SETS AND NUMBERS STRUCTURE 1 . 1 Introduction

Objectives

1.2 Sets and Functions Sets, Functions

1.3 'system of Real Numbers ,Natural Numbers Integers Rational Numbers

1.4' The Real Line

1.5 The Complex Numbers

1.6 Mathematical Induction

1.7 Summary

1.1 INTRODUCTION

One of the main features of Mathematics is the identification of the subject matter, its analysis and its presentation in a satisfactory manner. For this, we necd a simple language- a language that admits minimum vocabulary and an easy grammar- a language that is precise and has clear meanings. In other words, thc language should be a vehicle which carries ideas through the mind without affecting their meaning in any way. Set Theory comes closest t o being such a language. Introduced bctween 1873 and 1895 by a famous German mathematician, George Cantor (1 845-1918), Set Theory became the foundation of almost all the branches of Mathematics, Besides its universal appeal, it is quite amazing in its simplicity ?nd elegance.

A rigorous presentation of Set Theory is not the purpose here be~ause we believe that you are already familiar with it. Yet, we shall briefly recal! some of its basic concepts which are essential for a systematic study of Real Analysis. Closely linked with the sets, is the notion of a function, which also you have learnt in your previous studies. In this unit, we shall review this as well as other related concepts which are necessary for our discussion.

'Real Analysis' is an important branch of Mathematics which mainly deals with the study of real numbers. What is, then, the system of the real numbers'? We shall try t o find an answer to this cluestion as well as some other related questions in this unit. *

Also, we shall give the geometrical representation of the real numbers. This will help us In d1:cussing the algebraic structure of real numbers in Unit 2 and some relatcd . ,pects 111 Unit 3 and Unit 4.

In this unit, therefore, you should be able to

+recall the basic concepts of sets and functions,

*discuss the development of the system of real numbers,

-describe the geometrical representation of real numbers.

Most of modern Mathematics is based on the ideas that are expressed in .the language of sets and.functions. In this section, we shall give a brief review of certain basic . concepts of Set Theory which are quite familiar to you. These concepts will also serve an important purpose of recalling certain notations and terms that will be used

,throughout our discussion with' you. Also, this will be useful as a background material for what is going t o be discussed in the subsequent units and blocks.

1.2.1 SETS You are used to.the phrases like the 'team' of cricket players, the 'army' of a country, , the 'committee' on the education policy, the 'panchayat' of a village,, etC. The terms

IVumbers and Functions 'team', 'army', 'committee', panchayat', etc. indicate the notion of a 'collection' or I

'totality' or 'aggregate' of objects. These are well-known examples of a set. I

Therefore, our starting point is an informal description of the term 'set'. A set is treated as an undefmed term just as a point in Geometry is undefined. However, for our purpose 1 we say that a set is a well-defined collection of objects. A collection os well-defined of it '

is possible to say whether a given object belongs to the collection or not. I

The following are some examples of sets: 1 i) The collection of the students registered in Indira Gandhi National Open University.

ii) The collection of the planets namely jupiter, saturn, Earth, Pluto, Venus, Mercury, Mars, Urmus and Neptune.

iii) The collection of all the countries in the world. (Do you know how many countries I

are there in the world?)

iv) The collection of u~lmbers, 1, 2, 3, 4, ......

If we consider the collection of tall persons or beautiful ladies or popular leaders, then these collections are not well-defined and hence none of them forms a set. The reason is that the words 'tal' 'beautiful' or 'popular' are not well-defined. The objects constituting a set are called its elements or members or points of the set. Generally, sets are denoted by the capital letters A, B, C etc. and the elements are denoted by thesmall letters a, b, G,

etc. If S is any set and x is an element of S, we express it by writing that XES, where the symbol Erneans 'belongs to' or 'is a member of.If x is not an element an element of a set S, we write xc$ S. For example , if S is the set containing 1, 2, 3, 4 only, then 2 ES and 54 s.

You know that there are two method of describing a set. One is known as the Tabular method and the other is the Set-Builder method. in the tabular method we describe a set by actually listing all the elements belonging to it. For example, if S is the set consisting of all small letters of English allphabet, then we write

S = {a, b, c ,.... x, y, 2).

If N is the set of all natural numbers, then we write - N = {1, 2, 3 .... 1.

This is also called an explicit representation of a set.

In the set-builder method, a set is described by stating the property which determines the set as a well-defined collection . Suppose pdenotes this property and x is an element of a set S. Then

S = (x:x satisfies p).

For example, the two sets S and N can be writen as

S = {x: x is a small letter of English alphabet}

N = {n: n is a natural number).

This is also called an implicit representation of a set. I

SOLUTION : Tabular form is S = (7, 8, 9, 10, 1 1, 12, ). ? Set-Builder form is S = ( n EN: 75 n I 12, 1, i l

1 1

EXERCISE 1)

Note that in the representation of sets, the elements of a set are not repeated. Also, the 1

(i) Write the following in the set-builder from : A = (2, 4, 6, .... ). A = {I, 3, 5, ,.., ).

elements may be listed in any manner.

EXAMPLE 1: Write the set S whose elements are all natural numbers between 7 and Y

I

12 including both 7 and 12 in the tabular as well as in the set- builder forms. 1

(ii) Write the following in the tabular form: A= {x:x is a factor of 15). A= {x:x is a natural numbel- between 20 and 30). A= {x:x is a negative integer}.

The fofollowing standard notations are used for the sets of numbers:

N = Set of all natural numbers = { 1, 2, 3. . . . ) = {n:n is a natural number) = Set of all positive integers.

Z = Set of all integers = { .... -3,-2,-1, 0, 1,2, 3, .... ) = {p:p i$ an integer).

Q= Set of all rational numbers P

= { x : x = -: p E Z , ~ E z, q st 0). a

R= Set of real'numbel-s = {x :x is a real number).

We shall, however, discuss the development of the systeln of real numbers in Section 1.3.

A set is said to be finite if it has a finitc nuber of elements. A set is snid to be infinite if it is not finite. We shall, however, give a nathmatical dcfi~iitisn of finite and inti~iile sets in Unit 2.

Note that an element of a set must be carefully distinguished from thc set consisti~ig of this element. Thus, for instance, you must distinguish

from each other

We talk of equality of numbers, eq~~ality of objects etc,

The question, therefore, arises: What is the notion of t.he ecluality of sets'?

DEFlMTION 1 : EQUALITY OF SETS

Any two sets sre equal if tllet arc identical. Thus the two sets S and T are equal, written as S = T if they consist of exactly thc same elenlents. When the two sets S and T are unequal, we write

S # T .

It follows from the definition that S = T if any or~ly of x ES implies x ET and y ET implies YES. Also S is different from T (S +T) if there is at least one element in one of them which is not in the other.

If every member of a given set S is also a member of T, the11 we say that S is a subset of T or " S is contained in Tf ' , and write:

or equivaleiltly we say thet "T contains S" or T is a superset of S, alcl write

T z S

The relation

means that S is not a subset of T i.e. tlrere is at least one element in T which is not in S. .

Thus , you can easily see that any-two sets S and T are equal if and only if S is a subset of T and T is a subset of S i.e.

Ren! Numbers and Functions If S c T but Tc S, then we say that S is a proper subset of T. Note that S c S i.e. every set is a subset of itself.

Another important concept is that of a set having no elements Such a set, as you know, is called an empty set or a null set or a void set and is denoted by 0.

You can easily see that there is only one empty set i.e. 0 is unique. Futher 0 is a subset of every set.

Now why don't you try an exercise? I - -

EXERCISE 2)

Justify the following statements:

(i) The set N is a proper subset of 7,.

(ii) The set R is not a subset of Q.

(iii) I f A, B, C are any three sets such that A c B, and B c C, then A c C.

So far, we have talked about the elements and subsets of a given set. Let us now recall the method of constructing new sets from the given sets.

While studying subsets, \lire generally fix a set and consider the subsets of this set throughout our discussion. This set i:. usually &led the Universal set. This Universal set may vary from situations to situations. Fgr example, when we con-ider the subsets of R, then R is the Universal set. When we consider the set of points in the Eucliclean plane, then the set o f all points in the Euclidean plane is the Universal set. We shall denote the Univeral set by X.

Now, suppose that thc Universal set X is given as

x = (.1,2,3,4,5) and

S = { l , 2 ,3 ) is a subset of X. Conisder a subset of X whose elements do not belong to S. This set is {4,2). Such, a set, as you know is called the complement of S.

We define the complement of a set as follows:

DEFINITION 2: COMPLEMENT OF A SET

Let X be the Universal set and S be a subset of X. The complement of the set S is the set of all those elements of the Universal set X which do not belong to S. It is denoted by S. '

1

Thus, if S is an.arbitarary set contained in the Universal Set X, then the complement of 1

S is the set I i

Associated with each set S is the Power set P(S) of S consisting pf all the subsets of S. It is written as

P(S) = {A : A c S).

. NO" try the following exercise. 1 i

~ R C L S E 3)

Let X be universal set and let S be a subset of X. Prove that

(i) P(0) = {a). (ii) (St)' = S.

Let us consider the sets S and T given as

S = { l , 2 , 3 , 4 , 5 ) , T = { 3 , 4 , 5 , 6 , 7 ) . Construct a new set (1, 2, 3, 4, 5, 6, 7 ).Note that all the elements of this set have been taken from S or T such that no element of S and T is left out. This new set is called the union of the sets S and T and is denoted by S u T.

Thus

S u T = {1 ,2 ,3 ,4, 5,6, 7).

Again let us construct another set (3, 4, 5). This set consists of the elements that are common to both S and T i.e. .a set whose elements are in both S and T. This set is called the intersection of S and T. It is denoted by S n T. Thus S n T = (3, 4, 5).

These notions of Union and Intersection of 'two sets can be generalized for any ab: sets in the following way: Note the all the sets under discussion will be treated as subsets of the Universal set X.

DEFINITION 3 : UNION O F SETS

Let S and T be given sets. The collection of all elements which belong to S o r T is called the unio~n of S and T. I t is expressed as

Note that when way we say that x E S or x ET, then it means that x belong to S or x belongs to T or x belongs to both S and T.

DEFINITION 4: INTERSECTION O F SETS

The intersection S n T of the sets S and T is defined to be the set of all those elements which belong to both S and T i.e.

S n T = {x:x ES and xeT).

Note that the sets sre disjoint or murually exclusive when SnT=W i.e., when their intersection is empty.

You can now verify (or even prove ) by means of examples the following laws of u ~ ~ i o l ~ and intersection of sets given in the next exercise.

EXERCISE 4)

Let A, B and C be any three sets. Then prove the following:

(i) AuB=BuA,AnB=Bn A (Commutative laws).

(ii) Au(BuC)=(A uB) u C , An(B nC)=(A n B ) nC(Assciative laws).

(iii) Au@ nC)=(AuB)n(AuC)

An(BuC)=(AnB)u(A n C ) (Distributive laws).

(iv) ( A u B ) ~ = A ~ ~ B ~ , (A nB)C=ACuBC (De Morgan laws).

Also, you can easily sec that

Given any two sets S and T, we can construct a new set in such a way that it contains only those elements of one of the sets which do not belong to the other. Such a set is called the difference of the given sets. There will be two such sets denoted by S-T and T--3. For example, let

S ={2,4, 8, 10, 111, T={l, 2, 3.4). Then S--T={8, 10, 11), T-S={1, 3).

Thus, we can define the difference of two sets in the following way:

DEFINITION 5: DIFFRENCE O F TWO SETS Given two sets S and T, the difference S-T is a set consisting of precisely those members of S which are not in T. Thus

S-T={x:x E S and x eT).

Similarly, we can define T-S.

Sets and MUIIPI~IPCI'Y

' Real Numbers and Functions Consider a collection of sets S,, where i varies over some index set J. This simply means that to each element i EJ, there is a corresponding set Sin For example, the collection {S,, S,, S ,,... )could be expressed as EN, where N is the index set.

With the introduction of an index set, the notions of the union and the intersection of sets can be extended to an arbitrary collection of sets. For example

(i) . u Si = u {x : x €Si for at least one i EJ). 1 E J 1 EJ

(i) n Si = n {x : x ES; for ati i EJ). i a 1 i EJ

(iii) (u S,)f ., = n q. I €1

1.2.2 FUNCTIONS

Let S be the set of all books in IGNOU library and let N be the set of all natural numbers. Assign to each book the number of pages the book contains. Here each book corresponds to a unique natural number. In other words, there is a correspondence between the books and the natural numbers i.e., there is a rule or a mechanism by which we can associate to each book one and only one nahlral number. Such a rule o r correspondence is named as a function or a mapping.

DEFINITION 6: FUNCTION

Let S and T be any two non-empty sets. A function f from S to T denoted as f:S-+T is a rule which asigns to each element of the set S, a unique element in the set T.

The set S is called the domain of the function f and T is called its co-domain. If an elements x in S corresponds to an element y in T under the function f, then y is called the image,of x under f. This is expressed by writing y = f (x). The set {f(x): x ES) wliich is a subset of T is called the range of f . If range of f = co-domain of f, then f is called onto or surjective function; otherwise f is called an into function.

Thus, a function f: S+T is said to be onto if the range of S is equal to its codomain T.

Suppose S = ( 1 , 2, 3, 4) and T = { I , 2, 3, 4, 5, 6) and f : S t T is defined by f(n) = n+l,V n ES. Then the range o f f ={2, 3, 4, 5). This shows that f is an into function. On the other hand, if S = { l , 2, 3, 41, T = (1, 4, 9,161 and if f:S+T is defined by f(n) = nZ, then f is onto. You can verify that the range of f is, in fact, equal to T.

Refer back to the example on the books in IGNOU. It just possible that two books may have the same number of pages. If it is so, then under this function, two different books ' shall have the same natural number as their image. However if for a hnction any two distinct elements in the domain have distinct images in the co-domain, then the hnction is called one-one or injective. ,

Thus a function f is said to be one-one if distinct elements in the domain o f f have distinct image or in other words, if f(x,) = f(xz) 3 x,=x,, for any x,, x, in the domain of f.

A function which is one-one and onto, is called a bijection o r a 1-1 correspondence.

EXAMPLE 1: (i) Let S'= {1,2, 3) and T = {a, b, c) and let f: S-+T be defined as f(1) = a, f(2) = by f(3) = c. Then f is one-one and onto.

(ii) Let N = (1, 2,3, 4, ...) and f : N+N be defined as f(n) = n+l. As 1 does not belong to the range of f, therefor f is not onto. However, f is one-one . (iii) Let S = (1, -1, 2,3, -3) and let T = (1, 4, 9). Define f: S+T by f(n) = n2

I

V n ES. Then f is not one-one as f(1) = f(-1) = 1. However, f is onto. I

DEFlNlTION 7: IDENTITY FlNC'I'ION

Let S be any non-empty set. A function f: S+S defined by f(x)= x for each x in S is called the identity function.

It is generally denoted by Is. It is easy to see that IS is one-one and onto.

DEFINITION 8: CONSTANT FUNCTION Let S and T be any two non-empty sets. A function CS+T defined by f(x) = c, for each x in S, where c is fixed element of T, is called a constant function.

For example f: S+R defined as f(x)=2, for every x in S, is a constant function. Is this function one-one and onto? Verify it.

DEFINITION 9 : EQUALITY OF FUNCTIONS

Any two functions with the same domain are said to be equal if for each point of their domain, they have the same image. Thus i f f and g are any two functions defined on an non-empty set S, then

f = g if f(x) = g(x), b' x E S. In other words, f = g if f and g are identical.

Let us now discuss another important concept in this section. This is about the composition or cot~lbinatiori of two function. Consider the followilig situation:

Let S =: {I, 2, 3, 41, T ={I, 4, 9, 161, N ={I, 2, 3, 4 .... } be any three sets Let C S+T be defined by f(x)= x2, b' x xS and g:T+N be defined by g(x)= x +I, b' x xT. Then, by the function f, an elernent x ES is mapped to f(x) = x2. Further by the hnction g the element f(x) is mapped to f(x)+ I = x2 + 1. Denote this as g(f(x)). Define a function h:S+N by h(x) = g(f(x)). This f~lnction h maps each x in S to some unique elements g(f(x))=x2+1 of N. The function h is called the composition or the composite of tlie functions f and g. Thus, we have tlie following defintion:

DEFINITION 10: COMPOSITE OF FUNCTIONS

Let f: S+T and g: T+V be any two functions. A function 11: S+V denoted as h=gof and defined by

h(x)=(gof) (x) = g(f(x)), b'x ES is called the composite o f f and g.

Note tliat the domain of the composite tunction is the set S and its co-domain is the set V. The set T which contains the range of f is equal to the domain of g.

But in general, the composition of the two functions is meaningful whenever the range of the first is contained the domain of the second.

EXAMPLE 3: Let S = T = 11, 2, 3, 4 .... ). Define f(x) = 2x and g(x) = x + 5. Tl~en "gof is defined as (goo (x) = g(f(x)) = g(2x) = 2x + 5.

Note tliat wc can also define fog the composite of g and f. Here (fog) (x) = f(g(x)) = f(x + 5) = 2 (x + 5) = 2x = 10. Also (fog) (I) = 12 and (gof) ('I) = 7. This shows that 'fog' need not be equal to 'gof.

Lei S = { I , 2, 3)and T = {a, b, c). Let f:S+T be f(l) = a, f(2) = b, f(3) = c. Define a function g:T+S as g(a) = 1, g(b) = 2 and g(c) = 3. Under the function g, the element f(x) in T is taken back to the element x in S. This mapping g is called the inverse of f and is given by g(f(x)) = x, for each in S. You may note that f(g(a)) = a, f(g(b)) = b and f(g(c))=c. Thus, we have the following definition:

DEFINITION 11 : W E W E OF A FUNCTION

Let S and T be two non-empty sets. A function f:S+T is said to be invertible if there exists a function g:T+S such that

(gof)(x)=x for each x in S, and

(fog)(x)=x for each x in T. In this case, g is said to be the inverse of f and we write it as g = f-'.

You may ask: Do all function possess inverses?

No, all functions do not possess inverses. For example, let S = (1, 2, 3)and T ={a, b). If f:S+T is defined as f(1) = f(2) = a and f(3) = b, then f is not invertible. For, if g:S+ T is inverse o f f , then

(g f ) (I)= g(f(l)) = g(a)

Sets and Numbel-s

and (g f ) (2) = g(f(2)) = g(a). Therefore, 1 = 2, which is absurd.

' Real Numbers and Functions Consider a collection of sets Si, where i varies over some index set 9. This simply means that to each element i EJ, there is a corresponding set S,. For example, the collection {S,, S,, S ,,... )could be expressed as {S,), EN, where N is the index set.

With the introduction of an index set, the notions of the union and the intersection of sets can be extended to an arbitrary collection of sets. For example

(i) . US, = u {x : x €Si for atleast one i EJ). 1 EI 1 €1

(ii) n S i = n { x : x ~ S i f o r a l i i ~ J ) . i el 1 sl

(iii) (U S,); ,, = n Sf. i E J

1.2.2 FUNCTIONS

Let S be the set of all books in IGNOU library and let N be the set of all natural numbers. Assign to each book the number of pages the book contains. Here each book corresponds to a unique natural number. In other words, there is a co~~espondence between the books and the natural numbers i.e., there is a rule or a mechanism by which we can associate to each book one and only one natural number. Such a rule or correspondence is named as a function or a mapping.

Let S and T be any two non-empty sets. A function f from S to T denoted as CS+T is a rule which asigns to each element of the set S, a unique element in the set T.

The set S is called the domain of the function f and T is called its co-domain. If an elements x in S corresponds to an element y in T under the function f', then y is called the image.of x under f. This is expressed by writing y = f (x). The set {f(x): x ESJ which is a subset of T is called the range of f. If range of f = co-domain of f, then f is called onto or surjective function; otherwise f is called an into function.

Thus, a function f: S+T is said to be onto if the range of S is equal to its codomain T,

Suppose S = { l ,2 , 3, 4) and T = (1, 2, 3, 4, 5, 6) and f:S+ T is defined by f(n) = n+l,V n ES. Then the range of f ={2, 3, 4, 5). This shows that f is an into function. On the other hand, if S = {I, 2, 3, 41, T = (1, 4, 9,161 and if f:S+T is defined by f(n) = n2, then f is onto. You can verify that the range of f is, in fact, equal to T.

Refer back to the example on the books in IGNOU. It just possible that two books may have the same number of pages. If it is so, then under this function, two different books shall have the same natural number as their image. However if for a function any two distinct elements in the domain have distinct images in the co-domain, then the function is called one-one or injective.

Thus a function f is said to be one-one if distinct elements in the domain of f have distinct image or in other words, if f(x,) = f(x,) x,=x,, for any x,, x, in the domain of f.

A function which is one-one and onto, is called a bijection or a 1-1 correspondence.

EXAMPLE 1: (i) Let s'= (1, 2, 3) and T = {a, b, c ) and let f: S+T be defined as f(1) = a, f(2) = b, f(3) = c. Then f is one-one and onto.

(ii) Let N = {1,2, 3, 4, ...) and f : N+N be defined as f(n) = n+l. As 1 does not belong to the range of f , therefor f is not onto. However, f is one-one . (iii) Let S = (1, -1, 2, 3, -3) and let T = (1, 4, 9). Define f: S+T by f(n) = n2 V n ES. Then f is not one-one as f(1) = f(-1) = 1. However, f is onto.

DEF'INrrION 7: IDENTITY FINCIION

Let S be any non-empty set. A function f: S+S defined by f(x)= x for each x in S is called the identity function.

It is generally denoted by Is. It is easy to see that IS is one-one and onto,

DEFINlTION 8: CONSTANT FUNCTION Let S and T be any two non-empty sets. A function f:S+T defined by f(x) = c, for each x in S, where c is fixed element of T, is called a constant function.

For example f: S+R defined as f(x)=2, for every x in S , is a constant function. Is this bnction one-one and onto? Verify it.

DEFINITION 9 : EQUALITY OF FUNCTIONS

Any two functions with the same domain are said to be equal if for each point of their domain, they have the same image. Thus i f f and g are any two functions defined on an non-empty set S, then

f = g if f(x) = g(x), V x ES. In other words, f = g i f f and g are identical.

Let us now discuss another important concept in this section. This is about the composition or combination of two function. Consider the following situation:

Let S = { I , 2, 3, 4), T ={I, 4, 9, 16), N ={l, 2, 3, 4 ....I be any three sets Let f: S+T be defined by f(x)= x2, V x E S and g:T+N be defined by g(x)= x +1, V x ET. Then, by the function f, an element x E S is mapped to f(x) = xZ. Further by the function g the element f(x) is mapped to f(x)+ 1 = x2 + 1. Denote this as g(f(x)). Define a function h:S+N by h(x) = g(f(x)). This function h maps each x in S to some unique elements g(f(x))=x2+1 of N. The function h is called the composition or the composite of the functions f and g. Thus, we have the following defintion:

DEFINITION 10: COMPOSITE OF FUNCTIONS

Let f: S+T and g: T+V be any two functions. A function h: S+V denoted as h=gof and defined by

h(x)=(gof) (x) = g(f(x)), Vx ES is called the composite of f and g.

Sets and Numbers

Note that the domain of the colnposite tunction is the set S and its co-domain is the set V. The set T whicli contains the range of f is equal lo tlie domain of g.

But in general, the compositioli of the two functions is meaningful whenever the range of the first is contained the domain of the second.

EXAMPLE 3: Let S = T = { I , 2, 3, 4 .... }. Define f(x) = 2x and g(x) = x + 5. Tlien "gof is defined as (go9 (x) = g(f(x)) = g(2x) = 2x -t 5.

Note that we can also define fog the composite of g and f, Here (fog) (x) = f(g(x)) = f(x + 5) = 2 (x + 5) = 2x = 10. Also (fog) (1) = 12 and (gof) (1) = 7. This shows that 'fog' need not be equal to 'gof.

Let S = { I , 2, 3)and T = {a, b, c}. Let f:S+T be f(1) = a, f(2) = b, f(3) = c. Define a function g:T+S as g(a) = 1, g(b) = 2 and g(c) = 3. Under the function g, the element f(x) in T is taken back to the element x in S. This niapping g is called the inverse o f f and is given by g(f(x)) = x, for each in S. You may note that f(g(a)) = a, f(g(b)) = b and f(g(c))=c. Thus, we have the following definition:

DEFIMTION 11: INVERSE OF A FUNCTION

Let S and T be two non-empty scts. A function I:S+T is said to be invertible if there exists a function g:T+S such that

(gof)(x)=x for each x in S, and

(fog)(x)=x for each x in T. In this case, g is said to be the inverse of f and we write it as g = f-'.

You may ask: Do all function possess inverses?

No, all functions do not possess inverses. For example, let S = {I, 2, 3)and T ={a, b). If f:S+T is defined as f(1) = f(2) = a and f(3) = b, then f is not invertible. For, if g:S+ T is inverse o f f , then

(g o f ) (1 )= g(f(1)) = g(a)

and (g o f)(2) = g(f(2)) = g(a). Therefore, 1 = 2, which is absurd,

Red Numbem and Functioil:, '

Leopold Kroneckcr 18

This raises another rl%restion : Under what conditions a? fuaaac~ow Baas an inverse? If a function k S " T i s one-one and onto, then it is invertible* Conversely, i f f is invertible, then f is bath om;-one and onto. Thus if sa function is one-one and onto, then It must'hnve an Inverse.

1.3 SYSTEM OF REAL NUMBERS --- You are quite familiar with some number sjstems and some of their properties. You will, perhaps rccaif the following properties:

(i) Any number mui!iplied by zero is equal to zero,

(ii) the product of a positivz: number with a negative number is negative,

(iii) the praduct of a nqative number with a negative nundber is positive among atkers.

To illustrate these properties, you d most $ikely use the natural numbers or integers or even rational numbers. The questions, which begin to arise are: What k e these various types of numbers? Wlaat properiies characterise the distinction between these various sets of numbers?

In this section, we shall try to provide answers to these and many other related questions. Since we are dealing with the course on Real Analysis, therefore we confine our discussioll to the systcm of real numbers. Nevertheless, we shall make you peep into the realm of a still larger class of numbers, the so called complex numbers.

The system of real numbers has been evolved in different ways by different mathematicians. In the late 19th Century, the two famous German mathematicians Richard Dedekind [1815-18971 and George Cantor [1845-19181 gave two independent approaches for the construciion of real numbers. During the same time, an Italian mathematician, G. Peano [1858-19321 defined the natural numbers by the well-known Peano Axioms. However, we start with the set of natural numbers as the foundation and build up the integers. From integers, we construct the rational numbers and then through the set of rational numbers, we reach the stage of real numbers. This developmeilt of number system culniinates into the set of complex numbers. A detailed study of the system of numbers leads us to a beautiful branch of Mathematics

_namely The Number Theory, which is beyond the scope of this course. However, we shall outline the general development of the system of the real numbers in this section. This is crucial to understand the characterization of the real numbev in terms of the algebraic structure to be discussed in Unit 2. Let us start our discFssion with the natural numbers.

-1.3.1 NATURAL NUMBERS The notion of a number and its counting is so old that it ishifficult to trace its origin. I

It developed much before the time of even the recorded hiitory that its manner of development is based on conjectures and guesses. The mykind, even'in the most primitive times had some number sense. The man, at least, had the sense of recognizing 'more' and 'less', when some objects were added to or taken out from a s 1 3 collection. Studies have shown that even some animals possess such a sense. 1 With the gradual evolution of society, simple counting became imperative. A tribe had to count how many members it had, how many enemies and how many friends. A shephard or a cowboy found it necessary to know if his flock of sheep or cows was decreasing or increasing in size. Various ways were evolved to keep such a count. Stones, pebbles, scratches on the ground, notches on a big piece of wood, small sticks, knots in a string or the fingers of hands were used for this purpose. As a result of

1 several refinements of these counting methods, the numbers were expressed in the ,writtensymbols of various types called the digits. These digits were written differently according to the different languagues and cultures of the societies. In the ultimate development, the numbers denoted by the digits 1,2, 3, .... became universally acceptable and were named as natural numbers.

Different theories have been advanced about the origin and evolution of natural numbers. An axiomatic approach, as evolved by G. Peano, is often used to define the

1- natural numbers. Some mathematicians like L. Kronecker [1823-18911 have remarked that the natural numbers are a creation of God while all else is the work of man.

However, we shall not go into the origin of the natural numbers. In fact, we accept that the natural numbers are a gift of nature to the mankind.

Sets and Numbers

We denote the set of all natural numbers as N = {I, 2, 3, .... 1.

