69
Math 201 (Introduction to Analysis) Fall 2004-2005 Instructor: Dr. Kin Y. Li Office: Room 3471 Office Phone: 2358 7420 e-mail address: [email protected] Office Hours: Tu. 3:30pm - 4:30pm (or by appointments) Prerequisite: A-level Math or One Variable Calculus Website for Lecture Notes and Practice Exercises: http://www.math.ust.hk/ makyli/UG.html Grade Scheme: Homeworks (4%), Tutorial Presentations (6%), Midterm (40%), Final Examination (50%) All records of grades will be put on the website http://grading.math.ust.hk/checkgrade/ as soon as they are available. This course is essentially graded by curve with one exception, namely students who achieve 40% or less of the overall grade will fail the course. Students who take the same midterm will be grouped under the same curve. Students should make copies of homeworks before submitting the originals. In case homeworks are not received, students will be required to resubmit copies within a short period of time (may be less than a day). For tutorial presentations, students will form groups (of 1 to 3 students) and present solutions to assigned problems in the tutorial sessions. All members of a group must attend the sessions to assist in answering possible questions from presentations. Marks will be deducted for failure to present solutions or for absence in supporting his/her group. Course Description: This is the first of two required courses on analysis for Math majors. It is to be followed by Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, as appeared in one variable calculus. Along the way to establish the proofs, many new concepts will be introduced. These include countability, sequences and series of numbers, of functions, supremum/infimum, Cauchy condition, Riemann integrals and improper integrals. Understanding them and their properties are important for the development of the present and further courses. Textbooks: Kenneth A. Ross, Elementary Analysis : The Theory of Calculus , Springer-Verlag, 1980. Robert Bartle and Donald Sherbert, Introduction to Real Analysis , 3rd ed., Wiley, 2000. Ross is for students who are taking this course for the first time. Students who are repeating may use either Ross or Bartle and Sherbert. References: 1. J. A. Fridy, Introductory Analysis 2nd ed., Academic Press, 2000. 2. Manfred Stoll, Introduction to Real Analysis , Addison Wesley, 1997. 3. Walter Rudin, Principles of Mathematical Analysis , 3rd ed., McGraw-Hill, 1976. 4. Tom Apostol, Mathematical Analysis , 2nd ed., Addison-Wesley, 1974. *5. Chinese Solution Manual to Tom Apostol’s Mathematical Analysis , 2nd ed. Reminders: Students are highly encouraged to come to office hours for consultation. This is a difficult course for many, but not all students. Although there are lecture notes, students should attend all lectures and tutorials as lectures notes are only brief records of materials covered in class, which may contain typographical errors. Of course, questions from students and answers from instructors or other digressions will not be recorded. Students are advised to take your own notes. All materials presented in lectures and tutorials as well as proper class conduct are the students’ responsibility. The instructor reserves the right to make any changes to the course throughout the semester. The only way to succeed in this course is to do the work. 1

Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Embed Size (px)

Citation preview

Page 1: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Math 201 (Introduction to Analysis) Fall 2004-2005

Instructor: Dr. Kin Y. LiOffice: Room 3471 Office Phone: 2358 7420 e-mail address: [email protected] Hours: Tu. 3:30pm - 4:30pm (or by appointments)

Prerequisite: A-level Math or One Variable Calculus

Website for Lecture Notes and Practice Exercises: http://www.math.ust.hk/ � makyli/UG.html

Grade Scheme: Homeworks (4%), Tutorial Presentations (6%), Midterm (40%), Final Examination (50%)

All records of grades will be put on the website http://grading.math.ust.hk/checkgrade/ as soon as they areavailable. This course is essentially graded by curve with one exception, namely students who achieve 40% orless of the overall grade will fail the course. Students who take the same midterm will be grouped under thesame curve.

Students should make copies of homeworks before submitting the originals. In case homeworks are notreceived, students will be required to resubmit copies within a short period of time (may be less than a day).

For tutorial presentations, students will form groups (of 1 to 3 students) and present solutions to assigned problemsin the tutorial sessions. All members of a group must attend the sessions to assist in answering possible questions frompresentations. Marks will be deducted for failure to present solutions or for absence in supporting his/her group.

Course Description: This is the first of two required courses on analysis for Math majors. It is to be followed byMath 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, as appeared in onevariable calculus. Along the way to establish the proofs, many new concepts will be introduced. These includecountability, sequences and series of numbers, of functions, supremum/infimum, Cauchy condition, Riemann integralsand improper integrals. Understanding them and their properties are important for the development of the present andfurther courses.

Textbooks:Kenneth A. Ross, Elementary Analysis : The Theory of Calculus, Springer-Verlag, 1980.Robert Bartle and Donald Sherbert, Introduction to Real Analysis, 3rd ed., Wiley, 2000.

Ross is for students who are taking this course for the first time. Students who are repeating may use either Rossor Bartle and Sherbert.

References:

1. J. A. Fridy, Introductory Analysis � 2nd ed., Academic Press, 2000.2. Manfred Stoll, Introduction to Real Analysis, Addison Wesley, 1997.3. Walter Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, 1976.4. Tom Apostol, Mathematical Analysis, 2nd ed., Addison-Wesley, 1974.

*5. Chinese Solution Manual to Tom Apostol’s Mathematical Analysis, 2nd ed.

Reminders: Students are highly encouraged to come to office hours for consultation. This is a difficult course formany, but not all students. Although there are lecture notes, students should attend all lectures and tutorials aslectures notes are only brief records of materials covered in class, which may contain typographical errors. Of course,questions from students and answers from instructors or other digressions will not be recorded. Students are advisedto take your own notes. All materials presented in lectures and tutorials as well as proper class conduct are thestudents’ responsibility. The instructor reserves the right to make any changes to the course throughout thesemester. The only way to succeed in this course is to do the work.

1

Page 2: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Objectives of the Course

The objectives of the course are to learn analysis and learn proofs.

Questions: What is analysis? How is it different from other branches of mathematics?

Analysis is the branch of mathematics that studies limit and concepts derived from limit, such as continuity,differentiation and integration, while geometry deals with figures, algebra deals with equations and inequalitiesinvolving addition, subtraction, multiplication, division and number theory studies integers.

When we try to solve problems involving real or complex numbers, such as finding roots of polynomials orsolving differential equations, we may not get the right answers the first time. However, we can get approximationsand the limits of these approximations, we hope, will give us the right answers. At least, we can know solutions existeven though we may not be able to write them explicitly.

Problems involving integers can be much harder since integers are discrete, i.e. there are minimum distanceseparating distinct integers so that one cannot find integers arbitrarily close to another integer.

Try to see if there is a real solution to the equation2x � x2

4x4 �1

� 987654321 � Then try to see if there is an integer

solution. What are the differences in the way you solve these two problems?

Question: Why should we learn proofs?

A statement is true not because your teacher tells you it is true. A teacher can make mistakes! There were famousmathematicians who made conjectures that were discovered to be wrong years later. How can we be certain the factswe learned are true? How can we judge when more than one proposed solutions are given, which is correct?

Suppose we want to find limx � 0

x2 sin 1x

sin x� Since the numerator is between � x 2 and x2

� the numerator has limit 0.

The denominator also has limit 0. So, by l’Hopital’s rule, limx � 0

x2 sin 1x

sin x� lim

x � 0

2x sin 1x

� cos 1x

cos x� However, the new

numerator does not have a limit because cos 1x has no limit as x � 0 � while the new denominator has limit 1. So the

limit of the original problem does not exist. Is this reasoning correct? No. Where is the mistake?

Sometimes we explain facts by examples or pictures. For instance, the statement that every odd degree polynomialwith real coefficients must have at least one root is often explained by some examples or some pictures. In our lifetime,we can only do finitely many examples and draw finitely many pictures. Should we believe something is true by seeinga few pictures or examples?

Draw the graphs of a few continuous functions on [0 � 1] � Do you think every continuous function on [0 � 1]is differentiable in at least one point on � 0 � 1 � ? Or do you think there exists a continuous function on [0 � 1] notdifferentiable at any point of � 0 � 1 � ?

Consider the function f � n � � n2 � n�

41 � Note f � 1 � � 41 � f � 2 � � 43 � f � 3 � � 47 � f � 4 � � 53 � f � 5 � �61 � f � 6 � � 71 � f � 7 � � 83 � f � 8 � � 97 � f � 9 � � 113 � f � 10 � � 131 are prime numbers. Should you believe that f � n �is a prime number for every positive integer n? What is the first n that f � n � is not prime?

In order to have confident, you have to be able to judge the facts you learned are absolutely correct. Almostcorrect is not good enough in mathematics.

2

Page 3: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 1. Logic

To reason correctly, we have to follow some rules. These rules of reasoning are what we called logic. We willonly need a few of these rules, mainly to deal with taking opposite of statements and to handle conditional statements.

We will use the symbol � (or � ) to denote the word “not”. Also, we will use the symbol�

to denote “for all”,“for any”, “for every”. Similarly, the symbol � will denote “there is (at least one)”, “there exists”, “there are (some)”and usually followed by “such that”. The symbols

�and � are called quantifiers.

Negation. Below we will look at rules of negation (i.e. taking opposite). They are needed when we do indirectproofs (or proofs by contradiction). For any expression p � we have � � � p � � p �

Examples. (1)expression :

p� ��� �x � 0 and

q� ��� �x 1

opposite expression : x 0 or x � 1

rule : � � p and q � � � � p � or � � q �

(2) expression : x 0 or x � 1

opposite expression : x � 0 and x 1

rule : � � p or q � � � � p � and � � q �

(3) statement : For every x � 0 � x has a square root. (True)

quantified statement :�

x � 0 � x has a square root � �opposite statement : There exists x � 0 such that x does not have a square root. (False)

quantified opposite statement : � x � 0 � � x has a square root� �

(4) statement : For every x � 0 � there is y � 0 such that y2 � x � (True)

quantified statement :�

x � 0 � � y � 0 � y2 � x � �opposite statement : There exists x � 0 such that for every y � 0 � y2 �� x � (False)

quantified opposite statement : � x � 0 �

�y � 0 � � y2 � x � �

From examples (3) and (4), we see that the rule for negating statements with quantifiers is first switch every�to � and every � to

�, then negate the remaining part of the statement.

If-then Statements. If-then statements occur frequently in mathematics. We will need to know some equivalentways of expressing an if-then statement to do proofs. The statement “if p � then q” may also be stated as “p implies q”,“p only if q”, “p is sufficient for q”, “q is necessary for p” and is commonly denoted by “p �� q”. For example,the statement “if x � 3 and y � 4 � then x 2 �

y2 � 25” may also be stated as “x � 3 and y � 4 are sufficient forx2 �

y2 � 25” or “x2 �y2 � 25 is necessary for x � 3 and y � 4”.

Example. (5) statement : If x � 0 � then � x � � x � (True)

opposite statement : x � 0 and � x � �� x � (False)

rule : � � p �� q � � p and � � q �Remark. Note

p �� q � � � � � p �� q � �� � � p and � � q � �� � � p � or � � � q �� � � p � or q �

3

Page 4: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

For the statement “if p � then q” � p �� q � � there are two related statements: the converse of the statement is “ifq, then p” (q �� p) and the contrapositive of the statement is “if � � q � � then � � p � ” ( � q �� � p).

Examples. (6) statement : If x � � 3 � then x 2 � 9 � (True)

converse : If x2 � 9 � then x � � 3 � (False, as x may be 3.)

contrapositive : If x 2 �� 9 � then x �� � 3 � (True)

(7) statement : x � � 3 �� 2x � � 6 (True)

converse : 2x � � 6 �� x � � 3 (True)

contrapositive : 2x �� � 6 �� x �� � 3 (True)

(8) statement : If � x � � 3 � then x � � 3 � (False, as x may be 3.)

converse : If x � � 3 � then � x � � 3 � (True)

contrapositive : If x �� � 3 � then � x � �� 3 � (False, as x may be 3.)

Remarks. Examples (6) and (7) showed that the converse of an if-then statement is not the same as the statementnor the opposite of the statement in general. Examples (6), (7) and (8) showed that an if-then statement and itscontrapositive are either both true or both false. In fact, this is always the case because by the remark on the last page,

� � q � �� � � p � � � � � q � or � � p �� q or � � p �� � � p � or q� p �� q �

So an if-then statement and its contrapositive statement are equivalent.

Finally, we introduce the terminology “p if and only if q” to mean “if p � then q” and “if q � then p”. The statement“p if and only if q” is the same as “p is necessary and sufficient for q”. We abbreviate “p if and only if q” by “p

� �q”. So p

� � q means p �� q and q �� p”. The phrase “if and only if” is often abbreviated as “iff”.

Caution! Note��� ��� � �������

and � � � � � � � � � � but��� � � �� � ����� � For example, “every student is assigned a

number” is the same as “�

student, � number such that the student is assigned the number.” This statement impliesdifferent students may be assigned possibly different numbers. However, if we switch the order of the quantifiers, thestatement becomes “ � number such that

�student, the student is assigned the number.” This statement implies there

is a number and every student is assigned that same number!

4

Page 5: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 2. Sets

To read and write mathematical expressions accurately and concisely, we will introduce the language of sets. Aset is a collection of “objects” (usually numbers, ordered pairs, functions, etc.) If object x is in a set S, then we say x isan element (or a member) of S and write x � S � If x is not an element of S, then we write x �� S � A set having finitelymany elements is called a finite set, otherwise it is called an infinite set. The empty set is the set having no objects andis denoted by

� �A set may be shown by listing its elements enclosed in braces (eg. � 1 � 2 � 3 � is a set containing the objects 1 � 2 � 3 � the

positive integer � � � 1 � 2 � 3 � � � � � , the integer � � � � � � �� 2 �

� 1 � 0 � 1 � 2 � � � � � , the empty set� � ��� ) or by description

enclosed in braces (eg. the rational numbers � � � mn : m ��� � n ��� � the real numbers � � x : x is a real number �

and the complex numbers � � � x �iy : x � y ��� � ) In describing sets, the usual convention is to put the form of the

objects on the left side of the colon and to state the conditions on the objects on the right side of the colon. It is alsocommon to use a vertical bar in place of colon in set descriptions.

Examples. (i) The closed interval with endpoints a � b is [a � b] � � x : x �� and a x b � �(ii) The set of square numbers is � 1 � 4 � 9 � 16 � 25 � � � � � � � n2 : n ��� �

(iii) The set of all positive real numbers is � � � x : x �� and x � 0 � � (If we want to emphasize this is a subsetof � we may stress x is real in the form of the objects and write � � � x �� : x � 0 � � If numbers are alwaystaken to mean real numbers, then we may write simply � � � x : x � 0 � � )

(iv) The set of points (or ordered pairs) on the line � m with equation y � mx is � � x � mx � : x ��� �

For sets A � B � we say A is a subset of B (or B contains A) iff every element of A is also an element of B � Inthat case, we write A � B � (For the case of the empty set, we have

� � S for every set S � ) Two sets A and B areequal if and only if they have the same elements (i.e. A � B means A � B and B � A � ) So A � B if and only if(x � A

� � x � B). If A � B and A �� B � then we say A is a proper subset of B and write A � B � (For example,if A � � 1 � 2 � � B � � 1 � 2 � 3 � � C � � 1 � 1 � 2 � 3 � � then A � B is true, but B � C is false. In fact, B � C � Repeatedelements are counted only one time so that C has 3 elements, not 4 elements.)

For a set S � we can collect all its subsets. This is called the power set of S and is denoted by P � S � or 2S � Forexamples, P � � � � � � � � P ��� 0 � � � � � ��� 0 ��� and P ��� 0 � 1 � � � � � ��� 0 � ��� 1 � ��� 0 � 1 ��� � For a set with n elements, its powerset will have 2n element. This is the reason for the alternative notation 2S for the power set of S � Power set is oneoperation of a set. There are a few other common operations of sets.

Definitions. For sets A1 � A2 � � � � � An �

(i) their union is A1 � A2 ��������� An� � x : x � A1 or x � A2 or ����� or x � An � �

(ii) their intersection is A1 � A2 ��������� An� � x : x � A1 and x � A2 and ����� and x � An � �

(iii) their Cartesian product is

A1 � A2 ��������� An� � � x1 � x2 � � � � � xn � : x1 � A1 and x2 � A2 and ����� and xn � An � �

(iv) the complement of A2 in A1 is A1 � A2� � x : x � A1 and x �� A2 � �

Examples. (i) � 1 � 2 � 3 � � � 3 � 4 � � � 1 � 2 � 3 � 4 � ��� 1 � 2 � 3 � � � 2 � 3 � 4 � � � 2 � 3 � ��� 1 � 2 � 3 � � � 2 � 3 � 4 � � � 1 � �(ii) [ � 2 � 4] � � � � 1 � 2 � 3 � 4 � � [0 � 2] � [1 � 5] � [4 � 6] � [0 � 6] �

(iii) � [0 � 7] � � � � � n2 : n ��� � � 0 � 1 � 2 � 3 � 4 � 5 � 6 � 7 � � � 1 � 4 � 9 � 16 � 25 � � � � � � � 0 � 2 � 3 � 5 � 6 � 7 � �(iv) � � � � � x � y � z � : x � y � z ��� ��� � �� � � � � � � a � b � : a is rational and b is irrational � �Remarks. (i) For the case of the empty set, we have

A �� � A � �

� A � A �� � � � �

� A � A �� � � � �

� A � A �� � A and

�� A � � �

5

Page 6: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(ii) The notions of union, intersection and Cartesian product may be extended to infinitely many sets similarly. Theunion is the set of objects in at least one of the sets. The intersection is the set of objects in every one of the sets.The Cartesian product is the set of ordered tuples such that the i-th coordinate must belong to the i-th set.

(iii) The set A1 � A2 � ������� An may be written asn�

k � 1

Ak � If for every positive integer k � there is a set Ak � then the

notation A1 � A2 � A3 � ����� may be abbreviated as ��k � 1

Ak or�

k � � Ak � If for every x � S � there is a set Ax � then

the union of all the sets Ax ’s for all x � S is denoted by�x � S Ax � Similar abbreviations exist for intersection and

Cartesian product.

Examples. (i) � [1 � 2] � [2 � 3] � [3 � 4] � [4 � 5] ������� � � � � [1 �

��� � � � � � �

(ii) �n � �

�0 � 1

� 1n � [0 � 2 � �

�0 � 1

12 � � 0 � 1

13 � � 0 � 1

14 ������� � [0 � 1] �

(iii) For every k �� � let Ak� � 0 � 1 � � then

A1 � A2 � A3 ������� � � � x1 � x2 � x3 � � � � � : each xk is 0 or 1 for k � 1 � 2 � 3 � � � � � �

(iv) For each m �� � let � m be the line with equation y � mx on the plane, then�

m � � m� 2 � � � 0 � y � : y �� � y �� 0 �

and �m � � m

� � � 0 � 0 � � �

We shall say that sets are disjoint iff their intersection is the empty set. Also, we say they are mutually disjoint iffthe intersection of every pair of them is the empty set. A relation on a set E is any subset of E � E � The following isan important concept that is needed in almost all branches of mathematics. It is a tool to divide (or partition) the set ofobjects we like to study into mutually disjoint subsets.

Definition. An equivalence relation R on a set E is a subset R of E � E such that

(a) (reflexive property) for every x � E � � x � x � � R �

(b) (symmetric property) if � x � y � � R � then � y � x � � R �

(c) (transitive property) if � x � y � � � y � z � � R � then � x � z � � R �We write x � y if � x � y � � R � For each x � E � let [x] � � y : x � y � � This is called the equivalence classcontaining x � Note that every x � [x] by (a) so that

�x � E

[x] � E � If x � y � then [x] � [y] because by (b) and (c),

z � [x]� � z � x

� � z � y� � z � [y] � If x �� y � then [x] � [y] � �

because assuming z � [x] � [y] willlead to x � z and z � y � which imply x � y � a contradiction. So every pair of equivalence classes are either the sameor disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes.

Examples. (1) (Geometry) For triangles T1 and T2 � define T1� T2 if and only if T1 is similar to T2 � This is an

equivalence relation on the set of all triangles as the three properties above are satisfied. For a triangle T � [T ] is the setof all triangles similar to T �(2) (Arithmetic) For integers m and n � define m � n if and only if m � n is even. Again, properties (a), (b), (c) caneasily be verified. So this is also an equivalence relation on � � There are exactly two equivalence classes, namely[0] � � � � � �

� 4 �� 2 � 0 � 2 � 4 � � � � � (even integers) and [1] � � � � � �

� 5 �� 3 �

� 1 � 1 � 3 � 5 � � � � � (odd integers). Two integersin the same equivalence class is said to be of the same parity.

(3) Some people think that properties (b) and (c) imply property (a) by using (b), then letting z � x in (c) to conclude� x � x � � R � This is false as shown by the counterexample that E � � 0 � 1 � and R � � � 1 � 1 � � � which satisfies properties(b) and (c), but not property (a). R fails property (a) because 0 � E � but � 0 � 0 � �� R as 0 is not in any ordered pair in R �

6

Page 7: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

A function (or map or mapping) f from a set A to a set B (denoted by f : A � B) is a method of assigning toevery a � A exactly one b � B � This b is denoted by f � a � and is called the value of f at a � Thus, a function must bewell-defined in the sense that if a � a � � then f � a � � f � a � � � The set A is called the domain of f (denoted by dom f )and the set B is called the codomain of f (denoted by codom f ). We say f is a B-valued function (eg. if B � � thenwe say f is a real-valued function.) When the codomain B is not emphasized, then we may simply say f is a functionon A � The image or range of f (denoted by f � A � or im f or ran f ) is the set � f � x � : x � A � � (To emphasize this is asubset of B � we also write it as � f � x � � B : x � A � � ) The set G � �

�x � f � x ��� : x � A � is called the graph of f � Two

functions are equal if and only if they have the same graphs. In particular, the domains of equal functions are the sameset.

Examples. The function f : � � given by f � x � � x 2 has dom f � � � codom f � � Also, ran f �� 0 � 1 � 4 � 9 � 16 � � � � � � This is different from the function g : � given by g � x � � x 2 because dom g � �� dom f �Also, a function may have more than one parts in its definition, eg. the absolute value function h : � defined by

h � x � ��� x if x � 0� x if x 0

� Be careful in defining functions. The following is bad: let xn� � � 1 � n and i � xn � � n � The

rule is not well-defined because x1� � 1 � x3 � but i � x1 � � 1 �� 3 � i � x3 � �

Definitions. (i) The identity function on a set S is IS : S � S given by IS � x � � x for all x � S �(ii) Let f : A � B � g : B � � C be functions and f � A � � B � � The composition of g by f is the function

g � f : A � C defined by � g � f � � x � � g � f � x � � for all x � A �(iii) Let f : A � B be a function and C � A � The function f � C : C � B defined by f � C � x � � f � x � for every x � C

is called the restriction of f to C �(iv) A function f : A � B is surjective (or onto) iff f � A � � B �(v) A function f : A � B is injective (or one-to-one) iff f � x � � f � y � implies x � y �

(vi) A function f : A � B is a bijection (or a one-to-one correspondence) iff it is injective and surjective.

(vii) For an injective function f : A � B � the inverse function of f is the function f � 1 : f � A � � A defined byf � 1 � y � � x

� � f � x � � y �

Remarks. A function f : A � B is surjective means f � A � � B � which is the same as saying every b � B is an f � a �for at least one a � A � In this sense, the values of f do not omit anything in B � We will loosely say f does not omitany element of B for convenience. However, there may possibly be more than one a � A that are assigned the sameb � B � Hence, the range of f may repeat some elements of B � If A and B are finite sets, then f surjective implies thenumber of elements in A is greater than or equal to the number of elements in B �

Next, a function f : A � B is injective means, in the contrapositive sense, that x �� y implies f � x � �� f � y � �

which we may loosely say f does not repeat any element of B � However, f may omit elements of B as there maypossibly be elements in B that are not in the range of f � So if A and B are finite sets, then f injective implies thenumber of elements in A is less than or equal to the number of elements in B �

Therefore, a bijection from A to B is a function whose values do not omit nor repeat any element of B � If A andB are finite sets, then f bijective implies the number of elements in A and B are the same.

Remarks (Exercises). (a) Let f : A � B be a function. We have f is a bijection if and only if there is a functiong : B � A such that g � f � IA and f � g � IB � (In fact, for f bijective, we have g � f � 1 is bijective.)

(b) If f : A � B and h : B � C are bijections, then h � f : A � C is a bijection.

(c) Let A � B be subsets of and f : A � B be a function. If for every b � B � the horizontal line y � b intersects thegraph of f exactly once, then f is a bijection.

To deal with the number of elements in a set, we introduce the following concept. For sets S1 and S2 � we willdefine S1

� S2 and say they have the same cardinality (or the same cardinal number) if and only if there exists abijection from S1 to S2 � This is easily checked to be an equivalence relation on the collection of all sets. For a set S � theequivalence class [S] is often called the cardinal number of S and is denoted by card S or � S � � This is a way to assigna symbol for the number of elements in a set. It is common to denote, for a positive integer n � card � 1 � 2 � � � � � n � � n �

card � ��0 (read aleph-naught) and card � c (often called the cardinality of the continuum).

7

Page 8: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 3. Countability

Often we compare two sets to see if they are different. In case both are infinite sets, then the concept of countablesets may help to distinguish these infinite sets.

Definitions. A set S is countably infinite iff there exists a bijection f : � � S (i.e. � and S have the same cardinalnumber � 0 � ) A set is countable iff it is a finite or countably infinite set. A set is uncountable iff it is not countable.

Remarks. Suppose f : � � S is a bijection. Then f is injective means f � 1 � � f � 2 � � f � 3 � � � � � are all distinct and fis surjective means � f � 1 � � f � 2 � � f � 3 � � � � ��� � S � So n � ��� f � n �� S is a one-to-one correspondence between �and S � Therefore, the elements of S can be listed in an “orderly” way (as f � 1 � � f � 2 � � f � 3 � � � � � ) without repetition oromission. Conversely, if the elements of S can be listed as s1 � s2 � � � � without repetition or omission, then f : � � Sdefined by f � n � � sn will be a bijection as no repetition implies injectivity and no omission implies surjectivity.

Bijection Theorem. Let g: S � T be a bijection. S is countable if and only if T is countable.

(Reasons. The theorem is true because S countable implies there is a bijective function f : � � S � which impliesh � g � f : � � T is bijective, i.e. T is countable. For the converse, h is bijective implies f � g � 1 � h is bijective.)

Remarks. Similarly, taking contrapositive, S is uncountable if and only if T is uncountable.

Basic Examples. (1) � is countably infinite (because the identity function I � � n � � n is a bijection).

(2) � is countably infinite because the following function is a bijection (one-to-one correspondence):

� � � 1 � 2 � 3 � 4 � 5 � 6 � 7 � 8 � 9 � � � � �f� � � � � � � � � � � � �� � � 0 � 1 �

� 1 � 2 �� 2 � 3 �

� 3 � 4 �� 4 � � � � � �

The function f : � � � is given by f � n � �� n

2 if n is even� � n � 1

2 � if n is oddand its inverse function g : � � � is given

by g � m � ��

2m if m � 01 � 2m if m 0

� Just check g � f � I � and f � g � I � �

(3) � � � � � � m � n � : m � n ���� is countably infinite.

(Diagonal Counting Scheme) Using the diagram on the right, de-fine f : � � � � � by f � 1 � � � 1 � 1 � , f � 2 � � � 2 � 1 � , f � 3 � � � 1 � 2 � ,f � 4 � � � 3 � 1 � , f � 5 � � � 2 � 2 � , f � 6 � � � 1 � 3 � , � � � , then f is injectivebecause no ordered pair is repeated. Also, f is surjective because

� m � n � � f � m n � 2�k � 0

k�

n � � f � � m �n � 2 � � m �

n � 1 �2

�n � �

� 1 � 1 � � 1 � 2 � � 1 � 3 � � 1 � 4 �

� 2 � 1 � � 2 � 2 � � 2 � 3 �������

� 3 � 1 � � 3 � 2 � ��� ������

���� 4 � 1 �

������

������ ��� ��� ���

(4) The open interval � 0 � 1 � � � x : x �� and 0 x 1 � is uncountable. Also, is uncountable.

f � 1 � � 0 � a11a12a13a14 � � �f � 2 � � 0 � a21a22a23a24 � � �f � 3 � � 0 � a31a32a33a34 � � �f � 4 � � 0 � a41a42a43a44 � � �

��� ���

Suppose � 0 � 1 � is countably infinite and f : � � � 0 � 1 � is a bijection asshown on the left. Consider the number x whose decimal representation is

0 � b1b2b3b4 � � � , where bn��

2 if ann� 1

1 if ann�� 1

. Then 0 x 1 and x �� f � n �for all n because bn

�� ann . So f cannot be surjective, a contradiction. Next is uncountable because tan � x � 1

2 � provides a bijection from � 0 � 1 � onto �

To determine the countability of more complicated sets, we will need the theorems below.

8

Page 9: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Countable Subset Theorem. Let A � B � If B is countable, then A is countable. (Taking contrapositive, if A isuncountable, then B is uncountable.)

Countable Union Theorem. If An is countable for every n ��� � then�

n � � An is countable. In general, if S is countable

(say f : � � S is a bijection) and As is countable for every s � S � then�s � S As

� �n � � A f � n � is countable. (Briefly,

countable union of countable sets is countable.)

