Mathematical Analysis - Hong Kong Baptist Universityzeng/Teaching/2017math1006/wu.pdf · Mathematical Analysis Hongci Huang, Xiaonan Wu and Tieyong Zeng Department of Mathematics

Mathematical Analysis

Hongci Huang, Xiaonan Wu and Tieyong Zeng

Department of Mathematics

Hong Kong Baptist University

January 13, 2017

2

Contents

1 Sets and Numbers 9

1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 9

1.2 Real Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 10

1.3 Inequalities and Algebraic Identities . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 17

1.4 Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 20

1.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 21

2 Limit Theorey 25

2.1 Convergence and Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 25

2.1.1 Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 25

2.1.2 Convergence and Limit . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 26

2.1.3 Uniqueness of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 30

2.1.4 Boundedness of Convergent Sequence . . . . . . . . . . . . . . .. . . . . . . . . 30

2.1.5 Operations on Convergent Sequences . . . . . . . . . . . . . . .. . . . . . . . . 32

2.1.6 Squeezing Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 34

2.2 Some Important Theorems in Limit Theory . . . . . . . . . . . . . .. . . . . . . . . . . 36

2.2.1 Dedekind Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 36

2.2.2 Least Upper Bound Theorem . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 37

2.2.3 Monotone Convergence Theorem . . . . . . . . . . . . . . . . . . . .. . . . . . 39

2.2.4 Nested Interval Theorem . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 42

2.2.5 Heine-Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 43

2.2.6 Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . . . . . . .. . . . . . . . 44

2.2.7 Cauchy Convergence Criterion . . . . . . . . . . . . . . . . . . . .. . . . . . . . 45

2.3 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 49

3

4

3 Continuity 55

3.1 Limit of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 55

3.1.1 Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.1.2 Operations and properties . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 59

3.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 61

3.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 61

3.2.2 Operations and composition on continuous functions .. . . . . . . . . . . . . . . 61

3.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 62

3.3 Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 64

3.3.1 Maximizer and minimizer . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 64

3.3.2 Extreme value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 64

3.3.3 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 66

3.4 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 66

3.5 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 67

3.6 Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 69

3.7 Continuity of Elementary Functions . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 70

3.7.1 Power Function with Rational Exponential . . . . . . . . . .. . . . . . . . . . . 70

3.7.2 Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 70

3.7.3 Logarithmic Function . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 72

3.7.4 Power Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 73

3.8 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 73

4 Differentiation 83

4.1 Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 83

4.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 83

4.1.2 Physical and geometric interpretation . . . . . . . . . . . .. . . . . . . . . . . . 85

4.1.3 Algebra of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 85

4.2 Differentiating Inverses and Compositions . . . . . . . . . .. . . . . . . . . . . . . . . . 86

4.2.1 Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 86

4.2.2 Composition and Chain Rule . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 87

4.3 Derivatives of Elementary Function . . . . . . . . . . . . . . . . .. . . . . . . . . . . . 87

4.4 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 88

4.5 High (Order) Derivatives and Differential . . . . . . . . . . .. . . . . . . . . . . . . . . 92

4.6 Applications of Derivative . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 94

4.6.1 L’Hopital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 94

5

4.6.2 Extreme values of functions . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 96

4.6.3 Concavity and graph of functions . . . . . . . . . . . . . . . . . .. . . . . . . . 98

4.7 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 100

5 Indefinite Integral 107

5.1 Definition and Some Basic Formulas . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 107

5.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 107

5.2 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 108

5.3 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 110

5.4 Integration of Rational Functions . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 112

5.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 114

6 Integration 117

6.1 Darboux Sums and Definition of Integral . . . . . . . . . . . . . . .. . . . . . . . . . . . 117

6.1.1 Darboux Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 117

6.1.2 Refinement Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 117

6.1.3 Definition of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 119

6.1.4 Interpretation in Physics and Geometry . . . . . . . . . . . .. . . . . . . . . . . 120

6.2 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 120

6.2.1 Integrability Criterion . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 120

6.2.2 Two Kinds of Integrable Functions . . . . . . . . . . . . . . . . .. . . . . . . . . 122

6.2.3 Convergence of Darboux Sums . . . . . . . . . . . . . . . . . . . . . .. . . . . 123

6.2.4 Convergence of Riemann Sums . . . . . . . . . . . . . . . . . . . . . .. . . . . 126

6.3 Linearity, Monotonicity and Additivity . . . . . . . . . . . . .. . . . . . . . . . . . . . . 128

6.3.1 Linearity of Integral . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 128

6.3.2 Monotonicity of Integral . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 128

6.3.3 Additivity over Intervals . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 130

6.4 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 131

6.4.1 First Fundamental Theorem of Calculus . . . . . . . . . . . . .. . . . . . . . . . 131

6.4.2 Mean Value Theorem for Integral . . . . . . . . . . . . . . . . . . .. . . . . . . 133

6.4.3 Second Fundamental Theorem of Calculus . . . . . . . . . . . .. . . . . . . . . 134

6.4.4 Calculation of Integral . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 136

6.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 139

7 Series of Numbers 145

7.1 Definition and Basic Properties . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . 145

6

7.2 Series with Nonnegative Terms . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 146

7.3 Convergence of General Series . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 151

7.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 155

8 Series of Functions 157

8.1 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 157

8.1.1 Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . 157

8.1.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 158

8.1.3 Test for Uniform Convergence . . . . . . . . . . . . . . . . . . . . .. . . . . . . 159

8.2 Uniform Limit of Functions . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 163

8.2.1 Uniform Limit of Continuous Functions . . . . . . . . . . . . .. . . . . . . . . . 164

8.2.2 Uniform Limit of Integrable Functions . . . . . . . . . . . . .. . . . . . . . . . . 164

8.2.3 Uniform Limit of Differentiable Functions . . . . . . . . .. . . . . . . . . . . . 166

8.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 167

8.3.1 Radius of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 167

8.3.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 170

8.4 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 173

8.4.1 Taylor Polynomials and Order of Contact . . . . . . . . . . . .. . . . . . . . . . 173

8.4.2 Convergence of Taylor Series . . . . . . . . . . . . . . . . . . . . .. . . . . . . 176

8.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 180

9 Fourier Series 185

9.1 Fourier Series and Fourier Coefficients . . . . . . . . . . . . . .. . . . . . . . . . . . . . 185

9.2 Convergence of Fourier Series . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 187

9.3 Integration and Differentiation of Fourier Series . . . .. . . . . . . . . . . . . . . . . . . 194

9.4 Fourier Expansion on Finite Interval . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 197

9.5 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 198

10 Improper Integral 201

10.1 Integration on an Unbounded Interval . . . . . . . . . . . . . . .. . . . . . . . . . . . . 201

10.1.1 Nonnegative Integrand . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 202

10.1.2 Absolute Convergence Test . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 204

10.1.3 Improper Integral on(−∞, b] and(−∞,+∞) . . . . . . . . . . . . . . . . . . . 204

10.2 Integral with an Unbounded Integrand . . . . . . . . . . . . . . .. . . . . . . . . . . . . 205

10.2.1 Nonnegative Integrand . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 206

10.2.2 Absolute Convergence Test . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 207

Chapter 1: Sets and Numbers 7

10.2.3 Other Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 207

10.3 Gamma Function and Beta Function . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 208

10.3.1 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 208

10.3.2 Beta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 208

10.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 209

8 Mathematical Analysis

Chapter 1

Sets and Numbers

1.1 Sets

Set theory is the foundation in which virtually all mathematics is constructed. Set theory is quite easy

to learn and use.

Set. A collection of well-defined objects is a set. ”object” is also called ”element” or ”member”. ”Well-

defined” means that an object is distinguishable by some testwhether it is in the set or not. The sets

discussed in our course are mainly real number sets.

Notation: Capital letters A,B,C are usually used to denote Sets, and small letters a, b, c, x, y, z are used

to denote elements.

• x ∈ A means element x in A.

• x 6∈ A (or x∈A) means x not in A.

• A = {x | conditions on x}.

• The set of natural numbersN = {1, 2, 3, . . .}.

• The set of integers isZ = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .}.

• The set of rational numbers:Q = {mn |m ∈ Z, n ∈ N}.

Examples.

1. A = {x | x2 + 3x+ 1 = 0}⇔ A =

{

−3+√5

2 , −3−√5

2

}

.

2. A = {n ∈ N | n can be divided by 3} ⇔ A = {3, 6, 9, · · · } orA = {n | n = 3k, k ∈ N}, whereN

is the set of natural numbers.

9


Empty set. A set that has no member, denoted by∅.

A = {x ∈ N | x2 + 3x+ 1 = 0} is an empty set.

Subset.A is called a subset ofB if each element inA must be inB, denoted byA ⊆ B orB ⊇ A.

A = B meansA ⊆ B andB ⊇ A.

Example. Let

A = {m, a, t, h, e,m, a, t, i, c, s}, B = {m, a, t, h, e, i, c, s}.

If x ∈ B, thenx ∈ A. Therefore,B ⊆ A. Also, if x ∈ A, thenx ∈ B. Therefore,A ⊆ B, i.e.,A = B.

Union. Union of setsA andB, denoted byA ∪B is the set of all elements which belong toA or toB, i.e.,

A ∪B = {x | x ∈ A or x ∈ B}.

Intersection. Intersection of setsA andB, denoted byA ∩ B is the set of all elements which belong to

bothA andB, i.e.

A ∩B = {x | x ∈ A andx ∈ B}.

Complement. Complement ofA in B, denoted byB\A is the set of all elements which belong toB and

not belong toA, i.e.

B\A = {x | x ∈ B andx 6∈ A}.

1.2 Real Number

Ratio of the lengths of any two line segments.In ancient Greece, once mathematicians believed that any

quantity can be described by rational numbers. Especially they believed that for any two line segments,

there must be a third line segment such that the former two line segments arem andn (m,n are integers)

times of the length of the third line segment. Hence the ratioof the lengths of any two line segments is a

rational numberm/n. The proof of many geometric theorems were based on this assumption at that time.

Meanwhile some mathematicians announced that they had proved this basic assumption. But unfortunately

all the proofs were wrong.

By Pythagorean theorem, for a right angled triangle there isa2 + b2 = c2.

Let a = b = 1, thenc2 = 2.

That means the ratio of lengthsc/a = γ satisfiesγ2 = 2.

We now prove by algebraic method thatγ can not be a rational number.


Proposition 1.2.1No rational number satisfies the equationγ2 = 2.

Proof: We argue by contradiction. Assume that the proposition is false, then there is a rational number

γ = m/n that satisfies(m

n

)2

= 2,

wherem,n are integers and have no common factors other than 1. Sincem2 = 2n2, m is even. Let

m = 2k, k is an integer. Since(2k)2 = 2n2, n2 = 2k2, n is even. Therefore,2 is a common factor ofm

andn, contradicted with the assumption of no common factors other than1. Hence, the proposition is true.

Suppose that we divide all rational numbers into two sets (Dedekind cut), sayA andA′, such that

1. Any rational number belongs only one of the two sets.

2. If x ∈ A andx′ ∈ A′, thenx < x′.

There are three cases:

1. A has no largest number, butA′ has the smallest number.

2. A has the largest number, butA′ has no smallest number.

3. A has no largest number, andA′ has no smallest number.

Example. (case 1.)

A = {x | x < 1 andx is a rational numver.},

A′ = {x′ | x′ ≥ 1 andx is a rational numver.}.

Example. (case 3.)

A = {x | x ≤ 0 or x2 < 2, x is a rational numver.},

A′ = {x′ | x′ > 0 andx′2 > 2, x′ is a rational numver.}.

If a > 0, a ∈ A, then there exists integern, such thatn > 2a+12−a2 , or 2a+1

n < 2 − a2. Then we have

2an + 1

n2 <2a+1n < 2 − a2, or (a + 1

n )2 < 2, i.e.,A has no largest number. Similarly,A′ has no smallest

number.

In case 1 and 2, the largest of A, or the smallest ofA′ is a rational number. In case 3, we define an

irrational number. Case 1 and 3 are called normal Dedekind cuts. The union of all rational and irrational

numbers i.e., the normal Dedekind cuts, are calledreal numbers. The set of real numbers is denoted byR.

To define the order of the real numbers, let the normal Dedekind cutsA|A′ andB|B′ define the real

numbersα andβ. If B ⊆ A, and∃x ∈ A s.t. x /∈ B, then we callα > β, or β < α. If A = B and


A′ = B′, then we callα = β. It is easy to see that for anyα andβ, α = β, α > β, orα < β. Also, the real

number defined by the normal Dedekind cut is unique.

Let the normal Dedekind cutsA|A′ andB|B′ define the real numbersα andβ. LetC = {a + b|a ∈A, b ∈ B} andC′ = Q \ C whereQ is the set of rational numbers. The the normal Dedekind cutC|C′

defines a new real numberγ := α+ β. Similarly, other operations can be defined.

Real number axis.On a straight line, a pointO is selected as the Origin and a line segmentOE (E is on

the right ofO) is selected as the unit of length. Then, the set of all real numbers and the set of all the points

on the straight line can be established a one-to-one corresponding by the following rule:

• the numberx = 0 corresponds to the originO;

• the positive (negative) numberx corresponds to the pointP on the right (left) ofO with |x| = OPOE .

The straight line is called a Real Number Axis. Proposition 1.2.1 demonstrates that the set of rational points

cannot full fill the real number axis. The point on the axis that doesn’t correspond to a rational number

corresponds to an irrational number.

Order and operations. Using the normal Dedekind cut, we can also define the order (”<,>”) of two real

numbers, and the operations (”+,−,×,÷”) of real numbers.

Real number set is a field.We know that for any two real numbersx andy, the operations onx andy such

as ”addition” and ”multiplication” and their inverse ”subtraction” and ”division” are closed. That means

the results ofx+ y, x− y, x− y andx/y (y 6= 0) are still real numbers.

In addition, the operations have the following properties:

1. Commutative rule:a+ b = b+ a

2. Commutative rule:a · b = b · a

3. Associative rule:(a+ b) + c = a+ (b+ c)

4. Associative rule:(a · b) · c = a · (b · c)

5. Distributive rule:a · (b + c) = a · b+ a · c

Therefore, real number set is a Field. (The definition of ”Field” will be learnt in the subject ”Algebra”.

In fact, the ”real number set” is the prototype of the concept”Field”.

Real number set is an ordered field.For any two different real numbersx andy, there is an ”Ordering”

between them. That is, one of the following relations

x < y, or y < x


must hold, i.e.y bigger thanx or x bigger thany. On the real number axis,x < y means the pointy is on

the right of pointx.

The relationship ”Ordering” has the following properties:

1. x < y andy < z, thenx < z.

2. x1 < y1 andx2 < y2, thenx1 + x2 < y1 + y2.

3. 0 < x andy < z, thenxy < xz.

4. x < y, then−x > −y.

5. 0 < x < y, then 1x >

1y .

Symbolx ≤ y meansx < y or x = y.

Density of the rationals and irrationals. SymbolR denotes the set of all real numbers.

Interval notations (a < b):

• The set{x ∈ R | a < x < b} is called an open intervaldenoted by(a, b).

• The set{x ∈ R | a ≤ x ≤ b} is called a closed interval, denoted by[a, b].

• [a, b) = {x ∈ R | a ≤ x < b} is called an interval but not open and not closed.

• (a, b] = {x ∈ R | a < x <≤ b}. is also called an interval but not open and not closed.

• (−∞,+∞) = R is the whole axis of real numbers, it is open and closed at the same time.

• [a,+∞) = {x ∈ R | x ≥ a} and(−∞, b] = {x ∈ R | x ≤ b} are closed intervals.

• (a,+∞) = {x ∈ R | x > a} and(−∞, b) = {x ∈ R | x < b} are open intervals.

Definition 1.2.2A setS of real numbers is said to be Dense inR, provided that any intervalI = (a, b),

contains at least one member of S.

Archimedean property. For any real numberC, there is a natural numbern such thatC < n. We take this

property for granted.

By this property, for any given positive numberǫ, we can choose a natural numbern such that1ǫ < n.

Hence1n < ǫ.

Theorem 1.2.3 (Density theorem)The set of rational numbers and the set of irrational numbersare both

dense inR.

Proof Firstly we consider the case when0 ≤ a < b. Choose a natural numberm such that

1

m< b− a.


Let

A =

{

k | k ∈ N,k

m< b

}

.

SinceA has finite members, there must be a maximum, sayk0. We now showa < k0

m < b.

Sincek0 ∈ A, k0

m < b. Sincek0 is the maximum ofA, k0+1m ≥ b. Hence

k0m

≥ b− 1

m> b− (b− a) = a.

That means the rational numberk0

m satisfiesa < k0

m < b.

By a similar way we can find an irrational number in(a, b). Choose a natural numberm such that

1m < b−a√

2, i.e.

√2

m< b− a.

Let

B =

{

k | k ∈ N, k

√2

m< b

}

andk0 be the maximum ofB. The same reasoning shows that the irrational numberk0√2

m satisfiesa <

k0√2

m < b. If a < 0, 0 < b, then(0, b) ⊆ (a, b). From above proof(0, b) contains rational and irrational

numbers, so interval(a, b) does.

If a < b ≤ 0. From above proof, there are a rational numberr and an irrational numberx that satisfy

−b < r < −a

and

−b < x < −a.

Hence

a < −r < b

and

a < −x < b.

Hence we complete the proof.

It is easy to see from Theorem 1.2.3 that there is at least a rational number between any two irrational

numbers, and there is at least an irrational number between any two rational numbers.

Given a normal Dedekind cutA|A′,A′ is uniquely determined byA sinceA′ = Q \A. Hence, we can

also simply callA as normal Dedekind cut.


Normal Dedekind cuts

The main purpose of the normal Dedekind cuts is to build a reasonable logical foundation for the real

numbers system. Notice that any real numberα, intuitively, is completely decided by the rational numbers

strictly smaller thanα and strictly bigger thanα. This is the main motivation.

Definition A Normal Dedekind Cutis a non-empty proper subset of the rational numbersQ that satisfies

these properties:

1. α contains no greatest element.

2. Forx, y ∈ Q, if x ∈ α andy < x, theny ∈ α as well.

For two reasons, the normal Dedekind cuts are extremely powerful. Indeed, it is very easy to prove the

continuity of the real line, or completeness. Moreover, it is rather direct to distinct the rational numbers

from the irrational numbers on the real line and set up a firm logical foundation for the real number system.

Definition A real numberis a normal Dedekind cut. We denote the set of all real numbersby R and

we order the real numbers by set-theoretic inclusion, that is to say, for anyα, β ∈ R,

α < β, if and only if α ( β,

where the inclusion is strict. Ifα andβ are equal, we defineα = β as real numbers. We writeα ≤ β if

α < β orα = β. Furthermore, a real numberα is said to beirrational if Q \ α contains no least element.

Theorem (Least upper bound theorem)Every nonempty subset of real numbers that is bounded above

has a least upper bound.

Proof LetA be a nonempty set of real numbers, such that for eachα ∈ A we have thatα ≤ γ for some real

numberγ. Now define the set

supA =⋃

α∈A

α.

We show that this set is a real number by checking the conditions of the normal Dedekind cut.

1. ClearlysupA is nonempty. Asγ is a normal Dedekind cut, there exists some rationalx /∈ γ. But

∀α ∈ A, α ⊆ γ. Thusx /∈ α, andx /∈ supA. Hence,supA is a proper subset ofQ.

2. If supA has a greatest elementg, theng ∈ α for someαA. Theng is the greatest element ofα. This

is impossible sinceα is real. Hence,supA has no greatest element.

3. Finally, if x ∈ supA, thenx ∈ α for someα. Given any rationaly < x, becauseα is real,y ∈ α,

whencey ∈ supA.

Thus,supA is a real number. Evidently,supA is an upper bound ofA as for everyα ∈ A, α ⊆ supA.

Now it suffices to verify thatsupA ≤ γ whereγ is an arbitrary upper bound. This is easy since every


x ∈ supA is an element of someα ∈ A, andα ⊆ γ, x ∈ γ. ThussupA is the least upper bound ofA. We

call this real number thesupremumof A.

To finish the construction of the real numbers, we must endow them with algebraic operations, define

the additive and multiplicative identity elements, prove that these denitions give a field, and prove further

results about the order of the reals (such as the totality of this order) in short, build a complete ordered field.

This task is somewhat laborious, but we include here the appropriate denitions. Verifying their correctness

can be an instructive, albeit tiresome, exercise. We use thesame symbols for the operations on the reals as

for the rational numbers; this should cause no confusion in context.

Definition Given two real numbersα andβ, we dene

• Theadditive identity, denoted0, is

0 := {x ∈ Q : x < 0}.

• Themultiplicative identity, denoted1, is

1 := {x ∈ Q : x < 1}.

• Additionof α andβ denotedα+ β is

α+ β := {x+ y : x ∈ α, y ∈ β}.

• Theoppositeof α, denotedα, is

α := {−x : x ∈ Q \ α, x is not the least element ofQ \ α}.

• Theabsolute valueof α, denoted|α|, is

|α| :=

α, if α ≥ 0,

−α, otherwise.

• If α, β > 0, then multiplication ofα andβ, denotedα · β, is

α · β := {z ∈ Q : z ≤ 0 or z = xy for somex ∈ α, y ∈ β with x, y > 0}.

In general,

α · β :=

0, if α = 0, or β = 0,

|α| · |β|, if α > 0, β > 0 orα < 0, β < 0,

−(|α| · |β|), otherwise.


• The inverse ofα > 0, denotedα−1, is

α−1 := {x ∈ Q : x ≤ 0 or x > 0 and1

x∈ α, but

1

xis not the least element ofQ \ α}.

If α < 0,

α−1 := −(|α|)−1.

All that remains is to check that the above denitions do indeed define a complete ordered field, and

that all the sets implied to be real numbers are indeed so. Theproperties ofR as an ordered field

follow from these denitions and the properties ofQ as an ordered field. It is important to point out

that in two steps, in showing that inverses and opposites areproperly dened, we require an extra

property ofQ, not merely in its capacity as an ordered field. This requirement is the Archimedean

property. Moreover, becauseR is a field of characteristic 0, it contains an isomorphic copyof Q.

The rational numbers correspond to the normal Dedekind cutsα for whichQ \ α contains a least

member.

1.3 Inequalities and Algebraic Identities

At the heart of many arguments in analysis lies the problem ofestimating the sizes of various quotients,

differences and sums. In order to do so the following inequalities and identities will be used frequently.

Absolute value and basic inequalities.Absolute Value of a real numberx, denoted by|x| is defined by

|x| =

x, if x ≥ 0,

−x, if x < 0.

On the real number axis,|x| is the distance from the pointx to the origin0. |x − y| is the distance

between the pointx and the pointy.

By the definition we have the inequality

−|x| ≤ x ≤ |x|

and the following basic proposition.

Proposition 1.3.1Suppose thatc andd are any real numbers andd ≥ 0, then

|c| ≤ d⇔ −d ≤ c ≤ d.

Proof. ”⇒”: By the definition,|c| ≤ d means

c ≤ d and − c ≤ d.


From the second inequality it follows. Hence we obtain

−d ≤ c ≤ d.

”⇒”: Assume−d ≤ c ≤ d, i.e. −d ≤ c andc ≤ d. From−d ≤ c we get−c ≤ d. Combine−c ≤ d

andc ≤ d we obtain|c| ≤ d.

Theorem 1.3.2 (Triangle inequality)For any numbersa andb,

|a+ b| ≤ |a|+ |b|.

Proof. From inequalities

−|a| ≤ a ≤ |a|, −|b| ≤ b ≤ |b|

we have

−(|a|+ |b|) ≤ a+ b ≤ (|a|+ |b|).

Let c = a+ b, andd = |a|+ |b|, then−d ≤ c ≤ d. From Proposition 3.1.1 it follows|c| ≤ d, i.e.

|a+ b| ≤ |a|+ |b|.

From this basic form of Triangle Inequality, it is easy to getthe following inequalities:

|a| − |b| ≤ |a+ b|,

|a| − |b| ≤ |a− b|,

and

|x− z| ≤ |x− y|+ |y − z|.

We may see these inequalities as different forms of TriangleInequalities.

In many cases, we need to use mathematical induction to provetheorems. To introduce the principle of

mathematical induction, we first give the following definition.

Definition 1.3.1A setS of real numbers is said to be inductive if the number1 ∈ S, and ifx ∈ S, then

x+ 1 ∈ S.

The set of natural numbers is defined as the intersection of all inductive subsets ofR, denoted byA.

Let N = {1, 2, 3, · · · }. Clearly,N is inductive. Then,A ⊆ N. On the other hand,1 ∈ N, and1 belongs

to every inductive set. Ifk ∈ N, thenk belongs to every inductive set. Thus,N ⊆ A. Therefore, we have

A = N.

Principle of Mathematical Induction. For each natural numbern, letS(n) be some mathematical asser-

tion. Suppose thatS(1) is true. Also suppose that wheneverk is a natural number such thatS(k) is true,

thenS(k + 1) is also true. ThenS(n) is true for every natural numbern.


Proof. Let A ≡ {k ∈ N | S(k) is true.}. It is enough to show thatA = N. A ⊆ N by definition. The

assumption means thatA is an inductive subset ofN. SinceA is inductive, thenN ⊆ A. ThenA = N, i.e.,

S(n) is true for every natural numbern.

Theorem 1.3.3 (Bernoulli’s inequality)For any natural numbern and real numbera ≥ −1,

(1 + a)n ≥ 1 + na.

Proof. Prove by mathematical induction. Forn = 1, the proposition is obviously true. Assume that the

proposition is true forn = k, i.e.,

(1 + a)k ≥ 1 + ka,

we need to prove

(1 + a)k+1 ≥ 1 + (k + 1)a.

Since

(1 + a)k+1 = (1 + a)k(1 + a)

≥ (1 + ka)(1 + a)

= 1 + (k + 1)a+ ka2

≥ 1 + (k + 1)a.

which completes the proof.

Binomial formula.

(a+ b)n =

n

0

an +

n

1

an−1b+

n

2

an−2b2 + · · ·+

n

n− 1

abn−1 +

n

n

bn

=n∑

k=0

n

k

an−kbk,

where

n

k

=n!

k!(n− k)!, k! = k(k − 1) · · · 2 · 1, and0! = 1.

Difference of powers formula.

an − bn = (a− b)

n−1∑

k=0

an−1−kbk.

Geometric sum formula.n−1∑

k=0

γk =1− γn

1− γ, if γ 6= 1.


1.4 Function

In the real world, all things are varying. Variables are usedto describe the variation in some quantity

aspects of the matters we concern. In general, two or more variables for describing a same object or

phenomenon have some relationship. For example, the volumeV, temperature T and the pressure P of

certain gas satisfyV = cT/P . The concept ”function” is just for this application requirement to introduce

into mathematics. It is one of the most basic concept in Mathematical Analysis.

Definition 1.4.1. Let D andR be sets. A functionfrom D into R, denoted byf : D → R, is a rule

that assigns to each element ofD a unique element ofR. The setD is called the definition domainof the

function. Functions are usually denoted by letters such asf, g, F,G, and so on. Iff is a function fromD

into R, then the elementy ∈ R that is assigned to the elementx ∈ D by f is denoted byf(x) (read ”f

of x”), and is called the value off at x (or function value atx) or the image ofx underf . The elements

x ∈ D andy ∈ R are called variables, withx the independent variableandy the dependent variable. The

set{y | y = f(x), x ∈ D} is called the rangeof f, or the image of D under f.

The definition domain and range of function f are denoted byD(f) andR(f) or f(D) respectively.

Throughout most of this course, the setsD andR are real number sets. In this context, a function from

D into R is said to be a real-valued function of a real variable. In general, the rule for a function will be

specified by an equation, or a set of equations.

Surjective, injective and bijective.Let f be a function from setA to setB.

• f is surjective ifR(f) = B (or f(A) = B).

• f is injective if forx1 6= x2 in A, f(x1) 6= f(x2) in B.

• f is bijective iff is surjective as well as injective.

Example 1: f : A→ B, f(x) = x2.

(a) If A = B = R, f is neither surjective nor injective.

(b) If A = [0,∞), B = R, f is injective.

(c) If A = R, B = [0,∞), f is surjective.

(d) If A = B = [0,∞), f is bijective.

Example 2: f : A→ B, f(x) = sinx.

(a) If A = B = R, f is neither surjective nor injective.

(b) If A = R, B = [−1, 1], f is surjective.

(c) If A = [−π2 ,

π2 ],B = R, f is injective.

(d) If A = [−π2 ,

π2 ], B = [−1, 1], f is bijective.


Inverse and composition.

• If the functionf : A → B is bijective, i.e. for every pointy ∈ B there is exactly one pointx ∈ A

such thatf(x) = y. Then, functionf−1 : B → A is called the inverseof f , if for eachy ∈ B,

f−1(y) = x, with f(x) = y.

Example 1: f : [0,∞) → [0,∞), f(x) = x2. The inversef−1 : [0,∞) → [0,∞) is f−1(y) =√y.

Example 2: f : [−π2 ,

π2 ] → [−1, 1], f(x) = sinx. The inversef−1 : [−1, 1] → [−π

2 ,π2 ] is

f−1(y) = sin−1 y. .

• Let f andg be functions andR(g) ⊆ D(f). The compositionof f andg denoted byf ◦ g is defined

in D(g) by

(f ◦ g)(x) = f(g(x)).

Example: Let f(y) = sin y, g(x) = x2, then

(f ◦ g)(x) = f(g(x)) = sinx2.

1.5 Exercise

1. Use mathematical induction to prove the following identities:

(a) 12 + 22 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6.

(b) 13 + 23 + · · ·+ n3 = (1 + 2 + · · ·+ n)2.

(c) 20 + 21 + · · ·+ 2n−1 = 2n − 1.

2. Find a formula forn∑

j=1

j(j + 1).

3. Prove that there is at least one irrational number betweenany two rational numbers.

4. Prove that there are infinitely many irrational numbers between any two rational numbers.

Hint. By mathematical induction.

5. Determine whether the following sets are dense inR and justify your answer.

(a) S ={

x | x = n+1

m, n,m, are integers

}

.

(b) S ={

x | x = q +√2, q ∈ Q

}

.

Hint. For anya < b, consider the interval(a −√2, b−

√2)


(c) S ={

x | x =√2q, q ∈ Q

}

, Q denotes the set of all rational numbers.

6. Prove that if the setS is dense inR, then any intervalI = (a, b) contains infinitely many numbers

of the setS.

7. Suppose that the numbera has the property that for every natural numbern, a ≤ 1/n. Prove that

a ≤ 0.

8. Suppose that the numbera has the property that for every natural numbern, a ≥ 1 − 1n . Prove that

a ≥ 1.

9. Prove that ifa > 0, x satisfies|x− a| < a

2, thenx >

a

2.

10. Solve the inequality:|x+ 2| − |x| > 1.

Hint. The zeros of|x+ 2| and|x| are:x = −2 andx = 0, then we discuss three cases:x < −2, −2 ≤ x < 0andx ≥ 0, in order to remove the absolute value signs.

11. Prove the inequality:1

2· 34· · · 2n− 1

2n<

1√2n+ 1

.

12. Show that if|a− b| ≤ 1, then|a| ≤ |b|+ 1.

13. Solve the inequality:|2x+ 3| > |x− 2|.

14. (a) Prove the Cauchy’s Inequality: for any real numbera andb, ab ≤ 1

2(a2 + b2).

(b) Prove that ifa andb are real numbers, andn is a natural number, thenab ≤ 1

2

(

na2 +1

nb2)

.

Hint. The inequality is equivalent to:(√

na− 1√nb)2

≥ 0.

15. Prove that ifa1, · · · , an are of same sign andai > −1, i = 1, · · · , n, then

(1 + a1)(1 + a2) · · · (1 + an) ≥ 1 + a1 + a2 + · · ·+ an.

16. Determine whether the following functions are surjective, injective, bijective, or none of them:

(a) f : R → R, f(x) = cosx.

(b) f : (−π2,π

2) → R, f(x) = tanx.

(c) f : R → R, f(x) =

−1, −∞ < x < 0,

1, 0 ≤ x <∞..

(d) f : (−1, 1) → (−1, 1), f(x) =

x, −1 < x < 0,

0, 0 ≤ x < 12

2x− 1, 12 ≤ x < 1.

.

(e) f : (−1, 1) → (−1, 2), f(x) =

x, −1 < x < 0,

x+ 1, 0 ≤ x < 1..

Chapter 2: Limit Theory 23

17. Are the following functions bijective? If so, find their inverses.

(a) f : R → R, f(x) =

x, x is rational,

x+ 1, x is irrational..

(b) f : (−1, 1) → (−2, 0) ∪ [1, 5), f(x) =

2x, −1 < x < 0,

4x+ 1, 0 ≤ x < 1..

(c) f : (−1, 1) → (−1, 1), f(x) =

x2, −1 < x < 0,

−x3, 0 ≤ x < 1..

18. Find the compositionf ◦ g andg ◦ f :

f(x) =

−x− 1, x ≤ 0,

x, x > 0,g(x) =

x, x ≤ 0,

−x2, x > 0.


Chapter 2

Limit Theorey

2.1 Convergence and Limit

2.1.1 Sequence

A sequence of real number is a real-valued functionf(n) : N → R, i.e., a function from the set of

natural numbers to real numbers. It is used to replacef(n) with an and denoting a sequence by{an}. The

numberan associated with the indexn is called thenth term of the sequence.

Note:The collection of the terms from a sequence is a set. But different terms of a sequence may be a same

number.

Countable and uncountable set.A set is called countableprovided it can be counted, i.e., a one-to-one

correspondence can be established between the set and the finite set{1, 2, · · · , n} or the infinite setN.

Otherwise, the set is called uncountable. The rational number setQ is countable; but the real number setR

or its subset{x ∈ R | a < x < b} is uncountable.

The set consisted of the terms from a sequence is a countable set. An uncountable set such as{x ∈R | a < x < b} can not be arranged into a sequence which includes all the numbers of the set.

Examples 2.1.1

1. an = 1n , {an} =

{

1,1

2,1

3,1

4, · · · ,

}

.

2. an = 1 +

(

1

2

)n

, {an} =

{

1 +1

2, 1 +

1

4, 1 +

1

8, 1 +

1

16, · · · ,

}

.

3. a1 = 1, an+1 =

an + 1n , if a2n ≤ 2,

an − 1n , if a2n > 2,

{an} =

{

1, 2,3

2,7

6, · · · ,

}

.

4. an = 1 + (−1)n, {an} = {0, 2, 0, 2, · · · , }.

25


5. an =

(

1 +1

n

)n

, {an} =

{

2,9

4,64

27,625

256, · · · ,

}

.

2.1.2 Convergence and Limit

Observing the example 1, we find that whenn becomes larger and larger, the numberan = 1n becomes

closer and closer to zero. The sequence{an} has infinitely many terms andan never equals zero. Butan

can be arbitrarily close to zero while the indexn is sufficiently large. Similarly, the sequence in example 2

the termsan = 1+(

12

)nalthough never equal 1, butan can be arbitrarily close to 1 whilen is sufficiently

large.

Definition 2.1.1A sequence{an} is said to be convergentand converge to the numbera if for any positive

numberǫ (or say arbitrarily small numberǫ > 0) there exists a natural numberN (∃N ) such that

|an − a| < ǫ ∀n ≥ N.

The numbera is called the limit of the sequence{an}, where ”∀” means ”for all”, or we write as

Givenǫ > 0, ∃N > 0, s.t. |an − a| < ǫ ∀n ≥ N.

Notation:A sequence{an} converges to numbera is denote by

limn→∞

an = a, or an → a.

Geometrical explanation.The inequality

|an − a| < ǫ ∀n ≥ N.

means that all the termsan with indexn ≥ N are inside the interval(a− ǫ, a+ ǫ), equivalently, there are

only a finite number of termsan outside the interval.

Proposition 2.1.2A sequence{an} converges to a numbera if and only if for anyǫ > 0, only a finite

number of terms ofan are outside the interval(a− ǫ, a+ ǫ).

A sequence is said to be divergentif the sequence does not converge. The sequence in example 4 is diver-

gent.

Definition 2.1.3A sequence{xn} is said to tend to+∞ (or −∞) if for any positive numberM (or negative

number−M ) there exists a natural numberN such that

xn > M (or xn < −M), ∀n ≥ N


and denoted by

limn→∞

xn = +∞, (or limn→∞

xn = −∞).

If there exists a natural numberN such that

|xn| > M, ∀n ≥ N,

then we say that

limn→∞

xn = ∞.

Examples 2.1.2Prove limn→∞

1

n= 0.

Proof. According to the definition, for any givenǫ > 0 we need to findN such that

∣

∣

∣

∣

1

n− 0

∣

∣

∣

∣

< ǫ ∀n ≥ N.

⇔ 1

n< ǫ ∀n ≥ N.

⇔ 1

ǫ< n ∀n ≥ N.

From this, we see that when we chooseN =[

1ǫ

]

+ 1 ([x] means the integer part ofx.), then

∣

∣

∣

∣

1

n− 0

∣

∣

∣

∣

< ǫ ∀n ≥ N.

Thus we complete the proof.

Note:The numberN that we are going to find is a function ofǫ. It is not unique. In the example,N =[

1ǫ

]

+k

is a right choice with any positive integerk.

Examples 2.1.3Prove limn→∞

[

1 +

(

1

2

)n]

= 1.

Proof. For anyǫ > 0, we need to findN such that

∣

∣

∣

∣

[

1 +

(

1

2

)n]

− 1

∣

∣

∣

∣

< ǫ, ∀n ≥ N.

⇔(

1

2

)n

< ǫ ∀n ≥ N.

⇔ n ln

(

1

2

)

< ln ǫ ∀n ≥ N.

⇔ −n ln(

1

2

)

> − ln ǫ ∀n ≥ N.

⇔ n >− ln ǫ

− ln(

12

) =ln ǫ

ln(

12

) , ∀n ≥ N.


From this, we see that when we chooseN =

[

ln ǫ

ln( 12 )

]

+ 1 then

∣

∣

∣

∣

[

1 +

(

1

2

)n]

− 1

∣

∣

∣

∣

< ǫ, ∀n ≥ N.

Examples 2.1.4Let sequence{an} be defined as

a1 = 1, an+1 =

an + 1n , if a2n ≤ 2,

an − 1n , if a2n > 2,

Prove limn→∞

an =√2.

Proof. First we prove by mathematical induction the inequality

∣

∣

∣an −√2∣

∣

∣ ≤ 1

n− 1, n ≥ 2.

Whenn = 2, the inequality is true due to

∣

∣

∣a2 −√2∣

∣

∣ =∣

∣

∣2−√2∣

∣

∣ ≤ 1.

Assume that the inequality is true forn = k, i.e.,

∣

∣

∣ak −

√2∣

∣

∣≤ 1

k − 1. (2.1.1)

We need to show that the inequality is also true forn = k + 1. From (2.1.1) we get

− 1

k − 1< ak −

√2 <

1

k − 1.

There are two cases:

(i) − 1

k − 1< ak −

√2 ≤ 0, (2.1.2)

(ii) 0 < ak −√2 <

1

k − 1. (2.1.3)

In case (i), by the definition of the sequence,

ak+1 = ak +1

k.

Hence

ak+1 −√2 = ak −

√2 +

1

k.

Adding 1k to (2.1.2) we get

1

k− 1

k − 1< ak+1 −

√2 ≤ 1

k,

or,

− 1

k(k − 1)< ak+1 −

√2 ≤ 1

k.


Sincek ≥ 2, we have

− 1

k< ak+1 −

√2 ≤ 1

k,

i.e.,∣

∣

∣ak+1 −√2∣

∣

∣ ≤ 1

k.

In case (ii), by the definition of the sequence,

ak+1 = ak −1

k.

Hence

ak+1 −√2 = ak −

√2− 1

k.

Subtracting1k from (2.1.3) we get

− 1

k< ak+1 −

√2 <

1

k − 1− 1

k,

or,

− 1

k< ak+1 −

√2 <

1

k(k − 1).

Sincek ≥ 2, we have

− 1

k< ak+1 −

√2 <

1

k,

i.e.,∣

∣

∣ak+1 −√2∣

∣

∣ ≤ 1

k.

We have proved the inequality∣

∣

∣an −√2∣

∣

∣ ≤ 1

n− 1.

Thus for anyǫ > 0, we chooseN =[

1ǫ

]

+ 2, then∣

∣

∣an −√2∣

∣

∣ ≤ ǫ, ∀n ≥ N.

Examples 2.1.5Prove that the sequence{0, 2, 0, 2, · · · } is divergent.

Proof. We argue by contradiction. Assume thatlimn→∞ an = a. According to the definition of conver-

gence, we chooseǫ = 12 , then there exists a natural numberN such that

|an − a| < 1

2∀n ≥ N.

Therefore, the neighboring two termsaN andaN+1 satisfy

|aN − aN+1| ≤ |aN − a|+ |aN+1 − a| < 1

2+

1

2= 1.

But, according to the definition of the sequence

|aN − aN+1| = 2.

The contradiction shows that the sequence can’t converge.


2.1.3 Uniqueness of Limit

Proposition 2.1.4A sequence can not converge to two limits.

Proof. We argue by contradiction. Letlimn→∞

an = a and also limn→∞

an = a′, a′ 6= a. Takeǫ = 12 |a − a′|.

According to the definition of convergence,

an → a ⇒ ∃N1 s.t. |an − a| < ǫ, ∀n ≥ N1,

an → a′ ⇒ ∃N2 s.t. |an − a′| < ǫ, ∀n ≥ N2.

DefineN = max{N1, N2}, then we have

|an − a| < ǫ and|an − a′| < ǫ, ∀n ≥ N.

Thus,

|a− a′| ≤ |an − a|+ |an − a′| < 2ǫ = |a− a′| < ǫ,

which is impossible, hence a sequence can’t converge to two limits.

2.1.4 Boundedness of Convergent Sequence

Bounded above.A sequence{an} is said to be bounded above if for allan there is a numberB such that

an ≤ B.

The numberB is called an upper boundof the sequence. Obviously, ifB is an upper bound of a sequence,

then any numberB′ > B is also an upper bound of the sequence.

Bounded below.A sequence{an} is said to be bounded below, if for allan there is a numberA such that

A ≤ an.

The numberA is called a lower boundof the sequence. IfA is a lower bound of a sequence, then any

numberA′ < A is also a lower bound of the sequence.

Bounded sequence.A sequence{an} is said to be bounded if it is bounded above as well as bounded

below. Therefore there are numbersA andB such that for allan

A ≤ an ≤ B.

DefineM = max{|A|, |B|}, then

|an| ≤M,


or

−M ≤ an ≤M.

Lemma 2.1.5Every convergent sequence is bounded.

Proof. Let {an} be a convergent sequence and converges toa. Takingǫ = 1, it follows from the definition

of convergence that there exists a natural numberN such that

|an − a| < 1, ∀n ≥ N.

Then we have

|an| − |a| ≤ |an − a| < 1, ∀n ≥ N,

or

|an| < 1 + |a|, ∀n ≥ N.

Let

M = max{1 + |a|, |a1|, |a2|, · · · , |aN−1|},

Then

|an| < M, ∀n ∈ N.

Lemma 2.1.6Suppose that

limn→∞

bn = b 6= 0.

Then there is a natural numberN such that

|bn| >|b|2, ∀n ≥ N.

Proof. Takeǫ = |b|2 . Then there exists a natural numberN such that

|bn − b| < |b|2, ∀n ≥ N,

then we have

|b| − |bn| ≤ |bn − b| < |b|2, ∀n ≥ N,

or

|bn| >|b|2, ∀n ≥ N.


2.1.5 Operations on Convergent Sequences

For some simple sequences, we can use the ”ǫ−N ” definition to obtain their limits. For example, it is

easy to obtain the limits of the following sequences

an =

(

3

4

)n

, bn =1√n

by ”ǫ−N ” definition, but it is difficult to obtain the limit of the sequence

cn =1

1 +(

34

)n +

(

1 +1√n

)2

by ”ǫ−N ” definition.

Theorem 2.1.7Suppose that

limn→∞

an = a, and limn→∞

bn = b.

Then

(i) limn→∞

(an + bn) = limn→∞

an + limn→∞

bn = a+ b.

(ii) limn→∞

(an · bn) = limn→∞

an · limn→∞

bn = a · b.

(iii) If bn 6= 0, ∀n, andb 6= 0, then, limn→∞

anbn

=limn→∞

an

limn→∞

bn=a

b.

Proof of (i). Let ǫ > 0. We need to find a natural numberN such that

|(an + bn)− (a+ b)| < ǫ, ∀n ≥ N.

Sincean → a, we can choose a natural numberN1 such that

|an − a| < ǫ

2, ∀n ≥ N1.

Sincebn → b, we can choose a natural numberN2 such that

|bn − b| < ǫ

2, ∀n ≥ N2.

LetN = max{N1, N2}, then

|(an + bn)− (a+ b)| ≤ |an − a|+ |bn − b| < ǫ

2+ǫ

2= ǫ, ∀n ≥ N.


Proof of (ii). Let ǫ > 0. We need to find a natural numberN such that

|anbn − ab| < ǫ, ∀n ≥ N.


Since

anbn − ab = anbn − anb+ anb− ab

= an(bn − b) + (an − a)b,

then

|anbn − ab| < |an||bn − b|+ |b||an − a|.

Since{an} converges, by Lemma 2.1.5 there is a numberM such that

|an| < M, ∀n.

Hence,

|anbn − ab| < M |bn − b|+ |b||an − a|.

Sincebn → b, we can choose a natural numberN1 such that

|bn − b| < ǫ

2M∀n ≥ N1.

If b = 0, then

|anbn − ab| < M |bn − b| < ǫ

2< ǫ ∀n ≥ N1,

which gives the proof. Ifb 6= 0, sincean → a, we can choose a natural numberN2 such that

|an − a| < ǫ

2|b| ∀n ≥ N2.

LetN = max{N1, N2}, then

|anbn − ab| < Mǫ

2M+ |b| ǫ

2|b| = ǫ ∀n ≥ N.

We complete the proof.

Proof of (iii). Using (ii), it suffices to prove that

limn→∞

1

bn=

1

b.

Since1

bn− 1

b=b− bnbbn

, .

according to Lemma 2.1.6, we can chooseN1 such that

|bn| >|b|2, ∀n ≥ N1.

Thus,∣

∣

∣

∣

1

bn− 1

b

∣

∣

∣

∣

=

∣

∣

∣

∣

b− bnbbn

∣

∣

∣

∣

<2

|b|2 |bn − b| ∀n ≥ N1.


Sincebn → b, we can chooseN2 such that

|bn − b| < |b|22ǫ ∀n ≥ N2.