One of the basic properties of these numbers is that there is a starting number i. Then for each number there is a next number. This nextness property is an important idea that you may find fascinating with the natural numbers. You may think of any big natural number. Yet, you can always tell its next number. What's the next number after,forty nine? After seventy seven? After one hundred twenty three? After three thousand and ninety nine? Thus you have an endless chain of natural numbers.

Some of the basic properties of the natural numbers are concerning the well-known fundamental operations of addition, multiplication, subtraction and division. You know that the symbol 'f' is used for addition and the symbol '.' is used for multiplication. If we add o r multiply any two natural numbers, we again get natural numbers. We express it by saying that the set of natural numbers is closed with respect t o these operations.

However, if you subtract 2 from 2, then what you get is not a natural number. It is a number which we call zero denoted as '0'. The word zero in fact is a translation of the Sanskrit 'shunyr'. It is universally accepted that the concept of the number zero was given by the ancient Hindu mathematicians. You come across with certain concrete situations indicating the meaning of zero. For example the temperature of zero degrees is certainly not an absence of temperature.

After having fixed the idea of the number zero, it should not be difficult for you to understand the notion of negative natural numbcrs. You must have heard the weather experts saying that the temperature on the top of the hills is minus 5 degrees written as - 5 O . What does it mean? The simple and straight explanation is that -5 is the negative of 5 i.e. -5 is a number such that 5+ (-5) = 0. Hence -5 is a negative natural number. Thus for each natural n, there is a unique number -n, called the negative of n such that

n + (-n) = 0.

1.3.2 INTEGERS You have seen that in the set N of natural numbers, i f we subtract 2 from 2 or 3 from 2, we do not get back natural numbers. Thus set ol' natural numbers is not closed with respect to the operation of subtraction. After the operation of subtraction is introduced, the nced to include 0 and negatkcnumbcrs becomes apparent. To make this operation valid, we must enlarge the system ol' natural numbers, by including in it the numberD and all the negative natural numb6rs. This enlarged set consisting of all the natural numbers, zero and the negatives of natural numbcrs, is called the set of integers. It is denoted as

Z = { .... -3 , -2 , - l , 0 ,1 ,2 ,3 ,.... 1. Now you can easily verify that the set of integers is closed with respect to the operations of addition, multiplication and subts;lction.

The integers 1. 2, 3 .... are also called positive integers which are in fact natural numbers. The integers -1, -2, -3,.... are c;~lled negative integers which ;Ire ac t~~a l ly the negative natural numbers. The number 0, however, is ncither,a positive integer nor a negative one. The set consisting of all thc positivc integcrs and 0 is called the set of non-negative integers. Similarly we talk of the set oS non-positive integers. Can you describe it?

1.3.3 RATIONAL NUMBERS If you add or multiply the integers 2 and 3, then the result is, of course, iln integer in each case. Also if you subtract 2 1'1qorn 2 or 2 from 3 , the result once again in each case, is an integer. What d o you get, when you divide 2 by 3? Obviously, the result is not an integer. Thus if you divide an integer by a non-zero integer, you may not get an integer always. You may get the numbers of the form

Such numbers are called rational nu-mbers.

R r ~ l Numbers and Functions Thus the set Z of integers is inadequate when the operation of division is introduced. Therefore, we enlarge the set Z to that of all rational numbers. Accordingly, we get a bigger set which includes all integers and in which division by non-zero integers is

A rational number of pdssible. Such a set is called the set of rational numbers. Thus a rational number is a

--P P I fkrm - or - is equivalently number of the form x , q # 0, where p and q are integers. We shall denote the set qf

9 -9 9 -P written as - where p and q all rational numbers by Q . Thus, 0

are both positive integers.

, EXERCISE 5)

Justify that

(i)' N is a proper subset of Z.

(ii) Z is a proper subset of Q .

Now if you add or multiply any two rational numbers you again get a rational number. Also if you subtract one rational number from another or if you divide one rational number by a non-zero rational, you again get a rational numbers in each case. Thus the set Q of rational numbers looks to be a highly satisfactory system of numbers in the sense that the basic operations of addition, multiplication, subtraction and division are defined on it. However, Q is also inadequate in many ways. Let us now examine this aspect of Q .

Consider the equation x2 = 2. We shall show that there is no rational number which satisfy this equation. In other words, we have to show that there is no rational number whose square is 2. We discuss this question in the form of the following example:

EXAMPLE 4 : Prove that there is no rational number whose square is 2.

SOLUTION : If possible, suppose that there is arational number xsuch that x2 = 2. S ~ n c . x is a rational number, therefore x must be of the form

x = J y p ~ ~ , q ~ ~ , q # ~ . 9

Note that the integers p and q may or may not have a common factor. We assume that p and q have no common factor except 1.

Squaring both sides, we get 2

P = 2 . q2

Then we have

Indeed, if p were odd, then p would This means that p2 is even and henctp is even (verify it). Therefore, we can write be ofthe form P = 2k-t-1 lor some p = 2k for some integer k. Accordingly, we will have integer k. ~ c c o r d i n ~ l ~ . then p2=4k2 . + 4k+l which is obvitausly odd. P2 = 4k2 = 29 2 This contradicts the fact that p2 is even. Hence p must be even. or

q2 = 2k2.

Thus p and q are both even. In other words, p and q have 2 as a common factor:This contradicts our supposition that p and q have no common factor.

Hence there is no rational number whose square is 2. Why don't you try the following similar exercises?

EXERCISE 6) Show that there is no rational number whose square is 3.

EXERCISE 7) 20 There is no rational number x such that x2.=;5 I

I

EXERCISE 8) Use the arguement of Exarnpe 6 to show that there is no rational number whose square is 4 and show where does the proof fail.

~ h u s you have seen that there are numbers which are not rationals. Such numbers are called irrational. In other words, a number is irrational if it cannot bc expressed as plq, p EZ, q EZ, q + 0. In this way, 112, 113, 45, etc. are irrational numbers. In fact, such numbers are infinite. Rather, you will see in Unit 2 that such numbers are even uncountable. Also you should not conclude that all irrational numbers can be obtained in this way. For example, the irrational numbers e and n are not of this form. We denote by I, the set of all irrational numbers.

Thus we have seen that the set Q is inadequate in the sense that there are nu : , which are not rationals.

A number which is either ratinal o r irrational is called a real number. The set of 1.ea1 numbers is. denoted by R. Thus the set R is the disjoint union of the sets of rational and irrational numbers i.e. R = Q u I, Q n X = 0.

Now in order to visualise a clear picture of the relationship between the rationals and . irrationals, their geometrical representation as points on a line is of gredt help. We discuss this in the next section.

1.4 THE REAL LINE

Draw a straight line L as shown in the Figure 1.

Fig. i

Choose a point 0 on L and another point P, to the right of 0. Associate the number 0 (zero)lto the point 0 and the number 1 to the point PI. We take the distance between the points p and' PI as a unit length. We mark a succession of points P,, P,, ...... to the right of PI each at a unit distance from the preceding one. Then associate the integers 2, 3, .... to the points P,, P ,,..., respectively. Similarly, mark the points P-,, P -,,..,, to the left of the point 0. Associate the integers -1, -2 ,..... to the points P-,, P -,...... Thus corresponding to each integer, we have associated a unique point of the line L.

Now associate every rational number to a unique;point of L. Suppose you want to

2 2 I associate the rational number - to a point on the line L. Then - = 2 x - i.e., one 7 7 7

unit is divided into seven parts, out of which 2 are to be taken. ~ e t us sek how you ca i do it geometrically.

Real Numbers and Functions Through 0, draw a line OM inclined to the line L. Mark the points A,, A,, ..., A7 on the line OM at equal distances. Join PI A,. Now if you draw a line through A, parallel to P,A, to meet the line L in H. Then H corresponds to the rational

2 . 2 number - i.e., OH = -.

7 7

You can do likewise for a negative rational number. Such points, then, will be to the left of 0.

EXERCISE 9) 3

Give the geometrical representation of' - . 7

By having any line through 0 , you can see that the point P does not depend upon chosen line OM. Thus you have associated every rational number to a unique point on the line L.

Now arises the important question :

Have you used all the points of the line L while representing rational numbers on it?

The answer to this question is NO. Rut how? Let us examine this.

At the point P, draw a line perpendicular to the line and mark A such that P,A = 1 unit. Cut off OB = OA on the line-as shown in the Figure 3.

A

Fig. 3

Then B is a poi'nt which correspond to a number whose square is 2. You have already I seen that there is no rational number whose square is 2. In fact, the length OA = f i b y

I Pythagorean Theorem. In other words, the irrational number a i s associated with the point B on the ljne L. In this way, it can be shown that every irrational number can be associated to a unique point on the line L.

Thus, it can be shown that to every real n"mber, thefe correspon'ds a unique point on the line L. In other words, all the real numbers are represented as points on a line. Is the converse h e ? That is to say, does every point on the line correspond to a unique real number? This is true but we are not going to prove it here. Therefore, hence onwards, we shall say that every real number corresponds to a unique point on the line and conversely every point on the line corresponds to a unique real number. In this sense, the line L is called the Real Line.

Now let L be the real like. '

We may define addition ($) and multiplicatio~l (.) of real ni1mbel.s geometrically as follows:

Suppose A represents a real number r and B reps.csents a real number s so that OA = r and OB = s. Shift OB so that 0 coincides with A. The point C ivhich is the new position of B is defined to.represent r+s. See the Figure 4.

The construction is valid for positive as well as negative values of r snd s. A real number r is said to be positive if r corresponds to a point on the line l, on the right of the point 0. It is written as r > 0. Similarly, r is said to be negalivc if it corresponds to a point on the left of the point 0 and is written as r < 0. Thus if r is a real numl- then either r is zero or r is positive or r is negative i.e. either -I. = 0 or r 0 or r < 0. You should try the following exercise:

EXERCISE 10) Construct r-s, r-i- s and -r-s on the real line.

What about t l ~ c product r s of two seal numbers r and s ? We shall consider the case when r and s are both positive real numbers.

Though 0 draw some other line OM. On L, let A reprcsci,t the real nurrlber s. On OM take a point D so that OD = r. Let Q be a point on I , sn that OQ -r 1 unit. Joiv~ Q D. Through A draw a straight line parallel t o Q D to nlect OM at C. Cut off 0 P on the line equal to OC. The11 P represents the real nunaber r.b.

Suppose s is a positive real number rind r is a negative real rlu~nher. 'f l le~i, there exists a number r such that r = -rf whcre r' is a positivc real nutnbers. Ihcreforc, the '

product rs can be defined on L as rs = (- rf)s = -(rPs).

Similarly you can state that 1.s = r(-s') = -(rsl) where s is negative and s =; - -s ' fur some positive s', while r is positive.

If both r and s are negative and r -- -rf and s -. ---s' where r' iund s' :ire positive rcai numbers, then we define I

rs = r's' = (-r) (-s). I

We can also sirniiarly define 0, r - r! 0 = 0 for each real number r,

11.5 COMPLEX NUMBERS

So far, we have discussed the system of real numbers. We have. yet, another system of numbers. For example, if you ha've to find the square root of a negative real number say -5, then you will write at as G. J i i i ~ o u know that f i i s a renl number but what about $-=T? Again you call verify that a simple equation x2-I- I = 0 docs not have a solution in the set of real numbers because the solution involvcs the square root of a negative renl number. As a matter of fact, the problen~ is to investigate the nature of the number &I which we denotq by sucll that i'! --- -1. Let us discuss the following example to know the nature oC i.

EXAMPLE 5: Show that i is not a real number. \ Real Numbers and Functions I -- We claim that i is not a re,al number. If it is so, then either i = 0 or i > 0 or i < 0.

, If i = 0, then i2 = 0. This implies that -1 = 0 which is absurd. If i > 0, then i2 > 0 which implies that - 1 > 0. This is also absurd. Finally, if i < 0, then again i 2 > 0 which implies that - 1 > 0. This again is certainly absurd. Thus i is not a real number. This number 'i' is called the imaginary number. The symbol 'i' is called 'iota' in Greek language. This generates another class of numbers, the so called complex numbers. 1 The basic idea of extending the system of real numbers to the system of complex numbers arose due to the necessity of finding the solutions of the equations. like x2 t 1 = 0 or x2 f 2 = 0 and so on. The first contribution to the notion of such a number was made by the most celebrated Indian Mathematician of the 9th century, Mahavira, I

who showed that a negative real number does not have a square root in the set of real numbers. But it was an Italian mathematician, G. Cardon [1501-15761 who used imaginary numbers in his work without bothering about their existence. Due to notable contributions made by a large number of mathematicians, the system of complex numbers came into existence in the 18th century. Since we are dealing with real numbers, therefore, we shall not go into the detailed discussion of complex numbers. However, We shall give a brief introduction to the system of complex numbers. We denote the set of complex numbers as

C=[z=a+i b, a and b real numbers)

In a compiex~umber, z = a f i b, a is called its real part and b is called its imaginary part.

Any two complex numbers z, = a, -k i bl and z2= a2 + i b2 are equal if only if their corresponding real and imaginary parts are equal.

If z~ = a1 + i bl and zz = a2 + i bz are any two complex numbers, then we define addition (+) and multiplication (.) as follows:

ZI 4- 22 = (al + az) + i (bl f b2)

~ 1 . 2 2 = (ala2 - blbz) + i (a lb~ + a2b2).

The real numbers represent points on a line while complex numbers are identified as points on the plane.

- -- - - - -

E X E R ~ E - ~ Justify that R is a subset of C.

-

Before concluding this section, we would like to mention yet another classification o i numbers as enunciated by some mathematicians. Consider the number J2. This is an example of what is called an Algebraic Number because it satisfies the equation -

x2 -2 = 0.

A number is called an Algebraic Number if it satisfies a polynomial equation

a0xn + alxn-l t .,..I + an-] x + a,, + an = 0

where the coefficients a", a,, a ~ , .... a, are integers, a, # 0 and n > 1. The rational numbers are always algebraic numbers. The numbers defined in terms of the square root etc, are also treated as algebraic numbers. But there are some real numbers which are not algebraic. Such numbers are called the Transcendental numbers. The numbers n and e are transcendental numbers.

You may think th'at the operations of algebraic operations viz, addition, multiplication, etc. are the only aspects to be discussed about the set of real numbors. But certainly there are some more important aspects of the set of real numbers as points on the real line. We shall discuss these aspects in Unit 3 namely the point sets of the real line called also the topology of the real line. But prior to that, we shall discuss the structure of real numbers in Unit 2.

We conclude chis unit by calkin8 briefly about an important hypothesis-closely linked with the system of natural numbers. This is called the Principle of Induction.

MATHEMATICAL INDUCTION

The Principle of Induction and the natural numbers are inseparable. In Mathematics. we often deal with the proofs of various theorems and formulas. Some of these are derived by the direct proofs, while some others can be proved by certain indirec't methods. Consider, for example, the validity of the following two statements:

(i) The number 4 divides 5" - 1 for every natural number n.

n(n f 1) (ii) The sum of the first n natural numbers is i.e.

In fact, you can provide k o s t of the verifications for both statements in the following way:

For (i), if n = 1, then 5n - 1 = 5 -1 = 4 which is obviously divisible by ., if n = 2, then 52 -1 = 24, which is also divisible by 4; if n = 6 , then S6-fl = 15624, which is indeed divisible by 4.

Similarly for (ii) if n = 10 then 14-24- .... 4-10 = 55, while the formula

"("+ = 55 when n = 10. 2

Again, if n = 100; then also you can verify that in each way, the sum of the first hundred natural numbers is 5050 i.e.

"("'l) = 5050 for n = 100. 2

What do these statements have in common and what do they indicate? The answer is obvious that each statement is valid for every natural number.

Thus to a great extent, a large number of theorems, formulas, results etc. whose statement involves the phrase, " for every natural number n" are those for which an indirect proof is most appropriate. In such indirect proofs, clearly a criterion giving a general approach is applied. One such criterion is known as the criterion of Mathematical Induction. The principle of Mathematical Induction is-stated (without proof) as follows:

Principle of Mathematical Induction

Suppose that, for each n E N, P (n) is a statement about the natural number n, Also, Guppose that

(i) P(1) is true,

(ii) if P(n) is true, then P(n4-1) is also true.

Then ~ ( n j is true for every n EN. Let us illustrate this principle by an example:

EXAMPLE 6: The sum of the first n natural numbers is n(n4- I) 2

SOLUTION: In other words, we have to show that for each n E N,

n

= r, k. k= l

Let P(n) be the statement that

We, then, have S I = 1 and '(l+') - - 1. Hence P(1) is true. L

This proves part (i) of the Principle of Mathematical Induction. Now for (ii), we have

Real Numbers and Functions to verify that if P(n) is true, then P(n+l) is also true. For this, let us assume that P(n) is true and establish that P(n+l) is also true. Indeed, if we assume that

n(n+l) Sn = -

2 '

then we claim that (n+ 1) (n+2) -

S"+l - 2

Indeed

I = - n(n+l) + (n+l)

2

- - (n+ 1) (n+2) 2

Thus P(n+ 1) is also true.

Similarly, by using the Principle of Induction, you can prove that (i) the sum of the squares of the first n natural numbers is

1 - n(n+l) (2n + 1); and 6

(ii) the sum of the cubes of the first n natural numbers is

1.7 SUMMARY

We have recalled some of the basic concepts of sets and fknctions in section 1.2. A set is a well-defined collection of objects. Each object is an element or a member of the set. Sets are generally designated by capital letters and the members by small letters enclosed with braces. There are two ways to indicate the members of a set. The tabular method or listing method in which we list each element of a set individually and the set-builder method which gives a verbal description of the elements or a property that is common t o all the elements of a set.

A set with a limited number of elements is a Finite set. A set with an unlimited number of elements is an infinite set. A set with no elements is a null-set. A set S is a subset o f a set T if every element of S is in T. The set S is said to be a proper subset of T if every element of S is in T and there is at-least one element of T which does not belong to S . The sets S and T are equal if S is a subset of T and T is a subset of S. The null set is a subset o f every set and every set is a subset of itself.

The union of two sets S and T, written as S u T, includes elements of S and T without repetitions. The intersection of S and T, written as S n T, includes all those elements that are common to both S and T. The complement of a set S in a Universal set X is denoted as S' and it consists of all those elements of X which do not belong to S. The

I

laws with respect to union. intersection and complement have been asked in the form o f ( exercises. Also, these notions have been extended to an arbitrary family of sets.

A function f: S+ T is a rule by which you can associate to each element of S, a unique , element of T . The set S is the domain and the set T is the co-domain of f. The set {f(x):x ES) is the Range of f, where f(x) is an image of x under f. The function f is one- 1 one if f(x,) = f(x,) a x, = x,, for any x,, x, in the domain of f. It is said to be onto if the range of f is equal to the domain of f. A function f is said to be a one-one correspondence if it is both one-one and onto. A function i: S + S defined by i(x) = x, V x ES is called an identity function, while a function f:S+ T is said to be constant if

I

f(x) = c, V x ES, c being a fixed element of T. I

Any two functions with the same domain are said to be equal if they have the same image for each point of the domain. The composite of the functions f:S + T and g: T -+ V is a function denoted as 'g o f : S + V and defined by (g o f ) (x) = g(f(x)). The function f:S + T is said to be invertible if there exists a function g:T -+ S such

that both 'g o f and 'f o g9 are identity functions. Also, a function is invertible if it is both one-one and onto. The inverse off exists if f is invertible and it is denoted as f '. In Section 1.3, we have discussed the development of the system of numbers starting from the set of natural numbers. These are the following:

Nafkal Numbers (Positive Integers): N = {1, 2,4 ....}

Integers: Z = {.. . . 3, -2, - 1, 0, 1, 2, 3 . . . .}

Rational Numbers: Q = (1 : p E z, qEIZ, q1 Z 0} q

Irrational Numbers I

Real Numbers: R = Disjoint Union of Rational and Irrational Numbers

R = Q U I , Q n I = +

ComplexNumbers C=(z=x+iy:xER,yER},i=J-1.

A mathematical development of the number systems is depicted in Figure 6:

A mathematical development of numbers systems Fig. 6

Irl Section 1.4, we have discussed the geometrical representation of the real numbers and stated the continpum Hypothesis. According to this, every real number can be represented bfi a unique point on the line and every point on the line corresponds to a unique real number. In view of this, we call this line as the Real Line.

E 1) (i) A = {n: n EN , n is even}. A = {n: n E N, n is odd}. A = (x: x is a president of India}.

Sets and Numbers

Real Numbers and Functions (ii) A = ( l , 3 , 5 ) . A = { 21,22, 23, 24, ...., 291. A={ -1,-2,- 3 ,...... }.

E 2) (i) Each n EN. 3 n EZ but not conversely. i There exist at least one x ER such that x does not belong to Q e.g.,

x = 5/2. (iii) A c B implies that x E A 3 x E B.

B c C implies that x E B 3 x E C, which shows that x E A 3 x E C and hence A c C.

E 3) (0 Since 0 is a subset of every set, therefore 0 is a subset of itself and hence P (0) = (0).

(ii) Let x E S. Then x tir SC 3 x E (S)" i.e., S C ( S ' ) ~ . Similarly, reverse inclusion follows. Thus S = (S')".

/'

E 4) Using definitions of operations ' u ' and 'n' on set of subsets of a set X, say, it is easy to see that (i), (ii) and (iii) holds. For (iv), IL' x E (AuB)". Then, x tir AuB. This implies, by definition of 'us, that x ei A and x 65 B. Then x E A" and x E Bc, which gives, x E Ac nBc. So, x E (AuB)" 3 x E A' n Bc i.e., (AuB)' c Ac n Bc. Similarly, you can see that A' n B c (AuB)'. Thus (AuB)" = A' n B'. This proves first part of (iv), of course, its second part is analogue to first one and you can easily do that.

E 5) It is true because every natural is an integer, every integer is a rational but not conversely i.e., every rational may not be an integer and every integer may not be a positive integer. Give an example in each case.

E 6 ) (i) If possible, suppose that there ia a rational x such that x2 = 3.

pL Then x2 = -, where p E Z, p E Z, q + 0. Suppose p and q have no q2

common factor. Then

p2 = 3qZr

which implies that p2 is divisible by 3 and hence p is divisible by 3. indeed if p = 3k+l , then p2 = 9k2 + 6kH which is not divisible by 3. Hence p must be divisible by 3 i.e. p = 3k. Then

Thus both p and q are divisible by 3 which contradicts th,e assumption that p and q have no common factors. Hence the result. .

E 7) Prove it in the same way.

P' E 8) In this case, - = 4 a p 2 = 4q2 3 p2 is even 3 p = 2k (say) for

9' some integer k 4q2 = 4k7 3 q2 = k2. Proof fails. Give suitable arguments.

E 9)

< -r-S 2

Fig. 8

E 11) Every real number x can be expressed as a complex number e.g. z = x+iy. But every complex number z need not be a real number. Give an example. Think of

, i != 4-1.

UNIT 2 STRUCTURE OF REAL

STRUCTURE 2.1 Introduction

Objectives,

2.2 Order Relations in'Real Numbers Intervals Extended ~ i a l ~urnders

2.3 Algebraic Structure Ordered Field Complete Ordered Field

2.4 Countability Countable Sets Countability of Real Numbers

2.5 summary

2.6 Answers/ Hints/ Solutions

2.1 INTRODUCTION

In Unit 1 we have discussed the construction of real numbers from the rational numbers which, in turn, were constructed from integers. In this unit, we show that the set of real numbers has an additional property which the set of rational numbers does not have, namely it is a complete ordered field. The questions, therefore, that arise are: What is a field? What is an ordered field? What does it mean for an ordered field to be complete? In order to answers these questions we need a few concepts and definitions e.g. those of order inequalities and intervals in R. We shall discuss these concepts in Section 2.2. Also in this section, we shall explain the extended real number system.

You know that a given set is either finite or infinite. In fact a set is finite, if it contains just n ele ents where n is some natural number. A set which is not finite is called an infinite set. The elements of a finite set can be counted as one, two, three and so on, while those \ an infinite set can not be counted in this way. Can you count the elements of tbe set of natural numbers? Try it. In Section 2.4, we shall show that this notion of counting can be extended in certain sense to even infinite sets.

-OBJECTIVES In this unit, you should, therefore, be able to +identify the order relation in the set of real numbers and extended real number

system,

+describe the field structure of the set of real numbers,

+discuss the order-completeness of the set of real numbers apply the notion of countability to various infinite sets.

2.2' ORDEg RELATIONS IN REAL NUF'yXBERS-

In Section 1.3, we have demonstrated that every real number can be represented as a unique point on a line and every p$nt on a line represents-%unique real number. This helps us to introduce the notion of inequalities and intervals on'tlie real .line which we ' shall frequently use in our subsequent discussion through out the course.

You know that a real number x is s ~ d to be positive if it lies on the right side of 0, the point which corresponds to the nukber.0 (zero) on tlie real line. We write it as x > 0. Similarly, a real number x is negative, if it lies on the left side of 0. This is written as x <: 0. If x 1 0, then x is a non-negative real number. The real numlier x is sai& to be non-positive if x S 0.

Let x and y be any two real numbers. Then, we say that x-is greater than y if x-y 5 0. Structure of Real Numbers

We this by writing x > y. Similarly x is less than y if x - y < 0 and we write X-'< y. Also x is greater than eraqual to y (x 1 y) if x - y 1 0. Accordingly, x is less than or equal to y (x 5 y) if x - y 5 0. Given any two real numbers x'and y, exactly one of the following can hold:

either (9 X > Y or (ii) x < y or (5) x = y.

This is called the law of trichotomy. ~ h k order relation 5 has the following

PROPERTY 1: For any x, y, z in R, (i) If x I y and y I x, then x = y, jii) If x 5 y an y I z, then x I z, (iii) If x I y then x 4- z I y + z, (iv) If x I y and o 5 z, then x z 5 y z.

The relation satisfying (i) is called anG-symmetric. I t is called transitive if it satisfies (ii). The property (iii), shows that the inequality remains unchanged under addition of a real number. The property (iv) implies that the inequality also remains unchanged under multiplication by a non-negative real number. However, in this case the inequality gets reversed under multiplication by a non-positive real number. Thus, if x I y and z 5 0, then x z 2 yz. For instance, if z = -1, we see that

EXERCISE 1)

State the properties of order relation in the set R qf real numbers with respect to the relation 3 (is greater than or equal to) and illustratq the inequality under multiplication by a negative real number.

We state the following results without proof:

There lie an infinite number of rational numbers behveeh any two distinct rational numbers. -.

As a matter of fact, something more is true.

Between any two real numbers, there lie infinitely many rational (irrational) numbers. Thus there lie an infinite number or real numbers between any two given real numbers.

2.2.; INTERVALS Now that the notion of an order has been introduced in R, we can talk of some special subsets of R defined in terms of the order relation. Before we formally define subset, we first introduce the notion of 'betweenness', which we have already used intuitively in the previous results. If 1, 2 ,3 are three real numbers, then we say that 2 lies between 1 and 3. Thus, in general, if a, b and c are any three real numbers such that a 5 b % c then we say that b lies 'between' a and c. Closely related to notion of betweeness is the concept of an interval.

DEFINITION 1: INTERVAL

An interval in R is an nonempty subset of R which has the property that, whenever two numbers a and b belong to it, all numbers between a and b also belong to it. '

The set N of natural numbers is not an interval because while 1 and 2 belong to N, but 1.5 which lies between 1 and 2, does not belong to N.

We now discuss various forms of an interval.

(i) Consider'the set (x ER: a I x I b).' This set is denoted by [a, b], and is called a closed interval. Note that the-end a and b are included in it.

(ii) Consider the set {XE R : a < x < b}, This set is denoted by ]a, b[, and is called an open interval. In this case the end points a and b are not included in it,

Real Numbers and Functions (iii) The interval {x E R: a 5 x < b) is denoted by [a, b[.

(iv) The interval {x E R : a < x 5 b} is denoted by ]a, b]. You can see the graph of all the four intervals in the Figure 1.

a -.I

Fig. 1

Intervals of these types are called bounded intervals. Some authors also call them Ilnite intervals. But remember that these are not finite sets. In fact these are infinite sets except for the case [a, a] = {a).

You can ea_sily verify that an open interval ]a, b[ as well as ]a, b] and [a, b[ are alwayr contained in the closed interval I [a, b].

EXAMPLE 1: kest whether or not the union of any two intervals is an interval. I

SOLUTION : Let [2, 51 and [j, 121 be two intervals. Then [2,5] U 17, 121 is not an interval as can be seen on the line in Figure 2.