Product Theorem. If A, B are countable, then A � B � � � a � b � : a � A � b � B � is countable. In fact, if A1 � A2 � � � � � An

are countable, then A1 � A2 ��������� An is countable (by mathematical induction).

(Sketch of Reasons. For the countable subset theorem, if B is countable, then we can list the elements of B and tocount the elements of A, we can skip over those elements of B that are not in A � For the countable union theorem, ifwe list the elements of A1 in the first row, the elements of A2 in the second row, � � � � then we can count all the elementsby using the diagonal counting scheme. As for the product theorem, we can imitate the example of � � � and alsouse the diagonal counting scheme.)

Examples. (5) � � � �n � 1Sn � where Sn

��� m

n: m ���

�� For every n ��� � the function fn : � � Sn given by fn � m � � m

nis a bijection (with f � 1

n � mn � m), so Sn is countable by the bijection theorem. Therefore, � is countable by the

countable union theorem. (Then subsets of � like � � � 0 � � � � � 0 � ��� � � 0 � 1 � are also countable.)

(6) � � is uncountable. (In fact, if A is uncountable and B is countable, then A � B is uncountable as A � Bcountable implies � A � B � � � A � B � � A countable by the countable union theorem, which is a contradiction).

(7) � � � x �iy : x � y �� contains and is uncountable, so by the countable subset theorem, � is uncountable.

(8) Show that the set A � � r � m : m ��� � r � � 0 � 1 � � is uncountable, but the set B � � r � m : m ��� � r ��� � � 0 � 1 � �is countable.

Solution. Taking m � 1 � we see that � 0 � 1 � � A � Since � 0 � 1 � is uncountable, A is uncountable. Next we willobserve that B � �

m � � Bm � where Bm� � r � m : r � � � � 0 � 1 � � � �

r � ����� 0 � 1 � � r � m � for each m ��� � Since

� � � 0 � 1 � is countable and � r � m � has 1 element for every r ��� � � 0 � 1 � � Bm is countable by the countable uniontheorem. Finally, since � is countable and Bm is countable for every m � � � B is countable by the countableunion theorem.

(9) Show that the set L of all lines with equation y � mx�

b � where m � b ��� � is countable.

Solution. Note that for each pair m � b of rational numbers, there is a unique line y � mx�

b in the set L � So thefunction f : � � � � L defined by letting f � m � b � be the line y � mx

�b (with f � 1 sending the line back to

� m � b � ) is a bijection. Since � � � is countable by the product theorem, so the set L is countable by the bijectiontheorem.

(10) Show that if An� � 0 � 1 � for every n ��� � then A1 � A2 � A3 ������� is uncountable. (In particular, this shows that

the product theorem is not true for infintely many countable sets.)

Solution. Assume A1 � A2 � A3 � ����� � � � a1 � a2 � a3 � � � � � : each ai� 0 or 1 � is countable and f : � �

A1 � A2 � A3 � ����� is a bijection. Following example (4), we can change the n-th coordinate of f � n � (from 0 to1 or from 1 to 0) to produce an element of A1 � A2 � A3 ������� not equal to any f � n � � which is a contradiction.So it must be uncountable.

(11) Show that the power set P ��� � of all subsets of � is uncountable.

Solution. As in example (10), let An� � 0 � 1 � for every n ��� � Define g : P ��� � � A1 � A2 � A3 � ����� by

g � S � � � a1 � a2 � a3 � � � � � � where am��

1 if m � S0 if m �� S

� (For example, g ��� 1 � 3 � 5 � � � � � � � � 1 � 0 � 1 � 0 � 1 � � � � � � ) Note

g has the inverse function g � 1�

� a1 � a2 � a3 � � � � ��� � � m : am� 1 � � Hence g is a bijection. Since A1 � A2 � A3 �������

is uncountable, so P ��� � is uncountable by the bijection theorem.

9

Page 10: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(12) Show that the set S of all polynomials with integer coefficients is countable.

Solution. Let S0� � � For n ��� � the set of Sn of all polynomials of degree n with integer coefficients is countable

because the function f : Sn � � � � � 0 � � � � � ����� � � defined by f � anxn �an � 1xn � 1 � ����� �

a0 � � � an � an � 1 � � � � � a0 �is a bijection and � � � � 0 � � � � ��������� � is countable by the product theorem. So, S � S0 �

�n � � Sn is countable

by the countable union theorem.

(13) Show that there exists a real number, which is not a root of any nonconstant polynomial with integer coefficients.

Solution. For each polynomial f with integer coefficients, let R f denotes the set of roots of f � Then R f hasat most deg f elements, hence R f is countable. Let S � be the set of all nonconstant polynomials with integercoefficients, which is the subset

�n � � Sn of S in the last example. Then

�f � S � R f is the set of all roots of nonconstant

polynomials with integer coefficients. It is countable by the countable union theorem. Since is uncountable, �

�f � S � R f is uncountable by the fact in example (6). So there exist uncountably many real numbers, which are

not roots of any nonconstant polynomial with integer coefficients.

Remarks. Any number which is a root of a nonconstant polynomial with integer coefficients is called an algebraicnumber. A number which is not a root of any nonconstant polynomialwith integer coefficients is called a transcendentalnumber. Are there any algebraic numbers? Are there any transcendental numbers? If so, are there finitely many orcountably many such numbers?

Since every rational number ab is the root of the polynomial bx � a � every rational number is algebraic. There

are irrational numbers like� � 2 � which are algebraic because they are the roots of x 2 � 2 � Using the identity

cos 3 � � 4 cos3 � � 3 cos � � the irrational number cos 20 � is easily seen to be algebraic as it is a root of 8x 3 � 6x � 1 �Example (12) and the fact that every nonconstant polynomial has finitely many roots showed there are only countablymany algebraic numbers. Example (13) showed that there are uncountably many transcendental real numbers. It isquite difficult to prove a particular number is transcendental. In a number theory course, it will be shown that and eare transcendental.

The following are additional useful facts concerning countability.

Theorem.

(1) (Injection Theorem) Let f : A � B be injective. If B is countable, then A is countable. (Taking contrapositive,if A is uncountable, then B is uncountable.)

(2) (Surjection Theorem) Let g : A � B be surjective. If A is countable, then B is countable. (Taking contrapositive,if B is uncountable, then A is uncountable.)

(Reasons. For the first statement, observe that the function h : A � f � A � defined by h � x � � f � x � is injective(because f is injective) and surjective (because h � A � � f � A � ). So h is a bijection. If B is countable, then f � A � iscountable by the countable subset theorem, which implies A is countable by the bijection theorem.

For the second statement, observe that B � g � A � � �x � A

� g � x � � � If A is countable, then it is a countable union of

countable sets. By the countable union theorem, B is countable.)

A Famous Open Problem in Mathematics

For two sets A and B � it is common to define card A card B if and only if there exists an injective functionf : A � B � This is a way to indicate B has at least as many elements as A �Continuum Hypothesis. If S is uncountable, then card card S � (This means every uncountable set has at least asmany elements as the real numbers.)

In 1940, Kurt Godel showed that the opposite statement would not lead to any contradiction. In 1966, Paul Cohenwon the Fields’ Medal for showing the statement also would not lead to any contradiction. So proof by contradictionmay not be applied to every statement.

10

Page 11: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 4. Series

Definitions. A series is the summation of a countable set of numbers in a specific order. If there are finitely manynumbers, then the series is a finite series, otherwise it is an infinite series. The numbers are called terms. The sum ofthe first n terms is called the n-th partial sum of the series.

An infinite series is of the form a1�������1st term

�a2�������

2nd term

�a3�������

3rd term

� � � � or we may write it as ��k � 1

ak �

The first partial sum is S1� a1. The second partial sum is S2

� a1�

a2. The nth partial sum is Sn� a1

�a2

� � � � �an .

Series are used frequently in science and engineering to solve problems or approximate solutions. (E.g. trigono-metric or logarithm tables were computed using series in the old days.)

Examples.(1) 1� 1

2� 1

4� 1

8� 1

16� � � � � ? (Sn

� 1� 1

2� 1

4� � � � � 1

2n� 2 � 1

2n � 1� 1

2� 1

4� 1

8� 1

16� � � � � lim

n � ��2 � 1

2n� � 2.)

We say the series converges to 2, which is called the sum of the series.

(2) 1�

1�

1�

1�

1�

1� � � � � �

(Sn� 1

�1

� � � � �1� ��� �

n

� n, limn � � Sn

� �.) We say the series diverges (to

�).

(3) 1 � 1�

1 � 1�

1 � 1�

1 � 1� � � � . (Sn

��� 1 if n is odd0 if n is even

, limn � � Sn doesn’t exist.) We say the series diverges.

Definitions. A series ��k � 1

ak� a1

�a2

�a3

� � � � converges to a number S iff limn � � � a1

�a2

� � � � �an � � lim

n � � Sn� S �

In that case, we may write ��k � 1

ak� S and say S is the sum of the series. A series diverges to

�iff the partial sum Sn

tends to infinity as n tends to infinity. A series diverges iff it does not converge to any number.

Remarks. (1) For every series ��k � 1

ak , there is a sequence (of partial sums) � Sn � . Conversely, if the partial sum sequence

� Sn � is given, we can find the terms an as follows: a1� S1, a2

� S2� S1, � � � , ak

� Sk� Sk � 1 for k � 1. Then

a1� � � � �

an� S1

� � S2� S1 � � � � � � � Sn

� Sn � 1 � � Sn. So � Sn � is the partial sum sequence of ��k � 1

ak . Conceptually,

series and sequences are equivalent. So to study series, we can use facts about sequences.

(2) Let N be a positive integer. ��k � 1

ak converges to A if and only if ��k � N

ak converges to B � A � � a1� ����� �

aN � 1 �because

B � limn � � � aN

� ����� �an � � lim

n � � � a1�

a2� ����� �

an � � � a1� ����� �

aN � 1 � � A � � a1� ����� �

aN � 1 � �

So to see if a series converges, we may ignore finitely many terms.

Theorem. If ��k � 1

ak converges to A and ��k � 1

bk converges to B � then

��k � 1

� ak�

bk � � A�

B � ��k � 1

ak� ��

k � 1

bk � ��k � 1

� ak� bk � � A � B � ��

k � 1

ak� ��

k � 1

bk � ��k � 1

cak� cA � c ��

k � 1

ak

for any constant c �

For simple series such as geometric or telescoping series, we can find their sums.

11

Page 12: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Theorem (Geometric Series Test). We have

��k � 0

rk � limn � � � 1 �

r�

r2 � � � � �rn � � lim

n � �1 � rn 1

1 � r��

1

1 � rif � r � 1

doesn’t exist otherwise�

Example. 0 � 999 ����� � 9

10� 9

100� 9

1000� ����� � 9

10� 1

1 � 110

� � 1 � 1 � 000 ����� � So 1 has two decimal representa-

tions!

Theorem (Telescoping Series Test). We have ��k � 1

� bk� bk 1 � � lim

n � ��

� b1� b2 � � � b2

� b3 � � ����� � � bn� bn 1 ���

� limn � � � b1

� bn 1 � � b1� lim

n � � bn 1 converges if and only if limn � � bn is a number.

Examples. (1) ��k � 1

1

k � k �1 �

� ��k � 1

� 1

k� 1

k�

1� � �

1 � 1

2� � � 1

2� 1

3� � � 1

3� 1

4� � ����� � 1 � lim

n � �1

n�

1� 1 �

(2) ��k � 1

� 51 � k � 51 � � k 1 � � � � 5 � � 5 � � � � 5 � 3� 5 � � ����� � 5 � limk � � 51 � � k 1 � � 5 � 50 � 4 �

If a series is not geometric or telescoping, we can only determine if it converges or diverges. This can be donemost of the time by applying some standard tests. If the series converges, it may be extremely difficult to find the sum!

Theorem (Term Test). If ��k � 1

ak converges, then limk � � ak

� 0. (If limk � � ak

�� 0, then the series ��k � 1

ak diverges.) If

limk � � ak

� 0, the series ��k � 1

ak may or may not converge.

(Reason. Suppose ��k � 1

ak converges to S � Then limn � � Sn

� S and limk � � ak

� limk � � � Sk

� Sk � 1 � � S � S � 0.)

Term test is only good for series that are suspected to be divergent!

Examples. (1) 1�

1�

1�

1� � � � . Here ak

� 1 for all k, so limk � � ak

� 1. Series diverges.

(2) ��k � 1

cos � 1k

� � cos 1�

cos12

�cos

13

� � � � diverges because limk � � cos � 1

k� � cos 0 � 1 �� 0.

(3) ��k � 1

cos k � cos 1�

cos 2�

cos 3� � � � diverges because lim

k � � cos k �� 0 � (Otherwise, limk � � cos k � 0 � Then

limk � � � sin k � � lim

k � ��� 1 � cos2 k � 1 and 0 � limk � � � cos � k �

1 � � � limk � � � cos k cos 1 � sin k sin 1 � � sin 1 �� 0 � a

contradiction.)

(4) 1 � 12

� 14

� 18

� � � � . Here ak� � � 1

2 � k � 1 for all k, so limk � � ak

� 0 � (Term test doesn’t apply!) Series converges

by the geometric series test.

(5) 1� 1

2� 1

2� ��� �2 times

� 14

� 14

� 14

� 14� ��� �

4 times

� 18

� � � � � 18� ��� �

8 times

� � � � . We have limk � � ak

� 0 � (Term test doesn’t apply.) Series

diverges to�

because S1 S2 S3 ����� and S2n � 1� n has limit

� �

12

Page 13: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

For a nonnegative series ��k � 1

ak (i.e. ak � 0 for every k), we have S1 S2 S3 � � � and limn � � Sn must exist as a

number or equal to���

. So either ��k � 1

ak converges to a number or ��k � 1

ak diverges to���

. (In short, either ��k � 1

ak� S

or ��k � 1

ak� ���

.) For nonnegative series, we have the following tests.

Theorem (Integral Test). Let f : [1 �

��� � � decrease to 0 as x � ���. Then ��

k � 1

f � k � converges if and only if

� �1

f � x � dx � . (Note in general, ��k � 1

f � k � ��� �

1f � x � dx � )

(Reason. This follows from f � 2 � �f � 3 � � � � � �

f � n � � ����� � �

1f � x � dx f � 1 � �

f � 2 � � � � � �f � n � 1 � � �����

as shown in the figures below.)

...

1 2 3 4 n...

(2)

(3)

1 2 3 4 ... n

...

(2)

(1)f

f

f

f

Examples. (1) Consider the convergence or divergence of ��k � 1

1

1�

k2�

As x � �� 1

�x2 � �

� so1

1�

x2 � 0 � Now� �

1

1

1�

x2dx � arctan x ����

�1

� 2

� 4

� � So ��k � 1

1

1�

k2

converges.

(2) Consider the convergence or divergence of ��k � 2

1k ln k

and ��k � 2

1k � ln k � 2

.

As x � �� x ln x and x � ln x � 2 � �

� so their reciprocals decrease to 0 � Now� �

2

dx

x ln x� ln � ln x ������

�2

� � � So

��k � 2

1k ln k

diverges. Next� �

2

dx

x � ln x � 2� � 1

ln x�����2

� 1ln 2

� � So ��k � 2

1k � ln k � 2

converges.

Theorem (p-test). For a real number p ��� � p � � ��k � 1

1

k p� 1

� 1

2p

� 1

3p

� 1

4p

� � � � converges if and only if p � 1.

(Reason. For p 0 � the terms are at least 1, so the series diverges by term test. For p � 0 � f � x � � 1x p decreases

to 0 as x � ���. Since

� �1

1

x pdx � x � p 1

� p�

1�����1

� 1

p � 1if p � 1 �

� �1

1

x pdx � � ln x � � �1 � �

if p � 1 and� �

1

1

x pdx � x � p 1

� p�

1�����1

� �if p 1, the integral test gives the conclusion.)

Remarks. For even positive integer p � the value of � � p � was computed by Euler back in 1736. He got

� � 2 � � 2

6��� � 4 � � 4

90� � � � ��� � 2n � � � � 1 � n 1 � 2 � 2n B2n

2 � 2n � !� � � � �

13

Page 14: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

where B0� 1 and � k �

1 � Bk� � k � 1�

m � 0

� k �1

m� Bm for k � 1 � The values of � � 3 � � � � 5 � � � � � are unknown. Only in the

1980’s, R. Apery was able to show � � 3 � was irrational.

Theorem (Comparison Test). Given �k � uk � 0 for every k. If ��

k � 1

�k converges, then ��

k � 1

uk converges. If ��k � 1

uk

diverges, then ��k � 1

�k diverges.

(Reason. �k � uk � 0 � ��

k � 1

�k � ��

k � 1

uk � 0. If ��k � 1

�k is a number, then ��

k � 1

uk is a number. If ��k � 1

uk� ���

, then

��k � 1

�k

� ���.)

Theorem (Limit Comparison Test). Given uk , �k � 0 for every k � If lim

k � ��

k

ukis a positive number L � then either (both

��k � 1

uk and ��k � 1

�k converge) or (both diverge to

���). If lim

k � ��

k

uk

� 0 � then ��k � 1

uk converges � ��k � 1

�k converges. If

limk � �

�k

uk

� �� then ��

k � 1

uk diverges � ��k � 1

�k diverges.

(Sketch of Reason. For k large,�

k

uk

�L . For L � 0 �

� �k

� �Luk

� L�

uk . If one series converges, then the

other also converges. If one diverges (to���

), so does the other. For L � 0 �

�k uk eventually. For L � �

�k � uk

eventually. So the last two statements follow from the comparison test.)

Examples. Consider the convergence or divergence of the following series:

(1) ��k � 1

1

k2cos� 1k� (2) ��

k � 2

3k

k2 � 1(3) ��

k � 1

� k�

1

k2 �5k

(4) ��k � 1

sin� 1k� .

Solutions. (1) Since 0 1

k2cos� 1k� 1

k2and ��

k � 1

1

k2converges by p-test, ��

k � 1

1

k2cos� 1k� converges.

(2) Since 0 � 32� k 3k

k2 � 1for k � 2 and ��

k � 2

� 32� k diverges by the geometric series test, ��

k � 2

3k

k2 � 1diverges.

(3) When k is large,� k

�1

k2 �5k

� � k

k2� 1

k3 � 2 . We compute limk � �

�k 1

k2 5k�k

k2

� limk � �

�k

�1

k

k2

k2 �5k

� 1 � Since ��k � 1

1

k3 � 2converges by p-test, ��

k � 1

� k�

1

k2 �5k

converges by the limit comparison test.

(4) When k is large,1

kis close to 0, so sin

� 1k� is close to

1

kbecause lim� � 0

sin ��

� 1 (i.e. sin � � � as � � 0).

We compute limk � �

sin � 1k �

1k

� lim� � 0

sin ��

� 1. Since ��k � 1

1k

diverges by p-test, ��k � 1

sin� 1k� diverges by the limit

comparison test.

For series with alternate positive and negative terms, we have the following test.

Theorem (Alternating Series Test). If ck decreases to 0 as k � �(i.e. c1 � c2 � c3 � � � ��� 0 and lim

k � � ck� 0),

then ��k � 1

� � 1 � k 1ck� c1

� c2�

c3� c4

�c5

� � � � converges.

14

Page 15: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(Reason. Since c1 � c2 � c3 � � � � � 0, we have 0 S2 S4 S6 � � � S5 S3 S1 � Since limn � � � Sn

� Sn � 1 � �lim

n � � cn� 0, the distances between the partial sums decrease to 0 and so lim

n � � Sn must exist.)

Examples. Both ��k � 2

� � 1 � k

k ln kand ��

k � 1

e � k cos k converge by the alternating series test because as k � �� k ln k � �

and ek � �� so 1 � � k ln k � � 0 and e � k � 0 and cos k � � � 1 � k �

For series with arbitrary positive or negative term, we have the following tests.

Theorem (Absolute Convergence Test). If ��k � 1

�ak � converges, then ��k � 1

ak converges.

(Reason. From � �ak � ak �ak � � we get 0 ak� �ak � 2 �ak � � Since ��

k � 1

2 �ak � converges, so by the comparison test,

��k � 1

� ak� �ak � � converges. Then ��

k � 1

ak� ��

k � 1

� ak� �ak � � � ��

k � 1

�ak � converges.)

Definition. We say ��k � 1

ak converges absolutely iff ��k � 1

�ak � converges. We say ��k � 1

ak converges conditionally iff ��k � 1

ak

converges, but ��k � 1

�ak � diverges.

Examples. Determine if the following series converge absolutely or conditionally

(a) ��k � 1

cos k

k3(b) ��

k � 1

cos k 1

�k

.

Solutions. (a) ��k � 1

����cos k

k3���� ��

k � 1

1k3

. Since ��k � 1

1k3

converges by p-test, it follows that ��k � 1

����cos k

k3����

converges by the

comparison test. So ��k � 1

cos k

k3converges absolutely by the absolute convergence test.

(b) ��k � 1

����cos k 1

�k����

� ��k � 1

11

�k

because cos k � � � 1 � k .� �

1

dx

1�

x� ln � 1 �

x � �����1

� � � ��k � 1

11

�k

diverges.

However,1

1�

kdecreases to 0 as k � ���

. So by the alternating series test, ��k � 1

cos k 1

�k

� ��k � 1

� � 1 � k 11

�k

converges. Therefore ��k � 1

cos k 1

�k

converges conditionally.

Theorem (Ratio Test). If ak�� 0 for every k and lim

k � � �ak 1 � ak � exists, then

limk � � ����

ak 1

ak����

���������� ���������

1 � ��k � 1

ak converges absolutely

� 1 � ��k � 1

ak may converge � e � g � ��k � 1

1

k2� or diverge � e � g � ��

k � 1

1

k�

� 1 � ��k � 1

ak diverges

15

Page 16: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(Sketch of reason. Let r � limk � � ����

ak 1

ak����, then for k large, ����

ak 1

ak����, ����

ak 2

ak 1����, � � � , ����

ak n

ak n � 1����

�r, so �ak n � � �ak � rn and

�ak � � �ak 1 � � �ak 2 � � � � � � �ak � � 1 �r

�r2 �

r3 � � � � � which converges if r 1 by the geometric series test and�ak n

� � �akrn diverges if r � 1 by the term test.)

Theorem (Root Test). If limk � �

k

� �ak � exists, then

limk � �

k

� �ak �

���������� ���������

1 � ��k � 1

ak converges absolutely

� 1 � ��k � 1

ak may converge � e � g � ��k � 1

1k2� or diverge � e � g � ��

k � 1

1k�

� 1 � ��k � 1

ak diverges

(Sketch of reason. Let r � limk � �

k

� �ak � , then for k large, k� �ak � �r � So �ak � �

rk ,� �ak � � �

rk .)

Examples. Consider the convergence or divergence of the following series:

(1) ��k � 1

13k � 2k

(2) ��k � 1

k!kk

.

Solutions. (1) Since limk � �

13k � 1 � 2k � 1

13k � 2k

� limk � �

3k � 2k

3k 1 � 2k 1� lim

k � �13

� � 23 � k 1

3

1 � � 23 � k 1

� 1

3 1 � by the ratio test, ��

k � 1

1

3k � 2k

converges. Alternatively, since limk � �

k

�1

3k � 2k� lim

k � �1

k� 3k � 2k

� limk � �

1

3 k�

1 � � 23 � k

� 1

3 1, by the root

test, ��k � 1

1

3k � 2kconverges.

(2) Since limk � �

� k �1 � !

� k �1 � k 1

kk

k!� lim

k � �1

� 1 � 1k � k

� 1

e 1 � by the ratio test, ��

k � 1

k!

kkconverges.

Remarks. You may have observed that in example (1), the limit you got for applying the root test was the same as thelimit you got for applying the ratio test. This was not an accident!

Theorem. If ak � 0 for all k and limk � �

ak 1

ak

� r � � then limk � �

k� ak� r � (This implies that the root test can be

applied to more series than the ratio test.)

Examples. (1) Let ak� k � then lim

k � �ak 1

ak

� limk � �

k�

1

k� 1 � So, lim

k � �k� k � 1 �

(2) Let ak� k!

kk� then lim

k � �ak 1

ak

� 1

eas above. So lim

k � �k� ak

� limk � �

k� k!

k� 1

e� i.e. when k is large, k!

� � k

e� k

which is a simple version of what is called Stirling’s formula. It is useful in estimating n! when n is large. For

example, since log10100

e�

1 � 566 � so100

e�

101 � 566� then we get 100!

�10156 � 6

� which has about 157 digits.

Theorem (Summation by Parts). Let S j�

j�k � 1

ak� a1

�a2

� � � � �aj and � bk

� bk 1� bk

� k �1 � � k

� bk 1� bk � then

n�k � 1

akbk� Snbn

� n � 1�k � 1

Sk � bk �

16

Page 17: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(Reason. Note a1� S1 and ak

� Sk� Sk � 1 for k � 1. So,

n�k � 1

akbk� S1b1

� � S2� S1 � b2

� � � � � � Sn� Sn � 1 � bn

� Snbn� S1 � b2

� b1 � � � � � � Sn � 1 � bn� bn � 1 � � �

Example. Show that ��k � 1

sin k

kconverges.

Let ak� sin k and bk

� 1

k� Using the identity sin m sin

1

2� 1

2� cos � m � 1

2� � cos � m � 1

2� � � we have

Sk� sin 1

�sin 2

� ����� �sin k � cos 1

2� cos � k � 1

2 �2 sin 1

2

This implies � Sk � 1

sin � 1 � 2 � for every k � Applying summation by parts and noting that limn � �

Sn

n� 0 � we get

��k � 1

sin k

k� lim

n � �n�

k � 1

sin k

k� lim

n � � �Sn

n� n � 1�

k � 1

Sk � 1k

�1

� 1k� � � ��

k � 1

Sk � 1k

� 1k

�1� �

Now ��k � 1

����Sk � 1

k� 1

k�

1� ����

1sin � 1 � 2 � ��

k � 1

� 1k

� 1k

�1� � 1

sin � 1 � 2 � by the telescoping series test. So by the

absolute convergence test, ��k � 1

sin k

k� ��

k � 1

Sk � 1

k� 1

k � 1� converges.

Complex Series

Complex numbers S1 � S2 � S3 � � � � with Sn� un

�i �

n are said to have limit limn � � Sn

� u�

i � iff limn � � un

� u

and limn � �

�n

� � � A complex series is a series where the terms are complex numbers. The definitions of convergent,

absolutely convergent and conditional convergent are the same. The remarks and the basic properties following thedefinitions of convergent and divergent series are also true for complex series.

The geometric series test, telescoping series test, term test, absolute convergence test, ratio test and root test are

also true for complex series. For zk� xk

�iyk � we have ��

k � 1

zk converges to z � x�

iy if and only if ��k � 1

xk converges

to x and ��k � 1

yk converges to y � So complex series can be reduced to real series for study if necessary.

Examples. (1) Note limn � � in �� 0 (otherwise 0 � lim

n � � � in � � limn � � 1 is a contradiction). So ��

k � 1

ik diverges by term test.

(2) If � z � 1 � then ���zk

k2 ��� 1

k2and ��

k � 1

1k2

converges by p-test implies ��k � 1

zk

k2converges absolutely. However, if

� z � � 1 � then limk � � ���

zk 1

� k �1 � 2

k2

zk ���� lim

k � �k2

� k �1 � 2

� z � � � z � � 1 implies ��k � 1

zk

k2diverges by the ratio test.

17

Page 18: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Inserting Parentheses and Rearrangements of Series.

Definition. We say ��k � 1

bk is obtained from ��k � 1

ak by inserting parentheses iff there is a strictly increasing function

p : � � � 0 � � � � � 0 � such that p � 0 � � 0 � b1� a1

� ����� �ap � 1 � � b2

� ap � 1 � 1� ����� �

ap � 2 � � b3� ap � 2 � 1

� ����� �ap � 3 � � � � � �

(Note bn is the sum of kn� p � n � � p � n � 1 � terms.)

Grouping Theorem. Let ��k � 1

bk be obtained from ��k � 1

ak by inserting parentheses. If ��k � 1

ak converges to s � then ��k � 1

bk

will converge to s � Next, if limn � � an

� 0 � kn is bounded and ��k � 1

bk converges to s � then ��k � 1

ak will converge to s �

(Reason. Let sn� n�

k � 1

ak and tn� n�

k � 1

bk � For the first part, ��k � 1

bk� lim

n � � tn� lim

n � �p � n ��k � 1

ak� s � For the second part,

let p � n � � p � n � 1 � be bounded by M � For a positive integer j � let p � i � j p � i �1 � � For r � 1 � 2 � � � � � M � define

cr � j ��

ap � i � r if p � i � �r j

0 if p � i � �r � j

� Then ��k � 1

ak� lim

j � � sj� lim

i � � ti�

limj � � � c1 � j � ����� �

cM � j � � s�

0� ����� �

0 � s � )

Examples. (1) Since ��k � 1

12k

� 12

� 14

� 18

� 116

� ����� converges to 1, so by the theorem,

1

2� � 1

4� 1

8� � � 1

16� 1

32� 1

64� � � 1

128� 1

256� 1

512� 1

1024� � ����� � 1 �

(2) � 1 � 1 � � � 1 � 1 � � ����� converges to 0, but 1 � 1�

1 � 1� ����� diverges by term test. So lim

n � � an� 0 is important.

