. If a sequence {x n } converges, the unique point to which it converges is called the limit of the sequence and is denoted by lim x n or by lim n x n . A sequence {x n } in a metric space (S,d) can converge to at most one
n
得当 n N 时有
x
则称数列
n
n
A ,
n
x 收敛于 A , 定数 A 称为数列 x 的极限 . x ,当 n 无限增大时,x
n n
这个极限过程表达了一个很直观的观点:对于数列

பைடு நூலகம்
无限地接近某一个常数 A ,这样的数列叫做收敛数列 .
2.度量空间里的收敛数列:
定义 1 设
x 为度量空间 S , d 中的一个点列, p 为空间 S 中的一个点 .
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