- - - 1

1 I I , I 1 I I

,4 I I 1

- 1

16 ~i 12

\

Fig. 2

However,'if you take the intervals which are not disjoint, then the union is an interval , For example, the union of [2, $1 and [3,6] is [2,6] which is an interval. Thus the union of m y two intcrvrls is an interval provided the i n te~a ls are not disjoint.

32 Now try the following exercise:

h I

- .3

-

EXERCISE 2) Structure of Real Numbers

Give examples to sllow that the intersection of any two intervals may not be an interval. What happens, if the two intervals a r e not disjoint? Justify your answer by an example.

2.2.2 EXTENDED REAL NUMBERS The notion of the extended real number system is ,important since we need it in this unit as well as in the subsequent units.

You are quite familiar with the symbols + oo and - co. You often call these symbols are 'plus infinity' and 'minus infinity', respectively. The symbols +. w and - m are extremely useful. Note that these are not real numbers.,

Let us construct a new set R* by adjoining - m and + m to the set R and write it as

Let us extend the order structure to R* by a $lation < as - w < x < + oo, for every x E R. Since the symbols - oo and + a do not represent any real numbers, you should, therefore, not apply any result stated for real numbers, to the symbols + w and - w. The only purpose of using these symbols is that, it becomes convenient to extend the notion of (bounded) intervals to unbounded intervals which are as follows :

Let a and b be any two real numbers. Then we adopt the following notations :

[a, co] = { X ER: x 2 a) [a, do] = { X ER: x > a ) [-a, b] = {x ER: x 5 b ) [-a, b] = {x ER: x < b ) [ -a, m] = { X ER: - w < x < a ) .

You can see the geometric representation of these intervah. in Figure 3, l--

Fig. 3 All these unbounded intervals are also sometimes called infinite intervals.

You can perform the operations of addition and multiplication involving -m and +m in the following way: For any x ER, we have

Real Numbcrs and Functions X . ( - W ) = ) ~ . ~ ifx.<O ec* = +oo, -00 -03 = ,-m

w Note that the operations UJ -my O.", - are not defined.

w -, -

2.3 A L G E B R ~ c STRUCTURE

During the 19th Century, a new trend emerged in Mathematics to use algebraic structures in order to provide a solid foundation for Calculus and Analysis. In this . quest, several methods were used to characterise the red numbers. One of the methods was related to the least upper bound principle used by Richarc) Dedekind. which we discuss in this section.

This leads us to the description of the real numbers as a complete ordered field. In order to define a complete ordered field. we need some definitions and concepts.

You are quite familiar with the operations of addition and multiplication on numbers, I union and intersection on the subsets of a universal set. For example if you add or

multiply any two natural numbers, the sum or the product is a natural number..These operations of addition or multiplications on the sets of numbers are examples of a binary operation on a set. In general, we can define a binary operation on a set in the following way:

Richnrd Dedekind DEFINITION 2: BINARY OPERATION

Given a non-empty set §, a binary operation on S is a rule which associates with each pair of elements of S, a unique element of S.

We denote this rule by symbols such as ., *, +. etc.

By an Algebraic Structure, we mean a non-empty set together with one or more binary operations defiqed on it. A field is an algebraic structure which we define as follows:

DEFINITION 3: FIELD STRUCTURE

A field consists of a non-empty set F together with two binary operations defined on it, denoted by the symbols '+'(addition) and '.?(multiplication) and satisfying the following axiomh for any elements x, y, z of the set F.,

A,: x f y E F * (Additive Closure) Az: x f (y-i-zj = (x+y) f z , (Addition is Associative) A3: x+y = y+x (Addition is ~ommutat ive) Ad: Therq kxists an element in F, denoted by '0' and (Additive Identity)

cal,le& the zero or the zero element of F sdch t h a t x + O = O + x = x t f x E F

As: I For each x E F , there exists an element -x GF with (Additive Inverse) the property I

x+(-x) = (-x)+x = 0 The element -x is called additive inverse of x.

MI x . y E F (Multiplicative Closure) M2 (x.y).z = X. (y.~) (Multiplication is Associative) I

Mn x.y = y.x (Multiplication is Cowutat ive) I

M4 There exists an elenient 1 different from 0 called the unity of F,:such that (Multiplicative Identity) ? l . x s x . 1 = x Y x E F

I

Ms For each x E F, x # 0, there (Multiplicative Inverse) exists an element x-' E F such that . x.x-' = .x-l X = 1.

The, eiemenf x-' is called the multiplicative inverse of x. D: x.(y+z) = x.y +x.z

( ~ u l t i ~ l i c a i o n is distributive over Addition). (x+y) .z = x.z+y.z.

Since the unity is not equal to the zero i.e. 1 # 0 in a field, therefore any field must contain at least two elements. Note that the axioms A1 (closure under addition) and

M1 (closure under multiplicanon) are unnecessary because the closures are implied in Structure of Real Numbers

the definition of a binary operation. However, we include them, for the sake of emphasis. Now try the following exercises:

EXERCISE 3) Show that the set {0, 1) forms a field under the operations '-by and '.' defined by the following tables:

1 1

1 0 1

EXERCISE 4) '

Show that the zero and the unity are unique in a field.

Now, you can easily verify that all the eleven axioms are satisfied by the set of rational numbers with respect to the ordinary addition and multiplication. Thus, the set Q forms a field under the operations of addition and multiplication, and so does, the set R of all the .real numbers.

EXERCISE 5) . Do the sets N (of natural numbers) and Z (set of integers) form fields? Justify your answers. Also verify that the set C of complex numbers is a field.

We state (without proof) some important properties satisfied by a field. They follow from the field axioms. Can you try?

PROPERTY 1

For any x, y, z in F, 1. x t z = y t z * x = y , 2. x.0 = 0 = O.x, 3. (-x). y = -x,y = x. (-y), 4. (3). (_y) = x.y, 5. x.z=y.z, z # 0 0 x = y, 6. x.y = 0 + either x = 0 or y = 0.

Thus by now you know that the sets Q, R apd C form fields under the operations of addition and multiplication.

2.3.1 ORDERED FIELD In Section 2:2, we defined *e order nlatipn 5 in R. It is easy to see that this order relation satisfies the following properties:

PROPERTIES 2

Let x, y, z be any elements of R. Then

01: For any two elements x and y of R, one and only of the following holds: (i) x < y, (ii) y < x, (iii) x = y,

o h x 5 y , y ~ x * x ~ z , 0;: x I y --/ x t z 5 y+z, 04: X 5 y, 0 < z * x.z 5 y,z

We express this observation by saying that the field R is an ordered field (i.e. it satisfies the properties 01 --04). I t is easy to see that these properties are also satisfied ' by the field of rational numbers. Therefore, Q is also an ordered field.. What about . the field C of Complex numbers? ~ r y ' i t yourself as an exercise,

EXERCISE 6)

Show that the field C of Complex numbers is not an ordered field.

2.3.2 COMPLETE ORDERED FIELD Although R and Q are both ordeped fields, ~et ' there is a prope;ty associated with the order relation which is satisfied by R but not by Q. This property is known a's the Order-Completeness, introduced for,the first time by Richard Dedekind. To explain

Rest Numbers nnd Pullctions this situation more precisely, we need a few more mathematical concepts which are discussed as folIows:

Consider set S = { 1, 3, 5, 7). You can see that each element of S is less than or equal to 7. That is x s 7, for each x ES. Take another set S, where S ={x ER : x I 17). Once again, you see that each element of S is less than 18. That is, x < 18, for each x ES. In both the examples, the sets have a special property namely that every element of the set is less than or equal lto some number. This number is called an upper bound of the corresponding set and such a set is said to be bounded above. Thus, we have the following definition:

DEFINITION 4: UPPER BOLINLI OF A SET

Let S cR. If there is a number 11 E R such that X < U , for every x ES, then S is said to b. bounded above. The number u is called an upper bound of S.

EXAMPLE 2: Verify whether the following sets are bounded above. Find an upper bound of the set, if it exists.

(i) The set of negative integers (-1, -2, -3, ...... }.

(ii) The set N of natural numbers.

(iii) The sets 2, Q and R.

SOLUTION : (i) The set is bounded above with -1 as an upper bound, (i) The set N is not bounded above. (iii) All these sets are not bounded above.

EXERCISE 7) (i) Define a set which is bounded below. Also define a lower bound of a set. (ii) Give a t least two examples of a set (one of an infinite set) which is bounded

below and mention a lower bound in each case. (iii) Is the set of negative integers bounded below? Justify your answers.

Now consider a set S = {2, 3, 4, 5, 6, 7). You can easily see that this set is bounded above because 7 is an upper bound of S. Again this set is also bounded below because 2 is a lower bound of S. Thus S is both bounded above as well as bounded below. Such a set is called a bounded set. Consider the following sets:

s , = I... -3, -2,-1, 0, 1,2, ....... ), S,= {O, 1,2, .......... ), S,= (0,-1, -2, ..... ).

You can easily see that S, is neither bounded above nor b~unded below. The set S, is not bounded above while S, is not bounded below. Such sets are known as Unbounded Sets.

Thus, we can have the following definition.

DEFLNITION 5: BOUNDED SETS

A set S is bounded if it is both bounded above and bounded below.

I11 other words, S has an upper bound as well as a lower bound. Thus, if S is bounded, then there exist nulnbers u (an upper bound) and v (a lower bound) such that v S x S u, for every x ES.

If a set S is not bounded then S is called an unbounded set. Thus S is unbounded if either it is not bounded above or it is not bounded below.

EXAMPLE 3 : (i) Any finite set is bounded. (ii) The set Q of rationat numbers is unbounded. (iii) The set R of real numbers is unbounded (iv) The set P = {sin x, sin Zx, sin3x ,......, sin nx, ..... } is bounded

because -1 5 sin nx 5 1, for every n and x.

EXERCISE 8) Test which of the fallowing sets are bounded above, bounded below, bounded and unbounded. (i) The intervals la, b[, [a, b], la, b] and [a, b[, where a and b are any two real

numbers.

(U) The intenrls [2, q, ]-m,q,]5,m [ and 1-m, 41.

(iii) TIW set ~COS e, COS 2 e, 3 e, ......... j. (iv) S = ( x E R : - a I x I a } f o r s o m e a E R .

Structure of Real Nurnhers

' You can easily verify that a subset of a bounded set is always bounded since the bounds of the given set will become the bounds of the subset.

Now consider any two bounded sets say S = (1,2, 5, 7) and T = {2,3,4, 6,7, 8). Their uriion and intersection are given by

S U T = { l , 2 , 3 , 4 , 5 , 6 , 7 , 8 ] and

s n T = {2,7j.

Obviously S U T and S fl T are both bounded sets. You can prove this assertion ir, general for any two bounded sets.

EXERCISE 9) Prove that the union and the intersection of any two bounded sets qre bounded.

Now consider the set of negative integers namely

s = {-I, -3, -2, -4, .... j. You know that -1 is an upper bound of S. Is it the only upper bound of St Can you think of some other upper bound of S? Yes, certainly, you can. What about 01 The number 0 is also an upper bound of S. Rather, any real number greater than -1 is an upper bound of S. You can find infinitely many upper bounds of S. However, you can not find an upper bound less than -1. Thus -1 is the least upper bound of S.

It is quite obvious that if a set S is bounded above, then it has an infinite number of upper bounds. Choose the least of these upper bounds. This is called the least upper bound of the set S and is known as the Supremum of the set S . (The word 'Supremum' is a Latin word). We formulate t h definition of the Supremum of a set 'in the following way:

DEFINITION 6: THE SUPREMUM OF A SET

Let S be a set bounded above. The least of all the upper bounds of S is called the least upper bound or the Supremum of S. Thus, if a set S is bounded above, then a real number m is the supremum of S if the following two conditions are satisfied:

(i) m is an upper bound of S, (ii) if k is another upper bourid of S, then m 5 k.

EXERCISE 10)

Give an example of an infinite set which is bounded below. Show that it has an infinite number of lower boudds and hence develop the concept of the greatest lawer bound of the set.

-

I The greatest lower bound, in Latin terminology, is called the Infimum of a set.

Let us now discuss a few examples of sets having the supremum and the infimum:

EXAMRE 4: Each of the intervals ]a, b[, [a, b] ]a, b], [a, b[ has both the supremum and the infimum. The number a is the infimum and b is the'supremum in each case. In case of [a, b] the supremum and the infimum both belong to the set whereas this is not fhe'case for the set ]a, b[. In case of the set ]a, b], the i~tfimum does not belong to it and the supremum belongs to it. Similarly, the infimum belongs ta [a, b[ but the supremum does not belong to it.

Very often in our discussion, we have used the expressions 'the supremum', rather than a .supremum. What does it mean? It simply means that the supremum of a set, if it exists, is unique i.e. a set can not have more than one supremum. Let us prove it in the form of the following theorem:

Real ~urnbers acd Functions . THEOREM 1: Prove that the supremum of a set, if it exists, is unique. , PROOF : If possible, let there be two supremums (Suprema) say m and rn' of a set $.

'since m is the least upper bound of S, therefore by definition, we have

. m 5 m'.

( ~ i m i l a r l ~ , since m' the least upper bound of S, therefore, we must have

This shows that m = rn: which proves the theorem. You can now similarly prove the following result:

EXERCISE 11) Prove that the infimbm of a set, if it exists, Is unique.

In example 3, you have seen that supremum or the infimum of'a set may or may not belong to the set. If the supremum of a set belongs to the set, then it is called the ' greatest member of the set. Similarly, if the infimum of a set belongs to it, then it is called the least member of the set.

I

EXAMPLE 5: (1) Every f h t e set has the greateat as well ur tbe k t member.

(M) The set N.has the least member but not the -test. Determine that number.

(iii) The set of negative integers has the grenteet member but ~ i o t the , least member. What is that number?

Try the following exercise:

EXERCISE 12)

I Check which of the following sets have the greatest and the least member:

(i) {x: a l x l b}.

(ii) {x: a < x l b}.

(iii) {x: a 5 x < b}. .

(iv) {x: a < x.< b}.

(v) [a, m[, I a, *[. (vi) I-=, b], 1-0, b[. .

You have seen that whenever a set S is bounded above, then S has the supremum. In fact. this is true in general. Thus, we have the following property of R without proof:

PROPERTY 3: COMPLETENESS PROPERTY

Every non empty subset S of R which is bounded above, has the supremum.

Similarly, we have

1 Every non-empty subset S of R that is bounded below, has the infimum

In fact, it can be easily shown that the above two statements are equivalent.

Now, if you consider a non-empty subset S of 9, then S considered'as a subset of R must have, by property 2, a supremum. However, this supremum may not be in Q. This fact is expressed by saying that Q considered as a field in its dwn right is not

. Order-Complete. We illustrate this observation as follows;

Construct a subset S of Q consisting of all those positive rational numbers whose -squares are less than 2 i.e.

S={xEQ:x>0, ,xZ<2} .

Sincethe nuibe,rl ES, therefore S is non empty. Also, 2 is an upper boynd of S because every element of S is less than 2. Thus the set S is non-empty arid bounded, above. According to the Axiom of Completeness of R, the subset S must have (he supremum in R. We claim that this supremum does not belong to Q.

mppose rn is the supremum of the set S. If possible, let m belong to Q. Obvibusly, , 38 .

' then > O./%w either mZ < 2 or mZ = 2 or mZ > 2.

Case (i) When m2 < 2. Then a number y defined as Structure of Keal N u I ~ ~ c : ~

is a positive rational number and

Since rn2 < 2, therefore 2 - m2 > 0. Hence

nhich'implies that y > m. Again,

Y2 -2 =

Since m2 < 2, therefore

his shows that y E S and also it is greater than m (the supreInum of S). This is absurd. Thus the case m2 < 2 is not possible. -

Case (ii) When m2 = 2.

This means there exists a rational number whose square is equal to 2 which is again not possible;-since you have already proved this in Section 1.3.

Case (iii).When m2 > 2

In this case consider the positive rational number y defirled in case (i). Accordinglj, we have

Also 2 -y2 = 2- 4 + 3 m 2 _m2

(3 + 2m12

i.e. 2 - y2 < 0 or Y2 > 2,

which sl~ows that y is an upper bounh of S.

Thus y is an upper bound of S which d o ~ s not belong to S. At the same time y is less than the supremum of S. This is 'again absurd. Thus mi > 2 is also not possible. Hence none of three possibilities is true. This means there is something wrong with our suppOsition. I n other words, our supposition is false and therefore the set, S does pot possess the suprernum in Q.

This justifies that the fie1d.Q of rational numbers is not order-'complete.

Now you can also try a similar exercise.

EXERCISE .13) Let S be a subsdt of all those positive rationil numbers whose squares are less tinn 3

Show that S is nonempty and boi~nded above but it does not have the least upper bound in Q.

P

In Section 1.2, we recalled the notioil of a set and certain related concepts: Subsequently, we discussed certain properties of tlle sets of nllmhers N, Z, Q, R and

Real Numbem and Functions'

George Cantor. 1

'C. A few more imporiant properties and related aspects concerning these sets are yrt to be examined. One such significant aspect is the countability of these sets: The concept of Countability of sets was htroduced by George Cantor which, forms a corner stone of Modern Mathematics.

2.4.1 COUNTABLE SETS You can easily count the elements of a finite set. For example, you very frequently use the term 'one hundred rupees' or 'fifty boxm', 'two dozen eggs', etc. These figures pertain to the number of elements of a set. Denote the number of elements in a finite set S by n (S). If S = {a, b, c, dl, then n (S) = 4. Similarly n (S) = 26, if S is the set of the letters of English alphabet. Obviously, then n (4) = 0, where 4 is the null set.

You can make another interesting observation when you count the number of elements of a finite set. While you are counting these elements, you are indirectly and perhaps unconsciously, using a very important concept of the one-one correspondence between two sets. Rekall the concept of one-orre correspondence from Section 1.2. Here one of the^ sets is a finiie subset of the set of natural numbers and the other set is the set consisting of the articles/ objects like rupees, boxes, eggs, etc. Suppose you have a basket of oranges. While counting the oranges, you are associating a natural number to each of the oranges. This, s you know, is a one-one correspondence between'the set of oranges and a subs d t of natural members. Similarly, when you , count the fingers of your hands, you-are in fact showing a one-one correspondence between the set of the fingers with a subset, say Nlo = (1, 2, .... 10) of N.

Although, we have an intuitive idea of fi~iite and infinite sets, yet we give a mathematical definition of these sets in the following way:

DEFINITION 7: FINITE AND INFINITE SETS

A set S is said to be finite if it is empty or if there is a positive integer k such that there is one-one correspondence between the elements of the set S and the set NK = {I, 2,3 ...., k]. A set is said to be Mnite if it is not finite.

The advantage of using the concept of one-one correspondence is that it helps in studying the countability of infinite sets. Let E = {2,4,6, . . . . I be the set of even natural numbers. If we define a mapping f: N -. E as

f ( n )=2n ,VnEN,

then we find that f is a one-one correspondence between N and E.

Consider another examplk. Suppose S = {1,2, .... n) and T = {x,, xz, .... x,). Define a mapping f: S - T as

f (n)=x,Vn E S. *

Then again f is a one-one correspondence between S and T.

S y ~ h sets are known as equivalent sets. We define the equivalent sets in the following - w 8y:

D E F I N I ~ O ~ 8: 3QUIVALENT SETS

Any two sets are equivalent if there is one-one correspondence between them.

Thus if two sets S and T are equivalent, we write, as S - T

You can easily shop that S, T and P are any three sets such that S - T and T - P, then S - P. i

The notion of the equiralent sets is very important because it f ~ r m s the basis of the 'counting' of the infinite sets. ,

Now, consider any two line segments AB and CD.

C F

Let M denote the set of points on AB and N the set of points on CD. Let us check i

whether M and N are equivalent.

Join CA and D B to meet in the point P. Let a line through P meet AB.in E and CD in F. Define f:M+ N as f(x) = y where x is any point on AB and y is any point on CD. The construction shows that f i s a one-one correspo*dence. Thus M and N are equivalent sets.

. .

The follawing are some examples bf equivalent sets : Let I be an interval with end points a and b, and J be an interval with end pbinrs c and d. Also, we assume that I and J are. intervals of the same type. Define f:I -+ J, by

'- - a a ) d + (e) c, fo r t &I. b - a

Then, it is not difficult to see that f is a one-to-orre c0,rrespondence between intervals 1 and J. Hence, all the intervals of same type are equivalent to each another.

Now, we introduce the following definition:

DEFINITION 9: DENUMERABLE AND COUNTABLE SETS

A set which is equivalent to the set of natural numbers is call'ed a denumerable set. Any set which is either finite.or denumerable, is called a Countable set.

Any set whic11 is not countable is said to be an uncountable set.

EXAMPLE 6 : (i) A mapping f:Z+ N defined by

- 2n, if n is a negative integer 2n + 1, if n is no.-~~egative integer

is a one-to-one correspondence. Hence Z - N. 'Thus the set of integers is a denumerable set and hence a counhb~e set.

(ii) Let E denote the set of all even natural numbers. Then the mapping f:N -+ E defined as f(n) = 2n is a one-one correspondence. Hence the set E of even natural numbers is a denumerable set and hence a countable set.

(iii) Let D denote the set of all odd integers and E the set of even integers. Then the, mapping f: E+ D, defined as f(n) = 11 + 1 is a one-one correspondence.Thus E - D. But, E - N, therefore D - N. Hence D is a denumerable set and hence a countable set.

We observe that a set S is denumerable if and only if it is of the form {a,, a,, a, .....I for distinct elements a,, a,, a, ..... For, in this case the mapping f(all) = n is one-one mapping of S onto N i.e. the sets {a,, a,, a ,,.....I and the set N are equivalent.

If we consider the set S, = 12, 3, 4, .....}, we find that the mapping fiN j 's , defined as f(n) = n+l is one-one and onto. Thus S, is denumerable. Similarly if we consider S, = {3, 4 .....) or S, = {k, k+l , ...... }, then we find that all these are denumerable sets and hence are countable sets.

We have seen that the set of integers is countable.

Now' we discuss the countability of the rational and real numbers. Here is an interesting theorem.

THEOREM 2: Every infinite subset of a denu~nerable set is denumerable.

PROOF : Let S be a denumerable set. Then S can be written as

S = {a,, a,, a ,,..... ). Let A be an infinite subset of S. We want to show that A is also denumerable.

You.can see that the elements of S are designated by subscripts 1, 2, 3, .... Let n, be the smallest subscript for which a EA. Then consider the set A - {a ). Again, in this new

1 " I set, let n, be the smallest subscript such that an, E A - {a,,,}.

Let ni be.the sinallest subscript such that

allk E A - {anl, an, ,......, a }. "k-l Note that such an element a always exists for each k EN as A is infinite. For, then

"k

Structure of Real Numbers

.

Renll Numbers nud Functions A - {an1, an2, ...., # 0

for each k EN. Thus, we can write

A = {a , a , an], ...., an ,,.... }. " I "2

Define f: N +A by f(k) = a . Then it can be verified that f is a one-one correspondence. Hence A is denumerable. hi\ completes the proof of the theorem.

EXERCISE 14) Every subset of a countable set is countable.

Now consider the sets S = (6, 8, 10, 12 ,.... } and T = (3, 5, 7, 9, 1 1, ...), which are both denumerable. Their union S u T = {3, 5, 6, 7, 8, 9, ...) is an infinite subset of N and hence is denumerable. Again, if S = (-1, 0, 1, 2) and T = (20, 40, 60, 80, .... ), then we see that S u T = (-1, 0, 1, 2, 20, 40, 60, ..,..) is a denumerable set. Note that in each case S n T = 0 . In fact, you can prove a general result in the following exercise. -. - - - - - --- -- - - -

EXERCISE 15) (i) If S and T arc two denumerable sets, such that S n T = 0, then S u T is

denumera ble.

(P) If S is denumerable and T is finite such that S n T = 0, then also S u T is denumerable.

(iii) The condition S n T = 0 can be relaxed in (1) and (ii).

Thus, it follows that the union of any two countable sets is countable.

Indeed, let S and T be any two countable sets. Then S and T are either finite or denumerable.

If S and T are both finite, then S u T is also a finite set and hence S u T is countable.

If S is denumerable and T is finite, then also we know that S u T is denumerable. Hence S u T is countable. Again, if S is finite and T is denumerable, then again S u T is denumerable and countable.

Finally, if both S and T are denumerable, then S v T is also denumerable1 and hence countable. In fact, this result can be extended to countably many countable sets. We prove this in the following theorem.

THEOREM 3 : The union of a countable number of countable sets is countable.

PROOF : Let the given sets be A,, A,, A, ,.... Denote the elements of these sets, using double subscripts, as follows :

A, = {a1 ,, alp ,.-.. 1 A, = {az1, a,,, a,,, .... 1 A3 = {a3l, aI2, -**I 3

and so on. Note that the double subscripts have been used for the sake of convenience only. Thus aij is the jth element in the set A,. Now let us try to form a single list of all elements of the union of these given sets.

One method of doing this is by using Cantor's diagonalised counting as indicated by arrows in the following table.

Diagonalised Counting of 2 , ~ ~ .

:list the elements as indicated through the arrows. This is a scheme for making a single Structure of Real Numbers

;t of all the elements.

,llowing the arrows in above table, you can easily arrive at the new single list:

a,,, a,,, a,,, a,,, a,,, a,,, a,,, a,,, ......

ate that while doing so, you must omit the duplicates, if any.

ow, if any of the sets A,, A,, ........ are finite, then this will merely shorten the final list. hus, we have

...... ui Ai =q {a,,, a,, .... .), i = 1, 2, 3, e

which each element appears only once. This set is countable and, so, complete! the of of the theorem.

re are now in a position to discuss the countability lof the sets of rational and real ~mbers.

-4.2 COUNTABILITY OF REAL NUMBERS. re have already established that the sets N and Z are countable. Let us now consider le case of the set Q of rational numbers. For this we need the following theorems: .

HEOREM 4 : The set of all rational numbers between [0, I] is countable.

ROOF : Make a systematic scheme in an order for listing the rational numbers x where 5 x 1, (without duplicates) of the following sets

........I ...........................................................................................................................................................

........ ............................................................................................................................................................. 'ou can see that each of the above sets is countable. Their union is given by

1 1 2 1 3 1 2 3 4 1 ' A;={() ,- - - - - - - - - - ...} = [0, l]n Q, . . . I 2 ' 3 ' 3 ' 4 ' 4 ' 5 ' 5 ' 5 ' 5 ' 6 , I

lhich is countable by Theorem 3.

'HEOREM 5 : The set of all positive rational numbers is countable.

ROOF: Let Q, denote the set of all positive rational numbers. To prove that Q, is ountable, consider the following sets :

elist the elements of these sets in a manner as you have done in Theorem 3 or as '

lown below:

Real Numbers a n d Functions

You may follow the method of indicating by arrows for making a single list or you may follow another path as indicated here. Accordingly, write down the elements.of Q+ as they appear in the figure by the arrows, while omitting those numbers which are already listed to avoid the duplicates. We will have the following list:

which is countable by Theorem 3. Thus Q+ is countable.

Now let Q- denote the set of all negative rational numbers. But Q+ and Q- are equivalent sets because there is one-one correspondence between Q+ and Q-, f ' ~ + + Q-, given by

f(x)= -x, V x EQ+.

Therefore Q- is also countable. Further (0) being a finite set is countable. Hence,

Q = Q+ c/ (0) u Q- is a countable set. Tbus, in fact, we have proved the following theorem: I

THEOREM 6: The set Q of all rational numbers is countable. 1

You may start thinking that perhaps every finite set is denumerable. This is not true. We have not yet discussed the countability of the set of real numbers or of the set of irrational numbers. To do so, we first discuss the countability o f the set of real numbers 1

i in an interval with end points 0 and I , which may be closed or open or semi-closed.

Consider the real numbers in the interval 10, I[. ! : , I

Each real number in 10, 1 [ can be expressed in the decimal expansion. This expansion may be non-terminating or may be terminating e.g. I 1

I

: is an example of non-terminating decimal expansion, whereas i

are terminating decimal expansions. Even the terminating expansion can also be expressed ; h as non-terminating expansion in the sense that you, can write . C

I S

Thus, w e agree to say that each real number (rational of irrational) in the 10, . I [ can be I expressed as a non-terminating decim'al expansion in terms of the digits from 0 to 9.