Also, � 1 � 1 � � � 12

� 12

� 12

� 12 � � 1

3� 1

3� 1

3� 1

3� 1

3� 1

3 � ����� converges to 0. However, the series

without parentheses diverges (as Sn2� 1 and Sn2 n

� 0) even though the terms have limit 0. So kn bounded isimportant.

(3) Since � 1 � 1

2� � � 1

3� 1

4� � ����� � ��

j � 1

� 1

2 j � 1� 1

2 j� � ��

j � 1

1

2 j � 2 j � 1 � converges (by the limit compari-

son test with ��j � 1

1j 2

), so by the theorem, 1 � 12

� 13

� 14

� ����� � ��k � 1

� � 1 � k 1

kconverges to the same sum.

Definition. ��k � 1

bk is a rearrangement of ��k � 1

ak iff there is a bijection � : � � � such that bk� a � � k � �

Example. Given ln 2 � 1 � 12

� 13

� 14

� 15

� 16

� � � � (which converges conditionally). Consider the rearrangement

1� 1

3� ��� �2

� 1

2�������1 �

� 1

5� 1

7� ��� �2

� 1

4�������1 �

� 1

9� 1

11� ��� �2

� 1

6�������1 �

� � � � . Observe that

� 1 � 12 � � � 1

3� 1

4 � � � 15

� 16 � � � 1

7� 1

8 � � � � � � ln 2� 12

� 14

� 16

� 18 � � � � 1

2 ln 2

1� � 1

3� 1

2 � � 15

� � 17

� 14 � � � � � � 3

2 ln 2 �

Riemann’s Rearrangement Theorem. Let ak � and ��k � 1

ak converge conditionally. For any x �� or x � � �,

there is a rearrangement ��k � 1

a � � k � of ��k � 1

ak such that ��k � 1

a � � k � � x �

18

Page 19: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(Sketch of reason. Let pk��

ak if ak � 00 if ak 0

and qk��

0 if ak � 0�ak � if ak 0

� Then ak� pk

� qk and �ak � � pk�

qk �

Now both ��k � 1

pk � ��k � 1

qk must diverge to��� � (If both converges, then their sum ��

k � 1

�ak � will be finite, a contradiction.

If one converges and the other diverges to���

� then ��k � 1

ak� ��

k � 1

pk� ��

k � 1

qk will diverges to� �

� a contradiction

also.) Let un �

�n be sequences of real numbers having limits x and un �

n �

�1 � 0 � Now let P1 � P2 � � � � be the

nonnegative terms of ��k � 1

ak in the order they occur and Q1 � Q2 � � � � be the absolute value of the negative terms in

the order they occur. Since ��k � 1

Pk � ��k � 1

Qk differ from ��k � 1

pk � ��k � 1

qk only by zero terms, they also diverges to��� �

Let m1 � k1 be the smallest integers such that P1� ����� �

Pm1 � �1 and P1

� ����� �Pm1

� Q1� ����� � Qk1 u1 �

Let m2 � k2 be the smallest integers such that P1� ����� �

Pm1� Q1

� ����� � Qk1

�Pm1 1

� ����� �Pm2 � �

2 andP1

� ����� �Pm1

� Q1� ����� � Qk1

�Pm1 1

� ����� �Pm2

� Qk1 1� ����� � Qk2 u2 and continue this way. This is possible

since the sums of Pk and Qk are��� � Now if sn � tn are the partial sums of this series P1

� ����� �Pm1

� Q1� ����� � Qk1

� �����whose last terms are Pmn � Qkn � respectively, then � sn

� �n � Pmn and � tn

� un � Qkn by the choices of mn � kn � SincePn � Qn have limit 0, so sn � tn must have limit x � As all other partial sums are squeezed by sn and tn � the series weconstructed must have limit x � )

Dirichlet’s Rearrangement Theorem. If ak � and ��k � 1

ak converges absolutely, then every rearrangement ��k � 1

a � � k �converges to the same sum as ��

k � 1

ak .

(Reason. Define pk � qk as in the last proof. Since pk � qk �ak � � ��k � 1

pk � ��k � 1

qk converge, say to p and q � respectively.

Since a � � k � � p � � k � � q � � k � � we may view ��k � 1

p � � k � as a rearrangement of the nonnegative terms of ��k � 1

ak and inserting

zeros where a � � k � 0 � For any positive integer m � the partial sum sm� m�

k � 1

p � � k � ��k � 1

pk� p � Since pk � 0 � the

partial sum sm is also increasing, hence ��k � 1

p � � k � converges. Now, for every positive integer n �

n�k � 1

pk ��k � 1

p � � k � p �

As n � �� we get ��

k � 1

p � � k � � p � Similarly, ��k � 1

q � � k � � q � Then ��k � 1

a � � k � � p � q � ��k � 1

ak � )

Example. ��k � 1

� � 1

2� k � � 1

2� 1

22� 1

23

� 1

24� 1

25

� � � � converges (absolutely) to� 1

2

1 � � � 12 �

� � 1

3.

� 12

� 122

� 124

� 123� ��� �

2 terms

� 128

� 127

� 126

� 125� ��� �

4 terms

� 1216

� 1215

� 1214

� 1213

� 1212

� 1211

� 1210

� 129� ��� �

8 terms

� � � �

is a rearrangement of ��k � 1

� � 12

� k , so it also converges to � 13

.

Remarks. As a consequence of the rearrangement theorem, the sum of a nonnegative series is the same no matter howthe terms are rearranged.

19

Page 20: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 5. Real Numbers

Decimal representations and points on a line are possible ways of introducing real numbers, but they are not tooconvenient for proving many theorems. Instead we will introduce real numbers by its important properties.

Axiomatic Formulation. There exists a set (called real numbers) satisfying the following four axioms:

(1) (Field Axiom) is a field (i.e. has two operations�

and � such that for any a, b, c �� ,

(i) a�

b, a � b �� , (ii) a�

b � b�

a, a � b � b � a, (iii) � a �b � �

c � a� � b �

c � , � a � b � � c � a � � b � c � ,

(iv) there are unique elements 0, 1 � with 1 �� 0 such that a�

0 � a, a � 1 � a,

(v) for each x �� , there is a unique element � x �� such that x� � � x � � 0; if x �� 0 � then there is a unique

element x � 1 such that x � � x � 1 � � 1 �(vi) a � � b �

c � � a � b �a � c.)

(This axiom allows us to do algebra with equations. Define a � b to mean a� � � b ��� ab to mean a � b � a

b tomean a � � b � 1 � � Also, define 2 � 1

�1 � 3 � 2

�1 � � � � � )

(2) (Order Axiom) has an (ordering) relation such that for any a, b ��(i) exactly one of the following a b, a � b, b a is true,

(ii) if a b, b c, then a c,

(iii) if a b, then a�

c b�

c,

(iv) if a b and 0 c, then ac bc.

(This axiom allows us to work with inequalities. For example, using (ii) and (iii), we can see that if a band c d � then a

�c b

�d because a

�c b

�c b

�d � Also, we can get 0 1 (for otherwise 1 0

would imply by (iii) that 0 � 1� � � 1 � 0

� � � 1 � � � 1 � which implies by (iv) that 0 � � 1 � � � 1 � � 1 �

a contradiction). Now define a � b to mean b a � a b to mean a b or a � b � etc. Also, defineclosed interval [a � b] � � x : a x b ��� open interval � a � b � � � x : a x b ��� etc. Part (i) of theorder axiom implies any two real numbers can be compared. We define max � a1 � � � � � an � to be the maximumof a1 � � � � � an and similarly for minimum. Also, define � x � � max � x �

� x � � Then x � x � and � x � x � �

i.e. � � x �� x � x � � Next � x �� a if and only if x a and � x a � i.e. � a x a � Finally, adding� � x � x � x � and � � y � y � y � � we get � � x � � � y � x

�y � x � � � y � � which is the triangle inequality

� x �y � � x � � � y � � )

(3) (Well-ordering Axiom) � � � 1 � 2 � 3 � � � � � is a well-ordered subset of (i.e. for any nonempty subset S of � , thereis m � S such that m x for all x � S. This m is the least element (or the minimum) of S).

(This axiom allows us to formulate the principle of mathematical induction later.)

Definitions. For a nonempty subset S of , S is bounded above iff there is some M � such that x M for allx � S. Such an M is called an upper bound of S. The supremum or least upper bound (denoted by sup S or lub S)of S is an upper bound

�M of S such that

�M M for all upper bounds M of S.

Examples. (1) For S ��� 1n : n ���� ��� 1 �

12 �

13 � � � �� � the upper bounds of S are all M � 1 � So sup S � 1 � S �

(2) For S � � x �� : x 0 � , the upper bounds of S are all M � 0 � So sup S � 0 �� S �(4) (Completeness Axiom) Every nonempty subset of which is bounded above has a supremum in .

(This axiom allows us to prove results that have to do with the existence of certain numbers with specificproperties, as in the intermediate value theorem.)

Definitions. � � � 1 � 2 � 3 � 4 � � � � � is the natural numbers (or positive integers), � � � � � � �� 3 �

� 2 �� 1 � 0 � 1 � 2 � 3 � � � � �

is the integers, � � � mn : m � � and n ���� is the rational numbers and � � � � x �� : x �� ��� is the irrational

numbers.

Remarks (Exercises). The first three axioms are also true if is replaced by � � However, the completeness axiom isfalse for � � For example, S � � x : x � � � x 2 2 � � � � � 2 � � 2 � � � is bounded above in � � but it does not have asupremum in � �

20

Page 21: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

As above, we define S to be bounded below if there is some m �� such that m x for all x � S. Such an m iscalled a lower bound of S. The infimum or greatest lower bound (denoted by inf S or glb S) of S is a lower bound

�m

of S such that m �m for all lower bounds m of S.

lower boundsare here

upper boundsare hereS

inf S sup S

If S is bounded above and below, then S is bounded. (Note sup S, inf S may or may not be in S. Also, if S isbounded, then for all x � S � � x � M � max � � sup S � � � inf S � � � )Remarks (Exercises). (1) (Completeness Axiom for Infimum) Every nonempty subset of which is bounded belowhas an infimum in . This follows from considering � B � � � x : x � B � � (This is the reflection of B about 0.) If Bis bounded below, then � B is bounded above and inf B � � sup � � B � � Similarly, if B is bounded above, then � B isbounded below and sup B � � inf � � B � �

( )-B B

0inf(-B) sup(-B) supBinfA supA

AinfB

(2) For a set B, if it is bounded above and c � 0 � then let cB � � cx : x � B � � (This is the scaling of B by a factor ofc � ) We have sup cB � c sup B � If

� �� A � B, then inf B inf A whenever B is bounded below and sup A sup Bwhenever B is bounded above.

infB supB sup(c+B)inf(c+B)

c units

B c+B

(3) For c �� � let c�

B � � c �x : x � B � � (This is a translation of B by c units.) It follows that B has a supremum

if and only if c�

B has a supremum, in which case sup � c �B � � c

�sup B � The infimum statement is similar, i.e.

inf � c �B � � c

�inf B � More generally, if A and B are bounded, then letting A

�B � � x �

y : x � A � y � B � � wehave sup � A

�B � � sup A

�sup B and inf � A

�B � � inf A

�inf B �

Simple Consequences of the Axioms.

Theorem (Infinitesimal Principle). For x, y �� , x y���

for all� � 0 if and only if x y � (Similarly, y � � x

for all� � 0 if and only if y x.)

Proof. If x y, then for all� � 0, x y � y

�0 y

���by (iv) of the field axiom and (iii) of the order axiom.

Conversely, if x y���

for all� � 0 � then assuming x � y � we get x � y � 0 by (iii) of the order axiom. Let

0� x � y � then x � y

���0 � Since

0 � 0 � we also have x y���

0 � These contradict (i) of the order axiom. Sox y � The other statement follows from the first statement since y � � x is the same as y x

��� �Remarks. Taking y � 0, we see that � x � � for all

� � 0 if and only if x � 0. This is used when it is difficult toshow two expressions a � b are equal, but it may be easier to show �a � b � � for every

� � 0 �

Theorem (Mathematical Induction). For every n ��� , A � n � is a (true or false) statement such that A � 1 � is true andfor every k ��� , A � k � is true implies A � k �

1 � is also true. Then A � n � is true for all n ��� .

Proof. Suppose A � n � is false for some n � � . Then S � � n ��� : A � n � is false � is a nonempty subset of � . By thewell-ordering axiom, there is a least element m in S � Then A � m � is false. Also, if A � n � is false, then m n. Takingcontrapositive, this means that if n m � then A � n � is true.

21

Page 22: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Now A � 1 � is true, so m �� 1 and m ��� imply m � 2. So m � 1 � 1 � Let k � m � 1 ��� � then k � m � 1 mimplies A � k � is true. By hypothesis, A � k �

1 � � A � m � is true, a contradiction.

Theorem (Supremum Property). If a set S has a supremum in and� � 0 � then there is x � S such that

sup S � � x sup S.

Proof. Since sup S � � sup S, sup S � � is not an upper bound of S. Then there is x � S such that sup S � � x .Since sup S is an upper bound of S, x sup S. Therefore sup S � � x sup S.

Theorem (Infimum Property). If a set S has an infimum in and� � 0 � then there is x � S such that inf S

��� �x � inf S.

Proof. Since inf S� � � inf S, inf S

� �is not a lower bound of S. Then there is x � S such that inf S

� � � x . Sinceinf S is a lower bound of S, x � inf S. Therefore inf S

��� � x � inf S.

Theorem (Archimedean Principle). For any x �� , there is n ��� such that n � x .

Proof. Assume there exists x �� such that for all n ��� � we have n x � Then � � � n: n ���� has an upperbound x . By the completeness axiom, � has a supremum in . By the supremum property, there is n ��� such thatsup � � 1 n, which yields the contradiction sup � n

�1 �� .

Question. How is � contained in ? How is � � contained in ?

Below we will show that � is “dense” in in the sense that between any two distinct real numbers x � y, no matterhow close, there is a rational number. Similarly, � � is “dense” in � First we need a lemma.

Lemma. For every x �� � there exists a least integer greater than or equal to x � (In computer science, this is calledthe ceiling of x and is denoted by

�x � � ) Similarly, there exists a greatest integer less than or equal to x � (This is denoted

by [x] � In computer science, this is also called the floor of x and is denoted by � x � � )Proof. By the Archimedean principle, there is n ��� such that n � � x � � Then � n x n � By (iii) of the order axiom,0 x

�n 2n � The set S � � k ��� : k � x

�n � is a nonempty subset of � because 2n � S � By the well-ordering

axiom, there is a least positive integer m � x�

n � Then m � n is the least integer greater than or equal to x � So theceiling of every real number always exist.

Next, to find the floor of x � let k be the least integer greater than or equal to � x � then � k is the greatest integerless than or equal to x �Theorem (Density of Rational Numbers). If x y, then there is m

n ��� such that x mn y.

Proof. By the Archimedean principle, there is n ��� such that n � 1 � � y � x � � So ny � nx � 1 and hence nx�

1 ny �Let m � [nx]

�1 � then m � 1 � [nx] nx [nx]

�1 � m � So nx m nx

�1 ny � i.e. x m

n y �

Theorem (Density of Irrational Numbers). If x y, then there is � �� � � such that x �� y.

Proof. Let � 0 � � � . By the density of rational numbers, there is mn ��� such that x� �

0� m

n y� �0� � (If m

n� 0 �

then pick another rational number between 0 and y� �0� � So we may take m

n�� 0 � ) Let � � m

n � � 0 � � then � �� � � andx �� y �

Examples. (1) Let S � � � �� 3 � � � 4 � 5] � then S is not bounded below and so S has no infimum. On the other hand,

S is bounded above by 5 and every upper bound of S is greater than or equal to 5 � S � So sup S � 5 �(2) Let S � � 1

n : n ���� � � 1 �12 �

13 �

14 � � � ��� � In the examples following the definition of supremum, we saw sup S � 1 �

Here we will show inf S � 0 � (Note 0 �� S.) Since 1n � 0 for all n � � � 0 is a lower bound of S. So by the

completeness axiom for infimum, inf S must exist. Assume S has a lower bound t � 0 � By the Archimedeanprinciple, there is n �� such that n � 1 � t � Then t � 1 � n � S � a contradiction to t being a lower bound of S � So0 is the greatest lower bound of S �

(3) Let S � [2 � 6 � � � � Since 2 x 6 for every x � S � S has 2 as a lower bound and 6 as an upper bound. Wewill show inf S � 2 and sup S � 6 � (Note 2 � S and 6 �� S.) Since 2 � S � so every lower bound t satisfy t 2 �Therefore inf S � 2 � For supremum, assume there is an upper bound u 6 � Since 2 � S � so 2 u � By thedensity of rational numbers, there is a r ��� such that u r 6 � Then r � [2 � 6 � � � � S � As u r contradictsu being an upper bound of S � so every upper bound u 6 � Therefore, sup S � 6 �

22

Page 23: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 6. Limits

Limit is the most important concept in analysis. We will first discuss limits of sequences, then limits of functions.

Definitions. An (infinite) sequence in a set S (e.g. S � or S � [0 � 1]) is a list x1 � x2 � x3 � � � � of elements of S in aspecific order. Briefly it is denoted by � xn � � (Mathematically it may be viewed as a function x : � � S with x � n � � xn

for n ��� � ) We say the sequence � xn � is bounded above iff the set � x1 � x2 � x3 � � � ��� is bounded above. (Bounded belowand bounded sequences are defined similarly.) We will also write sup � xn � for the supremum of the set � x1 � x2 � x3 � � � ���and inf � xn � for the infimum of the set � x1 � x2 � x3 � � � � � �

CAUTION: Since we seldom talk about a set with one element from now on, so notations like � xn � will denotesequences unless explicitly stated otherwise.

For x � y �� � the distance between x and y is commonly denoted by d � x � y � � which equals � x � y � � Below we willneed a quantitative measure of what it means to be “close” for a discussion of the concept of limit. For

� � 0 � the openinterval � c � �

� c��� � is called the

-neighborhood of c � Note x � � c � �� c

��� � if and only if d � x � c � � � x � c � � �

i.e. every number in � c � �� c

��� � has distance less than�

from c �

Limit of a sequence � xn � is often explained by saying it is the number the xn’s are closer and closer to as n getslarger and larger. There are two bad points about this explanations.

(1) Being close or large is a feeling! It is not a fact. It cannot be proved by a logical argument.

(2) The effect of being close can accumulate to yield large separation! If two numbers having a distance less than orequal to 1 are considered close, then 0 is close to 1 and 1 is close to 2 and 2 is close to 3, � � � � 99 is close to 100,but 0 is quite far from 100.

So what is the meaning of close? How can limit be defined so it can be checked? Intuitively, a sequence � xn � getsclose to a number x if and only if the distance d � xn � x � goes to 0. This happens if and only if for every positive

� thedistance d � xn � x � eventually becomes less than

� � The following example will try to make this more precise.

Example. As n gets large, intuitively we may think xn� 2n2 � 1

n2 �1

gets close to 2. For� � 0 � 1 � how soon (that is, for

what n) will the distance d � xn � 2 � be less than�

? (What if� � 0 � 01? What if

� � 0 � 001? What if�

is an arbitrarypositive number?)

Solution. Consider d � xn � 2 � � ����2n2 � 1n2 �

1� 2 ����

� 3n2 �

1 � � Solving for n � we get n2 � � 3 � � � � 1 � If

� � 0 � 1 � then

n � � 29 � So as soon as n � 6 � the distance between xn and 2 will be less than� � 0 � 1 �

(If� � 0 � 01 � then n � � 299 � So n � 18 will do. If

� � 0 � 001 � then n � � 2999 � So n � 55 will do.If 0 � 3 � then n � [ � � 3 � � � � 1]

�1 will do. If

� � 3 � then since 3n2 1 3 � for every n � � � so

n � 1 will do. So for every� � 0 � there is a K � � so that as soon as n � K � the distance d � xn � 2 � will be less

than� � ) Note the value of K depends on the value of

� � the smaller�

is, the larger K will be. (Some people write K �

to indicate K depends on� � )

Definition. A sequence � xn � converges to a number x (or has limit x) iff for every� � 0, there is K �� such that for

every n � K � it implies d � xn � x � � � xn� x � � (which means xK � xK 1 � xK 2 � � � � � � x � �

� x��� � � �

Let us now do a few more examples to illustrate how to show a sequence converges by checking the definition.Later, we will prove some theorems that will help in establishing convergence of sequences.

Examples. (1) Let �n

� c � For every� � 0, let K � 1, then n � K implies � �

n� c � � 0 � . So � �

n � converges to c �

(2) Let � n� c � 1

n� For every

� � 0, there exists an integer K � 1� (by the Archimedean principle). Then n � K

implies � � n� c � � 1

n 1

K � . So � � n � converges to c �

23

Page 24: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(3) Let xn� n

� cos n � � n� Show that � xn � converges to � 1 by checking the definition.

Solution. For every� � 0 � there exists an integer K � 1

� 1� by the Archimedian principle. Then n � K implies

��n

� cos n � � n� � � 1 � ��

� ��cos n

� cos n � � n �� 1

n � 1 1

K � 1 � � So � xn � converges to � 1 �

(4) Let zn� n1 � n � Show that � zn � converges to 1 by checking the definition.

Solution. (Let un� � zn

� 1 � � zn� 1 � By the binomial theorem,

n � znn

� � 1 �un � n � 1

�nun

� n � n � 1 �2

u2n

� ����� �un

n � n � n � 1 �2

u2n

so that un �

2n � 1

� ) For every� � 0 � there exists integer K � 1

� 2� 2

(by the Archimedean principle). Then n � K

implies � zn� 1 � � un

�2

n � 1

�2

K � 1 � � So � zn � converges to 1.

Remarks. (i) From the definition, we see that � xn � converges to x � � xn� x � converges to 0 and � � xn

� x � � convergesto 0 are equivalent because in the definition, � xn

� x � is the same as � � xn� x � � 0 � � � � xn

� x � � 0 � �(ii) To show � xn � converges to x means for every

� � 0 � we have to find a K as in the definition or show such a Kexists. On the other hand, if we are given that � xn � converges to x , then for every

� � 0 � (which we can evenchoose for our convenience,) there is a K as in the definition for us to use.

Theorem (Uniqueness of Limit). If � xn � converges to x and y, then x � y (and so we may write limn � � xn

� x).

Proof. For every� � 0, we will show � x � y � � . (By the infinitesimal principle,we will get x � y.) Let

0� � � 2 � 0 �

By the definitionof convergence, there are K1, K2 ��� such that n � K1 � � xn� x � � 0 and n � K2 � � xn

� y � � 0.Let K � max � K1 � K2 � . By the triangle inequality, � x � y � � � � x � x K � � � xK

� y � � � x � xK � � � xK� y � � 0 � �

0� � .

Boundedness Theorem. If � xn � converges, then � xn � is bounded.

Proof. Let limn � � xn

� x . For� � 1, there is K ��� such that n � K � � xn

� x � 1 � � xn � � � � xn� x � �

x � 1� � x � .

Let M � max � � x1 � , � � � , � xK � 1 � , 1� � x � � � then for every n ��� � � xn � M (i.e. xn � [ � M � M]).

Remarks. The converse is false. The sequence � � � 1 � n � is bounded, but not convergent. In general, bounded sequencesmay or may not converge.

Theorem (Computation Formulas for Limits). If � xn � converges to x and � yn � converges to y, then

(i) � xn�

yn � converges to x�

y, respectively, i.e. limn � � � xn

�yn � � lim

n � � xn�

limn � � yn,

(ii) � xn yn � converges to x y, i.e. limn � � � xn yn � � � lim

n � � xn � limn � � yn ,

(iii) � xn � yn � converges to x � y, provided yn�� 0 for all n and y �� 0.

Proof. (i) For every� � 0, there are K1 � K2 �� such that n � K1 � � xn

� x � � � 2 and n � K2 � � yn� y � � � 2.

Let K � max � K1 � K2 � � Then n � K implies n � K1 and n � K2 � So for these n’s,

� � xn�

yn � � � x �y � � � � � xn

� x � � � yn� y � � � xn

� x � � � yn� y � � � 2

��� � 2 � � �

(ii) We prove a lemma first.

Lemma. If � an � is bounded and limn � � bn

� 0, then limn � � anbn

� 0.

Proof. Since � an � is bounded, there is M such that �an � M for all n. For every� � 0, since

� � M � 0 and� bn � converges to 0, there is K ��� such that n � K � �bn

� 0 � � � M � �anbn� 0 � M �bn � � .

24

Page 25: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

To prove (ii), we write xn yn� x y � xn yn

� xn y�

xn y � x y � xn � yn� y � �

y � xn� x � . Since � xn � converges,

� xn � is bounded by the boundedness theorem. So by (i) and the lemma,

limn � � xn yn

� limn � � � xn yn

� x y � �lim

n � � x y � limn � � xn � yn

� y � �lim

n � � y � xn� x � �

x y � 0�

0�

x y � x y �

(iii) Note 12 � y � � 0 � Since � yn � converges to y � there is K0 � � such that n � K0 implies � yn

� y � 12 � y � � By

the triangle inequality, � y � � � yn � � yn� y � 1

2 � y � � 12 � y � � yn � for n � K0 � Then for every n � � �

� yn � � m � min � � y1 � � � � � � � yK0 � 1 � �12 � y � � � 0 �

Next we will show limn � �

1yn

� 1y

(then by (ii), limn � �

xn

yn

� limn � � � xn

1yn

� � x1y

� x

y). For every

� � 0 � let�

0� m � y � � � 0 � Since lim

n � � yn� y �� 0, there is K ��� such that n � K � � yn

� y � � 0 � Then

n � K � ����1yn

� 1y����

� � y � yn �� yn � � y �

0

m � y �� � �

Remarks. (1) As in the proofs of the uniqueness of limit and part (i) of the computation formulas, when wehave n � K1 � �an

� a � � 1 and n � K2 � �bn� b � � 2 � we may as well take K � max � K1 � K2 � to say

n � K � �an� a � � 1 and �bn

� b � � 2 from now on.

(2) By mathematical induction, we can show that the computation formulas also hold for finitely many sequences.However, the number of sequences must stay constant as the following example shows

1 � limn � � �

1

n� ����� � 1

n� ��� �n terms

�� limn � �

1

n� ����� �

limn � �

1

n� 0

� ����� �0 � 0 �

Sandwich Theorem (or Squeeze Limit Theorem). If xn �� n yn for all n ��� and limn � � xn

� limn � � yn

� z � then

limn � � � n

� z.

Proof. For any� � 0, there is K such that n � K � � xn

� z � � and � yn� z � � , i.e. xn , yn � � z � �

� z� � � . Since

xn �� n yn � so � n � � z � �� z

��� � , i.e. � � n� z � � .

Example. Let � n� [10n � 2]

10n��� for every n ��� � (Note � 1

� 1 � 4 � � 2� 1 � 41 � � 3

� 1 � 414 � � 4� 1 � 4142 � � � � � )

Then10n � 2 � 1

10n�� n 10n � 2

10n� � 2 � Since lim

n � �10n � 2 � 1

10n� � 2 � by the sandwich theorem, lim

n � � � n� � 2 �

Theorem (Limit Inequality). If an � 0 for all n �� and limn � � an

� a, then a � 0.

Proof. Assume a 0. Then for� � �a � , there is K ��� such that n � K � �an

� a � � � �a � , which impliesa � � an a

��� � a� � � a � � 0, a contradiction.

Remarks. By the limit inequality above, if xn yn , limn � � xn

� x , limn � � yn

� y, then taking an� yn

� xn � 0 � we get

the limit y � x � 0 � i.e. x y � Also, if a xn b and limn � � xn

� x � then limn � � a � a lim

n � � xn� x lim

n � � b � b �

i.e. xn � [a � b] and limn � � xn

� x imply x � [a � b] � (This is false for open intervals as 1n � � 0 � 2 � � lim

n � �1n

� 0 �� � 0 � 2 � � )

Supremum Limit Theorem. Let S be a nonempty set with an upper bound c � There is a sequence � � n � in S convergingto c if and only if c � sup S.

Proof. If c � sup S � then for n �� , by the supremum property, there is � n � S such that c � 1n

� sup S � 1n �� n

sup S � c. Since limn � � � c � 1

n� � lim

n � � c � c, the sandwich theorem implies limn � � � n

� c � sup S.

Conversely, if a sequence � � n � in S converges to c � then � n sup S implies c � limn � � � n sup S � Since c is an

upper bound of S � so sup S c � Therefore c � sup S �

25

Page 26: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Infimum Limit Theorem. Let S be a nonempty set with a lower bound c � There is a sequence � � n � in S convergingto c if and only if c � inf S.