LetN = max{N1, N2}, then∣

∣

∣

∣

1

bn− 1

b

∣

∣

∣

∣

<2

|b|2 |bn − b| < 2

|b|2|b|22ǫ = ǫ ∀n ≥ N,

which completes the proof.

Corollary 2.1.8

limn→∞

(αan + βbn) = α limn→∞

an + β limn→∞

bn,

limn→∞

m∑

k=0

ckxkn =

m∑

k=0

ck

(

limn→∞

xn

)k

.

Now it is easy to obtain the limit of the sequence

cn =1

1 +(

34

)n +

(

1 +1√n

)2

mentioned above:

limn→∞

cn =1

limn→∞[

1 +(

34

)n] +

(

1 + limn→∞

1√n

)2

= 1 + 1 = 2.

2.1.6 Squeezing Principle

Lemma 2.1.9Suppose thatdn ≥ 0 for all n ≥ N0, whereN0 is some natural number andlimn→∞

dn = d.

Thend ≥ 0.

Proof. If d < 0, sincedn → d, for ǫ = |d|/2 there is a numberN such that

d− |d|2< dn < d+

|d|2

=d

2< 0 ∀n ≥ N.

which contradicts with the assumptiondn ≥ 0 for all n ≥ N0.

Theorem 2.1.10Supposean → a, bn → b, cn → c and

an ≤ cn ≤ bn ∀n ≥ N0,

whereN0 is some natural number. Then

a ≤ c ≤ b.

Proof. Let dn = cn − an. Sincedn ≥ 0 for all n ≥ N0, by Lemma 2.1.9dn → d ≥ 0. Hence

limn→∞

(cn − an) = limn→∞

cn − limn→∞

an = c− a = d ≥ 0,


i.e.,c ≥ a. Similarly we prove,c ≤ b.

Corollary 2.1.11Suppose thata ≤ cn ≤ b for all n ≥ N0, whereN0 is some natural number andcn → c.

Thena ≤ c ≤ b.

Theorem 2.1.12Suppose that sequences{an}, {bn} and{cn} satisfy

an ≤ cn ≤ bn ∀n ≥ N0,

whereN0 is some natural number, and

limn→∞

an = limn→∞

bn = l,

then

limn→∞

cn = l.

Proof. By the assumptionsan → l andbn → l, for any givenǫ > 0 there existN1 andN2 such that

|bn − l| < ǫ ∀n ≥ N1,

|an − l| < ǫ ∀n ≥ N2.

It follows than

bn < l + ǫ ∀n ≥ N1,

an > l − ǫ ∀n ≥ N2.

LetN = max{N0, N1, N2}, from the assumptionan ≤ cn ≤ bn we get

l − ǫ < an ≤ cn ≤ bn < l + ǫ ∀n ≥ N.

That is

|cn − l| < ǫ ∀n ≥ N,

or,

limn→∞

cn = l.

Note:Here we do not need to assume that{cn} converges.

The Squeezing Principle will be used frequently onwards. Itis very useful to find the limits of a

sequence with complicated expression.

Examples 2.1.6Show that

limn→∞

[(n+ 1)k − nk] = 0, 0 < k < 1.


Proof.

0 < (n+ 1)k − nk = nk[(

1 +1

n

)k − 1]

< nk[(

1 +1

n

)

− 1]

=1

n1−k.

Since limn→∞

1n1−k = 0, by Squeezing Principlelim

n→∞[(n+ 1)k − nk] = 0.

Examples 2.1.7Let

cn =1√

n2 + 1+

1√n2 + 2

+ · · ·+ 1√n2 + n

,

show that limn→∞

cn = 1.

Proof.

an ≡ n√n2 + n

< cn <n√

n2 + 1≡ bn.

Since limn→∞

an = limn→∞

bn = 1, by Squeezing Principlelimn→∞

cn = 1.

2.2 Some Important Theorems in Limit Theory

We have mentioned some properties of the real number set in Chapter 1 such as the following: The real

number set is an ordered field; the real number set is dense in itself, namely, there is another real number

in between any two real numbers. But these properties are also shared by the rational number set. In this

section we will introduce the property of completeness (or continuity) of the real number set, which is not

owned by the rational number set.

The property of completeness can be expressed in any one of the following seven equivalent theorems

that we will mention. Among these theorems we can take any oneof them as an Axiom and then the

other six theorems can be proved accordingly. These seven theorems compose the theoretical basis of

mathematical analysis.

2.2.1 Dedekind Theorem

Theorem 2.2.1 (Dedekind Theorem)Suppose thatA andA′ is a cut for the real number set (Dedekind

cut), then this cut generates a real number, which is the largest number ofA, or the smallest number ofA′.

Proof. LetB andB′ be the sets containing all rational numbers ofA andA′ respectively. ThenB andB′

is a cut for the rational number set, which defines a real number a: a ∈ A or a ∈ A′. If a ∈ A, thena is the

largest number ofA, since otherwise, there exists a real numberb > a, then we can find a rational number

r such thata < r < b, which contradicts with the conditions of the cutB andB′. Similarly, if a /∈ A, then

a ∈ A′ anda is the smallest number ofA′.

This theorem shows that the real number axis is complete, i.e., the real number axis has no holes. If

we just admit this property of the real number set without proof, then the above theorem is usually called

Dedekind Continuous Axiom.


2.2.2 Least Upper Bound Theorem

It is known that there must be a maximum and minimum within a nonempty set of finite numbers.

But a set of infinitely many numbers may not have maximum and minimum. For examples, the setA ={

r | r = 1n , n ∈ N

}

has no minimum; the setB ={

r | r = 2− 1n , n ∈ N

}

has no maximum; the set

C = A ∪B has neither maximum nor minimum.

Definition 2.2.2A nonempty setS of real numbers is said to be bounded above provided that there is a

numberC having the property

x ≤ C, ∀x ∈ S.

Such a numberC is called an upper bound ofS.

Corollary 2.2.3 If numberC is an upper bound of setS, then any numberC′ ≥ C is also an upper bound

of S.

Consider the exampleA ={

r | r = 1n , n ∈ N

}

. The number1 is an upper bound and also the maxi-

mum of setA. Because1 ∈ A, any upper boundC of A satisfiesC ≥ 1. Hence1 is the minimum of the

set of all upper bounds, or,1 is the least upper bound ofA.

Now consider the exampleB ={

r | r = 2− 1n , n ∈ N

}

. The set has no maximum, but bounded

above. The number2 is an upper bound of setB. For any numberC < 2, because2− C > 0, we can find

a natural numbern such that1n < 2 − C. HenceC < 2 − 1n and thenC is not an upper bound of setB.

That means the number2 is the least upper bound of setB.

Both the examples give an affirmative answer to the question.In fact, we have the following Axiom for

general set bounded above.

Theorem 2.2.4 (Least Upper Bound Theorem)Suppose thatS is a nonempty set of real numbers that is

bounded above. Then among the set of upper bound forS there is a smallest, or least upper bound.

Proof. Let A′ be the set of all upper bounds ofS, andA be the set of all other numbers. ThenA andA′

is a cut for the real numbers. By Dedekind Theorem, it generates a real numberb. b is an upper bound of

S, b ∈ A′, andb is the smallest number ofA′.

Similarly, if we just admit this property of the real numberswithout proof, then this theorem can also

be used as the basis of the real number system, and the theoremcan be called theCompleteness Axiom.

Using this theorem, we can derive the Dedekind Theorem, i.e., the two theorems are equivalent.

Second Proof.LetS := {Aα|A′α}, α ∈ I be a non-empty set of real numbers, i.e., for everyα ∈ I,Aα|A′

α

is a normal Dedekind cut. We can assume that there exists another Dedekind cutC|C′ such that

Aα ⊂ C, ∀α ∈ I,

sinceS is bounded above.


Now define

B =⋃

α∈I

Aα, B′ = Q \B,

then one can easily verify thatB|B′ is a normal Dedekind cut:

1. Any rational number belongs only one of the two sets.

2. If x ∈ B andx′ ∈ B′, thenx < x′.

3. B has no largest number.

Moreover,B|B′ is the least upper bound forS since for any upper boundC|C′, we have

Aα ⊂ C, ∀α ∈ I.

We thus haveB ⊂ C.

Proof of Dedekind Theorem. Suppose thatA andA′ is a cut for the real numbers. IfA has the largest

number, then we have the proof. IfA has no largest number, then by Least Upper Bound Theorem,A′ has

the smallest number.

If b is the least upper bound of setS, we denote it by

b = l.u.b.S

Sometimes the least upper bound ofS is also called the supremumof S and is denoted by

b = supS

According to the definition of least upper bound, it is obvious that

(i) b is the least upper bound of setS if and only if b is an upper bound and for any upper boundd of S,

d ≥ b.

(ii) b is the least upper bound of setS if and only if b is an upper bound ofS and∀ǫ > 0, ∃x0 ∈ S such

thatx0 > b− ǫ.

Definition 2.2.5A nonempty setS of real numbers is said to be bounded below provided that there is a

numberC having the property that

x ≥ C, ∀x ∈ S.

Such a numberC is called a lower bound ofS.

From the definition, ifC is a lower bound ofS andC′ < C, thenC′ is also a lower bound ofS. By the

Completeness Axiom for the set bounded above, we have the following corollary.


Corollary 2.2.6 Suppose thatS is a nonempty set and bounded below. Then among the set of lower bounds

for S there is a greatest lower bound.

We denote the greatest lower bound b of S by

b = g.l.b.S

Sometimes we call the greatest lower bound ofS the infimumof S and denote it by

b = inf S

According to the definition of greatest lower bound, it is obvious that

(i) b is the greatest lower bound if and only ifb is a lower bound and for any lower boundd of S, b ≥ d.

(ii) b is the greatest lower bound if and only ifb is a lower bound and∀ǫ > 0, ∃x0 ∈ S such thatx0 < b+ǫ.

In summary, the completeness property of real numbers can also be expressed in the following theorem.

Theorem 2.2.7 (Supremum and Infimum Theorem).A nonempty set of real numbers bounded above has

the supremum; A nonempty set of real numbers bounded below has the infimum.

Rational number set is not complete:S is a subset ofQ and is bounded above, butS may not have

supremum inQ.

Examples 2.2.1Consider the setS = {r ∈ Q | r ≤√2}. S is bounded above, but it has no supremum

(least upper bound).

Proof. We argue by contradiction. Assume thatS has a least upper boundb = mn . We have proved that

x2 = 2 has no solution inQ, hencemn 6= 2. Sinceb is an upper bound ofS, b >

√2. By the density

theorem, there is a rational numberb′ in interval(√2, b), i.e.

√2 < b′ < b,

which meansb′ also an upper bound ofS andb′ < b. Henceb is not the least upper bound. This contradic-

tion shows no least upper bound ofS in Q.

2.2.3 Monotone Convergence Theorem

Definition 2.2.8A sequence{an} is said to be monotonically increasing provided

an+1 ≥ an, ∀n ∈ N,

{an} is said to be monotonically decreasing provided

an+1 ≤ an, ∀n ∈ N,


Both cases are called monotone.

Theorem 2.2.9A sequence{an} that is monotonically increasing and bounded above or is monotonically

decreasing and bounded below must converge.

Proof. Let’s first assume that{an} is monotonically increasing and bounded above. By the Completeness

Axiom, the sequence{an} has a least upper bounda = l.u.b. {an}. We claim thatan → a. It means that

for any givenǫ > 0, we need to find a natural numberN such that

a− ǫ < an < a+ ǫ, ∀n ≥ N.

Sincea is the least upper bound of{an}, by the definition of least upper bound, there is a term of{an}, say

aN , such that

a− ǫ < aN .

The indexN of this termaN is just what we want to find. Sincean is monotonically increasing,

an ≥ aN , ∀n ≥ N.

On the other hand,a is an upper bound. Hence we obtain

a− ǫ < aN ≤ an < a+ ǫ, ∀n ≥ N,

which completes the proof of

limn→∞

an = a = l.u.b. {an},

Similarly if {an} is monotonically decreasing and bounded below, then we have

limn→∞

an = a = g.l.b. {an}.

In section 1, we have proved that a convergent sequence must be bounded. Therefore, a theorem

connects the three concepts ”bounded” ”monotone” and ”convergent” can be expressed as follows.

Theorem 2.2.10 (Monotone Convergence Theorem)A monotone sequence converges if and only if it is

bounded.

Examples 2.2.2Show that the sequence

sn =

n∑

k=1

1

2kk, n = 1, 2, · · · ,

converges.

Proof. Clearly, sn is monotonically increasing, andsn ≤n∑

k=1

1

2k≤ 1, i.e., sn is bounded above. By

Monotone Convergence Theorem,sn converges.



sn =

n∑

k=1

1

k, n = 1, 2, · · · ,

diverges.

Proof. Clearly,sn is monotonically increasing. Also, we have

s2 = 1 +1

2≥ 1 +

1

2,

s4 = s2 +1

3+

1

4≥ s2 +

1

4+

1

4= 1 +

2

2,

s8 = s4 +1

5+ · · ·+ 1

8≥ s4 +

4

8≥ 1 +

3

2.

If s2n−1 ≥ 1 +n− 1

2, then

s2n = s2n−1 +1

2n−1 + 1+ · · ·+ 1

2n

≥ s2n−1 +1

2≥ 1 +

n− 1

2+

1

2= 1 +

n

2.

By Mathematical Induction, it is true for anyn, or

sn ≥ 1 +1

2log2 n, ∀n.

Since limn→∞

log2 n = +∞, {sn} diverges.


an =

(

1 +1

n

)n

, n = 1, 2, · · · ,

converges.

Proof. According to the Monotone Convergence Theorem, we only needto prove that the sequence is

bounded and monotone. Expandan:

an = 1 + n · 1n+n(n− 1)

2 · 11

n2+ · · ·+ n!

k!(n− k)!

1

nk+ · · ·+ n!

n!(n− n)!

1

nn

= 1 + 1 +1

2!

(

1− 1

n

)

+1

3!

(

1− 1

n

)(

1− 2

n

)

+ · · ·+

1

n!

(

1− 1

n

)

· · ·(

1− n− 1

n

)

Similarly, the expansion ofan+1 is:

an+1 =

(

1 +1

n+ 1

)n+1

= 1 + 1 +1

2!

(

1− 1

n+ 1

)

+1

3!

(

1− 1

n+ 1

)(

1− 2

n+ 1

)

+ · · ·+

1

(n+ 1)!

(

1− 1

n+ 1

)

· · ·(

1− n

n+ 1

)


Comparing term by term, we get

an < an+1.

Hence the sequence is monotonically increasing. Accordingto the above expansion ofan, we see that

2 < an < 1 + 1 +1

2+

1

22+ · · ·+ 1

2n−1< 3,

which means that{an} is bounded. Therefore the Monotone Convergence Theorem implies that the se-

quence{an} converges. The limit of this sequence is one of the importantconstants of mathematics and is

denoted by

e = limn→∞

(

1 +1

n

)n

.

Examples 2.2.5Let a1 = 1, and

an+1 =1 + an2 + an

, n = 1, 2, · · · ,

Show that{an} converges and find the limit.

Proof. It is easy to see that{an} > 0, ∀n, anda2 < a1. Since

an+2 − an+1 =1 + an+1

2 + an+1− 1 + an

2 + an

=an+1 − an

(2 + an)(2 + an+1)< 0, ∀n,

{an} is monotonically decreasing and bounded below.{an} converges by the Monotone Convergence

Theorem. Let limn→∞

an = a, then

a =1 + a

2 + a,

anda =−1 +

√5

2.

2.2.4 Nested Interval Theorem

Nested sequence of intervals.A sequence of intervals{In} that satisfies

In+1 ⊆ In

for every natural numbern is called nested sequence of intervals.

Closed Sets.A subsetS of R is said to be closed provided that whenever{an} is a sequence inS that

converges to a numbera, thena ∈ S.

Open Sets.A subsetS of R is said to be open provided that for anyx ∈ S, there exists aδ > 0 such that

(x − δ, x + δ) ⊆ S. (x − δ, x + δ) is called a neighborhood ofx, and a such pointx is called an interior

point.


Theorem 2.2.11 (Nested Interval Theorem)Suppose that a sequence of intervals{In} satisfies

(i) All intervalsIn = [anbn] are closed intervals.

(ii) The sequence of interval{In} is nested, i.e.an ≤ an+1 < bn+1 ≤ bn for everyn.

(iii) limn→∞

(bn − an) = 0.

Then there is exactly one pointC that belongs to all intervalsIn, and both the sequences{an} and{bn}converge toC.

Proof. From the assumption (ii), the sequences{an} and{bn} are both monotone and bounded. By the

Monotone Convergence Theorem,

limn→∞

an = l.u.b. {an},

and

limn→∞

bn = g.l.b. {bn}.

From the assumption (iii),

limn→∞

(bn − an) = limn→∞

bn − limn→∞

an = 0.

DenoteC = limn→∞

bn = limn→∞

an. Then

C = l.u.b.{an} = g.l.b. {bn}.

NamelyC is an upper bound ofan as well as a lower bound ofbn. Hence

an ≤ C ≤ bn ∀n.

Since allIn are closed intervals, the above inequality meansC ∈ In, ∀n.

Note: If the assumption (i) of the theorem is changed from closed intervals to open intervals. The conclu-

sion may not be true. For example, forIn =(

0, 1n

)

the nested sequence of intervals{In} satisfies the

assumptions (ii) and (iii). But there doesn’t exist any point that belongs to allIn, since

limn→∞

an = limn→∞

bn = 0,

and0 6∈ In, ∀n.

2.2.5 Heine-Borel Theorem

Theorem 2.2.12 (Heine-Borel Theorem)Let Iλ, λ ∈ Λ ⊆ R, be open intervals and the collection of sets

E = {Iλ | λ ∈ Λ} be an open cover of[a, b], i.e.,∀x ∈ [a, b], ∃λ ∈ Λ s.t. x ∈ Iλ. ThenE has a finite

subcover, i.e.,∃E1 ⊆ E,E1 is an open cover of[a, b] and the number of sets inE1 is finite.


Proof. Suppose that[a, b] can not be covered by a finite number of open intervals. Divide[a, b] into [a, a+b2 ]

and[a+b2 , b], then one of these two intervals can not be covered by a finite number of open intervals. Denote

this interval by[a1, b1] and divide[a1, b1] into two subintervals[a1, a1+b12 ] and[a1+b1

2 , b1]. Then one of

these two intervals can not be covered by a finite number of open intervals. Continue in this way, we get a

series of intervals[a1, b1], [a2, b2], [a3, b3], · · · . For anyn, [an, bn] can not be covered by a finite number

of open intervals, and the sequencean andbn satisfy

an ≤ an+1 ≤ bn+1 ≤ bn,

limn→∞

(bn − an) = limn→∞

1

2(bn−1 − an−1) = · · · = lim

n→∞1

2n(b − a) = 0.

From Nested Interval Theorem,∃c = limn→∞

an = limn→∞

bn. Thus,∃λ s.t. c ∈ Iλ. SinceIλ is open,∃ǫ s.t.

(c− ǫ, c+ ǫ) ⊆ Iλ.

Since limn→∞

an = limn→∞

bn = c, for thisǫ, ∃N s.t.

|an − c| < ǫ

2, |bn − c| < ǫ

2, ∀n ≥ N,

or,

[an, bn] ⊆ (c− ǫ, c+ ǫ), ∀n ≥ N,

i.e., [an, bn] can be covered by one open intervalIλ, which is a contradiction.

This theorem is also calledFinite Cover Theorem.

Examples 2.2.6Let In = (1

n, 2), andE = ∪∞

n=1In. Then,E is an open cover of(0, 1], butE has no finite

cover.

2.2.6 Bolzano-Weierstrass Theorem

Subsequence.Consider a sequence{an}. Let {nk} be a sequence of natural number and satisfies

n1 < n2 < n3 < · · ·

Then the sequence{ank} is called a subsequence of{an}.

Proposition 2.2.13Suppose that the sequence{an} converges to numbera, then any subsequence of{an}also converges toa.

Theorem 2.2.14 (Bolzano-Weierstrass Theorem)Every sequence in a bounded and closed interval[a, b]

has a subsequence that converges to a point in[a, b].

Proof. Let {xn} ⊆ [a, b]. We claim that there is a pointc ∈ [a, b] s.t. there are infinite many terms of

{xn} in any neighborhood ofc. Suppose that this is not true, then∀x ∈ [a, b], ∃δx s.t. there are only a


finite number of terms of{xn} in (x− δx, x+ δx) ≡ Ix. ThenE = {Ix | x ∈ [a, b]} forms an open cover

of [a, b]. From the Finite Cover Theorem, there exists a finite subcover E1. Since there are only a finite

number of terms in eachIx ⊆ E1, the total terms inE1 is also finite, i.e., only a finite number of terms in

[a, b], which is a contradiction. Now,∀k, take one term of{xn} from (c − 1/k, c+ 1/k) and denote it by

xnk. Then

|c− xnk| < 1

k, k = 1, 2, · · · ,

and

limk→∞

xnk= c.

2.2.7 Cauchy Convergence Criterion

Cauchy sequence.Sequence{an} is said to be a Cauchy sequence provided that for each positive number

ǫ, there is a natural numberN such that

|an − am| < ǫ, ∀n,m ≥ N.

Examples 2.2.7Sequence{ 1n} is a Cauchy sequence since

∣

∣

∣

∣

1

n− 1

m

∣

∣

∣

∣

<1

n, ∀n > m,

hence for anyǫ > 0, chooseN =[

1ǫ

]

+ 1 then 1n < ǫ if n ≥ N and

∣

∣

∣

∣

1

n− 1

m

∣

∣

∣

∣

< ǫ, ∀n,m ≥ N.

Examples 2.2.8Sequence{an} = {(−1)n} is not a Cauchy sequence.

Proof. Let ǫ0 = 1, then for anyN , letn0 = N ≥ N andm0 = N + 1 ≥ N , but

|an0 − am0 | = |(−1)N − (−1)N+1| = 2 > ǫ0,

i.e.,{an} is not a Cauchy sequence.

Theorem 2.2.15 (Cauchy Convergence Criterion Theorem)A sequence converges if and only if it is a

Cauchy sequence.

Proof. Necessity part:i.e., assume that{an} is a convergent sequence, we want to show that{an} is a

Cauchy sequence. Givenǫ > 0, we need to findN such that

|an − am| < ǫ, ∀n,m ≥ N.

Assumean → a, we may choose a natural numberN s.t.

|ak − a| < ǫ

2, ∀k ≥ N.


Thus, ifn,m ≥ N , the Triangle Inequality implies

|an − am| = |(an − a) + (a− am)|

≤ |an − a|+ |a− am| < ǫ

2+ǫ

2= ǫ.

which means that{an} is a Cauchy sequence.

Sufficiency part:i.e., assume that{an} is Cauchy sequence, we want to show that{an} is a convergent

sequence. We first show that Cauchy sequence is bounded. Takeǫ = 1, then there is aN such that

|an − am| < 1, ∀n,m ≥ N.

In particular

|an − aN | < 1, ∀n ≥ N.

Consequently, we see that

|an| ≤ |aN |+ 1, ∀n ≥ N.

Then for alln,

|an| ≤ max{|aN |+ 1, |a1|, |a2|, · · · , |aN−1|},

i.e.,{an} is a bounded sequence. Hence, by Bolzano-Weierstrass Theorem{an} has a subsequence{ank}

that converges toa. We now prove that the whole sequence{an} converges toa. Since{an} is a Cauchy

sequence, we can choose aN such that for any givenǫ > 0

|an − am| < ǫ

2, ∀n,m ≥ N.

Since{ank} → a, there is a natural numberK such that

|ank− a| < ǫ

2, ∀k ≥ K.

ChooseK such that whenk ≥ K,nk ≥ N . Then it follows that ifn ≥ N ,

|an − a| = |(an − ank) + (ank

− a)|

≤ |an − ank|+ |ank

− a| < ǫ

2+ǫ

2= ǫ.

which means that{an} converges toa.

Examples 2.2.9Let

an =

n∑

k=1

sin k

2k,

show that{an} converges.


Proof. Letm > n, then

|am − an| =∣

∣

∣

sin(n+ 1)

2n+1+ · · ·+ sinm

2m

∣

∣

∣

≤ 1

2n+1+ · · ·+ 1

2m<

1

2n.

For givenǫ > 0, letN = [log2 1/ǫ] + 1, then

|am − an| < ǫ, ∀m,n ≥ N,

i.e.,{an} is a Cauchy sequence and converges.

Example 2.2.10The sequence

Sn = 1 +1

2+

1

3+ · · ·+ 1

n, n = 1, 2, · · ·

diverges.

Proof. For m > n

|Sn − Sm| = 1

n+ 1+

1

n+ 2+ · · ·+ 1

m.

Let ǫ0 = 13 , then for anyN , letn0 = N ≥ N andm0 = 2N ≥ N , but

|Sm0 − Sn0 | =1

N + 1+

1

N + 2+ · · ·+ 1

2N> N · 1

2N=

1

2> ǫ0,

i.e.,{Sn} is not a Cauchy sequence, by Cauchy Convergence Criterion, it diverges.

So far we have introduced seven important theorems:

1. Dedekind Theorem

2. Least Upper Bound Theorem

3. Monotone Convergence Theorem

4. Nested Interval Theorem

5. Heine-Borel (Finite Cover) Theorem

6. Bolzano-Weierstrass Theorem

7. Cauchy Convergence Criterion Theorem

These seven theorems are all equivalent. To show this, we only need to prove Dedekind Theorem or Least

Upper Bound Theorem by using Cauchy Convergence Criterion Theorem, which is given below.

Proof of Least Upper Bound Theorem from Cauchy Convergence Criterion Theorem. Let the real

number set{xn} be nonempty and bounded above. We need to prove that{xn} has the least upper bound.


If {xn} has only finite elements or has maximum, then obviously it hasthe least upper bound. Hence, here

we assume that{xn} has infinite elements and no maximum.

Let b be an upper bound of{xn}. Takex1 to bea0 andb to beb0. Subdivide the interval[a0, b0] into

two subinterval[a0, a0+b02 ] and[a0+b0

2 , b0]. Because{xn} has no maximum, the interval[a0+b02 , b0] can’t

have finite but not zero elements of the set{xn}. That means there are only two possibilities:[a0+b02 , b0]

has none or infinite elements of{xn}. In former case, choosea1 = a0 andb1 = a0+b02 and in later case

choosea1 = a0+b02 , b1 = b0. Then the interval[a1, b1] contains infinite elements of{xn} andb1 is still the

upper bound of the set{xn}. Continuously to subdivide the interval[ak−1, bk−1] into two subintervals and

choose one of these two intervals to be[ak, bk] which has the properties:

(1) the interval[ak, bk] contains infinite elements of{xn}

(2) bk is still the upper bound of the set{xn}

By this way we get a sequence of intervals{Ik}, whereIk = [ak, bk], and satisfy:

(1) ak ≤ ak+1 < bk+1 ≤ bk for all k

(2) bk − ak =b0 − a0

2k, k = 0, 1, · · ·

(3) everybk is an upper bound of{xn}.

We now show that both the sequences{ak} and{bk} are Cauchy sequences. For allǫ > 0, we can choose

K such thatb0 − a0

2k< ǫ, ∀k ≥ K.

For thisK, if k′ > k ≥ K, due to

ak ≤ ak′ ≤ bk,

we get

(ak′ − ak) ≤ bk − ak =b0 − a0

2k< ǫ,

i.e.

|ak′ − ak| < ǫ, for k′ > k ≥ K.

Similarly

(bk − bk′) ≤ bk − ak =b0 − a0

2k< ǫ,

i.e.

|bk′ − bk| < ǫ, for k′ > k ≥ K.

By definition,{ak} and{bk} are Cauchy Sequences, and both converge. From the theorem about operations

on convergent sequences, we obtain

limk→∞

bk − limk→∞

ak = limk→∞

(bk − ak) = limk→∞

b0 − a02k

= 0,


i.e., limk→∞

ak = limk→∞

bk. Let numberC be the limit of both sequences. We show thatC is the least upper

bound of{xn}. Because everybk is an upper bound of{xn}, that means for every fixedxn, the inequality

xn < bk, for all k

holds. Ask → ∞, limk→∞

bk = C, hence for everyxn

xn ≤ C,

which means thatC is an upper bound of{xn}. Since limk→∞

ak = C, for anyǫ > 0, there must be someN

such that

aN > C − ǫ.

By the construction of the interval sequences{Ik}, the intervalIN = [aN , bN ] contains infinite elements

of {xn}, hence such elements are larger thanaN . Hence, the inequalityaN > C − ǫ implies(C − ǫ) is no

longer an upper bound of{xn}. Therefore, C is the least upper bound of{xn}.

2.3 Exercise

1. Give a direct′′ǫ−N ′′ verification of the following limits:

(a) limn→∞

1

n+ 5= 0.

(b) limn→∞

(

2√n+

1

n+ 3

)

= 3.

2. Suppose that the sequence{an} converges toa and thata > 0. Show that there is an indexN such

thatan > 0 for all indicesn ≥ N .

3. Prove that the sequence{cn} converges toc if and only if the sequence{cn − c} converges to0.

4. Prove that the Archimedean Property ofR is equivalent to the fact thatlimn→∞

1

n= 0.

5. Show that if{an} converges, thenlimn→∞

(an+1 − an) = 0.

6. Show that if limn→∞

bn = b < 0, then there is a natural numberN such thatbn <b

2∀n ≥ N.

7. Letan =1

n+ (−1)n. Show that lim

n→∞an does not exist.

8. Consider the sequencexn =2

n3, n = 1, 2, · · · ,

(a) Letǫk = 10−k with k = 1, 2, 3. FindNk such that|xn − 0| < ǫk, ∀n ≥ Nk.

(b) For anyǫ > 0, findN such that|xn − 0| < ǫ, ∀n ≥ N.

9. Consider the sequencexn = 3√n, n = 1, 2, · · · ,


(a) LetMk = 10k with k = 1, 2, 3. FindNk such thatxn > Mk, ∀n ≥ Nk.

(b) For anyM > 0, findN such thatxn > M, ∀n ≥ N.

10. Assume thatlimn→∞

an = a.

(a) Prove thatlimn→∞

|an| = |a|.

(b) If an ≥ 0 anda ≥ 0, prove that limn→∞

√an =

√a.

11. Prove thatlimn→∞

(

n2 +√n− 1

n− 1000

)

= ∞.

12. Discuss the convergence of each of the following sequences:

(a) limn→∞

(√n+ 1−√

n).

(b) limn→∞

[(n+ 1)1/3 − n1/3].

13. Find the limit:

(a) limn→∞

3√n2 sinn!

n+ 1.

(b) limn→∞

[

12

n3+

22

n3+ · · ·+ (n− 1)2

n3

]

.

(c) limn→∞

1 + a+ a2 + · · ·+ an

1 + b+ b2 + · · ·+ bn, (|a| < 1, |b| < 1).

(d) limn→∞

[

12

n3+

32

n3+ · · ·+ (2n− 1)2

n3

]

.

14. Prove the following:

(a) limn→∞

n

2n= 0.

(b) limn→∞

5n

n!= 0.

(c) limn→∞

n!

nn= 0. Hint. n!

nn ≤ (1/2)n/2 .

(d) limn→∞

nk

n!, wherek is a positive integer.

(e) limn→∞

n√n = 1. Hint. Defineαn = n1/n − 1, thenn = (1 + αn)

n ≥ 1 +n(n− 1)

2α2n.

15. LetSn =1

2 · 1 +1

3 · 2 + · · ·+ 1

(n+ 1) · n, n ≥ 1. Prove that limn→∞

Sn = 1.

16. Suppose that limn→∞

an = a. Define Gn =a1 + a2 + · · ·+ an

n. Prove that lim

n→∞Gn = a.

Hint. We may assumea = 0, otherwise, letbn = an − a. ∃N1, s.t. |an| < ǫ/2, ∀n ≥ N1. Then∣

∣

∣

1

n

n∑

i=1

ai

∣

∣

∣≤

∣

∣

∣

1

n

N1∑

i=1

ai

∣

∣

∣+

∣

∣

∣

1

n

n∑

i=N1+1

ai

∣

∣

∣≤

∣

∣

∣

1

n

N1∑

i=1

ai

∣

∣

∣+

ǫ

2. ∃N2, s.t.

∣

∣

∣

1

n

N1∑

i=1

ai

∣

∣

∣≤ ǫ

2.

17. Suppose that0 < a < 1.


(a) Prove thatlimn→∞

an = 0.

(b) If limn→∞

an = a, prove that limn→∞

ann = 0.

(c) Prove that limn→∞

nan = 0.


(a) limn→∞

nk

an= 0 (a > 1). Hint. (a1/k)n = (1 + δ)n ≥ n(n− 1)δ2/2, thenn/(a1/k)n → 0.

(b) limn→∞

n√a = 1 (a > 0). Hint. Let xn = n

√a− 1, thena = (xn + 1)n ≥ nxn, or,xn ≤ a/n → 0.

19. Show thatlimn→∞

(1

2· 34· · · 2n− 1

2n

)

= 0.


xn = a, andxn > 0, n = 1, 2, · · · , then limn→∞

n√x1x2 · · ·xn = a.


xn+1

xn= a, andxn > 0, n = 1, 2, · · · , then lim

n→∞n√xn = a.

Hint. Let an = xn+1/xn, thenan → a. n√xn = n

√x1

n√a1 · a2 · · · an/ n

√an. Let sn = n

√an − 1, then

an = (1 + sn)n ≥ nsn, or sn ≤ an/n → 0.


[

1 · 3 · · · (2n− 1)

2 · 4 · · · 2n

]1/n

= 1.


nn√n!

= limn→∞

(

1 +1

n

)n

= e.

24. Show that the set(−∞, 0] is closed.

25. Show thatb is the greatest lower bound ofS if and only if b is a lower bound ofS and∀ǫ > 0, ∃x0 ∈S such thatx0 < b+ ǫ.

26. Prove that for any irrational numberx, there exists a rational number sequence{rn} such that

limn→∞

rn = x.

27. Suppose that the sequence{an} is monotone. Prove that{an} converges if and only if{a2n} con-

verges. Show that this result does not hold without the monotonicity assumption.

28. Show that if the sequence{xn} converges, then any its subsequence{xpn} also converges and

converges to the same limit:limn→∞

xpn= lim

n→∞xn.

29. Show that if a subsequence of a monotonic sequence converges, then the monotonic sequence itself

converges.


xn = ∞, then there exists ak such thatxk = inf{xn}.

31. Prove the convergence of the following sequences by using the Monotone Convergence Theorem:

(a) xn = p0 +p110

+ · · ·+ pn10n

, n = 1, 2, · · ·, wherepi, i = 0, 1, 2, · · · , are nonnegative integers,

andpi ≤ 9, i = 1, 2, · · · .


(b) xn =(

1− 1

2

)(

1− 1

4

)

· · ·(

1− 1

2n

)

.

(c) xn =10

1· 113

· · · n+ 9

2n− 1.

(d) x1 =√2, x2 =

√

2 +√2, x3 =

√

2 +

√

2 +√2, · · · , xn =

√

2 +

√

2 + · · ·+√2, · · ·.

32. Determine whether the following sequences converge or not by using Cauchy Convergence Criterion.

(a) xn = a1q + a2q2 + · · ·+ anq

n, where|q| < 1 and|ai| < M .

(b) xn = 1 +1

2+

1

3+ · · ·+ 1

n.

(c) xn =sin θ

2+

sin 2θ

22+ · · ·+ sinnθ

2n, whereθ is a constant.

(d) xn = 1 +1

2+

1

22+ · · ·+ 1

2n.

(e) xn =cos 1!

1 · 2 +cos 2!

2 · 3 + · · ·+ cosn!

n · (n+ 1).

33. Prove that the sequencexn = 1 +1

2!+ · · ·+ 1

n!converges.

34. Letc > 0 andx1 > 0. Prove that the sequencexn+1 =√c+ xn, n ≥ 1 converges and find the

limit.

35. LetSn =

n∑

k=1

(−1)k−1

k, n ≥ 1. Show that{Sn} converges.

36. Prove the inequality:1

n+ 1< ln

(

1 +1

n

)

<1

n, n = 1, 2, · · · .

37. The sequence{xn} is defined by

x1 = a, x2 = b, xn =xn−1 + xn−2

2, n = 3, 4, · · · .

Find limn→∞

xn.

Hint. xn − xn−1 = − 12(xn−1 − xn−2), then,x3 − x2 = − 1

2(x2 − x1) = − 1

2(b − a),

x4 − x3 = (− 12)2(b− a), x5 − x4 = (− 1

2)3(b − a), · · · , xn − xn−1 = (− 1

2)n−2(b− a).

Then we havexn − x2 = (b− a)((−1/2) + (−1/2)2 + · · ·+(−1/2)n−2 = (b−a)(−1− (−1/2)n−2)/3,andxn → (a + 2b)/3.

38. Show that the sequence

xn = 1 +1

2+

1

3+ · · ·+ 1

n− lnn, n = 1, 2, · · ·

converges, and hence we have the formula

1 +1

2+

1

3+ · · ·+ 1

n= C + lnn+ ǫn,

where C = 0.577216 · · · is called the Euler Constant, andǫn → 0, as n → ∞.

Hint. ln(

1 +1

n

)

<1

n. xn =

n∑

i=1

1

n−

n∑

i=2

(ln i− ln(i− 1)) >1

n> 0.

xn − xn+1 = ln(

1 +1

n

)

− 1

n+ 1> 0.

Chapter 3: Continuity 53

39. Suppose that the sequences{xn} and{yn} are defined by

x1 = a, y1 = b, xn+1 =√xnyn, yn+1 =

xn + yn2

, n = 1, 2, · · · .

where0 ≤ a ≤ b. Show that{xn} and{yn} have the same limit

µ(a, b) = limn→∞

xn = limn→∞

yn,

whereµ(a, b) is the arithmetic-geometric mean ofa andb.

40. Let{x2k+1} and{x2k} be subsequences of{xn} and supposelimk→∞

x2k+1 = limk→∞

x2k = l. Prove

that limn→∞

xn = l.

41. Suppose that{an} is a sequence of nonnegative real numbers, and{an} has no converged subse-

quences. Prove that

limn→∞

an = +∞.

42. Suppose that the sequence{xn} is bounded. Prove that iflimn→∞

xn does not exist, then there exist

two subsequences{xmk} and{xnk

} of {xn} such that

limk→∞

xmk= a, lim

k→∞xnk

= b,

anda 6= b.

Hint. Since{xn} is bounded, from Bolzano-Weierstrass Theorem{xn} has a converged subsequencexnk→ b.

Since limn→∞

xn 6= b, then ∃ǫ0 > 0, s.t., for anyN , ∃xn(N), s.t., |xn(N) − a| ≥ ǫ0. The sequence

xn(N), N = 1, 2, · · · , is bounded, then it has a converged subsequencexmk→ a, and|xmk

−a| ≥ ǫ0. Takethe limit, we get|b− a| ≥ ǫ0, i.e.,a 6= b.


Chapter 3

Continuity

3.1 Limit of Function

3.1.1 Limit

Limit point. A point p is called a limit point of the setS if every neighborhood ofp contains a pointq 6= p

such thatq ∈ S.

Definition 3.1.1 (Limit of a function) Letf be a function defined in the domainD andx0 be a limit point

ofD. We say that the limit of functionf is l asx tends tox0, provided that for any givenǫ > 0, there exists

a δ > 0, such that forx ∈ D,

|f(x)− l| < ǫ, ∀ 0 < |x− x0| < δ.

In this case, we denote

limx→x0

f(x) = l.

Examples 3.1.1Prove that

limx→1

x2 − 1

−x2 + 3x− 2= 2.

Proof.∣

∣

∣

x2 − 1

−x2 + 3x− 2− 2∣

∣

∣ =3|x− 1||2− x| .

For givenǫ > 0, let δ1 = 1/2. Then, if|x− 1| < δ1, we have

|2− x| = |1 + 1− x| ≥ 1− |x− 1| > 1− δ1 =1

2,

55


and3|x− 1||2− x| < 6|x− 1|.

Let δ2 = ǫ/6, andδ = min{δ1, δ2}, then we have

∣

∣

∣

x2 − 1

−x2 + 3x− 2− 2∣

∣

∣ ≤ 6|x− 1| < ǫ, ∀|x− 1| < δ,

i.e., limx→1

x2 − 1

−x2 + 3x− 2= 2.

Left limit and right limit If x is restricted from left (right) hand side tending tox0, i.e.,x < x0(x > x0),

then the limitl is called left (right) limit, denoted by

limx→x0−0

f(x) = l

(

limx→x0+0

f(x) = l

)

It is easy to show that the limit exists if and only if both the left and right limits exist and equal.

Theorem 3.1.2Letf be a function defined inD andx0 is a limit point ofD. Then limx→x0

f(x) exists if and

only if limn→∞

f(xn) exists, where{xn} is an arbitrary sequence inD andxn → x0.

Proof. The necessary part:Assume limx→x0

f(x) = l, we want to show limn→∞

f(xn) = l, for any sequence

xn → x0. Since limx→x0

f(x) = l, then for givenǫ > 0, there exists aδ > 0, such that forx ∈ D,

|f(x)− l| < ǫ, if |x− x0| < δ.

Sincexn → x0, then forδ > 0, there exists aN > 0, such that,

|xn − x0| < δ ∀n ≥ N.

Thus we have

|f(xn)− l| < ǫ, ∀n ≥ N,

i.e., limn→∞

f(xn) = l.

The sufficient part:Assume limn→∞

f(xn) = l, for some sequencexn → x0, and letyn → x0 be any se-

quence. Since both{xn} and {yn} can be subsequences of another sequence{zn} (e.g., take terms

from {xn} and{yn} in turn.), limn→∞

f(zn) exists, andf(zn) → l, thenf(yn) → l. We want to show

limx→x0

f(x) = l. We argue by contradiction. Suppose the conclusion is not true, which means that for some

ǫ0 > 0, for anyδ > 0 we have

|f(x)− l| ≥ ǫ0, for somex satisfying|x− x0| < δ.

Hence forδn = 1n there is a pointan such that

|an − x0| < δn, but |f(an)− l| ≥ ǫ0.


Thus we obtain a sequence{an} in D that converges tox0 but {f(an)} does not converge tol. This

contradicts with the assumption.


limx→0

x sin1

x= 0.

Proof. Let {xn} be any sequence withxn 6= 0 and limn→∞

xn = 0. Since

0 ≤∣

∣

∣

∣

xn sin1

xn

∣

∣

∣

∣

≤ |xn| ,

By Squeezing Principle

limn→∞

∣

∣

∣

∣

xn sin1

xn

∣

∣

∣

∣

= 0.

Then from Theorem 3.1.2 we getlimx→0

x sin1

x= 0,


limx→∞

(

1 +1

x

)x

= e.

Proof. We have defined

limn→∞

(

1 +1

n

)n

= e,

wheren is a natural number. Let{kn} be any subsequence of{n}, then

limn→∞

(

1 +1

kn

)kn

= e.

Now for an arbitrary sequence{xn}, xn → ∞, we prove

limn→∞

(

1 +1

xn

)xn

= e.

Assume allxn > 1, let kn = [xn], we have

1 +1

kn + 1< 1 +

1

xn< 1 +

1

kn.

Then we have(

1 +1

kn + 1

)kn

<

(

1 +1

xn

)xn

<

(

1 +1

kn

)kn+1

.

Since

limn→∞

(

1 +1

kn + 1

)kn

= limn→∞

(

1 + 1kn+1

)kn+1

(

1 + 1kn+1

) =e

1= e,

limn→∞

(

1 +1

kn

)kn+1

= limn→∞

(

1 +1

kn

)kn

·(

1 +1

kn

)

= e · 1 = e.


By Squeezing Principle

limn→∞

(

1 +1

xn

)xn

= e.

Since{xn} is an arbitrary sequence such thatxn → ∞, therefore

limx→∞

(

1 +1

x

)x

= e.

Examples 3.1.4Let f(x) =√x, x ≥ 0. Show that

limx→0+0

√x = 0.

Proof. {xn} be any sequence withxn 6= 0 and limn→∞

xn = 0. Hence for any giveǫ > 0, let ǫ1 = ǫ2 and

chooseN such thatxn < ǫ1, ∀n ≥ N , then we have

√xn <

√ǫ1 = ǫ, ∀n ≥ N.

The result follows from Theorem 3.1.2.

Examples 3.1.5Prove that

limx→0

sinx

x= 1.

Proof. From figure 3.1.1, the areas of the triangle OCB, the sector OCB, and the triangle OCD satisfy

1

2|OC||AB| ≤ 1

2|OC|2|x| ≤ 1

2|OC||CD|, |x| < π

2.

x

O A

B

C

D

Figure 3.1.1

Let |OC| = 1, then|AB| = | sinx|, |CD| = | tanx|. Hence,

| sinx| ≤ |x| ≤ | tanx|, |x| < π

2.

Then we have

| cosx| = | sinx|| tanx| ≤

| sinx||x| ≤ 1, |x| < π

2.


Sincecos2 x =√

1− sin2 x and| sinx| < |x|, then

| cosx| ≥√

1− |x|2, |x| ≤ 1.

Sincesinx andx have a same sign wher|x| ≤ 1,

√

1− |x|2 ≤ sinx

x≤ 1, |x| ≤ 1.

Let {xn} be any sequence satisfyingxn 6= 0 and limn→∞

xn = 0. Then there is a natural numberN such that

|xn| ≤ 1, ∀n ≥ N.

Hence√

1− |xn|2 ≤ sinxnxn

≤ 1, n ≥ N.

By Squeezing Principle,

limn→∞

sinxnxn

= 1.

Therefore,

limx→0

sinx

x= 1.

Definition 3.1.3 (Limit at infinity) Let f(x) be defined on(a,+∞). We say thatf(x) converges tol asx

tends to+∞ if for givenǫ > 0, there exists a constantM such that

|f(x)− l| < ǫ, ∀x ≥M,

and denote by

limx→+∞

f(x) = l.

Similar to the limit of sequences, we can define other types oflimit.

3.1.2 Operations and properties

Similar to the limit of sequences, we have the following properties for the limit of functions.

Theorem 3.1.4Suppose that

limx→x0

f(x) = a, and limx→x0

g(x) = b.

Then

(i) limx→x0

[(f(x) ± g(x)] = limx→x0

f(x)± limx→x0

g(x) = a± b.

(ii) limx→x0

[f(x) · g(x)] = limx→x0

f(x) · limx→x0

g(x) = a · b.


(iii) If g(x) 6= 0, ∀x, andb 6= 0, then, limx→x0

f(x)

g(x)=

limx→x0

f(x)

limx→x0

g(x)=a

b.