I I

Suppose x E]O, I[. Then it can be written as I I

X = .c,c2c3 ..... I

where c,, c, ,.... take their values from the set (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) of ten digits. . Similarly, let y be another,real lnumber in (0, 1). Then y can also be exiressed as I I

UNIT 3 TOPOLOGY bF THE REAL L INE

STRUCTURE , 3:1 Introduction

Objcctivcs

3.2 Modulus of a Real Number ' ~ro$rties of the Moduiw of a RE^ Number .

3.3 Neighbaurhood of a Point

3.4 Open Sets

3.5 Limit Point of a Set Bulzuro-Weiertrm Thcorem

3.6 Closed Sets

3.7 Compact Sets Heine-Borcl iheorrm

t 3.8 Summary ' ,

' 3.9 AnswerslHintsl Solutions

' 3.1. ' INTRODUCTION

You are quite familiar with an elastic string or a rubber tube or a spring. Suppose you have an elastic string. If you first stretch it and then release the pressure, thdn the string will come back to its original length. This is a physical phenomenon but in Mathematics, we interpret it differently. According to Geometry, the unstreched string and the stretched string are different since there is a change in the length. But you will be surprised to know that according to another branch of Mathematics, the two positions of the string are identical and there is no change. This branch is known as Topology, one of the most exciting areas of Mathematics.

The word 'topology' is a combination of the two Grekk words 'topos' and 'logos'. The term 'topos' means the top or the surface of an object and '1ogo~'means the study. I

! Thus 'topology' means the study of surfaces. Since the surfaces are directly related to ; geometrical objects, therefore there is a close link between Geometry and Topology. In Geometry, we deal with shapes like lines, circles, spheres, cubes, cuboids etc. and their geometrical properties like lengths, areas, volumes, congruences etc. In

I

Topology, we study the surfaces of these geometrical objects and certain related I

properties which are called topological properties. What are these topological properties of the surfaces of a geometrical figure? We shall not answer this question at

I this stage. However, since our discussion is confined to the real line, therefo~e, we shall discuss this question pertaining to the topo\ogical properties of the real line. I

These properties are related to the points and subsets'of the real line such as I

neighbourhood of a point, open sets, closed sets, limit points of a set of the real line etc. We shall, therefore, discuss these notions and concepts in this u ~ t . Howver, I

prior to all these, we discuss the modulus of a real number and its relationship with the order relations or inequalities in Section 2.2.

\

WOBJECTIVES After reading this unit, you should, therefore, be able to

* define the modulus of a real number and its connecthn with the order relations in the real numbers

*describe the hotion of a neighbourhood of a point on%e line I

+ define an open set and give examples

-+ find the limit points of a set I

. ,* define a closed set and establish its relation with an open set

:- explain the meaning of an open covering of a subset of real numbers and that of a compact set. .

50

3.2 MODULUS OF A REAL NUMBER

You know that a real number x is said to be positive if x is greater than 0. Equivalently, if 0 represents a unique point 0 on the real line, then a positive real number x lies on the right side of 0. Accordingly, we defined the inequality x > y (in terms bf this positivity of real numbers) if x- y > 0. You will recall from Section 2.2 that for the validity of the properties of order relations or the inequalities. such as the one concerning the multiplication of inequalities, it is essential to specify that some of the numbers involved should be positive. For example, it is necessary that z > 0 so that x > y implies xz > yz. Again, the fractional power of a number will not be real if the number is negative, for instance x'" when x = -4. Many of the fundamental inequalities, which you may come across in higher Mathematics, will involve such fractional powers of numbers. In this context, the concept of the absolute value or the modulus of a real member is important to which you are already familiar. Nevertheless, in this section, we recall the notion of the modulus of a real number and its related properties which we need for our subsequent discussion.

DEFINITION 1 : MODULUS OF A REAL NUMBER Let x be any real number. The absolute value or the modulus of x denoted by I,xl. is defined as follows :

1x1 = x i f x > O = - x i f x < O =Oi fx=O,

You can easily see that I-xl = Ixl,IVx,ER.

Note that I - xl is different from - 1x1.

3.2.1 PROPERTIES OF THE MODULUS OF A REAL NUMBER Since the modulus of a real number is essentially a non-negative real number, therefore the operations of usual addition, subtraction, multiplication and division can be performed on these numbers. The properties of the modulus are mostly related to these operations.

PROPERTY 1 : For any real number x, I xl = Maximum of (x, - x) , i

PROOF: Since x is any real number, therefore either x 2 0 or x < 0. If t 1 0 , then by definition, we have /

1x1 = X. Also, x 1 0 implies that - x $0. herefo fore,

maximum of (x, -x) = x = 1x1) Again x < 0, implies that -x > 0. Therefore again maximum of {x, -x) = - x = I X I .

Thus Max. (x, -.XI= 1x1

Now you can solve the following exercise :

-

4 Now consider the numbers 15 I ', 1 - 4.5 1 , 1;). It is easy to see that

and 4 - 4 131 - - and -!!! = i.e.

5 151 5 4 - I41 1-1 - - 5 I 5 1 '

All this leads us to the following properties:

EXERCISE 1)

Prove that - 1 X I = Min. Ex, - x] for any x E R . Deduce that - 1x1 5 x, for every /x\€R. Illustrate it with an example.

PROPERTY 2: For any real number x l X i 2 = x 2 = 1- x12

PROOF: We know that I xl = .x f0r.x 2'0. TJhps lx12 = 1x1 1x1 '= x.x = x2,,for x 2 0

Again for x < 0, we know that I xl = -x. Therefore 2 lx21=JxI I x + = - x . - X = X . ,

Therefore, it follows that lx12 = x2 for any XE\R.

Now vou should try the other part as an exercise.

! I EXERCISE 2)

I Prove that I -xi2 = x2, for my xEB. I

PROPERTY 3 : For any two real numbers x and y, prove t h d Ix.yl = 1x1 . IYI.

PROOF: Since x and y are any two real numbers, therefore, either both are positive or one is positive and the other is negative or Both are negative i.e. e i t h e r x > O , y 1 O o r x 1 O , y 5 O o r x I 0 , y 2 O o r x I O , y I O . V J e discuss the proof for all the four possible cases separately.

i ' ) : ,.

i , .

, , 8

8 5

Case (i): When x 1 0, y 2 0. Since x 2 0, therefore, we have, by definition, 1x1 = x, I Y I = y

Also x 2 0, y 1 0 imply that xy 2 0 and hence lxyl = XY = 1x1 l ~ l

which proves the property. s

' . Case (ii): When x 1 0 , y 5 0. Then obviously x y 5 0. Consequently by definition, it , follows that .

Ix I=x, IYI =- Y, I X Y I = - X Y Hence

lxyl'= - XY = x (: Y) = 1x1 l ~ l

which proves the property. , Case (iii): When x 5 0, y 1 0 . Interchange x and y in (ii).

Case (iv): When x 5 0, y 5 0, Then. x y 2 0. Accordingly, we have 1x1 = - x, IyI = - y, lxyl = xy.

Hence lxyl =xy =(--x)(-y)= 1x1 I Y I

using the field properties stated in Section 3.3. This concludes the proof of the property.

Alternatively, the proof can be given by using property 2 in following way: 1 ~ ~ 1 2 = ( ~ ~ ) 2 = x 2 Y 2 = 1x1~. ly12

' Since Ixy 1, I xl and 1 y ( are non-negajjve, therefore we take the positive sign only and I

we have

I lxyl = 1x1 I Y I which proves the property.

You can use any of the two methods to try the following exercise. I

Fm anv two red numbers x urd g (y P 81, prove that ,

1; = #. .Topology of the Real Line

- - ---

The next property is related to the modulus of the sum of two real members. This is . one of the most important properties and is known as Triangular hequ.lHty:

PROPERTY 4: TRIANGULAR INEQUALITY

FOP any two real numbem x and y, prove that :..

I X + Y / 5 lxl+ I Y I . I L.

PROOF : For any two real numbers x and y, the number x + y 2 0 or x + y < 0. If x + y L 0, then by definition

I x + y ( = x + y . (1)

Also, we know that 1x1 L x Y X ~ R I Y I Z Y W y€,R.

Therefore

From (1) znd (2), it follows that lx + Y I 51x1 + I Y I

Now, if x + y < 0, then again by definition, we have . I X + Y I =- (x+y )

or Ix+yI =(--x)+(--y)

Also we know that (see property 1) 7 x 5 1x1 a n d - y 5 ( y ( .

Consequently, we get ( - - X ) + ( - - Y ) ~ J ~ I + I Y I

or ( - x ) + ( - Y ) ~ I x I + I Y I

From (3) and (4), we get

Ix + Y I 5 1x1 + IY l

.This concludes the proof of the property.

You can try the lollowing exercise similar to this property.

EXERCISE 4): Rove that I X ~ Y I 2 I 1x1 - I Y I I for any real numbers x and y.

Now let us see another interesting relationship between the inequalities and the modulus of a real number.

By definition, 1x1 is a non-negative real number for any x E R..Therefore, there .. always exists R non-negative real number u SULXC that

either 1x1 (: u or 1x1 > u or 1x1 = u.

Suppose 1x1 < u. Let us choose u = 2. Then 1 I x 1 < 2/ +Max. {-x, x} < 2 +- x<(2, x < 2

* x> -2 , x<2 ==3 - 2<x , x<2

I 23 -2<x<2.

i.e. 1x1 < 2 - - 2 < x < 2

; Real Numbers and Functions

i

Conversely, we have - 2 < x < 2 q - 2 < x . x < 2

j. 12>-x,x<2 + - x < 2 , x < 2 * Max. {-x, x} < 2 =+ 1x1 <2. '

i.e.

-2<x<2*1xy<f2 Thus, we have shown,'\hat

1x1 <2<*-2<x<2. - -

This can be generalised as the following property. I

/PROPERTY 5: Let x and w be any two real numbers.

1x1 5 u < = > - ~ < x 5 I u .

PROOF : I xl 5 u Max. {-x, x} 5 u * - x < u , x I u e 3 x 1 - u , x , S u ~ - u 5 x 9 x l u

- u I x . S u

which proves the desired property. I

The property 5 can be generalized in the form of the folowing exercise:

W A W R b I U J 4 r l )

For any real numbers x, a and d,

EXAMPLE l:I Write the inequalitS3 < x < 5 in the modulus form.

SOLUTION : ' Suppose that there exists real numbers a and b such that a- b = j , a f b=5 .

Solving these equations for a and b, we get a = 4 , b = 1.

Accordingly, 3 < x < 5 ~ 4 - 1 < ~ < 4 + 1

e - 1 < x - 4 < 1 -1x-41< 1

Now you can abo try the following exercise. - - - -

.EXERCISE 6) (i) Write the inequality 2 < x < 7 in the modulus form (il) Convert Ix - 21 < 3 into the corresponding inequality. .

3.3 NEIGHBOURHOODS

You are quite familiar with the word 'neighbourhood'. You use this word frequently in your daily life. Loosely speaking, a neighbourhood of a given point c on the real line is a set of all those points which are close to c. This is the notion which needs a precise meaning. The term 'close to'& subjectve and therefore must be quantified. We should clearly say how much 'close to9.To elaborate this, let us first discuss the notion of a neighbourhood of a point with respect to a (small) positive real number S.

k t c be any point on the real line and let 6 > 0 be a real number. A set consisting of all those points on the real line which are at a distance of 6 from c is called a neighbourhood of c. This set is given by

{xER: Ix-cl <6} = { ~ E : R : c - 6 < x < c f 6) =]c- 6, c + 6[

which is an open interval. Since this set depends upon the choice of the positive real number 6, we call it a S-neighbourhood of the point c.

T~US, a 6-neighboured of a point c on the real line is an open interval ]c - 6, c + 6[, 8 > 0 while c is the mid point of this neighbourhood. we can give the general definition of n e i g h b ~ ~ r h ~ ~ d of a point in the following way.

DEFINITION 2: NEIGHBOuRHmD OF A POINT

A set P is said to be a Neighbourhood (NBD) of a point 'c' if there exists an open interval which contains c and is contained in P.

This is equivalent to saying that there exists an open interval of the form ]c-6, c+6[, for some 6 > 0, such that

]C - 6, c + 6[ c P.

EXAMPLE 2: (i) Every open interval ]a, b[ is a NBD of each of its points.

(ii) A closed interval [a, b] is a NBD of each of its points except the end point i.e. [a, b] is not a NBD of the points a and b, because it is not possible to find an open interval containing a or b which is contained in [a, b]. For instance, consider the closed interval [o, I ] . It is a NBD of every point in 10, l[. But, it is not a NBD of 0 because for every 6 > 0, 1-6, 61 d [O, I]. Similarly [0, I ] is not a NBD of 1.

(iii) The null set 0 is a NBD of each of its point in the sense there is no point in 0 of which it is not a NBD.

(iv) The set R of real numbers is a NBD of each real number x because for every 6 > 0, the open interval ]x - 6, x + 6 [ is contained in R.

(v) The set Q of rational numbers is not a NBD of any of its points x because any open interval containing x will also contains an infinite number of irrational numbers and hence the open interval can not be a subset of Q.

Now try the following exercise.

Topology of the Real Line

EXERCISE 7)

Examine the validity of the following statements. Justify your answer in each case.

(i) The interval [a, b[ is a neighbourhood of each of its points.

(ii) The unit interval 10, 11 is the neighbourhood of only of its end points.

(iii) The set {x E R: x 2 a) is not a neighbourhood of any of its points.

( i ) The set {x E R: x i a) is a neighbourhood of each of its points.

(v) The singleton {x), for any x E R, is a neighbourhood of x.

(vi) A finite subset of R is not a neighbourhood of any of its points.

1 1 1 1 Now consider any two neighbourhoods of the point 0 say 1- 10, 10 [ and 1- 7, 3[ as

shown in the Figure 1.

Fig. 1 The intersection, of these two neighbourhood is

- 1 1 1 1 1 1 l - l o . ~ [ " l - y . ~ [ = l - ~ . ~ [ .

1 1 which is again a NBD of 0. The union of these two neighbourhoods is ] -7 , j [. which

is also a NBD of 0. Let us now examine these results in general.

EXAMPLE 3 : The intersection of any two neighbourhoods of a point is a neighbourhood Of the point.

SOLUTION : Let A and B be any two NBDS of a point c in R. Then there exist open

intervals ]c -6,, c + 6,] and ]c -a,, c+6,[ such that d.

1~-6,,c+6,]cA,forsome6,~0,and]c-6,,c+6,[cB,forsome6,~0.

Real Numbers and Functions Let 6 = Min. {61, 62) = minimum of 81, Sz. This implies that ]c -6, c + 6[ C A n B which shows that A n B is a NBD of c.

EXAMPLE 4: Show that the superset of a NBD of a point is also a NBD of the . ' point.

SOLUTION: Let A be a NBD of aspoint c. Then there exists an open inteival ]c -6, c+ 6[, for some 6 > 0 such that

]c - 6, c + 6[ C A.

Now let S be a super set which contains A. Then obviously

which shows that S is.also a NBD of c.

: For instance, if ] I [ is a NBD of the point 0. 10 10

1 1 Then ] - - - [ is also a NBD of 0 as can be seen 5 5 - - .

from the Figure 1 .'

. . -I . - 1 - - , I ' I ' -- 5 10 10 5

I Fig. 1 .

Is a subset of a NBD of a point also a NBD of the point? Justify your answer. I

! Now you can try the following exercise.

EXERCISE 8) \ , Prove that the Union of any two NBDS of a point is a NBD of the point.

The conclusion of the Exercise 8, in fact, can be extended to a finite or an infinite or an arbitrary number of the NBDS of a point. ~ However, the situation is not the same in the case of intersection of the NBDS. It is true that the intersection of a finite numbereof NBDS of a point is a NBD of the point. But the intersection of an infinit. collection of NBDS of a point may not be a NBD of the point. For example, consider the class of NBDS given by a family of open intervals of the form

n n I

which are NBDS of the point 0. Then you can easily verify that

II n IZ n 1 3 n n ..... n I, n .....

which .is not a NBD of 0.

3.4 OPEN SETS

You have seen from the pievious examples and exercises that a given set may o r may not be a NBD of a point. Also, a set may be a NBD of some of its points and not of its other points. A set may even be a NBD of each of its points as in the case of the intervatla, b[. Such a set is called an open set.

DEFINITION 3:\A SET S IS SAID TO BE OPEN IF IT IS A NEIGHBOURHOOD ) OF EACH OF ITS POINTS. Thus, a set S is open if for each x in S, there exists an open interval ]x - 6, x t. 6[ , '

6 > 0 such that x ~ ] x - - 6 , x + 6 [ C S .

~t follows at once that a set S is not open if it is not a NBD of even one of its points.

EXAMPLE 5: An open interval Is an open s d .

SOLUTION: Let ]a, b[ be an open interval. Then a < b. Let c E ]a, b[. Then a < c < b and therefore

c - a > O a n d b - c > 0 . Choose

6 = Minimum of (b -c, c - a) = Min (b - c, c - a).

Note'that b - c > 0, c - a > 0. Therefore 6 > 0. N o w 8 5 c - - a ~ ~ > a 5 c - 6 a n d 6 I b - c = = > c $ 6 5 b .

i.e. Thereioi.e, ]c - 6, c 4- 4 C ]a, q and hence ]a, b[ is a NBD of c.

EXAMPLE 6: (i) The sctt R of real numbers is an open set (ii) The laull set 4 is an open set (iii) A firmite set is not an open set (iv) The interval ]a, b] is not an open set.

You can solve the fo l lo~~ ing exercise easily:

EXERCISE 9) Test which of the follov oing are open sets:

(i) An interval [a, b] f o r a E R, b ER, a < b

(ii) The intervals [O, l [r~nd ]0,1[

(id) The set Q of ratio] ma1 numbers

(iv) The set N of natur rl numbers and the set Z of integers. 1 (v) The set {- : n EiN) n

(vi) The intervals ]a, a9 [and [a, 00 [for a ER.

EXAMPLE 7: Prove that the intersgction of any two open sets is an open set.

SOLUTION: Let A iind B be any tCo open sets. Then we have to show that A n B is also an open set. If A , 0 B = 4, then obviously A n B is an open set. Suppose A n B =# 4. Let x be an arbitrary element of A n th then x € A n B => x E A and x E B. Since A and B are o p n sets, therefore A and B are both NBDS of x. Hence A n B is a NBD of x. But x E n B is chosen arbitrarily. Therefore A n B is a NBD of each of its points and henr:e A n B is an open set. This proves the result. In fact, you can prove that the intersection of a finite number of open sets is an open set. However, the intersection of an infinite number of open sets may not be an open set. Try the following /exercises:

Topology of the Real Line

EXERCISE 10) Give an example to slnow that intersection of an infinite number of open sets need not be an open set. EXERCISE 11) Prove that the uniolh of any two open sets is an open set. In fact, you can show that the union of an arbitrary family of open sets is an open set.

3.5 LIMIT POINT OF A SET,

You have seen that the concept of an open set is linked with that of a neighdourhood of a point on the real line. Another closely related concept with the notion of neighbourhood is that of a limit point of a set. Before we explain the meaning of limit point of a set, let us study the following situations:

(i) Consider a set S = [I, 2[, Obviously the number 1 belongs to S. In any NBD of the point 1, we can always find points of S which are different from 1. For instance 16.5, 5 7

Real Numbers and Functions ' 1.:1[ is a NBD of 1. In this NBD, we can figd the point 1.05 which is in S but at the same time we note that 1.05 # 1.

l i

(ii) Consider another set S = { 1' : n € N.). The number 0 does not belong toCthis set. n

1 Take any NBD of 0 say, 1-0.1,O. I[. The number -- = 0.05 of S is in this NBD of 0. . Note that 0.05 Z 0. 20

( i ) Again considei the same set S of (ii) in which the number 1 obviously belongs to S. We can find a NBD of 1, say 10.9, 1.1[ in which we can not find a point of S . different from 1.

I In the light of the three situations, we'are in a position to define the.following:

DEFINITION 4: LIMIT POIhIT.OP A SET .

A number p is said to be a limit point o i a set S of real numbers if every ! neighbourhood of p contains at least one point of the set S different from p. . . . EXAMPLE 8: (i) In the set S = [I, 2[, the number 1 is a limit point of S. This limit

I 1' point belongs to S. The set S 311- : n E N) has only one limit pdnt 0. You may ~ io te that 0 does not belong to S. , n ' (ii) Every point in Q, (the set of rational numbers), is a limit point of Q, becauue.for

I

every rational number r and 8 > 0, i.e.1 r - 6, r + S ['has at least one rational number I

: different from r. This is because of the reason that there are infinite rationals between ] I any two real numbers. Now, you can easily see that every irrational number is 81~0 P

limit point of the set Q for the same reason.

(iii) The set N of natural numbers has no limit point because for every real ~\:~nlber a, you can always find 6 > 0 such that ]a - 6, a + @ does not oontain a point of the set N other than a.

(iv) Every point of the interval ]a, b] is its limit point. The end points a and b are also the limit poiats of ]a, b]. But the limit point a does not belong to it whereas the limit point b belongs to it.

(v) Every point of the set [a, =[ is a limit point of the sets. This is also true for l-m,a[.

Now try the following exercise and justify your answer.

EXERCISE 12) (i) Does the set Z possess a limit point?

(ii) Every point of R, the set of real numbers is a limit point of R. Is it true? '

(iii) Is every point of an open interval ]a, b[ a limit point of ]a, b[? what about the end points a and b? .

(iv) Is every point of a closed interval [a, b] is its limit point? what about the end points a and b?

(v) Is every point of the sets ]a, oo[ and 1- 00, a] a limit poiqtWthe set? -

EXERCISE 13) I

Show that a given 'point p is a limit point of a set S if .and only if every neighbourhood.of p contains an infinite number of members of S.

From the foregoing examples and exercises, you can easily observe that

(i) A limit point of set may or may not belong to the set, (ii) A set may have no limit point, (iii) A set may have only one limit point. (iv) A set may have more than one limit point. ,

The question, therefore, arises: "How to know whether or not a set has a limit point?" One obvious fact is that a finite set can not have a limit point. Can you givi a reason for it? Try it. But then there are examples where even an infinite set may not have a limit point e.g.'the sets N and Z do'not have a limit point even though they are infinite 'sets. However, it is certainly clear that a set which has a limit point, must necessarily be an infinite set. . Thus . our question takes the following form:

58 , "What are the conditions for a set to have a limit point?"

. This question was first studied by a Czechoslovakian Mathematician, Bernhard Bulzano , {178 1-1 8481 in 18 17 and he gave some ideas.

unfortunately, his ideas were so far ahead of their time that the world could not appreciatk the full significance of his work. It was only much later that Bulzano's work was extended by Karl Weierstrass [I 8 15- 18971, a great German Mathematician, who is known as the "father of analysis". It was in the year 1860 that Weierstrass proved a fundamtntal result, now known as Bulzano-Weierstrass Theorem for the existence of the limit points of a set. We state and prove this t h e ~ ~ e m as follows.

3.5.1. BULZANO-WEIERSTRASS THEOREM

THEOREM 1: Every infinite bounded subset of set K has a limit point (in K). . .

PROOF : Let S be an infinite and bounded subset of R. Since A is bounded, therefore A has both a. lower bound as well as an upper bound. (Recall the definition of a bounded set from Section 2.3.)

Let m be a lower bound and M be an upper bound of A. Then obviously m < x < M , V x e A .

Construct a,set S in the following way: '"S = (x E R: x exceeds.at most finite number of the elements of A}. Now, let us examine

the following two questions: (i) 1's S a non-empty set? (ii) Is S also a bounded set?

Indeed, S is non-empty because m S x , 'd x EA, implies that m ES. Also M is an upper bound of S because no number greater than or equal to M can belong to S. Note that M can not belong to S because it exceeds an infinite number of elements of A.

Since the set S is non-empty and bounded above, therefore, by the axiom of completeness (see Section 2.3), S has its supremum in R. Let p be the supremum of S. We claim that p is a limit point of the set A.

In order to show that p is a limit point of A, we must establish that every NBD of p has at least one point of the. set A other than p. In other words, we have to show that every NBD of p has an infinite number of.elements of A. For this, it is enough to show that any open interval ]p - 6 , p + 6[, for 6 > 0, contains an infinite number of members of set A. For this, y e pioceed as follows.

Since p is the supl'emum of S,. therefore, by the definition of the Supremum of a set'(see Section 2.3), there is at least one element y in S such that y > p - 6, for 6 > 0. Also y is a member of S, therefore, y exceeds at the most a finite number of the elements of A. In other words, if you visualise it on the line as shown in the Figure.2 below, the number of elements of A lying on the left of p - 6 is finite at the most. But

Fig. 3

certainly, the number of elements of A lying on the right side of the point p - 6 is infmite.

Again since p is the supremum of S, therefore, by definition p + 6 can not belong to S. In other Words, p + 6 exceeds an infinite number of elements of A. This means that there lie an kfinite number of elements of A on the 16ft side of the point p + 6.

Thus we have shown that there lies, an infmite number of elements of 4 on the right side of p - 6 and also there ia an infinite number of elements of A-on the left side of p + 6 .What do you conclude from this? In other woids, what is the number of elements of A in between (i.e., within) the interval ]p. - 6, p + 6[? Indeed, this number is infinite i.e., there is an infinite number of elements of A in the open interval

Topology of the R C ~ I Linr

]p - 6, p + I5 [.Hence the interval ] p - 8 , p 4-6 [contains an infinite number of elements of A for some 6 > 0. Since 6 > 0 is chosen arbitrarily, therefore every interval ]p - 8, p +6 [ has an infinite number of elements of A. Thus every NBEof p contains an i

nfi

nite number of elements of A. Hence p is a limit point of the set A.

This completes the proof of the theorem.

EXAMPLE 9 (i) The intervals [0, 11, 10, I[, ] 0, I], [ O , 1 [ are all infinite and bounded sets. Therefore each of these intervals has a limit point. In fact, each of these intervals has an bd i i te number of limit points because every point in eacR interval is a limit poiat of the interval.

(4) The set [a, oc[ is W i i t e and unbounded set but has every point as a limit point. This shows that the condition of boundedness of an infinite set is only sufficient in the theorem.

EXERCISE 14) Give examples of the following:

(i) At least four infmite bounded sets indicating their corresponding limit points.

(ii) At least three unbounded (and i n f i t e ) .sets each having a limit point.

(iii) An infinite and unbounded set haviag no Limit point.

From the previous examples and exercises, it is clear that it is not necessa - for an infinite set to be bounded to possess a limit point. In other words, a set ir.:*y be unbounded and still may have a limit point. However for a set to have a limit point, it is necessary that it is infinite.

Another obvious fact is that a limit poinf of a set may or may not belong to the set and a set may have more than one limit point. In the next section, we s'hall further study how sets can be characterized in terms of their limit points.

3.6 CLOSED SETS

In Section 3.5, you have seen that a limit point of a set may or may not belong to the set. For example, consider the set S = { x E R : 0 5 x < 11. In this set, 1 is a limit point of S but it does not belong to S. But if you take S = {x: 0 5 x 5 1}, then all the limit points of S belong to S. Such a set is called a closed set. We define a closed' set as follows:

DEFINITION 5: CLOSED SET

A set is said to be closed if it contains all its limit points.

EXAMPLE 10:

(i) Every closed and bounded interval such as [a, b] and [0, 11 is a closed set.

(ii) An open interval is not a closed set. Check Why?

(iii) The set R is a closed set because every real number is a limit point of R and it belongs to R.

(iv) The null set I$ is a closed set.

1 (v) The set S = { - : n E N } is not a closed set. why?

n

(vi) he set ]a, m[ is not a closed set, but 1-00, a] is a closed set.

You can try the following exercise: , /

EXERCISE 15) ~ h & k . ~ h e t h e r or not the following sets are closed sets:

(i) ~ h e ' s ' e t . ~ of rational numbers

(ii) The set N'& natural numqers

(iii) The set Z of integers.

) A finite set of real numbers.

(v) The set S = [x E R: a I x 5 b).

(vi) The sets [a, m[ and ]-a, a[.

Topology of the Real Line

you may be thinking that the word open and closed should be having some link. If you are guessing some relation belween the two terms, then you are hundred per cent correct. Indeed, there is a fundamental connection between open and closed sets. What exactly is the relation between the two? Can you try to find out? Consider, ,the following subsets of R:

(i) 10,4[

(ii) [-2,51

(iii) lo,*[

(iv) I-* $1.

The sets (i) and (iii) are open while (ii) and (iv) are closed. If you consider their complements, then the complements of the open sets are closed while those of the closed sets are open. In fact, we have the following concrete situation in the fonn of following theorem.

THEOREM 2: A set is closed if and only if its complement is open.

PROOF : We assume that S is a closed set. Then we prove that its complement Sc is open.