Examples. (1) Let S � � 1n : n ���� � � 1 �

12 �

13 � � � ��� � Since 0 1

n for all n ��� � 0 is a lower bound of the set S � Nowthe sequence � 1

n � in S converges to 0 � By the infimum limit theorem, inf S � 0 �(2) Let S � � x � 1

y : x ��� � � 0 � 1] � y � [1 � 2] � � Since 12 x � 1

y for all x ��� � � 0 � 1] and y � [1 � 2] �12 is a

lower bound of S � Now the sequence � 1n � 1

2 � in S converges to 12 � By the infimum limit theorem, inf S � 1

2 �(3) Let A and B be bounded in � Prove that if A � 2B � � a � 2b : a � A � b � B � � then sup � A � 2B � � sup A � 2 inf B �

Solution. Since A and B are bounded, sup A and inf B exist by the completeness axiom. For x � A � 2B � we havex � a � 2b for some a � A and b � B � So x � a � 2b sup A � 2 inf B � Hence sup A � 2 sup B is an upper boundfor A � 2B � By the supremum limit theorem, there is a sequence an � A such that � an � converges to sup A � By theinfimum limit theorem, there is a sequence bn � B such that � bn � converges to inf B � Then � an

� 2bn � is a sequence inA � 2B and � an

� 2bn � converges to sup A � 2 inf B by the computation formulas for limits. By the supremum limittheorem, therefore sup � A � 2B � � sup A � 2 inf B �

Definition. A subsequence of � xn � is a sequence � xnj � , where n j �� and n1 n2 n3 � � � .Examples. For the sequence x1 � x2 � x3 � x4 � x5 � x6 � � � � � if we set n j

� j 2� we get the subsequence x1 � x4 � x9 � x16 � � � � �

If n j� 2 j

�1 � then we get the subsequence x3 � x5 � x7 � x9 � � � � � If n j is the j -th prime number, then we get the

subsequence x2 � x3 � x5 � x7 � � � � �Remarks. (1) Taking n j

� j � we see that every sequence is a subsequence of itself. A subsequence can also bethought of as obtained from the original sequence by throwing away possibly some terms. Also, a subsequence of asubsequence of � xn � is a subsequence of � xn � �(2) By mathematical induction, we have n j � j for all j �� because n1 � 1 and n j 1 � n j � j implies n j 1 � j

�1 �

Subsequence Theorem. If limn � � xn

� x, then limj � � xnj

� x for every subsequence � xnj � of � xn � . (The converse is

trivially true because every sequence is a subsequence of itself.)

Proof. For every� � 0, there is K ��� such that n � K � � xn

� x � � . Then j � K � n j � K � � xnj� x � � .

Question. How can we tell if a sequence converges without knowing the limit (especially if the sequence is given bya recurrence relation)?

For certain types of sequences, the question has an easy answer.

Definitions. � xn � is

��� �� increasingdecreasing

strictly increasingstrictly decreasing

� ���� iff

��� �� x1 x2 x3 � � �x1 � x2 � x3 � � � �x1 x2 x3 � � �x1 � x2 � x3 � � � �

� ���� � respectively. � xn � is

�monotone

strictly monotone � iff

� xn � is

�increasing or decreasing

strictly increasing or decreasing � � respectively.

Monotone Sequence Theorem. If � xn � is increasing and bounded above, then limn � � xn

� sup � xn � . (Similarly, if � xn �is decreasing and bounded below, then lim

n � � xn� inf � xn � .)

Proof. Let M � sup � xn � , which exists by the completeness axiom. By the supremum property, for any� � 0, there is

xK such that M � � xK M. Then j � K � M � � xK x j M � � x j� M � � M � x j � �

Remark. Note the completeness axiom was used to show the limit of xn exists (without giving the value).

Examples. (1) Let 0 c 1 and xn� c1 � n � Then xn 1 and cn 1 cn � xn

� c1 � n c1 � � n 1 � � xn 1 � So by themonotone sequence theorem, � xn � has a limit x � Now x2

2n� � c1 � 2n � 2 � c1 � n � xn � Taking limits and using the

subsequence theorem, we get x 2 � x � So x � 0 or 1. Since 0 c � x1 x � the limit x is 1. Similarly, if c � 1 �

then cn will decrease to the limit 1.

26

Page 27: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(2) Does

�2

��

2� � 2

� ����� represent a real number?

Here we have a nested radical defined by x1� � 2 and xn 1

� � 2�

xn � The question is whether � xn �converges to a real number x . (Computing a few terms, we suspect that � xn � is increasing. To find an upper bound,observe that if lim

n � � xn� x � then x � � 2

�x implies x � 2 � ) Now by mathematical induction, we can show that

xn xn 1 2 � (If xn xn 1 2 � then 2�

xn 2�

xn 1 4 � so taking square roots, we get xn 1 xn 2 2 � )By the monotone sequence theorem, � xn � has a limit x � We have x2 � lim

n � � x2n 1

� limn � � 2

�xn

� 2�

x � Then

x � � 1 or 2. Since � 2 � x1 x � so x � 2 �

Another common type of sequences is obtained by mixing a decreasing sequence and an increasing sequence intoone of the form a1 � b1 � a2 � b2 � a3 � b3 � � � � � In the next example, we will have such a situation and we need two theoremsto handle these kind of sequences.

Nested Interval Theorem. If In� [an � bn] is such that I1 � I2 � I3 � � � � , then � �n � 1

In� [a � b] � where

a � limn � � an lim

n � � bn� b � If lim

n � � � bn� an � � 0, then � �n � 1

In contains exactly one number.

[ ][ [ ] ]a1 a2 a3 b3 b2 b1...

Proof. I1 � I2 � I3 � � � � implies � an � is increasing and bounded above by b1 and � bn � is decreasing and boundedbelow by a1. By the monotone sequence theorem, � an � converges to a � sup � an � and � bn � converges to b � inf � bn � .Since an bn for every n ��� , taking limits, we have an a b bn . Consequently, x � [an � bn] (i.e. an x bn)

for all n if and only if limn � � an

� a x b � limn � � bn � So � �n � 1

In� [a � b]. If 0 � lim

n � � � bn� an � � b � a � then a � b

and � �n � 1In

� � a � �

Remarks. Note in the proof, the monotone sequence theorem was used. So the nested interval theorem also implicitlydepended on the completeness axiom.

Intertwining Sequence Theorem. If � x2m � and � x2m � 1 � converge to x � then � xn � also converges to x �Proof. For every

� � 0 � since � x2m � converges to x � there is K0 ��� such that m � K0 � � x2m� x � � � Since � x2m � 1 �

also converges to x � there is K1 � � such that m � K1 � � x2m � 1� x � � � Now if n � K � max � 2K0 � 2K1

� 1 � �

then either n � 2m � 2K0 � � xn� x � � � x2m

� x � � or n � 2m � 1 � 2K1� 1 � � xn

� x � � � x2m � 1� x � � �

Example. Does1

1� 1

1� �����

represent a number?

Here we have a continued fraction defined by x1� 1 and xn 1

� 1 � � 1 �xn � � We have x1

� 1 � x2� 1 � 2 � x3

�2 � 3 � x4

� 3 � 5 � � � � � Plotting these on the real line suggests 1 � 2 x2n x2n 2 x2n 1 x2n � 1 1 for all n ��� �This can be easily established by mathematical induction. (If 1 � 2 x2n x2n 2 x2n 1 x2n � 1 1 � then1

�x2n 1

�x2n 2 1

�x2n 1 1

�x2n � 1 � Taking reciprocal and applying the recurrence relation, we have

x2n 1 � x2n 3 � x2n 2 � x2n � Repeating these steps once more, we get x2n 2 x2n 4 x2n 3 x2n 1 � )Let In

� [x2n � x2n � 1] � then I1 � I2 � I3 � ����� � Now

� xm� xm 1 � � ����

1

1�

xm � 1

� 1

1�

xm����

� � xm � 1� xm �

� 1 �xm � 1 � � 1 �

xm � � xm � 1� xm �

� 1 � 12 � � 1 � 1

2 �� 4

9� xm � 1

� xm � �

Using this, we get � x2n � 1� x2n �

49� x2n � 2

� x2n � 1 � 49

49� x2n � 3

� x2n � 2 � ����� 49 �����

49� ��� �

2n � 2

� x1� x2 � � � 4

9� 2n � 2 1

2�

By the sandwich theorem, limn � � � x2n � 1

� x2n � � 0 � So by the nested interval theorem, � �n � 1In

� � x � for some

27

Page 28: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

x and limn � � x2n

� x � limn � � x2n � 1 � By the intertwining sequence theorem, lim

n � � xn� x � So x � lim

n � � xn 1�

limn � � 1 � � 1 �

xn � � 1 � � 1 �x � � Then x � � � 1

� � 5 � � 2 � Since x � I1 � x � � � 1� � 5 � � 2 �

(Instead of estimating the lengths of the In’s and squeezing them to 0 to see their intersection is a single point,we can also do the following. Let In

� [x2n � x2n � 1] be as above so that I1 � I2 � I3 ������� � By the nested interval

theorem, � �n � 1In

� [a � b] � where a � limn � � x2n and b � lim

n � � x2n � 1 � Taking limits on both sides of x2n 1� 1

1�

x2nand

x2n� 1

1�

x2n � 1� we get b � 1

1�

aand a � 1

1�

b� Then b � 1 �

a � � 1 � a � 1 �b � � which yields b

�ab � a

�ab �

so a � b � Hence � �n � 1In is a single point.)

Back to answering the question above in general, French mathematician Augustine Cauchy (1789–1857) intro-duced the following condition.

Definition. � xn � is a Cauchy sequence iff for every� � 0, there is K ��� such that m, n � K implies � xm

� xn � � .

Remark. Roughly, the condition says that the terms of these sequences are getting closer and closer to each other.

Example. Let xn� 1

n2� (Note that if m � n � K � say m � n � then we have � xm

� xn � � 1

n2� 1

m2 1

n2 1

K 2� )

For every� � 0 � we can take an integer K � 1� � (by the Archimedean principle). Then m � n � K implies

� xm� xn �

1

K 2 � � So � xn � is a Cauchy sequence.

Theorem. If � xn � converges, then � xn � is a Cauchy sequence.

Proof. For every� � 0 � since lim

n � � xn� x � there is K ��� such that j � K � � x j

� x � � � 2. For m, n � K , we

have � xm� xn � � xm

� x � � � x � xn � � � 2��� � 2 � � . So � xn � is a Cauchy sequence.

The converse of the previous theorem is true, but it takes some work to prove that. The difficulty lies primarilyon how to come up with a limit of the sequence. The strategy of showing every Cauchy sequence in must convergeis first to find a subsequence that converges, then show that the original sequence also converge to the same limit.

Theorem. If � xn � is a Cauchy sequence, then � xn � is bounded.

Proof. Let� � 1 � Since � xn � is a Cauchy sequence, there is K ��� such that m � n � K � � xm

� xn � � � 1 �In particular, for n � K � � x K

� xn � 1 � � xn � � � � xn� xK � �

xK � � xn� xK � � � xK � 1

� � xK � � LetM � max � � x1 � � � � � � � xK � 1 � � 1

� � xK � � � then for all n �� � � xn � M (i.e. xn � [ � M � M] � �

Bolzano-Weierstrass Theorem. If � xn � is bounded, then � xn � has a subsequence � xnj � that converges.

Proof. (Bisection Method) Let a1� inf � xn � , b1

� sup � xn � and I1� [a1 � b1]. Let m1 be the midpoint of I1. If there

are infinitely many terms of � xn � in [a � m1], then let a2� a1, b2

� m1 and I2� [a2 � b2]. Otherwise, there will be

infinitely many terms of � xn � in [m1 � b1], then let a2� m1, b2

� b1 and I2� [a2 � b2]. For k � 2, 3, 4, � � � , repeat

this bisection on Ik to get Ik 1. We have Ij� [aj � bj] and I1 � I2 � I3 � � � � . By the nested interval theorem, since

limj � � � bj

� aj � � limj � �

b1� a1

2 j � 1� 0, � �n � 1

In contains exactly one number x .

Take n1� 1 � then xn1

� x1 � I1. Suppose n j is chosen with xnj � Ij . Since there are infinitely many terms xn inIj 1, choose n j 1 � n j and xnj � 1 � Ij 1. Then lim

j � � � xnj� x � lim

j � � � bj� aj � � 0. Therefore, lim

j � � xnj� x .

Remarks. In the proof, the nested interval theorem was used, so the Bolzano-Weierstrass theorem depended on thecompleteness axiom.

Alternate Proof. We will show every sequence � xn � has a monotone subsequence. (If � xn � is bounded, then themonotone sequence theorem will imply the subsequence converges.)

28

Page 29: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Call xm a peak of � xn � if xm � xk for all k � m � If � xn � has infinitely many peaks, then we order the peaks bystrictly increasing subscripts m1 m2 m3 ����� � By the definition of a peak, xm1 � xm2 � xm3 � ����� � So � xm j �is a decreasing subsequence of � xn � � On the other hand, if � xn � has only finitely many peaks xm1 � ����� � xmk � then letn1

� max � m1 � ����� � mk � �1 � Since xn1 is not a peak, there is n2 � n1 such that xn2 � xn1 � Inductively, if xnj is not a

peak, there is n j 1 � n j with xnj � 1 � xnj � So � xnj � is a strictly increasing subsequence of � xn � �Remarks. This alternate proof used the monotone sequence theorem, so it also depended on the completeness axiom.

Cauchy’s Theorem. � xn � converges if and only if � xn � is a Cauchy sequence.

Proof. The ‘only if’ part was proved. For the ‘if’ part, since � xn � is a Cauchy sequence, � xn � is bounded. By theBolzano-Weierstrass theorem, � xn � has a subsequence � xnj � that converges in , say lim

j � � xnj� x .

We will show limn � � xn

� x . For every� � 0, since � xn � is a Cauchy sequence, there is K1 � � such that m,

n � K1 � � xm� xn � � � 2. Since lim

j � � xnj� x , there is K2 � � such that j � K2 � � xnj

� x � � � 2. If

n � J � max � K1 � K2 � , then n J � J � K1, J � K2 and � xn� x � � xn

� xnJ �� � xnJ

� x � � � 2��� � 2 � � .

Example. Does the sequence � xn � converge, where x1� sin 1 and xk

� xk � 1� sin k

k2for k � 2 � 3 � 4 � � � � ?

We will check the Cauchy condition. For m � n � xm� xn

� m�k � n 1

� xk� xk � 1 � � m�

k � n 1

sin k

k2and

� xm� xn �

1

� n �1 � 2

� ����� � 1

m2 1

n � n �1 �

� ����� � 1

� m � 1 � m� � 1

n� 1

n�

1� � ����� � � 1

m � 1� 1

m� � 1

n� 1

m 1

n�

So for� � 0 � by the Archimedean principle, there is K � 1 � � � Then m � n � K � � xm

� xn � 1K � � Therefore, by

the Cauchy theorem, � xn � converges.

Limits of Functions

Let S be an interval (or more generally a set). Suppose f : S � is a function. We would like to define limx � x0

f � x � .

For this to be a meaningful expression, in the interval case, x0 must be a point on the interval or an endpoint. In thegeneral case, x0 must be “approachable” by the points of S.

CONVENTION. In discussing the limit of a function f : S � at x0 � x0 is always assumed to be the limit of somesequence in S � � x0 � so that x0 can be approached by points of S � (We say x0 is an accumulation (or limit or cluster)point of S iff x0 is the limit of a sequence � xn � in S � � x0 � � )

(x

L

S0

y = f x)(

()x0S S

y = f (x)

LLet f : S � be a function. To say lim

x � x0f � x � � L

roughly means the distance between f � x � and L is as smallas we please when x � S is sufficiently close to x0.(Again,small and close need to be clarified.)

Example. Let f � x � � � x3 � 3x2 � � � x � 3 � � If x gets close to 3 and x �� 3 � then f � x � � x 2 should be close to 9. Inother words, the distance d

�f � x � � 9 � � � f � x � � 9 � goes to 0 when the distance d � x � 3 � � � x � 3 � gets small. So for

every positive�

� the distance � f � x � � 9 � will soon or later be less than�

when the distance � x � 3 � becomes smallenough. For

� � 0 � 1 � how small d � x � 3 � be (that is, for what x) will make d�

f � x � � 9 � � � f � x � � 9 � � ? Equivalently,we are seeking a positive � so that 0 � x � 3 � �� � � f � x � � 9 � � �

Now � f � x � � 9 � ��� 8 � 9 f � x � 9 � 1 � So 2 � 9833� � 8 � 9 x � 9 � 1 �

3 � 0166 and � 0 � 0167 x � 3 0 � 0166 � If we take � � 0 � 016 � then 0 � x � 3 � �� � � f � x � � 9 � � �

29

Page 30: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(For small� � 0 � � f � x � � 9 � � � 9 � � f � x � 9

��� � � 9 � � � 3 x � 3 � 9��� � 3 � So we may

take � � min � � 9��� � 3 � 3 � � 9 � � � � then 0 � x � 3 � �� � � f � x � � 9 � � � )

Following the example, we are ready to state the precise definition of the limit of a function.

Definition. Let f : S � be a function. We say f � x � converges to L (or has limit L) as x tends to x0 in S iff forevery

� � 0, there is � � 0 such that for every x � S � 0 � x � x0 � � implies � f � x � � L � � . This is denoted bylim

x � x0x � S f � x � � L (or lim

x � x0f � x � � L in short.)

In the definition, � depends on�

and x0 � For different�

(or different x0), � will be different. If a limit value exists,then it is unique. The proof is similar to the sequential case and is left as an exercise for the readers.

Examples. (1) For g : [0 �

� � � defined by g � x � � � x � show that limx � 0

g � x � � 0 and limx � 4

g � x � � 2 by checking

the�

- � definition.

Solution. For every� � 0 � let � � � 2 � Then for every x � [0 �

� � � 0 � x � 0 � �� implies � g � x � � 0 � � � x � � � � �This takes care of the checking for the first limit.

For the second limit, note that � � x � 2 � � � x � 4 �� x�

2 � x � 4 �

2� For every

� � 0 � let � � 2� � Then for every

x � [0 �

� � � 0 � x � 4 � �� implies � g � x � � 2 � � x � 4 �2

�2

� � �

(2) Let f : � � 0 � � be defined by f � x � � 15x

� Show that limx � 2

f � x � � 110

by checking the�

- � definition.

Solution. (Note that ����1

5x� 1

10����

� � x � 2 �10x

� x � 2 �10

for x � � 1 � 3 � � ) For every� � 0 � take � � min � 1 � 10

� � � Then

� 1 and � 10� � For every x � � � 0 � � 0 � x � 2 � �� implies x � � 1 � 3 � � So ����

f � x � � 110���� � x � 2 �

10 �

10 �

and we are done.

Notation: We will write xn � x0 in S � � x0 � to mean sequence � xn � in S � � x0 � converges to x0 �

Sequential Limit Theorem. limx � x0x � S f � x � � L if and only if for every xn � x0 in S � � x0 � , lim

n � � f � xn � � L �

In quantifier symbols (� � for any, for every, for all, � � there is, there exists),

limx � x0x � S f � x � � L

� � � � � 0 � � � � 0 such that�

x � S � 0 � x � x0 � �� � � f � x � � L � � �

limx � x0x � S f � x � �� L

� � � � � 0 such that�� � 0 � � x � S � 0 � x � x0 � �� and � f � x � � L � � � �

Proof. If limx � x0x � S f � x � � L � then for every

� � 0, there is � � 0 such that for x � S, 0 � x � x0 � �� � � f � x � � L � � .

If limn � � xn

� x0 and xn�� x0, then there is K � � such that n � K � 0 � xn

� x0 � � � � � f � xn � � L � � � . So

limn � � f � xn � � L .

Conversely, if limx � x0x � S f � x � �� L � then there is

� � 0 such that for every � � 0 � there is x � S with 0 � x � x0 � ��

and � f � x � � L � � � . Now, setting � � 1n � there is xn � S with 0 � xn

� x0 � � � 1n and � f � xn � � L � � � . By the

sandwich theorem, xn � x0 in S � � x0 � � So limn � � f � xn � � L � Then 0 � lim

n � � � f � xn � � L � � � , a contradiction.

30

Page 31: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Remarks. If limn � � f � xn � exists for every xn � x0 in S � � x0 � � then all the limit values are the same. To see this, suppose

xn � x0 and � n � x0 in S � � x0 � � Then the intertwining sequence � zn � � � x1 � � 1 � x2 � � 2 � x3 � � 3 � � � ��� converges to x0

in S � � x0 � � Since � f � xn � � and � f � � n � � are subsequences of the convergent sequence � f � zn � � , they have the same limit.

Consequences of Sequential Limit Theorem.

(1) (Computation Formulas) If f � g : S � are functions, limx � x0x � S f � x � � L1 and lim

x � x0x � S g � x � � L2 � then

limx � x0x � S � f � x �

��� ���

���

� ���� g � x � � L1

��� ������

� ���� L2

� limx � x0x � S f � x �

��� ������

� ���� lim

x � x0x � S g � x �

respectively (in the case of division, provided g � x � �� 0 for every x � S and L 2�� 0).

Proof. Since f � x � and g � x � have limits L 1 and L2 � respectively, as x tends to x0 in S � by the sequential limittheorem, f � xn � and g � xn � will have limits L 1 and L2 � respectively, for every xn � x0 in S � � x0 � � By thecomputation formulas for sequences, the limit of f � xn � �

g � xn � is L1�

L2 for every xn � x0 in S � � x0 � � By thesequential limit theorem, f � x � �

g � x � has limit L 1�

L2 as x tends to x0 in S � Replacing�

by �� � � � � we get the

proofs for the other parts.

Alternate Proof. If limx � x0x � S f � x � � L1 and lim

x � x0x � S g � x � � L2 � then for every

� � 0 � there are � 1 � 0 such that for

every x � S � 0 � x � x0 � �� 1 implies � f � x � � L1 � �

2and � 2 � 0 such that for every x � S � 0 � x � x0 � �� 2

implies � g � x � � L2 � �

2� Now we take � � min � � 1 � � 2 � � 0 so that � � 1 and � � 2 � Then, for every x � S �

0 � x � x0 � �� implies both 0 � x � x0 � �� 1 and 0 � x � x0 � �� 2 so that

���

f � x � �g � x � � � � L1

�L2 � ��

� � � f � x � � L1 � � � g � x � � L2 � � � f � x � � L1 � � � g � x � � L2 � �

2� �

2� � �

The other parts of the computation formulas can be proved by adapting the arguments for the sequential case.

(2) (Sandwich Theorem or Squeeze Limit Theorem) If f � x � g � x � h � x � for every x � S and limx � x0x � S f � x � � L �

limx � x0x � S h � x � � then lim

x � x0x � S g � x � � L �

(3) (Limit Inequality) If f � x � � 0 for all x � S and limx � x0x � S f � x � � L � then L � 0 �

The proofs of (2) and (3) can be done by switching to sequences like the first proof of (1) or by adapting thearguments of the sequential cases and checking the

- � definition like the alternate proof of (1).

Next we will discuss one-sided limits.

Definitions. For f : � a � b � � and x0 � � a � b � � the left hand limit of f at x0 is f � x0� � � lim

x � x�

0

f � x � � limx � x0

x � � a � x0 � f � x � .

The right hand limit of f at x0 is f � x0� � � lim

x � x �0

f � x � � limx � x0

x � � x0 � b � f � x � �

Theorem. For x0 � � a � b � � limx � x0

x � � a � b � f � x � � L if and only if f � x0� � � L � f � x0

� � .

Proof. If limx � x0

f � x � � L � then for every� � 0 � there is � � 0 such that for x � � a � b � � 0 � x � x0 � � �

� f � x � � L � � � In particular, for x � � a � x0 � � 0 � x � x0 � � � � f � x � � L � � � So f � x0� � � L � Similarly, for

x � � x0 � b � � 0 � x � x0 � �� � � f � x � � L � � � So f � x0� � � L �

31

Page 32: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Conversely, if f � x0� � � L � f � x0

� � � then for every� � 0 � there is � 1 � 0 such that for x � � a � x0 � �

0 � x � x0 � �� 1 � � f � x � � L � � and there is � 2 � 0 such that for x � � x0 � b � � 0 � x � x0 � �� 2 � � f � x � � L � � �As � � min � � 1 � � 2 � �� 1 and � 2 � we have for x � � a � b � � 0 � x � x0 � �� � � f � x � � L � � � So lim

x � x0f � x � � L �

Definitions. A function f : S � is

��� �� increasingdecreasing

strictly increasingstrictly decreasing

� ���� on S iff for every x , y � S, x y �

��� �� f � x � f � y �f � x � � f � y �f � x � f � y �f � x � � f � y �

� ���� �

Also, f is

�monotone

strictly monotone � on S iff f is

�increasing or decreasing

strictly increasing or strictly decreasing � on S � respectively.

For a nonempty subset S0 of S � we say f is

�bounded abovebounded below

bounded

�on S0 iff � f � x � : x � S0 � is

�bounded abovebounded below

bounded

��

respectively. If The set S0 is not mentioned, then it is the domain S �

For monotone functions, the following theorem is analogous to the monotone sequence theorem. It will be used

in the next chapter to prove the continuous inverse theorem, which will be used to prove thedx

dy� 1 � dy

dxrule later.

Also, it will be used again in the chapter on integration.

Monotone Function Theorem. If f is increasing on � a � b � , then for every x0 � � a � b � , f � x0� � � sup � f � x � : a

x x0 � and f � x0� � � inf � f � x � : x0 x b � and f � x0

� � f � x0 � f � x0� � . If f is bounded below, then

f � a � � � inf � f � x � : a x b � � If f is bounded above, then f � b � � � sup � f � x � : a x b � � Also f has countablymany discontinuous points on � a � b � � i.e. J � � x0 : x0 � � a � b � � f � x0

� � �� f � x0� � � is countable. (The theorem is

similarly true for decreasing functions and all other kinds of intervals.)

Proof. If a x x0 b, then f � x � f � x0 � . So M � sup � f � x � : a x x0 � f � x0 � � By the supremumproperty, for every

� � 0 � there is c � � a � x0 � such that M � � f � c � M � If we let � � x0� c � then

�x � � a � x0 � � 0 � x � x0 � �� � c � x0

� � x x0 � f � c � f � x � M � � f � x � � M � M � f � c � � �

So limx � x

0

f � x � � M f � x0 � . Similarly, f � x0 � limx � x �

0

f � x � � inf � f � x � : x0 x b � . In the case f is bounded

below or above, the proof of the existence of f � a � � or f � b � � is similar.

q (x1)

q(x0)

x x0 1

Next let x0 � � a � b � with f � x0� � f � x0

� � . By the density of rationalnumbers, we may choose a q � x0 ��� � between f � x0

� � and f � x0� � . The

function q: J � � is injective because if f is discontinuous at x0, x1 withx0 x1, then

q � x0 � f � x0� � f

� x0�

x1

2� f � x1

� � q � x1 � �By the injection theorem, J is countable, i.e. f has countably many discontinuouspoints on � a � b � � The cases of decreasing functions or other kinds of intervals aresimilar.

Appendix: Infinite Limits and Limit at Infinity

We begin with a definition of a sequence having���

as limit. In this case, the sequence does not have any upperbound in � i.e. the sequence will pass any fixed r � eventually and keep on going. Sequences with � � as limitand functions with

� �limit are defined similarly.

Definitions. (1) A sequence � xn � diverges to���

(or has limit���

) iff for every r �� � there is K �� (depending onr) such that n � K implies xn � r � Similarly, a sequence � xn � diverges to � � (or has limit � � ) iff for every r �� �

there is K ��� (depending on r) such that n � K implies xn r �

32

Page 33: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(2) A function f : S � diverges to���

(or has limit���

) as x tends to x0 in S iff for every r �� � there is� � 0 (depending on r and x0) such that for every x � S � 0 � x � x0 � � implies f � x � � r � Similarly, a functionf : S � diverges to � � (or has limit � � ) as x tends to x0 in S iff for every r �� � there is � � 0 (depending on rand x0) such that for every x � S � 0 � x � x0 � �� implies f � x � r �

Limit at infinity for functions are defined similarly as for sequences as follow.

Definitions. (1) Let f : S � be a function such that���

is an accumulation point of S (i.e. there is a sequencein S diverges to

��� � ) We say f � x � converges to L (or has limit L) as x tends to���

in S iff for every� � 0 � there

is K �� such that for every x � S � x � K implies � f � x � � L � � � Similarly, let f : S � be a function suchthat � � is an accumulation point of S (i.e. there is a sequence in S diverges to � � � ) We say f � x � converges to L(or has limit L) as x tends to � � in S iff for every

� � 0 � there is K �� such that for every x � S � x K implies� f � x � � L � � �

(2) Let f : S � be a function such that���

is an accumulation point of S (i.e. there is a sequence in Sdiverges to

��� � ) We say f � x � diverges to���

(or has limit���

) as x tends to���

in S iff for every r �� � there isK �� such that for every x � S � x � K implies f � x � � r � Similarly, let f : S � be a function such that � � isan accumulation point of S (i.e. there is a sequence in S diverges to � � � ) We say f � x � diverges to

���(or has limit���

) as x tends to � � in S iff for every r �� � there is K �� such that for every x � S � x K implies f � x � � r �By replacing f � x � � r with f � x � r � we get the definitions of limit equal to � � as x tends to

� �in S �

It is good exercises for the readers to formulate the computation formulas and properties for these limits, whichcan be proved by checking these definitions just like the finite cases.

33

Page 34: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 7. Continuity

Definitions. A function f : S � is continuous at x0 � S iff limx � x0x � S f � x � � f � x0 � � i.e. for every

� � 0, there is � � 0

such that for all x � S, � x � x0 � � � � f � x � � f � x0 � � � . For E � S � we say f is continuous on E iff f iscontinuous at every element of E � Also, we say f is continuous iff f is continuous on the domain S �

Sequential Continuity Theorem. f : S � is continuous at x0 � S if and only if for every xn � x0 in S �

limn � � f � xn � � f � x0 � � f � lim

n � � xn � �

Proof. Just replace L by f � x0 � , 0 � x � x0 � �� by � x � x0 � �� and xn � x0 in S � � x0 � by xn � x0 in S (i.e. deletethe xn

�� x0 requirement) in the proof of the sequential limit theorem.