Theorem 3.1.5 (Squeezing Principle)Suppose thatf(x) ≤ g(x) ≤ h(x), and

limx→x0

f(x) = limx→x0

h(x) = a.

then limx→x0

g(x) = a.

Examples 3.1.6Let k be a positive integer, prove that

limx→+∞

xk

ax= 0 (a > 1).

Proof. We may assume thatx > 1, then

0 ≤ xk

ax≤ ([x] + 1)k

a[x].

Since limn→∞

nk

an= 0, then

limx→+∞

([x] + 1)k

a[x]= lim

n→∞(n+ 1)k

an= 0.

By Squeezing Principle,limx→+∞

xk

ax= 0.

Definition 3.1.6The functionf : D → R is called monotonically increasingprovided that

f(v) ≥ f(u), ∀u, v ∈ D s.t. v > u.

The functionf : D → R is called monotonically decreasingprovided that

f(v) ≤ f(u), ∀u, v ∈ D s.t. v > u.

In both cases, if the equality does not hold, then the function is called strictly monotone

(strictly increasing or strictly decreasing).

Theorem 3.1.7 (Monotone Convergence Theorem)Suppose thatf(x) is defined on some interval(a, b)

and monotonically increasing. Iff(x) is bounded above, i.e.,

f(x) ≤M ∀x ∈ (a, b),

then limx→b−

f(x) exists.

Theorem 3.1.8 (Cauchy Convergence Criterion)limx→x0

f(x) exists if and only if for givenǫ > 0, ∃δ > 0,

such that for anyx andx′ satisfying0 < |x− x0| < δ and0 < |x′ − x0| < δ, we have

|f(x)− f(x′)| < ǫ.


Proof. Supposelimx→x0

f(x) = l, then for givenǫ > 0, ∃δ > 0, s.t.,

|f(x)− l| < ǫ

2, ∀0 < |x− x0| < δ,

|f(x′)− l| < ǫ

2, ∀0 < |x′ − x0| < δ.

Therefore,

|f(x)− f(x′)| ≤ |f(x)− l|+ |f(x′)− l| < ǫ. ∀0 < |x− x0| < δ, 0 < |x′ − x0| < δ.

On the other hand, letxn → x0 be any sequence. By assumption, for givenǫ > 0, ∃δ > 0, s.t.,

|f(x)− f(x′)| < ǫ, ∀0 < |x− x0| < δ, 0 < |x′ − x0| < δ.

For thisδ > 0, sincexn → x0, ∃N , s.t.,

0 < |xm − x0| < δ, 0 < |xn − x0| < δ ∀m,n ≥ N.

Then,

|f(xm)− f(xn)| < ǫ, ∀m,n ≥ N,

i.e.,{f(xn)} is a Cauchy sequence, and converges. By Theorem 3.1.2,limx→x0

f(x) exists.

3.2 Continuity

3.2.1 Definition

Let f be a function defined in the domainD andx0 ∈ D is a limit point ofD.

Definition 3.2.1 (Continuity) The functionf : D → R is said to be continuous at the pointx0 ∈ D

provided that

limx→x0

f(x) = f(x0).

In terms ofǫ− δ: For any givenǫ > 0, there exists aδ > 0, such that forx ∈ D,

|f(x)− f(x0)| < ǫ, ∀|x− x0| < δ.

The functionf : D → R is said to be continuous inD provided that it is continuous at every point inD.

3.2.2 Operations and composition on continuous functions

Theorem 3.2.2Suppose that the functionsf : D → R andg : D → R are continuous inD. Then

(i) f(x) ± g(x) is continuous inD;


(ii) f(x) · g(x) is continuous inD;

(iii) if g(x) 6= 0 in D, thenf(x)

g(x)is continuous inD.

Theorem 3.2.3 (Continuity of composition)Suppose thatf : D → R is continuous at the pointx0 ∈ D

andg : U → R is continuous at the pointf(x0) ∈ f(D) ⊆ U . Then the compositiong ◦ f is continuous

at x0.

Proof. Let {xn} be a sequence inD andxn → x0. By the continuity off at pointx0, we getf(xn) →f(x0). Then{f(xn)} is a sequence inU and converges tof(x0), so by the continuity ofg at pointf(x0)

we get

g(f(xn)) → g(f(x0)),

which means thatg ◦ f is continuous atx0.

3.2.3 Examples

Examples 3.2.1The function

Pn(x) =

n∑

k=0

akxn−k

is continuous everywhere.

Proof. It is obvious thatf(x) = x is continuous in(−∞,∞). Hence, by the theorem of operations on

continuous functions,Pn(x) is continuous in(−∞,∞).

Examples 3.2.2f(x) = sinx is continuous everywhere.

Proof. For anyx0 ∈ (−∞,∞)

sinx− sinx0 = 2 sinx− x0

2cos

x− x02

.

It has been proved that| sinx| ≤ |x|, then

| sinx− sinx0| ≤ 2

∣

∣

∣

∣

sinx− x0

2

∣

∣

∣

∣

≤ 2|x− x0|

2= |x− x0|.

Hence,∀ǫ > 0, chooseδ = ǫ, then

| sinx− sinx0| ≤ |x− x0| < ǫ, if |x− x0| < δ,

which means thatf(x) is continuous atx0.

Examples 3.2.3f(x) =√x is continuous in[0,∞).

Proof. It has been provedf(x) is continuous at the point0. Forx0 > 0, we have

|√x−√x0| =

|x− x0||√x+

√x0|

≤ |x− x0|√x0

.


∀ǫ > 0, chooseδ =√x0ǫ, then

|√x−√x0| < ǫ, if |x− x0| < δ,

which means thatf(x) is continuous atx0.

Examples 3.2.4cosx is continuous everywhere;tanx is continuous in(

−π2 ,

π2

)

.

Proof. Proof: Letf(x) = 1− sin2 x andg(x) =√x. f(x) is continuous in(−∞,∞), g(x) is continuous

in [0,∞), andf((−∞,∞)) ⊆ D(g). Since

cosx =

√

1− sin2 x = g ◦ f(x), x ∈[

−π2 + 2kπ, π2 + 2kπ

]

−√

1− sin2 x = −g ◦ f(x), x ∈[

π2 + 2kπ, 3π2 + 2kπ

]

,

k = 0, 1, · · · . By the theorem of composition on continuous functions,cosx is continuous everywhere.

Since

tanx =sinx

cosx, x ∈

(

−π2,π

2

)

,

it is continuous in(

−π2 ,

π2

)

due to the continuity ofsinx andcosx.

Examples 3.2.5The function

f(x) =

sin xx , x 6= 0,

1, x = 0

is continuous everywhere.

Proof. By the theorem of operations on continuous functions,sin xx is continuous everywhere exceptx = 0.

We have proved that

limx→0

sinx

x= 1.

Sincef(0) = 1, we have

limx→0

f(x) = f(0),

i.e.,f(x) is continuous atx = 0.

Examples 3.2.6Let

f(x) = limn→∞

x2n − 1

x2n + 1,

thenf(x) is not continuous (discontinuous) atx = ±1.

Examples 3.2.7Let

f(x) =

1x3 , x 6= 0,

0, x = 0,

thenf(x) is not continuous atx = 0.


Examples 3.2.8Let X : R → R (Dirichlet function) be defined by

X (x) =

1, x is rational,

0, x is irrational,

thenX (x) is not continuous everywhere.

Proof. In casex0 is a rational number,X (x0) = 1. Let ǫ0 = 12 . By the density theorem, no matter what the

δ is chosen, the interval(x0 − δ, x0 + δ) always contains an irrational numberx. SinceX (x) = 0, we have

|X (x)− X (x0) = 1 > ǫ0,

i.e.,X (x) is not continuous atx0. Similarly, we can prove the case of an irrational number.

3.3 Extreme Value Theorem

In sections 2-4, we will apply the theory mentioned in the first two chapters to prove three important

theorems about continuous functions defined in a bounded closed interval.

3.3.1 Maximizer and minimizer

Definition 3.3.1We say that a functionf : D → R attains a maximum (minimum) provided that its image

f(D) has a maximum (minimum); that is, there is a pointx0 (not necessary only one) inD such that

f(x) ≤ f(x0) (f(x) ≥ f(x0)), ∀x ∈ D.

We call such a point inD a maximizer (minimizer) of the functionf .

In general, no assertion can be made concerning the existence of a maximum or minimum. But we have

the following important theorem about a continuous function defined on a bounded closed interval[a, b].

3.3.2 Extreme value theorem

Theorem 3.3.2 (Extreme Value Theorem)Suppose thatf : [a, b] → R is continuous. Thenf attains both

a minimum and a maximum in[a, b].

We first to prove a weaker result as the following lemma.

Lemma 3.3.3Suppose thatf : [a, b] → R is continuous. Then the image off : [a, b] → R is bounded;

that is, there is a numberM such that

|f(x)| ≤M, ∀x ∈ [a, b].


Proof. We will argue by contradiction. Iff([a, b]) is not bounded above, then for every natural numbern,

there must be a pointxn such that

f(xn) > n, n = 1, 2, 3, · · ·

Sincea ≤ xn ≤ b, by Bolzano-Weierstrass Theorem, there is a subsequence{xnk} of {xn} converges to a

pointx0 ∈ [a, b]. Hence on one hand,f(xnk) > nk i.e. {f(xn)} has no upper bound; on the other hand,

due to continuity off, f(xnk) → f(x0).

This contradiction show thatf([a, b]) is bounded above. Similarly we can prove thatf([a, b]) is

bounded below. If it is not true, then we may choose a sequence{xn} such that

f(xn) < −n, n = 1, 2, 3, · · ·

Then we have a convergent subsequence{xnk} that on one hand{f(xn)} has no lower bound and on the

other handf(xnk) → f(x0) asxnk

→ x0.

Proof of Theorem 3.3.2Sincef([a, b]) is bounded above, according to the Completeness Axiom,f([a, b])

has the least upper bound. DefineC = sup f([a, b]). Now, we want to find a pointx0 ∈ [a, b] such that

f(x0) = C. Then

f(x0) = max f [a, b],

or,

f(x0) ≥ f(x), ∀x ∈ [a, b].

SinceC is the least upper bound off([a, b]), for any natural numbern,C− 1n is not an upper bound. Hence

there is a sequence of pointsxn at which

f(xn) > C − 1

n, n = 1, 2, 3, · · ·

Then we have

C − 1

n< f(xn) ≤ C,

it follows that the sequence{f(xn)} converges toC. Again according to Bolzano-Weierstrass theorem,

there is a subsequence{xnk} of {xn} that converges to a pointx0 ∈ [a, b]. Sincef is continuous on[a, b],

{f(xnk)} converges tof(x0). But {f(xnk

)} is a subsequence of{f(xn)}, so the limit of{f(xn)} is also

the limit of {f(xnk)}, i.e. f(x0) = C. Thus we have proved thatx0 is a maximizer andf(x0) attains the

maximum.

Becausef([a, b]) is bounded below,f([a, b]) has a greatest low bound. In a similarly way by Bolzano-

Weierstrass theorem we can find a minimizer at whichf attains a minimum.


Note: If a continuous function defined in an open interval. It may not have maximum and minimum. The

reason is that iflimn→∞

xn = x0 anda < xn < b, the limitx0 may be the end pointa or b.

Example 3.3.1Let f(x) = x, 0 < x < 1. The functionf is continuous in(0, 1), but it has no maximum

and minimum in the interval.

3.3.3 Compactness

Definition 3.3.4A setS is said to be compact (sequentially compact) provided that every sequence inS has

a subsequence that converges to a point ofS.

In this new terminology, the Bolzano-Weierstrass Theorem is simply the assertion that the set of a

bounded closed interval[a, b] is compact.

Theorem 3.3.5Suppose thatS is a compact nonempty set of real numbers andf : K → R is continuous.

Thenf : S → R attains both a minimum and a maximum.

3.4 Intermediate Value Theorem

Theorem 3.4.1Suppose that the functionf : [a, b] → R is continuous, and

f(a) < 0, f(b) > 0.

Then there is at least a pointx0 in the open interval(a, b) at whichf(x0) = 0.

Proof. Let a1 = a, b1 = b andc1 = 12 (a1 + b1).

If f(c1) ≤ 0, definea2 = c1 andb2 = b1.

If f(c1) > 0, definea2 = a1 andb2 = c1.

In general, letcn = 12 (an + bn).

If f(cn) ≤ 0, definean+1 = cn andbn+1 = bn.

If f(cn) > 0, definean+1 = an andbn+1 = cn.

Hence we define a sequence of nested, closed intervals{[an, bn]} that satisfiesf(an) ≤ 0, f(bn) > 0,

a ≤ an ≤ an+1 < bn+1 ≤ bn ≤ b,

and

bn − an =1

2(bn−1 − an−1) = · · · = 1

2n−1(b− a).


Thus the sequences{an} and{bn} satisfy the assumptions of the Nested Interval Theorem, so there is a

pointx0 ∈ [a, b] such thatan → x0 andbn → x0. Sincef(x) is continuous on[a, b],

limn→∞

f(an) = limn→∞

f(bn) = f(x0).

Sincef(an) ≤ 0, ∀n, we havef(x0) ≤ 0, and on the other hand we havef(x0) ≥ 0 due tof(bn) ≥ 0, ∀n.

Hencef(x0) = 0.

Theorem 3.4.2 (Intermediate Value Theorem)Suppose that the functionf : [a, b] → R is continuous

andc is a number strictly betweenf(a) andf(b), i.e.,

f(a) < c < f(b), or f(a) > c > f(b).

Then there is a pointx0 in the open interval(a, b) at whichf(x0) = c.

Proof. First let us suppose thatf(a) < c < f(b). Define the functiong(x) = f(x)− c. Then

g(a) < 0 < g(b).

It follows from Theorem 3.3.1 that there is a pointx0 ∈ (a, b) at whichg(x0) = 0, i.e.

g(x0) = f(x0)− c = 0, or f(x0) = c.

In the case whenf(a) > c > f(b), defineg(x) = c− f(x) and follow the same argument.

Example 3.4.1Let the integerk > 0 be odd, then the equation

akxk + ak−1x

k−1 + · · ·+ a0 = 0

has at least one real root.

Proof. We may assume thatak > 0. Let

f(x) = akxk + ak−1x

k−1 + · · ·+ a0.

Then, limx→+∞

f(x) = +∞, limx→−∞

f(x) = −∞. There existsM > 0, s.t.,f(−M) < 0, f(M) > 0. By

Intermediate Value Theorem, there is a pointx0 ∈ (−M,M), s.t.,f(x0) = 0.

3.5 Uniform Continuity

Suppose that a functionf : D → R is continuous inD. According to theǫ− δ definition of continuity,

for x0 ∈ D and givenǫ > 0, there exists aδ > 0, such that forx ∈ D,

|f(x)− f(x0)| < ǫ, ∀|x− x0| < δ.


In general,δ depends not only onǫ, but also onx0, i.e., at different points, we need to choose differentδ

such that the above inequality is true. Sometimes theδ may need to tend to0 even for a fixedǫ, for example,

f(x) = 1/x, x ∈ (0, 1). Then we may ask the question: What extra conditions are needed in order to find

aδ which is good for anyx0 ∈ D?

Definition 3.5.1 (Uniformly continuity) Suppose that the functionf(x) is defined on an intervalD. f(x)

is said to be uniformly continuous inD provided that for any givenǫ > 0, ∃δ > 0, s.t.

|f(u)− f(v)| < ǫ, ∀u, v ∈ D, |u− v| < δ.

If D is an open interval,f : D → R is continuous inD, butf may not be uniformly continuous inD.

Example 3.5.1Let f(x) = 1x , D = (0, 1), thenf(x) is continuous inD, but not uniformly continuous in

D. Since for anyu, v ∈ D,∣

∣

∣

∣

1

u− 1

v

∣

∣

∣

∣

=

∣

∣

∣

∣

u− v

uv

∣

∣

∣

∣

.

Takeǫ0 = 1, then for anyδ > 0, let 0 < u0 < δ andv0 = u0

2 , we have|u0 − v0| = u0

2 < δ, but

|f(u0)− f(v0)| =∣

∣

∣

∣

1

u0− 1

v0

∣

∣

∣

∣

=1

u0> 1 = ǫ0.

If D is a bounded closed interval, we have the following theorem.

Theorem 3.5.2A continuous function on a closed bounded interval,f : [a, b] → R, is uniformly continu-

ous.

Proof. We will argue by contradiction. LetD = [a, b]. Suppose thatf : D → R is not uniformly

continuous. Then there is someǫ0 > 0 such that∀δ > 0

|f(u)− f(v)| ≥ ǫ0, for someu, v ∈ D, |u− v| < δ.

Let δ = 1n , then there exist pointsun andvn in [a, b] such that|un − vn| < 1

n , but

|f(un)− f(vn)| ≥ ǫ0, ∀n.

This defines two sequences{un} and{vn} in [a, b]. By using Bolzano-Weierstrass Theorem, we can choose

a subsequence{unk} of {un} that converges to a pointu∗ ∈ [a, b]. But for each natural numberk,

|unk− vnk

| < 1

nk≤ 1

k,

or

unk− 1

k< vnk

< unk+

1

k.

By Squeeze Principle,{vnk} also converges tou∗. The assumption of continuity off at pointu∗ implies

that both{f(unk)} and{f(vnk

)} converge tof(u∗). Thus their difference{f(unk) − f(vnk

)} converges


to 0. This contradicts with the inequality|f(unk)− f(vnk

)| ≥ ǫ0. It follows thatf : D → R is uniformly

continuous.

3.6 Inverse Function

Definition 3.6.1A functionf : D → R is said to be one-to-one provided that for each pointy in its image

f(D), there is exactly one pointx in its domainD such thatf(x) = y. In other words, iff(x1) = f(x2),

thenx1 = x2. For a one-to-one functionf : D → R, we define its inverse, denoted by

f−1 : f(D) → R,

as following:

f−1(y) = x, if f(x) = y.

It is clear that iff : D → R is strictly monotone, it is one-to-one and its inversef−1 : f(D) → R is

also strictly monotone. In fact, iff is strictly increasing, thenf−1 is also strictly increasing; iff is strictly

decreasing, thenf−1 is also strictly decreasing.

Theorem 3.6.2LetD be an interval and suppose that the functionf : D → R is monotone. If its image

f(D) is an interval, then the functionf is continuous.

Proof. Since−f is increasing iff is decreasing, we only need to consider the case whenf is increasing.

For anyx0 ∈ D, it is easy to show that limx→x0−0

f(x) = f(x0 − 0) and limx→x0+0

f(x) = f(x0 + 0) exist.

Suppose thatf(x) is not continuous atx0, for example,f(x0− 0) < f(x0). Sincef(x) ≤ f(x0− 0) when

x < x0 andf(x) ≥ f(x0) whenx > x0, thenf(x) can not take the values betweenf(x0 − 0) andf(x0),

which contradicts with the assumption thatf(D) is an interval.

Theorem 3.6.3Let I be an interval and suppose that the functionf : I → R is strictly monotone. Iff(I)

is an interval, then the inverse functionf−1 : f(I) → R is continuous.

Proof. Suppose thatf is strictly increasing. LetD = f(I). Then the functionf−1 : D → R is a

monotonically increasing function that has the intervalI as its image. From Theorem 3.6.3,f−1 : D → R

is continuous.

Theorem 3.6.4Let I be an interval and the functionf : I → R is strictly monotone and continuous. Then

the inverse functionf−1 : f(I) → R is continuous and the definition domainf(I) is an interval.

Proof. Let y1 = f(x1), y2 = f(x2) ∈ f(I), x1 < x2, andc be any number in betweeny1 andy2.

Then from the Intermediate Value Theorem there exits a pointx0 ∈ (x1, x2) such thatf(x0) = c. Hence

c ∈ f(I), i.e.,f(I) is an interval. Sincef(I) is an interval, by Theorem 3.6.3.f−1 is continuous.


Example 3.6.2Definef : [0,∞) → R, asy = f(x) = x2. The inversex = f−1(y) =√y. We have

proved that the function√y is continuous in[0,∞) by the definition of continuity. Now, we apply Theorem

3.6.4 to prove this. Becausef(x) is strictly increasing and continuous in[0,∞) and the image off is

f([0,∞)) = [0,∞). Hence by the theorem,f−1(y) =√y is continuous in[0,∞).

Example 3.6.2Inverse functionssin−1 x, cos−1 x andtan−1 x.

We have proved the continuity ofsinx, cosx andtanx. Sincesinx is strictly increasing in[

−π2 ,

π2

]

and the image is[−1, 1], hence its inversesin−1 x is continuous in[−1, 1] and also strictly increasing.

Becausecosx is strictly decreasing in[0, π] and the image is[−1, 1], hence its inversecos−1 x is

continuous in[−1, 1] and also strictly decreasing.

Becausetanx is strictly increasing in[

−π2 ,

π2

]

and the image is(−∞,∞). Hence its inversetan−1 x

is continuous on(−∞,∞) and also strictly increasing.

3.7 Continuity of Elementary Functions

In section 1 and 6, we have seen the continuity of polynomialsPn(x), the trigonometric functionssinx,

cosx, andtanx and their inverses. Using the properties of the continuous functions, we can easily show

that the rational functions and other trigonometric functions are all continuous in their domains. In this

section, we discuss the continuity of some other elementaryfunctions, such as power functions, logarithmic

functions, and exponential functions. Then from the theorems for operations on continuous functions, for

inverse functions, and for compositions, we know that any combinations and operations (including inverse

and composition) on these functions result in also continuous functions.

3.7.1 Power Function with Rational Exponential

• y = x1n , x ≥ 0, n is a natural number.

Sincex = yn is a strictly increasing function in interval[0,∞) and its image is also interval[0,∞),

its inverse functiony = x1n is continuous in[0,∞).

• y = xmn , x ≥ 0,m,n are natural numbers.

This function can be treated as a composite function ofz = x1n with y = zm. Hencey = x

mn is

continuous in[0,∞).

3.7.2 Exponential Function

• y = ax, a > 0.

Functiony = aγ has been defined whenγ is rational. Let’s temporarily assumea > 1. We consider

f(γ) = aγ in the rational setQ. Thenf(γ) has the following properties.


(i) aγ1+γ2 = aγ1 · aγ2

(ii) f(γ) is strictly increasing, i.e., aγ1 < aγ2 , if γ1 < γ2.

(iii) For any sequence{γn} in Q that converges to0, the sequence{f(γn)} converges toa0 = 1.

(iv) If {γn} is a convergent sequence,{aγn} is a Cauchy sequence.

• Definition of ax. For any real numberx, let {γn} be a rational sequence and converges tox. It is

known from the above property (iv) that{aγn} is a Cauchy sequence. We define

ax = limn→∞

aγn .

For this definition, we need to prove that for any other rational sequence{γ′n} that converges tox,

limn→∞

aγ′

n = limn→∞

aγn .

Indeed

limn→∞

aγ′

n = limn→∞

(

aγ′

n − aγn

)

+ limn→∞

aγn

= limn→∞

aγn

(

aγ′

n−γn − 1)

+ limn→∞

aγn

= limn→∞

aγn · limn→∞

(

aγ′

n−γn − 1)

+ limn→∞

aγn

= limn→∞

aγn .

Hence, fora > 1 we have defineax on the whole real axis, i.e. the interval(−∞,∞).

• Properties of Functionax

(i) For any real numbersx1 andx2, ax1+x2 = ax1 · ax2 .

(ii) For any real numberx, ax > 0.

(iii) y = ax (a > 1) is strictly increasing.

Lemma 3.6.1y = ax is continuous atx = 0.

Proof. We need to show that for any real sequence{xn} that converges to0, {axn} converges toa0 = 1.

By the density property of rational numbers, we can choose two rational sequences{γn} and{γ′n} that

satisfy

xn − 1

n< γn < xn < γ′n < xn +

1

n.

whena > 1, ax is strictly increasing, hence

aγn < axn < aγ′

n .


Sinceγn → 0 andγ′n → 0 thenaγn → 1 andaγ′

n → 1. By using Squeezing Principle

limxn→0

axn = 1.

Up to now, we have assumed thata > 1. If 0 < a < 1, we define

ax = (a−1)−x,

and ifa = 1, we define

ax = 1.

Obviously,y = ax is strictly decreasing, continuous atx = 0, and satisfies (i), (ii).

Theorem 3.6.2The exponential functiony = ax (a > 0) is continuous on(−∞,∞) and strictly increasing

(decreasing) whena > 1 (a < 1). Its image is the interval(0,∞)

Proof. Proof For any pointx0 ∈ (−∞,∞)

ax − ax0 = ax0(ax−x0 − 1).

Since limx→0

ax = 1, we get

limx→x0

(ax − ax0) = limy→0

ax0(ay − 1) = 0,

which means thatax is continuous atx = x0.

3.7.3 Logarithmic Function

y = loga x andy = lnx: It has been known that the functionx = ay is a strictly monotone and

continuous function from(−∞,∞) to (0,∞). Now we define its inverse byy = loga x (y = lnx for

a = e), which is from(0,∞) to (−∞,∞). Sinceax (ex) is strictly monotone, its inverseloga x (ln x) is

continuous and also strictly monotone on(0,∞).

Example 3.7.1Show thatlimx→0

ln(1 + x)

x= 1.

Proof. By the definition of Logarithm,

1

xln(1 + x) = ln(1 + x)

1x .

Hence, letf(x) = (1 + x)1x andg(y) = ln y. Since

limx→0

(1 + x)1x = lim

y→∞

(

1 +1

y

)y

= e,

then

limx→0

g(f(x) = g(

limx→0

f(x))

= g(e) = ln e = 1.


3.7.4 Power Function

y = xα (α is a real number.) : This function can be defined as a composition of y = f(x) = α ln x

with g(y) = ey, i.e., (g ◦ f)(x) = g(f(x)) = eα ln x. Sincelnx is continuous in(0,∞) and the image

is (−∞,∞), ex is continuous in(−∞,∞). Hence by the theorem of continuity on composite functions,

y = xα = eα lnx is continuous in(0,∞).

3.8 Exercise

1. Letf(x) be defined by

f(x) =

n, if x = mn , wherem andn > 0 are integers and have no common factors besides1,

0, if x is irrational.

Show thatf(x) is finite for any givenx, but not bounded, i.e.,f(x) is unbounded in any interval

containingx.

2. Show that the functionf(x) =1 + x2

1 + x4is bounded in(−∞,+∞).

3. Show that the functionf(x) =1

xcos

1

xis unbounded in any interval containing0, butf(x) does not

tend to∞ asx→ 0.

4. Discuss the boundedness of the functionf(x) = lnx sin2π

xfor 0 < x < ǫ.

5. Show that the functionf(x) =x

1 + xhas theinf f(x) = 0 andsup f(x) = 1 in [0,+∞).

6. Find theinf andsup of the following functions:

(a) f(x) = x2, x ∈ (−2, 5).

(b) f(x) =1

1 + x2, x ∈ (−∞,+∞).

(c) f(x) =2x

1 + x2, x ∈ (0,+∞).

(d) f(x) = x+1

x, x ∈ (0,+∞).

(e) f(x) = 2x, x ∈ (−1, 2).

(f) f(x) = sinx, x ∈ (0,+∞).

(g) f(x) = sinx+ cosx, x ∈ [0, 2π].

(h) f(x) = [x], x ∈ (0, 2), or x ∈ [0, 2].

(i) f(x) = x− [x], x ∈ [0, 1].

7. Letm[f ] andM [f ] denote the greatest lower bound and the smallest upper boundof f(x). Prove

that if f1(x) andf2(x) are functions defined on(a, b), then

m[f1 + f2] ≥ m[f1] +m[f2], M [f1 + f2] ≤M [f1] +M [f2].

Give examples for the cases of equality and strict inequality.

8. Letf(x) =√1− x2, |x| ≤ 1. Show thatlim

x→0f(x) = 1.


9. Letf(x) =

1, x is irrational,

1n , x is rational andx = m

n ,

wherem andn are integers with no common factors other than1. Prove thatlimx→1

f(x) does not

exist.

10. Letf(x) = x3, x ∈ R. Verify theǫ− δ criterion for limx→x0

f(x) = x30 at

(a)x0 = 2, (b)x0 = 50, (c) any pointx0.

11. Letf(x) =1

x, x 6= 0. Verify theǫ− δ criterion for lim

x→x0

f(x) =1

x0at

(a)x0 = 10, (b)x0 = 1/10, (c) any pointx0 6= 0.

12. Give a directǫ− δ verification of limx→2

x2 = 4, Find δ for ǫ = 0.1, 0.01, 0.001.

13. Give a directM − δ verification of limx→1

1

(1 − x)2= +∞, Find δ for M = 10, 100, 1000.

14. Prove thatlimx→0

|x|2x

= 0.

15. LetP (x) = a0xn + a1x

n−1 + · · · + an, whereai, i = 0, 1, · · · , n, are real numbers. Show that

limx→∞

|P (x)| = +∞.

16. LetR(x) =a0x

n + a1xn−1 + · · ·+ an

b0xm + b1xm−1 + · · ·+ bm,

wherea0 6= 0, b0 6= 0. Show that limx→∞

R(x) =

∞, if n > m,

a0

b0, if n = m,

0, if n < m.

17. Find the limits:

(a) limx→1

x2 − 1√x− 1

. (b) limx→2

x4 − 16

x− 2.

18. Find the limits: (m andn are natural numbers.)

(a) limx→0

(1 + x)5 − (1 + 5x)

x2 + x5.

(b) limx→0

(1 +mx)n − (1 + nx)m

x2.

(c) limx→∞

(x+ 1)(x2 + 1) · · · (xn + 1)

[(nx)n + 1](n+1)/2.

(d) limx→3

x2 − 5x+ 6

x2 − 8x+ 15.

(e) limx→2

x3 − 2x2 − 4x+ 8

x4 − 8x2 + 16.

(f) limx→1

x+ x2 + · · ·+ xn − n

x− 1.

(g) limx→1

√x− 1

x− 1.

(h) limx→1

xm − 1

xn − 1.

(i) limx→a

(xn − an)− nan−1(x− a)

(x− a)2.

(j) limx→1

xn+1 − (n+ 1)x+ n

(x− 1)2.

(k) limx→1

(

m

1− xm− n

1− xn

)

.

(l) limn→∞

(

13 + 23 · · ·+ n3

n3− n

4

)

.

(m) limn→∞

13 + 43 + 73 + · · ·+ (3n− 2)3

[1 + 4 + 7 + · · ·+ (3n− 2)]2.



(a) limx→∞

1

n

[

(

x+a

n

)

+(

x+2a

n

)

+ · · ·+(

x+(n− 1)a

n

)

]

.

(b) limx→∞

1

n

[

(

x+a

n

)2

+(

x+2a

n

)2

+ · · ·+(

x+(n− 1)a

n

)2]

.


(a) limx→+∞

√

x+√

x+√x

√x+ 1

.

(b) limx→+∞

√x+ 3

√x+ 4

√x√

2x+ 1.

(c) limx→4

√1 + 2x− 3√x− 2

.

(d) limx→−8

√1− x− 3

2 + 3√x

.

(e) limx→a

√x−√

a+√x− a√

x2 − a2.

(f) limx→3

√x+ 13− 2

√x+ 1

x2 − 9.

21. Find the limits: (n is an integer.)

(a) limx→−2

3√x− 6 + 2

x3 + 8.

(b) limx→16

4√x− 2√x− 4

.

(c) limx→∞

√9 + 2x− 5

3√x− 2

.

(d) limx→0

n√1 + x− 1

x.

(e) limx→0

√1− 2x− x2 − (1 + x)

x.

(f) limx→0

3√8 + 3x− x2 − 2

x+ x2.

22. Find the limits: (m andn are integers.)

(a) limx→0

3√27 + x− 3

√27− x

x+ 23√x4

.

(b) limx→0

√1 + x−

√1− x

3√1 + x− 3

√1− x

.

(c) limx→7

√x+ 2− 3

√x+ 20

4√x+ 9− 2

.

(d) limx→0

m√1 + αx n

√1 + βx− 1

x.

(d) limx→0

3√

1 + x3 − 4

√

1 + x4

1−√

1− x2

.

(e) limx→0

x2

5√1 + 5x− (1 + x)

.

(f) limx→0

m√1 + αx− n

√1 + βx

x.


(a) limx→0

1− cosx

x2. (b) lim

x→0

tanx− sinx

x3.

24. Prove that limx→+∞

ax

x= +∞ (a > 1).

25. (Monotone convergence theorem)Suppose thatf(x) is defined on some interval(a, b) and mono-

tonically increasing. Iff(x) is bounded above, i.e.,f(x) ≤ M ∀x ∈ (a, b), prove that limx→b−

f(x)

exists.

26. LetP (x) = a1x+ a2x2 + · · ·+ anx

n andm is an integer. Show thatlimx→0

m√

1 + P (x) − 1

x=a1m.

27. Useǫ− δ criterion to prove the continuity of the following functions:


(a) f(x) = ax+ b, −∞ < x < +∞.

(b) f(x) = x2, −∞ < x < +∞.

(c) f(x) = x3, −∞ < x < +∞.

(d) f(x) =√x, 0 ≤ x < +∞.

(e) f(x) = 3√x, −∞ < x < +∞.

(f) f(x) = sinx, −∞ < x < +∞.

(g) f(x) = cosx, −∞ < x < +∞.

(h) f(x) = tan−1 x, −π2< x < +

π

2.

28. Definef(x) =√x, ∀x ≥ 0. Verify theǫ− δ criterion for continuity atx = 4 andx = 100.

29. Definef(x) =

x+ 1, if x ≤ 3/4,

2, if x > 3/4.

Use theǫ− δ criterion to show thatf is not continuous atx = 3/4.

30. Defineh(x) = 1/(1 + x2), ∀x. Verify theǫ− δ criterion for continuity at each pointx0.

31. A functionf : D → R is called a Lipschitz function if there is a constantc ≥ 0 such that

|f(u)− f(v)| ≤ c|u− v| ∀u, v ∈ D. Prove that a Lipschitz function is continuous inD.

32. Letf : D → R be continuous. Prove that the functiong(x) = |f(x)| is continuous inD.

33. Suppose thatf(x) is continuous andf(x) ≥ 0 in D. Prove that the functiong(x) =√

f(x) is

continuous inD.

34. Letf(x) =

x2, x ≥ 0,

x, x < 0.Show thatf : R → R is continuous.

35. Letg(x) =

x2, if x is rational,

−x2, if x is irrational.At what points is the function continuous? Justify your answer.

36. Suppose that the functionf : R → R is continuous at the pointx0 and thatf(x0) > 0. Prove that

there is an intervalI ≡ (x0−1/n, x0+1/n), wheren is a natural number, such thatf(x) > 0 ∀x ∈ I.

37. Suppose that the functiong : R → R is continuous and thatg(x) = 0 if x is rational. Prove that

g(x) = 0 ∀x ∈ R.

38. Letf(x) =

1 + x2, x > 0,

0, x = 0,

−(1 + x2), x < 0.

Show thatf : R → R is not continuous, but it has a continuous inverse.

39. Prove that limx→+∞

loga x

x= 0.

40. Find the limit limn→+∞

cosx

2cos

x

22· · · cos x

2n.

41. Suppose that the functionf : R → R is continuous at the pointx0. Prove that there is an interval

I ≡ (x0 − 1/n, x0 + 1/n), wheren is a natural number, such thatf(x) < n ∀x ∈ I.


42. Define the functionh : [1, 2] → R as follows:h(x) = 0 if x ∈ [1, 2] is irrational;h(x) = 1/n if

x ∈ [1, 2] is rational andx = m/n, wherem andn are natural numbers having no common factors

other than1.

(a) Prove thath : [1, 2] → R fails to be continuous at each rational number in[1, 2].

(b) Prove that ifǫ > 0, then the set{x ∈ [1, 2] | h(x) > ǫ} has only a finite number of points.

(c) Use part (b) to prove thath : [1, 2] → R is continuous at each irrational number in[1, 2].

43. Discuss the continuity of the following functions:

(a) f(x) = |x|.

(b) f(x) =x2 − 4

x− 2if x 6= 2, andf(2) = A.

(c) If x 6= −1, f(x) =1

(1 + x)2, f(−1) arbrary.

(d) If x 6= 0, f(x) =∣

∣

∣

sinx

x

∣

∣

∣, f(0) = 1.

(e) If x 6= 0, f(x) =sinx

|x| , f(0) = 1.

(f) If x 6= 0, f(x) = sin1

x, f(0) arbrary.


(a) If x 6= 0, f(x) = x sin1

x, f(0) = 0.

(b) If x 6= 0, f(x) = e−1x2 , f(0) = 0.

(c) If x 6= 1, f(x) =1

1 + e1

x−1

, f(1) arbrary.

(d) If x 6= 0, f(x) = x lnx2, f(0) = a.

(e) f(x) = [x].

(f) f(x) =√x− [

√x].

45. Find the discontinuous points of the following functions:

(a) f(x) =x

(1 + x)2.

(b) f(x) =1 + x

1 + x3.

(c) f(x) =x2 − 1

x3 − 3x+ 2.

(d) f(x) =1x − 1

x+11

x−1 − 1x

.

(e) f(x) =x

sinx.

(f) f(x) =

√

1− cosπx

4− x2.

46. Find the discontinuous points of the following functions:

(a) f(x) = cos21

x.

(b) f(x) = sgn(

sinπ

x

)

.

(c) f(x) = tan−1 1

x.

(d) f(x) =1

1− ex

1−x

.

(e) f(x) =√x tan−1 1

x.

(f) f(x) = ex+1x .

(g) f(x) =1

lnx.


(a) f(x) = sgn(sinx).

(b) f(x) = x− [x].

(c) f(x) = x[x].

(d) [x] sinπx.

(e) f(x) = x2 − [x2].

(f) f(x) =[ 1

x

]

.


48. Suppose that the functionf : R → R has the properties that (1)f(u+ v) = f(u)+ f(v) ∀u, v ∈ R,

(2) f(1) = m. Prove that (1)f(x) = mx for all rational numberx. (2) if f is continuous inR, then

f(x) = mx ∀x ∈ R.

49. Prove thatPn(x) = xn + a1xn−1 + · · ·+ an = 0 has at least one real root ifn is an odd number.

50. Letf : [0,∞) → R be continuous and satisfy (1)f(0) = 0, (2) f(x) ≥ √x ∀x ≥ 0. Show that for

c > 0, there is somex0 such thatf(x0) = c.

51. Suppose thatS is a nonempty set of real numbers that is not sequentially compact. Prove that either

(i) there is an unbounded sequence inS or (ii) there is a sequence inS that converges to a point

x0 /∈ S.

52. If a setS contains an unbounded sequence, show that the functionf : S → R, defined byf(x) =

x, ∀x ∈ S, is continuous but unbounded. If a setS contains a sequence that converges to a point

x0 /∈ S, show that the functionf : S → R, defined byf(x) = 1/|x− x0|, ∀x ∈ S, is continuous

but unbounded.

53. Show that ifS is a nonempty subset ofR that fails to be sequentially compact, then there is a function

f : S → R that is continuous but unbounded.

54. Suppose that the functionf : (a, b) → R is continuous and thatlimx→a+

f(x) and limx→b−

f(x) exist.

Prove thatf(x) is bounded on(a, b).

55. Prove that there is a solution of the equationx9 + x2 + 4 = 0, x ∈ R.

56. Prove that there is a solution of the equation1√

x+ x2+ x2 − 2x = 0, x > 0.

57. Consider the equationx = f(x),−1 < x < 1. If f(x) is continuous in[−1, 1] andf(−1) >

−1, f(1) < 1, show that the equation has a solution in(−1, 1).

58. For a functionf : D → R, a solution of the equationf(x) = x, x ∈ D, is called afixed point of

f . If f : [−1, 1] → R is continuous,f(−1) > −1, andf(1) < 1, show thatf : [−1, 1] → R has a

fixed point.

59. Suppose that the functionh : [a, b] → R and g : [a, b] → R are continuous. Show that if

h(a) ≤ g(a) andh(b) ≥ g(b), then the equationh(x) = g(x), x ∈ [a, b], has a solution.

60. Suppose thatf : R → R is continuous and its imagef(R) is bounded. Prove that there is a solution

of the equationf(x) = x, x ∈ R.


61. Suppose that the functionf(x) is continuous on[0, 1] and satisfies the conditionsf(0) = f(1). Show

that for any natural numbern, there exists at least one pointξ ∈ [0, 1] such thatf(ξ) = f(

ξ + 1n

)

.

Hint. By contradiction. If it is not true, then∃N > 1, s.t. f(x) 6= f(x + 1N), ∀x ∈ [0, 1 − 1

N]. Let

g(x) = f(x) − f(x+ 1N), theng(x) > 0 or g(x) < 0, ∀x ∈ [0, 1 − 1

N], sinceg(x) is continuous. Suppose

g(x) > 0, thenf(0) − f( 1

N

)

> 0, f( 1

N

)

− f( 2

N

)

> 0, · · · f(N − 1

N

)

− f(1) > 0. Adding up we get

f(0) − f(1) > 0, which is a contradiction.

62. Suppose thatf : [a, b] → R is continuous, and for everyx ∈ [a, b], there existsy ∈ [a, b]

such that|f(y)| ≤ 12 |f(x)|. Prove that there exists a pointξ ∈ [a, b] such thatf(ξ) = 0.

Hint. Let x1 = a, then|f(xn)| ≤1

2n−1|f(a)| → 0. ∃ subsequence{xnk

} → ξ, f(xnk) → f(ξ) = 0.

63. Suppose thatf(x) is continuous in[a, b]. Letx1, x2, · · · , xn be points in[a, b]. Prove that there is a

pointc ∈ [a, b] at whichf(c) =f(x1) + f(x2) + · · ·+ f(xn)

n.

64. Suppose that the functionf : [0, 1] → R is continuous and that its image consists entirely of rational

numbers. Prove thatf(x) is a constant function.

65. Suppose thatf : [0,+∞) → R is continuous. LetS be the set of real numbers defined by

S = {x | x = limn→+∞

f(xn) for some sequence{xn} ⊂ [0,+∞) such that limn→+∞

xn = +∞.}

Prove that ifa ∈ S andb ∈ S, then[a, b] ⊂ S.

Hint. Let c ∈ (a, b), anda ∈ S, b ∈ S. By assumption,∃{xn} and{yn} s.t.,lim

n→+∞

xn = +∞, limn→+∞

yn = +∞, and limn→+∞

f(xn) = a, limn→+∞

f(yn) = b.

Let ǫ =1

2min{(c− a), (b− c)}, then∃N1, s.t.,f(xn) < a+ ǫ < a+ 1

2(c+ a) < c, ∀n ≥ N1.

∃N2, s.t.,f(yn) > b− ǫ > b− 12(b− c) > c, ∀n ≥ N2. LetN = max{N1, N2},

thenf(xn) < c < f(yn), ∀n ≥ N . Sincef(x) is continuous on(xn, yn), from Intermediate Value Theorem,∃zn ∈ (xn, yn), s.t.,f(zn) = c. Therefore,c ∈ S, i.e., [a, b] ⊆ S.

66. Show that it is not necessary the case that iff : D → R andg : D → R are each uniformly

continuous, then so is the productfg : D → R.

67. Suppose that the functionsf : D → R andg : D → R are uniformly continuous and bounded.

Prove thatfg : D → R also is uniformly continuous.

68. A functionf : D → R is called a Lipschitz function if there is a constantc ≥ 0 such that|f(u)−f(v)| ≤ c|u− v| ∀u, v ∈ D. Prove that a Lipschitz function is uniformly continuous inD.

69. Prove that each of the following functions is uniformly continuous in its definition domain byǫ − δ

verification:

(a) f(x) =1

1 + x2, −∞ < x < +∞.

(b) f(x) = x2, −M < x < M, M > 0.

(c) f(x) =1

x, a ≤ x < +∞, a > 0.

70. Letf(x) =√x, x ∈ [0, 1]. Prove thatf(x) is not a Lipschitz function but uniformly continuous in

[0, 1].


71. Prove that each of the following functions is not uniformly continuous in its definition domain:

(a) f(x) = x2, 0 ≤ x < +∞.

(b) f(x) =1

x, 0 < x < 1.

(c) f(x) = sin1

x, 0 < x < 1.

72. Suppose that a continuous functionf : R → R is periodic, that is, there is numberp > 0 such that

f(x+ p) = f(x) ∀x. Show thatf : R → R is uniformly continuous.

73. Prove that the functionf(x) = sinπ

xis continuous and bounded in(0, 1), but not uniformly contin-

uous.

74. Prove that the functionf(x) = sinx2 is continuous and bounded in(−∞,+∞), but not uniformly

continuous.

75. Show that iff(x) is continuous in[a,+∞) and limx→+∞

f(x) exists, thenf(x) is uniformly continuous

in [a,+∞).

76. Show thatf(x) = x+ sinx is uniformly continuous in(−∞,+∞).

77. Discuss whether the following functions are uniformly continuous:

(a) f(x) =x

4− x2, −1 ≤ x ≤ 1.

(b) f(x) = lnx, 0 < x < 1.

(c) f(x) =sinx

x, 0 < x < π.

(d) f(x) = ex cos1

x, 0 < x < 1.

(e) f(x) = tan−1 x, −∞ < x < +∞.

(f) f(x) =√x, 1 ≤ x < +∞.

(g) f(x) = x sinx, 0 ≤ x < +∞.

78. Prove that the functionf(x) =| sinx|x

is uniformly continuous in each of the intervalsJ1 = (−1, 0)

andJ2 = (0, 1), but not uniformly continuous in the unionJ1 ∪ J2 (0 < |x| < 1).

79. For givenǫ > 0, find theδ(ǫ), for which the functionf(x) is uniformly continuous:

(a) f(x) = 5x− 3, −∞ < x < +∞.

(b) f(x) = x2 − 2x− 1, −2 ≤ x ≤ 5.

(c) f(x) =1

x, 0.1 ≤ x ≤ 1.

(e) f(x) =√x, 0 ≤ x < +∞.

(f) f(x) = 2 sinx− cosx, −∞ < x < +∞.

(g) f(x) = x sin1

x, x 6= 0, andf(0) = 0, 0 ≤ x ≤ π.

80. Suppose thatf(x) is continuous and monotone in a finite or infinite open interval (a, b). Prove that

f(x) is uniformly continuous in(a, b) if and only if f(x) is bounded in(a, b).

81. Letf(x) be continuous in a finite interval(a, b). Prove thatf(x) can be continuously extended to

[a, b] if and only if f(x) is uniformly continuous in(a, b).

82. Show thatf(x) = ax, −∞ < x < +∞, wherea = f(1) is an arbitrary constant, is the unique

continuous function that satisfies the equationf(x+ y) = f(x) + f(y), ∀x, y ∈ R.