To show that Sc is open, we have to prove that Sc is a NBD of each of its points. Let x E SC. Then, x E Sc d x E S. This means x is not a limit point of S because S is given to be a closed set. Therefore, there exists a 6 > 0 such that ]x - 6, x + 6 [ contains no points of S. This means that ]x - 6, x + 6[ is contained in Sc, This isrther implies that Sc

is a NBD of x. In other words, Sc is an open set, which proves the assertion.

Conversely, let a set $ be such that its complement Sc is open. Then we prove that S is closed.

To show that S is closed, we have to prove that every limit point x of S belongs to S. Suppose x ES. Then x E SC.

This implies that SC is a NBD of x because Sc is open. This means that there exists an open interval ]x- 6, x + 6[, for some 6 > 0, such that

In other words, ]x - 6, x + 6[ contains no point of S. Thus x is no1 a limit point of S, which is a contradiction. Thus our supposition is wrong and hence, x c S is not possible. In other words, the (limit) point x belongs to S and thus S is a closed set.

Note that the notions of open and closed sets are not inutually exclusive. In other words, if a set is open, then it is not necessary that it can not be closed. Similarly, if a set is closed, then it does not exclude the possibility of its being open. In fact, there are sets which are both open and closed and there are sets which are neither open nor closed as you must have noticed in the various examples we have given in our discussion. For example the set R of all the real numbers is both an open sets as well as a closed set. Can you give another example? What about the null set. Again Q, the set of rational numbers is neither open nor closed.

EXERCISE 16) Give examples of two sets which are neither closed nor open.

In Section 3.4, 'we have discussed the behaviour of the union and intersection of open sets. Since closed sets are closely connected with open sets, therefore, it is quite natural that we should say something about the union and intersection of closed sets. In fact, we have the following results.

EXAMPLE 11 : Prove that the union of two closed sets is a closed set.

SOLUTION : Let A and B be any two closed sets. Let S = A u B . we have to show

Real Numbers and Fu t~c t~on \ that S is a closed set. For tills, i l IS enough to prove that the co'rnplement S' is open. Now . .

s c = ( ~ ~ ~ ) c = ~ c n ~ L = ~ c n ~ L

Since A and B are closed sets, therefore A" and Bc arp. open sets. Also, we have proved in Section 3.4 that the intersection of any two bpen sets is open. Therefore A' n Bc is an open set and hence Sc is open.

This result can be extended to a finite number of closed sets. You can easily yerify that the union of a finite number of closed sets is a closed set. But, note that the union of an arbitrary family of closed sets may not be closed.

For example, consider the family of closed sets given as 1 1

'S1=[1,2],Sz= [ - , 2 . ] , S3= [ - , 2 ] ,.... and in general 2 3

1 S n = [ - ,2 ] .... f o r n Z 1 , 2 , 3 ,....

n Then, 00

U s,=s,us2us3 .... US,U .... n = l

= I 0321 which is not a closed set.

Now try the following exercise:

' EXERCISE 17) Prove that the intersection of an arbitrary family of closed sets is closed.

DEFINITION 6 : DERIVED SET

The set of all limit points of a given set S is called the derived set and iq denoted by S'.

EXAMPLE 12 : (i) Let S be a finite set. Then S'= C#J

1 (ii) S = ( - : n EN), the derived set S' = (0) n

(iii) The derived set of R is given by R' = R

(iv) The derived set of Q is given by Q' = R

We define another set connected with the notion of the limit point of a set. This is called the closure of a set.

DEFINITION 7: CLOSURE OF A SET

Let S be any set of real numbers (S C R). The closure of S is defined as the union of the set S and its derived set S. It is denoted by S. Thus *

-- s=sus1 In other words, the closure of a set is obtained by the conlbination of the elements of a given set S and its derived set S'.

1 For example, 3 of S = { - : n E N) is given by.

Similarly, you can verify that

3.7 COMPACT .SETS \

We discuss yet another concept of the so called compactness of a set. The concept of , cornpaqtness is formulated in terms of the notion bf an open cover of a set.

DEFINITION 8: OPEN COVER OF'A SE'F Topology of the Real Line I Let's be set and {Gal be a collection of some open subsets of R such that S c u Gas Then {Go} is called an open cover of S.

EXAMPLE 13 : Verify that the collection Gn = {G}:, , where Gn = 1 - n, n[ is an open cover of the set R.

As shown in the Figure 3 above, we see that every real number belongs to some G,,. Hence

EXAMPLE 14 : Examine whether o r not the following collections are open covers of the interval [l, 21.

SOLUTION : (i) Plot the subsets of GI on the real line as shown in the Figure 4.

From above figure, it follows that every element of the set S = [1,'2] = {x : 1 5 x I; 2) i

I belongs to at least one of the subsets of G,. Since each of the subsets in G, is an open 1 set, therefore G, is an open cover of S. I

63

Real Numbers and Functions (ii) Again plot the subsets of G, on the real line as done in the case of (i).

5 3 You will find that none of the points in the interval [;i-, ,] belongs to any of the subsets of G,. Therefore G, is not an open cover of S.

Now try the following exercise.

EXERCISE 18) Verify whether the following collections are open covers of the corresponding sets mentioned in each case: (i) GI = {In, n + 2 [ : n EZ) of R; (ii) G , = { ] n , n + l [ : n ~ Z ) o f R ;

Now consider the set [O, 11 and two classes of open covers of this set namely G, and G, given as

You can see that G ,cG, . In this case, we say that G, is a subcover of G,. In general, we have the following definition.

DEFINITION 9: SUBCOVER AND FINITE SUBCOVER O F A SET

Let G be an open cover of a set S. A subcollection E of G is called a subcover of S if E too is a cover of S. Further, if there are only a finite number of sets in E, then we say tliat E is a finite subcover of tlie open cover G of S. Thus if G is an open cover of a set S, then a collection E is a finite subcover of the open cover G of S provided the following three conditions hold.

(i) E is contained in G.

(ii) E is a finite collection.

(iii) E is itself a cover of S.

EXERCISE 19) Give an example of an infinite set S such tliat there is an open cover G of S which admits of a finite subcover of G.

From the forgoing example and exercise, it follows that an open cover of a set may or may not admit of a finite subcover. Also, there may be a set whose every open cover contains a finite subcover. Such a set is called a compact set. We define a compact set in the following way.

DEFINITION 10: COMPACT SET

A set is said to be compact if every open cover of it admits of a finite subcover of the set.

For example, consoder the finite set S = (1, 2, 3 ) and an open cover {Gal of S. Let GI, GZ, G3, be the sets in G which contain 1, 2, 3 respectively. Then {GI, G2, G3} is a finite subcover of S. Thus S is a compact set. In fact, you can show that every finite set in R is a compact set.

The collection G = { I -n, n[ : n EN} is an open cover of R but does not admit of a il

finite subcover of R. Thei-efore the set R is not a compact set. :I ;I

Thus you have seen that every finite set is always compact. But an infinite set may or may not be a compact set. The question, therefore, arises, " What is the criteria to determine when a given set is compact?" This question has been settled by a beautiful theorem known as Heine-Borel Theorem namtd $n the honour of the 64 I

I

German mathematician H.E. Heine [1821-18811 and the French mathematician F. E.E. Bore1 [1871-19561, both of whom were pioneers in the development of Mathematical Analysis.

We state this theorem without proof.

THEOREM 3: Heine-Bore1 Theorem -- Every closed and bounded subset of R is compact. The immediate consequence of this theorem is that every bounded and closed interval is compact.

3.8 SUMMARY -

I . In Section 3.2, we have defined the absolute value or the modulus of a real number and discussed certain related properties. The modulus of real number x is defined as

I x l = x i f x 2 0 =-x i f x<O.

Also, we have shown that Ix-a1 < d a - d < x < a + d

2. In Section 3.3, we have discussed the fundamental notion of NBD of a point on the real line i.e. first we have defined it as a 8 - neighbourhood and then, in general, as a set containing,an open interval with the point in it,

3. With the help of NBD of a point we have defined, in Section 3.4, an open set in the sense that a set is open if it is a NBD of each of its points.

4. We have introduced the notion of the limit point of a set in Section 3.4. A point p is said to be a limit point of a set S if every NBD of p contains a point of S different from p. This is equivalent to saying that a point p is a limit point of S if every NBD of p contains an infinite number of the members of S. Also, we have discussed Bulzano-Weiresstrass theorem which gives a sufficient condition for a set to possess a limit point. It states that a,, infinite and bounded set must have a limit point. This condition is not necessary in the sense that an unbounded set may have a limit point.

5. The limit points of a set may or may not belong to the set. However, if a set is such that every limit point of the set belongs to it, then the set is said to be a closed set. The concept of a closed set has been discussed in Section 3.6. Here, we have also shown a relationship between a closed set and an open set in the sense that a set is closed if and only if its complement is open. Further, we have also defined the Derived set of a set S as the set which consists of all the limit points of the set S. The Union of a given set and its Deived set is cqlled the closure of the set. Note the distinction between a closed set and the closure of a set S.

6. Finally, we have introduced another topological notion in Section 3.7. It is about the open cover of a given set. Given a set S, a collection of open sets such that their Union contains the set S is said to an open cover of S. A set S is said t d b e compact if every open cover of S admits of a finite subcover. The criteria to determine whether a given set is compsct or not, is given by a theorem named Heine-Bore1 Theorem wh'ich states that every closed and bounded subset of R is compact. An immediate consequence of this theorem is that every bounded and closed interval is compact.

E 1) If x 2 0 , then -x 5 0 Min (x, -x) = -X

Also 1x1 = x. Hence -1 X I = -X = Min (x,- x). If x < 0, -x > 0. Hence Min (x, -x) = x. Also

r I xl = -x. Therefore - 1x1 = x = Min (x, - x).

Topology of the ~ e a i l . i n c

- /

Real Numbers and Functions ~ 2 ) Follow the method as explained in Property 2.

I 2 = ($)' (by Property 2).

x' I* - - -- = ' - (again by Property 2) Y2 IYlZ

' * +lxl = J i , (why?) Y - 1yJ IYI

E 4) I X - yI2 = (X - Y)' = x2 + y2 - ~ X Y

1 Ixl2 t /YIZ f 2(- JxYI) = Ixl2 f IylZ - 21x1 I Y I = (1x1 - IYOZ = I 1x1 - IY I l 2 Therefore, lx - Y I 2 i (1x1 - IYI) Hence Ix - yJ 2 1x1 - lyl. (Why?)

E 5) Follow the procedure explained in the Property 5. J x - a ( 5 d e - d - < ( x - a ) S d

# a - d l x a + d

E 6) (i) Choose a and b such that a t b = 7, a - b = 2. 9 5 Then 2a = 9 or a =- and 2b = 5 or b = T . 2

Hence

E 7) ( i ) [a, b[ is a NBD of each of its points except of the point b.

(ii) Easy to solve.

(iii) Each of these is a NBD of each of its points except the point a.

(iv) Sameas(iii).

(v) It is not a NBD. Explain why?

(vi) It is also not a NBD of any of its points. (Why?)

E 8) Let S and T be any two NBDs of a point c. We have to show that S U T is a NBD of c. Now c ES u T =$ c ET. If c ES, then since S is a NBD of c and S c S u T, therefore S u T is also a NBD of c.

E 9) (i) [a, b] is not an open set. (Why?)

(ii) 10, 1[ is open while [0, 1[ is not open. Give reasons.

(iii) It is not an open set because it is not a NBD of any of its points.

(iv) Neither N nor Z is an open set.

(v) It is not open.

(vi) ]a, a[ is open but [a, co[ is not open.

E 10) Exmaple is given after E 8). Look for it.

E 11) Use the method of E 8) I

E 12) (i) No. Give reasons.

(ii) Every NBD of an arbitrary real number contains an infinite number of real numbers.

7-

(iii) Yes the end points are also the limit points but they do not belong to it. ~ o p o l o g y of ths Real Liiie

(iv) Yes. The end points, in this case, are also the limit points but they belong to the interval.

(v) Yes. Elaborate it.

E 13) Let p be a limit point of a set S. We have to prove that

(i) If every NBD of p contains at least one member of S different from p, then evely NBD of p contains an infinite number of the member of S, and

(i) If every NBD of p contains an infinite number of members of S, then it must have at least one member of S other than p.

The result (ii) is obvious. Therefore it is enough to prove (i).

Since every NBD of p contains atleast one member of S different from p, so, let ]p - 6,, p + 6, [ be a NBD such that it has a member x, of S and x, # p. Suppose Ix, - pl = S,, where 6, < 6,. Consider the NBD ]p - 6,, p + 6, [ of p. Then by definition, ]p - 6,, p + 6, [ must have an element say x,, of S such that x, # p. But since 6, < 6,, therefore ]p - 6,, p + 6, [ co~ltains two elements x,, x, of S which are different from p. Continuing like this, yo11 can show that the NBD ]p -6,, p + 6,[ contains an infinite number of elements of S.

E 14)(i) Any four bounded intervals.

1 (ii) The sets Q, R, and the set {-: n EN}

n (iii) Z

E 15)(i) Not closed.

(ii) Closed.

(iii) Closed.

( i ) Closed.

(v) Closed.

(vii First is closed and the second is not closed.

E 17) Consider an arbitrary family of closed sets such that their intersection is nonempty. Let x be a limit point of this intersection. Then every NBD ]x -6, x + 6[ , 6 > 0 , of x contains an infinite numbers of elements of this intersection and hence of each member of the given family. Therefore x is a limit point of each member of this family of closed sets. Hence x belongs to each member of the family and therefore x belongs to the intersection. Hence the intersection is also a closed set.

E18) ( i ) G , = ] { ....., ] -3,-I[,]-2,0[,]-1, 1[,10,2[ ,..... 1 . Since every x ER belongs to at least one of the open interval in G,, therefore, GI is an open cover of R.

Similarly verify (ii) and (iii).

E 19) Let S be an infinite open subset of R. Let G = {G,} be an open cover of S such that S EG. Then G admits of a finite subcover of S. (HOW?)

UNIT 4 REAL FUNCTIONS

STRUCTURE 4.1 Introduction

Objectives

4.2 Algebraic Functions Algebraic Combinations bf unctions Notion of an Algebraic Function Polynomial Functions Rational Functions Irrational Functions

I 4.3 ~ranscendental Functions Trigonometric Functions Logarithmic Functions Exponential Functions

4.4 Some Special Functions

I I Bounded Functions

I

I 4.5 Summary

I

I

4.1 INTRODUCTION

Real Analysis is often referred to as the Theory of Real ~unctions. The word 'function' was first introduced in 1694 by L.G. Leibniz [1646-17161, a famoll German mathematician, who is also credited along with Isacc Newton for the invention. of Calculus, Leibniz used the term functian to denote a quantity connected with a curve. A Swiss mathematician, L. Euler [1707-17831 treated function as an expression made up of a variablt and some constants. Euler's idea of a function was later generalized by an eminent French mathematician J. Fourier 11768-18301. Another German mathematician, L. Dirichlet (1805-1859) defined function as a.relationship between a variable (called an independent variable) and another variable (called the dependent variable). This is the definition which, you know, is now used in Calculus.

The concept of a function has undergone many refinements. With the advent of Set Theory in 1895, this concept was modified as a correspondence between any two non- empty sets. Given any twa non-empty sets S and T, a function f'from S into T, denoted as f: S 4 T, defines a rule which assigns to each x E S, a unique element

' Leonard Eulcr y-ET. This is expressed by writing is y =.f (x). This definition, as you will recall, was ' given in Section 1.2. A function fi S -4- T is said to be a

(i) Complex-valued function of a complex variable if both S and T are sets of complex numbers;

(ii) Complex-valued functiop of a real variable if S is a set of real numbers and T is a set of compler numbers;

. (iii) Real-valued function of a comp(lex variable if S is a set of complex numbers and T i a set of real numbers;

(iv) .Re f, -valued function of a real variable if both S and T are some sets of real numbers,

.Since we are deabgv&h the course on Real Analysis, we shall confine our discussion t'o those functions whose donlains'as well as codomains are s6me subsets of the set of real numbers. We shall call such, functions as Real Functions. ,

In this unit, we shall deal with the algebraic and transcendental functions. .Amorig the transcandental functions, we shall define the trigonometric functions, the exponential and logarithmic functions. Also, we shall talk about some special real functions including the bounded and monotonic functions. We shall frequently use these

68 functions to illustrate various concepts in Blocks 3 and 4.

Real Functions

After going through this unit, you should be able to

+identify various types of algebraic functions

-+define the trigonometric and the inverse trigonometric functions

-+describe the exponential and logarithmic functions

-+discuss some special functions including ths bounded and monotonic functions.

4.2 ALGEBRAIC FUNCTIONS

In Unit 1, we identified the set of natural numbers and built up various sets of numbers with the help of the algebraic operations of addition, subtraction, multiplication, division etc. In the same way, let us construct new functions from the real functions which we have chosen for our discussion. Before we do so, let us review the algebraic combinations of the functions under the operations of addition, subtraction, multiplication and division on the real-functions.

4.2.1 ALGEBRAIC COMBINATIONS OF FUNCTIONS Let f and g b e any two real functions with tlie same domain S C R and their co- domain as the set R of real numbers. Then we have the following definitions:

DEFINITION 1: SUM AND DIFFERENCE O F TWO FUNCTIONS

(i) The Sum of f and g, denoted as f + g, is a function defined from S into R such that

(f + g) (x) = f(x) -I- g(x), If x E S.

(ii) The Difference o f f i nd g, denoted as f - g, is a func t ih defined from S to R such that

(f - g) (x) = f(x) - g(x), IfxG S. Note that both f(x) and g (x) are elements of R. Hence each of their sum and difference is again a unique member of R.

DEFINITION 2: PRODUCT O F TWO FUNCTIONS

Let f: S -+ R and g: S ' E be any two functions. The product o f f and g, denoted as fag, is defined as a function f. g: S ' R by

(f . g) (x) = f(x) . g(x), it x ES.

DEFINITION 3: SCALAR MULTIPLE OF A FUNCTION

Let f: S -+ R be a function and k be same fixed real number. Then the scalar multiple of 'f' is a function k f S -+ R defined by

(kf (x) = k. f(x), 9 x E S.

This is also called the scalar multiplication.

DEFINITION 4: QUOTIENT OF TWO FUNCTIONS

Let f: S -+ R and g: S ' R be any two functions such that g(x) # 0 for each - x in S. Then s function!: S + R defined by

g f(x) (t) ( x ) = ----& V x E S

is called the quotient of the two functions. -

EX'ERCISE 1) Let f, g, h be any three functions, defined on S and taking values in R, as f (x) = ax2, g(x) = bx for every x in S, where a, b, are fixed real numbers. Find f + g, f - g, f , g, f /g and kf, when k is a constant.

4.2.2 NOTION O F AN ALGEBRAIC FUNCTION You are quite familiar with the equations ax + b = 0 and ax2 + bx -t c = 0, where a, b, c E R, a Z 0. These equations, as you know are, called linear (or first degree) and

l ea l Numbers and Functions quadratic (or second degree) equations, respectively. The expressions ax + b and ax2 + bx + c are, respectively, called the first and second degree polynomials in x. In the same way an expression of the form ax' + bx2 cx d (a # 0, a, b, c, d ER) is called a third degree polynomial (cubic polynomial) in x. In general, an expression of the form ao xn + a, x"l + a2 x " -~ + ..., 4- a,, where a, P 0, ar ER, i = 0, 1,2, ...., n, is called an nth degree polynomial in x.

A function which is expressed in the form of such a polynomial is called a polynomial function. Thus, we have the following definition:

DEFINITION 5:*POLYNOMIAL FUNCTION

Let 81 (i = 0,1, ...., n) be fixed real numbers where n is some fixed non-negative integer. Let S be a subset of R. A function f: S - R defmed by

is called a polynomial function of degree n.

Let us consider some particular cases of a polynomial function on R:

Suppose f: S " R is such that

(i) f(x) = k, Vx ES (k is a fixed real number). This is a polynomial function. This is generally called a constant function on S.

For example, f(x) = 2, f(x) = - 3,f(x)= T, tf xE R, are all constant functions.

(ii) One special case of a constant function is, obtained by taking k = 0 i.e. when f(x) = 0, v x E S.

This is called the zero function on S

EXERCISE 2) Draw the.graph of a constant function. Draw the graph of the zero function.

Let f: S ' R be such that (iii) f(x) = a, x + a,, f f xE S, a, # 0.

This is a polynomial function and is called a linear function on S. For example, qx) = 2x f 3, f(x) = - 2 x+3, f(x) = 2x - 3, f(x) = -.2x - 3, f(x) = 2x' for every x E S are all linear functions

(iv) The function f: S R defined by

f(x) = x , f f x E S 1 s called the identity function on S.,

(v) fi S ' R given as .

f(x),=ao.t2+at x + a2,VxER,ao#O.

is a polynomial function of degree two and.is called a quadratic function on S.

Fpr example, f(x) = 2x 2 + 3x - 4, f(x) = x2 + 3, f(x) = x2 f 2x, f(x) = - 3x2, for every x E S are all quadratic functions.

DEiilNITION 6: RATIONAL FUNCTION A function which can be expressed as the quotient of two polynomial functions is called a rational function.

Thus n function f: S ' R defined by

is called a rational fbnction.

Here ao # O bo # 0, ai, bj E R where i, j are some fixed seal numbers and the polynomial function in the denominator is never zero.

EXAMPLE 1: The following are all rational functions on R.

The functions which are not rational are known as irrational functions. A typical example of an irrational function is the square root function which we define as follows:

DEFINITION 7: SQUARE ROOT FUNCTION

Let S be the set of non-negative real numbers. A function f: S - R defined by

f(x) = 6 , - v x E S is called the square root function.

You may recall that Jx is the non-negative real number whose square is x. Also it is defined for all x L 0.

EXERCISE 3) Draw the graph of the function f(x) = &for x L 0.

- .

Polynomial functions, rational functions and the square root function are some of the examples of what are known as algebraic functions. An algebraic function, in general, is defined as follows

DEFINITION 8: ALGEBRAIC FUNCTION

An algebraic function f: S " R is a function defined by y = f(x) if it satisfies identically an equation of the form

po (~ )y "+p , (X )yn - '+ .... +p.-,(x) y+p,(x)=O where p (x), pl (x), ....p.-I (x), p. (x) are Polynomials in x for all x in S and n is a positive integer.

EXAMPLE 2: Show that f: R ' R defined by

is an algebraic function.

Solution

Let y = f(x) = d x 2 - 3, + 2 . JiFT

Then (4 x-1) y2- (x2-3x + 2) = 0

Hence f(x) is an algebraic function.

In fact, any function constructed tiy a finite number of algebraic operations (addition, subtraction, multiplication, division and root extraction) on the identity function and the constant function, is an algebraic function.

EXAMPLE 3: The functions f : R -+ R defined by

EXAMPLE 4: Rove that every rational function is an algebraic function.

S O L ~ T O N : Let f: R ' R be given as

- Real Funcrinns

Red Numbers and Functions bhere p(x) and q(x) are some polynomial functions such that q (x) # 0 for any x E R. Then we have

which shaws.that y = f(x) can be obtained by solving the equation 4tx) Y - P(X) = 0.

Hence f(x) is an algebraic function.

EXERCISE 4) Verify that a function f: R " R defined by f(x) = m x i s an algebraic function.

A function which is not algebraic is cdled a Transcendental Function. Examples of elementary transcendental functions are the trigonometric functions, the exponential functions and the logarithmic functions, which we discuss in the next section.

4.3 TRANSCENDENTAL FUNCTIONS

In Unit 1, we gave a brief introduction to the algebraic and transcendental numbers. Recall that a number is said to be an algebraic if it is a root of an equation of the form

with integral coeffioients and ~o # 0, where n is a positive integer. A number which is not algebrkc is called a transcendental number. For example the numbers e and n are transcendental numbers. In fact, the set of transcendental numbers is uncountable.

'Based on the same analogy, we have the transcendental functions. In Section 4.2, we have discussed algebraic functions. The functions that are non-algebraic are called transcendental functions. In this section, we discuss some of these functions.

I

4.3.1 TRIGONOMETRIC FUNCTIONS

You are quite familiar with the trigonometric functions from the study of Geometry and Trigonometry. The study of Trigonometry is concerned with the measurement of the angles and the ratio of the measures of the sides of a triangle. In Calculus, the trigonometric functions have an importance much greater than sirpply their use in relating sides and angles of a triangle. Let us review the definitions of the trigonometric functions sin x, cos x and some of their properties. These functions form an important class of real functions.

Consider a circle x2 t y2 = r2 with radius r and centre at 0. Let P be a poii~t on the circumference of this circle. If 8 is the radian measure of a central angle at the centre of the circle as shown in the Figure 1.

then you know that the lengths of the arc AP is given by s = 8 r .

You already know how the trigonometric ratios sin 0, cos 0, etc., are defined for an angle .Q measured in degrees or radians. We now define sinx, cos x, etc., for x ER.

If we put r = 1 in above relation, then we get 0 = s. Also the equation of circle beconies x1 + y2 = I . This, as you know,, is the Unit Circle. Let C represents this circle with centre 0 and radius 1. Suppose the circle meets the x-axis at a point A as shown in the Figure 2.

Fig. 2

' Through the point A = (1,O); we draw a verticle line labeled as t-axis with origin at A and positive direction upwards. Now, let t be any real number and we will think of this as a point on this verticle number line i.e., t-axis.

Imagine this t-axis as a line af thread that can be wrapped around the circle C, Let p(t) 7 (x, y) be the point where 't' ends up when this wrapping takes place. In other words, the line segment from A to point (t, 0) becomes the arc from A to P, pdsitive or negative i.e., counterclockwise or clockwise, depending on whether t > 0 or t c: 0. Of course, when t = 0, P = A. Then, the trigonometric functions 'sine' and 'cosine', for arbitrary t ER, are defined by

s i n t= s i n e = y, and c o s t = c o s e = x,

where '8' is the radian measure of the angle subtended by the arc AP at the centre of the circle C. More generally, if t is any real number, we may take (0 I 8 5 27c) to be the angle (rotation) whose radian measure is t. It is then clear that

sin (t + 27c) = sin t and cos (t + 2 x ) = cost.

You can easily see that as 6 increases from '0' to d2 , PQ increases from 0 to 1 and OQ decreases from 1 to 0: Further, as 6

Rea l Bunct ions

R P ~ B Yclmbtrs and Functions a increases from - to n; P Q decreases from 1 to 0 and OQ decreases from 0 to - 1,

2 3x

Again as B increases from n to , PQ decreases from 0 to - 1 and O Q 2

3a increases from - 1 to 0. As B increases kom - to 217. OQ increases 2

from 0 to 1 and PQ increases from - 1 to 0. The graphs of these functions take the shapes as shown in Figure 3.

L !- = s i n s

b

y 7 cos x

Frg. J I

Thus, we define sin x and cos x as follows :

DEFJNITION 9: SINE FUNCTlON

A function f: R -+ R dkfined by I

f(x) = sin x, V x ER I

I is called the sine of x. We often write y = sin x. I

I DEFINITION 10: COSINE FUNCIION

A function fi R -+ R defined by I

f(x) = cos x, V x ER I

I

is called the cosine of x and we write y = cos x.

74 i I i I

Note that the range of each of the sine and cosine, is [- 1, 11. In terms of the real fLinctions sine and cosine, the other four trigonometric functions can be defined as follows:

Kca! F~~nctions

(i) A function f: S --+ M defined by

sin x f(x) = tan x = - , cos x # 0, +Y' x E S = R - {(2n + 1) T] is called the COS 2

Tangent Function. The range of the tangent function is ] - m, + m [ = R and the

domain is S = R - {(2n + 1) E] , where n is a non-negative integer. 2

:(ii) A function f: S -+ R defined by

COS X f ( x ) = c o t x = -, sir1 X # 0 , U X E S = R - [n T } ,

sin x

is said to be the Cotangent Function. lts'range is also same as its co-domain i.e. range = ] - m, m [ = R and the domain is S = R - { n r ) , where n is a non- negative integer.

(iii) A function f: S -* R defined by

f ( x ) = a e c x = -I ~ o s ~ # o , v ~ E s = s - ( ~ ~ + l jz]. cos x 2

is called the Secant Function. Its range is the set

S=]-=,-I]. U[I,m[anddomain i s S = R - { 2 n + I)T}, 2

(iv) A function f: S * R defined by 1 -

f(x) = cosec x = ,--, sin x# 0, x E S = R - {n r } , sin x

is called the Cosecant function. Its range is also the set S = ] - m, - I ] U [I, oo[ and d o m a i n i $ S = K - { n T } .