Example. It is easy to give examples of continuous functions, such as polynomials. Here is an example of a function

not continuous at any point. Let f � x � ��

1 if x ���0 if x ���� � then f is discontinuous at every x �� !

Reason. For every x0 � � n ��� , by the density of rational numbers and irrational numbers, there are rn ��� , sn����

in � x0� 1

n � x0 � � Now rn � x0, sn � x0, but limn � � f � rn � � 1 and lim

n � � f � sn � � 0. So limx � x0

f � x � cannot exist.

Theorem. If f � g : S � are continuous at x0 � S � then f�

g � f g � f � g (provided g � x0 � �� 0) are continuous at x0 �Proof. Since f � g are continuous at x0 � lim

x � x0� f

�g � � x � � lim

x � x0f � x � �

limx � x0

g � x � � f � x0 � �g � x0 � � � f

�g � � x0 � �

So f�

g is continous at x0 by definition. The subtraction, multiplication and division cases are similar.

Theorem. If f : S � is continuous at x0, f � S � � S � � g : S � � is continuous at f � x0 � , then g � f is continuousat x0.

Proof. By the sequential continuity theorem, all we need to show is that limn � � � g � f � � xn � � � g � f � � x0 � for every

sequence � xn � converging to x0 � Since f is continuous at x0, by the sequential continuity theorem, the limit of f � xn �is f � x0 � � Since g is continuous at f � x0 � � so lim

n � � g � f � xn � � � g � limn � � f � xn � � � g � f � x0 � � �

In the discussions below, S will denote an interval of positive length.

Theorem (Sign Preserving Property). If g: S � is continuous and g � x0 � � 0, then there is an interval I �� x0

� � � x0� � � with � � 0 such that g � x � � 0 for every x � S � I � (The case g � x0 � 0 is similar by considering � g.)

Proof. Let� � g � x0 � � 0 � Since g is continuous at x0 � there is � � 0 such that for x � S � � x � x0 � � �

� g � x � � g � x0 � � � � So x � S � � x0� � � x0

� � � � 0 � g � x0 � � � g � x � �Intermediate Value Theorem. If f : [a � b] � is continuous and y0 is between f � a � and f � b � inclusive, then thereis (at least one) x0 � [a � b] such that f � x0 � � y0.

Proof. The cases y0� f � a � or y0

� f � b � are trivial as x0� a or b will do. We assume f � a � y0 f � b � . (The

other possibility f � a � � y0 � f � b � is similar.) The set S � � x � [a � b]: f � x � y0 � is nonempty (a � S) and has b asan upper bound. Then x0

� sup S � [a � b].

By the supremum limit theorem, there is a sequence � xn � in S such that limn � � xn

� x0. Note xn � [a � b] implies

x0 � [a � b] � Then f � x0 � � limn � � f � xn � y0 by continuity of f at x0. Assume f � x0 � y0. Then x0

�� b as y0 f � b � �Define g � x � � y0

� f � x � on [a � b]. Since g � x0 � � 0, by the sign preserving property, there is x1 � x0 such thatg � x1 � � 0 � Then f � x1 � y0. So x0 x1 � S � which contradicts x0

� sup S. Therefore, f � x0 � � y0.

Examples. (1) The equation x 5 �3x

�sin x � cos x

�10 has a solution. To see this, let f � x � � x 5 �

3x�

sin x �cos x � 10 � then f � 0 � � � 11 and 26 f � 2 � 30 � Since f is continuous, the intermediate value theorem impliesf � x � � 0 for some x � � 0 � 2 � �

34

Page 35: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(2) Suppose p � x � � xn �a1xn � 1 � ����� �

an with n odd. Let x0� 1

� �a1 � � ����� � �an � � 1 � then we havep � x0 � � xn

0�

a1xn � 10

� ����� �an and p � � x0 � � � xn

0�

a1xn � 10

� ����� �an � So xn

0� p � x0 � � � a1xn � 1

0� ����� � an

and xn0

�p � � x0 � � a1xn � 1

0� ����� �

an � By the triangle inequality,

xn0

� p � x0 �xn

0�

p � � x0 � � �a1xn � 10 � � ����� � �an � �a1 � xn � 1

0� ����� � �an � xn � 1

0� � �a1 � � ����� � �an � � xn � 1

0 xn0 �

Then p � x0 � � 0 and p � � x0 � 0 � By the intermediate value theorem, there is a root of p � x � between � x0 and x0 �

a x w b0 0

Extreme Value Theorem. If f : [a � b] � is continuous, then there are x0,� 0 � [a � b] such that f � � 0 � f � x � f � x0 � for every x � [a � b] � So the rangeof f is f

�[a � b] � � [ f � � 0 � � f � x0 � ] � In particular, f is bounded on [a � b]

In this case, we write f � x0 � � sup � f � x � : x � [a � b] � � maxx � [a � b]

f � x � and

f � � 0 � � inf � f � x � : x � [a � b] � � minx � [a � b]

f � x � �

Proof. We first show f � [a � b] � � � f � x � : x � [a � b] � is bounded above. Assume it is not bounded above. Then eachn ��� is not an upper bound. So there is zn � [a � b] such that f � zn � � n � By the Bolzano-Weierstrass theorem, � zn �in [a � b] has a subsequence � znj � converging to some z0 � [a � b] � Since f is continuous at z0 � lim

j � � f � znj � � f � z0 � �

which implies � f � znj � � is bounded by the boundedness theorem. However, f � znj � � n j � j implies � f � znj � � is notbounded, a contradiction.

By the completeness axiom, M � sup � f � x � : x � [a � b] � exists. By the supremum limit theorem, there is asequence � xn � in [a � b] such that lim

n � � f � xn � � M. By the Bolzano-Weierstrass theorem, � xn � has a subsequence � xnk �such that lim

k � � xnk� x0 in [a � b]. Since f is continuous at x0 � [a � b], M � lim

k � � f � xnk � � f � x0 � by the subsequence

theorem and the sequential continuity theorem, respectively.

Similarly, inf � f � x � : x � [a � b] � � f � � 0 � for some � 0 � [a � b].

The following two theorems are for explaining thedx

dy� 1 � dy

dxrule in the next chapter.

Continuous Injection Theorem. If f is continuous and injective on [a � b], then f is strictly monotone on [a � b] andf�[a � b] � � [ f � a � � f � b � ] or [ f � b � � f � a � ]. (This is true for any nonempty interval in place of [a � b] � The range is an

interval with f � a � � f � b � as endpoints.)

f (y)

f (b)

f (a)

a y b

Proof. Since f is injective, either f � a � f � b � or f � a � � f � b � . Supposef � a � f � b � . Let y � � a � b � . Then f � y � cannot be greater than f � b � , otherwiseby the intermediate value theorem, there is ��� � a � y � such that f � � � � f � b � ,contradicting injectivity. Similarly, f � y � cannot be less than f � a � . Therefore,f � a � f � y � f � b � . If a x y b, then similarly f � a � f � x � f � y � f � b � , i.e. f is strictly increasing on [a � b] and f

�[a � b] � � [ f � a � � f � b � ]

by the intermediate value theorem.The case f � a � � f � b � leads to f strictly decreasing on [a � b].

Continuous Inverse Theorem. If f is continuous and injective on [a � b], then f � 1 is continuous on f�[a � b] � .

Proof. By the last theorem, f is strictly monotone on [a � b]. We first suppose f is strictly increasing. Then f � 1 isalso strictly increasing on f

�[a � b] � � [ f � a � � f � b � ].

For y0 � � f � a � � f � b � ], let r � f � 1 � y0� � � lim

y � y�

0

f � 1 � y � � which exists by the monotone function theorem.

Since f � 1 � y ��� [a � b] � r � [a � b] � Let � yn � be a sequence in � f � a � � y0 � converging to y0 � then � n� f � 1 � yn �

will converge to r by sequential limit theorem. Since f is continuous at r � by the sequential continuity theorem,f � r � � lim

n � � f � � n � � limn � � f � f � 1 � yn � � � y0 � Then f � 1 � y0 � � r � f � 1 � y0

� � . Similarly, if y0 � [ f � a � � f � b � � , then

f � 1 � y0� � � f � 1 � y0 � . So f � 1 is continuous on [ f � a � � f � b � ] � f

�[a � b] � . The case f is strictly decreasing is similar.

35

Page 36: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Appendix: Fundamental Theorem of Algebra

The extreme value theorem is often used in estimating integrals. More precisely, if f : [a � b] � is continuous,

then for m � minx � [a � b]

f � x � and M � maxx � [a � b]

f � x � � we have m f � x � M and m � b � a � � b

af � x � dx M � b � a � �

Here we will mention that there is a version of the extreme value theorem for continuous functions defined onclosed disks of finite radius on the plane. As an application of this fact, we can sketch a proof of the fundamentaltheorem of algebra, which asserts that every nonconstant polynomial with complex coefficients must have a root.

Let p � z � � zn �a1zn � 1 � ����� �

an and m � inf � � p � z � � : z � � � � We have

� zn � � ���p � z � � n�

k � 1

akzn � k ��� � p � z � � � n�

k � 1

�ak � � z � n � k � � p � z � � � � z � n � n�k � 1

�ak � � z � n � k � � z � n � 1 � n�k � 1

�ak � � z � � k

� ��� �� 1 � as � z � � �

So for � z � large, 1 � n�k � 1

�ak � � z � � k � 1

2� For a very large K � n� 2m � we have � z � � K implies

� p � z � � � � z � n � 1 � n�k � 1

�ak � � z � � k � 12� z � n � 1

2K n � m �

Let DK be the closed disk � z : � z � K � � We have m � inf � � p � z � � : z � � � � inf � � p � z � � : z � DK � � � p � z0 � � for somez0 � DK by the extreme value theorem, since � p � z � � is continuous on DK �

We claim m � � p � z0 � � � 0 � Suppose m �� 0 � Then f � z � � p � z �z0 � � p � z0 � is a polynomial of degree n and

f � 0 � � 1 � So f � z � � 1�

b1z� ����� �

bnzn for some b1 � � � � � bn with bn�� 0 � Let k be the smallest positive integer such

that bk�� 0 � Then f � z � � 1

�bkzk � ����� �

bnzn � Since � p � z �z0 � � � m � � p � z0 � � � we get � � � � f � z � � � 1 for all z �

Introduce the notation ei � � cos� �

i sin�

� which describes the points on the unit circle. Since the absolute valueof � �bk � � bk is 1, so � �bk � � bk

� ei � for some� � Let � � � � k � then eik

�bk

� ei � bk� � �bk � � Considering � � rei

�with r �bk � � 1 � k � � 1 � rk �bk � � 0 � � we have �1 �

bk � k � � �1 �bkrk eik

�� � ��

1 � rk �bk � ��� 1 � rk �bk � and

� f � � � � � �1 �bk � k � ����� �

bn � n � �1 �bk � k � � �bk 1 � k 1 � � ����� � �bn � n �

� 1 � rk �bk � � �bk 1 � rk 1 � ����� � �bn � rn

� 1 � rk��bk � � �bk 1 � r � ����� � �bn � rn � k� ��� �

� �bk� � 1

2

�bk� �

0 as r � 0 � �

Taking � � rei�

with a very small positive r � we have � f � � � � 1 � r k � 12 �bk � � 1 � a contradiction to � � � � Therefore

m � � p � z0 � � � 0 and z0 is a root of p � z � �

36

Page 37: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 8. Differentiation.

Definitions. Let S be an interval of positive length. A function f : S � is differentiable at x 0 � S iff f � � x0 � �

limx � x0x � S f � x � � f � x0 �

x � x0exists in . Also, f is differentiable iff f is differentiable at every element of S �

Theorem. If f : S � is differentiable at x0, then f is continuous at x0.

Proof. By the computation formulas for limits,

limx � x0

f � x � � limx � x0

� � f � x � � f � x0 �x � x0

� � x � x0 � �f � x0 ��� � f � � x0 � � 0 �

f � x0 � � f � x0 � �

Theorem (Differentiation Formulas). If f � g : S � are differentiable at x0, then f�

g, f � g, fg, f � g (wheng � x0 � �� 0) are differentiable at x0. In fact, � f

�g � � � x0 � � f � � x0 � �

g � � x0 � , � f g ��� � x0 � � f � � x0 � g � x0 � �f � x0 � g � � x0 �

and � f

g� � � x0 � � f � � x0 � g � x0 � � f � x0 � g � � x0 �

g � x0 � 2.

Proof. By the computation formulas for limits,

limx � x0

� f�

g � � x � � � f�

g � � x0 �x � x0

� limx � x0

�f � x � � f � x0 �

x � x0

� g � x � � g � x0 �x � x0

� � f � � x0 � �g � � x0 � �

limx � x0

� f g � � x � � � f g � � x0 �x � x0

� limx � x0

�f � x � � f � x0 �

x � x0g � x � �

f � x0 � g � x � � g � x0 �x � x0

� � f � � x0 � g � x0 � �f � x0 � g � � x0 � �

limx � x0

� f

g� � x � � � f

g� � x0 �

x � x0

� limx � x0

1g � x � g � x0 �

�f � x � � f � x0 �

x � x0g � x0 � � g � x � � g � x0 �

x � x0f � x0 ���

� f � � x0 � g � x0 � � f � x0 � g � � x0 �g � x0 � 2

Theorem (Chain Rule). If f : S � is differentiable at x0 � f � S � � S � and g : S � � is differentiable at f � x0 � ,then g � f is differentiable at x0 and � g � f ��� � x0 � � g � � f � x0 � � f � � x0 � .

Proof. The function

h � y � ��� � g � y � � g � f � x0 � �

y � f � x0 � if y �� f � x0 �g � � f � x0 � � if y � f � x0 �

is continuous at f � x0 � because limy � f � x0 � h � y � � g � � f � x0 � � � h � f � x0 � � . So g � y � � g � f � x0 � � � h � y � � y � f � x0 � � holds

for every y in the domain of g. Then

limx � x0

g � f � x � � g � f � x0 �x � x0

� limx � x0

h � f � x � � f � x � � f � x0 �x � x0

� h � f � x0 � � f � � x0 � � g � � f � x0 � � f � � x0 � �

Remarks. Note f differentiable at x0 does not imply f � is continuous at x0 � In fact, the function f � x � � x 2 sin 1x for

x �� 0 and f � 0 � � limx � 0

x2 sin1x

� 0 is continuous and differentiable with f � � x � � 2x sin 1x

� cos 1x for x �� 0 and

f � � 0 � � limx � 0

� x2 sin1

x� 0 � � x � 0 � However, f � is not continuous at 0 because lim

x � 0f � � x � � lim

x � 02x sin

1

x� cos

1

xdoes

not exist, hence not equal to f � � 0 � � In particular, f is not twice differentiable. (Also, the function g � x � � x 2 sin1

x2

for x �� 0 and g � 0 � � 0 is differentiable on � but g � � x � is not continuous at 0 and g � � x � is unbounded on every openinterval containing 0.)

37

Page 38: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Notations. C0 � S � � C � S � is the set of all continuous functions on S � For n � � � Cn � S � is the set of all functionsf : S � such that the n-th derivative f � n � is continuous on S. C � � S � is the set of all functions having n-th derivativesfor all n �� . Functions in C1 � S � are said to be continuously differentiable on S �

Inverse Function Theorem. If f is continuous and injective on � a � b � and f � � x0 � �� 0 for some x0 � � a � b � , then f � 1

is differentiable at y0� f � x0 � and � f � 1 ��� � y0 � � 1

f � � x0 � , i.e.dx

dy� 1 � dy

dx.

Proof. The function g � x � �

��� ��x � x0

f � x � � f � x0 � if x �� x0

1

f � � x0 � if x � x0

is continuous at x0 because limx � x0

g � x � � 1

f � � x0 �� g � x0 � �

Since f is continuous and injective on � a � b � � f � 1 is continuous by the continuous inverse theorem. So limy � y0

f � 1 � y � �

f � 1 � y0 � � x0 and � f � 1 � � � y0 � � limy � y0

f � 1 � y � � f � 1 � y0 �y � y0

� limy � y0

g�f � 1 � y ��� � g � x0 � � 1

f � � x0 � �

Local Extremum Theorem. Let f : � a � b � � be differentiable. If f � x0 � � minx � � a � b � f � x � or f � x0 � � max

x � � a � b � f � x � �

then f � � x0 � � 0 �

Proof. If f � x0 � � minx � � a � b � f � x � � then 0 lim

x � x �0

f � x � � f � x0 �x � x0

� f � � x0 � � limx � x

0

f � x � � f � x0 �x � x0

0 � So f � � x0 � � 0 �The other case is similar.

w0

x0

Rolle’s Theorem. Let f be continuous on [a � b] and differentiable on � a � b � . Iff � a � � f � b � then there is (at least one) z0 � � a � b � such that f � � z0 � � 0.

Proof. This is trivial if f is constant on [a � b]. Otherwise, by the extremevalue theorem, there are x0, � 0 � [a � b] such that f � x0 � � max

x � [a � b]f � x � �

minx � [a � b]

f � x � � f � � 0 � . Either f � x0 � �� f � a � or f � � 0 � �� f � a � . So either x0 or

� 0 is in � a � b � � By the last theorem, f � � x0 � � 0 or f � � � 0 � � 0 �

a x0 b

Mean-Value Theorem. Let f be continuous on [a � b] and differentiable on� a � b � . Then there exists x0 � � a � b � such that f � b � � f � a � � f � � x0 � � b � a � .

Proof. Define F � x � � f � x � � � f � b � � f � a �b � a

� x � a � �f � a ��� . Then F � a � �

0 � F � b � . By Rolle’s theorem, there is x0 � � a � b � such that 0 � F � � x0 � �

f � � x0 � � f � b � � f � a �b � a

, which is equivalent to f � b � � f � a � � f � � x0 � � b � a � �

Examples. (1) For a b � show � sin b � sin a � �b � a � � (By the mean-value theorem, there is x0 � � a � b � such thatsin b � sin a � � cos x0 � � b � a � � so � sin b � sin a � �b � a � � )

(2) Show � 1 �x � � � 1

� �x for x � � 1 and

� � 1 � (Let f � x � � � 1 �x � � � 1 � �

x � then there is x0 � � 0 � x � ifx � 0 or x0 � � x � 0 � if � 1 x 0 such that

� 1 �x � � � 1 � �

x � f � x � � f � 0 � � f � � x0 � � x � 0 � � � � � 1 �x0 � � � 1 � 1 � x � 0 � �

(3) Show ln x x � 1 for x � 0 � (Let f � x � � ln x � x�

1 � then f � 1 � � 0 � If x � 1 � then there is x0 � � 1 � x � suchthat

ln x � x�

1 � f � x � � f � x � � f � 1 � � f � � x0 � � x � 1 � � � 1x0

� 1 � � x � 1 � 0 �

The case 0 x 1 is similar.)

38

Page 39: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(4) To approximate � 16 � 1 � we can let f � x � � � x � Then f � 16 � 1 � � f � 16 � � f � � c � � 16 � 1 � 16 � for some cbetween 16 and 16 � 1. Now c

�16 and f � 16 � 1 � � f � 16 � �

f � � 16 � � 16 � 1 � 16 � � 0 � 0125 � which gives� 16 � 1 � f � 16 � 1 � �4 � 0125 �

Theorem. If

������� ������

f � � 0f � � 0f � 0f � 0f � �� 0f ��� 0

� ������������ everywhere on � a � b � , then f is

������� ������

increasingstrictly increasing

decreasingstrictly decreasing

injectiveconstant

� ������������ on � a � b � � respectively.

Proof. If x , y � � a � b � and x y, then by the mean value theorem, there is x0 � � x � y � such that f � y � � f � x � �

f � � x0 � � y � x �

������� ������

� 0� 0 0 0�� 0

� 0

� ������������ . Therefore,

������� ������

f � x � f � y �f � x � f � y �f � x � � f � y �f � x � � f � y �f � x � �� f � y �f � x � � f � y �

� ������������ �

Generalized Mean-Value Theorem. Let f , g be continuous on [a � b] and differentiable on � a � b � . Then there isx0 � � a � b � such that g � � x0 � � f � b � � f � a � � � f � � x0 � � g � b � � g � a � � . (Note the case g � x � � x is the mean-value theorem.)

Proof. Let F � x � � f � x � � g � b � � g � a � � � � f � b � � f � a � � � g � x � � g � a � � , then F � a � � f � a � � g � b � � g � a � � � F � b � . ByRolle’s theorem, there is x0 � � a � b � such that 0 � F � � x0 � � f � � x0 � � g � b � � g � a � � � g � � x0 � � f � b � � f � a � � �

Theorem (L’Hopital’s Rule). Let f , g be differentiable on � a � b � and g � x � , g � � x � �� 0 for x � � a � b � � Let limx � b �

f � x � �

0 � limx � b �

g � x � � If limx � b �

f � � x �g � � x �

� L � then limx � b �

f � x �g � x �

� L. (The rule is also true if x � a or L � � � � )

Sketch of Proof. Define f � b � � 0 and g � b � � 0 � Then f and g are continuous at b � So by the generalized mean value

theorem, there is x0 between x and b such thatf � x �g � x �

� f � x � � f � b �g � x � � g � b �

� f � � x0 �g � � x0 � � As x � b � � x0 will be squeezed to

b � andf � x �g � x �

� f � � x0 �g � � x0 � will be squeezed to L � The case x � a is similar.

Example. (Even if limx � a

f � x � � 0 � limx � a

g � x � and limx � a

f � � x �g � � x � does not exist, lim

x � a

f � x �g � x � may still exist as the following

example will show.) Let f � x � � x 2 sin1

x, g � x � � sin x , a � 0, then � x2 x2 sin

1

x x2 implies lim

x � 0f � x � � 0

by sandwich theorem, limx � 0

g � x � � 0, limx � 0

f � � x �g � � x �

� limx � 0

2x sin 1x

� cos 1x

cos x� lim

x � 0

0 � cos 1x

1does not exist. However,

limx � 0

f � x �g � x �

� limx � 0 � x

sin x � x sin1x

� � 1 � 0 � 0.

Taylor’s Theorem. Let f : � a � b � � be n times differentiable on � a � b � . For every x, c � � a � b � , there is x0 betweenx and c such that

f � x � � f � c � � f � � c �1!

� x � c � � f � � � c �2!

� x � c � 2 � � � � � f � n � 1 � � c �� n � 1 � !

� x � c � n � 1 � f � n � � x0 �n!

� x � c � n �

(This is called the n-th Taylor expansion of f about c � The term Rn � x � � f � n � � x0 �n!

� x � c � n is called the Lagrange

form of the remainder.)

Proof. Let I be the closed interval with endpoints x and c. For t � I , define g � t � � � n � 1 � !n � 1�k � 0

f � k � � t �k!

� x � t � k�

where f � 0 � � f � and p � t � � � � x � t � n

n� We have g � � t � � f � n � � t � � x � t � n � 1 and p � � t � � � x � t � n � 1 � By the generalized

39

Page 40: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

mean value theorem, there is x0 between x and c such that g � � x0 �� ��� �f

�n � � x0 � � x � x0 � n � 1

[p � x � � p � c �� ��� �� x � c � n � n ] � p � � x0 �� ��� �� x � x0 � n � 1

[ g � x �� ��� �� n � 1 � ! f � x �� g � c � ].

Solving for f � x � � we get f � x � � g � c �� n � 1 � !

� f � n � � x0 �n!

� x � c � n �

Appendix 1: L’Hopital’s Rule

The following is a more comprehensive version of L’Hopital’s rule.

L’Hopital’s Rule. Let f , g be differentiable on � a � b � and g � x � � g � � x � �� 0 for x � � a � b � � where � � a b ��� �Let lim

x � a �f � � x �g � � x �

� L � where � � L ��� � Then

� limx � a � f � x � � 0 � lim

x � a � g � x � or � limx � a � g � x � � ��� �� lim

x � a �f � x �g � x �

� L �

(Similarly, the rule is also true if x � b � or g � x � � � � in the latter case).

Proof. For the case � � L ���� consider r �� be such that L r � (If L �� � then consider r � L

��� � ) Let

s � such that L s r � Since limx � a �

f � � x �g � � x �

� L � there is c � � a � b � such that a x c impliesf � � x �g � � x � s � (If

a �� � then c may be taken to be a� � 0 � ) Now, for a x y c � by the generalized mean value theorem, there is

t � � x � y � such that � � � f � x � � f � y �g � x � � g � y �

� f � � t �g � � t � s �

If limx � a � f � x � � 0 � lim

x � a � g � x � � then letting x � a � we seef � y �g � y � s r for all y � � a � c � �

If limx � a � g � x � � ���

� then keeping y fixed, g � x � � g � y � for all x on some interval � a � c � � � Multiplyingg � x � � g � y �

g � x �on both sides of � � � � we get

f � x �g � x � s � s

g � y �g � x �

� f � y �g � x � for all x � � a � c � � � As x tends to a � the limit of the right side

is s � which is less than r � So,f � x �g � x � r on some interval � a � c � � � �

So in both cases, there exists d1 � a (d1� c in the former case and d1

� c � � in the latter case) such that

a � d1 impliesf � � �g � � � r � Similarly, for the case � � L ���

� consider r � �� such that r � L � (If

L �� � then consider r � � L � � � ) There is d2 such that a �� d2 implies r � f � � �g � � � � Taking d � min � d1 � d2 � (and

� � d � a in case a �� ), the definition of limx � a �

f � x �g � x �

� L is successfully verified.

Examples. (1) Show that limx � �

xr

ex� 0 for every r �� �

Solution. There is an integer n � � r � (by the Archimedian principle). We have x r xn on [1 �

��� � � So 0 xr

ex xn

ex

on [1 �

� � � Sincedn

dxnxn � n! and lim

x � �n!ex

� 0 � applying L’Hopital’s rule n times, we see limx � �

xn

ex� 0 � which

implies limx � �

xr

ex� 0 by the sandwich theorem.

(2) Let f : � a �

��� � � be differentiable. Show that limx � � f � � x � �

f � x � � 0 if and only if limx � � f � x � � 0 and

limx � � f � � x � � 0 � (This type of result is often needed in the studies of differential equations. Here if lim

x � � g � x � � 0 �

then every solution y � f � x � of the equationdy

dx�

y � g � x � will have limit 0 as x � ��� � )

40

Page 41: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Solution. The if part follows by the computation formulas for limit. For the only if part, consider f � x � � f � x � ex

ex� Note

the denominator on the right side tend to���

as x tend to��� � Since lim

x � ��

f � x � ex � �� ex � � � lim

x � � f � � x � �f � x � � 0 �

by L’Hopital’s rule, we get limx � � f � x � � lim

x � �f � x � ex

ex� 0 and lim

x � � f � � x � � limx � �

�f � � x � �

f � x � � � f � x � � 0 �

Appendix 2: Convex and Concave Functions.

tf (a) 1(+ - t ) f (b)

ta+(1- t ) b ba

f (b)

f(a)

convex function

y = f (x)

Definitions. Let I be an interval and f : I � , we say f is a convexfunction on I iff for every a, b � I , 0 t 1,

f�ta

� � 1 � t � b � t f � a � � � 1 � t � f � b � �We say f is a concave function on I if

f�ta

� � 1 � t � b � � t f � a � � � 1 � t � f � b � �

Remarks. As t ranges from 0 to 1, the point�ta

� � 1 � t � b � f � ta�

� 1 � t � b ��� traces the graph of y � f � x � for a x b, while the point�ta

� � 1 � t � b � t f � a � � � 1 � t � f � b � � traces the chord joining � a � f � a � �to � b � f � b � � . So f is convex on I if and only if the chord joining twopoints on the graph always lies above or on the graph.

(x, f (x))

(a, f ))a(

(b, f (b))

a x b

Theorem. f is convex on I if and only if the slope of the chords always

increase, i.e. a, x , b � I , a x b � f � x � � f � a �x � a

f � b � � f � x �b � x

.

Proof. x � ta� � 1 � t � b for some t � [0 � 1]

� � 0 t � b � x

b � a 1.

f�ta

� � 1 � t � b � t f � a � � � 1 � t � f � b � � � f � x � b � x

b � af � a � � x � a

b � af � b � � � f � x � � f � a �

x � a f � b � � f � x �

b � x�

Theorem. For f differentiable on I � f is convex on I� � f � is increasing on I . (For f twice differentiable on I �

f is convex on I� � f � � � 0 on I .)

Proof. ( � ) If a, b � I , a b, then f � � a � � limx � a �

f � x � � f � a �x � a

limx � a �

f � b � � f � x �b � x

� f � b � � f � a �b � a

limx � b �

f � x � � f � a �x � a

limx � b �

f � b � � f � x �b � x

� f � � b � .

(�

) If a, x , b � I , a x b, then by the mean value theorem, there are r, s such that a r x s b andf � x � � f � a �

x � a� f � � r � f � � s � � f � b � � f � x �

b � x.

Theorem. If f is convex on � a � b � , then f is continuous on � a � b � .