Chapter 4: Differentiation 81

83. Show thatf(x) = ax, −∞ < x < +∞, wherea = f(1) is a positive constant, is the unique contin-

uous function that is not identically zero and satisfies the equationf(x+y) = f(x)f(y), ∀x, y ∈ R.

84. Show thatf(x) = loga x, x > 0, wherea is a positive constant, is the unique continuous function

that is not identically zero and satisfies the equationf(xy) = f(x)f(y), ∀x, y > 0.

85. Show thatf(x) = xa, x > 0, wherea is a constant, is the unique continuous function that is not

identically zero and satisfies the equationf(xy) = f(x)f(y), ∀x, y > 0.

86. Find all continuous functionsf(x), −∞ < x < +∞, that satisfies the equationf(x+ y) + f(x−y) = 2f(x)f(y), ∀x, y ∈ R.

87. Find all bounded continuous functionsf(x) andg(x), −∞ < x < +∞, wheref(0) = 1 and

g(0) = 0, that satisfies the equationsf(x+ y) = f(x)f(y)− g(x)g(y) andg(x+ y) = f(x)g(y) +

f(y)g(x), ∀x, y ∈ R. Hint: Consider the functionF (x) = f2(x) + g2(x).

88. Let∆f(x) = f(x+∆x) − f(x) and∆2f(x) = ∆(∆f(x)) denote the first and second order finite

difference off(x). Show that iff(x), −∞ < x < +∞, is continuous and∆2f(x) ≡ 0, thenf(x)

is a linear function, i.e.,f(x) = ax+ b, wherea andb are constants.

89. Suppose that the functionf : [0, 1] → R is continuous,f(0) > 0, andf(1) = 0. Prove that there is

a numberx0 ∈ (0, 1] such thatf(x0) = 0 andf(x) > 0 for 0 ≤ x ≤ x0; that is, there is a smallest

point in the interval[0, 1] at which the functionf attains the value0.

90. Letωf (δ) = sup |f(x1) − f(x2)|, wherex1, x2 ∈ (a, b) satisfying the condition|x1 − x2| ≤ δ.

Prove thatf(x) is uniformly continuous in(a, b) if and only if limδ→+0

ωf (δ) = 0.

91. Suppose thatf(x) is uniformly continuous in an open interval(a, b). Prove thatf(x) is bounded in

(a, b).

92. Suppose that the functionf : R → R is uniformly continuous andf(0) = 0. Prove that

|f(x)| ≤ 1 +M |x|, ∀x ∈ (−∞,+∞),

whereM > 0 is a constant.

Hint. Supposex ≥ 0. Sincef(x) is uniformly continuous on[0, x], then for givenǫ = 1, ∃δ > 0, s.t.,|f(x′)−f(x′′)| < ǫ, ∀|x′−x′′| < δ, x′, x′′ ∈ [0, x]. Divide [0, x] into subintervals[0, δ], [δ, 2δ], · · · , [kδ, x], wherekδ < x andx− kδ ≤ δ. Then we have|f(x)| = |f(x)− f(0)| ≤ |f(x)− f(kδ)|+ |f(kδ)− f((k − 1)δ)| + · · ·+ |f(δ) − f(0)|

≤ 1 + k = 1 +1

δkδ < 1 +

1

δx.

LetM = 1δ

, then|f(x)| ≤ 1 +M |x|. Similarly, if x < 0, then|f(x)| ≤ 1 +M |x|.


Chapter 4

Differentiation

4.1 Derivative

4.1.1 Definition

Definition 4.1.1Let f(x) be a function defined in interval(a, b) andx0 ∈ (a, b). Consider the following

ratiof(x)− f(x0)

x− x0or

f(x0 +∆x) − f(x0)

∆x.

If the ratio has a limitl asx tends tox0 (or ∆x → 0). Then we say thatf(x) is differentiable at the point

x0 andl is the derivative off at x0, denote byl = f ′(x0). Namely

limx→x0

f(x)− f(x0)

x− x0= f ′(x0),

or

lim∆x→0

f(x0 +∆x) − f(x0)

∆x= f ′(x0).

The one side limits

lim∆x→0−

f(x0 +∆x)− f(x0)

∆xand lim

∆x→0+

f(x0 +∆x)− f(x0)

∆x.

are calledleft andright derivative off(x), and denoted byf ′−(x0) andf ′

+(x0).

Obviously,f(x) is differentiable at thex0 if and only if bothf ′−(x0) andf ′

+(x0) exist and equal.

By the definition, if f(x) is differentiable atx0, we have

f(x)− f(x0) = f ′(x0)(x − x0) + α(x)(x − x0),

where

α(x) =f(x)− f(x0)

x− x0− f ′(x0)

83


and satisfies

limx→x0

α(x) = 0.

Hence

limx→x0

[f(x)− f(x0)] = 0.

It means that we have the following proposition.

Proposition 4.1.2If f(x) is differentiable atx0, thenf(x) is continuous atx0.

The opposite assertion may not be true. Consider the following example:

Example 4.1.1f(x) = |x| is continuous atx = 0 but not differentiable atx = 0.

Proof: Since

limx→0x>0

f(x)− f(0)

x− 0= lim

x→0x>0

x

x= 1,

limx→0x<0

f(x)− f(0)

x− 0= lim

x→0x<0

−xx

= −1.

Hence, the ratiof(x)−f(0)x−0 has no limit asx→ 0.

Example 4.1.2f(x) = xn , f ′(x) = nxn−1

Proof: Using Binomial Formula we have

(x+∆x)n = xn + nxn−1∆x+n∑

j=2

(

n

j

)

xn−j∆xj ,

(x+∆x)n − xn

∆x= nxn−1 +

n∑

j=2

(

n

j

)

xn−j∆xj−1.

Hence

lim∆x→0

(x +∆x)n − xn

∆x= nxn−1.

Example 4.1.3f(x) = sinx , f ′(x) = cosx

Proof: sin(x+∆x)− sinx = 2 sin ∆x2 cos(x + ∆x

2 ). Sincecosx is continuous everywhere and

lim∆x→0

sin ∆x2

∆x2

= 1.

We have

f ′(x) = lim∆x→0

f(x+∆x)− f(x)

∆x= lim

∆x→0

sin∆x2

∆x2

· cos(x+∆x

2)

= lim∆x→0

sin∆x2

∆x2

· lim∆x→0

cos(x+∆x

2) = cosx.


4.1.2 Physical and geometric interpretation

1. Velocity in Mechanics. Let S = S(t) be the function of displacement. ThenS(t0+∆t)−S(t0)∆t is the

average velocity during the time period fromt0 to t0 +∆t, and

V (t0) = lim∆t→0

S(t0 +∆t)− S(t0)

∆t= S′(t0)

is the instant velocity at timet0.

2. Slope of a tangent line.

The tangent line at pointx0 is y = f(x0) +m0(x− x0) wherem0 = tan θ0, the slope of the line

m0 = limx→x0

f(x)− f(x0)

x− x0= lim

x→x0

tan θ = tan θ0 =⇒ Tangent Line:y = f(x0) + f ′(x0)(x − x0)

4.1.3 Algebra of Derivatives

Theorem 4.1.3Suppose thatf(x) andg(x) are differentiable atx0 ∈ (a, b). Then

1. (f + g)′(x0) = f ′(x0) + g′(x0).

2. (f · g)′(x0) = f(x0)g′(x0) + f ′(x0)g(x0).

3. if g(x) 6= 0 for all x in (a, b), then(f/g)′(x0) =g(x0)f

′(x0)− f(x0)g′(x0)

[g(x0)]2.

Proof of 2: Since

f(x0 +∆x)g(x0 +∆x)− f(x0)g(x0) = f(x0 +∆x)g(x0 +∆x)

−f(x0 +∆x)g(x0) + f(x0 +∆x)g(x0)− f(x0)g(x0),

we have

f(x0 +∆x)g(x0 +∆x)− f(x0)g(x0)

∆x=

f(x0 +∆x)g(x0 +∆x)− g(x0)

∆x+f(x0 +∆x)− f(x0)

∆xg(x0)


Take limits on both sides as∆x→ 0 we get

(fg)′(x0) = f(x0)g′(x0) + f ′(x0)g(x0).

Proof of 3: Since

f(x0 +∆x)

g(x0 +∆x)− f(x0)

g(x0)=f(x0 +∆x)g(x0)− f(x0)g(x0 +∆x)

g(x0 +∆x)g(x0)

=[f(x0 +∆x)− f(x0)]g(x0)− f(x0)[g(x0 +∆x) − g(x0)]

g(x0 +∆x)g(x0)

Dividing both sides by∆x and taking limits we get

(f/g)′(x0) =g(x0)f

′(x0)− f(x0)g′(x0)

[g(x0)]2.

4.2 Differentiating Inverses and Compositions

4.2.1 Inverse

Theorem 4.2.1Let I be an open interval containing the pointx0. Suppose that the functionf : I → R is

strictly monotone and continuous,f is differentiable atx0, andf ′(x0) 6= 0. Then the inversef−1 : f(I) →R is differentiable at the pointy0 = f(x0) and

(f−1)′(y0) =1

f ′(x0)

Proof. It follows from Theorem 3.6.4 thatf(I) is an open interval containing the pointy0 = f(x0). For a

pointy ∈ f(I), there is a pointx ∈ I such thaty = f(x) andx = f−1(y). Then we have

f−1(y)− f−1(y0)

y − y0=

1

f(x)− f(x0)

x− x0

.

Since the inverse function is continuous and monotone, theny → y0 if and only if x → x0. Sincef is

strictly monotone,f−1(y) 6= f−1(y0) if and only if y 6= y0. Hence,

limy→y0

f−1(y)− f−1(y0)

y − y0= lim

x→x0

1

f(x)− f(x0)

x− x0

=1

f ′(x0).

That is

(f−1)′(y0) =1

f ′(x0).

Corollary 4.2.2 Suppose thatf is strictly monotone and differentiable withf ′(x) 6= 0 in an open interval

I. Then the inverse functionf−1 is differentiable in the intervalf(I).


4.2.2 Composition and Chain Rule

Theorem 4.2.3Let I be an open interval containing a pointx0 and suppose that the functionf : I → R

is differentiable atx0. Let J be an open interval for whichf(I) ⊆ J , and suppose thatg : J → R is

differentiable aty0 = f(x0). Then the compositiongof : I → R is differentiable atx0 and

(gof)′(x0) = g′(y0)f′(x0).

Proof. Let y = f(x) with x 6= x0. If there is an open interval containingx0 in which f(x) 6= f(x0) if

x 6= x0, then(gof)(x) − (gof)(x0)

x− x0=g(y)− g(y0)

y − y0

f(x)− f(x0)

x− x0.

By taking limits in both sides, we get the required result. Toaccount for the possibility that no such interval

exists, we introduce an auxiliary functionh : J → R by defining

h(y) =

[g(y)− g(y0)]/(y − y0), if y 6= y0,

g′(y0), if y = y0.

Then

g(y)− g(y0) = h(y)(y − y0)

and(gof)(x)− (gof)(x0)

x− x0= h(y)

[f(x)− f(x0)

x− x0

]

.

From the definition ofh(y), h is continuous aty0. Because differentiability implies continuity,f is contin-

uous atx0. Therefore, the composite functionh(f(x)) is continuous atx0. Hence,

limx→x0

g(f(x))− g(f(x0))

x− x0= h(f(x0))f

′(x0) = g′(y0)f′(x0).

The formula

(gof)′(x0) = g′(f(x0))f′(x0)

is called the Chain Rule for differentiating composite function.

4.3 Derivatives of Elementary Function

1. f(x) = xn, f ′(x) = nxn−1, x ∈ (−∞,∞).

2. f(x) = ex, f ′(x) = ex, x ∈ (−∞,∞).

Proof. It’s known that lim∆x→0

e∆x − 1

∆x= 1. Therefore,

lim∆x→0

ex+∆x − ex

∆x= lim

∆x→0ex

e∆x − 1

∆x= ex.


3. f(x) = lnx, f ′(x) = 1x , x ∈ (0,∞).

Proof. It’s known that by definitionx = f−1(y) = ey. By the property of differentiating inverse

f ′(x) =1

(f−1)′(y)=

1

ey=

1

x..

4. f(x) = ax, f ′(x) = ax · ln a, a > 0, x ∈ (−∞,∞).

Proof. f(x) = ax = ex ln a. Let y = g(x) = x ln a, andh(y) = ey. Thenf(x) = h(g(x)) =

hog(x). By the property of differentiating compositionf ′(x) = (hog)′(x) = h′(y)g′(x) = ey ·ln a = ex ln a · ln a = ax ln a.

5. f(x) = xα, f ′(x) = αxα−1, x ∈ (0,∞).

6. f(x) = sinx, f ′(x) = cosx, x ∈ (−∞,∞).

7. f(x) = cosx, f ′(x) = − sinx, x ∈ (−∞,∞).

8. f(x) = tanx, f ′(x) = 1

cos2 x= sec2 x, x ∈ (−π

2 ,π2 ).

Proof. Sincetanx =sinx

cosx, we have

f ′(x) =cosx · cosx− sinx(− sinx)

cos2 x=

cos2 x+ sin2 x

cos2 x=

1

cos2 x.

9. (sin−1 x)′ = 1/√1− x2, x ∈ (−1, 1),

(cos−1 x)′ = −1/√1− x2, x ∈ (−1, 1),

(tan−1 x)′ = 11+x2 , x ∈ (−∞,∞).

4.4 Mean Value Theorems

Let f be a continuous function defined on(a, b). For a pointx0 ∈ (a, b), if there is aδ > 0 such that

f(x0) ≥ f(x), (or f(x0) ≤ f(x) ), for x0 − δ < x < x0 + δ,

then we sayf(x0) is a local maximum (or minimum ) andx0 is a local maximizer (or minimizer). In both

cases,f(x0) is called a local extreme.

Lemma 4.4.1Let I be a neighborhood ofx0 and suppose that the functionf : I → R is differentiable at

x0. If the pointx0 is either a local minimizer or local maximizer, thenf ′(x0) = 0.

Proof. By the definition of derivative

limx→x0x<x0

f(x)− f(x0)

x− x0= lim

x→x0x>x0

f(x)− f(x0)

x− x0= f ′(x0).


First suppose thatx0 is a local maximizer, then there is aδ > 0 such that

f(x)− f(x0)

x− x0≥ 0 for x0 − δ < x < x0

and hence

f ′(x0) = limx→x0x<x0

f(x)− f(x0)

x− x0≥ 0.

On the other hand,f(x)− f(x0)

x− x0≤ 0 for x0 < x < x0 + δ

and hence

f ′(x0) = limx→x0x>x0

f(x)− f(x0)

x− x0≤ 0.

Therefore

f ′(x0) = 0.

In the case whenx0 is a local minimizer, the proof is similar.

x

y

a b

Theorem 4.4.2 (Rolle’s Theorem)Suppose that the functionf : [a, b] → R is continuous andf is differ-

entiable in(a, b). Assume, moreover, that

f(a) = f(b).

Then there is a pointx0 in (a, b) at whichf ′(x0) = 0.

Proof. Sincef(x) is continuous in[a, b], according to the Extreme Value Theorem, it attains both a mini-

mum value and a maximum value on[a, b]. Sincef(a) = f(b), if both the maximizer and minimizer occur

at the end points, thenf(x) is a constant, sof ′(x) = 0 in whole(a, b). Otherwise, the function has either a

maximizer or a minimizer at some pointx0 ∈ (a, b). Hence, by Lemma 4.4.1, at this pointf ′(x0) = 0.

Rolle’s Theorem is a special case of the following Lagrange Mean Value Theorem, but in fact, it has

some important applications by itself.

Theorem 4.4.3 (Lagrange Mean Value Theorem)Suppose that the functionf(x) is continuous in[a, b]

and differentiable in(a, b). Then there is a pointx0 ∈ (a, b) at which

f ′(x0) =f(b)− f(a)

b− a.


x

y

a b

b a-

f(b) f(a)-

( , ( ))x f x0 0

Proof. Let y = g(x) be the line passing through the points(a, f(a)) and(b, f(b)), namely

y = g(x) = f(a) +f(b)− f(a)

b− a(x− a).

Define the auxiliary function

d(x) = f(x)− g(x) = f(x)− f(a)− f(b)− f(a)

b− a(x− a).

Thend(x) is continuous on[a, b] and differentiable in(a, b), andd(a) = d(b) = 0. We apply Rolle’s

theorem to select a pointx0 ∈ (a, b) at whichd′(x0) = 0.

d′(x0) = f ′(x0)−f(b)− f(a)

b− a= 0,

or,

f ′(x0) =f(b)− f(a)

b− a.

Example 4.4.1Show that iff ′(x) exists for allx ∈ (a, b), thenf ′(x) cannot have jump discontinuity.

Proof. Let x0 ∈ (a, b), and take∆x > 0 such thatx0 + ∆x ∈ (a, b). Then,f(x) is continuous on

[x0, x0 +∆x], and differentiable on(x0, x0 +∆x). Using Lagrange Mean Value Theorem we get

f(x0 +∆x) − f(x0)

∆x= f ′(x0 + θ∆x), 0 < θ < 1.

If f ′(x) has a jump discontinuity atx0, then limx→x0

f ′(x) exists. Then we have

f ′(x0) = f ′+(x0) = lim

∆x→0

f(x0 +∆x) − f(x0)

∆x= lim

∆x→0f ′(x0 + θ∆x).

Similarly,

f ′(x0) = f ′−(x0) = lim

∆x→0f ′(x0 − θ∆x),

i.e.,f ′(x) is continuous atx0, which is a contradiction.

Example 4.4.2Let f : (−1, 1) → R be defined by

f(x) =

x2 sin(

1x

)

, x 6= 0,

0 x = 0.


Show that iff ′(x) exists for allx ∈ (−1, 1), butf ′(x) is not continuous atx = 0.

Proof. It is easy to see thatf ′(x) exists for allx ∈ (−1, 1). If x 6= 0, then

f ′(x) = 2x sin( 1

x

)

− cos( 1

x

)

.

Since limx→0

f ′(x) does not exist, thenf ′(x) is not continuous atx = 0.

The Lagrange Mean Value Theorem is one of the most important result in calculus. The following

propositions are some Corollaries from this theorem.

Corollary 4.4.4 Suppose thatf(x) is differentiable in(a, b). Thenf(x) is constant if and only iff ′(x) = 0

in whole(a, b).

Proof. f(x) = C then obviouslyf ′(x) = 0 in (a, b). To prove the converse, letu andv be any two points

in (a, b) with u < v. Sincef(x) is continuous on[u, v] and differentiable in(u, v). According to the Mean

Value Theorem, there is a pointx0 ∈ (u, v) such that

f ′(x0) =f(v)− f(u)

v − u.

But f ′(x0) = 0, and thusf(u) = f(v). Consequently,f(x) is a constant.

From Corollary 1, we get the following corollary immediately.

Corollary 4.4.5 Suppose thatf(x) andg(x) are differentiable in(a, b). Then they differ by a constant, i.e.,

f(x) = g(x) + C, if and only ifg′(x) = f ′(x).

Corollary 4.4.6 Suppose thatf(x) is differentiable andf ′(x) > 0 on the interval(a, b). Thenf(x) is

strictly increasing on(a, b). Moreover, iff(x) is continuous on[a, b], thenf(x) is strictly increasing on

[a, b].

Proof. Let u and v be any two points withu < v in (a, b). Sincef(x) is continuous on[u, v] and

differentiable in(u, v), we apply the Mean Value Theorem to choose a pointx0 ∈ (u, v) at which

f ′(x0) =f(v)− f(u)

v − u.

Sincef ′(x0) > 0 andv − u > 0, it follows thatf(v) > f(u).

Replacing the assumptionf ′(x) > 0 by f ′(x) < 0, the conclusion isf(x) strictly decreasing in(a, b).

Theorem 4.4.7 (Cauchy Mean Value Theorem)Suppose that the functionsf(x) andg(x) are continuous

on [a, b] and differentiable in(a, b). Moreover, assume thatg′(x) 6= 0 ∀x ∈ (a, b). Then there is a point

x0 ∈ (a, b) at which

f(b)− f(a)

g(b)− g(a)=f ′(x0)

g′(x0).


Proof. Sinceg′(x) 6= 0 in whole (a, b), it follows from the Lagrange Mean Value Theorem thatg(b) −g(a) 6= 0. Now define an auxiliary functionψ on [a, b] by

ψ(x) = f(x)− f(a)−[

f(b)− f(a)

g(b)− g(a)

]

[g(x)− g(a)].

ψ(x) is continuous on[a, b] and differentiable in(a, b). Furthermore,

ψ(a) = ψ(b) = 0.

Therefore, we can apply Rolle’s Theorem to choose a pointx0 ∈ (a, b) at whichψ′(x0) = 0, namely

f ′(x0)−f(b)− f(a)

b− ag′(x0) = 0 =⇒ f(b)− f(a)

b− a=f ′(x0)

g′(x0).

4.5 High (Order) Derivatives and Differential

Definition 4.5.1 Let f ′(x) be the derivative off(x) in (a, b) and also differentiable in(a, b). Then,

the derivative off ′(x) is called the second (two) derivative off(x), denoted byf ′′(x) or f (2)(x), i.e.,

(f ′(x))′ = f ′′(x). Iteratively, iff (k)(x) is differentiable in(a, b) then the derivative off (k)(x) is called

the(k + 1) (or (k + 1)th) derivative off(x) denoted byf (k+1)(x), i.e.,[f (k)(x)]′ = f (k+1)(x).

Theorem 4.5.2 (Leibniz’s Formula)Let I be an open interval andn be a natural number. Suppose that

bothf : I → R andg : I → R haven derivatives, thenfg : I → R hasn derivatives, and

(fg)(n)(x) =

n∑

k=0

( n

k

)

f (k)(x)g(n−k)(x) ∀x ∈ I.

Notation of Leibnitz. An alternate notation, due to Leibnitz, for derivativef ′(x) is as following

f ′(x) =df

dxor

d

dx(f(x))

ordy

dx= y′(x) if y = y(x) or f(x)

Advantages. For numbersa, b we haveab = 1/ b

a . If y = f(x) has its inversex = f−1(y) and f(x) is

differentiable, then according to the related theorem we have

(f−1)′(y) =1

f ′(x)

Using the notation of Leibnitz, we can write

dx

dy= 1/

dy

dx


For composite functionh(x) = f(u(x)), we have the Chain Rule

h′(x) = f ′(u)u′(x)

Now we can write

df

dx=

df

du

du

dx

that is just like for numbersa, b 6= 0, c 6= 0

a

b=a

c· cb

For second derivative or high derivative, we denotef ′′(x) or f (k)(x) by

f ′′(x) =d2f

dx2or f (k)(x) =

dkf

dxk.

Definition 4.5.3Suppose thaty = f(x) is defined in some neighborhood ofx. Let∆y = f(x+∆x)−f(x).If

∆y = A∆x+ α(∆x),

whereA is independent of∆x, and

lim∆x→0

α(∆x)

∆x= 0,

thenA∆x is called thedifferential of y, and denoted by

dy = A∆x.

Clearly, if f ′(x) exists, thenA = f ′(x). The reverse is also true: if∃A s.t.∆y = A∆x+α(∆x), then

f ′(x) = A. Sincedx = ∆x, then

dy = Adx = f ′(x)dx.

All the formulas for derivatives can be transferred to differentials, e.g.,

d(u ± v) = du± dv

d(uv) = vdu + udv

d(u

v

)

=vdu− udv

v2

df(g(x)) = f ′(g(x))g′(x)dx, or, df(u) = f ′(u)du = f ′(u)g′(x)dx


4.6 Applications of Derivative

4.6.1 L’Hopital’s Rule

Theorem 4.6.1Suppose thatf(x) andg(x) are differentiable in a neighborhood ofa, except possibly ata,

g′(x) 6= 0, and

limx→a

f ′(x)

g′(x)= A (or ±∞).

If limx→a

f(x) = limx→a

g(x) = 0, then

limx→a

f(x)

g(x)= A.

Proof. Definef(a) = g(a) = 0, thenf(x) andg(x) are continuous in the neighborhood. Ifx < a, then

from Cauchy Mean Value Theorem there exists aξ ∈ (x, a) such that

limx→a−0

f(x)

g(x)= lim

x→a−0

f(a)− f(x)

g(a)− g(x)= lim

x→a−0

f ′(ξ)

g′(ξ).

Sincex→ a− 0, thenξ → a− 0. Thus, we have

limx→a−0

f(x)

g(x)= A.

Similarly, we can prove

limx→a+0

f(x)

g(x)= A.

Theorem 4.6.1’Suppose thatf(x) andg(x) are differentiable in a neighborhood ofa, except possibly at

a, g′(x) 6= 0, and

limx→a

f ′(x)

g′(x)= A (or ±∞).

If limx→a

g(x) = +∞, then

limx→a

f(x)

g(x)= A.

Proof. Sincelimx→a

g(x) = +∞, we may assume thatg(x) > 0 in a small interval(a, b]. Givenǫ > 0, ∃δ1,

s.t.,∣

∣

∣

f ′(x)

g′(x)−A

∣

∣

∣ <ǫ

2, ∀a < x < a+ δ1 ≡ x0.

Using Cauchy Mean Value Theorem on[x, x0], we get

f(x)− f(x0)

g(x)− g(x0)=f ′(c)

g′(c), x < c < x0.

Then we have∣

∣

∣

f(x)− f(x0)

g(x)− g(x0)− A

∣

∣

∣ <ǫ

2, ∀a < x < x0.


Since

f(x)

g(x)−A =

f(x)− f(x0)

g(x)− g(x0)· g(x)− g(x0)

g(x)+f(x0)

g(x)−A

=[f(x)− f(x0)

g(x)− g(x0)−A

]

·[g(x)− g(x0)

g(x)

]

+A[g(x)− g(x0)

g(x)

]

−A+f(x0)

g(x)

=[

1− g(x0)

g(x)

]

·[f(x) − f(x0)

g(x) − g(x0)−A

]

+f(x0)−Ag(x0)

g(x),

then we have

∣

∣

∣

f(x)

g(x)−A

∣

∣

∣≤ ǫ

2·∣

∣

∣1− g(x0)

g(x)

∣

∣

∣+∣

∣

∣

f(x0)−Ag(x0)

g(x)

∣

∣

∣.

Sincelimx→a

g(x) = +∞, ∃δ2, s.t.,∀a < x < a+ δ2,

∣

∣

∣

f(x0)−Ag(x0)

g(x)

∣

∣

∣ <ǫ

4,

and∣

∣

∣

g(x0)

g(x)

∣

∣

∣ ≤ 1

2.

Therefore,

∣

∣

∣

f(x)

g(x)−A

∣

∣

∣ ≤(

1 +1

2

)

· ǫ2+ǫ

4= ǫ,

i.e.,

limx→a+

f(x)

g(x)= A.

Similarly,

limx→a−

f(x)

g(x)= A.

For the casex→ ∞, we have

Theorem 4.6.2Suppose thatf(x) andg(x) are differentiable for|x| > a > 0, g′(x) 6= 0, and

limx→∞

f ′(x)

g′(x)= A (or ±∞).

If limx→∞

f(x) = limx→∞

g(x) = 0, then

limx→∞

f(x)

g(x)= A.

Proof. LetF (t) = f(1/t) andG(t) = g(1/t). Then

limt→0

F ′(t)

G′(t)= lim

t→0

f ′(1/t)(−1/t2)

g′(1/t)(−1/t2)= A.

Therefore, from Theorem 4.6.1 we get

limx→∞

f(t)

g(t)= lim

t→0

F (t)

G(t)= A.


Theorem 4.6.2’Suppose thatf(x) andg(x) are differentiable for|x| > a > 0, g′(x) 6= 0, and

limx→∞

f ′(x)

g′(x)= A (or ±∞).

If limx→∞

f(x) = +∞, and limx→∞

g(x) = +∞, then

limx→∞

f(x)

g(x)= A.

Using L’Hopital’s Rule, we can find the following limits easily:

limx→0

sinx

x, lim

x→0

ln(1 + x)

x, lim

x→0

ex − 1

x.

Example 4.6.1(00 ) limx→0

sinx

x= 1.

Example 4.6.2(∞∞ ) limx→+∞

lnx

x2= 0.

Example 4.6.3(0 · ∞) limx→0+

x ln x = 0.

Example 4.6.4(∞−∞) limx→0

( 1

x2− cos2 x

sin2 x

)

=1

3.

Example 4.6.5(1∞) limx→0

(sinx

x

)1

1−cos x

= e−13 .

Example 4.6.6 limx→+∞

x− sinx

x+ sinx= 1.

4.6.2 Extreme values of functions

By using high derivatives, we have the following important theorem related to local minimizer and

local maximizer.

Theorem 4.6.3Suppose that functionf(x) has second derivative in(a, b) and at the pointx0 ∈ (a, b),

f ′(x0) = 0 (stationary point). Then

if f ′′(x0) > 0, x0 is a local minimizer;

if f ′′(x0) < 0, x0 is a local maximizer.

Proof. First suppose thatf ′′(x0) > 0. Since

f ′′(x0) = limx→x0

f ′(x) − f ′(x0)

x− x0> 0.

it follows that there is aδ > 0 such that

f ′(x) − f ′(x0)

x− x0> 0 if 0 < |x− x0| < δ.


Butf ′(x0) = 0, hence

f ′(x) > 0 if x0 < x < x0 + δ,

and

f ′(x) < 0 if x0 − δ < x < x0.

From Corollary 4.4.6 we have

f(x) > f(x0) if 0 < |x− x0| < δ.

A similar argument applies whenf ′′(x0) < 0.

Lemma 4.6.4Suppose that functionf hasn derivatives in(a, b) and at the pointx0 ∈ (a, b)

f (k)(x0) = 0, k = 0, 1, · · · , n− 1.

Then for each pointx 6= x0 in (a, b), there is a pointξ strictly betweenx andx0 at which

f(x) =f (n)(ξ)

n!(x− x0)

n.

Proof. Defineg(x) = (x − x0)n Then

g(k)(x0) = 0, k = 0, 1, · · · , n− 1.

Thenf andg haven derivatives and satisfy

f (k)(x0) = g(k)(x0) = 0, k = 0, 1, · · · , n− 1.

From the proof of Theorem 4.6.1 (repeatedly) we see that there exists aξ such that

f(x)

g(x)=f (n)(ξ)

g(n)(ξ),

ξ is betweenx andx0. Sinceg(n)(x) = n!, we obtain

f(x) =f (n)(ξ)

n!(x− x0)

n.

Theorem 4.6.5Suppose that functionf(x) has continuousn derivatives in(a, b) and at the pointx0 ∈(a, b),

f (k)(x0) = 0, k = 0, 1, · · · , n− 1, f (n)(x0) 6= 0.

Then ifn is odd,x0 is not a local maximizer or minimizer; ifn is even,x0 is a local local maximizer or

minimizer, and

1. if f (n)(x0) > 0, x0 is a local minimizer;


2. if f (n)(x0) < 0, x0 is a local maximizer.

Proof. Leth(x) = f(x)− f(x0). Then

h(k)(x0) = 0, k = 0, 1, · · · , n− 1.

From the above theorem, we know that in intervalx0 − δ < x < x0 + δ with (x0 − δ, x0 + δ) ⊂ (a, b)

h(x) =h(n)(ξ)

n!(x− x0)

n, x0 − δ < ξ < x0 + δ,

or,

f(x)− f(x0) =f (n)(ξ)

n!(x − x0)

n, x0 − δ < ξ < x0 + δ.

Becausef (n)(x) is continuous atx = x0, we may chooseδ such thatf (n)(x) has a same sign withf (n)(x0)

in the intervalx0 − δ < x < x0 + δ. Therefore, Ifn is odd, in the interval(x0 − δ, x0 + δ)

f(x)− f(x0) =f (n)(ξ)

n!(x− x0)

n ξ ∈ (x0 − δ, x0 + δ),

f (n)(ξ) has a same sign withf (n)(x0). f(x)− f(x0) will change its sign fromx < x0 to x > x0. Thusx0

is not a local local maximizer or minimizer. If n is even, due to (x−x0)n

n! ≥ 0, f(x)− f(x0) has a same sign

with f (n)(x0) in (x0 − δ, x0 + δ).

1. If f (n)(x0) > 0, f(x) − f(x0) =f(n)(ξ)

n! (x − x0)n ≥ 0 in (x0 − δ, x0 + δ), hencex0 is a local

minimizer;

2. If f (n)(x0) < 0, f(x) − f(x0) =f(n)(ξ)

n! (x − x0)n ≤ 0 in (x0 − δ, x0 + δ), hencex0 is a local

maximizer.

4.6.3 Concavity and graph of functions

Lemma 4.6.6Suppose that the functionf(x) has continuous second derivatives on(a, b). Then∀x0, x1 ∈(a, b), ∃η, betweenx0 andx1, s.t.

f(x1) = f(x0) + (x1 − x0)f′(x0) +

f ′′(η)

2(x1 − x0)

2.

Proof. Let

ψ(x) = f(x)− f(x0)− (x− x0)f′(x0),

φ(x) =1

2(x− x0)

2.

Then, using Cauchy Mean Value Theorem we get

f(x1)− f(x0)− (x1 − x0)f′(x0)

12 (x1 − x0)2

=f ′(ξ)− f ′(x0)

ξ − x0= f ′′(η),


whereη is betweenx0 andx1.

Lemma 4.6.7Suppose that the functionf(x) has continuous second derivatives on(a, b), andf ′′(x) ≥0, ∀x ∈ (a, b). Then for anyx0 ∈ (a, b), the tangent line off(x) at (x0, f(x0)) is below the graph off(x).

Proof. The tangent line at(x0, f(x0)) is

y = f(x0) + (x− x0)f′(x0).

Let (x1, y1) be any point on the tangent line and(x1, f(x1)) be the corresponding point onf(x). Then

from Lemma 4.6.6, we get

f(x1) = f(x0) + (x1 − x0)f′(x0) +

f ′′(η)

2(x1 − x0)

2

≥ f(x0) + (x1 − x0)f′(x0) = y1.

Definition 4.6.8Suppose that the functionf(x) is defined onI. If ∀x0, x1 ∈ I, x0 < x1, and∀t ∈ (0, 1)

f [(1− t)x0 + tx1] ≤ (1− t)f(x0) + tf(x1),

thenf(x) is called concave up on(a, b). If ∀x0, x1 ∈ I, x0 < x1, and∀t ∈ (0, 1)

f [(1− t)x0 + tx1] ≥ (1− t)f(x0) + tf(x1),

thenf(x) is called concave down on(a, b).

Theorem 4.6.9Suppose that functionf(x) has continuous second derivatives on(a, b) and f ′′(x) ≥0, ∀x ∈ (a, b). Thenf(x) is concave up on(a, b). If f ′′(x) ≤ 0, ∀x ∈ (a, b). Thenf(x) is concave

down on(a, b).

Proof. ∀x ∈ (a, b), from Lemma 4.6.7, we have

f(x) ≥ f(u) + (x− u)f ′(u), ∀u ∈ (a, b).

Then∀x0, x1 ∈ (a, b), we get

f(x0) ≥ f(xt) + (x0 − xt)f′(xt),

f(x1) ≥ f(xt) + (x1 − xt)f′(xt).

Multiplying the first inequality by(1− t), the second inequality byt, and lettingxt = (1− t)x0 + tx1, we

get the result.

Example 4.6.7Sketch the graph off(x) =1

32(3x5 − 20x3).


4.7 Exercise

1. Letf(x) = x3 + 2x+ 1. Find the equation of the tangent line to the graph off at the point(2, 13).

2. Letf(x) =

m1x+ 4, x ≤ 0,

m2x+ 4, x > 0,wherem1 6= m2 are constants. Show thatf : R → R is

continuous but not differentiable atx = 0.

3. Use the definition of derivative to compute the derivativeof the following functions atx = 1:

(a) f(x) =√x+ 1.

(b) f(x) = x3 + 2x.

(c) f(x) =1

1 + x2.

4. LetI andJ be open intervals, and the functionf : I → R andh : J → R have the property that

h(J) ⊆ I, so the compositionf ◦ h : J → R is defined. Show that ifx0 is in J , h : J → R is

continuous atx0, h(x) 6= h(x0) if x 6= x0, andf : I → R is differentiable ath(x0), then

limx→x0

f(h(x)) − f(h(x0))

h(x) − h(x0)= f ′(h(x0)).

5. Letf(x) =

0, x ≤ 0,

xn, x > 0,Prove thatf : R → R is differentiable forn ≥ 2.

6. Suppose that the functionf : R → R has the property that−x2 ≤ f(x) ≤ x2, ∀x. Prove thatf is

differentiable atx = 0 and thatf ′(0) = 0.

7. Letg(x) =

3x2, x ≤ 1,

a+ bx, x > 1,For what values ofa andb is the functiong : R → R differentiable

at x = 1.

8. Suppose that the functiong : R → R is differentiable atx = 0. Also, suppose that for each natural

numbern, g(1/n) = 0. Prove thatg(0) = 0 andg′(0) = 0.

9. Suppose that the functionf : R → R is differentiable and monotonically increasing. Show that

f ′(x) ≥ 0 for all x.

10. Suppose that the functionf : R → R is differentiable and that there is a bounded sequence{xn}with xn 6= xm, if n 6= m, such thatf(xn) = 0 for every indexn. Show that there is a pointx0 at

whichf(x0) = 0 andf ′(x0) = 0. Hint. Use the Sequential Compactness Theorem.

11. Suppose that the functionf : R → R is differentiable atx0.

Prove that limx→x0

xf(x0)− x0f(x)

x− x0= f(x0)− x0f

′(x0).

12. Suppose that the functionf : R → R is differentiable atx = 0. Prove thatlimx→0

f(x2)− f(0)

x= 0.


13. Suppose that the functionf : R → R is differentiable atx = 0. For real numbersa, b, andc, with

c 6= 0, Prove thatlimx→0

f(ax)− f(bx)

cx=[a− b

c

]

f ′(0).

14. Let the functionh : R → R be bounded. Define the functionf : R → R byf(x) = 1+4x+x2h(x).

Prove thatf(0) = 1 andf ′(0) = 4. Note: There is no assumption about the differentiability of the

functionh.

15. For a natural numbern, the Geometric Sum Formula asserts that

1 + x+ · · ·+ xn =1− xn+1

1− x, x 6= 1.

By differentiating, find a formula for1+x+2x2+ · · ·+nxn and then for12+22x+ · · ·+n2xn−1.

16. Letf(x) = x2 sin1

x, x 6= 0, andf(0) = 0. Show thatf ′(x) is not continuous.

17. Letf(x) = xn sin1

x, x 6= 0, andf(0) = 0. Under what conditions, (1)f(x) is continuous atx = 0;

(2) f ′(x) exists atx = 0; (3) f ′(x) is continuous atx = 0.

18. Letf(x) = |x|n sin1

|x|m , x 6= 0, and f(0) = 0 (m > 0). Under what conditions, (1)f ′(x) is

bounded in some neighborhood ofx = 0; (2) f ′(x) is not bounded in any neighborhood ofx = 0.

19. Letf(x) = (x− a)ϕ(x), whereϕ(x) is continuous atx = a, Find f ′(a).

20. Letf(x) = |x − a|ϕ(x), whereϕ(x) is a continuous function and atϕ(a) 6= 0. Show thatf ′(a)

does not exist. What aref ′−(a) andf ′

+(a).

21. Letf(x) = x2∣

∣

∣ cosπ

x

∣

∣

∣, x 6= 0, andf(0) = 0. Show thatf ′(0) exists, but in any neighborhood of0,

there exists a pointx0, such thatf ′(x0) does not exist.

22. Letf(x) = x2, if x is rational, andf(x) = 0, if x is irrational, Show thatf(x) is differentiable only

at x = 0.

23. Discuss whether the following functions are differentiable:

(a) y = |(x− 1)(x− 2)2(x− 3)3|.

(b) y = | cosx|.

(c) y = |π2 − x2| sin2 x.

(d) y = sin−1(cos x).

(e) y =

x−14 (x+ 1)2, |x| ≤ 1,

|x| − 1, |x| > 1,

24. Find the left and right derivativef ′−(x) andf ′

+(x):

(a) f(x) = |x|.

(b) f(x) = [x] sinπx.

(c) f(x) = x∣

∣

∣ cos πx

∣

∣

∣.

(d) f(x) = x

1+e1x

, x 6= 0; f(0) = 0.

(e) f(x) =√1− e−x2.

(f) f(x) = | ln |x||, x 6= 0.

(g) f(x) = sin−1 2x1+x2 .

(h) f(x) = (x− 2) tan−1 1x−2 , x 6= 2; f(2) = 0.


25. Letf(x) = x sin1

x, x 6= 0, andf(0) = 0. Show thatf(x) is continuous atx = 0, butf(x) has no

left and right derivatives.

26. Letf(x) = x2, x ≤ x0, andf(x) = ax + b, x > x0. For whata and b f(x) is continuous and

differentiable atx = x0?

27. LetF (x) = f(x), x ≤ x0, andf(x) = ax+ b, x > x0, wheref(x) has left derivative. For whata

andb F (x) is continuous and differentiable atx = x0?

28. Find the formula for the sum:

Pn = 1 + 2x+ 3x2 + · · ·+ nxn−1, Qn = 12 + 22x+ 32x2 + · · ·+ n2xn−1.

29. Find the formula for the sum:

Sn = sinx+ sin 2x+ · · ·+ sinnx, Tn = cosx+ 2 cos 2x+ · · ·+ n cosnx.

30. Use the identitycosx

2cos

x

4· · · cos x

2n=

sinx

2n sin x2n

, to find the formula for the sum:

1

2tan

x

2+

1

4tan

x

4+ · · ·+ 1

2ntan

x

2n.

31. Show that the derivative of an even function is an odd function, and the derivative of an odd function

is an even function.

32. Show that the derivative of a periodic function is still aperiodic function with the same period.

33. Find the derivative:

(a)(1− x)p

(1− x)q, x 6= −1. (b) 2tan(1/x).

34. Letf(x) =1√

1 + x2∀x > 0. Find (f−1)′(

√

1/5).

35. LetI be a neighborhood ofx0 and letf : I → R be continuous, strictly monotone, and differentiable

at x0. Assume thatf ′(x0) = 0. Use the characteristic property of inverse,f−1(f(x)) = x, ∀x, and

the Chain Rule to prove that the inverse functionf−1 : f(I) → R is not differentiable atf(x0).

36. Suppose that the functionf : R → R is differentiable and that{xn} is a strictly increasing bounded

sequence withf(xn) ≤ f(xn+1) ∀n ∈ N. Prove that there is a numberx0 at whichf ′(x0) ≥ 0.

Hint. Apply the Monotone Convergence Theorem.

37. For real numbera, b, c, andd, defineQ = {x | cx+ d 6= 0}. Then definef(x) =ax+ b

cx+ d, ∀x ∈ Q.

Show that if the functionf : Q → R is not contant, then it fails to have any local maximizer or

minimizer. Sketch the graph.

38. For c > 0, prove that the following equation does not have two solutions: x3 − 3x + c = 0, 0 <

x < 1.

39. Prove that the following equation has exactly one solution: x5 + 5x+ 1 = 0, −1 < x < 0.


40. Prove that the following equation has exactly two solutions:x4 + 2x2 − 6x+ 2 = 0, x ∈ R.

41. Suppose that the functionf : [a, b] → R is continuous and that the restriction off to the open

interval (a, b) is differentiable. Also suppose thatf(a) > a, f(b) < b, and f ′(x) 6= 1 for all

x ∈ (a, b). Prove that there exists a unique pointξ ∈ (a, b) such thatf(ξ) = ξ.

42. Suppose that the functionf : (0, 1) → R is differentiable, but unbounded. Prove thatf ′(x) is also

unbounded on(0, 1). Hint. If not, |f(x)| ≤ |f(x0)|+ |f ′(ξ)| · |x− x0|.

43. For any numbersa and b and an even natural numbern, show that the following equation has at

most two solutions:xn + ax+ b = 0, x ∈ R. Is this true ifn is odd?

44. For numbersa and b, prove that the following equation has exactly three solutions if and only if

4a3 + 27b2 < 0: x3 + ax+ b = 0, x ∈ R.

45. Letn be a number. Suppose that the functionf : R → R is differentiable and that the equation

f ′(x) = 0, x ∈ R, has at mostn − 1 solutions. Prove that the equationf(x) = 0, x ∈ R, has at

mostn solutions.

46. Use an induction argument to prove that ifp : R → R is a polynomial of degreen, then the equation

p(x) = 0, x ∈ R, has at mostn solutions.

Hint: use the result of the previous problem.

47. Letg : R → R and f : R → R be differentiable functions and suppose thatg(x)f ′(x) =

f(x)g′(x), ∀x. If g(x) 6= 0, ∀x ∈ R, show that there is somec ∈ R such thatf(x) = cg(x), ∀x ∈ R.

48. Definef(x) =

x− x2, if x is rational,

x+ x2, if x is irrational.Show thatf ′(0) = 1 and yet there is no neighbor-

hoodI of the point0 on which this function is monotonically increasing.

49. Let the functionf : R → R have the property that there is a positive numberc such that|f(u) −f(v)| ≤ c(u− v)2, ∀u, v ∈ R. Prove thatf is constant.

50. Suppose that the functionf(x) is differentiable on(a,+∞), and limx→+∞

f ′(x) = 0. Prove that

limx→+∞

f(x)

x= 0.

Hint. Using mean value theorem we get∣

∣

∣

f(x)

x− f(x0)

x

∣

∣

∣=

∣

∣

∣

f(x)− f(x0)

x

∣

∣

∣=

∣

∣

∣

f ′(ξ)(x− x0)

x

∣

∣

∣. For given

ǫ > 0, since limx→+∞

f ′(x) = 0, ∃M > 0, s.t.,|f ′(x)| < ǫ/2, ∀x > M . Takex0 > M , then ifx > x0, we

have∣

∣

∣

f(x)

x− f(x0)

x

∣

∣

∣<

ǫ

2

∣

∣

∣

x− x0

x

∣

∣

∣<

ǫ

2. Therefore,

∣

∣

∣

f(x)

x

∣

∣

∣≤

∣

∣

∣

f(x)

x− f(x0)

x

∣

∣

∣+∣

∣

∣

f(x0)

x

∣

∣

∣<

ǫ

2+∣

∣

∣

f(x0)

x

∣

∣

∣.

Since limx→+∞

f(x0)

x= 0, ∃M1, s.t.,

∣

∣

∣

f(x0)

x

∣

∣

∣<

ǫ

2, ∀x > M1. Therefore,

∣

∣

∣

f(x)

x

∣

∣

∣≤ ǫ

2+

ǫ

2= ǫ, ∀x >

max{x0,M1}, i.e., limx→+∞

f(x)

x= 0.