The graphs of these functions are shown in the Figure 4 on pages 76-77.

EXAMPLE 5: Let 5 = - h TI]. Show fhat the function f: S- R defined by 2 ' 2

f(x) = sin x, V x E S

is one-one. When is f only onto? Under what conditions f is both one-one and onto?

SOLUTION: .Recall from Unit 1 that a function i is one-one if f(x1) = f(x2) =+ XI = X2

for every XI, x2 in the domain of f .

Therefore, here we have for any XI , XZES, -

XI) = f(x2) =+ sin X I = sin x2 ,, =+ sin X I - sin x2 = 0

Recnll the trigonometric identities \khich you have lenrnt in your prevlous study of Trigonometry.

x1+x2 = o X'-X2 COS - + 2sin - 2 2

XI + x2 = 0. + Either sin = 0, or cos - 2 2

X 1-X: XI-X? If sin - = 0, then - = O , f 7r, + 2 ~ , .... 2 2

X I + X: If cos - x,+x. Tr 3Tr = O , then- = * -, f - , .,,. 2 2 2 2

~erel'ore we can only have 2 2-

Real Numbers and Functions R X - X x -- < L 2 < - 2 - 2. 2

and

i s , then x, + x, = + x .

X Since x,, x, E[ - - ,

2

X 'IC therefore, x, = x2 = - or x, = x, = - -

2 2

Hence (x,) = f(x,) 3 x, = x,, which proves that f is one-one. Then function f(x) = sin x dsfined as such, is not onto because you know that the range of sin x is [- 1, 11 # R.

y = tan x

I Fig. 4t i )

Fig .4- (ii )

Fig. 4 (iii)

If you dzfipe f: R----3. [-I, I] as

f(x) = sin x, ft x ER.

Then f is certainly onto. But then it is not one-one. However the function.

f: [-T. I d [ - I , l ldefinedby

f(x) = sin. x, V x E

is both one-one and onto.

Rcal Functions '

-- Real Numbers and Functions EXERCISE 3

Two functions g and h are defined as follows : cia g: S -+ R def i~~ed by

g(x) = 60.5 X, X E S = 10, 7C]

(ii) h: S -+ R defined by

Show that the finctiol~s are one-one. Under what conditions the function are one-one and onto?

4.3.2 INVERSE TRIGQNOMETWIC FUNCTIONS

In Section 1.2 we discussed inverse functions. You Icnow that if a function is one-one and onto, then it will have an inverse. If a functioli is not one-one and onto, then sometimes it is possible lo restrict its domain in some suitable manner such that the restricted function is one-one and onto. Let us use these ideas to define the inverse trigonometric functions. We begin with the inverse of the sine function.

Refer to the graph of f(x) = sin x in Fig. 3. The x-axis cuts the curve y = sin x at the points x = 0, x = x , x = 2.x, ....... This shows that function f(x) = sin x is not one-one. However, we have already shown in Example 5 that if we restrict the domain of

f(x) = sinx to the interval [-n/2, 71/21, then the functio~l

7t x f [-- -1 -+ [- 1, 11 defined by

2 ' 2

71 71 f(x) = sinx, - -

2 1 x 5 - 2

is one-to-one as well as onto. Hence it will have the inverse. The inverse function is Be careful about the notation used.The superscript - 1 that called the inverse sine of x and is denoted as sin x. In other words, appears in y = sin-' is not y = sin-' x o x = sin y, an exponent, but is Lhe

71 7t symbol f' used to denote the where - - 5 y 2 - and -1 2 x 2 1. inverse of a function f. Two 2 2

I avoid this, notation y = I ' arcsin, illstead of ,, = Thus, we have the following definition: ' 1 x is used sometimes.

DEFINlTION 11 : INVERSE SllVE FUNCTION

'rt TK A function g : [-I, 1) -> [-- , -1 defined by

2 2

g(x) = sin-'x, V x E [-I, I ]

is called the inverse sine function. -

Again refer back to the graph of f(x) = cos x in Figure 3. You can easily see that cosine function is also not one-one. However, if you ~~estrict the domain of f(x) = cos x to the interval [0, x ] , then the function f [O, 711 -> [ -1, 11 defined by

f(x) = cosx, 0 I x 1 ?I,

is one-one and onto. Hence it will have the inverse. The inverse function is called the inverse cosine of x and is denoted by cos-I x (or by arc cosx). In other words,

y = cos-'x G X = cosy, !

where 0 I y 5 7~ and -1 5 x I I .

Thus, we have the following definition:

DEFINITION 12: A function g: 1-1,1] --A 10, n] defined by g(x) = cos-' x, v x E[-1, 11,

is called the inverse cosine function.

You can easily see from Figure 4 that the tangent function, in general, is not one-one. However, again if we restrict the domain of f(x) = tan x to the interval 1-~12, ~12 [ , then the hnction

n n f: ] - - , -[ -+ R defined hy

2 2

n: .n f(x) = tan x, - - < x < -

2 2

Real Punctinna

is one-one and onto. Hence it has an inverse. The inverse function is called the inverse tangent of x and is denoted by tan-, x (or by arctan x). In other words,

y = tan-' x a x = tan y,

Thus, we have the following definition:

DEFINITION 13: Inverse Tangent Function

71 n: A function g: R -+ ] - - , - ( defined by

2 2

g(x) = tan-' x, v x E R I is called the inverse tangent function.

Now you can t1-y the following exercise to define the remaining three inverse trigonometric functions:

EXERCISE 6) Define the inverse cotangent, inverse secant and inverse cosecant function. Specify their domain and range.

Now, before we proceed to define the logarithmic and exponential functions, we need the concept of the monotonic functions. We discuss these fullctions as follows:

4.3.3 MONOTONIC FUNCTIONS Consider the following functions:

(i) f ( x ) = x , b ' x ~ R .

(ii) f(x) = sin x, V x E [-7212, ~ 1 2 1 .

(iii) f(x) = - x2, 'd x E [0, ca[.

(iv) f(x) = cos x, 'v' x E [O, n:].

Out of these functions, (i) and (ii) are such that for any x,, x, in their domains,

XI < x, f(x,) 2 f(x,),

whereas (iii) and (iv) are such that for any x,, x, in their domains,

x, < x, =$ f(x,) 2 f(x,).

The functiolls given in (i) and (ii) are called illonotonically increasing while those of (iii) and (iv) are called monotonically decreasing. We define these functions as follows:

Let f: S + R (S c R) be a function

(i) I t is said to be a monotonically increasing function on S if

x, < x, a f(x,) I f(x,) for any x,, x, E S

(ii) I t is said to be a monotonically decreasing function on S if

X, < x2 3 f(x,) 2 f(x2) for any x,, x, ES.

(iii) The function f is said to be a monotonic function on S if it is either monotonically increasing o r monotonically decreasing.

(iv) The function f is said tb be strictly increasing on S if

X, < x2 3 f(x,) < f(x2), for x,, x2 ES.

(v) It is said to be strictly decreasing on S if

X, < X, 3 f(xI) > f(x,), for x,, xz ES.

Red Numbem and Functions You can notice immediately that i f f is monotonically increasing then - f i.e. - f: R ' R defined by (- f) (x) = -f(x), tf x ER is monotonically decreasing and vice-versa.

EXAMPLE 6: Test the monotonic character of the function f: R ' R defined as

SOLUTION: For any XI, XZER, XI h 0, x2 5 0 XI < x2 => x21 > x22 => f(x1) > f(x2)

which shows that f is strictly decreasing.

Again if X I > 0, x2 > 0, then xl<x2=>~:<~:=+-~:>-x:=>f(xl)>f(x2) which shows that Z is strictly decreasing for x > 0. .Thus f is strictly decreasing for every x E R.

Now, we discuss an interesting property of a strictly increasing function in the form of the following theorem:

THEOREM 1: Prove that a strictly increasing function is always one-one.

PROOF: Let f: S " T be a strictly increasing function. Since f is strictly increasing, therefore, X I < xz XI) < f(x2) for any XI, xz f S.

Now to show that f: S " T is one-one, it is enough to show that f(x1) = f(x2) => X I = X2.

- Equivalently, it is enough to show that distinct elements in S have distinct images in T i.e. XI # x2 XI) # f (xz), for X I , xz E S.

Indeed, x l # x2 =>XI < xz or X I > xz

=> f(x1) < f(x2) or f(xl)> f(x2) => XI) # ~ ( x z )

which proves the theorem.

EXAMPLE 7: Let f: S + T be a strictly increasing function such that f(S) = T. Then prove that f is invertible and fl: T + S is also strictly increasing.

SOLUTION: Since fi S ' T is strictly increasing, therefore, f is one-one. 'Further, , I since f (S) = T, therefore f is onto. Thus f is one-one and onto. Hence f is invertible. In other words, f - I : T 'S exists.

Now, for any y ~ , y 2 E T , we have yl= f(xl), y2 = f(xz). If y~ < yz, then we claim X I < XZ.

Indeed if XI 2 x2, then f (xl) 2 f(x2) (why?). This implies that y, 2 y2 which contradicts that yl < y2. Hence yl < yz =>XI < xz =>f (yl)< f- ' (y2) which shows that f -' is strictly increasing.

You can similarly solve the following exercise for a strictly decreasing function:

EXERCISE 7) Let f: S ' T be a strictly decreasing function such that f(S) = T. Show that f is invertible and f -': T ' S is also strictly decreasing.

4.3.4 LOGARITHMIC FUNCTION You know that a definite integral of a function represents the area enclosed between the curve of the function, X-Axis and the two Ordinates:.You will now see that this can be used to define logarithmic function and then the exponential function.

1 We consider the function f(x) = - for x > 0. We find that the graph of the X

function is as shown in the Figure 5,

Pig. 5 - - DEFINITION 14: LOG,4IPITiWMIC FUNCTION

For x 2 1, we define thc! arraturitai logarithulic function log x as

a P log x = J - dt I t

1 In the Figure 5, log x represents the area between the curve f(t) = - 9 X - Axis and

t

the two ordinates.at 1 ; ~ s d at ;r. F,..; 0 < x < 1, we define

I 1ogx=-J- dt x t

This means that for O < x < 1, log x is the negative of the area under the graph of

1 f(t) = - X - Axis and the.twc3 ordinates at x and at.1

?P X.

Fig 6 '

We also see by this definition that l o g x < o i f o < x < l log 1 = 0

and l ogx>O i f x> 1. It 'also follows by definition that if .

xt > x2 > 0, then log xl > log x2. This shows that log x is qtrictly increasin

shown in the Figure 7. f' The reason for this is quite clear if we realise by log XI as the area under the gf ph as

The logarithmic function defined here is called the Natural logarithmic function. For any x > 0, and for any positive real number a + 1, we can define

! log x log.,* = - - log a This function is called the logarith,mic function with respect to the base a,.

If a = 10, then this function is ca l l~d the common logarithmic funktion. 811

Fig.

Logarithmic function to the base a has the following properties

( i ) log, ( ~ 1 ~ x 2 ) = log, XI + log, xz

(iii) log, xm = m loga x for every integer rn.

(iv) lo&. = I.

(v) log,,' = 0 1

By the definition of log x, we see that log 1 = 0 and as xbecomes larger and larger, 1

the area covered by the curve f(t) = X - axis and :he ordinates at 1 and x, t

becomes larger and larger. Its graph is as shown in the Figure 8,

You already know what is meant, by inverse of ;l funcion. You had also seen in Unit 1 that if f is 1 - 1 and onto, then f is invertible. Let us ?ply that study to logarithmic function. I

4.3.5 EXPONENTIAL FUNCTION We now come to define exponential function. We hak seen that

log x: 10, w[ " R is strictly increasing function. The graph of the logaribmic function also showsjthat

log x: 10; =[ R .

is also onto. Therefore this function' admits of invers function. Its inverse function, called the Exponential function, Exp (x) has domair qs the set R of all red numbers and range as ]0,%. If

log x = y, then Exp (y) = x. 1 1

The graph of this function is the mirror image of logarithmic function as shown in the Figure 9.

I (ii) a > 1

Fig. 9

- The Exp (x) satisfies the following properties

(i) Erp (x + y) = Exp x Exp y

(ii) Exp (x - y) = Exp x/Exp y

(iii) (Exp x)"= Exp (nx)

(iv) Exp (0) = 1 I.*

We now come to define ax for a > 0 and x any real number. We write

ax =-E.xp (x log a)

If x is any raGonai number, thenwe know that log a' = x log. a. ~ e n c e Exp (x log a) = Exp (log ax) = ax. Thus our definition agrees with the already known definition of a in cace x is a rational number. The function ax satisfies the following 'properties

I (iii) (ax)Y = axY

, (iv) ax bx = (ab)x, a > 0, b > 0.

Denote E (I) = e, so that log e = 1. The number e is an irrational number and its approximation say upto five places of decimals is 2.71828. Thus -

ex = Exp (x log e) = Exp (x). I Thus Exp (x) is also den,oted as ex and ye~write for each a > 0, a" ex log a

EXAMPLE 8: Plot the graph of the function I: R -+ R defined by f(x) A 2'.

I

' The required graph takes the shape as shown in the Figure 10.

Real Functions

Fig. 10

r-

EXERCISE 8) 1 Show the graph off: R ' R defined by F(x) = ( - )" 2

4.4 SOME SPECIAL FUNCTIONS

So far, we have discussed two main classes of real functions-Algebraic and Transcendental. Some functions have been designated as special functions because of their special nature and behaviour. Some of these special functions are of great interest to us. We shall frequently use these functions in our discussion in the subsequent units and blocks.

1. Identity Function

We have already discussed some of the special functions in Section 4.2. For example, the Identity function i : R + R, defined as i (x) = x, V x ER has already been discussed as an algebraic function. However, this function is sometimes, referred to as a special function because of its special characteristics, which are as follows:

(i) domain of i = Rangc of i = Codomain of i

(ii) The function i is one-one and onto. Hence it has an inverse i-' which is also one- one and onto.

I

(iii) ;The function i is invertible and its inverse i-' = !

(iv) The graph of the identity function is a straight line through the origin which forms an angle of 45' along the positive direction of X-Axis as shown in the Figure 11.

84

i l

Real F ~ ~ n c l i o ~ ~ s

Fig. 11

2. .periodic Function

You know that sin (27r + x) = sin (4n + x) = sin x, tan (IT + x ) = tan (27r-t x) = tanx.

This leads us to define a special class of functions, known as Periodic functions. All trigonometric functions belong to this class.

A function f: S - R is said to be periodic if there exlsts a positive real number k such that

f(x + k) = f(x), tf x ES where S C R.

The smallest such positive number k is called the period of the function.

You can verify that the functions sine, cosine, secant and cosecant are periodic each with a period 2n while tangent and cotangent are periodic functions each with a period IT.

EXERCISE 9) F'ind the period of the function f where f(x) = I sin3 xl

-<.

3. Modulus Function

The modulus or the absolute (numerical) value of a real number has already been defined in Unit 1. Here we define the modulus (absolute value) function as follows:

Let S be a subset of R. A function f: S R defined by f(x)= 1x1, v xES

is called the modulus function.

In short, it is written as ~ d d function.

You can easily see the following properties of this function:

(i) The domain of the Modulus function may be a subset of R or the set R itself. ,

(ii) The range of this function is a subset of the set of non-negative real numbers. I

(iii) The ~odu l ' us function I: R R is not an onto #unction. (Check why?).

(iv) The Modulus function f: R " R is not one-one. For instance, both 2 and - 2 in the domain have the same image 2 in the range.

(v) The modulus function fi R - R does not have an inverse function (why)?

(vi) The graph of the Modulus function is R -+ R given in the Figure 12.

Reel Nurnkrs and Functions

It consists of two straight lines:

(i) y = x (y 2 0)

and (ii) y = -x (y 2 0) through 0, the origin, making an angle of rr/4 and 3n/4*with the pojitive direction of X-axis:

4. Signum Function A fuqction'k R ' R defined by

x when x # 0 x when x = 0:

or equivalently by: - l i f x<O

1 if x > 0.

is called the signum function. It is generally mitten as sgn (x).

Its range set is {-I, 0, I}. Obviously sgn x is neither one-one nor onto. The graph of sgn x is shown in the Figure 13,

5. Greatest Integer Function

Consider the number 4.01. Can you find the greatest integer which is less than or equal to this number? Obviously, the required integer is 4 and we write it as [4.01] = 4.

Similar.ly, ifthe syrnbol [x] denotes the greatest integer contained in x then we have [3/4] = 0, [5.01] -5,

[-.OO:i] = -1 and [-3.961 = -4.

Based on these, thc greatest integer function is defined as follows: A function f: R -+ R defined by

f(x) = 1x1, V x ER,

where [x ] is the largest integer less than or equal to x is called the greatest integer function: The following properties of this function are quite obvious:

(i) The domain is R and the range is the set Z of all integers.

(ii) The function is neither one-one nor onto.

(iii) If n is any integer and x is any real number such that x is greater than or equal to n but less than n + 1 i.e., if n S x < n + 1 (for some integer n), then [x] = n i.e.,

The graph of tile greatest integer function is shown in the Figure 14.

Real Functioris

I I I I I I ( I I I I I

Fig. 14

EXAMPLE 9: Prove that [ x + m] = 1x1 + m, V x e R a n d ~ E Z .

SOLUTION: You know that for every x ER, there exists an integer n such that n S x < n + l .

Therefore, n + m < x + r n < n + I + m ,

and hence [x+m] = n + m = [ x ]+m,

which proves the result. 87

I

--C o , I I I I I

C -9 -8 -7 -6 -5 -4 -3 -2 - I I I I I I I I I 1 2 3 4 5 6 7 8 9 x-axis

EXERCISE 10) Test whetlher or not allre function f: R + R defined by f(x) = x-[x] M x E R, is psddic. PB i h m9 h d its psimi.

6. Even md Odd Funstions

Consider a function C R + R defined as f(x) = 2x, V x E R.

If you change x to -x, then you have f(-X) := 2 (-x) = -2x -f (x).

Suck a function is called an add function. Now, consider a function f:R - R defined as

f(x) = x2 V x E R Then changing x to -x we get

q- x) = (-x)~ = x2 = f(x) Such a function is called an even function.

The definitions of even and odd functions are as follows: A function fi R + R is called even if f.( - x) = f(x), tf x ER. It is callcd odd Pf f(- x) = - f(x), tf x E R

EXAMPLE 7: Verify whether the function P: R " R defined by

(i) f(x) = s i d x + em3 2x

qx) = \I- - JZ are even or odd.

SOLUTION: (1 f(x) = sin2 x + cos3 2x. V. x € R - I => (-X) = sin (-x) + cos3 2 (-x) = sinZ x + cos3 2x = f(x), tf x E R => f is an even function.

(ii) fix) =

=> P(-x) =

f is an odd function.

EXERCISE 11) Dct wbkboftbefdlowiug

(iii) a l n x , c ~ x , ~ x , x- 4 (iv) f(x) = - ~ f t xER,xC( -3 ,3 ) x2- 9

In Unit 2, you were introduced to the notion of a bounded set, upper and lower bounds of a set. Let us now extend these notions to a function.

You know that if I: S -+ R is a function, (S C R), then {Kx) : x €s],

-

is called the range set or simply the range of the function f. -

A function 18 mid io be bounded if iti I*age-h bounded.

Let f: S R be a function. It is said to be bounded above #$here kxists a real number K such that

f ( x ) l K. Vx ES

The number K is called an upper bound of It. The function f is said to be bounded below if there exists a number.k such that

I 1

a R X ) ~ ~ ~ X E S ' I

I The number k je called a lower bound of f. . -

- - .- - I

1 4 s

A function I: 'S + R, which is bounded above as well as bounded below, is said to be bounded. This implies that there exist two real numbers k and K such that k S f ( x ) S K V x E S .

This is equivalent to say that a function f: S + R is bounded if there exists a real number M such that

If(x)l S M , V x E S .

A function may be bounded above only or may be bounded below only or neither bounded above nor bounded below.

RecalI that sin x and cos x are both bounded functions. Can you say why? It is because of the reason that the range of each of these functions is [-I, 11.

EXAMPLE 8: A function fi R - R defined by * 2 - -

(i) f(x) = - x , tr x E R is bounded above with 0 as an upper bound

(Ii) f(x) = x ,q x 1 0 is bounded below with 0 as a lower bound

( i ) f ) = for 1x1 5 1 is bounded because If(r)( 5 l for 1 rl 5 1 .

Try the following 'exercise.

EXERCISE 12) Test which of the following functions with domain and co-domain as R are bounded and unbounded:

(i) f(x) =: tan x

(W f(x) = 1x1 (iii) f(x).= ex

(iv) f(x) = log x

EXERCISE 18) Suppose t: S 4 -R and g: S -+- R are any bounded functions on S. h a v e that f + g aqd'f. g are also bounded functions on S.

4.5 SUMMARY

In this unit, we have discussed various types of real functions. We shall frequently use these functions in the concepts and examples to be discussed in the subsequent units throughout the course).

In Section 4.2, we haveroduced the notion of an algebraic - function --- and its varioua types. A function f: S ---+ R ( S C R) defrned as y = f(x), Y x E S is said to be algebraic if it satisfies identically an equation of the form PO (x) y" + PI (x) Y"" + p2 (x) yn-2 + .... + P(X) Y + pn (x) = 0, where p, (x), pl (x), ...., p, (x) are polynomials in x for all x € S and n is a positive integer. In fact, any function constructed by a finite number of algebraic operations-addition, subtraction, multiplication, division and root extraction-is an algebraic function. Some of the examples of algebraic functions are the polynomial functions, rational functions and irrational functions.

But not all functions are algebraic. The functions which are not algebraic, are called transcendental functions. These have been discussed in Section 4.3. Some important examples of the transcendental functions are trigonometric functions, logarithmic functions and exponential functions which have been defined in this section. We have defined the monotonic functions also in this section.

In Section 4.4, we have discussed some special functions. These are the identity I

function, the periodic functions, the modulus function, the signum function, the matest integer function, even and odd functions. Lastly, we have introduced the bounded functions and discussed a few examples. . 8 9

Relei Numbers and Functions 4.6 ANSWERS/IEPINTS/SOLUTIONS

El) (f + g) (x) ax2 + bx (f - g) (x) = ax2 - bx (fg) (x) = ax2.bx = abx3

ax2 ax (flg) (x) =- = - provided b + 0, x s 0.

bx b

constant function Zero function (x-axis) I

I E3) y = f(x) = d x y2 = x. Now draw the graph.

E4) y = Rx) = y =

i = y 2 = x + ,LC / / = > y 2 - x = , E i ' 1 <

=? (y2 - x ) ~ = x

which shows that y = f(x) is an algebraic function.

E5) (i) Let x,, x, €10, x[. Then f(xl) = f (x2)= COS XI = COS X2

=? cos x, - cos x, = 0

X I + x2 (x2 - x,) 2 sin (-) sin - = 0 2 2

Cx, - x,) either sin (-1 = 0 or sin - = 0 2 2

X - X NOW, sin 51.5 = 0 a 3 = 2 2

0, + n, 2 2n, ...., and

x + % sin = O 3- = o , + X , + 23, ......

2 2

X - X Now, ~ = 0 ~ x I = x 2

2

X - X --1---! = t n n : ( n 2 I ) a x , = t ; 2 n n : + x 1 ,

2

which, is not possible.

X - X Hence 2 = +- n n: is not possible.

2 - Thus, the only possibility is

X2 -- X,

2 = 0 which means x, = x,.

Since, x,, x, E]O, A[, we cannot have the case

X t x . 1 2 , 2 0, 2 n:, i2n,. .*

Hence - qx,) = f(X2) 3 XI = Xr

90 In other words, f ( x ) = cos x is .one-one in.[O, n].

a

Now, the range of cos x is [-I, 11 = R. Therefore cos x, defined from [0, 7c] to Real Functions

R is not onto. But, if cos x is defiend fiom [O, ] to [-I, I], then it is certainly one-one and onto.

(i) Do it yourself.

Ed) (i) Contangent Inverse. y = COP x o x = cot y where 0 < y < n and - ca < x < + m,

(ii) Secant !nverse y = sec-' x o x = sec y,

n where 0 I y < n, y # - andlxl L 1.

2 (iii) Cosecant inverse

y = cosec-I x e x = cosec y,

n n where - - 5 y s - , y z O a n d I x l 2 1 .

2 2

E7) (i) Let f:S + T be a strictly decreasing function. Let x,, x, be any two distinct elements of S. Then

X, + x2 3 X , < X, or X, > X,

= f(x,) > f(x,) or f(x,) < f(x,) 3 f(x,) # f(x2).

This shows that f is one-one. since f(S) = T, f is onto. Thus f is one-one and onto and hence f is invertible. In other words f-I exists i.e. f-I : T + S is defined.

For y,, y, ET, we have y, = f(x,), y, = f(x2), for some x,, x, in S. If y, < y,, we claim that x, x,. If not, then x, 5 x, which implies that f(x,) > f(x2) i.e., y, > y,. This is'a contradiction. Hence y, < y, a d l > x, 3 El (y,) > f - I (y,).

Thus, f' is also strictly increasing.

I E8) Follow the method of Example 8. ! ' I

( E9) Since f(n + x) = ]sin3 (TK + x) 1 = 1 - sin' x 1 = I sin3 x I, I therefore n is the period of f. You may note that n is the least such number I , . satisfying the above relation.

E10) The function f(x) = x - [x] is periodic with period 1 because 1 is the least , number sudh that

f(x+ l ) = ( x + 1)- [x+ I ] = ( x + 1) - [XI - 1 = x - [XI = fIx).

1 Ell) (i) odd (ii) even.

(iii) sin x is odd, cos x is even, tan x is odd.

-x-4 f(-x) = (-x1-4 , which shows that f is neither even

(-X)~ - 9 . x2 - 9

nor odd.

E12) (i) It is unbounded because its range is ] - a3, + co[.

(ii) 1x1 is bounded below with 0 as,a lower bound.

(ki) ex is bounded below because its range is 10, cof.

(iv) log x is unbounded.

E13) Since f and g are given to be bounded functions, therefore there exist numbers 1 k,, K, and \, 5 such that

k, I f(x) 5 K ,, b' x ES; I i k, S g(x) s K,, b' x ES. I

(i) Since (f+g) (x) = f(x) + g(x), V x ES,

therefore, k, + k, 5 f(x) + g(x) r K, + &, V x ES. I 91

Rerl Numkn and Functions '

I !

+ ' k L ( f + g ) ( x ) 5 K W x E S where k = k l+ kz, K = KI + Kz are some real nimbers Thus f(x) + g(x) is a bounded function.

(ii) We know (f.g) (x) = f(x). g(x) _x_ E S. Since f and g are bounded, therefore, we can find ml, mz such that (f(x), 1 5 r n ~ I, Ig(x)l 5 mz, x E S. Thep ' I(f.s) (XI I = l f(x).

= I f(x) l Ig(x1I P m l . m z t t x E S

which shows that f.g is bounded.

REVIEW Real Functions-

Attempt the following self-assessment questions and verify your answers given at the end:

I. Test whether the .following are rational numbers: (i) d l 7 (ii) d g (iii) 3 4- d2

2. The inequality x2 - 5x 4- 6 < 0 holds for (i) x < 2, x < 3 (ii) x > 2, x < 3 (ii) x < 2, x > 3 (iii) x > 2, x > 3

3. If a, b, c, d are real numbers duch that a 2 + b 2 = 1 , c 2 + d 2 = 1, then show that ac+bd 5 1.

C,

4. P r o v e t h a t l a + b + c l 5 l a l + ( b l + I c l for all a, b, c, E R.

5. Show that (a, + az+ .... + anl 5 (all + la21 + .... + la,l

.for al, az, .... a, ER. - . - - 6. Which of the'following sets are bounded above? Write the supremum of the set

if it exists. 01)

(i) '(n, e) (ii) U [2n, 2n + I ] n= 1

1 (v) ( x E R : x < O } (vi)(- :nENandnispr ime) n

!

(vii) ( x 2 : x E R } (viii)

7. Find which of the sets in question 6 are bounded below. Write the infemum if it exists.

8. Which of the sets in question 6 are bounded and unbounded.

9. Test whether the following statements are true or i l s e : (i) The set Z of integers is not a NBD of any of its points. (ii) The interval 10, 11 is a NBD of each of its points (iii) The set ] 1, 3[ U ] 4, 5 [ is open. (iv) The set [a,=[ U 1-00, a ] is not open. (v) N is a closed set. (vi) The derived set of Z is nonempty. (vii) Every real number is a limit point of the set Q of rational numbers. (viii) A finite bounded set has a limit point. (ix) [4,5] U [7, 81 is a closed set. (x) Every infinite set is closed.