Proof. For x0 � � a � b � , consider u, � , � � � a � b � such that u x0 � � . Thenf � x0 � � f � u �

x0� u

f � � � � f � x0 �� � x0

f � � � � f � � �� � � . Solving for f � � � � we get

f � x0 � � f � u �x0

� u� � � x0 � �

f � x0 � f � � � f � � � � f � � �� � � � � � x0 � �

f � x0 � �Taking limit as � � x 0 , we have f � x0 � f � x0

� � f � x0 � , i.e. f � x0� � � f � x0 � . Similarly, we can show

f � x0� � � f � x0 � by taking u � x0 �� � Therefore, lim

x � x0f � x � � f � x0 � for every x0 � � a � b � .

Example. The function f � x � � � 0 if 0 x 11 if x � 1

is convex on [0 � 1] by checking the slopes of the chords over [0 � 1],

but f is not continuous at 1 because limx � 1 �

f � x � � 0 �� 1 � f � 1 � .

Remark. The proof of the last theorem uses the fact that on an open interval, at any point x0, there are points on itsleft and points on its right, which is not true for endpoints of a closed interval.

41

Page 42: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Chapter 9. Riemann Integral

For a � b � � let f be a bounded function on [a � b] � say � f � x � � K for every x � [a � b]. Let P � � x0 � x1 � � � � � xn �be a partition of [a � b], i.e. a � x0 x1 � � � xn � 1 xn

� b. The length of [x j � 1 � x j] is � x j� x j

� x j � 1

and the mesh of P is�P� � max � � x1 � � � � � � xn � � Also, on [x j � 1 � x j] � let m j

� inf � f � x � : x � [x j � 1 � x j] � andMj

� sup � f � x � : x � [x j � 1 � x j] � �

a = x0 x1 x2 . . . xn - 1 xn = b

Definitions. For the partition P � a Riemann sum is a sum of the form S �n�

j � 1

f � tj � � x j � where every tj is in [x j � 1 � x j] � The lower Riemann sum is L � f � P � �

n�j � 1

m j � x j and the upper Riemann sum is U � f � P � � n�j � 1

Mj � x j �

(Note � K m j f � tj � Mj K implies � K � b � a � L � f � P � S U � f � P � K � b � a � � )

In 1854, George B. Riemann defined the integral of f � x � on [a � b] to be lim�P� � 0

n�j � 1

f � tj � � x j � There are two

immediate technical problems with such a definition.

(1) As the choices of the t j’s may vary, it is hard to say how close the Riemann sum is to the actual integral. Inparticular, it is not clear if the Riemann sum is greater than or less than the integral at any time.

(2) The type of limit involved is not the limit of a sequence nor the limit of a function. In fact, there are many variablesin a Riemann sum!

Instead of dealing with these technicalities, we introduce integral following the approach of J. Gaston Darboux in 1875.

Definition. We say P � is a refinement of P or P � is finer than P iff P and P � are partitions of [a � b] and P � P � �

Refinement Theorem. If P � is a finer partition of [a � b] than P, then L � f � P � L � f � P � � U � f � P � � U � f � P � .

x xw

w

jj - 1

j

Proof. P � can be obtained from P by adding one point at a time. It suffices to considerthe case P � � P � � � � with � �� P � � x0 � x1 � � � � � xn � . Suppose x j � 1 �� x j �

Let m j� inf � f � x � : x � [x j � 1 � x j] � � m �j � inf � f � x � : x � [x j � 1 � � ] � and m � �j �

inf � f � x � : x � [ � � x j] � � Since [x j � 1 � � ] � [ � � x j] � [x j � 1 � x j] � we have m j m �j andm j m � �j � Then m j � x j

� m j � � � x j � 1 � �m j � x j

� � � m �j � � � x j � 1 � �m � �j � x j

� � � �So L � f � P � L � f � P � � . Similarly, U � f � P � � U � f � P � .

Observe that L � f � P � “underestimates” the area under the curve; U � f � P � “overestimates” the area under thecurve.

Definitions. The lower integral of f on [a � b] is

� L �� b

af � x � dx � sup � L � f � P � : P is a partition of [a � b] � (the largest underestimate �

and the upper integral of f on [a � b] is

� U �� b

af � x � dx � inf � U � f � P � : P is a partition of [a � b] � (the smallest overestimate) �

(So for every partition P of [a � b] � L � f � P � � L ��� ba f � x � dx � U ��� b

a f � x � dx U � f � P � � )We say f is (Riemann) integrable on [a � b] iff � L � � b

a f � x � dx � � U � � ba f � x � dx � In that case, we denote the

common value by � ba f � x � dx . (For b a, define � b

a f � x � dx � � � ab f � x � dx . In particular, � a

a f � x � dx � 0.)

42

Page 43: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

U( f, P) - L ( f, P)= sum of areas

of shaded regions

Theorem (Integral Criterion). Let f : [a � b] � be a bounded function. Thefunction f is (Riemann) integrable on [a � b] if and only if for every

� � 0, thereis a partition P of [a � b] such that U � f � P � � L � f � P � � �

Proof. If for every� � 0 � there is a partition P of [a � b] such that U � f � P � � L � f � P � � , then 0 � U � � b

a f � x � dx �� L ��� b

a f � x � dx U � f � P � � L � f � P � � � By the infinitesimal principle, we get � L � � ba f � x � dx � � U ��� b

a f � x � dx �

i.e. f is (Riemann) integrable on [a � b] �Conversely, if f is (Riemann) integrable on [a � b] � then for every

� � 0 � by the supremum property, we have� L ��� b

a f � x � dx � � � 2 L � f � P1 � � L ��� ba f � x � dx for some partition P1 of [a � b] � Similarly, � U � � b

a f � x � dx U � f � P2 � � U ��� b

a f � x � dx� � � 2 for some partition P2 of [a � b] � Let P � P1 � P2 � then by the refinement

theorem, L � f � P1 � L � f � P � U � f � P � U � f � P2 � � Since U � f � P2 � � L � f � P1 � [ � U � � ba f � x � dx

� � � 2] �[ � L ��� b

a f � x � dx � � � 2] � �� so U � f � P � � L � f � P � U � f � P2 � � L � f � P1 � � �

Question. Are there integrable functions? Are there non-integrable functions?

Below we will show that constant functions and continuous functions are integrable.

Example. Recall the function f � x � ��

1 if x ���0 if x ���� is not continuous anywhere. Partitioning an interval [a � b] into

subintervals [x j � 1 � x j], we get by the density of rational numbers and irrational numbers that m j� 0 and Mj

� 1 � Then

L � f � P � � 0 and U � f � P � � b � a for every partition P of [a � b] � So � L �� b

af � x � dx � 0, � U �

� b

af � x � dx � b � a.

Therefore, f � x � is not integrable on every interval [a � b] with a b.

Example. If f � x � � c for every x � [a � b] � then L � f � P � � c � b � a � � U � f � P � for every partition P of [a � b] � So

� L �� b

af � x � dx � c � b � a � � � U �

� b

af � x � dx � Therefore, f is integrable on [a � b] and

� b

af � x � dx � c � b � a � �

Uniform Continuity Theorem. If f : [a � b] � is continuous, then f is uniformly continuous (in the sense that forevery

� � 0 � there exists a � � 0 such that for all x � t � [a � b] � � x � t � �� implies � f � x � � f � t � � � � )Proof. Suppose f is not uniform continuous. Then there is an

� � 0 such that for every � � 1n � n � � � there are

xn � tn � [a � b] such that � xn� tn � � � 1

n and � f � xn � � f � tn � � � � � By the Bolzano-Weierstrass theorem, � xn � has

a subsequence � xnj � converging to some � � [a � b] � Since � tnj� � � � tnj

� xnj �� � xnj

� � � 1

n j

� � xnj� � � � by

the sandwich theorem, � tnj � converges to � � Then 0 � � f � � � � f � � � � � limj � � � f � xnj � � f � tnj � � � � � 0 � which is a

contradiction.

Theorem. If f is continuous on [a � b], then f is integrable on [a � b].

Proof. For� � 0, since f is uniformly continuous on [a � b], there is � � 0 such that x , ��� [a � b], � x � � � � �

� f � x � � f � � � � � � � b � a � . Let P � � a � x0 � x1 � � � � � xn� b � be a partition of [a � b] with max

1�

j�

n� x j

� x j � 1 � �� . By

the extreme value theorem, M j� f � � j � and m j

� f � u j � for some � j � u j � [x j � 1 � x j] � Then

U � f � P � � L � f � P � � n�j � 1

�f � � j � � f � u j � � � x j

b � a

n�j � 1

� x j� x j � 1 � �

� � xn� x0 �

b � a� � �

By the integral criterion, f is integrable on [a � b] �

43

Page 44: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Exercises. (1) Let f : [a � b] � be a function and c � [a � b] � Show that f is integrable on [a � b] if and only if f isintegrable on [a � c] and [c � b] �(2) If f : [a � b] � is bounded and discontinuous only at x1 � � � � � xn � [a � b] � show that f is integrable on [a � b] �

(Hint: This can be done directly or by using (1) to reduce the problem to intervals having only one discontinuity.)

Questions. How bad can an integrable function be discontinuous? Which functions are integrable?

Definitions.(i) A set S � is of measure 0 (or has zero-length) iff for every� � 0, there are intervals � a1 � b1 � � � a2 � b2 � �

� a3 � b3 � � � � � such that S � � �n � 1� an � bn � and ��

n � 1

�an� bn � � .

(ii) We say a property holds almost everywhere iff the property holds except on a set of measure 0. (It is common toabbreviate almost everywhere by a.e. in advanced courses. In probability, almost surely is used instead of almosteverywhere.)

Lebesgue’s Theorem(1902). For a bounded function f : [a � b] � � f is integrable on [a � b] if and only if the setSf

� � x � [a � b]: f is discontinuous at x � is of measure 0 (i.e. f is continuous almost everywhere).

For a proof of Lebesgue’s theorem, please go to appendix 1 of this chapter.

Examples. (1) The empty set�

is of measure 0 because� � � �n � 1

� 0 � 0 � and ��n � 1

�0 � 0 � � 0 � � So every continuous

function f : [a � b] � is integrable on [a � b] � since the set of discontinuities S is�.

(2) A countable set � x1 � x2 � � � ��� is of measure 0 because

� x1 � x2 � � � ��� � � �n � 1� xn

��

4n� xn

� �

4n� and ��

n � 1

2�

4n� 2�

3 � �

( ) ( ) ( ( () ) )

1x41ε−x 41

ε+x 2x 3x 4x 5x162ε−x 643

ε−x162ε+x 643

ε+x

So every monotone function on [a � b] is integrable on [a � b] � since the set of discontinuities S is countable, henceof measure 0.

(3) There exist uncountable sets, which are of measure 0 (eg. the Cantor set).

(4) A countable union of sets of measure 0 is of measure 0. (If S1 � S2 � S3 � � � � are of measure 0 and S is their union,

then for every� � 0 � since Sk is of measure 0, there are intervals � ak � n � bk � n � such that Sk � � �n � 1

� ak � n � bk � n � and

��n � 1

�ak � n � bk � n � �4k

� Then S � � �k � 1 � �n � 1� ak � n � bk � n � and ��

k � 1

��n � 1

�ak � n � bk � n � ��k � 1

4k��

3 � � )

(5) If a set S is of measure 0 and S � � S � then S � is also of measure 0. (For S � � use the same intervals � an � bn � for S � )

(6) Arrange � � [0 � 1] in a sequence r1 � r2 � r3 � � � � and define fn � x � ��� 1 if x � r1 or r2 or � � � or rn

0 otherwise� Then fn � x �

is integrable on [0 � 1] � since the set of discontinuities S f n� � r1 � r2 � � � � � rn � is finite, hence of measure 0. Now

limn � � fn � x � � f � x � �

�1 if x ���0 if x ���� is not integrable on [0 � 1] since Sf

� � x � [0 � 1] : f is discontinuous at x � �

[0 � 1] � which is not of measure 0. So the limit of a sequence of Riemann integrable functions may not be Riemannintegrable.

44

Page 45: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Remarks. In B. R. Gelbaum and J. M. H. Olmsted’s book Counterexamples in Analysis, p. 106, there is anexample of the limit of a sequence of

continuous functions on [0 � 1] � which is not Riemann integrable.

Theorem. For c � [a � b] � f is integrable on [a � b] if and only if f is integrable on [a � c] and [c � b] �Proof. Let S � S1 � S2 be the set of discontinuous points of f on [a � b] � [a � c] � [c � b] � respectively. If f is integrableon [a � b] � then by Lebesgue’s theorem, S is of measure 0. Since S1 � S2 � S � so both S1 � S2 are of measure 0. ByLebesgue’s theorem, f is integrable on [a � c] and [c � b] �

Conversely, if f is integrable on [a � c] and [c � b] � then S1 � S2 are of measure 0. Since S � S1 � S2 � � c � � byexamples (4) and (5), S is of measure 0. By Lebesgue’s theorem, f is integrable on [a � b] �

Theorem. If f � g : [a � b] � are integrable on [a � b] � then f�

g � f � g and fg are integrable on [a � b] �Proof. If f and g are integrable on [a � b] � then f and g are bounded on [a � b] � So f

�g � f � g and f g are bounded

on [a � b] �Observe that if f and g are continuous at x � then f

�g is continuous at x � Taking contrapositive, we see that if

f�

g is discontinuous at x � then f or g is discontinuous at x � So if x � S f g � then x � S f or x � Sg � which impliesSf g � S f � Sg � Since f � g are integrable on [a � b] � by Lebesgue’s theorem, S f � Sg are of measure 0. By example(4), Sf � Sg is of measure 0. By example (5), S f g is also of measure 0. Therefore, by Lebesgue’s theorem f

�g is

integrable on [a � b] � Using a similar argument, we can see that f � g � f g are integrable on [a � b] �

Remarks. By a similar argument, we can show that if f : [a � b] � is integrable on [a � b] and g is bounded andcontinuous on f � [a � b] � � then Sg � f � Sf � (This is because if f is continuous at x � then g � f is continuous at x � Takingcontrapositive, we get if x � Sg � f � then x � Sf � So Sg � f � S f � ) Since S f is of measure 0, by example (5), Sg � f isof measure 0. So g � f : [a � b] � is integrable on [a � b] � In particular, if f is integrable on [a � b] � then takingg � x � � � x � � x2

� ex� cos x � � � � � respectively, we get � f � � f 2

� e f� cos f � � � � are integrable on [a � b] �

However, even if f : [a � b] � [c � d] is integrable on [a � b] and g : [c � d] � is integrable on [c � d] � g � f maynot be integrable on [a � b] � (For example, define f : [0 � 1] � [0 � 1] by f � 0 � � 1 � f � m � n � � 1 � n � where m � n arepositive integers with no common prime factors, and f � x � � 0 for every x � [0 � 1] � � � Next, define g : [0 � 1] � [0 � 1]by g � 0 � � 0 and g � x � � 1 for every x � � 0 � 1] � As an exercise, it can be showed that S f

� [0 � 1] � � and Sg� � 0 � � So

f and g are integrable on [0 � 1] � However, g � f is the nonintegrable function that is 1 on [0 � 1] � � and 0 on [0 � 1] � � � )There is even an example of a continuous function f : [0 � 1] � [0 � 1] and an integrable function g : [0 � 1] �

such that g � f is not integrable (see B. R. Gelbaum and J. M. H. Olmstad’s book Counterexample in Analysis,pp. 106-107).

Up to now, we have been trying to determine which functions are integrable. Below we will look at how theintegrals of functions can be computed.

Theorem (Simple Properties of Riemann Integrals). Let f and g be integrable on [a � b] �

(1)� b

a

�f � x � �

g � x ��� dx �� b

af � x � dx

�� b

ag � x � dx �

� b

ac f � x � dx � c

� b

af � x � dx for every c � .

(2) If f � x � g � x � for all x � [a � b], then� b

af � x � dx

� b

ag � x � dx . Also, ����

� b

af � x � dx ����

� b

a� f � x � � dx .

(3)� b

af � x � dx �

� c

af � x � dx

� � b

cf � x � dx for c � [a � b] �

Proof. To get (1), by the supremum and infimum properties, for every� � 0 � there is a partition P of [a � b] such that

� b

af � x � dx

� � b

ag � x � dx � � L � f � P � �

L � g � P � L � f�

g � P �

U � f�

g � P � U � f � P � �U � g � P �

� b

af � x � dx

� � b

ag � x � dx

��� �

45

Page 46: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Letting� � 0 � we get

� b

a

�f � x � �

g � x � � dx �� b

af � x � dx

� � b

ag � x � dx � Next, from U � � f � P � � � L � f � P � and

inf � � S � � � sup S � we get� b

a

� f � x � dx � �� b

af � x � dx � Then

� b

a

�f � x � � g � x ��� dx �

� b

a

�f � x � � � � g � x � ��� dx �

� b

af � x � dx

� � b

a

� g � x � dx �� b

af � x � dx �

� b

ag � x � dx �

For the second statement, the case c � 0 is clear since� b

a0 f � x � dx � 0 � 0

� b

af � x � dx � If c � 0 � then

� b

ac f � x � dx � sup � cL � f � P � : P is a partition of [a � b] �

� c sup � L � f � P � : P is a partition of [a � b] � � c� b

af � x � dx �

If c 0 � then

� b

ac f � x � dx �

� b

a

� � � c � f � x � dx � �� b

a� � c � f � x � dx � � � � c �

� b

af � x � dx � c

� b

af � x � dx �

For (2), observe that g � f � 0 on [a � b] implies L � g � f � P � � 0 for every partition P � So

� b

a

�g � x � � f � x ��� dx � sup � L � g � f � P � : P is a partition of [a � b] � � 0 implies

� b

af � x � dx

� b

ag � x � dx �

For the second statement, since � � f � f � f � on [a � b] � we get �� b

a� f � x � � dx

� b

af � x � dx

� b

a� f � x � � dx �

For (3), let P be a partition of [a � b] � P � � P � � c � � P1� P � � [a � c] and P2

� P � � [c � b] � Then P � is a finerpartition of [a � b] than P � P1 is a partition of [a � c] � P2 is a partition of [c � b] and L � f � P � � � L � f � P1 � �

L � f � P2 � � Let

A � � L � f � P � : P is a partition of [a � b] � and B � � L � f � P � � : P is a partition of [a � b] and P � � P � � c � � �

Since P � is also a partition of [a � b] � we see B � A and so sup B sup A � By the refinement theorem, L � f � P � L � f � P � � sup B � Hence sup A sup B � Therefore, sup A � sup B and

� b

af � x � dx � sup A � sup B

� sup � L � f � P1 � �L � f � P2 � : P1 � P2 are partitions of [a � c] � [c � b] � respectively �

� sup � L � f � P1 � : P1 is a partition of [a � c] � �sup � L � f � P2 � : P2 is a partition of [c � b] �

�� c

af � x � dx

� � b

cf � x � dx �

The most important tool for computing an integral is to find an antiderivative or primitive function of the integrablefunction, which is a function whose derivative is the integrable function. What can we say about such a function?

Example. For x � [ � 1 � 1] � define f � x � � � 1 if x � 0� 1 if x 0

� Since f � x � has only one point of discontinuity at 0, f � x �is integrable on [ � 1 � 1] � Now the function

F � x � �� x

0f � t � dt ��� x if x � 0

� x if x 0� � x �

46

Page 47: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

is continuous on [ � 1 � 1] � hence uniformly continuous there by the uniform continuity theorem, but F � x � is notdifferentiable at 0, the same point where f � x � is discontinuous!

Theorem. If f is integrable on [a � b] and c � [a � b] � then F � x � �� x

cf � t � dt is uniformly continuous on [a � b] �

Proof. Recall there is K � 0 such that � f � t � � K for every t � [a � b] � For every� � 0 � let � � �

K � Then � x � x0 � ��implies � F � x � � F � x0 � � � ����

� x

x0

f � t � dt ���� K � x � x0 � � �

The next theorem is the most important theorem in calculus. It not only tells us how to compute an integral,but also the deep fact that differentiation and integration are inverse operations on functions. Roughly, it may be

summarized by the formulasd

dx

� x

ch � t � dt � h � x � and

� x

c

d

dth � t � dt � h � t � ���

x

c�

Fundamental Theorem of Calculus. Let c � x0 � [a � b] �(1) If f is integrable on [a � b], continuous at x0 � [a � b] and F � x � �

� x

cf � t � dt , then F � � x0 � � f � x0 � .

(2) If G is differentiable on [a � b] with G � � g integrable on [a � b], then� b

ag � x � dx � G � b � � G � a � .

(Note G � need not be continuous by an example in the chapter on differentiation.)

Proof. (1) We will check limx � x0

F � x � � F � x0 �x � x0

� f � x0 � using the definition of limit. For every� � 0 � since f is

continuous at x0 � there exists a � � 0 such that for every x � [a � b] � � x � x0 � � � � f � x � � f � x0 � � � � With thesame � � we get

����F � x � � F � x0 �

x � x0

� f � x0 ������� �����

� xc f � t � dt � � x0

c f � t � dt

x � x0

� � xx0

f � x0 � dt

x � x0

������ �����

� xx0

� f � t � � f � x0 � � dt

x � x0

����� �����

� xx0

dt

x � x0

������ � �

(2) The conclusion will follow from the infinitesimal principle if we can show ���

� b

ag � x � dx � � G � b � � G � a ��� ���

for every� � 0 � Now for every

� � 0 � since g is integrable on [a � b] � by the integral criterion, there is a partitionP � � a � x0 � x1 � � � � � xn

� b � such that U � g � P � � L � g � P � � � For j � 1 � 2 � � � � � n � by the mean value theorem,there exists tj � � x j � 1 � x j � such that G � x j � � G � x j � 1 � � G � � tj � � x j

� x j � 1 � � g � tj � � x j � Then

L � g � P � n�

j � 1

g � ti � � x j� n�

j � 1

� G � x j � � G � x j � 1 � � � G � b � � G � a � U � g � P � �

Also, L � g � P � � b

ag � x � dx U � g � P � � Then ���

� b

ag � x � dx � � G � b � � G � a � � ���

U � g � P � � L � g � P � � �

Remarks. In (2) above, the condition G � is integrable is important. The function G � x � � x 2 sin1x2

for x �� 0 and

G � 0 � � 0 is differentiable on � but G � � x � is unbounded on [0 � 1] � hence not integrable there. In Y. Katznelson andK. Stromberg’s paper Everywhere Differentiable, Nowhere Monotone Functions, American Mathematical Monthly,vol. 81, pp. 349-354, there is even an example of a differentiable function on such that its derivative is bounded, butnot Riemann integrable on any interval [a � b] with a b �

Theorem (Integration by Parts). If f , g are differentiable on [a � b] with f � , g � integrable on [a � b], then

� b

af � x � g � � x � dx � f � b � g � b � � f � a � g � a � �

� b

af � � x � g � x � dx �

47

Page 48: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Proof. By part (2) of the fundamental theorem of calculus,� b

a� f g � � � x � dx � f � b � g � b � � f � a � g � a � � Since � f g � � � x � �

f � � x � g � x � �f � x � g � � x � � we can subtract the integral of f � � x � g � x � on both sides to get the integration by parts formula.

Theorem (Change of Variable Formula). If�

: [a � b] � is differentiable,� � is integrable on [a � b] and f is

continuous on� �

[a � b] � , then��� � b �� � a � f � t � dt �

� b

af� � � x ��� � � � x � dx �

Proof. Let g � x � ���� � x �� � a � f � t � dt � By part (1) of the fundamental theorem of calculus and chain rule, g � � x � �

f� � � x � � � � � x � � So

� b

af� � � x � � � � � x � dx �

� b

ag � � x � dx � g � b � � g � a � � g � b � �

� � � b �� � a � f � t � dt �

Appendix 1: Proof of Lebesgue’s Theorem

We first modify the uniform continuity theorem to obtain a lemma.

Lemma. Let f : [a � b] � be a function continuous on a subset K � [a � b] � � �j � 1� � j �

�j � � Then for every

� � 0 �

there exists a � � 0 such that for all x � K � t � [a � b] � � x � t � �� implies � f � x � � f � t � � � �Proof. Suppose the lemma is false. Then there is an

� � 0 such that for every � � 1n � n �� � there are xn � K � tn � [a � b]

such that � xn� tn � � � 1

n and � f � xn � � f � tn � � � � � By the Bolzano-Weierstrass theorem, � xn � has a subsequence� xnj � converging to some � � [a � b] � Since

� tnj� � � � tnj

� xnj �� � xnj

� � � 1n j

� � xnj� � � �

by the sandwich theorem, � tnj � converges to � �Next we will show � � K � Suppose � �� K � then � � � � i �

�i � for some i � Since lim

j � � xnj� � and � � � � i �

�i � �

by the definition of limit, there will be some xnp � � � i �

�i � � contradicting xnp � K � [a � b] � � �j � 1

� � j �

�j � � So � � K �

Since f is continuous at � � by the sequential continuity theorem,

0 � � f � � � � f � � � � � limj � � � f � xnj � � f � tnj � � � � � 0 �

which is a contradiction.

Proof of Lebesgue’s Theorem. First suppose f : [a � b] � is integrable. Note that if f is discontinuous at x � thenthere is an

� � 0 such that for every � � 0 � there is z � � x � � � x� � � � [a � b] such that � f � x � � f � z � � � � � Let Dk be the

set of all x � [a � b] such that for every open interval I with x � I � there is z � I � [a � b] such that � f � x � � f � z � � � 1

k�

Since every positive� � 1

kfor some positive integer k by the Archimedian principle and every open interval I with x � I

contains an interval � x � � � x� � � for some � � 0 � it follows that S f

� � x � [a � b] : f is discontinuous at x � � � �k � 1Dk �

We will show each Dk is of measure 0, which will imply S f is of measure 0.

For every� � 0 � by the integral criterion, there is a partition P � � x0 � x1 � � � � � xn � of [a � b] such that

U � f � P � � L � f � P � �

2k� If there is x � Dk � � x j � 1 � x j � � then Mj

� m j � � f � x � � f � z � � � 1

kfor some z �

� x j � 1 � x j � � Let J be the set of j such that Dk � � x j � 1 � x j � �� � � Then Dk � � x0 � x1 � � � � � xn � � �j � J

� x j � 1 � x j � and�j � J

� x j � 1� x j �

�j � J

k � Mj� m j �� ��� �

�1 � k � x j k

�U � f � P � � L � f � P � � �

2� Next around each x j � we take an open interval

48

Page 49: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Ij containing x j with length�

2 � n �1 � � Then the intervals � x j � 1 � x j � with j � J and Ij ’s together contain Dk and the

sum of their lengths is less than� � So each Dk (and hence Sf ) is of measure 0.

Conversely, suppose S f is of measure 0. For every� � 0 � let

0�

3 � N�

b � a � � where N is an upper

bound for � f � x � � on [a � b] � By the definition of measure 0, there are intervals � � i �

�i � such that Sf � � �i � 1

� � i �

�i �

and ��i � 1

� � i� �

i � � 0 � Then f is continuous on K � [a � b] � � �i � 1� � i �

�i � � Now take the � in the lemma for

0 and

a partition P � � x0 � x1 � � � � � xn � such that � x j � 1� x j � � for j � 1 � 2 � � � � � n � If K � [x j � 1 � x j] � �

� then

[x j � 1 � x j] � � �i � 1� � i �

�i � � If there is x � K � [x j � 1 � x j] � then Mj

� m j� sup � � f � t0 � � f � t1 � � : t0 � t1 � [x j � 1 � x j] �

sup � � f � t0 � � f � x � � � � f � x � � f � t1 � � : t0 � t1 � [x j � 1 � x j] � 2�

0 � So

U � f � P � � L � f � P � � �K � [xj � 1 � xj ] ��� � Mj

� m j �� ��� ��

2N

� x j� �

K � [xj � 1 � xj ] ���� � Mj� m j �� ��� �

�2 � 0

� x j 2N�

0�

2�

0 � b � a � � �

Therefore, by the integral criterion, f is integrable on [a � b] �

Appendix 2: Riemann’s Definition of the Integral

Recall the definition of limx � 0

f � x � � L is that for every� � 0 � there is � � 0 such that for every x with � x � � �

we have � f � x � � L � � � Based on this, Riemann’s approach of integral can be defined as follow.

Definition. Let f be a bounded function on [a � b] � We write lim�P� � 0

n�j � 1

f � tj � � x j� I iff for every

� � 0 � there is a

� � 0 such that for every partition P of [a � b] with�P� � and every choice t j � [x j � 1 � x j] � j � 1 � 2 � � � � � n � we

have ���n�

j � 1

f � tj � � x j� I ���

� � (Such I is unique as in the proof of uniqueness of limit.)

Below we will show Darboux’s approach is the same as Riemann’s by establishing the following theorem.

Darboux’s Theorem. Let f be a bounded function on [a � b] � Then the following are equivalent:

(a)� b

af � x � dx � I �

(b) lim�P� � 0

n�j � 1

f � tj � � x j� I �

(c) for every� � 0 � there is a partition P of [a � b] such that for every refinement Q � � x0 � x1 � � � � � xn � of P � every

choice tj � [x j � 1 � x j] � we have ���n�

j � 1

f � tj � � x j� I ���

� � (Such a number I is unique as in the proof of uniqueness of

limit.)