51. Suppose thatf : R → R is differentiable and that there is a positive numberc such thatf ′(x) ≥c, ∀x. Prove thatf(x) ≥ f(0) + cx if x ≥ 0 andf(x) ≤ f(0) + cx if x ≤ 0. Use these inequalities

to prove thatf(R) = R.

52. Let the functionf : R → R have two derivatives and suppose thatf(x) ≤ 0 andf ′′(x) ≥ 0 for all

x. Prove thatf is constant. Hint: Observe thatf ′ is increasing.

53. Suppose that the functionf : R → R has two derivatives, withf(0) = f ′(0) = 0 and |f ′′(x)| ≤ 1

if |x| ≤ 1. Prove thatf(x) ≤ 1/2 if |x| ≤ 1.

54. Letp : R → R be a polynomial of degree no greater than5. Suppose that at some pointx0 ∈ R,

p(x0) = p′(x0) = · · · = p(5)(x0) = 0. Prove thatp(x) = 0 ∀x ∈ R.

55. Suppose that the functionf : [a, b] → R and g : [a, b] → R are continuous and that their

restrictions to the open interval(a, b) are differentiable. Also suppose that|f ′(x)| ≥ |g′(x)| >0 ∀x ∈ (a, b). Prove that|f(u)− f(v)| ≥ |g(u)− g(v)| ∀u, v ∈ [a, b].

56. Suppose that the functionf : (−1, 1) → R hasn derivatives and that itsnth derivativef (n) :

(−1, 1) → R is bounded. Assume also thatf(0) = f ′(0) = · · · = f (n−1)(0) = 0. Prove that there

is a positive numberM such that|f(x) ≤M |x|n ∀x ∈ (−1, 1).

57. Suppose that the functionf : (−1, 1) → R hasn derivatives. Assume that there is a positive number

M such that|f(x)| ≤M |x|n ∀x ∈ (−1, 1). Prove thatf(0) = f ′(0) = · · · = f (n−1)(0) = 0.

58. Let I be a neighborhood ofx0 and suppose that the functionf : I → R has two continuous

derivatives. Prove that

limh→0

f(x0 + h)− 2f(x0) + f(x0 − h)

h2= f ′′(x0).

59. Suppose thatf(x) is differentiable in the finite or infinite interval(a, b), and

limx→a+

f(x) = limx→b−

f(x). Show that there exists ac ∈ (a, b) such thatf ′(c) = 0.

60. Suppose that the functionf(x) is continuous on[a, b] and differentiable on(a, b). If f(a) = 0 and

there exists a constantC > 0 such that|f ′(x)| ≤ C|f(x)|, ∀x ∈ (a, b). Show thatf(x) ≡ 0 on

[a, b]. Hint. Letx0 = inf{x | f(x) 6= 0, x ∈ [a, b], then it is easy to see thatf(x0) = 0. If x0 6= b, then letx1 > x0,

andM = maxx∈[x0,x1]

|f(x)|. Assume thatf(x) = M > 0, x ∈ [x0, x1]. Thenf(x)− f(x0)

x− x0= f ′(ξ), or

M = |f(x)| = |f ′(ξ)(x− x0)| ≤ C|f(ξ)|(x− x0) ≤ CM(x− x0), i.e.,C(x− x0) ≥ 1, or x− x0 ≥ 1C

Takex1 such thatx− x0 ≤ x1 − x0 < 1C

, we get a contradiction.

61. Suppose thatf (n−1)(x) is continuous in the closed interval[x0, xn], f (n)(x) exists in the open

interval (x0, xn), andf(x0) = f(x1) = · · · = f(xn), x0 < x1 < · · · < xn. Show that there exists

a c ∈ (x0, xn) such thatf (n)(c) = 0.


62. Suppose thatf (p+q)(x) is continuous in the closed interval[a, b], f (p+q+1)(x) exists in the open

interval (a, b), andf(a) = f ′(a) = · · · = f (p)(a) = 0, f(b) = f ′(b) = · · · = f (q)(b) = 0. Show

that there exists ac ∈ (a, b) such thatf (p+q+1)(c) = 0.

63. Suppose thatPn(x) = a0xn + a1x

n−1 + · · · + an, a0 6= 0, whereak, k = 0, 1, · · · , n are real

numbers, has only real roots. Show thatP ′n(x), P

′′n (x), · · · , P (n−1)

n (x) also have only real roots.

64. Prove that the following equation has exactly two solutions:

x4 − 4x3 + 8x2 − 14x+ 11 = 0, 0 ≤ x < +∞.

65. Prove that ifx ≥ 0, then√x+ 1−√

x =1

2√

x+ θ(x), where

1

4≤ θ(x) ≤ 1

2, and lim

x→0θ(x) =

1

4,

limx→+∞

θ(x) =1

2.

66. Prove the inequalities:

(a) | sinx− sin y| ≤ |x− y|. (b)a− b

a< ln

a

b<a− b

b, 0 < b < a.

67. Prove the inequalities:

(a) x− x2/2 < ln(1 + x) < x, x > 0. (b) x− x3/6 < sinx < x, x > 0.

68. Suppose thatf(x) is differentiable in the interval(x0,+∞) and limx→+∞

f ′(x) = 0.

Prove that limx→+∞

f(x)

x= 0.


(a) limx→0

tanx− x

x− sinx. (b) lim

x→+∞xn

eax, a > 0, n > 0.

70. Letf(x) = (x − a)nϕ(x), where the functionϕ(x) has continuousn − 1 derivatives in the neigh-

borhood ofa. Find f (n)(a).

71. Find the local maximizer , local minimizer or stationarypoints of the following functions.

(a) x3 − x+ 5. (b) x9 − x7 + 11.

72. If (a)f(x) is differentiable atx0, butg(x) is not differentiable atx0; (b) bothf(x) andg(x) are not

differentiable atx0, can you conclude thatF (x) = f(x) + g(x) is not differentiable atx0?

73. If (a)f(x) is differentiable atx0, butg(x) is not differentiable atx0; (b) bothf(x) andg(x) are not

differentiable atx0, can you conclude thatF (x) = f(x)g(x) is not differentiable atx0? Consider

the example (a)f(x) = x, g(x) = |x|; (b) f(x) = |x|, g(x) = |x|.

74. If (a)f(x) is differentiable atx = g(x0), butg(x) is not differentiable atx0; (b) f(x) is not differ-

entiable atx = g(x0), but g(x) is differentiable atx0; (c) f(x) is not differentiable atx = g(x0),

and g(x) is not differentiable atx0, can you conclude thatF (x) = f(g(x)) is not differentiable

at x0? Consider the example (a)f(x) = x2, g(x) = |x|; (b) f(x) = |x|, g(x) = x2; (c)

f(x) = 2x+ |x|, g(x) = 23x− 1

3 |x|.


75. Canf(x) have (a) finite derivative; (b) infinite derivative, at its discontinuous points? Consider the

examplef(x) = sgnx.

76. If f(x) is differentiable on the finite interval(a, b) and limx→a

f(x) = ∞. Is it true limx→a

f ′(x) = ∞?

Consider the examplef(x) =1

x+ cos

1

x, asx→ 0.

77. If f(x) is differentiable on the finite interval(a, b) and limx→a

f ′(x) = ∞. Is it true limx→a

f(x) = ∞?

Consider the examplef(x) = 3√x, asx→ 0.

78. If f(x) is differentiable on the interval(x0,+∞) and limx→+∞

f(x) exists. Is it true limx→+∞

f ′(x)

exists? Consider the examplef(x) =sin(x2)

x.

79. If f(x) is bounded, differentiable on the interval(x0,+∞), and limx→+∞

f ′(x) exists. Is it true

limx→+∞

f(x) exists? Consider the examplef(x) = cos(lnx).

Chapter 5

Indefinite Integral

5.1 Definition and Some Basic Formulas

5.1.1 Definition

Definition 5.1.1Letf(x) be a function defined on[a, b]. If F (x) is differential on(a, b) and satisfies

F ′(x) = f(x) x ∈ (a, b),

thenF (x) is called anantiderivative of f(x) and the set of all antiderivatives off(x) is called theindefi-

nite integral of f(x) and denoted by∫

f(x)dx.

If F (x) andF (x) are two antiderivatives off(x), then

[F (x) − F (x)]′ = F ′(x)− F ′(x) = f(x)− f(x) = 0.

By Corollary 4.4.5,

F (x) − F (x) = C, F (x) = F (x) + C.

Hence ifF (x) is an antiderivative off(x), the indefinite integral is∫

f(x)dx = F (x) + C

where C is an arbitrary constant. By the definition for any constantsα, β∫

[αf(x) + βg(x)]dx = α

∫

f(x)dx + β

∫

g(x)dx.

Existence of the antiderivative.The existence of the antiderivative will be discussed in thenext chapter.

The antiderivative may not exist for some functions.

107


Example 5.1.1Let

f(x) =

1, x ≤ 0,

0, x > 0.

Then there is no functionF (x) such thatF ′(x) = f(x), x ∈ (−∞,+∞).

Proof. By contradiction. Suppose that there exists aF (x), s.t.,F ′(x) = f(x), x ∈ (−∞,+∞). Then we

haveF ′(0) = f(0) = 1. Let∆x > 0 and consider the interval[0,∆x]. F (x) is continuous on[0,∆x] and

differentiable on(0,∆x). Using Lagrange Mean Value Theorem we get

F (0 + ∆x)− F (0)

∆x= F ′(ξ) = f(ξ), 0 < ξ < ∆x.

Therefore,

F ′+(0) = lim

∆x→0+

F (0 + ∆x)− F (0)

∆xlim

ξ→0+f(ξ) = 0,

which contradicts withF ′(0) = 1.

Some Basic Formulas.

1.∫

xαdx =1

α+ 1xd+1 + C, α 6= 1.

2.∫

dx

x= ln |x|+ C.

3.∫

exdx = ex + C.

4.∫

cosxdx = sinx+ C.∫

sinxdx = − cosx+ C.

5∫

sec2 xdx = tanx+ C.∫

csc2 xdx = − cotx+ C.

6.∫

dx√a2 − x2

= sin−1(x

a

)

+ C.

7.∫

dx

a2 + x2=

1

atan−1

(x

a

)

.

8.∫

axdx =1

ln aax + C, (a > 0, a 6= 1).

5.2 Integration by Substitution

Suppose thatF ′(y) = f(y), y = ψ(x). By the Chain Rule of differentiating composite function

d

dxF (ψ(x)) = F ′(y)ψ′(x) = f(ψ(x))ψ′(x).

Hence∫

f(ψ(x))ψ′(x)dx = F (ψ(x)) + C,

or,∫

f(ψ(x))dψ(x) =

∫

f(y)dy = F (y) + C = F (ψ(x)) + C.

We will gain advantage from the formula if the antiderivative off(y) is known and the functiong(x) that

we are going to find its indefinite integral can be expressed as

g(x) = f(y(x))y′(x).

Chapter 5: Indefinite Integrals 109

Example 5.2.1

∫

tanxdx =

∫

sinx

cosxdx =

∫ −d(cosx)

cosx

= − ln | cosx|+ C.

Example 5.2.2Lety = x2, then

∫

xdx

1 + x4=

1

2

∫

dy

1 + y2=

1

2tan−1 y + C =

1

2tan−1 x2 + C

Example 5.2.3

∫

sinmx sinnxdx =1

2

∫

[cos(m− n)x− cos(m+ n)x]dx

=1

2

[

1

m− n

∫

cos(m− n)xd(m − n)x− 1

m+ n

∫

cos(m+ n)xd(m + n)x

]

=1

2

[

1

m− nsin(m− n)x− 1

m+ nsin(m+ n)x

]

+ C, m 6= n.

Similarly, we can find the following integrals:

∫

sinmx cosnxdx,

∫

cosmx cosnxdx.

Example 5.2.4

∫

dx

a2 − x2=

1

2a

[∫

dx

a+ x+

∫

dx

a− x

]

=1

2a[ln |a+ x| − ln |a− x|+ C]

=1

2aln

∣

∣

∣

∣

a+ x

a− x

∣

∣

∣

∣

+ C

Example 5.2.5

∫

dx

sinx=

∫

dx

2 sin x2 cos

x2

=

∫

sec2 x2

2 tan x2

dx

=

∫

d(tan x2 )

tan x2

= ln∣

∣

∣tanx

2

∣

∣

∣+ C


Example 5.2.6Letx = a sin t, thent = sin−1 xa , we have

∫

√

a2 − x2dx =

∫

a cos t · a cos tdt

= a2∫

cos2 dt = a2∫

1

2(1 + cos 2t)dt

=a2

2

[

t+1

2sin 2t

]

+ C

=a2

2[t+ sin t cos t] + C

=a2

2

[

sin−1 x

a+x

a

√

1− (x

a

2)

]

+ C

=a2

2sin−1 x

a+x

2

√

a2 − x2 + C

Example 5.2.7∫

dx√x2 + a2

= ln |x+√

x2 + a2|+ C

Example 5.2.8∫

dx√x2 − a2

= ln |x+√

x2 − a2|+ C

5.3 Integration by Parts

From the differentiation rule

[f(x) · g(x)]′ = f ′(x)g(x) + f(x)g′(x)

we get the formula for indefinite integral

∫

f(x)g′(x)dx =

∫

[f(x)g(x)]′dx−∫

f ′(x)g(x)dx,

or,∫

f(x)dg(x) = f(x)g(x) −∫

g(x)df(x).

This formula is calledintegration by parts. For some complicated integrals, the integral on the right hand

side may be a simplified one.

Example 5.3.1Let In =

∫

xnexdx, then we have

In =

∫

xndex = xnex −∫

exd(xn)

= xnex − n

∫

xn−1exdx = xnex − nIn−1


Therefore,

In = xnex − nIn−1

= xnex − nxn−1ex + n(n− 1)In−2

= · · ·

= ex[xn − nxn−1 + n(n− 1)xn−2 − · · ·+ (−1)nn!] + C

Example 5.3.2∫

xn lnxdx =

∫

lnxd

(

xn+1

n+ 1

)

=xn+1

n+ 1lnx−

∫

xn+1

n+ 1d(lnx)

=xn+1

n+ 1lnx− 1

n+ 1

∫

xn+1 · 1xdx

=xn+1

n+ 1lnx− 1

n+ 1

∫

xndx

=xn+1

n+ 1lnx− xn+1

(n+ 1)2+ C

Example 5.3.3∫

tan−1 xdx = x tan−1 x−∫

xd(tan−1 x)

= x tan−1 x−∫

x

1 + x2dx

= x tan−1 x− 1

2

∫

d(x2)

1 + x2

= x tan−1 x− 1

2ln(1 + x2) + C

Example 5.3.4(Example 5.2.6)∫

√

a2 − x2dx = x ·√

a2 − x2 −∫

xd√

a2 − x2

= x ·√

a2 − x2 +

∫

x2dx√a2 − x2

= x ·√

a2 − x2 −∫

a2 − x2√a2 − x2

dx+ a2dx√a2 − x2

= x ·√

a2 − x2 −∫

√

a2 − x2dx+ a2 sin−1 x

a+ C

=⇒∫

√

a2 − x2dx =x

2

√

a2 − x2 +a2

2sin−1 x

a+ C

In some case we can’t apply the method of integration by partsto obtain the result directly, but we can

obtain a recurrent formula and then by this formula to get theresult that we want.


Example 5.3.5Find∫

dx

(x2 + a2)4.

We considerIn =

∫

dx

(x2 + a2)n

In =x

(x2 + a2)n−∫

xd1

(x2 + a2)n

=x

(x2 + a2)n+ 2n

∫

x2

(x2 + a2)n+1dx

=x

(x2 + a2)n+ 2n In − 2na2 In+1

=⇒ In+1 =2n− 1

2na2In +

x

2na2(x2 + a2)n

It’s known that

I1 =

∫

dx

x2 + a2=

1

atan−1 x

a+ C

Hence by the recurrent formula we can getI2, I3 and then the required resultI4

5.4 Integration of Rational Functions


x+ 3

(x+ 1)(x− 2)2dx.

Letx+ 3

(x+ 1)(x− 2)2=

a

x+ 1+

b

x− 2+

c

(x − 2)2.

Then we have

x+ 3 = a(x− 2)2 + b(x+ 1)(x− 2) + c(x+ 1).

Compare the coefficients we geta = 2/9, b = −2/9, c = 5/3. Therefore,∫

x+ 3

(x + 1)(x− 2)2dx = a

∫

dxx+ 1

+ b

∫

dxx− 2

+ c

∫

dx(x− 2)2

=2

9ln |x+ 1| − 2

9ln |x− 2| − 5

3· 1

x− 2+ C


6x2 − 15x+ 22

(x+ 3)(x2 + 2)2dx.

Let6x2 − 15x+ 22

(x+ 3)(x2 + 2)2=

a

x+ 3+bx+ c

x2 + 2+

dx + e

(x2 + 2)2.

Then we have

6x2 − 15x+ 22 = a(x2 + 2)2 + (bx+ c)(x+ 3)(x2 + 2) + (dx+ e)(x+ 3).


Compare the coefficients we geta = 1, b = −1, c = 3, d = −5, e = 0. Therefore,

∫

6x2 − 15x+ 22

(x+ 3)(x2 + 2)2dx =

∫

dxx+ 3

+

∫ −x+ 3

x2 + 2dx+

∫ −5x

(x2 + 2)2dx

= ln |x+ 3| − 1

2ln |x2 + 2|+ 3√

2tan−1

( x√2

)

+5

2(x2 + 2)+ C

In general, to decompose a rational functionf(x) = p(x)/q(x) into partial fractions, we proceed as

follows:

1. If f(x) is improper, i.e.,p(x) is of degree at least that ofq(x), dividep(x) by q(x), obtaining

p(x)

q(x)= g(x) +

r(x)

q(x).

2. Factorq(x) into a product of linear and irreducible quadratic factors with real coefficients.

3. For each factor of the form(ax+ b)k, expect the decomposition to have the terms

A1

(ax+ b)+

A2

(ax+ b)2+ · · ·+ Ak

(ax+ b)k.

4. For each factor of the form(ax2 + bx+ c)m, expect the decomposition to have the terms

B1x+ C1

(ax2 + bx+ c)+

B2x+ C2

(ax2 + bx+ c)2+ · · ·+ Bmx+ Cm

(ax2 + bx+ c)m.

5. Setr(x)/q(x) equal to the sum of all the terms found in Step 3 and 4. The number of constants to be

determined should equal the degree of the denominatorq(x).

6. Multiply both sides of the equation found in Step 5 byq(x) and solve for the unknown constants.

This can be done by either of the two methods: (1) Equate coefficients of like-degree terms or (2)

assign convenient values to the variablex.

The above decompositions are based on the following two theorems:

Theorem 5.4.1If q(x) = (ax+ b)kq1(x), q1(−b/a) 6= 0, then

r(x)

q(x)=

Ak

(ax+ b)k+

r1(x)

(ax+ b)k−1q1(x),

whereAk is a constant, andr1(x) is a polynomial.

Proof. Only need to selectAk andr1(x) s.t.

r(x) = Akq1(x) + (ax+ b)r1(x).

Letx = −b/a, then we getAk. Sincer(−b/a)− Akq1(−b/a) = 0, r(x) − Akq1(x) has a factorax+ b.

Then we getr1(x).


Theorem 5.4.2If q(x) = (ax2 + bx+ c)mq1(x), andq1(x) has no factorax2 + bx+ c, then

r(x)

q(x)=

Bmx+ Cm

(ax2 + bx+ c)m+

r1(x)

(ax2 + bx+ c)m−1q1(x),

whereBm andCm are constants, andr1(x) is a polynomial.

Proof. Only need to selectBm, Cm, andr1(x) s.t.

r(x)

ax2 + bx+ c− (Bmx+ Cm)

q1(x)

ax2 + bx+ c= r1(x).

Suppose that

r(x)

ax2 + bx+ c= g(x) +

αx+ β

ax2 + bx+ c,

q1(x)

ax2 + bx+ c= h(x) +

γx+ δ

ax2 + bx+ c.

Then we have

r1(x) = g(x) + (Bmx+ Cm)h(x) − Bmγ

a

+

(

α−Bmδ − Cmγ + Bmγa b

)

x+ β − Cmδ +Bmγa c

ax2 + bx+ c.

There is a unique solutionBm andCm such thatr1(x) is a polynomial. .

In fact, it’s not true that we can find the indefinite integral for every elementary function. It can be

proved that some function looks quite simple such as

sinx

xor e−x2

but∫

sin xx dx or

∫

e−x2

dx can’t be expressed by some formula of elementary functions.

5.5 Exercise

1. Evaluate the following indefinite integrals using substitution:

(a)∫

x 3√x+ πdx.

(b)∫

dx√x2 + 2x+ 5

.

(c)∫

x2 + 3x√x+ 4

dx.

(d)∫

dx

x2√x2 − 1

.

2. Evaluate the following indefinite integrals using integration by parts:

(a)∫

x cos2 x sinxdx.

(b)∫

lnx

x2dx.

(c)∫

tan−1(1

x

)

dx.

(d)∫

x2 cosxdx.

(e)∫

eax cosxdx.

(f)∫

cos6 x sin2 xdx.

Chapter 6: Integration 115

3. Evaluate the following indefinite integrals using partial fractions:

(a)∫

3x3 − 8x+ 13

(x+ 3)(x− 1)2dx. (b)

∫

1

(x− 1)2(x2 + 2x+ 6)2dx.


Chapter 6

Integration

6.1 Darboux Sums and Definition of Integral

6.1.1 Darboux Sum

If n is a natural number and

a = x0 < x1 < · · · < xn = b

thenP = {x0, x1, · · · , xn} is called a Partitionof the interval[a, b], andxi is called a partition pointof

P , the interval[xi−1, xi] is called a subinterval of partitionP .

Suppose that functionf(x) is defined on[a, b] and bounded. For a partitionP we define

mi = inf{f(x) |x ∈ [xi−1, xi]} i = 1, · · · , n

Mi = sup{f(x) |x ∈ [xi−1, xi]} i = 1, · · · , n

and then define

L(f, P ) =n∑

i=1

mi(xi − xi−1)

U(f, P ) =

n∑

i=1

Mi(xi − xi−1)

We callU(f, P ) andL(f, P ) the Upper Darboux Sum and the Lower Darboux Sum for the function f

based on the partitionP .

6.1.2 Refinement Lemma

Lemma 6.1.1 Suppose that functionf(x) is bounded on[a, b] and satisfies

m ≤ f(x) ≤M, for all x ∈ [a, b]

117


Then for any partitionP of [a, b],

m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(b− a).

Proof. Since

m ≤ infa≤x≤b

{f(x)} ≤ infxi−1≤x≤xi

{f(x)} = mi

and

Mi = supxi−1≤x≤xi

{f(x)} ≤ supa≤x≤b

{f(x)} ≤M

and obviously

mi ≤Mi

Hence

n∑

i=1

m(xi − xi−1) ≤n∑

i=1

mi(xi − xi−1) ≤n∑

i=1

Mi(xi − xi−1) ≤n∑

i=1

M(xi − xi−1)

Namely,

m(b− a) ≤ L(f, P ) ≤ U(f, P ) ≤M(b− a).

Definition 6.1.2 Partition P ∗ of [a, b] is called a refinement ofP if each partition point ofP is also a

partition point ofP ∗.

Lemma 6.1.3 (Refinement Lemma) Suppose that the functionf(x) is bounded on[a, b]. Let P be a

partition of [a, b] andP ∗ be a refinement ofP . Then

L(f, P ) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P ).

Proof. LetPi be the partition of[xi−1, xi] that is induced byP ∗ and[xi−1, xi] is a subinterval ofP . By

the above lemma 1 in each subinterval[xi−1 xi],

mi(xi − xi−1) ≤ L(f, Pi) ≤ U(f, Pi) ≤Mi(xi − xi−1)

Hence, we have

n∑

i=1


i=1

L(f, Pi) ≤n∑

i=1

U(f, Pi) ≤n∑

i=1

Mi(xi − xi−1)

Butn∑

i=1

L(f, Pi) = L(f, P ∗) andn∑

i=1

U(f, Pi) = U(f, P ∗). Therefore

L(f, P ) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P ).


Lemma 6.1.4Suppose that functionf(x) is bounded on[a, b]. Then for any two partitionsP1 andP2 of

[a, b],

L(f, P1) ≤ U(f, P2)

Proof. LetP ∗ be a partitions formed by taking the union of partition points ofP1 andP2. HenceP ∗ is a

common refinement ofP1 andP2. From the Refinement Lemma

L(f, P1) ≤ L(f, P ∗) ≤ U(f, P ∗) ≤ U(f, P2).

6.1.3 Definition of Integral

Let L be the collection of all lower Darboux sums andU be the collection of all the upper Darboux

sums. From the Lemma 6.1.4, we know that any upper Darboux sumU(f, P ) is an upper bound ofL.

Hence by Completeness Axiom, the setL has the least upper bound (denoted byI∗ or I∗(f)) and

I∗ = supL ≤ U(f, P )

Similarly, any lower Darboux sum L(f,P) is a lower bound of the setU . HenceU is bounded below and has

the greatest lower bound (denoted byI∗ or I∗(f)) and satisfies

L(f, P ) ≤ inf U = I∗

for any partitionP . Therefore,

I∗ ≤ I∗.

I∗ is called the lower integralof f on [a, b], andI∗ is called the upper integralof f on [a, b].

Definition 6.1.5 Suppose that functionf(x) is defined on[a, b] and bounded. Thenf(x) is said to be

integrableprovided that

I∗ = I∗ = I,

and the numberI is called the integral(or definite integral) of the functionf(x) on the interval[a, b],

denoted by∫ b

a

f or∫ b

a

f(x)dx.

If I∗ 6= I∗, then we say thatf(x) is not integrable on[a, b]. This integral is also called Riemann integral.

If a ≥ b andf is integrable on[b, a], we define∫ b

a

f(x)dx = −∫ a

b

f(x)dx,

∫ a

a

f(x)dx = 0.

Note. From the definition, the integral is independent of the variable under the integral, i.e.,∫ b

a

f(x)dx =

∫ b

a

f(t)dt.


6.1.4 Interpretation in Physics and Geometry

• If v(t) is the speed function of a car, then∫ t1

t0

v(t)dt is the distance covered by the car from timet0

to t1.

• If ρ(x) is the line density of a rod with lengthl, then∫ l

0

ρ(x)dx is the mass of the rod.

•∫ b

a

f(x)dx is the area of the region between the curvey = f(x) and the x-axis

x

y

y=f(x)

6.2 Integrability

6.2.1 Integrability Criterion

Theorem 6.2.1Suppose that functionf(x) is bounded on[a, b]. Thenf(x) is integrable if and only if for

any positive numberε, there exists a partitionP ∗ of [a, b] such that

U(f, P ∗)− L(f, P ∗) < ε

Proof. Part of ”if” : If for any ε > 0, the inequality

U(f, P ∗)− L(f, P ∗) < ε

holds for some partitionP ∗, then we can prove

supL = inf U

That meansf(x) integrable. For partitionP ∗, we have the inequalities

L(f, P ∗) ≤ supL ≤ inf U ≤ U(f, P ∗)

Therefore, we obtain

0 ≤ inf U − supL ≤ U(f, P ∗)− L(f, P ∗) < ε


Becauseinf U andsupL are two constant andε is any positive number, the inequalities

0 ≤ (inf U − supL) < ε implies

inf U − supL = 0

i.e., inf U = supLPart of ”only if” : If inf U = supL = I, then we can prove for anyε > 0, there exists a partitionP ∗ such

that

U(f, P ∗)− L(f, P ∗) < ε

BecauseI is the greatest lower bound ofU , that means for a givenε > 0, there is an elementU(f, P1) in

U such that

U(f, P1)− I <ε

2

Similarly I is the least upper bound ofL, there is an elementL(f, P2) in L such that

I − L(f, P2) = I <ε

2

LetP ∗ be formed by taking the union of partition points ofP1 andP2. ThenP ∗ is common refinement of

P1 andP2. By Refinement Lemma,

U(f, P ∗) ≤ U(f, P1)

L(f, P ∗) ≥ L(f, P2)

Therefore

U(f, P ∗)− L(f, P ∗) ≤ U(f, P1)− L(f, P2)

= [U(f, P1)− I] + [I − L(f, P2)]

<ε

2+ε

2= ε


Example 6.2.1Functionf(x) is piecewise constant, meaning that there are pointsx0, x1, . . . , xn with

a = x0 < x1 < · · · < xn = b and

f(x) = ci for x ∈ (xi−1, xi)

and at pointsxi, f(x) may be any value. Prove thatf is integrable.

Proof. Because the number ofxi is finite, f(x) is bounded. Letm,M be two real numbers such that

m ≤ f(x) ≤M and leth = min1≤i≤n

(xi − xi−1). Take partitionP as follows,

P = {x0, x0 + δ, x1 − δ, x1 + δ, · · ·xn−1 − δ, xn+1 + δ, xn − δ, xn}


with δ ≤ h

3. (Note thatδ ≤ h

3assuresxi+1 − δ > xi + δ). For the partitionP

U(f, P )− L(f, P ) ≤ (M −m)(δ + (n− 1)2δ + δ)

= 2n(M −m)δ

Hence for anyε > 0, takeδ = min

{

h

3,

ε

1 + 2n(M −m)

}

, then for theP with thisδ,

U(f, P )− L(f, P ) < ε.

By the Integrability Criterion, the piecewise constant function is integrable.

6.2.2 Two Kinds of Integrable Functions

Theorem 6.2.2Suppose thatf(x) is monotone on[a, b]. Thenf is integrable.

Proof. LetPn = {x0, x1, . . . , xn} be a partition with evenly spaced points. That isxi − xi−1 =b− a

nfor

all i. Pn is called Regular Partition. Since

mi = infxi−1≤x≤xi

{f(x)} = f(xi−1)

Mi = supxi−1≤x≤xi

{f(x)} = f(xi)

U(f, Pn)− L(f, Pn) =n∑

i=1

(Mi −mi)(xi − xi−1)

=b− a

n

n∑

i=1

(f(xi)− f(xi−1))

=b− a

n(f(b)− f(a))

Hence, we can choose a sufficiently largen such that

U(f, Pn)− L(f, Pn) < ε.

Theorem 6.2.3Suppose thatf(x) is continuous on[a, b]. Thenf is integrable.

Proof. f(x) is continuous in[a, b], then it is uniformly continuous. Hence we may choose a number δ > 0

such that for any givenε > 0

|f(u)− f(v)| < ε

b− a, if |u− v| ≤ δ

Now we choose a regular partitionPn withn such thatb− a

n≤ δ. For thisPn, the length of any subinterval

[xi−1, xi] isb− a

n≤ δ. By Extreme Value Theorem of continuous function, there areui, vi ∈ [xi−1 xi]

such thatMi = f(ui),mi = f(vi). Then we obtain

Mi −mi = f(ui)− f(vi) <ε

b− a


due to

|ui − vi| ≤ (xi − xi−1) ≤ δ

Hence

U(f, Pn)− L(f, Pn) =

n∑

i=1


<ε

b− a

n∑

i=1

(xi − xi−1) = ε.

Corollary 6.2.4 Suppose that functionf(x) is bounded on[a, b] and continuous in(a, b). Thenf is inte-

grable.

Proof. Supposef(x) ≤ M, ∀x ∈ [a, b], and letδ > 0 be a small number. Sincef(x) is continuous on

[a+ δ, b− δ], thenf(x) is integrable on[a+ δ, b− δ]. For givenǫ > 0, there exists a partitionP ∗ (Theorem

6.2.1), s.t.,

U(f, P ∗)− L(f, P ∗) <ǫ

2.

By adding two subintervals[a, a+ δ] and[b− δ, b] to the partitionP ∗, we get a partitionP of [a, b]. Letma

andMa be the inf and sup off(x) on [a, a+ δ], and letmb andMb be the inf and sup off(x) on [b− δ, b],

then we have

U(f, P )− L(f, P ) = (Ma −ma)δ + U(f, P ∗)− L(f, P ∗) + (Mb −mb)δ

≤ 2Mδ +ǫ

2+ 2Mδ = 4Mδ +

ǫ

2.

Chooseδ such that4Mδ <ǫ

2, then we have

U(f, P )− L(f, P ) < ǫ.

From Theorem 6.2.1,f(x) is integrable on[a, b].

Example 6.2.2f(x) =

sinπ

x, if 0 < x ≤ 1

4, if x = 0is integrable.

Note.Thef(x) in this example is not continuous atx = 0.

6.2.3 Convergence of Darboux Sums

Definition 6.2.5 For a partition P = {x0, . . . xn} of interval [a, b], we define the Gap ofP , denoted by

||P ||, the length of the largest subinterval induced byP . That is

||P || = max1≤i≤n

[xi − xi−1].


Then||P || < α means every subinterval of partitionP has length less thanα.

Lemma 6.2.6Suppose that functionf(x) is bounded on[a, b]. Thenf is integrable if and only if for any

ε > 0, there is a positiveδ such that

U(f, P )− L(f, P ) < ε, if ||P || < δ.

Proof. By the Integrability Criterion, the proof of ”part if” has completed immediately. We now prove the

converse. Suppose thatf(x) is integrable. According to the Integrability Criterion, for givenε > 0, we can

choose a partitionP ∗ = {x∗0, . . . , x∗n} of [a, b] such that

U(f, P ∗)− L(f, P ∗) <ε

2.

ChooseM > 0 such that

−M ≤ f(x) ≤M, x ∈ [a, b].

We are now going to obtain the following estimate

U(f, P )− L(f, P ) ≤ 2nM ||P ||+ [U(f, P ∗)− L(f, P ∗)],

for any partitionP . Once this estimate is proven, we takeδ = ε/4nM . Then

U(f, P )− L(f, P ) < ε, if ||P || < δ.

Let P = {x0, . . . , xm}. Separate the set of indices{1, . . . ,m} into two subsetsA andB. DefineA to

be the set of indicesi such that the subinterval(xi−1, xi) contains a partition point ofP ∗. DefineB to

be the set of indicesi such that the subinterval(xi−1, xi) doesn’t contain any partition point ofP ∗. Since

Mi −mi ≤ 2M, xi − xi−1 ≤ ||P || and there are fewer thann indices inA,

∑

i∈A

(Mi −mi)(xi − xi−1) ≤ 2nM ||P ||.

On the other hand, if indexi ∈ B, then the subinterval[xi−1 xi] is contained in one of the subintervals

[x∗j−1 x∗j ] induced byP ∗. Because in such subintervals[xi−1 xi],

∑

[xi−1 xi]⊆[x∗

j−1 x∗

j]

(Mi −mi)(xi − xi−1) ≤

[

supx∗

j−1≤x≤x∗

j

{f(x)} − infx∗

j−1≤x≤x∗

j

{f(x)}]

(x∗j − x∗j−1)

Hence∑

i∈B

(Mi −mi)(xi − xi−1) ≤ U(f, P ∗)− L(f, P ∗)


Consequently

U(f, P )− L(f, P ) =∑

i∈A

(Mi −mi)(xi − xi−1) +∑

i∈B


≤ 2nM ||P ||+ [U(f, P ∗)− L(f, P ∗)].

Theorem 6.2.7 (Darboux Sum Convergence Criterion)Suppose that functionf(x) is bounded on[a, b].

Thenf is integrable and∫ b

a

f(x)dx = A if and only if

lim||Pn||→0

L(f, Pn) = lim||Pn||→0

U(f, Pn) = A

Proof. part of ”only if”: Let {Pn} be any sequence of partitions of[a, b] with limn→∞

||Pn|| = 0. Becausef

is integrable, according the Lemma 6.2.6, for anyε > 0, we can chooseδ > 0 such that for any partitionP

U(f, P )− L(f, P ) < ε, if ||P || < δ.

Since limn→∞

||Pn|| = 0, we can choose a natural numberN such that

||Pn|| < δ if n ≥ N.

Thus

0 ≤ U(f, Pn)− L(f, Pn) < ε, if n ≥ N.

By the definition of integral

L(f, Pn) ≤∫ b

a

f(x)dx ≤ U(f, Pn).

Consequently,

|∫ b

a

f(x)dx − L(f, Pn)| < ε,

|∫ b

a

f(x)dx− U(f, Pn)| < ε, n ≥ N.

That is

lim||Pn||→0


U(f, Pn) =

∫ b

a

f(x)dx.

part of ”if”: Because

lim||Pn||→0


U(f, Pn) = A,

for anyε > 0, there is a natural numberN such that

U(f, Pn)− L(f, Pn) < ε, if n ≥ N.


By the Integrable Criterion,f is integrable and satisfies

L(f, Pn) ≤∫ b

a

f(x)dx ≤ U(f, Pn).

Hence, by the Squeezing Principle,∫ b

a

f(x)dx = A.

6.2.4 Convergence of Riemann Sums

Definition 6.2.8 Suppose that functionf(x) is defined on[a, b] andP = {x0, . . . , xn} is a partition of

[a, b]. Letci ∈ [xi−1, xi]. Then the sum

n∑

i=1

f(ci)(xi − xi−1)

is called a Riemann sum for the functionf(x) based on the partitionP . Denote it byR(f, P ).

Note.Riemann sumR(f, P ) depends onf(x) and partitionP , and also depends on the choice of theci’s.

But the notationR(f, P ) does not exhibit this dependence.

From the definitions of Riemann sum and Darboux sums, for eachpartitionP ,

L(f, P ) ≤ R(f, P ) ≤ U(f, P ).

Lemma 6.2.9Suppose that functionf(x) is bounded on[a, b]. Then for each partitionP of [a, b] and each

positiveε, there are Riemann sumsR(f, P ) andR′(f, P ) (with suitableci’s) such that

0 ≤ U(f, P )−R(f, P ) < ε

and 0 ≤ R′(f, P )− L(f, P ) < ε.

Proof. It’s known that the upper Darboux sum

U(f, P ) =∑

Mi(xi − xi−1) where Mi = supxi−1≤x≤xi

{f(x)}.

By the meaning of least upper bound,

Mi −ε

(b − a)

is not an upper bound of the set{f(x) |x ∈ [xi−1 xi]}. Therefore, there is a pointci ∈ [xi−1, xi] such that

f(ci) > Mi −ε

(b− a).

Hencen∑

i=1

[

Mi −ε

(b − a)

]

(xi − xi−1) <

n∑

i=1

f(ci)(xi − xi−1),


and thenU(f, P )− ε < R(f, P ), or 0 ≤ U(f, P )−R(f, P ) < ε. Similarly we choosec′i ∈ [xi−1 xi] such

that

f(c′i) < mi +ε

(b − a),

and then obtain

0 ≤ R′(f, P )− L(f, P ) < ε,

where

R′(f, P ) =n∑

i=1

f(c′i)(xi − xi−1).

Theorem 6.2.10 (Riemann Sum Convergence Criterion)Suppose thatf(x) is bounded on[a, b] and

{Pn} is any sequence of partition of[a, b] with limn→∞

||Pn|| = 0. Thenf is integrable and∫ b

a

f(x)dx = A,

if and only if

limn→∞

R(f, Pn) = A.

Proof. part of ”only if”: Suppose thatf is integrable and∫ b

a

f(x)dx = A. From the Theorem of Darboux

Sum Convergence Criterion, we have

lim||Pn||→0


U(f, Pn) = A.

But it’s known that

L(f, Pn) ≤ R(f, Pn) ≤ U(f, Pn).

By Squeezing Principle,

lim||Pn||→0

R(f, Pn) = A.

part of ”if”: By the preceding Lemma 6.2.9, forε = 1/nwe choose Riemann sumsR(f, Pn) andR′(f, Pn)

such that

0 ≤ U(f, Pn)−R(f, Pn) <1

n,

0 ≤ R′(f, Pn)− L(f, Pn) <1

n.

Consequently, we have

R′(f, Pn)−1

n< L(f, Pn) ≤ U(f, Pn) < R(f, Pn) +

1

n.

By the assumption

limn→∞

R(f, Pn) = limn→∞

R′(f, Pn) = A.

Hence

limn→∞

L(f, Pn) = limn→∞

U(f, Pn) = A.

According to the Darboux Sum Convergence Criterion,f(x) is integrable and∫ b

a

f(x)dx = A.


6.3 Linearity, Monotonicity and Additivity

6.3.1 Linearity of Integral

Theorem 6.3.1Supposef(x) andg(x) are integrable on[a, b], then

∫ b

a

[αf(x) + βg(x)]dx = α

∫ b

a

f(x)dx + β

∫ b

a

g(x)dx.

Proof. By using the theorem on convergence of Riemann sums (or Darboux sums), we can prove the identity

immediately.

6.3.2 Monotonicity of Integral

Theorem 6.3.2Suppose thatf andg are integrable on[a, b] and

f(x) ≤ g(x), x ∈ [a, b]

then∫ b

a

f(x)dx ≤∫ b

a

g(x)dx.

Proof. For any partitionP , U(f, P ) ≤ U(g, P ). Choose a sequence{Pn} with limn→0

||Pn|| = 0. Then we

obtain

limn→0

U(f, Pn) ≤ limn→0

U(g, Pn), i.e.,∫ b

a

f(x)dx ≤∫ b

a

g(x)dx.

Corollary 6.3.3 Suppose thatf is integrable on[a, b], then∣

∣

∣

∣

∣

∫ b

a

f(x)dx

∣

∣

∣

∣

∣

≤∫ b

a

|f(x)|dx.

Proof. Sincef is integrable on[a, b], then from Theorem 6.2.1, for givenǫ > 0 there exists a partition

P = {x0, . . . , xn} of [a, b] such that

n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) < ǫ,

whereMi(f) is the sup off(x) on [xi−1, xi], andmi(f) is the inf off(x) on [xi−1, xi]. If mi(f) and

Mi(f) have the same sign, then

Mi(|f |)−mi(|f |) =Mi(f)−mi(f).


If Mi(f) ≥ 0,mi(f) ≤ 0, andMi(f) ≤ |mi(f)|, then we have

Mi(|f |)−mi(|f |) = −mi(f)−mi(|f |) ≤ −mi(f) ≤Mi(f)−mi(f).

If Mi(f) ≥ 0,mi(f) ≤ 0, andMi(f) ≥ |mi(f)|, then we have

Mi(|f |)−mi(|f |) =Mi(f)−mi(|f |) ≤Mi(f) ≤Mi(f)−mi(f).

Therefore, we have

n∑

i=1

(Mi(|f |)−mi(|f |))(xi − xi−1) ≤n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) < ǫ,

i.e., |f(x)| is integrable on[a, b]. Since

−|f(x)| ≤ f(x) ≤ |f(x)|, ∀x ∈ [a, b],

we have

−∫ b

a

|f(x)|dx ≤∫ b

a

f(x)dx ≤∫ b

a

|f(x)|dx,

or∣

∣

∣

∣

∣

∫ b

a

f(x)dx

∣

∣

∣

∣

∣

≤∫ b

a

|f(x)|dx.

Example 6.3.1Letf(x) be defined on[0, 1] and

f(x) =

1, x is rational,

−1, x is irrational,

then|f(x)| is integrable, butf(x) is not integrable.

Example 6.3.2Suppose thatf(x) andg(x) are integrable on[a, b]. Prove thatf2(x), g2(x), andf(x)g(x)

are integrable on[a, b].

Proof. Sincef(x) andg(x) are integrable on[a, b], then there existM > 0 such that

|f(x)| ≤M, |g(x)| ≤M, ∀x ∈ [a, b],

and a partitionP = {x0, . . . , xn} of [a, b] such that

n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) <ǫ

2M,

n∑

i=1

(Mi(g)−mi(g))(xi − xi−1) <ǫ

2M,


whereMi(f) is the sup off(x) on [xi−1, xi], andmi(f) is the inf off(x) on [xi−1, xi]. Givenx′, x′′ ∈[xi−1, xi], we have

|f(x′′)g(x′′)− f(x′)g(x′)| = |(f(x′′)− f(x′))g(x′′) + f(x′)(g(x′′)− g(x′))|

≤ M |f(x′′)− f(x′)|+M |g(x′′)− g(x′)|

≤ M(Mi(f)−mi(f) +Mi(g)−mi(g)).

For given integerk > 0, there existx′′ andx′ such that

Mi(fg)−1

k< f(x′′)g(x′′),

mi(fg) +1

k> f(x′)g(x′).

Then we have

Mi(fg)−mi(fg)−2

k< f(x′′)g(x′′)− f(x′)g(x′)

≤ M(Mi(f)−mi(f) +Mi(g)−mi(g)).

Sincek is arbitrary, we have

Mi(fg)−mi(fg) ≤M(Mi(f)−mi(f) +Mi(g)−mi(g)),

and

n∑

i=1

(Mi(fg)−mi(fg))(xi − xi−1)

≤n∑

i=1

(Mi(f)−mi(f))(xi − xi−1) +

n∑

i=1

(Mi(g)−mi(g))(xi − xi−1) < ǫ,

i.e.,f(x)g(x) is integrable on[a, b], andf2(x) andg2(x) are also integrable on[a, b].

6.3.3 Additivity over Intervals

Proposition 6.3.4If f is integrable on[a, b], and interval[c, d] ⊆ [a, b], thenf is integrable on[c, d].

Proof. Becausef is integrable on[a, b], we can choose a partitionP on [a, b] such that the pointc andd

are partition points ofP and

U(f, P )− L(f, P ) < ε,

whereε is any given positive number. LetP ′ be the partition on[c, d] induced fromP , i.e., the partition

points ofP ′ are the same as that ofP in interval [c, d]. Hence

U(f, P ′)− L(f, P ′) ≤ U(f, P )− L(f, P ) < ε.


Thus,f is integrable on[c, d].

Theorem 6.3.5If f is integrable on[a, b], thenf is integrable on[a, c] and[c, b], and furthermore,

∫ b

a

f(x)dx =

∫ c

a

f(x)dx +

∫ b

c

f(x)dx, a < c < b.

Conversely iff is integrable on[a, c] and[c, b], thenf is integrable on[a, b] and

∫ c

a

f(x)dx +

∫ b

c

f(x)dx =

∫ b

a

f(x)dx,

wherec may be anywhere.

Proof. From Proposition 6.3.4,f is integrable on[a, c] and[c, b]. LetP ′n andP ′′

n be the regular partitions

of [a, c] and[c, b], and letPn = P ′n ∪ P ′′

n . Then,R(f, Pn) = R(f, P ′n) +R(f, P ′′

n ), and

lim||Pn||→0

R(f, Pn) = lim||Pn||→0

R(f, P ′n) + lim

||Pn||→0R(f, P ′′

n ),

or,∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx a < c < b.