10. Justify the following statements: (i) The identity function is an odd function. (ii) The absolute value function is an even function. (iii) The greatest integer function is not onto.

'

(iv) The tangent function is periodic with period r. (v) The function f(x) = I xl for - 2 5 x 5 3 is bounded. (vi) The function f(x) = ex is not bounded n n (vii) The function f(x) = sin x; for x E [ - - -1 is monotonically increasing.

2 2 -(viii) The function f(x) = cos x for 0 5 x 5 n is monotonically decreasing.

7T (ix) The function f(x) = tan x is strictly inleasing for x E [0, : 1. L

(XI f(x) = is an algebraic function 3x-2

- - ANSWERS

1. None is a rational number

2. For (ii) only since2 < x < 3. .

4. , Use the triangle inequality.

5 . Use the principle of Induction.

6 . (i) ; (iv) 2

1 (v) 0 (vi) - (viii) 1 2

(ix) and (x) are unbounded.

7. (i) e (ii) 2 ( i ) 0 (vi) 0 (viii) - 1

8. All the sets are unbounded except the (i)

9. (i) True (ii) False (iii) 'True (iv) False (v) True (vi) False (vii) True (viii) False (ix) True (x) False.

NOTES

UNIT 5 SEQUENCES

Structure 5.1 Introduction

Objectives

5.2 Real Sequences Bounded Sequences Monotonic Sequences

5.3 Convergent Sequences 5.4 Criteria for the Convergence of Sequences

Cauchy Sequences

5.5 Algebra of Convergent Sequences 5.6 Summary 5.7 Answers/Hints/Solutions

5.1 INTRODUCTION

In Unjt 2, you were introduced to the structure of the real numbers. In Unit 3, some interesting properties of the system of real numbers were discussed. In addition to these properties, there are several other fascinating features of the real numbers. In this unit, we discuss one such feature. This is related to the problem of obtaining the sum of an infinite 'number of real humbew.

You kAow thnt it is easy to find the sum of a finite number of real numbers. The addition of an infinite number of real numbers, however, poses some problem. Apparently, you may conclude that it is not possible to add an infinite number of real numbers. But an infinite sum i.e. the sum of an infinite number of real numbers is not artificial. Under certain limiting processes, it is possible to give a meaning to an infinite sum of the form

Recall that this is the infinite sum of a Geometrical Progressi~n with first term 1 and 1

common ratio - 2 '

To obtain the infinite sums, we need the notion of a sequence of real numbers, and its convergence to a limit. What is, then, a sequence? What is the meaning of the convergence of a sequence? What is the criteria to determine the convergence of a sequence? We shall try to find answers for these questions. Also we shall discuss a few related concepts such as boundedness and monotoncity of sequences. We shall frequentby use these concepts in the Units 6 and 7 as well as later on in Blocks 3 and 5.

Objectives

After studying this wit, you shauld, therefore, be able to . * define a sequence and its subsequence * discuss a bounded and a monotonic sequence * find the limit of a sequenca, if it exists * verify whether a given sequence is convergent or not

,* use the criteria for the convergence of a sequence and define the Cauchy sequence. I

5*2 REAL SEQUENCES

Very often you use the word"ssquence1 in your daily life in hveral wnys. You talk of 'a . sequence of events' or 'arranging the library books in a sequence' and so on. Intuitively the

>

Sequences and Series idea of a sequence is that of a progression or succession of numbers, e.g. the first, the

second, the third and so on. For example, if you want to evaluate fi up to myiy decimal .... places, you can mange its approximate values as 1.4, 1.4 1, 1.414, 1.4142, and so on.

Thus, intuitively, a sequence of real numbers would mean a successiorn of real numbers x,, x, .........; where x, is the first element, x2 being the second element, ..... and so on. Hence, you may say that a sequence is an ordered collection of numbers. But in Mathematics, we define a sequence as a special type of function in the following way:

Recall from unit 1 that f is a function from a nonempty set A to a nonempty set B, if to each element x E A, there is assigned a unique element f(x) E B.

If a function s has its domain as the set N of natural numbers and range in R, then s is called a sequence. Let us study the following two examples:

EXAMPLE 1: Let a function s: N + R be defined as

Then s(1) = 1, s(2)=3, s(3) = 5 ,..........., and so on.

This ' function s: N+ R is called a sequence.

Let us write s(n) = s, V n E N. Then obviously,

s1 = I , s2 = 3. s3 = 5 ..... and so on.

The set {s, . s2. .......... s, ....... ) forms the range of the sequence s: N+R. The values s,, s,, ........ are called the terms of the sequence. The sequence is generally written as

(sn)nEN Or (s,),O~=, Or (s").

EXAMPLE 2: Let a function s: N+ R be defined as

1 1 1 1 Then s, = 5 ,Is2 = 5' s3 = 8' sa = ,........ and so on.

The sequence s, is given by

EXAMPLE 3: Let s: N+ R be defined by

......... In this case, sl = 1, s2 = 1, s3 = 1, where 1 is the first element of N, 2 is the second element of N, 3 is the third element of N, and so on. Thus in this case the sequence (s,,),,~ is such that sl = ~ ( 1 ) . = 1, is the first term of the sequence,' s, = s(2) = 1 is the second term of the sequence, s3 =s (3) 1 1 is the third term of the sequence, and so on.

In all these examples, we have taken the range as a subset of real numbers. Such sequences are called Real Sequences. A formal definition of a Real Sequence is as follows:

rC

DEFINITION 1: REAL SEQUENCE

A real sequence is a function s from the set N of natural numbers to the set R of real numbers whose values are denoted by Isl, s,, ....... ) o r by (s , , ) , ,~~ . o r by s(n), where s(n) = s, for n = 1, 2, 3, ....... The number s,

\ is called the nth term of the sequence. -.

We shall use the notation (s,) throughout our discussion. Thus, the sequence in Example 1

is (2n-l), the sequence in Exaqile 2 is (&) and the sequence in Example 3 is ( 13 or

(1, 1, I , ...... ). It is important to distinguish between a sequence and its set of values since the validity of many results depends on whether we are working with a sequence or a set, We shall always use parentheses ( ) to denate a sequence and the braces { ] to signify a set. The sequence (s,, s,, sf, ...- ) should not be confused with the set (s,, s2, s ,,..... }. For instance I

(1, 1, 1 ...... ) is a sequence whose first term is 1, second term is 1, Bird term is 1, and so

on, wheras the set { I , 1, 1 , ...} is just the singlctou { I ) . kience to mclke the distillctio~i clear, we some:imes write this sequence as (1") or

L.ei us look at a few more exa~nplcs of Real Sequellces.

EX/BM)EE 4 : (E) The erpsessioi: ( 8 , 2, 3, 4, ...I i s a sequence. This is the sequence (n),,. , or @I=

. (ii) k sequence such as (c, c, c, C, ... ), ~l1ht:re every term is LBle same na~n l~er c, is called a constant sequence. This is the sequence ( c ) ~ ,. (iii) The sequence (-1)" 21as terms (-I, 1, -1, 1 , --1, 1, ... ). EXAMPLE 5 : Let a seqalsalce (sn) be given as

In this case (sn) becomes (1, I, 2, 3, 5, 8 ...., nr, n, m+n, ...). This sequence is called the Fibonacci Sequence given by an eminent Italian mathematician L. Fibonacci 11 175-1250].

It has many fascinating mc! interesting properties. Also, it ha. lot of ~pplications particiilarly in puzzles and riddles in Matllematics. In fact, Fibonacci, was illspired by Hindu-Arabic methods uf calculation. 1-11: fol111d t11i.s scquellce when he was trying to solve the following problem:

"How many pairs of rabbits can be [~roduced fr-ona a single pair in a ycer i f d v t ~ y . . I hc nicll~oti 01' spccifjiir~g a

monil~ each pair bcgrt:; a new pair w h i c l ~ from the s c r o ~ ~ d month o u becomcs scqurllcc irl terms of prcvicjusly

producti~e'?" ~ I I I ) W I I tcr~ns of the sequerlce. is called a rrcorsiorr formula.

Now consider the scqucnce ( sl, ) gii'cn by ql =- ( - 1 )" n2. For111 a secguence ( t, 1 w!~ost:

terms are the positive terms of the sequence ( s,, ). Then the telnls of (s l l ) are (--I, 4, -4,

16, -25, 36, -49, 64; ...I slid the te:tnc of i t , ) are (4,16, 36, 64, ...I. Cbviocsly ( i , ) 1s

obpr' a ' by selecting, in ordcr, an infinite number of the dcrllls of (s"). 111 such a case, w t say hat (t,) is a subsequence of the sequerlce (sn ).

Let ( sn ) be a sequealcc. Let n,, n,, n3, ... he natorul numbers such that n, < n2 I I ~ <: ... Then a sequerice

(s,,,, Sn2, S,, ...I = (s,, 3 = ( t, I,., or ( t, 1, is called a subsequence of ( sn ).

In other wordb, a subsequence of a sequence is a sequence obtained by omining some teniis of the original sequence and not disturbing the relative positions of the remaining terms. For instance, (s,, s,, s,, ...) and (s;, s,, so, ... ) are subsequences of the sequence (s1, s2, s,, s4, ...). But (s2, sI, s4, s3.,. ) is not a subsequence of (s,, s,, s,, s,, ...). The sequence (-1)" has two subsequence namely (1, 1, 1, ...) and (-1, -1, -1, ...).

Note that n, < n, .: n, < ....... < nk-, .: n, n,,, .: .,, defines an infinite subset of N, namely, {n,, n,, n,, ...I. Conversely, you can say that every infinite subset of N can be described by

Thus, a subsequence of ( s n ) is a sequerlce obtairled by selecting, in order, an infinite subset (: the terns of the sequence. Now by the following exercise. -- - EXERCISE 1 Which of the fallowing sequences are subsequences of the sequcnce (1, 2, 3, 4, .,.j?

i) ( l , O , I , 0 , 1 , 0 ,...) ii) ( I , 3, 6, 10, 15, ... ) iii) ( 1 , 1 , 1 , 2 , 1 , 3 , 1 , 4 , 1 , 5 , . . . ) .

----

Sequences &Series Since a sequence is a function from N to R and N has the natural order, it makes sense to talk about a sequence being bounded and a sequence being monotonic. Recall from Unit 4, the definitions of bounded and monotonic functions.

DEFINITION 3: BOUNDED SEQUENCE A sequence s: N -+ R is said to be bounded if its range is bounded in R. In other words, a sequence ( s,, ) is bounded if there exists a number K > 0 such that

) s n ( < K f o r n = 1 , 2 , 3 ,... For instance, the sequence in Example 1 is not bounded whereas the sequence in Example 3 is bounded. What about the sequence in Example 2?. Is it bounded or not? Verify it yourself. Again, the sequence in Exercise 1 (i) is bounded, while the sequence in Exercise I (ii) is not bounded. What about the sequence in Exercise 1 (iii) 7

Just as, jn Unit 4, we defined a function which is bounded below only or bounded above only, you can similarly define a sequence which is bounded below only or bounded dbove only or both bounded below and bounded above i.e, bounded.

Let us consider the sequence

Since, for k' 2 2,

We have,

Thus, the sequance (sn) is bounded above by 2. Next, if we consider a sequence (sn), where

(s,Z - 2), then, in view of identity (s,+J2 - 2 = -

4s; , it follows easily that s n ' r 6 V n 2 2 i.e.,

the sequence (sn) is bounded below by 42.

Now, try the following exercise.

EXERCISE 2 Define a sequence which is bounded below ohly or bounded above only. Give an example for each. Verify whether an Arithmetical Sequence ( a, a+d, a+2d, ... ), d # 0 is bounded or bounded below only or bounded above only.

EXERCISE 3 Examine which of the sequence in Exercise 1 are bounded and unbounded.

Again, recall the notion of monotonic functions as discussed in Unit 4. Since a sequence is a special type of function; therefore, we can say something about a monotonic sequence. First, let us study the following examples:

1 2 - 3 4 1 2 Consider the sequence (-Z-, --;j; ,...). Here s, = $1) = 7, s, = s (2) =

3 $3) = r , and so on. You can see that s, < s, < s, c s, < ..., that is, s(1) < s(2) <

s(3) < ... . In other words, the sequence preserves the order in N. Such a sequence is called a monotonically incresing sequence.

1 1 1 Again, consider the sequence ( 1, - 9 - 9 - 9 ...). Here, the inequalities are reversed 2 3 4 1 1 1 1 > - > -j- > > ...; that is, 2

s(1) > $2) > s(3) > s(4) > ...

In this case the sequence reverses the order in N. Such a sequence is called a monotonically decreasing sequence.

1 Also look at the sequence (1, - a, +, - 7 ...-).

You can see that tl~is sequence neither preserves nor reverses the order in PI. Such a sequence is neither monotonically increasing nor monotonically decreasing.

A sequence which is either monotonically increasing or monotonically decreasing is called a monotonic sequence. We have the formal definition as follows:

DEFINITION 4: MONOTONIC SEQUENCE A sequence ( s,, s,, ..... ) is called a nionotonic sequence if either s, 5 s, 5 s, s , .., or s,> s, 2 s, ,... . In the first case, the sequence is called a nlonoto~~ically increasing sequence, wllile in the second case it is called a monotonically decreasing sequence.

s 2 + 2 Once again, let us look at the sequence (4) : s, = 1, sn,, = % , n 2 1. Since

and, so, in view of the fact that sn 2- 6, for n 2 2, it follows easily that sn - sntl 2 0 i.e., sntl I. sn, foi all n r 1. Hence tlie sequence (sn) is a mo~lotonically decreasing sequence.

I However, if you look at the sequence sn = (1 + +r, then we call show that it is an

monoto~iiciilly increasing sequence. Let us see how. By Binomial Theorem,

In the analogous formula for sntl, every term on the r.h.s. of last equality (except the first-term) increases in value and an additional tenn will appear on the right. Therefore, s,+~ > s,,, 'd n 21.

1 Thus (sn), where sn = (1 4- -) , is an strictly increasing sequence. In fact, this is bounded n as well.

For, observe that

(by (*) above)

As you will see below (Theorem 4), any such sequence has to be convergent. The limit

1 of the convergent sdquence sn = ( 1 i- ---)n is an important number denoted by e. Recall

I that the number e is same as the value of the exponential function e: for x = I I

Now try the following exercises.

EXERCISE 4 I I Which of the following sequences are monotonic? 1 i) ( sinn); ii) ( tann);

iv) (2n+ (-1)")).

EXERCISE 5

i), Show that a subsequence of a monotonic sequence is also monotonic.

ii) Do there exist sequences which are both monotonically increasing and monotonically decreasing?

Sequences

Sequences & Series

Since the number m depends upon the clloice of c > 0. Therefore, someti~nes the number m is dcnoted as mq.

We state a theorem (without proof! which we shall use i11 the next section:

THEOREM I :'Every sequence Ilas a monotonic subsequence. 1 For example, the sequence (T) has a subsequence

which is monotonic, is it monotonically increasing or decreasi~g? 7

5.3 CONVERGENT SEQUENCES - The basic tool of analysis is the notion of a limit and the simplest f om~ of a limit is that of the lirnit of a sequence. A real number s is called the limit of a sequei~ce ( sll ) if a large number of values of (s,,) are close to s.

1 For example, consider the sequence ti;-). It is intuitively clear that the tenns of the

sequence "approach" the number 0 as n becomes larger and larger. In other words, we can say that the n"' term of the sequence is 'as close to 0.' as we desire f ~ r

1 "sufficiently large n". See Fig. 1 for the limit of the sequence ($.

1 Alternatively, we can say that the sequence (+ approaches the limit 0 if

be made as small as possible for larger and larger values of n. Note that such a behaviour is not true of the seqrlence (n). Check why?

Thus, we have the following definition of the liniit of a sequence.

DEFINITION 5 : LIMIT OF A SEQUENCE A real number s is said to be a limit of a real sequence (sil) if, for every E > 0, there exists a number rn EN such thqt

Isn - SI < E , for all n > m.

The condition that a sequence (sn) has a limit s is often expressed by saying that the sequence (s,,) tends to a limit s. We say also that a sequence (sn) is convergent if there exists a number s (called the limit of the sequence) such that Isn - sl can be made "as small as possible" for sufficiently large values of n. Note, however, our intuition is not

1 sharp enough to tell if the sequence (n sin T) converges, we, therefore, need a precise definition of the convergence of a sequence, and also soine criteria to determine whether a given sequence converges or not. We shall first define converges of sequence in this section, and later on take up the criteria of convergence of seqiiences in the next section.

DEFINITION 6 : CONVEKGENCE OF A SEQUENCE A real sequence (sn) is said to converge to a real number s (called the limit of the sequence) if, for a given E > 0, there exists a positive integer m such that

IS,,- I] < E, for all n > rn.

We express the above fact in .several ways. We say that

(i) The sequence (sn) converges to a real number s;.or (ii) lim sn = s; or

n-tm

(iii) S,+S asn+oo.

The number s is called the limit of the sequence (sll). The convergence of (s,,) to s can be reviewed in the language of neighbourhood (Unit 3) as follows:

We say that $4; s,, = s if and only if except for finite number of terms, whole of the

sequence is in every neighbourhood of s.

GeC)n~etrir,al]y, sn -> s if, given E > 0, we are able to cut off an initial segment of the sequence, is,, s,, ... sm} such that, every member of the 'tail' sm+,, sn,+ ...a. is in the interval ]s-.E. SSE [. The initial segment that must be cut off depends on ~ . i .e . how close to s are the tail elements.

Figure 2 below illustrate the limit concept graphically. Consider the horizontal strip of width 2~ generated by the lines y = s - E and y = s+e. A given term sn of the sequence ( sn ) lies inside this strip exactly if the inequality Isn - sJ < E hold good. n u s , for a llumber s to be the litnit of the scquellce (s,,), we must be able to specify a point m on the x-axis such that, for all n lying to the right of nl, the corresponding term s,, lie wilhin the horizontal strip.

n>m

Fig. 2

Thus, if (s,,) is a sequence having the number s as a lirni!, we write /&sn= s

(or simply t&sn = s) and say the limit of sn is s as n tends t:. Iariinity or the sequence

(sn) converges to the limit s. Let us look at some examples.

1 EXAMPLE 6: Discuss tile convergence of the sequence (?).

SOLUTION : Intuitively, you can see that !5-$ - 0.

Let us use Definition 6 to justify our intuitive understanding. For this, cons'd i er an arbitrary E > 0 and let us by p find a number rn (depending upon E) such that

As you know that quite often (for example in the proof of the trigotlometric identities), we use the if method i.e. we initially work backward from our desired conclusion, but in the formal proof we must nuke sure that our steps ai-e reversible. In the present example, we want to know how big n must be so that

So, we will operate on this inequality algebraically and try to solve it for n. Thus, we

1 start with -- < E. n2

By multiplying both sides by n2 and dividing both sides by e, we find that we want

If our steps are reversible, then we can see that

1 This suggests that we should take 111 - -..? . Tllus, we proceed as follows: &

1 Let E > 0 and take m = -- . Then 6

Sequenccs & Series

I which proves that the sequence (-,) converges to the limit 0. n

A similar problem is given in the next exercise, which you should solve to check your understanding and computational skill.

EXERCISE 6 1 Use Definition 6 to prove that the sequence ($ converges to the limit 0.

EXAMPLE 7: Test the convergence of the following sequences i) (s,,), where sn = n;

ii) (I, -1, 1, -1, ... ). SOLUTION : (i) Look at the sequence sn = (n). This sequence can not converge. For, if

1 possible, suppose the sequence converges to a limit s. Taking E = , all but finitely

1 1 many elements of s,, should lie in the interval 1s - -, s + - [. But this is clearly not 2 2 so. Thus the sequence (n) does not converge.

1 ii) Suppose the sequence [ (-1)" converges to some number s. Taking E = 7, all

1 , I but finitely many elements of this sequence rrlust lie in the interval 1s - 7, s + -2- [. I 1 That means both 1 and -1 are in 1s - 7 , s It 7 [. But that is impossible because

1 1 the distance between 1 and -I is 2, while the length of the interval Is- 3, s + -2- [ is 1.

Thus the sequence (1, -1, 1, -1, ... ) does not converge.

In above example, we have come across two sequences which are not convergent. Such sequences are called divergent sequences.

Thus a sequence is said to be divergent if it is not convergent. Now we take up the question: Can a sequence converge to two different limits? The answer to this question is no and we prove this in the following form.

THEOREM 2: If a sequence is convergents, then its limit is unique.

PROOF : Suppose that a sequence (sn) has two distinct limits s and sf. Then s # sf. Let S - s'

us assume, without loss of generality, that s . sf. Take E = - 3 .

Sincr: sn+ s, all but iinitcly many elements of the sequence are in the interval IS-E, S+E [. For the same reason, all but finitely many elements of the seqence are in tlle interval ]st-E, s t + E [. Look at the Fig. 3.

The two intervals ]sf- E, S'+E [ and Is-&, s t& [ are disjoint (why?), and a tail of the sequence cannot be contained in both the intervals. Hence, it is not possible for the sequence (sn) to converge to two different limits. This proves the theorem.

Having defined convergence of sequence, the question that remains to be settled is : How to test the convergence of a ;+:en sequence? Let us, first, obtain necessary conditions for the convergence of a sequence. This condition is provided by the following theorem.

THEOREM 3: Every convergent sequence is boutadcd. Sequences

PROOF : Let (sn ) be a sequence which converges to s. Then, for E > 0, there is number rn E N such that 1 s,, -s / < 6 , for n > m.

With E = 1, we obtain m in N so that

Using triangle inequality (see Unit 3), we see that

Thus, n > r n = . I s , , / < l s / + 1. ChooseK=max { (sit I, ls,l, Is,I, ..... /sml}. Then,

we bave Isn 1 5 K, for all n s N. Hence ( s,, ) is bounded.

From this theorem, i t follows that if a sequence is convergent, then it is always bounded. Is the cowerse true? That is to say that if a sequence is bounded, then, is it convergent? The answer is No. For example the sequence (-1, 1, -1, 1, -1, I.....) is bounded but not convergent, Thus boundedness is a necessary condition for the convergence of a sequence. It is not a sufficient condition, However, for a monotonic sequence, boundedness is both necessary and sufficient, We prove this in the next theorem:

THEOREM 4: A tnonotonic sequence is convergent if and only if it is bounded.

PROOF : We already know that every convergent sequence is bounded. Thus, it is enough to prwe that a rn~notonic and bounded sequrnce is convergent. We shall prove this assertion for a monotonically increasing sequehce. Let ( sn) be a tnonotonically increasing sequence which is bounded above.

Let S denote the 6et (sn : n E N) = {s,, s ,,.,, ), where s, 5 s, S s, ... Since ( s n ) is bounded aboye, therefore, there exists an upper bound for the sequence ( s,, s,, s, ... ). Thus S hrls the lessf upper bound, say u. We slailn that the sequence (sll) converges to u,

For, let F > 0 be some real number. Since u - E is not an upper bound for the set S, therefore, u - E < U, This rneans that there exists some integer m such that sm > u - E. Also, since the sequence (s,,) is an iacreasing, we have

s,,, < s,,, for all n m.

But, sn < p, for all n. Therefore,

3 IS,, - UJ < e,

which shows that the sequel~ce (s,) converges to the limit u.

In case sf a monotonically decreasing sequence, the proof is similar to the one given above. 'So, we leave this as an easy exercise for you.

-- Sequei,ci.s & Sel its EXEBCISE 7

Let (sll) bc a monotonically decrensi~ag sequcnce which is bounded Seloav. Show that the sequence converges to its greatest lower bound.

EXERCISE 8 i) Suppose that the sequence (s,:) converges to s and sn 5 A, for every n. Show that

s 5 A. ii) Supposc the sequence (sn) converges to s and s,, 2 a, for every n. Show that s 2 a.

EXERCISE 9 i) Suppose that the sequence (st) ccnverges to s. Show that every subsequence of (sll)

also converges to s. ii) If a sequence docs not converge, then no subsequence of it caal converge. Is the

statement true? Justify your answer. ---- - - In fact, it is not difficult to see t!lat a sequlenca (fill) converges to s if and only if all its s~absequcnce converges to the mme limit s. However, if a sequence is such that its subsequences converges to different limits, then the sequence will not convergent. For example. the sequepce (s,,), where sn = (- I)", is ona s ~ ~ c h sequence. Because, here (- 1, -. 1, - 2 ,...) and (1, 1, 1 ,...) are its two subsequences converging to - 1 and 1, respecti\leiy. But, chc: sequence ((- 1)") itself is not convergent.

EXERCISE 10 Sappo'ie the sequence (sn) converges to s. Show that the sequence (!s,,() converges to Isl. Give an example to show that the converse of this is not true.

We now state a result (without proof) in tlie follouling theorem which is also referred to as rhe Bulzano-Weierstreas Theorem.

THEOREM 5 : Every bounded sequence of real numbers c o ~ ~ t a i n s a convtrgent subsequence.

We have already talked about the divagent sequences.

You have encountrred two examples of divergent sequences in Example 7, namely (1, 2, 3, 4 ,... ) and (1, -1, 1, -1, ...). The sequencc (1, 2, 3, 4, ....) is said to be divergent sequence because its terms become "hvo big", whereas (I, -1, 1, -1, ....) is divergent because it has no limit point.

We now give a formal definition of a divergent sequence.

DEFINITION 7: DIVERGENT SEQUENCE . A seqllence is said to be divergent if it is not convergent. A sequence (sn) diverges to +a if, given any real number-e > 0, there is a positivc integer rn such that sn E, fo r all n > m.

In this case, we also write lim sn - co or sn -+ m as 11 -+ co. Such a sequence is said to be divergent sequence. You can similarly define the divergence of a sequence (sn) to -m.

We car1 write this as 1% sn = -m.

Let us consider the sequence

Suppose ( sn ) is convergent and l i~n sn = s. It is clear, from the definition of n+m

convergence, that if sn -4 s, then s," -, s, as well.

But, for (s,,) as given above,

and so, we would have, Sequences

This contradiction establishes that ( sn ) does not converge, .

EXAMPLE 8: Show that the sequence ( dn ) diverges to +m.

'SOLUTION : Let E > 0 be given. Then > E for all n - > 2. This shows that thc sequence 6 + + w as n -+ w . Hence the sequence ( dn ) diverges to +m.

you can easily see that if a sequence ( s n ) diverges to +& or -m, then the sequence is unbounded. Some diverges to +m; while sqme other diverges to -a.

'

'TO end this section, let us look at a very important sequence, which will be frequently used in the larer sections:

EXAMPLE 9 : i ) If 0 5: x < I, then ( xu) converges to 0.

ii) If x = 1, then ( x" ) converges to I .

iii) Lf x 1, then ( x n ) diverges to +m.

SOLUTION: i) If x = 0, then tlie sequence (x n) is the constant sequence. (0, 0. 0 .....). Hence it converges to zero. Let 0 < x < I . Then

x"l = X. xn C xn,

implies that the sequence ( x n ) is a monotonic decreasing sequence, which is bounded below by 0. Hence the sequence converges to its grzatest lower bound. Therefore, it is enough to show that 0 is the greatest lower boultd of (XI, x2, x3, ,..).

clearly, 0 is a lower bound of the sequence. Therefore, it is enough to show tl'iat if r > 0, then r is not a lower bound of the sequence. Let r > 0. We wish to show that, for some n E N, x n < 1.. That is, for some n E N, nlogx < log r. (Recall from Unit 4 that .

1

I log x is an increasing function). This is equivalent to finding n EN such that n > - log r i log x '

Surely there are infinitely many such n's. This' sl?ows Lhat, when 0 < x < 1, liln x n = 0. Thus, when 0 5 x < 1, tlie sequence ( x n ), converges to 0.

ii) If x = 1, the sequence ( x") is die constant sequence (1, I , 1, ...) and thus converges 1 to 1. I

iii) Let x > 1. Then, silice xn*' > xn the sequence (x n) is a lnonotonic increasing sequence. To show that the sequence diverges to +w, it is sufficient to show .that the sequence is unbounded.

Let M > 0 be any number. Then

x" > M a logx" > logM (why?)

i a n I O ~ X > IogM

16g M a n > -

log x I which shows that the seqlience (x") is unbounded and hence diverges to +po. j

I Try some exercises now.

i I EXERCISE 11

! i) Show that, if -1 < x < 0, then ( x " converges to 0,

ii) Discuss the nature of the sequence ( x n) when x = -1, and x < -1.