Proof. � a � � � b � Suppose� b

af � x � dx � I � For every

� � 0 � by the proof of the integral criterion, there is a

partition P�� � � 0 � � 1 � � � � � � q � of [a � b] such that I �

2 L � f � P� � U � f � P� � I

� �

2� Let � �

4q K� where

K � sup � � f � x � � : x � [a � b] � � If P � � x0 � x1 � � � � � xn � is a partition of [a � b] with�P� �� � then U � f � P � � S1

�S2 �

where S1 is the sum of the terms from intervals not containing points in P� and S2 the sum of the remaining terms.

Note the rectangles for the terms of S1 are inside the rectangles for U � f � P� � � So we have S1 U � f � P� � I� �

2�

Since P�� � � 0 � � 1 � � � � � � q � � S2 has at most 2q terms (one for � 0 � � q each and at most two for � 1 � � � � � � q � 1

49

Page 50: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

each). Then S2 2q � K�P� � 2qk � �

2� Thus U � f � P � � S1

�S2 I

��� � Similarly, L � f � P� � � I � � � So

I � � L � f � P� n�

j � 1

f � tj � � x j U � f � P � I��� � Then ���

n�j � 1

f � tj � � x j� I ���

� �

� b � � � c � For every� � 0 � take � as the definition above. Let P be any partition of [a � b] with

�P� � � Then for

every refinement Q of P � we have�Q� �

P� �� and so (c) follows from the conclusion of (b).

� c � � � a � For� � 0 � let P be the partition in (c) for

3� For each j � 1 � 2 � � � � � n � by the supremum limit theorem, there

are sequences tj � k such that limk � � f � tj � k � � Mj

� sup � f � x � : x � [x j � 1 � x j] � � Since ���n�

j � 1

f � tj � k � � x j� I ���

3� so taking

limit, we get �U � f � P � � I � �

3� Similarly, we can also get � L � f � P � � I �

3� Then �U � f � P � � L � f � P � � 2

3 � �

By the integral criterion, it follows that� b

af � x � dx exists. By � a � � � b � and � b � � � c � � it must equal I by uniqueness.

Improper Riemann Integral – Integration of Unbounded Functions or on Unbounded Intervals

We would like to integrate unbounded functions on unbounded intervals by taking limit of integrals on boundedintervals. Since the functions may or may not be continuous, we have to make sure the functions are integrable onbounded intervals first.

Definition. Let I be an interval. We say f : I � is locally integrable on I iff f is integrable on every closed andbounded subintervals of I � We denote this by f � L loc � I � .

Roughly, there are three cases of improper integrals.

Case 1: (Unbounded at one endpoint) Let f be locally integrable on [a � b � � where a is real and b is real or��� � We

define the improper (Riemann) integral of f on [a � b � to be� b

af � x � dx � lim

d � b �

� d

af � x � dx � provided the limit exists

in � and we say f is improper integrable on [a � b � in that case. For f locally integrable on an interval of the form� a � b] � where a is real or � � and b is real, the definitions of the improper (Riemann) integral of f on � a � b] and f isimproper (Riemann) integrable on � a � b] are similar.

Case 2: (Unbounded at both endpoints) Let f be locally integrable on � a � b � � with a � b real or infinity, and x 0 � � a � b � .

We define the improper (Riemann) integral of f on � a � b � to be� b

af � x � dx � lim

c � a �� x0

cf � x � dx

�lim

d � b �

� d

x0

f � x � dx

provide both limits exist in � and say f is improper integrable on � a � b � in that case. (Note the integral is the sameregardless of the choices of x0 � � a � b � .)

Case 3: (Unbounded inside interval) Let f be improper integrable on intervals [a � c � and � c � b] � The improper integralof f on [a � b] is the sum of the improper integrals on the two intervals. The cases where a or b is excluded are similarlydefined.

In each case, if the improper integral is a number, then we say the improper integral converges to the number,otherwise we say it diverges.

Examples.(1) Consider the unbounded function f � x � � ln x on � 0 � 1]. Observe that f is continuous, hence integrable,

on every closed subinterval of � 0 � 1] � So f � L loc � 0 � 1] � Now limc � 0 �

� 1

cln x dx � lim

c � 0 � � � 1 � c ln c�

c � � � 1, so� 1

0ln x dx � � 1 and ln x is improper integrable on � 0 � 1] �

50

Page 51: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(2) Consider f � x � � 1

x2on the unbounded interval [1 �

��� � . Observe that f � L loc[1 �

��� � because of continuity.

Now limd � �

� d

1

1

x2dx � lim

d � � � 1 � 1

d� � 1, so

� �1

1

x2dx � 1 and

1

x2is improper integrable on [1 �

��� � �

(3) Consider f � x � � ex on � � � ��

��� � � Since f is continuous everywhere, f � L loc � � ��

��� � � Now take a

x0 � � � ��

��� � � say x0� 0 � Then lim

c � � �� 0

cex dx � lim

c � � � � 1 � ec � � 1 � but limd � �

� d

0ex dx � lim

d � � � ed � 1 �does not exist in � So ex is not improper integrable on � � �

��� � �

Remarks. By taking limits, many properties of Riemann integrals extend to improper Riemann integrals as well.There are also some helpful tests to determine if a function is improper integrable.

p-test. For a � 0 � we have

� �a

1x p

dx � if and only if p � 1 � Also,

� a

0

1x p

dx � if and only if p 1 �

Proof. In the proof of p-test for series, we saw� �

1

1x p

dx � if and only if p � 1 � Adding� 1

a

1x p

dx � � we

get the first statement. Next, by the change of variable y � 1

x� we have

� 1

c

1

x pdx �

� 1 � c1

1

y2 � pdy � Taking limit as

c � 0�

� we see that� 1

0

1

x pdx � if and only if

� �1

1

y2 � pdy � � By the first statement, this is the same as

2 � p � 1 � i.e. p 1 � Again adding� a

1

1x p

dx � � we get the second statement.

Theorem (Comparison Test). Suppose 0 f � x � g � x � on I and f � g � L loc � I � � If g is improper integrable on I ,then f is improper integrable on I . If f is not improper integrable on I , then g is not improper integrable on I .

Proof. For the case I � [a � b � � if 0 f g on I and g is improper integrable on I � then� d

af � x � dx is an increasing

function of d and is bounded above by� b

ag � x � dx � So

� b

af � x � dx � lim

d � b �

� d

af � x � dx exists in by the monotone

function theorem. The other cases I � � a � b] and � a � b � are similar.

Theorem (Limit Comparison Test). Suppose f � x � � g � x � � 0 on � a � b] and f � g � L loc � � a � b] � � If limx � a

g � x �f � x � is a

positive number L � then either (both

� b

af � x � dx and

� b

ag � x � dx converge) or (both diverge). If lim

x � a g � x �f � x �

� 0 �

then� b

af � x � dx converges implies

� b

ag � x � dx converges. If lim

x � a g � x �f � x �

� �� then

� b

af � x � dx diverges implies

� b

ag � x � dx diverges. In the case [a � b � � we take lim

x � b �g � x �f � x � and the results are similar.

Proof. If limx � a

g � x �f � x � is a positive number L � then for

� � L

2� 0 � there is � � 0 such that for every x � � a � a

� � � � we

haveL

2� L � � g � x �

f � x � L��� � 3L

2� Then

L

2

� a �

af � x � dx

� a �

ag � x � dx 3L

2

� a �

af � x � dx � So either

(both� b

af � x � dx and

� b

ag � x � dx converge) or (both diverge).

If limx � a

g � x �f � x �

� 0 � then there exists � � � 0 such that for every x � � a � a� � � � � we have 0 g � x �

f � x � 1 � which

implies 0 g � x � f � x � � Then 0 � a

���

ag � x � dx

� a ���

af � x � dx � So

� b

af � x � dx � �

� b

ag � x � dx � �

51

Page 52: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

If limx � a

g � x �f � x �

� �� then there is � � ��� 0 such that for every x � � a � a

� � � � � � we haveg � x �f � x � � 1 � which implies

g � x � � f � x � � Hence� a

� � �

ag � x � dx �

� a � � �

af � x � dx � So

� b

af � x � dx � � �

� b

ag � x � dx � � �

Theorem (Absolute Convergence Test). If f � L loc � I � and � f � is improper integrable on I , then f is improperintegrable on I .

Proof. We have � � f � f � f � on I � which implies 0 f� � f � 2 � f � on I � By the comparison test, f

� � f � isimproper integrable on I � Therefore, f � � f

� � f � � � � f � is improper integrable on I �

Examples. (4) Isln x

1�

x2improper integrable on � 0 � 1]? Observe first that the function is locally integrable on � 0 � 1]

by continuity. Also, ����ln x

1�

x2���� � ln x � on � 0 � 1] � Since � ln x � � � ln x is improper integrable on � 0 � 1] (cf.

example (1)), so ����ln x

1�

x2����

is improper integrable on � 0 � 1] by the comparison test. Thenln x

1�

x2is improper

integrable on � 0 � 1] by the absolute convergence test.

(5) Does� �

2

dx� x2 � 1converge? Observe that on [2 �

��� � � 0 1

x 1� x2 � 1

� Both1

xand

1� x2 � 1are

continuous, hence locally integrable, on [2 �

��� � � Now� �

2

1

xdx � �

by p-test. By the comparison test,� �2

dx� x2 � 1diverges.

(6) Does� �

1

sin x

xdx converge? Since

sin x

xis continuous on [1 �

��� � � it is locally integrable there. Integrating

by parts, � c

1

sin x

xdx � � cos c

c�

cos 1 �� c

1

cos x

x2dx �

Since � cos c � 1 � limc � �

cos c

c� 0 � Since ���

cos x

x2 ��� 1

x2on [1 �

��� � and� �

1

1

x2dx � by p-test, we have� �

1

cos x

x2dx converges by the comparison test and the absolute convergence test. Therefore,

� �1

sin x

xdx

converges.

(7) For the improper integral� 1

0

dx

1 � x3� the integrand is continuous (hence locally integrable) on [0 � 1 � � Observe that

11 � x3

� 11 � x � 1

1�

x�

x2 and the second factor on the right has a positive limit as x � 1 � � More precisely,

since limx � 1 �

1 � � 1 � x �1 � � 1 � x3 �

� limx � 1 � � 1 �

x�

x2 � � 3 and� 1

0

dx

1 � x� lim

d � 1 �� d

0

dx

1 � x� lim

d � 1 �� ln � 1 � d � � �

by the limit comparison test,� 1

0

dx

1 � x3diverges.

(8) For the improper integral� 5

0

dx3� 7x

�2x4

� the integrand is continuous (hence locally integrable) on � 0 � 5] � Observe

that1

3� 7x�

2x4� 1

3� x � 13� 7

�2x3 and the second factor on the right has a positive limit as x � 0

� � More

precisely, since limx � 0

1 � 3� x

1 � 3� 7x�

2x4� 3� 7 and

� 5

0

dx3� x

� by p-test as p � 13 1 � by the limit comparison

test,� 5

0

dx3� 7x

�2x4

converges.

52

Page 53: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Principal Value – Symmetric Integration about Singularities.

Definition. Let f � L loc �� � � then the principal value of� �� �

f � x � dx is P � V �� �� �

f � x � dx � limc � �

� c

� cf � x � dx .

Examples. (1) Because of continuity, f � x � � 11

�x2

� L loc �� � � Now P � V �� �� �

11

�x2

dx � limc � �

� c

� c

11

�x2

dx

� limc � � � 2 tan � 1 c � � exists and the improper integral

� �� �1

1�

x2dx � lim

c � � �� 0

c

11

�x2

dx�

limd � �

� d

0

11

�x2

dx � limc � � � � � tan � 1 c � �

limd � � � tan � 1 d � � �

(2) Because of continuity, f � x � � x � L loc �� � � Now P � V �� �� �

x dx � limc � �

� c

� cx dx � lim

c � � 0 � 0 exists, how-

ever the improper integral� �� �

x dx � limc � � �

� 0

cx dx

�lim

d � �� d

0x dx � lim

c � � � � � c2

2� �

limd � � � d2

2� does not

exist.

Theorem. If the improper integral

� �� �f � x � dx exists, then P � V �

� �� �f � x � dx exists and equals the improper integral� �� �

f � x � dx. (Note the converse is false by example (2).)

Proof. If the improper integral� �� �

f � x � dx exists, then it implies that both limd � � �

� 0

df � x � dx and lim

c � �� c

0f � x � dx

exist. So

P � V �� �� �

f � x � dx � limc � �

� c

� cf � x � dx � lim

c � �� � 0

� cf � x � dx

� � c

0f � x � dx �

� limd � � �

� 0

df � x � dx

�lim

c � �� �

0f � x � dx �

� �� �f � x � dx �

Definition. For a � b �� or�

� let f : [a � c � � � c � b] � be locally integrable on [a � c � and on � c � b] � Define

P � V �� b

af � x � dx � lim

� � 0 � � � c � �

af � x � dx

� � b

c �

f � x � dx � �

Example. Consider the improper sense and the principal value sense of� 1

� 1

1x

dx �

Because of continuity, 1x � L loc[ � 1 � 0 � and 1

x � L loc � 0 � 1] � In the improper sense,� 1

0

1

xdx � lim

c � 0 �� 1

c

1

xdx

� limc � 0 � � � ln c � does not exist. So the improper integral on [ � 1 � 0 � � � 0 � 1] does not exist. However,

P � V �� 1

� 1

1x

dx � lim� � 0 � � � � �

� 1

1x

dx� � 1

1x

dx � � lim� � 0 � � ln � � � � � ln � � � � � 0 �

Remarks. (1) It is incorrect to say� 1

� 1

1x

dx � ln � � 1 � � ln �1 � � 0 � which attempts to use part (2) of the fundamental

theorem of calculus, but here f � x � � ln � x � is not differentiable on [ � 1 � 1] failing the required condition.

(2) There is a similar theorem for this type of principal value integrals as the theorem above.

53

Page 54: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Taylor Series of Common Functions

Question. How do calculators compute sin x � cos x � ex� x y

� ln x � � � � ?

Recall Taylor’s theorem with the Lagrange form of the remainder is the following.

Taylor’s Theorem. Let f : � a � b � � be n times differentiable on � a � b � . For every x, c � � a � b � , there is x0 betweenx and c such that

f � x � � f � c � � f � � c �1!

� x � c � � f � � � c �2!

� x � c � 2 � � � � � f � n � 1 � � c �� n � 1 � !

� x � c � n � 1 � f � n � � x0 �n!

� x � c � n �

(This is called the n-th Taylor expansion of f about c � The term Rn � x � � f � n � � x0 �n!

� x � c � n is called the Lagrange

form of the remainder.)

There are two other forms of the remainder.

Theorem (Taylor’s Formula with Integral Remainder). Let f be n times differentiable on � a � b � � For everyx � c � � a � b � � if f � n � is integrable on the closed interval with endpoints x and c � then

f � x � � f � c � � f � � c �1!

� x � c � � ����� � f � n � 1 � � c �� n � 1 � !

� x � c � n � 1 �Rn � x � � where Rn � x � � 1

� n � 1 � !

� x

c� x � t � n � 1 f � n � � t � dt �

Proof. Noted

dt

� � � x � t � � � 1 � Applying integration by parts n � 1 times, we get

f � x � � f � c � �� x

cf � � t � � 1 dt �

� x

cf � � t �

� � � x � t ��� � dt

� � f � � t � � x � t � ���x

c

� � x

c� x � t � f � � � t � dt

� f � � c � � x � c � � � � f � � � t � � x � t � 2

2! ���x

c

� 12!

� x

c� x � t � 2 f � � � � t � dt � �����

� f � � c � � x � c � � ����� � f � n � 1 � � c �� n � 1 � !

� x � c � n � 1 � 1

� n � 1 � !

� x

c� x � t � n � 1 f � n � � t � dt �

Mean Value Theorem for Integrals. If f is continuous on [a � b], then

� b

af � x � dx � f � x1 � � b � a � for some

x1 � [a � b], (i.e. the mean value or average of f on [a � b] is1

b � a

� b

af � x � dx � f � x0 � ). More generally, if g is

integrable and g � 0 on [a � b] � then

� b

af � x � g � x � dx � f � x1 �

� b

ag � x � dx for some x1 � [a � b].

Proof. The first statement is the special case of the second statement when g � x � � 1 � So we only need to prove thesecond statement. Let M and m be the maximum and the minimum of f on [a � b] � respectively. Since m f � x � M

on [a � b] � we have m� b

ag � x � dx

� b

af � x � g � x � dx M

� b

ag � x � dx � If

� b

ag � x � dx � 0 � then the last sentence

implies� b

af � x � g � x � dx � 0 and so we may take x1 to be any element of [a � b] � If

� b

ag � x � dx � 0 � then dividing by

� b

ag � x � dx � we see that

� b

af � x � g � x � dx

� � b

ag � x � dx is between m and M � By the intermediate value theorem,

this expression equals f � x1 � for some x1 � [a � b] � which gives the second statement.

54

Page 55: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Theorem (Taylor’s Formula with Cauchy Form Remainder). Let f be n times differentiable on � a � b � � Forx � c � � a � b � � if f � n � is continuous on the closed interval with endpoints x and c � then there is x1 between x and c suchthat

f � x � � f � c � � f � � c �1!

� x � c � � ����� � f � n � 1 � � c �� n � 1 � !

� x � 1 � n � 1 �Rn � x � � where Rn � x � � � x � c � � x � x1 � n � 1 f � n � � x1 �

� n � 1 � !�

Proof. This follows by applying the mean value theorem for integrals to the integral remainder of Taylor’s formulaabove.

Remarks. For every x �� � the series ��k � 0

x k

k!converges. This follows from the ratio test because lim

k � � ����x k 1

� k �1 � !

k!

x k����� lim

k � �� x �

k�

1� 0 1 � By term test, we have lim

k � �x k

k!� 0 for every x � �

Examples.

(1) For f � x � � sin x , f � n � � x � ��

� � 1 � k cos x if n � 2k�

1� � 1 � k sin x if n � 2k

. So � f � n � � x � � 1 for every x �� �Taking c � 0 � we have f � 2k 1 � � 0 � � � � 1 � k and f � 2k � � 0 � � 0 for every k ��� � By Taylor’s theorem,

sin x � n � 1�k � 0

� � 1 � k x2k 1

� 2k�

1 � !�

R2n � x � for every x � . Now � R2n � x � � 1

� 2n � !� x � 2n � 0 as n � �

by the re-

marks above. Therefore,

sin x � x � x3

3!� x5

5!� x7

7!� ����� � ��

k � 0

� � 1 � k x2k 1

� 2k�

1 � !for � � x ��� �

Remarks. For 0 x 2

� � R18 � x � � � x � 18

18! � � 2 � 18

18! 6 � 10 � 13 � So x � x3

3!� ����� � x17

17!can be used to

compute sin x to 10 decimal places.

(2) For f � x � � cos x , f � n � � x � ��

� � 1 � k sin x if n � 2k � 1� � 1 � k cos x if n � 2k

. So � f � n � � x � � 1 for every x �� �Taking c � 0 � we have f � 2k � 1 � � 0 � � 0 and f � 2k � � 0 � � � � 1 � k for every k ��� � By Taylor’s theorem,

sin x � n � 1�k � 0

� � 1 � k x2k

� 2k � !�

R2n � 1 � x � for every x � . Now � R2n � 1 � x � � 1

� 2n � 1 � !� x � 2n � 1 � 0 as n � �

by the remarks above. Therefore,

cos x � 1 � x2

2!� x4

4!� x6

6!� ����� � ��

k � 0

� � 1 � k x2k

� 2k � !for � � x ��� �

(3) For f � x � � ex , f � n � � x � � ex � Taking c � 0 � we have f � n � � 0 � � 1 for every n � � � By Taylor’s theorem,

ex � n � 1�k � 0

x k

k!�

Rn � x � for every x � � Now � Rn � x � � � ex0

n!� x � n max � e0

� ex �n!

� x � n � 0 as n � �by the

remarks above. So

ex � 1�

x� x2

2!� x3

3!� ����� � ��

k � 0

x k

k!for � � x ��� �

Remarks. The Taylor series for sin x � cos x and ex also converge if x is a complex number! In fact, this is howthe sine, cosine, and exponential functions are defined for complex numbers. For every x � � we have

eix � ��k � 0

� ix � k

k!� ��

n � 0

� � 1 � n x2n

� 2n � !�

i ��n � 0

� � 1 � n x2n 1

� 2n�

1 � !� cos x

�i sin x �

55

Page 56: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

In particular, we have ei � � cos �i sin � � 1 so that ei � �

1 � 0 � This is known as Euler’s formula. It isthe most beautiful formula in mathematics because it connects the five most important constants 1 � 0 � � i � e ofmathematics in one single equation!

(4) (Binomial Theorem) For a �� � if � 1 x 1 � then

� 1 �x � a � 1

�ax

� a � a � 1 �2!

x2 � a � a � 1 � � a � 2 �3!

x3 � ����� � 1� ��

k � 1

a � a � 1 � ����� � a � k�

1 �k!

x k �

To see this, let f � x � � � 1 �x � a

� then f � n � � x � � a � a � 1 � ����� � a � n�

1 � � 1 �x � a � n � By Taylor’s formula with

integral formula,

Rn � x � � 1

� n � 1 � !

� x

0� x � t � n � 1 f � n � � t � dt � a � a � 1 � ����� � a � n

�1 �

� n � 1 � !

� x

0� x � t

1�

t n � 1� 1 �

t � a � 1 dt �

For x � [0 � 1 � � the function g � t � � x � t

1�

thas derivative g � � t � � � 1 � x

� 1 �t � 2

0 � On [0 � x] � g � t � g � 0 � � x � Let

k �� x

0� 1 �

t � a � 1 dt � then � Rn � x � � �a � a � 1 � ����� � a � n�

1 � kxn � 1 �� n � 1 � !� ��� �call this bn

� Since limn � �

bn 1

bn

� limn � �

�a � n � � x �n

� x � 1 � by ratio test, ��n � 1

bn converges. By term test, limn � � bn

� 0 � Then limn � � Rn � x � � 0 by the sandwich

theorem and the binomial formula is true for x � [0 � 1 � � The case x � � � 1 � 0] is similar.

Here are a few more common Taylor series. (Note the series only equal the functions on a small interval.) They

are obtained from the cases a � � 1 � a � � 1 (with x replaced by x 2) and a � � 1

2(with x replaced by � x 2) of the

binomial theorem by term-by-term integration.

ln � 1 �x � � x � x2

2� x3

3� x4

4� ����� � ��

k � 1

� � 1 � k � 1x k

kfor � 1 x 1

tan � 1 x � x � x3

3� x5

5� x7

7� ����� � ��

k � 0

� � 1 � k x2k 1

2k�

1for � 1 x 1

sin � 1 x � x� 1

2

x3

3� 1 � 3

2 � 4x5

5� ����� � x

� ��k � 1

1 � 3 � 5 ����� � 2k � 1 �2 � 4 � 6 ����� � 2k �

x2k 1

2k�

1for � 1 x 1

Remarks. Unfortunately, the Taylor series of a function does not always equal to the function away from the center.

For example, the function f � x � ��

e � 1 � x2if x �� 0

0 if x � 0can be shown to be infinitely differentiable and f � n � � 0 � � 0

for every n � � � So the Taylor series of f � x � about c � 0 is ��k � 0

0x k � 0 � i.e. the Taylor series is the zero function.

Therefore, the Taylor series of f � x � equals f � x � only at the center c � 0 �

56

Page 57: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

Practice Exercises for Math 201

The starred problems are difficult and involve more work or deeper thinking.

For exercises 1 to 7, negate each of the following expressions or statements.

1. (x � 0 and x 1) or x � � 1

2. x � 0 and (x 1 or x � � 1)

3. For every triangle ABC �

�A

� �B

� �C � 180 � �

4. There exists a man who does not have any wife.

5. For each x , there is a y such that x�

y � 0 �

6. � � ��� ��� such that � � � � � �� �

7. If x and y are positive, then x�

y � 0.

8. Give the contrapositive of the following statements. (Since the contrapositives are equivalent to the statements,they say the same thing.)

(a) If AB � AC in � ABC � then�

B � �C in � ABC �

(b) If a function is differentiable, then it is continuous.

(c) If limx � 0

f � x � � a and limx � 0

g � x � � b � then limx � 0

� f � x � �g � x � � � a

�b �

(d) If x2 �bx

�c � 0 � then x � � b

� � b2 � 4c

2or x � � b � � b2 � 4c

2�

9. Compute the following sets.

(a)�� x � y � z � � � � � z � � � � u �

�� � � � (Here u �

�� � � x � y � z are distinct objects.)

(b) � 1 � 2 � � � 3 � 4 � � � 5 � �(c) � � [0 � 10] � � n2 �

1 : n ���� �(d) � n ��� : 5 n 9 � � � 2m : m ��� �(e) � [0 � 2] � [1 � 3] � � � [1 � 3] � [0 � 2] � �

10. (i) Let A � [0 � 1] and B � [0 � 1] � [2 � 3] � Plot the graphs of A � A and B � B on the plane.

(ii) If A � B are sets that are not the empty set and A � B � B � A � what can be said about A and B?

11. (a) If B � C � then prove that A � B � A � C �(b) For sets X � Y � Z � prove that � X � Y � � Z � � X � Z � � Y �

12. (i) For all sets A � B � C � is it always true that � A � B � � C � A � � B � C � ?

(ii) For all sets A � B � C � is it always true that if A � B � A � C � then B � C?

(iii) For all sets A � B � C � is it always true that A � � B � C � � � A � B � � � A � C � ?

13. (i) Show that if A � B and C � D � then A � C � B � D �

57

Page 58: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(ii) Is it always true that if A � B and C � D � then A � C � B � D?

(iii) For a b � let

� a � b � Q� � x : x ��� and a x b � and [a � b � Q

� � x : x ��� and a x b � �

Does ��n � 1

[1

n� 2 � Q

� ��n � 1

� 1

n� 2 � Q?

14. Define functions f � g : � by f � x � ��

0 if x � 01 if x 0

and g � x � � 1 � 2x � For f and g � determine if each is

injective or surjective. Compute f � g and g � f �

15. (i) Let f : A � B be a function. Show that if there is a function g : B � A such that g � f � I A and f � g � IB �

then f is a bijection.

(ii) Show that if f : A � B and h : B � C are bijections, then h � f : A � C is a bijection.

16. Let A � B be subsets of and f : A � B be a function. If for every b � B � the horizontal line y � b intersectsthe graph of f at most once, then show that f is injective. If ‘at most once’ is replaced by ‘at least once’, whatcan be said about f ?

For each of the sets in exercises 17 to 29, determine if it is countable or uncountable:

17. intervals � a � b � and [a � b], where we assume a b �18. � � �� � � ���

19. A � � 1

2n

� 1

3m: n � m ��� � �

20. B � � x � � 2y : x � y ������21. the set C of all lines in 2 passing through the origin;

22. D � � x �� : x5 �x

�2 �������

23. the set E of all circles in 2 with centers at rational coordinate points and positive rational radius.

24. the set F � � a : x4 �ax � 5 � 0 has a rational root ���

25. the set G � � a3 �b3 : a � X � b � Y � � where X is a nonempty countable subet of and Y is an uncountable

subet of ;

26. the set H � � X � Y � � � Y � X � � where X is a countable set and Y is an uncountable set. (Remark: The set� X � Y � � � Y � X � is called the symmetric difference of X and Y and is usually denoted by X � Y � This conceptwill appear in other algebra and analysis courses later.)

27. (Bartle and Sherbert, p. 21) Show that the set F of all finite subsets of � is countable.

28. If S is a countable subset of 2� show that for any two points x � y �� 2 � S � there is a parallelogram in 2 � S

having x � y as opposite vertices.

*29. From K0� [0 � 1] � remove the middle thirds to get K1

� [0 � 1] � � 13 �

23 � � [0 �

13] � [ 2

3 � 1] � Then remove the middlethirds of the 2 subintervals of K1 to get K2

� [0 �19 ] � [ 2

9 �13 ] � [ 2

3 �79 ] � [ 8

9 � 1] � Inductively, remove the middle

58

Page 59: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

thirds of the 2n subintervals of Kn to get Kn 1 � The set K � K0 � K1 � K2 � K3 � ����� is called the Cantor set.Prove that K is uncountable.(Hint: Consider base 3 representations, i.e. representations of the form

� � a1a2a3 � � � � 3� a1

3� a2

32

� a3

33

� ����� �

where each ai� 0 � 1 or 2. What do the base 3 representations of numbers in Kn have in common? Note some

numbers have 2 representations, e.g. 13

� � � 1000 � � � � 3 and � � 0222 � � � � 3 � )

30. For each of the following series, determine if it converges or diverges.

(a) ��k � 1cos � sin

1k� (term test) (b) ��k � 1

1� k � k �1 � � k �

2 �

(c) ��k � 1ln � 1 � 1

k� (find sum) (d) ��k � 1

� 12

� 1k� k

(e) ��k � 2

ln k

k(f) ��k � 1

cos 2k

k2

(g) ��k � 1

k�

2

k�

1� 2

3� k

(h) ��k � 1

cos k � k

(i) ��k � 1ke � k2

(j) ��k � 1

k

� k �1 � !