Conversely, letP ′n andP ′′

n be the regular partitions of[a, c] and [c, b], and letPn = P ′n ∪ P ′′

n . Sincef is

integrable on[a, c] and[c, b], from Riemann Sum Convergence Theorem we get

lim||Pn||→0

R(f, Pn) = lim||Pn||→0

R(f, P ′n) + lim

||Pn||→0R(f, P ′′

n )

=

∫ c

a

f(x)dx+

∫ b

c

f(x)dx, a < c < b,

i.e.,f is integrable on[a, b].

6.4 Fundamental Theorem of Calculus

6.4.1 First Fundamental Theorem of Calculus

Theorem 6.4.1 (First Fundamental Theorem)Suppose thatf(x) is integrable on[a, b], and also suppose

thatF (x) is continuous on[a, b], differentiable on(a, b), and

F ′(x) = f(x), x ∈ (a, b).

Then∫ b

a

f(x)dx = F (b)− F (a).

Proof. Firstly we prove that for any partitionP of [a, b]

L(f, P ) ≤ F (b)− F (a) ≤ U(f, P ).


LetP = {x0, x1, . . . xn}. We apply the Lagrange Mean Value Theorem to functionF on every subinterval

[xi−1, xi],

F (xi)− F (xi−1) = F ′(ci)(xi − xi−1)

= f(ci)(xi − xi−1), xi−1 < ci < xi.

Since

mi = infxi−1≤x≤xi

{f(x)} ≤ f(ci) ≤ supxi−1≤x≤xi

{f(x)} =Mi,

we obtain

mi(xi − xi−1) ≤ F (xi)− F (xi−1) ≤Mi(xi − xi−1),

and thenn∑

i=1


i=1

(F (xi)− F (xi−1)) ≤n∑

i=1

Mi(xi − xi−1),

or

L(f, P ) ≤ F (b)− F (a) ≤ U(f, P ).

Therefore,

supL ≤ F (b)− F (a) ≤ inf U .

Sincef(x) is integrable,supL = inf U =

∫ b

a

f(x)dx. Hence,

∫ b

a

f(x)dx = F (b)− F (a).

Note.The First Fundamental Theorem of Calculus gives the relation betwween the definite integral and

indefinite integral (antiderivative): if∫

f(x)dx = F (x) + c,

then∫ b

a

f(x)dx =[

∫

f(x)dx]b

a.

LetS(t) be the displacement function of a particle. Then the distance covered by the particle from time

t = 0 to t = T is S(T )− S(0). One the other hand, the speed functionv(t) of the particle is

v(t) = S′(t).

By usingv(t) to calculate the distance covered by the particle in the timeperiod[0, T ] is∫ T

0

v(t)dt. Hence

we know∫ T

0

S′(t)dt = S(T )− S(0).


Example 6.4.1Evaluate the integral∫ π

4

π6

xdx

1 + x4.

Solution. It’s easy to find the antiderivative off(x) =x

1 + x4. That is

F (x) =1

2tan−1(x2).

By the First Fundamental Theorem we get,∫ π

4

π6

xdx

1 + x4= F (

π

4)− F (

π

6) =

1

2

[

tan−1(π

4)2 − tan−1(

π

6)2]

.

Example 6.4.2Find the limit of

limn→∞

√1 +

√2 + · · ·+√

n

n32

.

Solution. It’s the Riemann Sum of the functionf(x) =√x in interval [0, 1]. Hence the limit is

∫ 1

0

√xdx =

[

2

3x

32

]1

0

=2

3.

6.4.2 Mean Value Theorem for Integral

Lemma 6.4.2Suppose that the functionsf(x) and g(x) are bounded and integrable on[a, b], then the

functionf(x)g(x) is also integrable on[a, b].

Theorem 6.4.3 (Mean Value Theorem for Integral)Suppose that the functionf(x) is continuous on

[a, b] and that the functiong(x) is integrable and does not change the sign on[a, b]. Then there is a point

x0 ∈ [a, b] such that∫ b

a

f(x)g(x)dx = f(x0)

∫ b

a

g(x)dx.

Proof. Without loss of generality assume thatg(x) ≥ 0. Sincef(x) is continuous on[a, b], f(x) has a

maximumM and a minimumm. Then we have

mg(x) ≤ f(x)g(x) ≤Mg(x).

Sincef(x) andg(x) are integrable, thenf(x)g(x) is also integrable. Using Theorem 6.3.2 we get

m

∫ b

a

g(x)dx ≤∫ b

a

f(x)g(x)dx ≤M

∫ b

a

g(x)dx.

If∫ b

ag(x)dx = 0, then

∫ b

af(x)g(x)dx = 0 from the above inequality, and result follows. If

∫ b

ag(x)dx > 0,

then

m ≤∫ b

af(x)g(x)dx∫ b

a g(x)dx≤M.


Then from the Intermediate Value Theorem there is a pointx0 ∈ [a, b] such that

f(x0) =

∫ b

a f(x)g(x)dx∫ b

ag(x)dx

,

which is the result of the theorem.

In the special case wheng(x) ≡ 1 we get

∫ b

a

f(x)dx = f(x0)(b− a).

6.4.3 Second Fundamental Theorem of Calculus

Proposition 6.4.4Suppose that the functionf(x) is integrable on[a, b]. Define

F (x) =

∫ x

a

f(t)dt ∀x ∈ [a, b].

Then the functionF (x) is continuous on[a, b].

Proof. Sincef is integrable, it is bounded. ChooseM > 0 such that

|f(x)| ≤M, for x ∈ [a, b].

For any pointx0 ∈ [a, b]

F (x)− F (x0) =

∫ x

a

f(t)dt−∫ x0

a

f(t)dt =

∫ x

x0

f(t)dt.

Then we have,

|F (x)− F (x0)| =

∣

∣

∣

∣

∫ x

x0

f(t)dt

∣

∣

∣

∣

≤∣

∣

∣

∣

∫ x

x0

|f(t)|dt∣

∣

∣

∣

≤∣

∣

∣

∣

∫ x

x0

Mdt

∣

∣

∣

∣

=M |x− x0|.

Hence

limx→x0

{F (x)− F (x0)} = 0.

Note that here we have proved that the functionF (x) given above is Lipschitz continuous.

Theorem 6.4.5 (Second Fundamental Theorem)Suppose that the functionf(x) is continuous on[a, b].

Thend

dx

∫ x

a

f(t)dt = f(x) ∀x ∈ (a, b).

Proof. Let

F (x) =

∫ x

a

f(t)dt ∀x ∈ (a, b).


Using Mean Value Theorem we have

F (x+∆x)− F (x) =

∫ x+∆x

x

f(t)dt = f(ξ) ·∆x ∀x ∈ (a, b),

whereξ is betweenx andx+∆x. Hence as∆x→ 0, ξ → x. Therefore

F ′(x) = lim∆x→0

F (x +∆x)− F (x)

∆x= f(x).

Note.The Second Fundamental Theorem gives the existence of antiderivative (indefinite integral): A con-

tinuous function on[a, b] has an antiderivative. However, iff(x) is not continuous,F (x) may not be

differentiable on(a, b).

Example 6.4.3Functionf(x) define on[−1, 1] as following

f(x) =

−1, −1 ≤ x < 0,

0, x = 0,

1, 0 < x ≤ 1.

Then

F (x) =

∫ x

−1

f(x)dx =

−x− 1, −1 ≤ x < 0,

x− 1, 0 ≤ x ≤ 1,

or

F (x) = |x| − 1.

F is not differentiable atx = 0.

Corollary 6.4.6 Suppose thatf(x) is continuous on(a, b) andϕ(x) is differentiable on(c, d), and satisfies

ϕ((c, d)) ⊆ (a, b). Then for a fixed pointx0 ∈ (a, b) we have

d

dx

∫ ϕ(x)

x0

f(t)dt = f(ϕ(x))ϕ′(x) for x ∈ (c, d).

Proof. Let

G(x) =

∫ ϕ(x)

x0

f(t)dt ∀x ∈ (c, d),

and letF (x) be the antiderivative off(x). Then

G(x) =

∫ ϕ(x)

x0

f(t)dt = F (y)− F (x0),

wherey = ϕ(x). By the Chain Rule of composite function,

G′(x) =dF

dy· dydx

= f(y)ϕ′(x) = f(ϕ(x))ϕ′(x).


6.4.4 Calculation of Integral

In Chapter 5, there are two methods for calculating the antiderivative of a function: Integration by

Substitution and Integration by Parts. Through the First Fundamental Theorem, we can use these methods

for calculating integrals.

Theorem 6.4.7 (Integration by Parts)Suppose that the functionsh andg are continuous, and thatf ′ and

g′ exist and integrable on[a, b]. Then

∫ b

a

h(x)g′(x)dx = h(x)g(x)|ba −∫ b

a

g(x)h′(x)dx.

Proof. Since

(h(x)g(x))′ = h(x)g′(x) + g(x)h′(x) ∀x ∈ (a, b),

hg′ andgh′ are integrable on[a, b], then by the First Fundamental Theorem we have

∫ b

a

[h(x)g′(x) + g(x)h′(x)]dx =

∫ b

a

[h(x)g(x)]′dx = h(x)g(x)|ba.

On the other hand,

∫ b

a

[h(x)g′(x) + g(x)h′(x)]dx =

∫ b

a

h(x)g′(x)dx +

∫ b

a

g(x)h′(x)dx.

Combining the above two equations we get the result.

Example 6.4.4(Second Mean Value Theorem for Integrals) Suppose thatg(x) is continuous on[a, b],

f(x) is differentiable on[a, b]. Then we have the following results:

(a) If f ′(x) ≥ 0 andf(a) ≥ 0, then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.

(b) If f ′(x) ≤ 0 andf(b) ≥ 0, then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx.

(c) In general, iff ′(x) ≥ 0 or f ′(x) ≤ 0, then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx + f(b)

∫ b

ξ

g(x)dx.

Proof. Case (a). Let

G(x) =

∫ b

x

g(t)dt,


thenG′(x) = −g(x), and

∫ b

a

f(x)g(x)dx = −∫ b

a

f(x)dG(x)

= −f(x)G(x)|ba +

∫ b

a

G(x)f ′(x)dx

= f(a)G(a) +

∫ b

a

f ′(x)G(x)dx.

Letm andM be the minimum and maximum value ofG(x) on [a, b], then

m(f(b)− f(a)) ≤∫ b

a

f ′(x)G(x)dx ≤M(f(b)− f(a)).

Sincef(a) ≥ 0, we have

f(a)G(a) +

∫ b

a

f ′(x)G(x)dx ≤ f(a)M +M(f(b)− f(a)) = f(b)M.

Similarly,

f(a)G(a) +

∫ b

a

f ′(x)G(x)dx ≥ f(b)m.

Therefore,

mf(b) ≤∫ b

a

f(x)g(x)dx ≤Mf(b).

If f(b) = 0, take anyξ, we have the proof. Iff(b) 6= 0, thenf(b) > 0, we have

m ≤ 1

f(b)

∫ b

a

f(x)g(x)dx ≤M.

SinceG(x) is continuous, by the Intermediate Value Theorem, there exists aξ ∈ [a, b], such that

1

f(b)

∫ b

a

f(x)g(x)dx = G(ξ) =

∫ b

ξ

g(x)dx,

or∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.

Case (b) can be proved similarly by introducing

G(x) =

∫ x

a

g(t)dt.

For case (c), if f ′(x) ≥ 0, letF (x) = f(x)− f(a), thenF (x) satisfies the conditions in case (a).

The conditions in the above example can be relaxed, and we have the following more general

Second Mean Value Theorem for Integrals:

Suppose thatg(x) is integrable on[a, b], then we have the following results:

(a) If f(x) is monotonically increasing on[a, b] andf(x) ≥ 0, then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(b)

∫ b

ξ

g(x)dx.


(b) If f(x) is monotonically decreasing on[a, b] andf(x) ≥ 0, then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx.

(c) In general, iff(x) is monotonic on[a, b], then there exists aξ ∈ [a, b], s.t.,

∫ b

a

f(x)g(x)dx = f(a)

∫ ξ

a

g(x)dx + f(b)

∫ b

ξ

g(x)dx.

Theorem 6.4.8 (Integration by Substitution)Let the functionf : [a, b] → R be continuous. Suppose that

the functiong : [c, d] → R and its derivativeg′ are also continuous on[c, d], and, moreover, thatg(c) = a,

g(d) = b, anda ≤ g(t) ≤ b ∀t ∈ [c, d]. Then,

∫ d

c

f(g(t))g′(t)dt =

∫ b

a

f(x)dx.

Proof. DefineH : [c, d] → R by

H(t) =

∫ t

c

f(g(s))g′(s)ds−∫ g(t)

g(c)

f(x)dx ∀t ∈ [c, d].

Sincef(g(t)) is continuous on[c, d] by Continuity of Composition functions, thenf(g(t))g′(t) is continuous

on [c, d]. Then by Proposition 6.4.4H(t) is continuous on[c, d]. Moreover, from the Second Fundamental

Theorem and Corollary 6.4.6 we have

H ′(t) = f(g(t))g′(t)− f(g(t))g′(t) = 0 ∀t ∈ (c, d),

i.e.,H(t) is a constant. SinceH(t) is continuous andH(c) = 0, we getH(d) = 0.

Example 6.4.5∫ π

4

π6

xdx

1 + x4.

Solution.

∫ π4

π6

xdx

1 + x4=

∫ π4

π6

1

2

dx

1 + x4=

1

2

∫ (π4 )2

(π6 )2

dy

1 + y2

=1

2

[

tan−1y](π

4 )2

(π6 )2

=1

2

[

tan−1(π

4)2 − tan−1(

π

6)2]

Example 6.4.6∫ 1

0

x2exdx.


Solution. By the method of Integration by Substition, it can be found that the antiderivative off(x) = x2ex

is

F (x) = ex[x2 − 2x+ 2]

By the theorem,∫ 1

0

x2exdx = F (1)− F (0) = e− 2

You can also calculate the integral in the following way:

∫ 1

0

x2exdx =

∫ 1

0

x2dex = [x2ex]10 −∫ 1

0

ex2xdx

= e− 2

∫ 1

0

xdex = e− 2[xex]10 + 2

∫ 1

0

exdx

= e− 2e+ 2[ex]10 = −e+ 2e− 2

= e− 2

6.5 Exercise

1. Consider the partitionP = [0, 1/4, 1/2, 1] of the interval[0, 1]. ComputeL(f, P ) andU(f, P ) for

f(x) = x, x ∈ [0, 1].

2. Letf(x) = mx+ b, x ∈ [0, 1], wherem andb are positive.

(a) Show that the area under the graph ofy = f(x) and above thex axis isb+m/2.

(b) For any partitionP , show that L(f, P ) < b+m/2 < U(f, P ).

3. Suppose that the bounded functionf : [a, b] → R has the property that for each rational numberx

in the interval[a, b], f(x) = 0. Prove thatI∗ ≤ 0 ≤ I∗.

4. Suppose that the bounded functionf : [a, b] → R has the property thatf(x) ≥ 0 ∀x ∈ [a, b]. Prove

thatI∗ ≥ 0.

5. Suppose that the two bounded functionf : [a, b] → R andg : [a, b] → R have the property that

g(x) ≤ f(x) ∀x ∈ [a, b].

(a) For P a partition of[a, b], show thatL(g, P ) ≤ L(f, P ).

(b) Use part (a) to show thatI∗(g) ≤ I∗(f).

6. Define

f(x) =

x, if x ∈ [0, 1] is rational,

0, if x ∈ [0, 1] is irrational.

Prove thatI∗ = 0 andI∗ ≥ 1/2.


7. Let the functionf : [a, b] → R be monotonically decreasing and letPn be the regular partition of

[a, b] into n intervals of equal length(b− a)/n. Show that

U(f, Pn)− L(f, Pn) =[f(a)− f(b)](b− a)

n,

and then show thatf is integrable on[a, b].

8. Define

f(x) =

x, if 2 ≤ x ≤ 3,

2, if 3 ≤ x ≤ 4.

Prove thatf : [2, 4] → R is integrable.

9. Let

f(x) =

sgn(

sin πx

)

, x 6= 0,

0, x = 0.

Prove thatf(x) is integrable on[0, 1].

10. Use the Integrability Criterion to show thatf(x) = x2, x ∈ [0, 1] is integrable.

11. Define

f(x) =

x, if x ∈ [0, 1] is rational,

−x, if x ∈ [0, 1] is irrational.

Prove thatf : [0, 1] → R is not integrable.

12. Letf(x) = x+1, x ∈ [0, 1], thenf(x) is integrable sincef(x) is monotonically increasing. LetPn

be the regular partition of[0, 1] into n intervals of equal length1/n. ComputeU(f, Pn) of f , and

then find∫ 1

0f(x)dx by lettingn→ ∞.

13. Suppose that the functionsf, g, f2, g2 andfg are integrable on the closed bounded interval[a, b].

Prove that[f − g]2 also is integrable on[a, b] and that∫ b

a [f − g]2dx ≥ 0. Use this to prove that

∫ b

a

f(x)g(x)dx ≤ 1

2

[

∫ b

a

f2(x)dx +

∫ b

a

g2(x)dx

]

.

14. Suppose that the continuous functionf : [a, b] → R has∫ b

af(x)dx = 0. Prove that there is some

pointx0 ∈ [a, b] at whichf(x0) = 0. Hint. Use the Extreme Value Theorem and the Intermediate Value Theorem.

15. Suppose the continuous functionf : [a, b] → R has the property that∫ d

c f(x)dx ≤ 0 ∀a ≤ c ≤ d ≤b. Prove thatf(x) ≤ 0 ∀x ∈ [a, b]. Is this true if we require only integrability of the function?

16. Suppose that the functionf : [0, 1] → R is continuous and thatf(x) ≥ 0 ∀x ∈ [0, 1]. Prove that∫ 1

0 f(x)dx > 0 if and only if there is a pointx0 ∈ [0, 1] at whichf(x0) > 0.


17. Suppose that the functionsf : [a, b] → R andg : [a, b] → R are continuous. Prove that

∫ b

a

|f(x) + g(x)|dx ≤∫ b

a

|f(x)|dx +

∫ b

a

|g(x)|dx.

18. Suppose that the functionf : [a, b] → R is Lipschitz, that is, that there is a constantc ≥ 0

such that|f(u) − f(v)| ≤ c|u − v| ∀u, v ∈ [a, b]. For a partition P of [a, b], prove that0 ≤U(f, P )− L(f, P ) ≤ c|b− a|‖P‖.

19. Show that a Lipschitz function is integrable.

20. Suppose that the functionsf, g, f2, g2 andfg are integrable on the closed bounded interval[a, b].

prove that∫ b

a

f(x)g(x)dx ≤(

∫ b

a

f2(x)dx

)1/2(∫ b

a

g2(x)dx

)1/2

.

21. Suppose that the functionsf : [a, b] → R is bounded and that it is continuous except at a finite

number of points. Prove thatf : [a, b] → R is integrable.

22. Define

f(x) =

1, if x ∈ [0, 1] is rational,

−1, if x ∈ [0, 1] is irrational.

Prove thatf : [0, 1] → R is not integrable.

23. Suppose that the functionsf(x) is integrable on[a, b]. Prove that|f(x)| is also integrable on[a, b].

Is the reverse true? (Consider the above problem.)

24. Suppose that the functionsf : [a, b] → R is continuous and∫ b

a

f2(x)dx = 0. Prove thatf(x) ≡ 0

on [a, b]. Hint. Prove by contradiction.

25. Suppose that the functionsf : [−a, a] → R is continuous. Prove that∫ a

−a

f(x)dx = 0 if f(x) is an

odd function, and that∫ a

−a

f(x)dx = 2

∫ a

0

f(x)dx if f(x) is an even function.

26. Use the First Fundamental Theorem to evaluate each of thefollowing integrals:

(a)∫ 2

1

[ 1

x2+ x+ cosx

]

dx.

(b)∫ 1

0

x√

4− x2dx.

(c)∫ 1

0

xdx

1 +√1 + x

.

(d)∫ π/2

0

cosn xdx.

(e)∫ 3

1

x√10− xdx.

(f)∫ π

0

cos2 xdx.

(g)∫ 1

0

(1− x2)ndx.


27. Find the following limits by using Riemann sums:

(a)

limn→∞

[1β + 2β + · · ·+ nβ

nβ+1

]

.

(b) limn→∞

[

n∑

k=1

k

n2 + k2

]

.

(c) limn→∞

[ 1

n+ 1+

1

n+ 2+ · · ·+ 1

2n

]

.

(d) limn→∞

[ 1√n · n +

1√

n · (n+ 1)+ · · ·+ 1

√

n · (n+ n)

]

.

28. Find the area of the region bounded by:

(a) y = x2 andx+ y = 2.

(b)x2

a2+y2

b2, y = 0.

(c) y = x andy = x+ sin2 x, 0 ≤ x ≤ π.

29. A car runs on the road for one hour with the instant velocity

v(t) =

500t, 0 ≤ t ≤ 0.1,

50, 0.1 ≤ t ≤ 0.9,

500(1− t), 0.9 ≤ t ≤ 1,

where the unit is mile/hour. Find the distance that the car covers.

30. Calculate the following derivatives:

(a)d

dx

(

∫ x

0

x2t2dt)

.

(b)d

dx

(

∫ ex

1

ln tdt)

.

(c)d

dx

(

∫ x

−x

et2

dt)

.

(d)d

dx

(

∫ x

1

cos(x+ t)dt)

.

31. Suppose that the functionsf : R → R is differentiable. Define the functionH : R → R by

H(x) =

∫ x

−x

[f(t) + f(−t)]dt ∀x. FindH ′′(x).

32. Suppose that the functionf : R → R is continuous. DefineG(x) =∫ x

0

(x− t)f(t)dt ∀x. Prove

thatG′′(x) = f(x) ∀x.

33. For numbersa1, · · · , an, definep(x) = a1x+a2x2+ · · ·+anxn for all x. Suppose that

a12+a23+

· · ·+ ann+ 1

= 0. Prove that there is some pointx ∈ (0, 1) such thatp(x) = 0.

34. Suppose that the functionf : R → R has a continuous second derivative. Prove that

f(x) = f(0) + f ′(0)x+

∫ x

0

(x− t)f ′′(t)dt ∀x.

35. Suppose that the functionsf : [a, b] → R andg : [a, b] → R are bounded and integrable. Prove

thatfg is integrable.

36. Suppose thatf : [a, b] → R is continuous, and∫ b

a

f(x)g(x)dx = 0

for any continuous functiong(x) satisfyingg(a) = g(b) = 0. Prove thatf(x) ≡ 0.

Hint. If not, ∃x0 ∈ (a, b), s.t. (e.g.),f(x0) > 0, ∀x ∈ (x0 − δ, x0 + δ), then take a properg(x).

Chapter 7: Series of Numbers 143

37. Suppose thatf : [a, b] → R is continuous and satisfies the conditions

∫ b

a

xkf(x)dx = 0, k = 0, 1, · · · , n.

Prove that the equationf(x) = 0 has at leastn+ 1 solutions in(a, b). Hint. By induction.

38. Suppose thatf : [a, b] → R is integrable. Show that for any[c, d] ⊂ [a, b], there existsx0 ∈ [c, d],

such thatf(x) is continuous atx0.

Hint. Let ω(c, d) = supx∈[c,d]

f(x) − infx∈[c,d]

f(x), then for givenǫ > 0, ∃[c, d] ⊆ [a, b], s.t., ω(c, d) < ǫ.

Otherwise,f(x) is not integral on[a, b]. Let a1 = a, b1 = b, then∃[a2, b2] ⊆ [a1, b1], s.t., ω(a2, b2) <1/2, · · · , ω(an, bn) < 1/n, By Nested Interval Theorem,∃c ∈ [an, bn]. For givenǫ > 0, taken s.t.,1/n < ǫ,and takeδ s.t.,(c− δ, c+ δ) ⊆ [an, bn], then|f(x) − f(c)| ≤ ω(an, bn) < 1/n < ǫ, ∀x ∈ (c − δ, c+ δ), i.e.,f(x) is continuous atc.

39. (a). Suppose thatf(x) andf ′(x) are continuous on[a, b]. Prove that

limn→∞

∫ b

a

f(x) sinnxdx = 0.

(b). Suppose thatf(x) is integrable on[0, 2π]. Prove that

limn→∞

∫ 2π

0

f(x) sinnxdx = 0.

Hint. (a). Integration by parts. (b).∫ b

af(x) sinnxdx =

n∑

k=1

∫(2k−1)π

n

(2k−2)πn

f(x) sinnxdx+n∑

k=1

∫ 2kπn

(2k−1)πn

f(x) sinnxdx =n∑

k=1

2

nλk −

n∑

k=1

2

nµk ,

wheremk ≤ λk, µk ≤ Mk, andmk = minx∈[

(2k−2)πn

, 2kπn

]

f(x), Mk = maxx∈[

(2k−2)πn

, 2kπn

]

f(x).

Sincef(x) is integral, limn→∞

∫ 2π

0f(x) sinnxdx = lim

n→∞

1

π

[

n∑

k=1

λk2π

n−

n∑

k=1

µk2π

n

]

= 0.

40. Suppose thatf : [a, b] → R is integrable. Show that for everyǫ > 0, there exists a functiong,

continuous on[a, b] such that∫ b

a

|f(x)− g(x)|dx < ǫ.


Chapter 7

Series of Numbers

7.1 Definition and Basic Properties

Definition. Let {ak} be a sequence of real numbers. Then the sum∞∑

k=1

ak is called aSeries(or Number

Series),ak is called thekth term of the series. The sum

Sn =

n∑

k=1

ak

for any natural numbern is called a Partial Sum. If the sequence of partial sums{Sn} converges, i.e.,

limn→∞

Sn = S,

then we say that the seriesconvergesand callS the sum of the series, denoted by

∞∑

k=1

ak = S.

If {Sn} does not converge, then we say that the series∞∑

k=1

ak diverges.

Proposition 7.1.1 (necessary condition for convergence)Suppose that the series∞∑

k=1

ak converges. Then

limk→∞

ak = 0.

Proof. The convergence of∞∑

k=1

ak means thatlimn→0

Sn = S. Hence

limk→∞

ak = limn→∞

(Sk − Sk−1)

= limn→∞

Sk − limn→∞

Sk−1 = S − S = 0.

145


We have learnt the Cauchy Convergence Criterion for sequence. Apply to Series, we obtain the follow-

ing proposition.

Proposition 7.1.2 (Cauchy Convergence Criterion)A series∞∑

k=1

ak converges if and only if for every

ǫ > 0, there exists a natural numberN such that

|an+1 + an+2 + · · ·+ am| < ε, if m > n ≥ N.

Proof. Using Cauchy Convergence Criterion to the sequence of partial sums{Sn}, {Sn} converges if and

only if for everyε > 0, there exists a natural numberN such that

|Sm − Sn| < ε, if m > n ≥ N.

But

|Sm − Sn| = |an+1 + · · ·+ am|,

i.e.,

|an+1 + · · ·+ am| < ε, if m > n ≥ N.

Proposition 7.1.3 (Convergence of Geometric Series)If |r| < 1, then

∞∑

k=0

rk =1

1− r.

Proof. The partial sum of the series is

Sn =

n∑

k=0

rk =1− rn+1

1− r.

Hence∞∑

k=0

rk = limn→∞

Sn =1

1− r.

If |r| ≥ 1, the general termsak = rk don’t converge to 0. Hence the series∞∑

k=0

rk does not converge. The

Geometric Series is the simplest series but will play an important role in the following Comparison Test.

7.2 Series with Nonnegative Terms

Definition 7.2.1If ak ≥ 0 for all k, then∞∑

k=1

ak is called a series with nonnegative terms.


Sinceak ≥ 0 for all k, the partial sums satisfy

Sn = Sn−1 + an ≥ Sn−1 for all n.

The sequence{Sn} is monotonically increasing. Hence by the limit theory, we have the following proposi-

tion.

Proposition 7.2.2A series with nonnegative terms converge if and only if the sequence of partial sums{Sn}has an upper bound.

Proof. It’s known that{Sn} is monotonically increasing. By the Monotone Convergence Theorem,{Sn}converges if{Sn} has an upper bound. On the other hand, we know that a convergent sequence must be

bounded.

Theorem 7.2.3 (Comparison Test)Suppose that{ak} and{bk} are sequences of numbers such that0 ≤ak ≤ bk ∀k.

(i) If∞∑

k=1

bk converges, then∞∑

n=1

ak converges.

(ii) If∞∑

n=1

ak diverges, then∞∑

n=1

bk diverges.

Proof. Let {An} and {Bn} be sequences of partial sums of∞∑

k=1

ak and∞∑

k=1

bk respectively. From the

assumption0 ≤ ak ≤ bk ∀k, we know

An ≤ Bn for all n.

Hence, by the above Proposition, it follows

(i) If∞∑

k=1

bk converges, then{Bn} is bounded above. FromAn ≤ Bn, {An} is bounded above. There-

fore∞∑

k=1

ak converges.

(ii) If∞∑

k=1

ak diverges, then{An} has no upper bound. FromAn ≤ Bn, {Bn} has no upper bound also.

Therefore∞∑

k=1

bk diverges.

Corollary 7.2.4 LetN be a natural number andC is a positive real number. Suppose that

0 ≤ an ≤ Cbn for n ≥ N.

(i) If∞∑

k=1


k=1

ak converges.


(ii) If∞∑

k=1

ak diverges, then∞∑

k=1

bk diverges.

Proof. It’s obvious that∞∑

k=1

bk converges if and only if∞∑

k=N

bk converges, whereN is any natural number.

Also,∞∑

k=1

bk converges if and only if∞∑

k=1

Cbk converges, whereC is a nonzero constant. Hence∞∑

k=1

bk con-

verges,∞∑

k=1

Cbk converges and∞∑

k=N

Cbk converges.∞∑

k=N

ak and∞∑

k=N

Cbk are two series with nonnegative

terms, by Comparison Test,∞∑

k=N

ak converges due to convergence of∞∑

k=N

Cbk. Hence∞∑

k=1

ak converges.

The proof of part (ii) is left as exercise.

Corollary 7.2.4’ Suppose that

limn→∞

anbn

= l.

Then,

(i) If 0 < l < +∞, then∞∑

k=1

ak and∞∑

k=1

bk converge the the same time, or diverge the same time.

(ii) If l = 0 and∞∑

k=1


k=1

ak converges.

(iii) If l = +∞ and∞∑

k=1

bk diverges, then∞∑

k=1

ak diverges.

Theorem 7.2.5 (Ratio Test)Suppose that∞∑

k=1

uk is a series withuk > 0 and satisfies

limk→∞

uk+1

uk= l.

(i) If l < 1, then the series converges.

(ii) If l > 1, then the series diverges.

(iii) If l = 1, it is indefinite.

Proof.

(i) Take a constantq such thatl < q < 1. From the assumptionlimk→∞

uk+1

uk= l, there is a natural

numberN such that

uk+1

uk< q if k ≥ N.


It follows that

uN+1 < quN

uN+2 < quN+1 < q2uN

· · ·

uN+k < · · · < qkuN

Since∞∑

k=1

qkuN converges, by comparison test∞∑

k=1

uN+k converges and then∞∑

k=1

uk =N∑

k=1

uk +

∞∑

k=1

uN+k converges.

(ii) Since limk→∞

uk+1

uk= l > 1, there is a natural numberN such that

un+1

un> 1, if n ≥ N,

i.e. uk+1 > uk. It means that{uk}k≥N is an increasing sequence and can not converge to 0. Hence

the series∞∑

k=1

uk diverges.

(iii) Consider the series∞∑

k=1

1

kp, p is a constant. It is known that whenp = 2 the series converges and

p = 1 the series diverges. But for any real numberp

limn→∞

1(n+1)p

1np

= 1.

Therefore, whenl = 1, the ratio test is failed to give any conclusion.

Theorem 7.2.6 (Cauchy Test)Suppose that∞∑

k=1

uk is a series withuk > 0 and satisfies

limk→∞

k√uk = l.

(i) If l < 1, then the series converges.

(ii) If l > 1, then the series diverges.

(iii) If l = 1, it is indefinite.

Proof.

(i) Take a constantq such thatl < q < 1. From the assumptionlimk→∞

k√uk = l, there is a natural

numberN such that k√uk < q with k ≥ N , i.e.,

uk < qk, if k ≥ N.

By comparison test∞∑

k=N

uk converges due to the convergence of∞∑

k=N

qk, and then∞∑

k=1

uk converges.


(ii) From the assumptionlimk→∞

k√uk = l > 1, there is a natural numberN such that

uk > 1, if k ≥ N.

{uk} can’t converge to 0, hence∞∑

k=1

uk does not converge.

(iii) Also for example∞∑

k=1

uk with uk =1

kp, limk→∞

k√uk = 1. But the series converges whenp = 2 and

diverges whenp = 1.

Definition 7.2.7Suppose thatf(x) is defined on[a,+∞) and for anyA > a, f(x) is integrable in[a,A].

If the limit

limA→∞

∫ A

a

f(x)dx

exists, then we say that the integral∫ ∞

a

f(x)dx converges and denote by

∫ ∞

a

f(x)dx = limA→∞

∫ A

a

f(x)dx.

Proposition 7.2.8Suppose thatf(x) is nonnegative on interval[a,+∞) and for anyA > a, f(x) is

integrable on[a,A]. Then∫ ∞

a

f(x)dx converges if and only if the function

F (A) =

∫ A

a

f(x)dx

has an upper bound.

Proof. Sincef(x) is nonnegative, forA2 > A1 > a

F (A2)− F (A1) =

∫ A2

A1

f(x)dx ≥ 0.

HenceF (A) is monotonically increasing. From the limit theory,limA→∞

F (A) exist if and only if F(A) has an

upper bound.

Theorem 7.2.9 (Integral Test)Let∞∑

k=1

uk be a series with nonnegative terms. Suppose that the function

f(x) is monotonically decreasing and satisfies

f(k) = uk.

Then the series∞∑

k=1

uk converges if and only if∫ ∞

1

f(x)dx converges.

Proof. From the assumptions, we have

uk = f(k) ≥ f(x) ≥ f(k + 1) = uk+1 if k ≤ x ≤ k + 1.


Then we have

uk ≥∫ k+1

k

f(x)dx ≥ uk+1

andn∑

k=1

uk ≥∫ n+1

1

f(x)dx ≥n∑

k=1

uk+1,

or

Sn ≥∫ n+1

1

f(x)dx ≥ Sn+1 − u1.

If∫ ∞

1

f(x)dx converges,{Sn} has an upper bound∫ ∞

1

f(x)dx + u1. Therefore{Sn} converges. If∫ ∞

1

f(x)dx diverges, that means

{∫ n+1

1

f(x)dx

}∞

n=1

has no upper bound. By the inequality so{Sn}

has no upper bound, i.e.,∞∑

k=1

uk diverges.

Using the Integral Test we get

Theorem 7.2.10 (p-Test)For a positive numberp, the series∞∑

k=1

1

kp

converges if and only ifp > 1.

Proof. Consider the integral

∫ ∞

1

1

xpdx = lim

A→∞

∫ A

1

1

xpdx =

limA→∞

1

1− p(A1−p − 1), p 6= 1

limA→∞

lnA, p = 1

The integral converges only forp > 1.

Example 7.2.1Consider the convergence of the following series:

(a)∞∑

n=2

1

n lnn.

(b)∞∑

n=2

1

(lnn)lnn.

(c)∞∑

n=1

(1000)n

n!.

(d)∞∑

n=1

n1000

2n.

7.3 Convergence of General Series

Now we consider the general series∞∑

k=1

uk, i.e.,uk may be positive or negative. But first we discuss the

Alternating Series.

Theorem 7.3.1 (Alternating Series Test)Suppose that{uk} is a monotonically decreasing sequence of

nonnegative numbers andlimk→∞

uk = 0. Then the series

∞∑

n=1

(−1)n+1un


converges.

Proof. We first show that the subsequences{S2n} and {S2n+1} converge. Since{uk} is monotonically

decreasing,

S2n+2 − S2n = u2n+1 − u2n+2 ≥ 0

and

S2n = u1 −n−1∑

k=1

(u2k − u2k+1)− u2n ≤ u1.

We conclude that{S2n} is monotonically increasing and bounded byu1. Hence{S2n} converges. But

S2n+1 = S2n + u2n+1

and limn→∞

un = 0, then limn→∞

S2n+1 = limn→∞

S2n. Denote this limit byS. Now we show that the sequence

{Sn} converges toS. For givenǫ > 0, we choose natural numbersN1 andN2 such that

|S2n − S| < ε if n ≥ N1

and

|S2n+1 − S| < ε if n ≥ N2.

DefineN = max{2N1, 2N2 + 1}. Then

|Sn − S| < ε if n ≥ N.

Example 7.3.1Discuss the convergence of the following series:

1 +1

2+

1

3− 1

4− 1

5− 1

6+

1

7+

1

8+

1

9− · · · .

Definition 7.3.2A series∞∑

k=1

uk is said to converge absolutelyprovided that the series∞∑

k=1

|uk| converges.

Theorem 7.3.3 (Absolute Convergence Test)The series∞∑

k=1

uk converges if the series converges abso-

lutely, i.e.∞∑

k=1

|uk| converges.

Proof. Since∞∑

k=1

|uk| converges, by using Cauchy Convergence Criterion to series, for givenǫ > 0, ∃N

such that

|un+1|+ |un+2|+ · · ·+ |un| < ε, if m > n ≥ N,

and then

|un+1 + un+2 + · · ·+ un| ≤ |un+1|+ · · ·+ |um| < ǫ.


Hence again by Cauchy Convergence Criterion,∞∑

k=1

uk converge.

Note.∞∑

k=1

uk converges,∞∑

k=1

|uk| may not converge (Consider∞∑

k=1

(−1)k+1

k).

According to this theorem, all the convergence tests mentioned in Section 2 for series with nonnegative

terms can be used as sufficient conditions for general series.

Example 7.3.2Consider the series∞∑

n=1

sinn

n2. Since

∣

∣

∣

∣

sinn

n2

∣

∣

∣

∣

≤ 1

n2

and∞∑

n=1

1

n2converges. Then

∞∑

n=1

∣

∣

∣

∣

sinn

n2

∣

∣

∣

∣

converges, i.e.,∞∑

n=1

sinn

n2converges absolutely. By Theorem 7.3.3,

∞∑

n=1

sinn

n2converges.

Definition 7.3.4 (Conditional Convergence)A series that converges but does not converge absolutely is

said to converge conditionally.

Example 7.3.3∞∑

n=1

(−1)n+1

nconverges conditionally.

There are some interesting properties related to absolute convergence and conditional convergence.

For examples:

• If a series converges absolutely, then rearranging the terms in any order the new series still converges

absolutely and the sum does not change.

• If a series converges conditionally, then we can rearrange the terms in some order to make the partial

sums sequence converge to any specified number including+∞ and−∞.

Next, we introduce two tests: Abel and Dirichlet Tests, which are very useful for some series. Consider

the partial sum:

S =m∑

k=1

akbk.

Let

Bj =

j∑

k=1

bk, j = 1, 2, · · · ,m,

then

b1 = B1, bk = Bk −Bk−1, k = 2, 3, · · · ,m.


LetB0 = 0, then

S =

m∑

k=1

akbk =

m∑

k=1

ak(Bk −Bk−1)

=

m∑

k=1

akBk −m∑

k=2

akBk−1

= amBm +

m−1∑

k=1

akBk −m−1∑

k=2

ak+1Bk

= amBm +m−1∑

k=1

(ak − ak+1)Bk.

Lemma 7.3.5 (Abel)If {a1, a2, · · · , am} and{b1, b2, · · · , bm} satisfy the conditions:

(i) {ak} is monotone;

(ii) {Bk} is bounded, i.e.,∃M > 0, s.t.,

|Bk| ≤M, k = 1, 2, · · · ,m.

Then we have

|S| =∣

∣

∣

m∑

k=1

akbk

∣

∣

∣ ≤M(|a1|+ 2|am|).

Proof. From the expression ofS, we get

|S| ≤m−1∑

k=1

|ak − ak+1| · |Bk|+ |amBm|

≤ M[

m−1∑

k=1

|ak − ak+1|+ |am|]

.

Since{ak} is monotone, we have

|S| ≤ M[

|a1 − am|+ |am|]

≤ M(|a1|+ 2|am|).

Theorem 7.3.6 (Abel Test)Suppose the the series∞∑

n=1

anbn satisfies

(i)∞∑

n=1

bn converges;

(ii) {an} is monotone and bounded.

Then∞∑

n=1

anbn converges.


Proof. From (ii), there is aM such that|an| ≤ M, ∀n. Since∞∑

n=1

bn converges, then for givenǫ > 0, ∃N ,

s.t.,∣

∣

∣

n+p∑

k=n+1

bk

∣

∣

∣<

ǫ

3M, ∀p > 0, n > N. From Abel’s lemma we get

∣

∣

∣

n+p∑

k=n+1

akbk

∣

∣

∣ ≤ ǫ

3M(|an+1|+ 2|an+p|) ≤ ǫ.

By Cauchy Convergence Criterion,∞∑

n=1

anbn converges.

Theorem 7.3.7 (Dirichlet Test)Suppose the the series∞∑

n=1

anbn satisfies

(i) Bn =

n∑

k=1

bk is bounded;

(ii) {an} converges to zero monotonically.

Then∞∑

n=1

anbn converges.

Proof. Suppose that|Bn| ≤M, ∀n, then

|bn+1 + · · ·+ bn+p| = |Bn+p −Bn| ≤ 2M, ∀n, p.

Since{an} → 0, then for givenǫ > 0, ∃N , s.t.,|an| <ǫ

6M, ∀n > N . From Abel’s lemma we have

∣

∣

∣

n+p∑

k=n+1

akbk

∣

∣

∣ ≤ 2M(|an+1|+ 2|an+p|) ≤ ǫ, ∀p > 0, n > N.

By Cauchy Convergence Criterion,∞∑

n=1

anbn converges.

7.4 Exercise

1. Examine the following series for convergence:

(a)∞∑

n=1

sin2 nθ

n2 + 1

(b)∞∑

n=1

2 · 5 · 8 · · · (3n− 1)

1 · 5 · 9 · · · (4n− 3)

(c)∞∑

n=1

2n sinx

3n, x > 0

(d)∞∑

k=1

( k + 1

k2 + 1

)3

(e)∞∑

n=1

1000n

n!

(f)∞∑

k=2

1

(ln k)ln k

2. Examine the following series for convergence and decide whether the series converges absolutely:


(a)∞∑

k=1

(−1)kck

k, c > 1

(b)∞∑

k=2

(−1)k1

ln k

(c)∞∑

n=1

cosnx

np, p > 0, 0 < x < π.

(d)∞∑

k=1

(−1)kk5

k!

(e)∞∑

k=2

(ln k)(sin(kπ/3))

k2

3. For any positive numberα, prove that the series

∞∑

k=1

kα

ek

converges.

4. Fix a positive numberα and consider the series

∞∑

k=1

1

(k + 1)[ln(k + 1)]α.

For what values ofα does this series converge?

5. (The Cauchy Root Test) For the series∞∑

k=1

ak, suppose that there is a numberr with 0 ≤ r < 1 and

a naturalN such that

|ak|1/k < r ∀k ≥ N.

Prove that∞∑

k=1

ak converges absolutely.

6. Use the Cauchy Convergence Criterion for series to provide another proof of the Alternating Series

Test.

7. Let

u+n =

un, un ≥ 0,

0, un < 0,u−n =

0, un ≥ 0,

−un, un < 0.

Show that (i).∞∑

n=1

un converges absolutely if and only if both∞∑

n=1

u+n and∞∑

n=1

u−n converge. (ii). If

∞∑

n=1

un converges conditionally, then both∞∑

n=1

u+n and∞∑

n=1

u−n diverge.

8. Prove that if a series converges absolutely, then we can rearrange the terms of the series in any order,

the new series still converges absolutely and the sum is the same.

Chapter 8

Series of Functions

8.1 Uniform Convergence

8.1.1 Pointwise Convergence

Definition 8.1.1A sequence of functions{fn(x)} defined on the intervalI is said to converge pointwise(or

simply converge) tof(x) on I, if for eachx0 ∈ I,

limn→∞

fn(x0) = f(x0).

Definition 8.1.1′ A series of functions∞∑

n=1

un(x) defined on the intervalI is said to converge pointwiseto

S(x) on I, if for eachx0 ∈ I, the sequence of partial sums{Sn(x)}, whereSn(x) =

n∑

k=1

uk(x), satisfies

limn→∞

Sn(x0) = S(x0),

i.e. for every point ofI, Sn(x) converges toS(x).

In many problems of mathematics, we may need to know the properties of the limit function. In other

words, we may need the answers to the following questions:

(1) If {fn(x)} are all continuous on the intervalI, and converges pointwise tof(x). Isf(x) continuous?

(2) If {fn(x)} are all differentiable on the intervalI and converges pointwise tof(x). Is f(x) differen-

tiable and limn→∞

f ′n(x) = f ′(x)?

(3) If {fn(x)} are all integrable on the interval[a, b] and converges pointwise tof(x). Isf(x) integrable

157


and

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx.

Example 8.1.1Letfn(x) = xn, 0 ≤ x ≤ 1. Then

limn→∞

xn = f(x) =

0, x 6= 1,

1, x = 1.

For anyn, fn(x) are all continuous on[0, 1] and differentiable but the limit functionf(x) is not continuous

and not differentiable on[0, 1].

Example 8.1.2Letfn(x) = nxe−nx2

, 0 ≤ x ≤ 1. Then

limn→∞

fn(x) = 0 ∀x ∈ [0, 1].

The integral offn is

∫ 1

0

fn(x)dx =

∫ 1

0

e−nx2

dnx2

2

=1

2

(

−enx2)∣

∣

∣

1

0=

12 − e−n

2.

Then we have

limn→∞

∫ 1

0

fn(x)dx =1

2.

However, since∫ 1

0

[ limn→∞

fn(x)] = 0, then

limn→∞

∫ 1

0

fn(x)dx 6=∫ 1

0

[ limn→∞

fn(x)]dx.

The above examples show that the assumption of Pointwise Convergence can not guarantee affirmative

answers to the above three questions. To find the answers to these questions, we need the concept of uniform

convergence.

8.1.2 Uniform Convergence

Definition 8.1.2A sequence of functions{fn(x)} is said to converge uniformlyto f(x) on the intervalI, if

for any givenǫ > 0, there exists a natural numberN (which may depend onǫ but notx.) such that for all

x ∈ I

|fn(x) − f(x)| < ε, if n ≥ N.

Chapter 8: Series of Functions 159

The geometrical interpretation is that the graph of functionsfn(x), n ≥ N , lie between the graph of

functionf(x)− ǫ and functionf(x) + ǫ.

Example 8.1.3Letfn(x) = xn. Then{fn(x)} converges uniformly tof(x) = 0 on the interval[0, 1− δ],

0 < δ < 1, but does not converge uniformly tof(x) = 0 on the interval[0, 1].

Proof. For anyε > 0, chooseN =

[

ln ε

ln(1− δ)

]

+ 1, then for allx ∈ [0, 1− δ],

|fn(x) − 0| = |xn| ≤ (1 − δ)n < ε , if n ≥ N

Hence{fn(x)} converges uniformly tof(x) = 0 on [0, 1− δ]. On the interval[0, 1],

limn→∞

fn(x) = f(x) =

0, 0 ≤ x < 1,

1, x = 1,

We now prove that{fn(x)} does not converge uniformly on[0, 1]. We argue by contradiction. If{fn(x)}converges uniformly on[0, 1], then when forǫ = 1/2, there is a natural numberN such that

|fn(x)− f(x)| < 1

2if n ≥ N.

Thus for thisN , we have

|xN − 0| < 1

2for 0 ≤ x < 1

But it is obvious thatlimx→1

xN = 1, which means thatfN(x) can be arbitrarily close to1 asx sufficiently

close to1. For example, choosex0 = (2/3)1/N , then0 ≤ x0 < 1, and

xN0 =2

3>

1

2.

Thus we get a contradiction. It shows thatf(x) can not converge uniformly on[0, 1].

Definition 8.1.3A function series∞∑

n=1

un(x) defined on the intervalI is said to converge uniformlytoS(x)

on the intervalI, if for any givenǫ > 0, there exists a natural numberN such that for allx ∈ I, the partial

sumSn(x) =

n∑

k=1

uk(x) satisfies

|Sn(x)− S(x)| < ε , if n ≥ N.

8.1.3 Test for Uniform Convergence

Theorem 8.1.4 (Cauchy Convergence Criterion)A sequence of functions{fn(x)} converges uniformly

on the intervalI, if and only if for any givenǫ > 0, there exists a natural numberN such that for allx ∈ I

|fm(x)− fn(x)| < ε, if m > n ≥ N.


Proof. If {fn(x)} converges tof(x) uniformly, then for givenǫ > 0, ∃N , s.t.,

|fn(x)− f(x)| < ǫ

2, ∀n ≥ N, x ∈ I.

Then,

|fm(x)− fn(x)| < |fm(x) − f(x)|+ |fn(x)− f(x)| < ǫ, ∀m,n ≥ N, x ∈ I.

If {fn(x)} is a Cauchy sequence, then{fn(x)} converges pointwise tof(x), then for givenǫ > 0, ∃N , s.t.,

|fm(x)− fn(x)| < ǫ, ∀m,n ≥ N, x ∈ I.

Fix n, and letm→ +∞, we get

|fn(x)− f(x)| < ǫ, ∀n ≥ N, x ∈ I,

i.e.,{fn(x)} converges uniformly on the intervalI.

Theorem 8.1.4′ (Cauchy Convergence Criterion)A function series∞∑

n=1

un(x) converges uniformly on the

intervalI, if and only if for any givenǫ > 0 there exists a natural numberN , such that for allx ∈ I

|un+1(x) + · · ·+ um(x)| < ε, if m > n ≥ N.

Theorem 8.1.5 (Weierstrass Uniform Convergence Criterion) If the general termun(x) of∞∑

n=1

un(x)

satisfy

|un(x)| ≤ an, ∀x ∈ I

and the series∞∑

n=1

an converges. Then the function series∞∑

n=1

un(x) converges uniformly on the intervalI.

Proof. Since∞∑

n=1

an converges, by Cauchy Convergence Criterion, for givenǫ > 0, ∃N such that

|an+1 + · · ·+ am| < ǫ, if m > n ≥ N.

Butak are all nonnegative,

|an+1 + · · ·+ am| = an+1 + · · ·+ am < ǫ.

From the assumption|un(x)| ≤ an, ∀x ∈ I, if follows that for allx ∈ I,

|un+1(x) + · · ·+ um(x)| ≤ |un+1(x)| + · · ·+ |um(x)|

≤ an+1 + · · ·+ am < ǫ, if m > n ≥ N.


By Theorem 8.1.4,∞∑

n=1

un(x) converges uniformly onI.

According to this theorem, all the tests for number series with nonnegative terms can be used for the test of

uniform convergence of function series.

Example 8.1.4Consider the function series

∞∑

n=1

sin(nπx)

np, −∞ < x < +∞, p > 1.

Since∣

∣

∣

∣

sin(nπx)

np

∣

∣

∣

∣

≤ 1

np, −∞ < x < +∞

and∞∑

n=1

1

npconverges whenp > 1, the function series converges uniformly on(−∞,+∞).

Theorem 8.1.6 (Dini Test)Suppose the{fn(x)} converges pointwise tof(x) on [a, b], fn(x), ∀n, andf(x)

are continuous on[a, b], and for anyx ∈ [a, b] {fn(x)} is monotone. Then{fn(x)} converges uniformly to

f(x) on [a, b].

Proof. By contradiction. If{fn(x)} does not converge uniformly tof(x) on [a, b], then∃ǫ0 > 0, for any

nk, ∃xk, s.t.,

|fnk(xk)− f(xk)| ≥ ǫ0, k = 1, 2, · · · ,

Since{xk} is bounded, then{xk} has a converged subsequence{xki}, xki

→ x0 ∈ [a, b], as i → +∞.

Sincefn(x0) → f(x0), ∃N , s.t.,

|fN (x0)− f(x0)| < ǫ0.

SincefN (x)− f(x) is continuous atx0, we have

limi→+∞

|fN (xki)− f(xki

)| = |fN (x0)− f(x0)| < ǫ0,

and∃I, s.t.,

|fN(xki)− f(xki

)| < ǫ0, ∀i > I.

For eachx, {fn(x)} is monotone, then

|fnk(xki

)− f(xki)| ≤ |fN(xki

)− f(xki)| < ǫ0, ∀i > I, nk > N,

which is a contradiction.

Example 8.1.5Suppose thatf(x) is continuous on[0, 1] andf(1) = 0. Show thatgn(x) = f(x)xn, n =

1, 2, · · · , converges uniformly on[0, 1].

Proof. gn(x) andg(x) are continuous on[0, 1]. For eachx ∈ [0, 1], {xn} is monotone, then{gn(x)} is

monotone. By Dini’s test,{gn(x)} converges uniformly on[0, 1].


Theorem 8.1.6’ (Series form of Dini Test)Suppose the∞∑

n=1

un converges pointwise toS(x) on [a, b],

un(x), ∀n, andS(x) are continuous on[a, b], and for anyx ∈ [a, b] all un(x) have the same sign. Then∞∑

n=1

un converges uniformly toS(x) on [a, b].

Theorem 8.1.7 (Abel Test)Suppose that

(i) the function series∞∑

n=1

an(x) converges uniformly on the intervalI;

(ii) the sequence{bn(x)} is monotone for givenx ∈ I;

(iii) the sequence{bn(x)} is uniformly bounded onI, i.e.,∃M > 0, s.t.

|bn(x)| ≤M ∀x ∈ I, ∀n.

Then the function series∞∑

n=1

an(x)bn(x) converges uniformly onI.

Proof. From (i) we have, for givenǫ > 0, ∃N , s.t.,

∣

∣

∣

n+p∑

k=n+1

ak(x)∣

∣

∣ <ǫ

3M, ∀x ∈ I, ∀p > 0, n > N.

Since for eachx ∈ I, {bn(x)} is monotone, from Abel’s lemma and (iii) we get

∣

∣

∣

n+p∑

k=n+1

ak(x)bk(x)∣

∣

∣ ≤ ǫ

3M(|bn+1(x)| + 2|bn+p(x)|)

≤ ǫ

3M(M + 2M) = ǫ, ∀x ∈ I, ∀p > 0, n > N,

i.e.,∞∑

n=1

an(x)bn(x) converges uniformly onI by Cauchy Convergence Criterion.

Example 8.1.6Prove that the series∞∑

n=1

(−1)n√n

· 2 + xn

1 + xn

converges uniformly on[0,∞).

Proof. Letan(x) =(−1)n√

nandbn(x) =

2 + xn

1 + xn. Since

∞∑

n=1

an(x) converges, then it converges uniformly.

For x ∈ [0, 1], {bn(x)} is increasing. Forx ∈ (1,+∞), {bn(x)} is decreasing. Also,

2 + xn

1 + xn≤ 2 + 2xn

1 + xn= 2, ∀x ∈ [0,+∞), ∀n.

By Abel’s test,∞∑

n=1

an(x)bn(x) converges uniformly on[0,∞).

Theorem 8.1.8 (Dirichlet Test)Suppose that


(i) An(x) =

n∑

k=1

ak(x), n = 1, 2, · · · , is uniformly bounded onI, i.e.,∃M > 0, s.t.

|An(x)| ≤M ∀x ∈ I, ∀n;

(ii) for eachx ∈ I, the sequence{bn(x)} is monotone;

(iii) the sequence{bn(x)} converges to0 uniformly onI.

Then the function series∞∑

n=1

an(x)bn(x) converges uniformly onI.

Proof. From (i) we have,

∣

∣

∣

n+p∑

k=n+1

ak(x)∣

∣

∣ = |An+p(x)−An(x)| ≤ 2M, ∀x ∈ I, ∀p > 0, ∀n.

From (iii) we have, for givenǫ > 0, ∃N , s.t.,

|bn(x)| <ǫ

6M, ∀x ∈ I, ∀n > N.

From (ii) and Abel’s lemma we have,

∣

∣

∣

n+p∑

k=n+1

ak(x)bk(x)∣

∣

∣ ≤ 2M(|bn+1(x)| + 2|bn+p(x)|)

≤ 2M( ǫ

6M+

2ǫ

6M

)

= ǫ, ∀x ∈ I, ∀p > 0, ∀n > N,

i.e.,∞∑

n=1

an(x)bn(x) converges uniformly onI by Cauchy Convergence Criterion.

Example 8.1.7Prove that the series∞∑

n=1

(−1)n(1− x)xn

converges uniformly on[0, 1].

Proof. Let an(x) = (−1)n and bn(x) = (1 − x)xn. An(x) =

n∑

n=1

an(x) is uniformly bounded. Since

b′n(x) = nxn−1 − (n+ 1)xn, bn(0) = bn(1) = 0, thenbn(x) reaches its maximum atxn =n

n+ 1. Then,

bn(x) <1

n+ 1, i.e., {bn(x)} converges to0 uniformly on[0, 1]. Also, for eachx ∈ [0, 1], {bn(x)} is

decreasing. By Dirichlet’s test,∞∑

n=1

an(x)bn(x) converges uniformly on[0, 1].

8.2 Uniform Limit of Functions

With the concept of uniform convergence, we can now give answers to the three questions proposed in

Section 1.1.


8.2.1 Uniform Limit of Continuous Functions

Theorem 8.2.1Suppose that{fn(x)} is a sequence of continuous functions defined on the intervalI and

uniformly converges tof(x) in I, then the limit functionf(x) is continuous inI.

Proof. Letx0 be any pointI. We now prove∀ε > 0, ∃δ > 0 such that

|f(x)− f(x0)| < ε if |x− x0| < δ.

Sincefn(x) converges uniformly tof(x), there exists a natural numberN0 such that

|fN0(x) − f(x)| < ε

3, ∀x ∈ I.

BecausefN0(x) is continuous in intervalI, especially continuous at the pointx0, we can find aδ > 0 such

that

|fN0(x)− fN0(x0)| <ε

3, if |x− x0| < δ.

Hence

|f(x)− f(x0)| ≤ |f(x)− fN0(x)|+ |fN0(x)− fN0(x0)|+ |fN0(x0)− f(x0)|

<ε

3+ε

3+ε

3= ε if |x− x0| < δ.

Theorem 8.2.1′ Suppose that∞∑

n=1

un(x) is a series of continuous functionsun(x) defined on the intervalI

and converges uniformly toS(x). Then the sumS(x) is continuous onI.

8.2.2 Uniform Limit of Integrable Functions

Theorem 8.2.2Suppose that{fn(x)} is a sequence of integrable functions on[a, b] and converges uniformly

to f(x) on [a, b]. Then the limit functionf(x) is integrable, and moreover

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx =

∫ b

a

limn→∞

fn(x)dx.

Proof. We first prove thatf(x) is integrable in[a, b]. By the integrable criterion in Chapter 5, we need to

show that∀ǫ > 0, we can find a partitionP∗ on [a, b] such that

U(f, P∗)− L(f, P∗) < ǫ.

Sincefn(x) converges uniformly tof(x) on [a, b], there is a sufficiently largeN0 such that

fN0(x) −ǫ

3(b− a)≤ f(x) ≤ fN0(x) +

ǫ

3(b− a)∀x ∈ [a, b].


Hence for any partitionP of [a, b],

L(fN0 , P )−ǫ

3≤ L(f, P ) ≤ U(f, P ) ≤ U(fN0, P ) +

ǫ

3.

Then we have

U(f, P )− L(f, P ) ≤ U(fN0 , P )− L(fN0 , P ) +2

3ǫ.

SincefN0 is integrable in[a, b], we can find a partitionP∗ of [a, b] such that

U(fN0 , P∗)− L(fN0 , P∗) <1

3ǫ.

Therefore, for this partitionP∗, we have

U(f, P∗)− L(f, P∗) <1

3ǫ+

2

3ǫ = ǫ.

Hence,f(x) is integrable on[a, b]. Next, we prove:∀ǫ, ∃N such that

∣

∣

∣

∣

∣

∫ b

a

fn(x)dx −∫ b

a

f(x)dx

∣

∣

∣

∣

∣

< ǫ, if n ≥ N.

Since{fn(x)} converges uniformly tof(x), there existsN such that for allx ∈ [a, b]

|fn(x) − f(x)| < ε

b− aif n ≥ N.

Hence∣

∣

∣

∣

∣

∫ b

a

fn −∫ b

a

f

∣

∣

∣

∣

∣

≤∫ b

a

|fn − f | <∫ b

a

ε

b− a= ε, if n ≥ N,

which means that

limn→∞

∫ b

a

fn(x)dx =

∫ b

a

f(x)dx.


n=1

un(x) is a series of integrable functionsun(x) on interval [a, b] and

converges uniformly toS(x). Then the sumS(x) is integrable on[a, b], and

limk→∞

∫ b

a

n∑

k=1

uk(x)dx =

∫ b

a

S(x)dx,

or∫ b

a

∞∑

n=1

un(x)dx =

∞∑

n=1

∫ b

a

un(x)dx.

In this case we say that the function series is integrable termwise.


8.2.3 Uniform Limit of Differentiable Functions

Theorem 8.2.3Suppose that{fn(x)} is a sequence of continuously differentiable functions on the interval

[a, b] and satisfies

(i) The sequence{fn(x)} converges pointwise to the functionf(x) on [a, b].

(ii) The sequence{f ′n(x)} converges uniformly to the functiong(x) on [a, b].

Then the functionf(x) is continuously differentiable and

f ′(x) = g(x), ∀x ∈ [a, b],

or

[ limn→∞

fn(x)]′ = lim

n→∞f ′n(x).

Proof. According to the Fundamental Theorem of Calculus, for everyn and allx ∈ [a, b]∫ x

a

f ′n(t)dt = fn(x)− fn(a).

From assumption (ii) and using Theorem 8.2.2, we have

limn→∞

∫ x

a

f ′n(t)dt =

∫ x

a

g(t)dt for a ≤ x ≤ b,

i.e.,

limn→∞

[fn(x) − fn(a)] =

∫ x

a

g(t)dt.

By assumption (i),

f(x)− f(a) =

∫ x

a

g(t)dt, ∀x ∈ [a, b].

From assumption (ii) and using Theorem 8.2.1, we know thatg(x) is continuous on[a, b]. Hence, by the

Intermediate Value Theorem of Integral, at each pointx0 ∈ [a, b]

1

∆x[f(x0 +∆x)− f(x0)] =

1

∆x

∫ x0+∆x

x0

g(t)dt

=1

∆xg(ξ) ·∆x,

whereξ is betweenx0 andx0 +∆x. As∆x→ 0, we have

f ′(x0) = g(x0), for x0 ∈ [a, b].



n=1

un(x) is a function series of which eachun(x) is continuously differen-

tiable on the interval[a, b] and satisfies


(i)∞∑

n=1

un(x) converges pointwisely to the functionS(x) on [a, b]

(ii)∞∑

n=1

u′n(x) converges uniformly to the functionσ(x) on [a, b]

Then the functionS(x) is continuously differentiable and

S′(x) = σ(x), ∀x ∈ [a, b],

or( ∞∑

n=1

un(x)

)′

=

∞∑

n=1

u′n(x).

In this case, we say that the function series is differentiable termwise.

8.3 Power Series

8.3.1 Radius of Convergence

Definition 8.3.1A function series∞∑

k=0

uk(x) is called a Power Series if its terms have the form

uk(x) = ak(x− x0)k, k = 0, 1, . . . ,

whereak are real numbers.

An important property of a power series is that it has a Radiusof Convergence. In other words, the domain

of convergence of a power series∞∑

k=0

ak(x − x0)k is an interval with the centerx0 and a radiusR, i.e.,

one of the following cases:[x0 − R, x0 + R], (x0 − R, x0 + R), [x0 − R, x0 + R), (x0 − R, x0 + R] or

(−∞,+∞). The last case can be understood asR = +∞. WhenR = 0, it means that the series converges

only atx0.

For∞∑

k=0

ak(x−x0)k, a transformt = x−x0 will change it to∞∑

k=0

aktk. Hence, we only need to discuss

the convergence of power series in the form∞∑

k=0

akxk.

Lemma 8.3.2

(1) If∞∑

k=0

akxk converges at the pointx = x1 6= 0, then

∞∑

k=0

akxk converges absolutely for|x| < |x1|,

i.e.∞∑

k=0

|akxk| converges on the interval(−|x1|, |x1|)


2) If∞∑

k=0

akxk does not converge at the pointx = x2 6= 0, then

∞∑

k=0

akxk does not converge for

|x| > |x2|, i.e. the series does not converge on(−∞,−|x2|) and(|x2|,∞).

Proof. From the convergence of the number series∞∑

k=0

akxk1 , we know thatlim

k→∞akx

k1 = 0. Hence, there is

a constantM > 0 such that

|akxk1 | ≤M, k = 0, 1, . . . , n, . . . .

For anyx 6= 0 satisfying|x| < |x1|

|akxk| = |akxk1 |∣

∣

∣

∣

x

x1

∣

∣

∣

∣

k

≤M

∣

∣

∣

∣

x

x1

∣

∣

∣

∣

k

.

Since

∣

∣

∣

∣

x

x1

∣

∣

∣

∣

< 1,∞∑

k=0

∣

∣

∣

∣

x

x1

∣

∣

∣

∣

k

converges. By Comparison Test,∞∑

k=0

|akxk| converges when|x| < |x1|. There-

fore,∞∑

k=0

akxk converges absolutely in(−|x1|, |x1|).

We prove case (2) by contradiction. If there is a pointx, which satisfies|x| > |x2| and∞∑

k=0

akxk

converges. Then from case (1),∞∑

k=0

akxk2 converges absolutely. But it contradicts with the assumption, so

∞∑

k=0

akxk does not converge.

Using Lemma 8.3.2 and the Nested Interval Theorem, we can prove the following theorem.

Theorem 8.3.3For any power series∞∑

k=0

akxk, there is a nonnegative numberR (maybe 0 and+∞) such

that

(1) For |x| < R, the series converges absolutely.

(2) For |x| > R, the series diverges.

Proof. If the series converges everywhere, thenR = +∞. If the series converges only at the point zero,

thenR = 0. Now we need only to prove the case when the series converges at some pointx1 > 0 and

diverges at some pointx2 > x1. In this case, we prove that there exists a numberR > 0 such that the

series converges on(−R,R) and diverges on(−∞,−R) ∪ (R,+∞). Let u1 = x1, v1 = x2. Choose

c1 = (u1 + v1)/2. If the series converges atc1, let u2 = c1 andv2 = v1; otherwise letu2 = u1 and

v2 = c1. Step by step chooseck =uk + vk

2, and letuk+1 = ck andvk+1 = vk if the series converges

at ck; otherwise letuk+1 = uk andvk+1 = ck. Hence, we get two sequences{uk} and{vk}. The series∞∑

k=0

akxk converges at alluk and diverges at allvk and{uk}. {vk} satisfy:

(1) {uk} and{vk} are monotonically increasing and decreasing respectively.

(2) limk→∞

(vk − uk) = limk→∞

x2 − x12k

= 0.


Therefore by the Nested Interval Theorem, there is a unique pointR such that

uk ≤ R ≤ vk for all k

and

limk→∞

uk = limk→∞

vk = R.

Hence, for anyx with |x| < R, there exists some indexN such that|x| < uN . By the Lemma, the series∞∑

k=0

akxk converges due to the convergence of the series

∞∑

k=0

akukN . On the other hand, for anyx with

|x| > R, there exists some indexM such that|x| > vM . By the Lemma the series∞∑

k=0

akxk diverges due to

the divergence of the series∞∑

k=0

akvkM . Thus we complete the proof.

Note.The theorem does not give any conclusions on the two end pointsx = R andx = −R. The series

may converge or diverge at the end points. For example:∞∑

k=1

xk converges on(−1, 1),∞∑

k=1

xk

kconverges

on [−1, 1), and∞∑

k=1

xk

k2converges on[−1, 1]. Three series have a same radius of convergence. But the

convergence of end points are different. They should be discussed individually.

Theorem 8.3.4 (Ratio Test)If the coefficients of a power series∞∑

n=0

anxn satisfy

limn→∞

|an+1||an|

= l,

wherel may be+∞. Then the radius of convergence is1/l, or, we have

(1) if 0 < l < +∞,R = 1/l;

(2) if l = 0,R = +∞;

(3) if l = +∞,R = 0;

Proof. Letun(x) = anxn, and supposex 6= 0. Then

∣

∣

∣

∣

un+1(x)

un(x)

∣

∣

∣

∣

=|an+1||an|

|x|.

By the assumption,

limn→∞

∣

∣

∣

∣

un+1(x)

un(x)

∣

∣

∣

∣

= l |x|.

Hence,

(1) If 0 < l < +∞, by the Ratio Test for Series with Nonnegative Terms:l|x| < 1, then∞∑

n=0

|un(x)|

converges, i.e.|x| < 1/l, then∞∑

n=0

un(x) converges absolutely. At the same time, ifl|x| > 1, i.e.


|x| > 1/l, then{|un(x)|} doesn’t converge to zero, so{un(x)} doesn’t converge to zero also. Hence∞∑

n=0

uk(x) doesn’t converge. Thus, we get the conclusionR = 1/l.

(2) If l = 0, by the assumption, for anyx 6= 0

limn→∞

|un+1(x)||un(x)|

= 0.

Hence∞∑

n=0

un(x) converges absolutely for anyx 6= 0. Since∞∑

n=0

anxn converges obviously atx = 0,

hence∞∑

n=0

anxn converges on(−∞+∞), i.e.,R = +∞.

(3) If l = +∞, that means for anyx 6= 0

limn→∞

|un+1(x)||un(x)|

= limn→∞

|an+1||an|

|x| = +∞.

Then forn sufficiently large|un+1(x)||un(x)|

> 1,

i.e.,{un(x)} doesn’t converge to zero and∞∑

n=0

un(x) does not converge at any pointx 6= 0. Hence,

∞∑

n=0

anxn converges only atx = 0, i.e.,R = 0.

Theorem 8.3.5 (Root Test)If the coefficients of a power series∞∑

n=0

anxn satisfy

lim n√

|an| = l,

wherel may be+∞. Then the radius of convergence is1/l.

8.3.2 Uniform Convergence

Theorem 8.3.6Suppose that the radius of convergence of the power series∞∑

n=0

anxn isR > 0. Then

(1) For any positive numberb < R, the series∞∑

n=0

anxn converges uniformly on[−b, b].

(2) If∞∑

n=0

anxn converges atx = R, then

∞∑

n=0

anxn converges uniformly on[0, R].

(3) If∞∑

n=0

anxn converges atx = −R, then

∞∑

n=0

anxn converges uniformly on[−R, 0].

Proof.


(1) We have proved that if|un(x)| ≤ cn on intervalI and∞∑

n=0

cn converges, then∞∑

n=0

un(x) uniformly

converges inI. Now

|anxn| ≤ |an|bn on [−b, b]

and∞∑

n=1

|an|bn converges. Hence∑

anxn converges uniformly on[−b, b]

(2),(3) Use Abel Test.

Theorem 8.3.7If the radius of convergence of the power series∞∑

n=0

anxn isR, then the series is continuous

on (−R,R). Moreover, if the series converges atx = R (or x = −R), then∞∑

n=0

anxn is also continuous at

x = R (or x = −R).

Proof. For any fixed pointx0 ∈ (−R,R), there is a positive numberb < R such thatx0 ∈ (−b, b). From

Theorem 8.3.6, we know that∞∑

n=0

anxn converges uniformly on[−b, b]. Obviously, each termanxn of the

series is continuous on[−b, b], hence by Theorem 8.2.1’, the series is continuous in[−b, b], especially at

the pointx0. Becausex0 is any point in(−R,R),∞∑

n=0

anxn is continuous in(−R,R)

If∞∑

n=0

anxn also converges atx = R (or x = −R), from Theorem 8.3.6

∞∑

n=0

anxn converges uniformly

on [0, R] (or [−R, 0]). Hence it is continuous on[0, R] (or [−R, 0]), especially continuous atx = R (or

x = −R).

Theorem 8.3.8If the radius of convergence of power series∞∑

n=0

anxn isR, then

∫ x

0

∞∑

n=0

antndt =

∞∑

n=0

∫ x

0

antndt

=∞∑

n=0

ann+ 1

xn+1 for −R < x < R.

Proof. For x ∈ (−R,R),∞∑

n=0

anxn converges uniformly on[−|x|, |x|], especially on[−|x|, 0] or [0, |x|].

Obviously, each termanxn is integrable. Hence by Theorem 8.2.2’, the series can be integrated termwisely.

Theorem 8.3.8′ If the radius of convergence of the power series∞∑

n=0

anxn is R, and furthermore, if the

series converges atx = R, then∫ R

0

∞∑

n=0

antndt =

∞∑

n=0

∫ R

0

antndt

=

∞∑

n=0

anRn+1

n+ 1


Proof. The conclusion comes from Theorem 8.3.6 directly.

Theorem 8.3.9If the radius of convergence of the power series∞∑

n=0

anxn isR, then

( ∞∑

n=0

anxn

)′

=

∞∑

n=0

(anxn)′

=

∞∑

n=0

nanxn−1, for −R < x < R

Proof. For any fixed pointx ∈ (−R,R), there is a positive numberb such that|x| < b < R. It is known

that∞∑

n=0

anxn converges pointwise on[−b, b]. By the Theorem 8.2.3’, we need only to prove that the series

∞∑

n=1

nanxn−1 converges uniformly on[−b, b]. In order to find a convergent series

∞∑

n=1

cn which satisfies

cn > 0 and|nanxn−1| ≤ cn in [−b, b], we choose a numberr such thatb < r < R. Then forx ∈ [−b, b]

|nanxn−1| ≤ n|an|bn−1 =n

r|anrn|

(

b

r

)n−1

.

Since∞∑

n=1

|anrn| converges,{|anrn|} is bounded, i.e., there exists a constantM > 0 such that

|anrn| ≤M, n = 1, 2, . . . .

Hence

|nanxn−1| ≤ M

rn

(

b

r

)n−1

= Cn.

Due tob

r< 1, the series

∞∑

n=1

cn converges. Hence∞∑

n=1

nanxn−1 converges uniformly in[−b, b]. By the

Theorem 8.2.3’, the series can be differentiated termwisely on [−b, b], especially at the pointx which we

choose. Sincex can be any point in(−R,R), the formula holds on(−R,R).

In the above proof, we can see that the convergent radius of∞∑

n=1

nanxn−1 is also the same as

∞∑

n=1

anxn.

Iteratively apply Theorem 8.3.9, we get the following corollary.

Corollary 8.3.10Suppose that the radius of convergence of a power series isR. Then the power series can

be differentiated termwisely fork times,k is any natural number. That is

S(k)(x) =∞∑

n=0

(anxn)(k) =

∞∑

n=k

n(n− 1) · · · (n− k + 1)anxn−k k = 1, 2, . . . .

The convergent radius ofS(k)(x) is also the same asS(x).

Example 8.3.1Find the domain of convergence of the following power series:


(a)∞∑

n=1

xn

n2n.

(b)∞∑

n=1

n!

nnxn.

(c)∞∑

n=1

xn√n

.

(d)∞∑

n=1

n!

nn(x− 2)n.

8.4 Taylor Series

8.4.1 Taylor Polynomials and Order of Contact

Definition 8.4.1Let functionsf(x) andg(x) be defined on an open intervalI andx0 ∈ I. We say that

f(x) andg(x) have contact of ordern at x0 provided that

f (k)(x0) = g(k)(x0) for k = 0, 1, . . . , n.

Lemma 8.4.2LetI be an open interval andx0 ∈ I. Suppose that the functionf(x) has(n+1) derivatives

onI, and

f (k)(x0) = 0 for k = 0, . . . , n.

Then for each pointx ∈ I andx 6= x0, there is a pointξ strictly betweenx andx0 at which

f(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

Corollary 8.4.3 If f(x) andg(x) have contact of ordern at x0, then

f(x)− g(x) =f (n+1)(ξ)− g(n+1)(ξ)

(n+ 1)!(x− x0)

n+1 wherex0 < ξ < x or x < ξ < x0.

Corollary 8.4.3 shows thatf(x) andg(x) are very closed to each order in the neighborhood ofx0.

Polynomials are the simplest kind of functions. Hence, for ageneral functionf(x) it is natural to seek

a polynomial that is a good approximation off(x).

Proposition 8.4.4Suppose that the functionf(x) hasn derivatives on an open intervalI. Then there is a

unique polynomial of degree at mostn that has contact of ordern with the functionf at pointx0 ∈ I. This

polynomial is defined by the formula

pn(x) = f(x0) + f ′(x0)(x − x0) + · · ·+ f (n)(x0)

n!(x− x0)

n

=

n∑

k=0

f (k)(x0)

k!(x − x0)

k.


Proof. Let pn(x) =n∑

k=0

ck(x − x0)k be a polynomial that has contact of ordern with functionf(x). By

the definition,

p(k)n (x0) = f (k)(x0), k = 0, 1, . . . , n.

Butp(k)n (x0) = k!ck. Hence,ck = f(k)(x0)k! .

Definition 8.4.5For the functionf(x), the polynomial

pn(x) =

n∑

k=0

f (k)(x0)

k!(x− x0)

k

is called thenth Taylor polynomial at the pointx0

Theorem 8.4.6 (Lagrange Remainder Theorem)Suppose that the functionf(x) has(n+ 1) derivatives

on the open intervalI andpn(x) is thenth Taylor polynomial forf(x) at the pointx0 ∈ I, i.e.

pn(x) =

n∑

k=0

f (k)(x0)

k!(x− x0)

n.

Then for each pointx ∈ I, there is a pointξ strictly betweenx andx0 such that

f(x)− pn(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

Proof. Sincef(x) andpn(x) have contact of ordern, from Lemma 8.4.2 we obtain

f(x) − pn(x) =f (n+1)(ξ)− p

(n+1)n (ξ)

(n+ 1)!(x− x0)

n+1.

Butp(n+1)n (x) ≡ 0, it follows that

f(x)− pn(x) =f (n+1)(ξ)

(n+ 1)!(x− x0)

n+1.

The difference betweenf(x) andpn(x) is called theRemainder of the Taylor Polynomial approximation.

Definition 8.4.7Let I be a neighborhood ofx0. If f : I → R has derivatives of all orders. Let

pn(x) =n∑

k=0

f (k)(x0)

k!(x− x0)

k.

If

limn→∞

pn(x) = f(x) ∀x ∈ I,

we write

f(x) =

∞∑

k=0

f (k)(x0)

k!(x− x0)

k ∀x ∈ I.


This formula is called a Taylor Series expansionof f(x) aboutx0. If x0 = 0, then the series

∞∑

n=0

f (n)(0)

n!xn ∀x ∈ I

is called MaclurinSeries off(x).

Theorem 8.4.8Suppose that the power series∞∑

n=0

an(x − x0)n converges tof(x) on (x0 − R, x0 + R).

Then the series must be the Taylor Series off(x) expanding atx = x0. That is

an =1

n!f (n)(x0), n = 0, 1, 2, . . . .

Proof. For anyk = 0, 1, 2, . . . fixed, according to Corollary 8.3.10,f(x) hask-the derivative and

f (k)(x) =∞∑

n=0

(an(x− x0)n)(k) =

∞∑

n=k

n(n− 1) · · · (n− k + 1)an(x− x0)n−k.

Takingx = x0, we then get

ak =1

k!f (k)(x0).

Note:Sometime it is difficult to determine the domain of convergence analyzing the Remainder Term of the

Taylor Polynomial. Using this theorem, we may get the convergence by applying the theory of power series.

Theorem 8.4.9 (Cauchy Integral Remainder Formula)Suppose that the functionf(x) hasn+1 deriva-

tives on an open intervalI andf (n+1)(x) is continuous onI. Then for each pointx in I,

f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k =1

n!

∫ x

x0

f (n+1)(t)(x− t)ndt.

Proof. By the First Fundamental Theorem of Calculus,

f(x)− f(x0) =

∫ x

x0

f ′(t)dt.

Integrating by parts,∫ x

x0

f ′(t)dt = −∫ x

x0

f ′(t)d(x − t)

= −f ′(t)(x − t)

∣

∣

∣

∣

x

x0

+

∫ x

x0

f ′′(t)(x − t)dt

= f ′(x0)(x − x0) +

∫ x

x0

f ′′(t)(x − t)dt,

which is the result forn = 1. The general formula follows by induction. If the theorem istrue form, i.e.

f(x) =

m∑

k=0

f (k)(x0)

k!(x− x0)

k +1

m!

∫ x

x0

f (m+1)(t)(x − t)mdt,


then

1

m!

∫ x

x0

f (m+1)(t)(x − t)mdt =−1

(m+ 1)!

∫ x

x0

f (m+1)(t)d(x − t)m+1

=−1

(m+ 1)!f (m+1)(t)(x − t)m+1

∣

∣

∣

∣

x

x0

+1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x − t)m+1dt

=1

(m+ 1)!f (m+1)(x0)(x − x0)

m+1 +1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x− t)m+1dt

We obtain

f(x) =

m+1∑

k=0

f (k)(x0)

k!(x − x0)

k +1

(m+ 1)!

∫ x

x0

f (m+2)(t)(x− t)m+1dt

That means the theorem is also true form+ 1.

8.4.2 Convergence of Taylor Series

Example 8.4.1Leta > 0 and

f(x) =

e−1x2 , x 6= 0,

0, x = 0.

Show that the series∞∑

n=0

f (n)(0)

n!xn converges on(−a, a), but does not converge tof(x).

Proof. It is easy to see thatf (n)(x) exists on(−a, a) for anyn, andf (n)(0) = 0, ∀n. Then, we have

∞∑

n=0

f (n)(0)

n!xn = 0 6= f(x), ∀x ∈ (−a, a), exceptx = 0.

Lemma 8.4.9For any numberc

limn→∞

cn

n!= 0.

Proof. Choosek to be a natural number such thatk ≥ 2|c|. Then forn ≥ k

0 ≤∣

∣

∣

∣

cn

n!

∣

∣

∣

∣

=

( |c|1

· · · |c|k

)( |c|k + 1

· · · |c|n

)

≤ |c|k(

1

2

)n−k

= (2|c|)k(

1

2

)n

.

2|c|k is a constant andlimn→∞

(

1

2

)n

= 0. Hence by squeezing principle,lim

∣

∣

∣

∣

cn

n!

∣

∣

∣

∣

= 0 and so limn→∞

cn

n!= 0.


Theorem 8.4.10Suppose thatf(x) has derivatives of all order on the interval[x0 − r, x0 + r] and

|f (n)(x)| ≤Mn for x0 − r ≤ x ≤ x0 + r,

wherer andM are positive andn is any natural number. Then

∞∑

k=0

fk(x0)

k!(x− x0)

k = f(x) ∀x ∈ [x0 − r, x0 + r].

Proof. According to the Lagrange Remainder Theorem, for eachx ∈ [x0 − r, x0 + r]∣

∣

∣

∣

∣

f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k

∣

∣

∣

∣

∣

=|f (n+1)(ξ)|(n+ 1)!

|x− x0|n+1,

whereξ is strictly betweenx andx0. Hence,∣

∣

∣

∣

∣

f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k

∣

∣

∣

∣

∣

≤ Mn+1

(n+ 1)!|x− x0|n+1 ≤ cn+1

(n+ 1)!,

whereC =Mr. The conclusion follows immediately from Lemma 8.4.9.

Example 8.4.2Letf(x) = ex. Since

f (n)(x) = ex, f (n)(0) = 1, n = 0, 1, 2, · · · .

Then for anyr > 0 and for anyn ≥ 0 we have

|f (n)(x)| = ex ≤ er ∀x ∈ [−r, r].

Therefore,

ex =

∞∑

k=0

xk

k!∀x ∈ (−∞,+∞).

Example 8.4.3Letf(x) = sinx. Since

f (n)(x) = sin(

x+nπ

2

)

, f (n)(0) = sinnπ

2=

0, n even,

(−1)(n−1)/2, n odd.

Then for anyn ≥ 0 we have

|f (n)(x)| =∣

∣

∣ sin(

x+nπ

2

)∣

∣

∣ ≤ 1 ∀x ∈ (−∞,+∞).

Therefore,

sinx =

∞∑

k=0

(−1)k

(2k + 1)!x2k+1 ∀x ∈ (−∞,+∞).

Example 8.4.4Similarly, we have

cosx =

∞∑

k=0

(−1)k

(2k)!x2k ∀x ∈ (−∞,+∞).


Example 8.4.5Letf(x) = ln(1 + x), x > −1. Since

f (n)(x) =(−1)n−1(n− 1)!

(1 + x)n, f (n)(0) = (−1)n+1(n− 1)!, n ≥ 1.

Then we have the series∞∑

k=1

(−1)k−1

kxk. (8.4.1)

We need to determine the domain of convergence of the series.Consider the series

∞∑

k=0

(−1)kxk =1

1 + xx ∈ (−1, 1).

The radius of convergence of this power series isR = 1, then from Theorem 8.3.8 we have

ln(1 + x) =

∫ x

0

dx

1 + x

=

∫ x

0

∞∑

k=0

(−1)kxk

=∞∑

k=0

∫ x

0

(−1)kxk

=

∞∑

k=1

(−1)k−1

kxk, x ∈ (−1, 1).

Moreover, since the series (1) converges atx = 1, from Theorem 8.3.7

ln(1 + x) =

∞∑

k=1

(−1)k−1

kxk x ∈ (−1, 1].

Example 8.4.6Prove that∞∑

n=1

(−1)n−1

n= ln 2.

Example 8.4.7 (Newton’s Binomial Expansion)Letf(x) = (1 + x)α, whereα is a real number. Since

f (n)(x) = α(α − 1) · · · (α− n+ 1)(1 + x)α−n, f (n)(0) = α(α− 1) · · · (α− n+ 1).

Then we have the series

1 +

∞∑

k=1

α(α− 1) · · · (α− k + 1)

k!xk, −1 < x < 1.

Proof. Using Cauchy Integral Remainder we have

Rn(x) = f(x)−n∑

k=0

f (k)(x0)

k!(x− x0)

k

=1

n!

∫ x

x0

f (n+1)(t)(x− t)ndt

=1

n!f (n+1)(ξ)(x − ξ)n(x− x0),


whereξ is betweenx0 andx. Letξ = x0 + θ(x− x0), 0 < θ < 1, then

Rn(x) =1

n!f (n+1)(x0 + θ(x− x0))(1 − θ)n(x − x0)

n+1.

For x0 = 0 we have

Rn(x) =1

n!f (n+1)(θx)(1 − θ)nxn+1

=1

n!α(α− 1) · · · (α− n)(1 + θx)α−n−1(1− θ)nxn+1

=1

n!α(α− 1) · · · (α− n)xn+1(1 + θx)α−1

( 1− θ

1 + θx

)n

.

Consider the series∞∑

k=0

α(α− 1) · · · (α− k)

k!xk+1.

It is easy to see that the radius of convergence of this seriesisR = 1. Therefore, for|x| < 1

limn→+∞

α(α − 1) · · · (α− n)

n!xn+1 = 0.

Since0 < 1− |x| ≤ |1 + θx| ≤ 1 + |x| < 2, letM(x) = max{(1− |x|)α−1, 2α−1}, then

|1+ θx|α−1 ≤M(x), i.e., for fixedx, |1+ θx|α−1 is bounded. Also, for|x| < 1, 0 < 1− θ < 1+ θx, then

0 <( 1− θ

1 + θx

)n

< 1. Therefore,Rn(x) → 0, ∀|x| < 1.

Example 8.4.8Let f(x) = sin−1 x. It’s somewhat complicated to expand by directly differentiating f(x)

times by times. It’s known that

sin−1 x =

∫ x

0

dt√1− t2

for x ∈ (−1, 1).

By Newton’s Binomial Expansion, we know

1√1− t2

=∞∑

k=0

12

k

(−t2)k =∞∑

k=0

(2k − 1)!!

2k!!x2k.

By Applying Theorem 8.3.8 (Termwise integrate the uniformly convergent series), for anyx ∈ (−1, 1)

sin−1 x =

∫ x

0

dt√1− t2

=

∞∑

k=0

(2k − 1)!!

2k!!

x2k+1

2k + 1.

Example 8.4.9Letf(x) = tan−1 x. Expand1

1 + x2to a power series

1

1 + x2= 1− x2 + x4 − x6 + · · ·

=

∞∑

n=0

(−1)nx2n


The convergent radius of the series isR = 1. By Theorem 8.3.8, we have∫ x

0

1

1 + t2dt =

∞∑

n=0

∫ x

0

(−1)nt2ndt − 1 < x < 1

⇒ tan−1 x =

∞∑

n=0

(−1)nx2n+1

2n+ 1, −1 < x < 1

The series is convergent at the end pointsx = 1 as well asx = −1.

Note: The derivative of the series∞∑

n=0

(−1)nx2n+1

2n+ 1is

∞∑

n=0

(−1)nx2n which is not convergent at both end

points -1 and 1. This example shows that although the convergent radiusR will not change for differentia-

tion of a power series, the convergence at the end points willnot preserve.

8.5 Exercise

1. For each natural numbern and each numberx, define

fn(x) =1− |x|n1 + |x|n .

Find the functionf : R → R to which the sequence{fn : R → R} converges pointwise. Prove that

the convergence is not uniform.

Hint. fn continuous,fn(0) = 1 and fn(1) = 0. By Intermediate Value Theorem,∀n, ∃xn ∈ (0, 1), s.t.fn(xn) = 1/2. Sincef(xn) = 1, then,|f(xn)− fn(xn)| = 1/2.

2. For each natural numbern and each numberx ≥ 2, define

fn(x) =1

1 + xn.

Find the functionf : [2,∞) → R to which the sequence{fn : [2,∞) → R} converges pointwise.

Prove that the convergence is uniform.

3. For each natural numbern and each numberx ∈ (0, 1), define

fn(x) =1

nx+ 1.

Find the functionf : (0, 1) → R to which the sequence{fn : (0, 1) → R} converges pointwise.

Prove that the convergence is not uniform.

Hint. ∀n, let xn = 1/n, thenfn(xn) = 1/2, and|f(xn)− fn(xn)| = fn(xn) = 1/2.

4. For each natural numbern and each numberx ∈ [0, 1], define

fn(x) =x

nx+ 1.

Find the functionf : [0, 1] → R to which the sequence{fn : [0, 1] → R} converges pointwise.

Prove that the convergence is uniform.

Hint. |f(x)− fn(x)| = |0− x/(nx+ 1)| = 1/(n+ 1/x) < 1/n if x 6= 0, and|f(0) − fn(0)| = 0.


5. For each natural numbern and each numberx ∈ (−1, 1), define

pn(x) = x+ x(1− x2) + · · ·+ x(1 − x2)n.

Prove that the sequence{pn : (−1, 1) → R} converges pointwise.

6. Suppose that the sequence{fn : D → R} and{gn : D → R} converge uniformly to the functions

f : D → R andg : D → R respectively. For any two numbersα andβ, prove that the sequence

{αfn + βgn : D → R} converges uniformly to the functionαf + βg : D → R.

7. For each natural numbern, let the functionfn : R → R be bounded. Suppose that the sequence

{fn} converges uniformly tof onR. Prove that the limit functionf : R → R is also bounded.

8. Let{an} be a bounded sequence of numbers. For each natural numbern and each numberx, define

fn(x) = a0 + a1x+a2x

2

2!+ · · ·+ anx

n

n!.

Prove that for eachr > 0, the sequence of functions{fn : [−r, r] → R} is uniformly convergent.

9. Letfn(x) =√

x2 + 1/n andf(x) = |x|. Prove that the sequence{fn} converges uniformly on the

open interval(−1, 1) to the functionf . Check that each functionfn is continuously differentiable,

whereas the limit functionf is not differentiable atx = 0. Does this contradict Theorem 8.2.3?

10. Letfn(x) = nxe−nx2

. Prove that{fn(x)} converges pointwise tof(x) = 0 on the interval[0, 1],

but that the sequence{

∫ 1

0

fn(x)dx}

does not converge to0. Does this contradict Theorem 8.2.2?

11. Let fn(x) = nxe−nx2

. Prove that{fn(x)} converges uniformly tof(x) = 0 on the interval

[δ, 1], 0 < δ < 1, but does not converge uniformly tof(x) = 0 on the interval[0, 1].

12. Prove that if{fn : R → R} is a sequence of continuously differentiable functions such that the

sequence of derivatives{f ′n : R → R} is uniformly convergent and the sequence{fn(0)} is also

convergent, then{fn : R → R} is pointwise convergent. Is the assumption that the sequence

converges necessary? (Considerfn(x) = n ∀x ∈ R.)

13. Give an example of a sequence of differentiable functions fn : (−1, 1) → R that converges uni-

formly but for which{f ′n(0)} is unbounded. (fn(x) = (sinnx)/

√n.)

14. Let{an} and{bn} be bounded sequences. Prove that the series

∞∑

n=1

1

n2(an cosnx+ bn sinnx)

converges uniformly on(−∞,+∞).


15. Determine the domain for conditional convergence and absolute convergence of the following series:

(a)∞∑

n=1

(−1)n+1xn

n

(b)∞∑

n=1

(−1)n

2n− 1

(1− x

1 + x

)n

(c)∞∑

n=1

xn

n!