(k) ��k � 1

Arctan k

k2 �1

(l) ��k � 1

1

k1 1k

(compare with a p-series)

(m) ��k � 1tank � k

�1

k� (n) ��k � 1

� 1 � cos1k� (compare with a p-series)

(o) ��k � 1k2 sinp � 1

k� (depends on p) (p) ��k � 1

� � 1 � k 1 � � k�

1 � � k �

31. (Bartle and Sherbert, p. 95) Let a1 � a2 � a3 � � � � � 0 � Prove that ��k � 1ak converges if and only if ��k � 1

2ka2k

converges. (This is called Cauchy’s condensation test.) Use this test to determine if ��k � 3

1k ln k ln � ln k � converges.

32. Show that ��k � 2

k

2k � 1� 2

2� 3

22

� 423

� ����� converges and find the sum. (Hint: Compare the same series with

index from 1 to� � )

(Probabilistic interpretation: the sum is the expected number of births to get babies of both sex. The probability

of the k-th birth finally resulted in babies of both sex is2

2k� 1

2k � 1� )

*33. Let pn be the n-th prime number, i.e. p1� 2 � p2

� 3 � p3� 5 � p4

� 7 � ����� � Show that ��k � 1

1

pkdiverges. (Hint:

Suppose it converges to s � Then the partial sum sn has limit s � as n � � � So for some n � s � sn� ��k � n 1

1

pk 1

2� Let

Q � p1 p2 ����� pn � Show, by considering the prime factorization of 1�

mQ � thatN�

m � 1

1

1�

mQ ��j � 1

� ��k � n 1

1

pk� j

for every positive integer N � which will lead to a contradiction.)

59

Page 60: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

For exercises 34 to 36, use definitions, Archimedean principle, density of rationals, density of irrationals, etc. tosupport your reasonings.

34. For each of the following sets, if it is bounded above, give an upper bound and find its supremum with proof. Ifit is bounded below, give a lower bound and find its infimum with proof.

(a) A � � � m� � n : m � n ���� (b) B � � � �

� ] ��4 � 1

n: n �� �

(c) C ��

1n

� 12m

: m � n ��� � (d) D � � � � 0 � � 2]

35. Let A and B be nonempty subsets of � which are bounded above. Let

S � � x y : x � A � y � B � and T � � x � y : x � A � y � B � �

Must S or T be bounded above? Give a proof if you think the answer is ‘yes’ or give a counterexample if youthink the answer is ‘no’.

36. Let A and B be nonempty subsets of � which are bounded above. Let

C � � x �y : x � A � y � B � �

Show that C is bounded above and sup C � sup A�

sup B � (Hint: If sup C sup A�

sup B � then consider� � � sup A

�sup B � sup C � � 2 � Apply supremum property to get a contradiction.)

37. Let � n� 4n

�5

n3� then � � n � should converge to 0. For a given

� � 0 � show there is a positive integer K such

that if n � K � then � � n� 0 � � � If

� � 0 � 1 � give one such positive integer K �

38. Let x be positive and an� [x]

�[2x]

� ����� �[nx]

n2� Show that � an � converges to x

2 by the squeeze limit theorem.

(Here [y] is the greatest integer less than or equal to y � )

39. Show that if x is a real number, then there is a sequence of rational numbers converging to x �

40. If limn � � an

� A � then show that limn � � �an � � � A � � (Hint: Show ��

� x � � � y � �� � x � y � for x � y �� first.) Is the

converse true?

41. If � an � converges to A � then

�an

�an 1

2 � should converge toA

�A

2� A � Prove this by checking the definition.

42. Let x1� 4 and xn 1

� 4 � 1 �xn �

4�

xnfor n � 1 � 2 � 3 � � � � � Plot the first 3 terms on the real line. Then prove the

sequence � xn � converges.

43. Show that the sequence � � 1 � 1

n� n � is increasing and bounded above.

44. Let � xn � be a bounded sequence in � Mn� sup � xn � xn 1 � xn 2 � � � � � and mn

� inf � xn � xn 1 � xn 2 � � � ��� for n ��� �(a) Prove that both sequence � Mn � and � mn � converge. (The limit of Mn is called the limit superior of xn and isdenoted by limsup

n � �xn � while the limit of mn is called the limit inferior of xn and is denoted by liminf

n � � xn � )

(b) Prove that limn � � xn

� x if and only if limn � � Mn

� x � limn � � mn (i.e. limsup

n � �Mn

� x � liminfn � � xn).

60

Page 61: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

45. Let x1� 1 � x2

� 2 and xn 1� xn

�xn � 1

2for n � 2 � 3 � 4 � � � � � Plot the first 4 terms on the real line. Then prove

the sequence � xn � converges.

46. If ��k � 2� xk

� xk � 1 � converges, then show the sequence � xn � is a Cauchy sequence by checking the definition of

Cauchy sequence.

47. If an � 0 for all n ��� and � an � is a Cauchy sequence, then show that � � an � is also a Cauchy sequence bychecking the definition of Cauchy sequence.

48. Let 0 k 1 � If � xn 1� xn � k � xn

� xn � 1 � for n � 2 � 3 � 4 � � � � � then prove that � xn � is a Cauchy sequence.

49. Prove that if � an � converges to A � then � � n � converges to A � where�

n� a1

�a2

� ����� �an

n� Show that the

converse is false.

(Hint: Let bn� an

� A and�

n� b1

�b2

� ����� �bn

n� then the first part of the problem becomes showing � � n �

converges to 0. Here use � bn � converges to 0 and so �bK0 � � �bK0 1 � � � � � is small when K0 is large. For n � K0 �

write�

n� b1

�b2

� ����� �bK0 � 1

n� bK0

� ����� �bn

n� �

50. If � xn � is bounded and all its convergent subsequences have the same limit x � then prove that limn � � xn

� x �

51. Let S � � 2 � n : n ���� and f : � � S is injective. Show that limn � � f � n � � 0 �

*52. Let � an � be a sequence satisfying limn � � � an 1

� an

2� � 0 � Prove that lim

n � � an� 0 �

*53. Let � xn � be a sequence satisfying limn � � � xn

� xn � 2 � � 0 � Prove that limn � �

xn� xn � 1

n� 0 �

*54. Let � xn � be a sequence and let y1� 0 and yn

� xn � 1�

2xn for n � 2 � 3 � 4 � � � � � If � yn � converges, prove that � xn �also converges.

55. For a sequence � an � of nonzero numbers, we say the infinite product ��

n � 1

an converges to a nonzero number L (or

has value L) if limk � � a1a2 ����� ak

� L � We say it diverges if the limit is 0 or does not exist. Determine if each of the

following infinite products converges or diverges. Find the values of the infinite products that converge.

(a) ��

n � 2

� 1 � 2n � n �

1 � � (b) ��

n � 2� 1 � 1

n2 (c) ��

n � 2

n3 � 1

n3 �1

(d) ��

n � 0

� 1 �z2n � for � z � 1 �

Remarks: In Apostol’s book, the following theorem is proved: In the case every an � 0 � we have ��

n � 1

� 1 �an � �

��

n � 1

� 1 � an � � ��n � 1an all converge or all diverge. Try to do the above exercises without using this theorem.

56. Prove that every bounded infinite subset of has an accumulation point. (This is also often called the Bolzano-Weierstrass theorem.)

61

Page 62: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

57. Let f : � 0 �

��� � � be defined by f � x � � x

x�

1� Show that lim

x � 1f � x � � 1

2by checking the definition.

58. Define f : � by

f � x � ��� 8x if x is rational,2x2 �

8 if x is irrational.

For which x0 � does limx � x0

f � x � exist? (Hint: Sequential limit theorem.)

59. If f : � is continuous and f � x � � 0 for every rational number x � show that f � x � � 0 for all x �

60. (a) Find all functions f : � � such that f � x �y � � f � x � �

f � y � for all x � y ��� �(b) Find all strictly increasing functions f : � such that f � x �

y � � f � x � �f � y � for all x � y �� �

61. For a function f : � � we say f has a local (or relative) maximum at x0 if there exists an open interval� a � b � containing x0 such that f � x � f � x0 � for every x � � a � b � � Similarly, we say f has a local (or relative)minimum at x1 if there exists an open interval � c � d � such that f � x � � f � x1 � for every x � � c � d � � If f : � iscontinuous and has a local maximum or a local minimum at every real number, show that f is a constant function.

62. If f � x � � x3� then f � f � x � � � x9 � Is there a continuous function g : [ � 1 � 1] � [ � 1 � 1] such that g � g � x � � � � x 9

for all x � [ � 1 � 1]? (Hint: If such a function g exists, then it is injective.)

63. A fixed point of a function f is a number � such that f � � � � � � Show that if f : [0 � 1] � [0 � 1] is continuous,then f has at least one fixed point. (Hint: Consider g � x � � f � x � � x � )

64. Let f : [0 � 1] � [0 � 1] be an increasing function (perhaps discontinuous). Suppose 0 f � 0 � and f � 1 � 1 �

show that f has at least one fixed point. (Hint: Sketch the graph of f and consider the set � t � [0 � 1] : t f � t � � .Does it have a supremum?)

65. Let f : � be a continuous function such that � f � x � � f � y � ��� � x � y � for all x � y � � Show that f isbijective. (Hint: Easy to show f is injective. To show f is surjective, let � � and M � � � � f � 0 � � , show that� and f � 0 � are between f � � M � and f � M � � )

*66. Let f : [0 � 1] � � 0 �

��� � be continuous and M � sup � f � x � : x � [0 � 1] � � Show that

limn � � �

� 1

0f � x � ndx � 1

n � M

if the limit exists. (Hint: M � f � x0 � � For every k � � � use the sign preserving property to show thatf � x0 � � 1

k f � x � M on an interval containing x0. Squeeze the integral.)

67. Find the derivatives of the functions f � x � ��

x2 if x �� 0x if x � 0

and g � x � � � cos x � �

68. Let f : � be differentiable at c and In� [an � bn] be such that I1 � I2 � I3 � ����� and � �n � 1

[an � bn] � � c � �

Prove that if an bn for all n ��� � then f � � c � � limn � �

f � bn � � f � an �bn

� an�

69. Let f � x � � � x � 3 for every x �� � Show that f � C2 �� � � However, f � � � � 0 � does not exist.

70. Let f : � satisfies � f � a � � f � b � � �a � b � 2 for every a � b �� � Show that f is a constant function. Ifthe exponent 2 in the inequality is replaced by a number greater than 1, must f be a constant function? If 2 isreplaced by 1, must f be a constant function?

62

Page 63: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

71. Let n be a positive integer and f � x � � � x 2 � 1 � n � Show that the n-th derivative of f has n distinct roots.

72. Let f : [0 �

� � � be continuous and f � 0 � � 0 � If f � � x � f � x � for every x � 0 � then show that f � x � 0 forevery x � [0 �

� � � (Hint: What if f � � x � � f � x � ?)

73. If f : � 0 �

��� � � is differentiable and � f � � x � � 2 for all x � 0 � then show that the sequence xn� f

� 1n�

converges. Also, show limx � 0

f � x � exists. (Hint: Check � xn � is a Cauchy sequence. For the second part, consider

the sequential limit theorem and the remarks following it.)

74. For 0 x 2

� prove that � ln � cos x � � x tan x �

75. Let f : [0 �

� � � be continuous and f � 0 � � 0 � If � f � � x � � � f � x � � for every x � 0 � show that f � x � � 0 forevery x � [0 �

� � � (Hint: Let � f � has maximum value M on [0 �12] � Apply the mean value theorem to f on [0 �

12 ] � )

76. Let f : � be differentiable. If f � is differentiable at x0 � show that

limh � 0

f � x0�

h � �f � x0

� h � � 2 f � x0 �h2

� f � � � x0 � �

77. Prove that if 0 � �2 , then

1 � � 2

2 cos � 1 � � 2

2� � 4

24�

(Hint: Apply Taylor’s theorem to the four times differentiable function cos � � )

78. Let f : � 0 �

��� � � be twice differentiable, M0� sup � � f � x � � : x � 0 � � � M1

� sup � � f � � x � � : x � 0 � �and M2

� sup � � f � � � x � � : x � 0 � � � Show that M21 4M0M2 � (Hint: Let h � 0 � Apply Taylor’s theorem to

f � x � with x � c�

2h � then solve for f � � c � � )

79. (a) If f : � a � b � � is differentiable and � f � � x � � 2 for all x � � a � b � � then show that f is uniformlycontinuous.

(b) Show that f : � 0 �

��� � � defined by f � x � � sin1

xis not uniformly continuous.

80. (a) Prove that if the union of a collection of open intervals contains [a � b] � then there are finitely many of theseintervals, whose union also contains [a � b] � (Hint: Suppose this is false. Let m1

� � a �b � � 2 � Then one of [a � m1]

or [m1 � b] is not contained in the union of finitely many of these open intervals. Proceed as in the proof of theBolzano Weierstrass theorem.)

(b) Give a proof of the uniform continuity theorem using part (a).

81. If f is continuous on [a � b] � f � x � � 0 for all x � [a � b] and� b

af � x � dx � 0 � then prove that f � x � � 0 for all

x � [a � b] �

82. Let f : [a � d] be a bounded function and a b c d �(i) Use the integral criterion to show that if f is integrable on [a � b] and [b � c] � then f is integrable on [a � c] �(ii) Use the integral criterion to show that if f is integrable on [a � d] � then f is integrable on [b � c] �

83. Let f : [a � b] � be bounded and � x � [a � b] : f is discontinuous at x � � � x1 � x2 � � � � � xn � � where x1 x2 ����� xn � Use the integral criterion to show that f is integrable on [a � b] � (Hint: Divide [a � b] into subintervalswhere f is continuous except at one endpoint and note part (i) of problem above.)

63

Page 64: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

84. (i) Let f � g : [a � b] � be bounded and P is a partition of [a � b] � Show that

L � f � P � �L � g � P � L � f

�g � P � U � f

�g � P � U � f � P � �

U � g � P � �

(ii) If f and g are integrable on [a � b] � show that

� b

a� f � x � �

g � x � � dx �� b

af � x � dx

� � b

ag � x � dx �

(Hint: First show RHS � � L � f � P � �L � g � P � L � f

�g � P � for some P using supremum property.)

85. Determine if each of the following improper integrals exists or not.

(a)� �

0

dx� ex(b)� �

0sin x dx

(c)� 1

0

dx

x2 �5x

(d)� 1

� 1

dx3� x

(e)� 1

0

dx

x � x � 1 � (f)� �

0

cos x

1�

x2dx

86. Find the principal value of each of the following integrals if it exists.

(a) P.V.� �� �

x

ex2 dx (b) P.V.� 2

0

dx

x2 � 1

87. Prove that for 0 x �� the improper integral � � x � �

� �0

t x � 1e � t dt exists. This is called the gamma

function. (Hint: Consider the cases x � � 0 � 1 � and x � [1 �

� � separately.)

Past Exam Problems

88. Define the following terms:(a) S is a countably infinite set(b) S is a countable set

(c) a series ��k � 1ak converges to a number S

(d) a nonempty subset S of is bounded above(e) the supremum of a subset S of that is bounded above(f) a sequence � xn � converges to a number x(g) a sequence � xn � is a Cauchy sequence(h) x is an accumulation point of a set S(i) f : S � has a limit L at x0

(j) f : S � is continuous at x0 � S (the�

- � definition)

89. For each of the following sets S, determine if it is countable or uncountable. Be sure to give reasons to supportyour answer.(a) S is the set of all intersection points � x � y � of the line y � x with the graphs of all equations y � x 3 �

x�

m �

where m ��� �(b) S is the set of all intersection points � x � y � of the graph of y � x 3 �

x�

1 with all lines y � mx � wherem ��� �

(c) S is the set of all intersection points � x � y � of the circle x 2 �y2 � 1 with all hyperbolas x y � 1

m� where

m ��� �

64

Page 65: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(d) S � � a �b �� : �a ��� M � b ����� � where M is a uncountable subset of the positive real numbers.

(e) S � � a �b �� : �a ��� M � b ����� � where M is a countable subset of the positive real numbers.

(f) S � � � � 2 � ��� a�

b � 2

c�

d � 2: a � b � c � d ��� � c

�d � 2 �� 0

��

(g) S � � x2 �y2 �

z2 : x � A � B � y ��� � A � z � B � ��� � where A is a nonempty countable subset of and B is an uncountable subset of �

(h) S � � x � y : x � y � A � � where A is an uncountable subset of �(i) S � � x2 �

y2 : x � y � A � � where A is a nonempty countable subset of �(j) S � � x �

sin y : x � � A � y ��� � � where A is a nonempty countable subset of �(k) S � � � x � y � �� 2 : x � A � y �� � A � � where A is a nonempty countable subset of �(l) S � � x �

y � 2 : x ��� � y � A � � where A is a nonempty countable subset of �(m) S � � � a �

b � 2 � c � 3 : a � b � c � T � � where T � � r : r ����� �(n) S � T � U � where T � � � and U � � � � m

� � n : m � n � N � � (Hint: Consider � � T � U � � )(o) S is the set of all squares on the plane that can be circumscribed by circles with rational radii and centers

with rational coordinates.(p) S is the set of all nonconstant polynomials with coefficients in G � where i � � � 1 and G � � a �

bi : a � b �� � �

90. Determine if each of the following series converges or diverges. Be sure to give reasons to support your answer.

(a) ��k � 1

cos k k2 �

2kand ��k � 1

e�

k

� k

(b) ��k � 1

� 2k � !3kk4

and ��k � 1

� cos k � � sin 2k �2k

(c) ��k � 1

12 � cos

1k

�sin

1k and ��k � 1

sin� 1k

� 1k

�1�

(d) ��k � 1

2k �3k

1k �4k

and ��k � 1cos � sin

1k�

(e) ��k � 1

21 � k �31 � k

11 � k �41 � k and ��k � 1

� cos k � � sin1

k �(f) ��k � 1

� k! � 2

� k2 � !and ��k � 1

� cos1

k� � sin

1

k� � tan

1

k�

(g) ��k � 2

2k cos k

� k � 1 � !and ��k � 2

sin � 1k �

ln k

(h) ��k � 1

k � �cos k

3�

k4and ��k � 1

k � cos k 3k4

(i) ��k � 2

� 2k � !

� k �1 � ! � k � 1 � !

and ��k � 1k cos � 1

k2�

(j) ��k � 1

� 3k � !k! � 2k � !

and ��k � 2

cos � 1 � k �k2 � 1

(k) ��k � 1

k!� 2k � 1 � !

and ��k � 1

cos k � k�

1

(l) ��k � 1

2kk2

k!and ��k � 1

1� ksin � 1� k

�(m) ��k � 1

1k cos k and ��k � 1

k2 sin � 1 � k �� 2k

�1 � !

(n) ��k � 1cosk � 1 � 1

k� and ��k � 1

cos � sin � 1 � k � �sin � cos � 1 � k � �

91. Determine if each of the following set has an infimum and a supremum. If they exists, find them and give reasonsto support your answers.

(a) S ��� 1

m� 1

n: m � n ��

�� � 2

k: k ���

�65

Page 66: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

(b) S ��� x �y : x � y �

� 12

� 1 � � � � 2 � 1

n: n ���

�(c) S � � x � 1

n: x � [0 � 1] � � � n ���

��� � 1 �

1

2�

(d) S ��� x � x

� : x � �� � � � � [ �

� ��

(e) S ��� x � x

� : x ��� � [0 �

� ��

(f) S � � x3 �y3 : x ��� � [0 � 1] � y � � � 1 � 0] �

(g) S ��� � 2

m�

n� 1

k � 2: m � n � k ��

�(h) S � ��

k � 1

[1 � 1

2k � 1� 1 � 1

2k�

(i) S ��� � x�

y2 : x � y � � 0 � 1] � ��

(j) S � � 1

n�

x : x � [0 � 1] � � � n � 1 � 2 � 3 � � � ��

(k) S � � x �y : x � [0 � 1] � � � y � [0 � 1] � �� � � � �

(l) S � � x �� : x � x �1 � 0 and x �� � ���

(m) S � � kn! : k � n � N �

kn! � 2 �

(n) S � 10�n � 1

� � 1

n � 2� 2 � 1

n

�� �

(o) S � � x2 �y3 �

z4 : x � � � 1 � 0 � � � � y � � 0 � 1 � � � � z � � � 1 � 1 � �

92. For each of the following sequences � xn � � show it converges and find its limit.

(a) x1� 1 and xn 1

� xn

2� � xn for n � 1 � 2 � 3 � � � � �

(b) x1� 1 � x2

� 2 and xn 1� � xn

� � xn � 1 for n � 2 � 3 � 4 � ����� �(c) x1

� 1 and xn 1� 2 � xn

3�

xnfor n � 1 � 2 � 3 � � � � �

(d) x1� 15

16 and xn 1� 1 � � 1 � xn for n � 1 � 2 � 3 � ����� � Also, do the sequence � xn 1

xn

��

(e) xn� an 1

an� for n � 1 � 2 � 3 � � � � � where a1

� 1 � a2� 2 and an 1

� an�

an � 1 for n � 2 � 3 � 4 � ����� �

(f) x1� 1 and xn 1

� 1 � 1

4xnfor n � 1 � 2 � 3 � � � � �

(g) x1� 4 and xn 1

� 12� xn

� 4xn� for n � 1 � 2 � 3 � � � � � (Hint: Sketch the graph of f � x � � x

� 4x

for x � 2 � )

(h) x1� 5 and xn 1

� 3� 4

xnfor n � 1 � 2 � 3 � ����� �

(i) x1� 2 and xn 1

� 2 � 1

xnfor n � 1 � 2 � 3 � ����� �

(j) x1� 0 and xn 1

� x2n

�4

5for n � 1 � 2 � 3 � � � � �

(k) x1� 1 and xn 1

� � xn� 1

4for n � 1 � 2 � 3 � � � � �

(l) x1� 3 and xn 1

��� 1 � 1

xn�

1for n � 1 � 2 � 3 � � � � �

(m) 0 x1 1 and 7xn 1� x3

n�

6 for n � 1 � 2 � 3 � � � � � (Hint: x 3 � 7x�

6 � � x � 1 � � x � 2 � � x �3 � � )

(n) x1 � [1 �

� � and xn 1� � 3xn

� 2 for n � 1 � 2 � 3 � � � � �(o) x1

� 0 � x2 � [0 �12 ] and xn 1

� 13 � 1 �

xn�

x3n � 1 � for n � 2 � 3 � 4 � � � � �

93. If � an � is a decreasing sequence (of nonnegative real numbers) converging to 0, show that the sequence � xn �

66

Page 67: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

converges, where

xn� n�

k � 1� � 1 � k 1ak

� a1� a2

�a3

� ����� � � � 1 � n 1an �

94. Let 0 a b and a1� a � b1

� b � an 1� an

�bn

2� bn 1

��

a2n

�b2

n

2for n � 1 � 2 � 3 � � � � Show that � an �

and � bn � both converge and their limits are the same.

95. (i) If a b and 0 t 1, then show that a ta� � 1 � t � b b �

(ii) Let

x1� 1 � x2

� 2 and xn 1� 1

3xn

� 23

xn � 1 for n � 2 � 3 � 4 � ����� �

Show that the sequence � xn � converges.

96. Show that the sequence � xn � given by

x0� 0 � x1

� 1 and xn 1�

�1

4x2

n� 3

4x2

n � 1 for n ���

converges and find its limit.

97. Let S be a set of real numbers such that every sequence in S has a convergent subsequence, show that S is bounded.

98. Let A be a nonempty subset of the nonnegative real numbers. If A is bounded above and B � � x 2 �y2 : x � y � A � �

show B is bounded above and sup B � 2 � sup A � 2 �

99. Suppose An � � � �� 2 � and xn

� sup An for n � 1 � 2 � 3 � � � � � 10 � Prove that

sup� 10�k � 1

An � � max � x1 � x2 � � � � � x10 � �

100. Use the definitions of infimum and supremum to explain the statement: Let f : � � [0 � 1] be a function,g � x � � sup � f � x � y � : y ��� and h � y � � inf � f � x � y � : x ��� � Show that sup � h � y � : y ��� inf � g � x � : x �� �

101. Show that for every x �� � there is a strictly increasing sequence of irrational numbers � xn � converging to x �

102. Prove that limn � �

� 1

n2� � 2

n3� � 0 by checking the definition of limit.

103. Prove that limn � �

� 2

n�

1� 1

n2� � 0 by checking the definition of limit.

104. Given limn � � xn

� 0 � Prove that limn � �

�xn

� 1n� � 0 by checking the definition of limit.

105. Given limn � � xn

� 12

� Prove that limn � � xn

n� 0 by checking the definition of limit.

106. Given limn � � xn

� 8 � Prove that limn � �

3� xn� 2 by checking the definition of limit.

67

Page 68: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

107. Given sequences � xn � and � yn � both converge to A � For n ��� � let zn� max � xn � yn � � Show that � zn � converges

to A by checking the definition.

108. If � xn � is a sequence such that � xk 1� xk �

1

2kfor k � 1 � 2 � 3 � � � � � then show that � xn � is a Cauchy sequence.

109. (a) Let S be an open interval and x0 � S � For a function f : S � � state the definition of f � x � converges to L(or has limit L) as x tends to x0 in S �(b) Let f : � 1 � 3 � � be defined by f � x � � x 2 � 1

x� Prove that lim

x � 2f � x � � 9

2by checking the definition.

(c) Let f : � 1 � 4 � � be defined by f � x � � � x 2 � 9 � � Prove that limx � 2

f � x � � 5 by checking the definition.

110. If f � g : � 0 � 1 � � are increasing functions, show that h � x � � max � f � x � � g � x � � has countably many jumps onthe interval � 0 � 1 � �

111. Give an example of a function f : � that is continuous at x ��� � but discontinuous at every x ���� � Be sureto give reasons to support your example.

112. (a) State the intermediate value theorem.(b) Let f : [0 � 2] � be continuous and f � 0 � � f � 2 � � Show that there exists c � [0 � 1] such that f � c � � f � c �

1 � �(c) Show that there is a nonzero function g : � such that g � t � �

g � 2t � �g � 3t � � g � 4t � �

g � 5t � for everyt � � (Hint: Try g � t � � � t � r for some constant r � )

113. (a) Show that the set T � � x : sin x ����� is countable. (Hint: For a fixed rational r � how many solutions ofsin x � r are there in the interval [k � k �

2 � ?)(b) If f : [0 � 1] � is continuous and sin f � x � ��� for every x � [0 � 1] � then show that f is a constant function.

114. Show that there is no continuous function f : � such that for every c �� , f � x � � c has exactly 2 solutions.

115. Suppose f : � is a function such that f � x �y � � f � x � �

f � y � for every x � y �� and � f � x � � x 4 � � x � forevery x �� 0 �(a) Show that f is continuous at some x �� �(b) Show that f is continuous at every x � �(c) Give an example of such a function.

116. Suppose f � g : [1 � 2] � [3 � 4] are continuous functions and also � g � x � : x � [1 � 2] � � [3 � 4] � Show that there isc � [1 � 2] such that f � c � � g � c � �

117. Let f : � be continuous such that � f � x � � f � y � � 12 � x � y � for every x � y �� �

(a) Let � � � Define x1� � and xn 1

� f � xn � for n �� � Show that � xn � is a Cauchy sequence. (Hint: How is� xk 1

� xk � compared to � x2� x1 � ?)

(b) Show that there is x � such that f � x � � x �

118. Give an example of a function f : � 0 � 2 � � that is differentiable for every x � � 0 � 2 � � but f � � x � is notcontinuous at x � 1 � Be sure to give reasons to support your example.

119. Let f : � 0 � � � satisfies � � f � a � � f � b � � sin �a � b � for every a � b � � 0 � � � Show that f is a constantfunction.

120. (a) State the mean value theorem.(b) For the function f � x � � sin x � find the smallest constant K such that � f � b � � f � a � � K �b � a � holds forevery a � b �� �

121. If f : � is differentiable and limx � 0

f � � x � exists, then show that f � is continuous at 0.

68

Page 69: Math 201 (Introduction to Analysis)makyli/201/201_04-05Sp/201_SylLectNt.pdf · Math 301 (Real Analysis). This course will focus on the proofs of basic theorems of analysis, ... Try

122. Use the mean value theorem to prove that f � x � � sin 5x is uniformly continuous.

123. Prove that if f : � � 0 �

��� � is uniformly continuous, then the function g � x � � � f � x � is also uniformlycontinuous.

124. (a) State Lebesgue’s theorem.(b) Let f � g : [0 � 2] � [0 � 1] be Riemann integrable. Prove that the function h : [0 � 2] � [0 � 1] defined by

h � x � ��

f � x � if x � [0 � 1 �g � x � if x � [1 � 2]

is Riemann integrable.

125. If f � g : [0 � 1] � are Riemann integrable, show that the function h � x � � min � f � x � � g � x � � is Riemannintegrable on [0 � 1] �

126. For every positive integer n � give an example of a Riemann integrable function fn : [0 � 1] � [0 � 1] such thatlim

n � � fn � x � exists for every x � [0 � 1] � but the function f � x � � limn � � fn � x � is not Riemann integrable on [0 � 1] �

Be sure to give reasons to support your example.

127. (a) Determine if the improper integral� �� �

cos 3x

1�

x2dx exists or not.

(b) Determine if the principal value P.V.� �� �

cos 3x

1�

x2dx exists or not.

(c) Determine if the improper integral� 1

� 1

13� x

dx exists or not.

(d) Determine if the principal value P.V.� 1

� 1

13� x

dx exists or not.

(e) Determine if the improper integral� �� �

sin x dx exists or not.

(f) Determine if the principal value P.V.� �� �

sin x dx exists or not.

69