(d)∞∑

n=1

xn

1− xn

16. Prove that the series∞∑

n=1

(−1)n

n(xn + 1)

converges uniformly on[1,+∞). Hint. Use Abel Test.

17. Prove that the series∞∑

n=1

(−1)n

x+ n

converges uniformly on(0,+∞). Hint. Use Dirichlet Test.

18. Determine the domain of convergence of each of the following power series:

(a)∞∑

k=1

xk

k5k

(b)∞∑

k=1

k!xk

(c)∞∑

k=0

(−1)kx2k−1

(2k + 1)!


(a)1

1− x=

∞∑

k=0

xk, |x| < 1.

(b)1

1 + x=

∞∑

k=0

(−1)kxk, |x| < 1.

(c)1

(1 + x)2=

∞∑

k=1

(−1)k+1kxk−1, |x| < 1.

20. Use (b) in the above problem to obtain a series expansion for the integral∫ 1/2

0

1

1 + x4dx and justify

your calculation.

21. Prove that1

(1 + x2)2=

∞∑

k=0

(−1)k+1kx2k−2, |x| < 1.

22. Prove thatx =∞∑

k=0

(

1− 1

x

)k

, |1− x| < |x|.

23. Letf(x) =1

(1− x)3, |x| < 1. Find a power series expansion for the functionf : (−1, 1) → R.

24. Suppose that the domain of convergence of the power series∞∑

k=0

ckxk contains the interval(−r, r).

Definef(x) =∞∑

k=0

ckxk, |x| < r. Let the interval[a, b] be contained in the interval(−r, r). Prove

that∫ b

a

f(x)dx =

∞∑

k=0

ckk + 1

(bk+1 − ak+1).

Chapter 9: Fourier Series 183

25. Find the Taylor polynomial

p3(x) =

3∑

k=0

f (k)(1)

k!(x − 1)k

and the Lagrange Remainder for the functionf(x) = 1 + 3x+ 5x2 − 2x3.

26. Find the Taylor series expansion about0 for the functionf(x) = (1− x)/(1 + x).

27. Suppose thatf(x) is continuous on(−∞,∞). Let

fn(x) =

n−1∑

k=0

1

nf(x+ k/n).

Prove that the function sequence{fn(x)} converges uniformly on any finite interval.

Hint. Since limn→+∞

fn(x) = limn→+∞

n−1∑

k=0

1

nf(x+ k/n) =

∫ 1

0f(x+ t)dt, andf(x) is continuous, then{fn(x)}

converges forx ∈ (−∞,+∞). Rewritefn(x) and its limit as

fn(x) =

n−1∑

k=0

∫ (k+1)/n

k/nf(x+ k/n)dt,

∫ 1

0f(x+ t)dt =

n−1∑

k=0

∫ (k+1)/n

k/nf(x+ t)dt.

Then we have∣

∣

∣fn(x)−

∫ 1

0f(x+ t)dt

∣

∣

∣=

∣

∣

∣

n−1∑

k=0

∫ (k+1)/n

k/n[f(x+ k/n)− f(x+ t)]dt

∣

∣

∣

≤n−1∑

k=0

∫ (k+1)/n

k/n|f(x+ k/n)− f(x+ t)|dt.

Sincef(x) is uniformly continuous on any finite interval, then for given ǫ > 0,∃δ > 0, s.t.,|f(x1)− f(x2)| < ǫ,

∀x1, x2 ∈ [a − 1, b + 1], |x1 − x2| < δ. For x ∈ [a, b], takeN s.t.,1

N< δ, then for t ∈

[ k

N,k + 1

N

]

,

∣

∣x+k

n− (x+ t)

∣

∣ < δ, and|f(x+ k/n)− f(x+ t)| < ǫ, ∀n ≥ N . Therefore,∣

∣

∣fn(x)−

∫ 1

0f(x+ t)dt

∣

∣

∣< ǫ,

∀x ∈ [a, b],∀n ≥ N , i.e.,{fn(x)} converges uniformly on any finite interval.


Chapter 9

Fourier Series

9.1 Fourier Series and Fourier Coefficients

In physics and engineering, periodic phenomenon are frequently encountered. To describe these phe-

nomenon, Periodic Functions are required. The general periodic functions are defined as

f(t+ T ) = f(t).

The numberT is called theperiod. The well known periodic functions are Trigonometric functionssinx

andcosx which have the periodT = 2π. Now, we try to expand any periodic functionf(x) with periodT

to a series ofsin andcos function. That is

f(x) =a02

+∞∑

k=1

(ak cos kωx+ bk sin kωx),

whereω = 2π/T . For simplicity, we consider periodic functionsf(x) with period2π and then we have

f(x) =a02

+∞∑

k=1

(ak cos kx+ bk sin kx).

Definition 9.1.1A function set{φk(x)} is said to be orthogonal on[a, b] provided

∫ b

a

φm(x)φn(x)

= 0, if m 6= n,

> 0, if m = n.

Theorem 9.1.2The function set{1, cosx, sinx, cos 2x, sin 2x, · · · , cosnx, sinnx, · · · } is orthogonal

on [−π, π].According to the theory of function series, if

f(x) =a02

+

∞∑

k=1

(ak cos kx+ bk sin kx)

185


converges uniformly on[−π, π], then multiplyingcosnx on both sides and integrating from−π to π, we

get

an =1

π

∫ π

−π

f(x) cosnx, n = 0, 1, 2, · · · .

Multiplying sinnx on both sides and integrating from−π to π, we get

bn =1

π

∫ π

−π

f(x) sinnx, n = 1, 2, · · · .

In general, the assumption of ”uniform convergence” is not reasonable. In fact we often want to

use such kind of trigonometric series to approximate a discontinuous functionf(x). By the theory of

function series, because trigonometric functions are continuous, the sum of the Trigonometric series must

be continuous if the series is uniformly convergent. Hence the trigonometric series can’t converge uniformly

to a discontinuous function.

Definition 9.1.3 For a periodic functionf(x) with period2π defined on(−∞,+∞), the trigonometric

series

a02

+

∞∑

n=1

(an cosnx+ bn sinnx)

is called the Fourier Series off(x), provided that the coefficientsan, bn are calculated by the above formula.

We denote this by

f(x) ∼ a02

+

∞∑

n=1

(an cosnx+ bn sinnx).

The coefficientsan, bn are called the Fourier Coefficients off(x).

Note 1: The symbol ”∼” doesn’t mean equal, since we still do not know whether the series converges and

whether it converges tof(x).

Note 2: If the period off(x) is T , then

f(x) ∼ a02

+∞∑

n=1

(an cosnωx+ bn sinnωx),

whereω = 2π/T and

an =2

T

∫ T/2

−T/2

f(x) cosnωx, n = 0, 1, . . .

bn =2

T

∫ T/2

−T/2

f(x) sinnωx, n = 1, . . .


9.2 Convergence of Fourier Series

Theorem 9.2.1(Weierstrass Approximation Theorem) Suppose thatf(x) is continuous on[a, b], then for

givenǫ > 0, there exists a polynomialpn(x),

pn(x) =

n∑

k=0

f(k

n

)( k

n

)

xk(1− x)n−k,

such that

|f(x)− pn(x)| < ǫ, ∀x ∈ [a, b].

Proof. We only need to prove the case[a, b] = [0, 1]. Assume that|f(x)| ≤ M . Sincef(x) is uniformly

continuous, then for givenǫ > 0, there exists aδ > 0, such that

|f(x)− f(x′)| < ǫ/2, ∀x ∈ [0, 1].

Since

f(x) =n∑

k=0

f(x)( k

n

)

xk(1− x)n−k,

then we have

|f(x)− pn(x)| ≤n∑

k=0

∣

∣

∣f(x)− f(k

n

)∣

∣

∣

( k

n

)

xk(1− x)n−k.

Dividek = 0, 1, · · · , n into two sets:

k ∈ A∣

∣

∣

k

n− x∣

∣

∣ < δ,

k ∈ B∣

∣

∣

k

n− x∣

∣

∣ ≥ δ.

If k ∈ A, we have

∑

k∈A

∣

∣

∣f(x)− f(k

n

)∣

∣

∣

( k

n

)

xk(1− x)n−k <ǫ

2.

If k ∈ B, using the following lemma we get

∑

k∈B

∣

∣

∣f(x)− f(k

n

)∣

∣

∣

( k

n

)

xk(1 − x)n−k ≤ 2M

n2δ2

∑

k∈B

(k − nx)2( k

n

)

xk(1− x)n−k

≤ M

2nδ2<ǫ

2.


Lemma 9.2.2

n∑

k=0

(k − nx)2( k

n

)

xk(1 − x)n−k ≤ n

4, ∀x ∈ [0, 1].

Proof. Readily, we have,

n∑

k=0

k( k

n

)

xk(1− x)n−k = nx.

n∑

k=0

k(k − 1)( k

n

)

xk(1− x)n−k = n(n− 1)x2.

Using these facts, we can prove that

n∑

k=0

(k − nx)2( k

n

)

xk(1 − x)n−k = nx(1− x) ≤ n

4,

sincex ∈ [0, 1].

Note 1:Weierstrass Approximation Theorem is also called Bernstein Approximation Theorem.

Using Weierstrass Approximation Theorem, we get

Lemma 9.2.3 (Riemann’s Lemma)Suppose thatf(x) is integrable on[a, b]. Then we have

limλ→+∞

∫ b

a

f(x) sinλxdx = 0, limλ→+∞

∫ b

a

f(x) cosλxdx = 0.

Proof. The proof is divided into the following four steps:

(i) For givenǫ > 0, there exists a continuous functiong(x) such that

∫ b

a

|f(x)− g(x)|dx < ǫ

3.

Indeed, suppose thatPn = {I1, . . . , In} is a partition for[a, b] and denote

Mk = supx∈Ik

f(x), mk = infx∈Ik

f(x).

Takingg(xk) = f(pk) and linear between subinterval ends, the bounding off − g in Ik is simply:

|f(x)− g(x)| ≤Mk −mk.


Hence,∫ b

a

|f(x)− g(x)|dx ≤n∑

k=1

(Mk −mk)|Ik| <ǫ

3,

if the partition is reasonably chosen.

(ii) For givenǫ > 0, there exists a polynomialp(x) such that

|g(x)− p(x)| < ǫ

3(b− a), ∀x ∈ [a, b].

(iii)

limλ→+∞

∫ b

a

p(x) sinλxdx = 0.

Indeed, assume thatp(x), p′(x) are bounded byM on [a, b], then we have,

∣

∣

∣

∫ b

a

p(x) sinλxdx∣

∣

∣ =1

λ

∣

∣

∣

∫ b

a

p(x)d (cosλx)∣

∣

∣

=1

λ

∣

∣

∣p(b) cosλb− p(a) cosλa−∫ b

a

p′(x) cosλxdx∣

∣

∣

≤ 2M + (b− a)M

λ.

Takingλ→ +∞, we get the desired result.

(iv)

∣

∣

∣

∫ b

a

f(x) sin λxdx∣

∣

∣≤

∣

∣

∣

∫ b

a

[f(x)− g(x)] sin λxdx∣

∣

∣+

∣

∣

∣

∫ b

a

[g(x)− p(x)] sin λxdx∣

∣

∣ +∣

∣

∣

∫ b

a

p(x) sin λxdx∣

∣

∣

≤∫ b

a

|f(x)− g(x)|dx +

∫ b

a

|g(x)− p(x)|dx +ǫ

3< ǫ.

Suppose that the period functionf(x) is integrable on[−π, π], we consider the convergence of the

Fourier series off(x). For any given pointx, we have

Sn(x) =a02

+

n∑

k=1

(ak cos kx+ bk sin kx)

=1

π

∫ π

−π

f(t)[1

2+

n∑

k=1

(cos kt cos kx+ sin kt sin kx)]

dt

=1

π

∫ π

−π

f(t)[1

2+

n∑

k=1

(cos k(t− x)]

dt

=1

π

∫ π

−π

f(t)sin[(2n+ 1)(t− x)/2]

2 sin[(t− x)/2]dt.


Letu = t− x, we get

Sn(x) =1

π

∫ π−x

−π−x

f(u+ x)sin[(2n+ 1)u/2]

2 sin(u/2)du

=1

π

∫ π

−π

f(u+ x)sin[(2n+ 1)u/2]

2 sin(u/2)du

=1

π

∫ π

0

[f(u+ x) + f(x− u)]sin[(2n+ 1)u/2]

2 sin(u/2)du.

If we takef(x) ≡ 1, then we get

1

π

∫ π

0

sin[(2n+ 1)u/2]

sin(u/2)du = 1

LetS = [f(x+ 0) + f(x− 0)]/2, then we have

Sn(x)− S =1

π

∫ π

0

φ(u)sin[(2n+ 1)u/2]

2 sin(u/2)du,

where

φ(u) = f(x+ u) + f(x− u)− f(x+ 0)− f(x− 0).

Thus, we have the following lemma:

Lemma 9.2.4Suppose thatf(x) is periodic with period2π and integrable on[−π, π]. Also, suppose that

at the point x the left and right limitsf(x− 0) andf(x+0) exist. ThenSn(x) converges toS if and only if

limn→∞

∫ π

0

φ(u)sin[(2n+ 1)u/2]

sin(u/2)du = 0.

Since the function

φ(u)[ 1

sin(u/2)− 1

(u/2)

]

is integrable on[0, π], then from Riemann’s Lemma we get

limn→∞

∫ π

0

φ(u)[ 1

sin(u/2)− 1

(u/2)

]

sin[(2n+ 1)u/2]du = 0.

Therefore, Lemma 9.2.4 can be rewritten as

Lemma 9.2.5Suppose thatf(x) is periodic with period2π and integrable on[−π, π]. Also, suppose that

at the point x the left and right limitsf(x− 0) andf(x+0) exist. ThenSn(x) converges toS if and only if

limn→∞

∫ π

0

φ(u)sin[(2n+ 1)u/2]

udu = 0.

For any givenδ > 0, the functionφ(u)/u is integrable on[δ, π]. Then from Riemann Lemma

limn→∞

∫ π

δ

φ(u)sin[(2n+ 1)u/2]

udu = 0.


Using this we have

Theorem 9.2.6 (Riemann’s Localization Theorem)Suppose thatf(x) is periodic with period2π and

integrable on[−π, π]. Also, suppose that at the point x the left and right limitsf(x− 0) andf(x+0) exist.

ThenSn(x) converges toS if there exists aδ > 0 such that

limn→∞

∫ δ

0

φ(u)sin[(2n+ 1)u/2]

udu = 0,

i.e., the convergence of the Fourier series off(x) at the pointx depends only on the function values off(x)

in the neighborhood ofx.

Theorem 9.2.7 (Lipschitz Test)Suppose thatf(x) is periodic with period2π and integrable on[−π, π].Also, suppose that at the point x the left and right limitsf(x − 0) andf(x + 0) exist, and there exist a

constantL andδ0 such that

|f(x± u)− f(x± 0)| ≤ Lu, 0 < u < δ.

ThenSn(x) converges toS.

Proof.φ(u)

u=f(x+ u)− f(x+ 0)

u+f(x− u)− f(x− 0)

u.

Consider

g(u) =f(x+ u)− f(x+ 0)

u.

By assumption,g(u) is bounded on[0, δ0], i.e., |g(u)| ≤ L. Sincef(x) is integrable, then for anyη > 0,

g(u) is integrable on[η, δ0], andg(u) is integrable on[0, δ0]. Similarly,f(x− u)− f(x− 0)

uis integrable

on [0, δ0]. Therefore,φ(u)

uis integrable on[0, δ0]. From Riemann’s Lemma,

limn→∞

∫ δ

0

φ(u)sin[(2n+ 1)u/2]

udu = 0.

From Theorem 9.2.6,Sn(x) converges toS.

Corollary 9.2.8 Suppose thatf(x) is periodic with period2π and integrable on[−π, π]. Also, suppose that

at the point x,f(x) is differentiable, or the left and right derivatives off(x) exist. ThenSn(x) converges

to f(x).

Proof. Suppose

f ′+(x) = lim

u→0+

f(x+ u)− f(x+ 0)

u= A,

then,∃δ1 > 0, s.t.,∣

∣

∣

f(x+ u)− f(x+ 0)

u−A

∣

∣

∣ ≤ 1,


or,

|f(x+ u)− f(x+ 0)| ≤ (|A|+ 1)u = L1u, 0 < u < δ1.

Similarly,

|f(x− u)− f(x− 0)| ≤ L2u, 0 < u < δ2.

LetL = max{L1, L2}, then

|f(x± u)− f(x± 0)| ≤ L|u|, 0 < u < δ0 = min{δ1, δ2}.

From Theorem 9.2.7,Sn(x) converges tof(x).

Definition 9.2.9A functionf(x) is called piecewise continuous on[a, b], if there exist finite number of points

a = x0 < x1 < · · · < xn = b such thatf(x) is continuous on every subinterval(xi−1, xi), i = 1, 2, · · · , n,

and the limitsf(a+0), f(xi − 0), f(xi +0), i = 1, 2 · · · , n− 1, andf(b− 0) exist. Iff(x) andf ′(x) are

piecewise continuous on[a, b], we say thatf(x) is piecewise smooth on[a, b].

Lemma 9.2.10Suppose thatf(x) is piecewise smooth on[a, b], then

limx→c+

f(x)− f(c+ 0)

x− c= f ′(c+ 0), if c ∈ [a, b)

and

limx→c−

f(x)− f(c− 0)

x− c= f ′(c− 0), if c ∈ (a, b].

Proof. If c 6= xi, i = 0, 1, · · · , n, the lemma is obviously true. Ifc = xi, i = 0, 1, · · · , n, redefine

f(xi) = f(xi + 0), thenf(x) is continuous on[xi, x], for x ∈ (xi, xi+1). Using Lagrange Mean Value

Theorem, there existsξ ∈ (xi, x) such that,

f(x)− f(xi + 0)

x− xi= f ′(ξ),

and

limx→xi+

f(x)− f(xi + 0)

x− xi= f ′(xi + 0).

Theorem 9.2.11Suppose thatf(x) is periodic with period2π and piecewise smooth on[−π, π]. Then

Sn(x) converges toS.

Proof. Using Lemma 9.2.10,φ(u)

uis piecewise continuous on[0, δ] and integrable. By Riemann’s Lemma

and Theorem 9.2.6,Sn(x) converges toS.

Example 9.2.1Letf(x) be periodic with period2π and

f(x) = x, −π < x < π.


By the formula of calculating the Fourier series, we get

f(x) ∼ 2 ·∞∑

n=1

(−1)n+1 sinnx

n.

By Theorem 9.2.11, the Fourier series convergesf(x) everywhere except at the points(2k + 1)π, k =

0,±1,±2, . . . . The Fourier series converges to0 at these points.

Example 9.2.2Letf(x) be periodic with period2π and

f(x) = x2 − π < x < π.

By the formula of calculating the Fourier series we get

f(x) ∼ π2

3+ 4

∞∑

n=1

(−1)n cosnx

n2.

By Theorem 9.2.11, the Fourier series converges tof(x) everywhere.

Note 2:sinnx and cosnx are odd function and even function respectively. Iff(x) is odd function then

f(x) cosnx is still odd function, and

an =1

π

∫ π

−π

f(x) cosnx = 0.

Hence only the terms ofsinnx appear in the Fourier series. Similarly iff(x) is even function then

f(x) sinnx is odd function, and allbn = 0. Hence only the terms ofcosnx appear in the Fourier se-

ries. In above 2 examples, one is odd and one is even.

Example 9.2.3The interval[−π, π] is divided into−π = x0 < x1 < · · · < xn = π, andf(x) is defined

as follows

f(x) = ci xi−1 < x < xi, i = 1, 2, . . . , n

and

f(x+ 2π) = f(x).

This kind of functions often appear in physics. By Theorem 9.2.11, the Fourier series off(x) converges to

f(x) except at the pointsxi. Atxi, i = 1, . . . , n− 1, the Fourier series converges to

f(xi + 0) + f(xi − 0)

2=ci+1 + ci

2.

At the pointsx0 andxn, the Fourier series converges to

f(xn + 0) + f(x0 − 0)

2=cn + c1

2.


9.3 Integration and Differentiation of Fourier Series

For a general function series, we usually require the uniform convergence of the series for termwise

integration and differentiation of the series. However, for Fourier series, the requirement of uniform con-

vergence is not necessary for termwise integration and differentiation of the series. This is actually the

advantage of the Fourier series. First, we introduce some lemmas.

Lemma 9.3.1Suppose thatf(x) is continuous and piecewise smooth on[a, b], then the First Fundamental

Theorem holds, i.e.,∫ b

a

f ′(x)dx = f(b)− f(a).

Proof.∫ b

a

f ′(x)dx =n∑

k=1

∫ xk

xk−1

f ′(x)dx

=

n∑

k=1

f(xk)− f(xk−1) = f(b)− f(a).

Lemma 9.3.2Suppose thatf(x) is continuous and piecewise smooth on[a, b], g(x) is continuous on[a, b],

andG(x) is an antiderivative ofg(x). Then the Integration by Parts formula holds, i.e.,

∫ b

a

g(x)f(x)dx = G(x)f(x)|ba −∫ b

a

G(x)f ′(x)dx.

Proof. On [xk−1, xk], f(x) andG(x) are continuous,f ′(x) andg(x) are integrable. Then we have,

∫ b

a

g(x)f(x)dx =n∑

k=1

∫ xk

xk−1

f(x)g(x)dx

=

n∑

k=1

[

G(x)f(x)∣

∣

∣

xk

xk−1

−∫ xk

xk−1

g(x)f ′(x)dx]

= G(x)f(x)|ba −∫ b

a

G(x)f ′(x)dx.

Theorem 9.3.3Suppose thatf(x) is periodic with period2π and piecewise continuous on[−π, π]. If

f(x) ∼ a02

+

∞∑

n=1

(an cosnx+ bn sinnx),

then for anyx ∈ (−∞,+∞), we have

∫ x

0

f(t)dt =a0x

2+

∞∑

n=1

∫ x

0

(an cosnt+ bn sinnt)dt.


Proof. Let

F (x) =

∫ x

0

[

f(t)− a02

]

dt, x ∈ (−∞,+∞).

By Theorem 9.2.11, the Fourier series ofF (x) converges toF (x), i.e.,

F (x) =A0

2+

∞∑

n=1

(An cosnx+Bn sinnx), (9.3.1)

with

An =1

π

∫ π

−π

F (x) cosnxdx = −bnn, n = 1, 2, · · · ,

Bn =1

π

∫ π

−π

F (x) sinnxdx =ann, n = 1, 2, · · · .

Letx = 0 in (9.3.1), we get

A0

2=

∞∑

n=1

bnn.

Theorem 9.3.4Suppose thatf(x) is periodic with period2π and continuous, andf ′(x) is piecewise smooth

on [−π, π]. If

f(x) ∼ a02

+

∞∑

n=1

(an cosnx+ bn sinnx),

then for anyx ∈ (−∞,+∞), we have

1

2[f ′(x+ 0) + f ′(x− 0)] =

∞∑

n=1

(nbn cosnx− nan sinnx), (9.3.2)

i.e., the Fourier series off ′(x) can be obtained from the Fourier series by termwise differentiation.

Proof. Let

f ′(x) ∼ a′02

+

∞∑

n=1

(a′n cosnx+ b′n sinnx).

Then we have

a′0 = 0, a′n = nbn, b′n − nan, n = 1, 2, · · · ,

and

f ′(x) ∼∞∑

n=1

(nbn cosnx− nan sinnx).

Sincef ′(x) is piecewise smooth, we have (9.3.2).

Next, we consider the uniform convergence of the Fourier series. The following lemma will be used in the

proof of the theorem.


Lemma 9.3.5 (Bessel’s Inequality)Suppose thatf(x) is integrable on[−π, π], ak andbk are the Fourier

coefficients off(x). Then for any natural numbern, we have

a202

+

n∑

k=1

(a2k + b2k) ≤1

π

∫ π

−π

f2(x)dx.

Proof. Let

Sn(x) =a02

+

n∑

k=1

(ak cos kx+ bk sinkx).

Then we have,

0 ≤ 1

π

∫ π

−π

[f(x)− Sn(x)]2dx

=1

π

∫ π

−π

f2(x)dx − a202

−n∑

k=1

(a2k + b2k).

Theorem 9.3.6Suppose thatf(x) is periodic with period2π, continuous, and piecewise smooth on[−π, π].Then the Fourier series off(x) converges uniformly tof(x) on (−∞,+∞).

Proof. By assumption,

f(x) =a02

+∞∑

k=1

(an cosnx+ bn sinnx). (9.3.3)

Leta′n andb′n be the Fourier coefficients off ′(x), then we have,

f(x) =a02

+

∞∑

k=1

(−b′n

ncosnx+

a′nn

sinnx).

From Bessel’s Inequality, the series∞∑

k=1

(a2k + b2k) converges. Since

∣

∣

∣− b′nn

cosnx+a′nn

sinnx∣

∣

∣ ≤ 1

n2+ a2n + b2n,

(9.3.3) converges uniformly.

Theorem 9.3.7(Parseval’s Equality) Suppose that the Fourier series off(x) converges uniformly tof(x)

on [−π, π], then1

π

∫ π

−π

f2(x)dx =a202

+

∞∑

n=1

(a2n + b2n).

Proof. Since[Sn(x) − f(x)]2 converges to0 uniformly, then we have

limn→∞

1

π

∫ π

−π

[Sn(x)− f(x)]2dx = 0.


9.4 Fourier Expansion on Finite Interval

Suppose that we have a functionf(x) defined on the interval[a, b], we want to obtain the Fourier series

of f(x). For a functionf(x) defined in[a, b], it’s easy by a linear transform

x = ky + l

to obtain a functionϕ(y) = f(ky + l) such thatϕ(y) is defined on some standard interval such as

[0, 1], [−1, 1], [0, π], or [−π, π]. Hence, for simplicity, in the following discussion the definition domain

of function will be only[−π, π] and[0, π].

Fourier series of f(x) on [−π, π]: Let f(x) be defined on[−π, π]. We extendf(x) periodically to a

function defined on(−∞,+∞) through

f(x) = f(x+ 2π).

The original functionf(x) may not satisfy the condition

f(−π) = f(π).

If the function values at these two end points are not important, then when making the periodical extension

we may revise both the function values at−π andπ to be[f(−π+ 0)+ f(π− 0)]/2. This revision doesn’t

affect the calculation of Fourier coefficients. We calculate the coefficients of Fourier series by formula

an =1

π

∫ π

−π

f(x) cosnx, n = 0, 1, 2, . . .

bn =1

π

∫ π

−π

f(x) sinnx, n = 1, 2, . . .

and get

f(x) ∼ a02

+∞∑

n=1

(an cosnx+ bn sinnx).

However, we still need to check the convergence of the series. For example, iff(x) is piecewise smooth on

[−π, π]. Then we have

f(x+ 0) + f(x− 0)

2=

a02

+

∞∑

n=1

(an cosnx+ bn sinnx), ∀x ∈ (−π, π),

f(−π + 0) + f(π − 0)

2=

a02

+

∞∑

n=1

(an cosnx+ bn sinnx), x = −π, x = π.

Fourier series off(x) on [0, π]: Let f(x) be defined on[0, π]. In this case, it is possible to make an odd

or even extension off(x) to the interval[−π, 0] and then make a periodical extension to(−∞,+∞).


1. Even Extension:In this case, the extended function is even. As mentioned before, the Fourier series

only includes the terms ofcosnx, i.e.,

f(x) ∼ a02

+

∞∑

n=1

an cosnx.

The resulting series is called Fourier cosine series. Sincef(x) cosnx is even,

an =1

π

∫ π

−π

f(x) cosnxdx

=2

π

∫ π

0

f(x) cosnxdx

For even extension,f(x) is continuous at the point0,−π, andπ.

2. Odd Extension: In this case, the extended function is Odd. As mentioned before, the Fourier series

only includes the terms ofsinnx, i.e.,

f(x) ∼∞∑

n=1

bn sinnx.

The resulting series is called Fourier sine series. Sincef(x) sinnx is even,

bn =1

π

∫ π

−π

f(x) sinnxdx

=2

π

∫ π

0

f(x) sinnxdx.

For odd extension,f(x) may not be continuous at the point0, −π andπ.

9.5 Exercise

1. Letf(x) = f(x+ 2π) and

f(x) =

0, −π < x < 0,

1, 0 < x < π.

Find the Fourier series off(x) and discuss its convergence.

2. Letg(x) = g(x+ 2π) and

g(x) =

0, −π < x < −π2 ,

1, −π2 < x < π

2 ,

0, π2 < x < π.

Find the Fourier series ofg(x) and discuss its convergence.

3. Suppose that the Fourier series off(x) converges tof(x) uniformly on[−π, π]. Prove the following

Parseval’s equality:1

π

∫ π

−π

f2(x)dx =a202

+

∞∑

n=1

(a2n + b2n).

Chapter 10: Improper Integral 199

4. Use Riemann Lemma to find the following limits:

(a) limλ→+∞

∫ π

−π

sin2 λxdx. (b) limλ→+∞

∫ a

0

cos2 λx

1 + xdx, a > 0.

5. Find the Fourier series and its sum for the following functions:

(a) f(x) =

0, −π ≤ x < 0,

x, 0 ≤ x < π.

(b) f(x) =

x, −π ≤ x < 0,

x2, 0 ≤ x < π.

(c) f(x) = sinαx, −π ≤ x < π.

(d) f(x) = ex, −π ≤ x < π.

(e) f(x) =

0, −π ≤ x < 0,

cosx, 0 ≤ x < π.

(f) f(x) =

1, −π ≤ x < 0,

ex, 0 ≤ x < π.

(g) f(x) = max{2− |x|, 0}, −π ≤ x < π.

(h) f(x) = sgn(cosx), −π ≤ x < π.

6. Find the following sum by using a suitable Fourier series:

(a)∞∑

n=1

1

n2.

(b)∞∑

n=1

(−1)n+1

n2.

(c)∞∑

n=1

1

(2n− 1)2.

(d)∞∑

n=1

1

n4.

(e)∞∑

n=1

(−1)n+1

n4.

(f)∞∑

n=1

1

(2n− 1)4.

(g)∞∑

n=1

1

1 + n2.

(h)∞∑

n=1

(−1)n+1

1 + n2.

7. Suppose thatf(x) is an odd function on(0, π). Also suppose thatf(x) is non-negative and inte-

grable on(0, π). Prove that the Fourier coefficients off(x) satisfies|bn| ≤ nb1, n = 1, 2, · · · .Hint. By induction, using Mean Value Theorem andsinnx = sin(n− 1 + 1)x.

8. Suppose thatf(x) is integrable on[−π, π] and has the periodπ. Prove that the Fourier coefficients

of f(x) satisfiesa2n−1 = 0, b2n−1 = 0, n = 1, 2, · · · .

9. Suppose thatf(x) is integrable on[−π, π] andf(x+π) = −f(x). Prove that the Fourier coefficients

of f(x) satisfiesa2n = 0, b2n = 0, n = 1, 2, · · · .

10. Find thesin andcos series off(x) = π − x on [0, π].

11. Suppose thatf(x) is continuous on[−π, π] and all the Fourier coefficients off(x) are zero. Prove

thatf(x) ≡ 0.

12. Suppose thatf ′(x) is monotonically increasing and bounded on(0, 2π). Prove that the Fourier

coefficients off(x) satisfiesan ≥ 0, n = 1, 2, · · · .


Chapter 10

Improper Integral

10.1 Integration on an Unbounded Interval

In this section, we consider the first kind of improper integral and mainly the case with interval[a,+∞).

Definition 10.1.1Suppose thatf(x) is bounded on[a,+∞), and integrable on any interval[a, u], where

u > a. If

limu→+∞

∫ u

a

f(x)dx = A,

then we say that the improper integral off(x) on [a,+∞) converges and denote by

∫ +∞

a

f(x)dx = A.

Otherwise, we say that the improper integral off(x) on [a,+∞) diverges.

Example 10.1.1Let ǫ > 0, then

∫ +∞

0

e−ǫxdx = limu→+∞

∫ u

0

e−ǫxdx = limu→+∞

[−e−ǫx

ǫ

]u

0

=1

ǫ− lim

u→+∞e−ǫu =

1

ǫ.

Example 10.1.2Letp > 1, then

∫ +∞

1

x−pdx = limu→+∞

∫ u

1

x−pdx = limu→+∞

[ x1−p

1− p

]u

1

=1

p− 1− lim

u→+∞u1−p

p− 1=

1

p− 1.

201


10.1.1 Nonnegative Integrand

We first consider the case that the integrandf(x) is nonnegative.

Theorem 10.1.1Suppose thatf(x) ≥ 0 on [a,+∞) and integrable on any interval[a, u], whereu > a.

Let

F (u) =

∫ u

a

f(x)dx.

Then∫ +∞

a

f(x)dx converges if and only ifF (u) is bounded on[a,+∞).

Proof. Sincef(x) ≥ 0,

F (u2)− F (u1) =

∫ u2

u1

f(x)dx ≥ 0, ∀u2 > u1.

Hence,F (u) is monotonically increasing on[a,+∞). By the Monotone Convergence Theorem,∫ +∞

a

f(x)dx = limu→+∞

F (u)

exists if and only ifF (u) is bounded on[a,+∞).

Theorem 10.1.2 (Comparison Test)Suppose that there is a numberb, b > a, such that

0 ≤ f(x) ≤ Cg(x) ∀x ≥ b,

whereC is a positive constant. We have

1. If∫ ∞

a

g(x)dx converges, then∫ ∞

a

f(x)dx converges.

2. If∫ ∞

a

f(x)dx diverges, then∫ ∞

a

g(x)dx diverges.

Proof.

1.∫ ∞

a

g(x)dx converges implies that∫ ∞

a

Cg(x)dx converges. From the assumption we get∫ u

b

f(x)dx ≤∫ ∞

b

Cg(x)dx. By Theorem 10.1.1,∫ ∞

b

f(x)dx converges and then∫ ∞

a

f(x)dx

converges.

2.∫ ∞

a

f(x)dx diverges implies that∫ u

a

f(x)dx is unbounded ona < u < +∞. By assumption,∫ u

a

g(x)dx is also unbounded ona < u < +∞. By Theorem 10.1.1,∫ ∞

a

g(x)dx diverges.

Theorem 10.1.3Suppose thatf(x) ≥ 0 andg(x) > 0 on [a,+∞) and

limx→+∞

f(x)

g(x)= l > 0.

Then∫ ∞

a

f(x)dx and∫ ∞

a

g(x)dx converge or diverge at the same time.


Proof. Let ǫ = l/2. Then from the assumption, there exists aM > 0 such that

l− ǫ <f(x)

g(x)< l + ǫ ∀x ≥M,

or,

l

2g(x) < f(x) <

3l

2g(x) ∀x ≥M.

By Theorem 10.1.2,∫ ∞

a

f(x)dx and∫ ∞

a


Theorem 10.1.4Suppose thatf(x) ≥ 0 on [a,+∞), a > 0.

1. If for α > 1,

limx→+∞

xαf(x) = l ≥ 0,

then∫ ∞

a

f(x)dx converges.

2. If for α ≤ 1,

limx→+∞

xαf(x) = l > 0, or, limx→+∞

xαf(x) = +∞,

then∫ ∞

a

f(x)dx diverges.

Proof.

1. From the assumption,

limx→+∞

f(x)

x−α= l,

and∫ ∞

a

x−αdx converges forα > 1. Then∫ ∞

a

f(x)dx converges from Theorem 10.1.3.

2. From the assumption,

limx→+∞

f(x)

x−α= l > 0,

and∫ ∞

a

x−αdx diverges forα ≤ 1. Then∫ ∞

a

f(x)dx diverges from Theorem 10.1.3. If

limx→+∞

f(x)

x−α= +∞, α ≤ 1,

then there exists a numberM such that

f(x) > x−α ∀x ≥M.

By Comparison Test,∫ ∞

a

f(x)dx diverges due to the divergence of∫ ∞

a

x−αdx whenα ≤ 1.


10.1.2 Absolute Convergence Test

Consider a general improper integral∫ ∞

a

f(x)dx and letF (u) =

∫ u

a

f(x)dx. By Cauchy Conver-

gence Criterion, limu→+∞

F (u) exists if and only if for givenǫ > 0, there exists aM > 0 such that

|F (u2)− F (u1)| < ǫ ∀u2 > u1 ≥M.

In other words, the following theorem holds.

Theorem 10.1.5∫ ∞

a

f(x)dx converges if and only if for givenǫ > 0, there exists aM > 0 such that

∣

∣

∣

∫ u2

u1

f(x)dx∣

∣

∣ < ǫ ∀u2 > u1 ≥M.

Definition 10.1.6 If∫ ∞

a

|f(x)|dx converges, then we say that∫ ∞

a

f(x)dx converges absolutely. If∫ ∞

a

|f(x)|dx diverges, but∫ ∞

a

f(x)dx converges, then we say that∫ ∞

a

f(x)dx converges conditionally.

Theorem 10.1.7If∫ ∞

a

f(x)dx converges absolutely, then∫ ∞

a

f(x)dx converges.

Theorem 10.1.8 (Absolute Convergence Test)Suppose thatg(x) ≥ 0 on [a,+∞) and∫ ∞

a

g(x)dx con-

verges. If

|f(x)| ≤ Cg(x), x ≥M,

whereC > 0 is a constant andM ≥ a, then∫ ∞

a

f(x)dx converges absolutely.

Theorem 10.1.9 (Abel Test)Suppose that∫ ∞

a

f(x)dx converges,g(x) is monotone and bounded on

[a,+∞). Then∫ ∞

a

f(x)g(x)dx converges.

Theorem 10.1.10 (Dirichlet Test)Suppose that∫ A

a

f(x)dx is bounded for anyA ∈ [a,+∞), g(x) is

monotone on[a,+∞), and limx→+∞

g(x) = 0. Then∫ ∞

a

f(x)g(x)dx converges.

10.1.3 Improper Integral on (−∞, b] and (−∞,+∞)

Definition 10.1.11Suppose thatf(x) is bounded on(−∞, b] and integrable on any interval[u, b], u < b.

If

limu→−infty

∫ b

u

f(x)dx = A.

Then we say that the improper integral off(x) on (−∞, b] converges and denote it by∫ b

−∞f(x)dx = A.


A simple transformy = −x will change the integration interval from(−∞, b] to [−b,+∞), i.e.,

∫ b

−∞f(x)dx =

∫ +∞

−b

f(−y)dy.

Hence, all the theory above can be used for∫ b

−∞f(x)dx.

Definition 10.1.12Suppose thatf(x) is bounded on(−∞,+∞) and integrable on any interval[u, v]. If

limu→−∞,v→+∞

∫ v

u

f(x)dx = A,

or equivalently,

limu→−∞

∫ 0

u

f(x)dx+ limv→+∞

∫ v

0

f(x)dx = A.

Then we say that the improper integral off(x) on (−∞,+∞) converges and denote it by

∫ +∞

−∞f(x)dx = A.

By the definition,∫ +∞

−∞f(x)dx converges if and only if both

∫ 0

−∞f(x)dx and

∫ +∞

0

f(x)dx converge and

∫ +∞

−∞f(x)dx =

∫ 0

−∞f(x)dx +

∫ +∞

0

f(x)dx.

10.2 Integral with an Unbounded Integrand

In this section, we discuss another kind of improper integral: the interval is finite, but the function is

unbounded.

Definition 10.2.1Suppose thatf(x) is unbounded on[a, b] and limx→b

f(x) = +∞ (or −∞), but f(x) is

bounded and integrable on any interval[a, u], a < u < b, and

limu→b

∫ u

a

f(x)dx = A.

Then we say that the improper integral off(x) on [a, b] converges and denote it by

∫ b

a

f(x)dx = A.

If the limit does not exist, we say that the improper integralof f(x) on [a, b] diverges.


10.2.1 Nonnegative Integrand

Parallel to Theorem 10.1.1 to Theorem 10.1.4, we have the following theorems.

Theorem 10.2.1Suppose thatf(x) ≥ 0 on [a, b] and integrable on any interval[a, u], wherea < u < b.

Let

F (u) =

∫ u

a

f(x)dx.

Then∫ b

a

f(x)dx converges if and only ifF (u) is bounded on[a, b).

Theorem 10.2.2 (Comparison Test)Suppose that

0 ≤ f(x) ≤ Cg(x) ∀a ≤ x < b,

whereC is a positive constant. We have

1. If∫ b

a

g(x)dx converges, then∫ b

a

f(x)dx converges.

2. If∫ b

a

f(x)dx diverges, then∫ b

a

g(x)dx diverges.

Theorem 10.2.3Suppose thatf(x) ≥ 0 andg(x) > 0 on [a, b), and

limx→b

f(x) = +∞, limx→b

g(x) = +∞.

If

limx→b

f(x)

g(x)= l > 0,

then∫ b

a

f(x)dx and∫ b

a


Theorem 10.2.4Suppose thatf(x) ≥ 0 on [a, b).

1. If for α < 1,

limx→b

(b − x)αf(x) = l ≥ 0,

then∫ b

a

f(x)dx converges.

2. If for α ≥ 1,

limx→b

(b− x)αf(x) = l > 0, or, limx→b

(b− x)αf(x) = +∞,

then∫ b

a

f(x)dx diverges.


10.2.2 Absolute Convergence Test

For general improper integrals on[a, b], we have the following theorems which are parallel to Theorem

10.1.5, Theorem 10.1.7, and Theorem 10.1.8.

Theorem 10.2.5Suppose thatlimx→b

f(x) = +∞ (or − ∞). Then∫ b

a

f(x)dx converges if and only if for

givenǫ > 0, there exists aδ > 0 such that

∣

∣

∣

∫ u2

u1

f(x)dx∣

∣

∣< ǫ ∀b− δ < u1 < u2 < b.

Definition 10.2.6 If∫ b

a

|f(x)|dx converges, then we say that∫ b

a

f(x)dx converges absolutely. If∫ b

a

|f(x)|dx diverges, but∫ b

a

f(x)dx converges, then we say that∫ b

a

f(x)dx converges conditionally.

Theorem 10.2.7If∫ b

a

f(x)dx converges absolutely, then∫ b

a

f(x)dx converges.

Theorem 10.2.8 (Absolute Convergence Test)Suppose thatg(x) ≥ 0 on [a, b) and∫ b

a

g(x)dx converges.

If

|f(x)| ≤ Cg(x), a′ ≤ x < b,

whereC > 0 is a constant anda ≤ a′ < b, then∫ b

a

f(x)dx converges absolutely.

10.2.3 Other Cases

Definition 10.2.9Suppose thatf(x) is unbounded on[a, b] and limx→a

f(x) = +∞ (or −∞), but f(x) is

bounded and integrable on any interval[u, b], a < u < b. If

limu→a

∫ b

u

f(x)dx = A,

then we say that the improper integral off(x) on [a, b] converges and denote it by

∫ b

a

f(x)dx = A.

If the limit does not exist, we say that the improper integralof f(x) on [a, b] diverges.

Definition 10.2.10Suppose thatf(x) is unbounded on[a, b] and

limx→a

f(x) = ±∞, limx→b

f(x) = ±∞,

butf(x) is bounded and integrable on any interval[u, v], a < u < v < b. If

limu→a,v→b

∫ v

u

f(x)dx = A,


or equivalently,

limu→a

∫ c

u

f(x)dx+ limv→b

∫ v

c

f(x)dx = A, a < c < b,

then we say that the improper integral off(x) on [a, b] converges and denote it by

∫ b

a

f(x)dx = A.

By the definition,∫ b

a

f(x)dx converges if and only if both∫ c

a

f(x)dx and∫ b

c

f(x)dx converge (a < c <

b) and∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

10.3 Gamma Function and Beta Function

In this section, we recommend two important functions that are defined by improper integrals. They are

widely used in physics, statistics, as well as other branches of mathematics.

10.3.1 Gamma Function

The following improper integral that includes a parameters

Γ(s) =

∫ ∞

0

xs−1e−xdx, s > 0

is calledGamma Function. Whens < 1, limx→0

xs−1e−x = +∞. Hence it is of two kinds of improper

integrals. However, both integrals∫ 1

0

xs−1e−xdx and∫ ∞

1

xs−1e−xdx converge fors > 0. Therefore,

Gamma Function is well defined on the interval(0,+∞).

Theorem 10.3.1For s > 0,

Γ(s+ 1) = sΓ(s).

Theorem 10.3.2For natural numbern,

Γ(n+ 1) = n!.

10.3.2 Beta Function

Beta functionβ(p, q) involves two parameters, and is defined by the improper integral

β(p, q) =

∫ 1

0

xp−1(1 − x)q−1dx, p > 0, q > 0.


When0 < p < 1 and0 < q < 1, the integrand is unbounded on(0, 1) and

limx→1

xp−1(1− x)q−1 = +∞, limx→0

xp−1(1 − x)q−1 = +∞.

We can write the improper integral in form of

β(p, q) =

∫ 1/2

0

xp−1(1− x)q−1dx +

∫ 1

1/2

xp−1(1− x)q−1dx.

Since

xp−1(1− x)q−1 ≤(1

2

)q−1

xp−1, 0 < x ≤ 1

2,

xp−1(1− x)q−1 ≤(1

2

)p−1

xq−1,1

2≤ x < 1.

By Comparison Test, when0 < p < 1 and0 < q < 1, both integrals converge. Hence the Beta function is

defined forp > 0, q > 0.

Theorem 10.3.3β(p, q) = β(q, p).

Theorem 10.3.4β(p+ 1, q + 1) =q

p+ q + 1β(q + 1, p) =

p

p+ q + 1β(p, q + 1).

Theorem 10.3.5β(p, q) =Γ(p)Γ(q)

Γ(p+ q), p > 0, q > 0.

10.4 Exercise

1. Evaluate the following integrals:

(a)∫ +∞

0

dx

1 + x3. (b)

∫ +∞

0

e−ax sinxdx, a > 0.

2. Evaluate the following integrals:

(a)∫ 1

0

dx

(2 − x)√1− x

. (b)∫ 1

0

sin−1 x

xdx.

3. Discuss the convergence of the following integrals:

(a)∫ +∞

1

dx√1 + x2 ln2(1 + x)

. (b)∫ +∞

0

xp

1 + xqdx, p > 0, q > 0.

4. Discuss the convergence of the following integrals:

(a)∫ 1

0

lnx

1− xdx. (b)

∫ 1

0

xp

lnq(1 + x)dx, p > 0, q > 0.


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Mathematical Analysis - Hong Kong Baptist Universityzeng/Teaching/2017math1006/wu.pdf · Mathematical Analysis Hongci Huang, Xiaonan Wu and Tieyong Zeng Department of Mathematics