1 EXERCISE 12

Suppose (s t ) ) and ( t n ) are convergent sequences such that sn I tn, for each n. Prove 1 17

! I *

Scqucnees ti Serier that Iitn sn I liln t,,. I1 -r. I, *

- --- ---"-I-

- Wc ~onclucle tills section by drsc1rs5ins a very 1111portant erta~l-lplc of a convergent sequence.

EXAMPI I.- 10 : I.et s., = nk. Show that i i l i l sn = I , tha t is. !jr?; 11'= I . "-,.I

1 - SOLU'FION: Writr: sn = tin = I + xn. Then x,, 0, tor n , 1

Ey Bino~n~sl 'Theorem. we Iiave

--- 2 : 2

'This Iniplies ~ l la t O 4,: x,: .: -. , fcl~ 11 2 2. Hence 0 i >i,, <: y , , For n 2 2. n-. 1

Lei c 0 bc given. *Then

7 Let m E N such that m :a --T -1 1. 'l'heii. x,) < 6, h r all n , rn.

E-

Helice 1 % - I = i X" 1 C: e, fur ill1 ir :. n(. That is, ( s,, ) coilverges to l

5.4 CRITERIA FOR THE CONVERGENCE OF SEQUENCES

In section 5.3, we have delincd a sequence ! sn ) lo be convergent if we arc able to find a nu~nber s such that sll -S s as n + ~m. 111 soti~e cases, it may be easy enough to guess the existence of such a nu~nber s. Hut. quite often, i t is not easy to do so. For exa~iiple,

I ' . consider thc saque~cc sinn). Does this scquenee converge? We cannot say

I1

anytliing unless we know the limit of this sequence. What we really need is a criterion for the convergencr: of' a sequence ( sn ) which uses only the terrns of the glven sequelicc - and 170t to search for a possible limit of the sequence frum among the infinite set R. 'That is precisely what we intend to do in this section I n fact, we shall obtain a necessary

\ ' > * . 'I r r ' i~~y le Incqrral~ty as and sufficient condition for the coavcrgence of a sequence in this section. .IIVLI\~C~ 111 [;IIIL 3 We state, first, a necessary condition for a sequence to converge. You will see later that

the condition is also sufficient for a sequence to converge but under certain aclditio~ial res~r~ct ions

THEOREM 6: I f ' a sequence ( s,, ) is convcrgcnt, then for a given E > 0, there esists m E N SI IC~ I that Isn - sL[ C: E, whenever n > li 1n.

PROOF: Let the sequence ( s n ) converge to a nunalsar s (say). Let E 0 be given.

Since sl, --+ s as n -+ cc , tlierc exists a natiiral number m such that

Thus isn - ski <'E. for aB n, k > 111 (n > k).

The theorem says that if the terms of the sequence (sen) get close to some number, then they are already close to each other. Motivated by the above theorem, we have the following definition due to A.L. Cauchy, an eminent French mathematician.

DEFINITION 9: CAUCHY SEQUENCE A sequence ( s " ) of real numbers is called a Cauchy sejuence or a fundamental sequence. if, for each E > 0, there exists a natural number m such that

, Isn - s,, I < E , for ,all n > k > m.

i n Theorem 3, we have seen that every convergent sequence is bounded. And, in Theorem 6 , it is shown that every convergent sequence is Cauclly. In the next two results we will prove that class of convergent sequences is same as the class of cauchy sequences and is properly contained in the class of bounded sequences. This inclusion is proper because there exists bounded sequence which are not monotone.

We state and pr,ove the following tileorem.

THEOREM 7: Evcry cauchy sequence is bounded.

PROOF: Let ( srl ) be a cauchy sequence. Then, by definition, it follc. .::s that for E = 1. there is a positive integer m sucll that

Isn - q( < 1 whenever n > k > m.

In particular, Isn - sm+,l < 1, for all n > m. In other words (see Unit 3)

I Sn t = Its,, - snl*,) + SII1+, I Isn - srn+,l -1- I Sn,+, I < I Sm+l I + 1- n > m.

Thus if

A = { I s, I, I s2 I , ..., I snl I, I sm+l i + 11,

then j'sn I I: A,' for all n E N.

This proves that the Sequence (sn) is bounded if it is Cauchy.

THEOREM 8: Every cauchy sequence is convergent.

PROOF': Let ( 5 ) be a cauchy sequence. Then by Theore111 7, it follows that ( sn ) is bounded. Using Bulzano-Weirstrass Theorem ('Theorem S), we can conclude that the sequence ( s n ) must have a convergent subsequence, say ( snr ). Suppose that

.snr + p as r -+ w.

Our claim is that sn -+ p as 11 + w.

Let E > 0 be given. Then there exists a natural number q such that, for any r > q, w e have

Again since ( s,, ) is a cauchy sequence, therefore, tht:e cxists a natural number k such . that for n > k > rn, we have

E lSn-SkI< 7.

Now, for an n > m, clloose r so large thal n, I. nl and r > q. 'Tben

Sequences

Sequences % Series E I s n r - P I ' T s

E is satisfied. Also 1 sn - s, ,~ / is satisfied with k = n,. TIILIS, tbr n > ~ n ,

(Triangle Inequality)

I s,, - P I = I Sn - Sn, -' Sn, - P I

5 Is,, - snIrl + ISt,[ - PI

E 1 < - - + - E = E . 2 2

Thus, given an E > 0, we have found an m such that, for any n > In, I s,, - p I < E.

This shows that s,, -> p as n -+ a, which concludes the proof.

Theorems 6 and 8 are solneti~nes combined into one theorem wl;.cli is popularly known as Cauchy's General Principle of Convergence o r Cauchy's criteria for the convergence of sequence.

THEOREM 9: Cauchy's General Principle of Convel.gence A necessary and sufficient condition for the convergence of a sequence'( st ' ) is that, for every E > 0, there exists a positive integer rn suc l~ that

( sn - s,, ( < E, whenever n > k > m.

Thus a sequence of real numbers converges if and only if it is a cauclly sequence

Let us illustrate this by the following example:

I EXAMPLE 11 : Sl~ow that the sequence -;;- is a cauchy sequence. What about ( nl)?

SOLUTION : Let E > 0 be given. Then by taking m = III (E) > $ , we see that

whenever n, k 2 m. Thus, fLr n > k > rn,

In other words,

1 whenever n > In and k >m, which shows that the sequence (x) is convergent.

The sequ ce (n2), however, is not a caucliy sequence.

For, if n and k are any two integers ( n > k), then

'n2 - kl' = l(n-k) (n+k), > 2k1 > 1,

whatever m may be. Thus, if we choose E = I , then by above inequality

inZ - k 2 ! > E , for any m (n > k > m).

Thus the sequence (nZ) is not convergent.

The Cauchy's Criteria is sometimes described in the following way:

A sequence (s,,) is said to be a cauchy srquence, if $or any E > 0, there is a positive integer m such that

Is, - s,,1 4 E, whenever k > (I.

AS you can observe, in Theorein 9, there is no mention of iimit. Thus the advantage of the Cauchy Criteria is that we are able to test the convergence of a sequence without necessarily knowing the value of its limit. Example 10 has justified the utility of Cauchy Criteria. To further elaborate this assertion, c0nsider.a sequence ( rn ) of rational nurtlbers

which converges to 4 (e.g., sequence (8,) defined below Definition 4 in this unit). If (rn)

is treated as a sequence of real numbers, then we have a real number fi, as the limit of

the sequence ( rn ). Thus the sequence ( rn) satisfies the definition of convergence. However, if we treat ( rn ) as a sequence of rational numbers and if our definition of convergence requires us to find a rational number which is the limit of ( rn) , then ( r n ) is

no longer convergent since fi is not a rational number. Thus lack of convergence has

arisen not due to the change of the sequence in any way, In fact, we have merely modified the context in which we are considering the sequence 'by changing the underlying field in which the scquence is being discussed. By changing the context in which the sequence is considered has no effect on whether the sequence is cauchy or not. ~ h u s : it also implies that there is no difference between convergent sequences of real numbers and cauchy sequences of real numbers. This is true because of the axiom of completeness of R, the set. of real numbers, But then this is not the case if we confine our sequences to the field of rational numbers. In view of this, the completeness of R is also described by saying that the system'of real numbers is complete if every caucl~y sequence in R has a limit in R.

5.5 ALGEBRA OF CONVERGENT SEQUENCES

In this section, we shall discuss the behaviour of convergent sequences with respect to algebraic operations like addition, niultiplicat~on, and so on. Recall that a sequence of real numbers is a function s: N -+ R. Since, the sum,, product, etc, of real-valued functions are defined, you can easily define the sum, product etc. of sequences.

DEFINITION 10: COMPOSITION OF SEQUENCES

Let (s,,) and ( t n ) be two sequences of real numbers and cc E R. Then,

i) ( sn + ( trt ) = ( Sn + tt, (slim of two sequences)

ii) ( sn') ( tQ 1 = ( sn tt, ) (product of two seqiences)

iii) a ( sn ) = (a S; ) .

(scalar multiple ut' ii sequence)

We'shall sllow that convergence and limits are preserved under above operations.

THEOREM 10: ' I f lirn sn = s and fim.tn = t, then n->m !I.-)=

lirn (q, + tn ) = s + t, n-tr

PROOF: Let E > 0 be given. Then ~ 1 2 , 0. Since lirn sn = s, there is.m, E b4 such that n->n

E . ' I sn - s J < - , whenever n 2 m,. 2

Since lirn tn = t, there is m, E N such that n 4 r

E Itn - t J -, whenever n r rn,.

2

Let m = max (m,, m,). Then, for all n 2; m, we have

Thus ( s,, + tn j converges to s + t.

The next theorem is easikr to prove. I

Sequences & Scrirs THEOREM 11: I f liii s,, = s and a E R, then

lim ( a sn) a s n- r

PROOF: The theorem is obvious if u = 0. So, let us consider the case when a t 0.

Let E > 0 be given. Then > 0, because / a ( 0. Since (s,) converges to s. there 1s la I

m E N such that

E / sn - s I < -, whenever.n 2 m. la I

Thus, if n 2 m, then we have E

~ 1 s ~ - a s I = l a ( I s ~ - s ( < I c ~ I - - - = E . la I

That is, lirn ( a sn) = a s . n-+%

Now you can easily solve following exercise. .

EXERCISE 13

Let linl sn = s and !i,mP t,, = t. Let a, p E 'R. Show that lirn (.a sn + P t n ) = cc s + [Jt. 1142 n-r=

Deduce that lirn ( sn - tn ) = lirn sn .- lirn 'tn. n - l r n->z n-+x . .

Now to prove that the limit of the productGof twb convergent sequences is the prod;~ct 01' their limits, we need the following theorem:

THEOREM 12: &f lirn ( sn )' = s, then !$ s: = sz. n-+E

PROOF: Let E > 0 be given. We have to find m E N such that

1 s ; - s 2 1 ' < : ~ , f o r - a l l n > . m , t h a t i s , '

1s"- s J ( s n + S I C & , for a l l n 3 m.

Since (sn) converges, therefore it is bounded. Hence, there exists a'ieal ,nu~~lber ti' such

that 1 sn I < K, for all n. Since 1i1n s,, = s, we have Is I s K. Hence Il--IX

i I s n + s I < I s n I - I - ( s ( < 2 K fo ra l tn .

E , Since lirn sn = s, there is an m E N such that I sn - s I < -- . for all 11 2 ni.

n+r , 2K

Hence, whenever n m, we have

'This proves that l.im s: = s2. 1 1 4 5

Now we are ready to prove the following theorem:

THEOR.EM 13: Let lim s,, = s and lirn tn = t. Then lirn (s,, , tn ) = s.t. '

n+x n-r r

1 PROOF: We use the wcll-known algebraic identity ab = - [(a t- b)l - (a - b)?].

4 ' Now, by Theorem 10, we know that sn + ln -+ s 4- t as n-+m. Ti~erefore. using Tl~eorem I

12. we get

(so + tJ2 + ( ~ + t ) ~ .

-

~ 1 ~ 0 , by E 13, sn-- tn + s - t as n+m. Therefore. ligalli using Theorem 12, we have

Ifence, (sl, + tn)' - (sn - tJ2 -> ( s -1- l )' - ( S - t)'. Finally, using the algebraic, idcnl i :~,

we get

verify t.fiat all the steps are justif'ied. Nniice that this proof' uses no E. 'T*l~e tcuhlliqut, of ~!:;i!lg'the'iilgebraic ideritity to deal with Ihe product is c,alled polsrizi~tioa.

I'inally. we turn our attention to the qe~otient of'convery,ent seqr.ri.liccs. I:nr l.hi.r;. \rife ?sail.\ 1 I )reim: ilr7ed ttie fi.~llocv~n,b , I( (

i I'WBEOF: '1'6: p~.ove tlic ~i.leul.em, rse nccd s i i! f,(> that ;:- c;in bc rl*:iint:J. Bur: w l ~ : 3

1 : ' I : :abot,lt 7-- ! if SOIIIC g,':= 0. then --- IS 1101 ilefi;iccj. "To ovclecumc tliii; difficulty \.vt! 112:iy !> 3,)

\,Ye shall discuss tlio proof for the casc s 0 The case \ O call he discurstd 1.y' applyilig the case for s :-. O to the sequence ( --s,, )

Let c ' 0 be given. We must find rn E N such tliat

Since iim sn - s. tlierelblr tl'iere exists m, e N such that n ->I

S , This implies that sn > for all n , m,.

2

Similarly, there exists m, E N such that

s2 E .. Isn - s I < - , whencvcr n > 111,.

2

Let m = niax (m,, m,), If 11 > rn, then we have

1 1 This proves that I,ic ,

Now, it should not hc tl;ffic~llt t b r ; \NI to prow the

'I'HEOREM 1% I F lim sn = s :(II(J liiu t , , 'I, then 11 +T 11 .

Sequences & Scrics 2n3 + 5n 1 EXAMPLE 12: Prove that iim, (dn3 + n z ) = i .

. .

SOLUTION: You have seen earlier that liif = 0. Consequently, .

5 lirn (2 + -)

n+m n 2 . 2 - - 1 - -- 1

lim (4 + -) 4 2 '

n-+= n .

What we have proved is that, if lim sn = s and Iim t = t, then it is true that n+= n + s

lirn (sn i- tn) = s + t. In other words, convergence of the sequences (sn) and (tn) is n + r

sufficient for the convkrgence of (sn -+ tn).'It is possible for (sn + tn) to be convergent even if ( s,, ) and ( t, ) do not converge.

Similar reniarks are true for sequence (a sn ) and (sn tn). Now try the following exercises.

'EXERCISE 14

i) Given an example of divergent sequence ( sn ) and ( tn ) such that ( sn + tn ) converges.

ii) Given an example of a divergent sequence (s,,) and a codvergent sequence ( t n ) such that'( s. tn ) converges.

EXERCISE IS

Show that if (sn ) is a bounded sequence and if ( tn ) converges to 0, then (sntn) converges to 0.

We have discussed the algebra of convergent sequences. Is there an algebra of divergent sequences? The following re'sults do justify that there is algebra of divergent sequences also.

If lim sn = +. oo and lirn tn = +oo, then n-bm n-+oc

111. If ( sn ) diverges to +OD and if ( tn ) converges, then (sn +- tn) diverges to 4-m.

You can similarly try to formulate some similar results for the sequences.diverging to minus infinity.

5.6 SUMMARY

In this unit, we have initiated the study of the limiting procebs by introducing the norion . of a sequence and other related concepts. In section 5.2, we have defined a sequence, a subsequence and a few types of sequences such as bounded and monotonic sequences .

etc. We have confined our discussion to the real sequences. A real sequences is a special type of real function whose domain is the set N of natural numbers and the range is a subset of the set R of real numbers. If s: N + R is a sequence, then its values are . ,

denoted. by s,, s,, .... The sequence is generally denoted by ( sn ) where the values s,, s,, .... are known as its terms. A sequence'( tn) is called a subsequence of the sequence ( s,, ) if all terms of tn are taken in order from *those of ( sn). A sequence ( x ) is said to be bounded if there exists a real number K such that ( s,, I I K, for' every n E N. A seiuerice (s, ) is said to be monotonicall;r illcreasing if sn I snr,, for, every n E N and it is said to be monotonically decreasing if sn 2 sn+, for every n EN. The sequence (sm) is said to be

strictly increasing if sn < sn+, (strict inequality), for every n EN and strictly decreasing if

Sn ' Sn+l' for every n EN. A sequence which is either increasing or decreasing is said to be a monotonic sequence.

Section 5.3 deals with the convergence of a sequence. When a sequence ( s n ) possesses a limit as n-+m, then it is said to be convergent. In other words, we say that a sequence (s,,) .converges to a limit s if for a given E > 0, there exists a positive integer m such that

A sequence which is not convergent is said to be a divergent sequence. This is due to the reason that ( s, ) is unbounded or because ( sn ) does not have a unique limiting value.

We have proved that a convergent sequence is always bounded. According to Bulzano- Weirstrass Theorem, every bounded sequence has a convergent subsequence. Similarly a bounded and monotonic sequence is convergent.

In Section 5.4, we have discussed Cauchy's Criteria to test the convergence of a sequence without taking the botheration of finding the limit of the sequence. Tl1i.s criteria states that a sequence ( s n ) is convergent if and only if, for an E > 0, there exists a positive integer n1 such that

1 s, - sk I < E, for a11 n, k > m ( n > k ).

Finally in section 5.5, we have discussed the algebra of convergent sequences i.e., the sum, difference, product and the quotient of two convergent sequences is a sequence which is convergent under certain necessary restrictions.

E 1 ) ii) is a subsequence o r (n) while (i) and (iii) are not subsequence of (n). .

E 2) Since d r 0, either d >-0 or d < 0. Let us consider the case d > 0. The case d < 0 , % .

is si~nilar and can be proved in an analogous way.

We have to show that ( a, a+d, a+2d, ....,) is unbounded.

Clearly it is bounded bclow by a. We show that it cannot have an upper bound.

Suppose m > 0 is any nu~liber. Then, there is a number a+nd such that a+nd > In for some n EN. TIILIS there are infinitely such ,positive integers. Hence (a, a+d, a+2d ....) is unbounded.

E 3) i) (1, 0, 1, 0, ....) is bounded, 1 is an upper bound and 0 is a lower bound.

ii) (1, 3, 6, 10, 15, ....) is not bounded above.

iii) (1, 1, 2, 1, 3, 1, 4, 1, 5, .....) is not bounded above.

E 4) i) is not monotonic.

ii) is also not monotonic.

1 1 1 iii) is monotonically decreasing since - > - > - 1+12 1+22. 1+32 > """'

- iv) The sequence is monotonically increasing.

The sequence is (1, 5, 5, 7, 7, 9, 9, .....)

E 5) i) Suppose ( sn ) is a monotonic increasing sequence. That is,

s, 5 s2 5 s3 5 ..", Let ( s,,,) be a subsequence. This means that

nI < n2 < no ... Hence s,, 5 sni I sn3 < ...

25

Sequences &Series [n other words, (s,,) is n~onotonically increasing.

ii) Yes. Suppose ( s n ) is both an increasing slid a decreasing sequence. This means that s, 2 s: 2 s, 2 .... and s , I. s, 5 s, s ..... By the law of trichotollly ( do you recall it), this can happen it and only if s, = s, = s, = ..... In other words constant sequences are the orfly sequences which are both increasing and decreasing.

E 6) Proof is similar to that of Exampie 6.

E 7 ) We have s, 2 s, 2 s, .... and we are given that the sequence is bounded. Hence, by the completeness axiom, the set is,. s,, ..,..I has the greatest lower bound. Let a be the greatest lower bound. Hence sn 2 a, for eveiy n EN.

Let E > 0 be given. Then a -t E is not a lower bound of the sequence. Hence sill L= a + E , for somr: m EN. But then a + E =- sn, 2 s ,,,? , > ..... But then a - E z sill 2 sm, .... 2 a. That is, s,, lies i n the interval ]a - E, a + E [, for all n > m. In other words, lim sll = a.

n->z

E 8) i) We prove this by contradiction.

If possible, let s > A.

Take c = s-A.

Since ( sn ) converges to s, there exists m EN such that / s - s,, 1 < E , for all n In.

Hence sn > s - E = A, for all n > m, which is a contradiction to the hypothesis that: sn 5 A for all n.

ii) This is entirely analogous to Part (i). You have nlerely to reverse tlle inequalities.

E 9) i) Given that ( sl, ) converges to s.

~ e t ( s,,~ ) = (s,,, s , , ~ , sIl3, .......) be a subsequence of ( s,, ).

This means, by definition, that n, n, < n, < ...... To show that (snr) converges to s, let E > 0 be, given.

Since ( sn ) converges to s, then exists m EN sucll that 1s - sll / < E, wilenever n 2 m.

Since n, < 11, < n, < ..... is an increasing sequence of natural numbers, there is an integer i EN such that n,, nit,, nit,, .... are all 2 m.

Hence I s - snrl < E, whenever r 2 i .

That is, ( snr) converges to s.

ii) The sL7tement is false. For instance, consider tlle sequence (1, 0, 1. 0, ...). This sequence does not converge. However (1, 1, 1, ...) is a subsequence ~dl ich converges to 1.

E 10) Given that ( s,, ) converges to s.

To show that ( I sn 1 ) coilverges to 1 s I.

Let E > 0 be given.

Since ( Is,! - J S 1 ( E, whenever 11 > rn,

then. I I sn 1 - 1 s I I " sn - s I . [Refer to Unit 31.

Hence, 1 i s,, 1 - ( s 1 I < E, whenever n > m.

Hence ( / s, I ) converges to I s 1 . Sequences

The converse is not true, as you can see by considering the non-convergent sequence (1, -1, 1, -1 .... ).

E 11) A sequence (sn ) is said to diverge to -w if, given any real number k < 0, there is ni E N such that sn k, for all n 2 m.

L,et -1 < x < 0. Then, 0 - 1 ?: < 1.

Let E > 0 be given. Then

( 0 - xn I = I x I".' NOW,

I x l " ~ i f n log 1x1 ' log&.

Remembering that log 1 x I < 0 sinco 0 < ( x 1 < 1. Take

log E , , m > ---

log 1x1 log E

Hence 10 - xn I < E , for all n > - 1% 1x1 '

Hence ( x") converges to '0.

ii) If x = -1, then the sequence becomes

(-1, 1, -1, +I, ..... ), which we know, oscillates finitely.

If x < -1, then (x" oscillates infinitely.

E 12) Let.lini sn = s and l i ~ n t,, = t. Then, for E > 0, there exists integers m,, rn, EN n+T n+m

such that Isn - S I < E, M n > m,, and Itn - tl < E, V n > m,. Let m = max (m,, m,). Then, for n > m, both Isn - sJ < e and Itn - t( .< E holds.

This, in term, implies

s - sn < E and t,, - t < 'E, V n > m.

Consequently, for n > m,

s ' - t = (S - sn) + (tn - t) 4 (sn - tn)

< 2&, since sn - tn 5 0,

That is, s - t < 2 ~ , V E > 0. Thus, S - t I 0 or

E 13) . lirn (asn) = a s , by Theorem 11. n - r a

Similarly, lirn (Pt,) = Pt, by h he or em 11. n+=

lirn (asn + Ptn ) = lirn (asn ) + lirn (Pt, ), by Theorem 10. n+x n+x n+d

= a s + p t .

lirn ( -tn ) = -lim sn = -t, by Theorem 1 1. n+m n+w

Hence, lirn ( sn - tn ) = lirn sn + lim ( -tn ) = s-t. n + r n + s n+m

E 14) i) Let sn = n and tn = -n. .

Then (sn ) diverges, ( tn ) diverges, but

(sn+ tn ) is the consiant sequence (0, 0, 0, .....) which converges to 0.

27

: I Sequcnccs & Series

Then (sn) diverges, while ( sn t n ) is the constant sequence ( 1, 1, 1, .....) which converges to 1.

E 15) Let I s n ( K , f o r a l l n.

let.^ < 0 be given. E

Since ( tn ) canverges to 0; there is m E N such that

for all n > m.

Hence, for all n > m,

Hence ( sn tn ) converges to 0.

1 ii) ~ e t sn = n and tn = - n .

UNIT 6 . POSITIVE TERM SERIES

Structure 6.1 Introduction

Objectives 6.2 Infinite Series 6.3 Series of Positive Terms! 6.4 General Tests of Convergence

Co~nparison Tests

p-test 6.5 Some Special Tests of Convergence

D'Alembertls Ratio Tesl Cauchy's Root Test

Cauchy's Integral Test

Raabe's Tcst Ciauss's Test

6.6 Summary 6.7 Answers/Hints/Solutions

6.1 INTRODUCTION

In the Unit 5, you were introduced to the notion of a real sequence and its convergence to a limit. It was also stated that one of the main aims of discussing the real sequences and its convergence'was to find a method of obtaining the sum of an infinite number of real numbers. In other words, we have to give a meaning to the infinite sums of the forms

1 t 2 + 3 + 4 + 5 + .............

where the ' .................. ' is interpreted to indicate the remainiae infkite number of additions which have to be performed. The clear explanation of this concept will, then, lead us to conclude that it is possible to achieve the addition of an infinite number of real numbers, by using the limiting pracess of the real sequences.

To give a satisfactory meaning to-the summation of the infinite number of the terms of a sequence, we have to define a summation which is popularly known as an infinite series. The infinite series have been classified mainly into two categories - the positive term series and the general Series. What are, then, the positive term series and the general series? We shall try to find answers for these questions. The summation of an infinite series of real numbers is directly connected with the convergence of the associated real sequences. We shall, therefore, give a meaning to the term associated sequence for an infinite series and hence its convergence which will lead us ultimately. to find the sum of an infinite series.

Although the famous Greek philosopher and mathematician, Archimedes had summed up the well-known Geometric Series, yet oqer results on infinite series did not appear in Europe until the 14th century when Nicole Oresme [1330-13821 showed that the Harmonic series diverges. Since then, a lbt of work has been going on in thiS direction. There is evidence that this type of work was known in 1ndia also as early as in 1550. Indeed, even modem work has shown evidence of the discovery of a number of mathematical ideas pertaining to the infinite series in China, India and Persia much before they came to be known in Europe. In the 17th century, there seemed to be little concern for the convergence of the infinite series. But during the 18th century, two ~rench$mathematicians D' Aleinbert and Cauchy devised

+ remarkable tests for the convergence of infinite series under certain conditions which we shall discuss in this unit. Also, we shall discuss, in this unit, a few more tests for the

UNIT 6 . POSITIVE TERM SERIES

Structure 6.1 Introduction

Objectives 6.2 Infinite Series 6.3 Series of Positive Terms1 6.4 General Tests of Convergence

Comparison Tests p-test

6.5 Some Special Tests of Convergence D'Alembert's Ratio Test

Cauchy's Root Test Cauchy's Integral Test Raabe's Tcst Ciauss's Test

6.6 Summary 6.7 Answers/Hints/Solutions

6.1 INTRODUCTION

In the Unit 5, you were introduced to the notion of a real sequence and its convergence to a limit. It was also stated that one of the main aims of discussing the real sequences and its convergenceewas to find a method of obtaining the sum of an infinite number of real numbers. In other words, we have to give a meaning to the infinite sums of the forms

1 + - 2 + 3 + 4 + 5 + .............

where the ' ....,............. ' is interpreted to indicate the remainiiig infifiite number of additions which have to be performed. The clear explanation of this concept will, then, lead us to conclude that it is possible to achieve the addition of an infinite number of real numbers, by using the limiting process of the real sequences.

To give a satisfactory meaning toathe sllmmation of the infinite number of the terms of a sequence, we have to d@ne a summation which is popularly known as an infinite series. The infinite series have been classified mainly into two categories - the positive term series and the general Series. What are, then, the positive term series and the general series? We shall try to find answers for these questions, The summation of an infinite series of real numbers is directly connected with the convergence of the associated real sequences. We shall, therefore, give a meaning to the term associated sequence for an infinite series and hence its convergence which will lead us ultimately. to find the sum of an infinite series.

L

Although the famous Greek philosopher and mathematician, Archimedes had summed up the well-known Geometric Series, yet other results on infinite series did not appear in Europe until the 14th century when Nicole Oresme [ I 330-1382] showed that the Harmonic series diverges. Since then, a lbt of work has been going on in this direction. There is evidence that this type of work was known in lndia also as early as in 1550. Indeed, even modem work has shown evidence of the discovery of a number of mathematical ideas pertaining to the infinite series in China, India and Persia much before they came to be known in Europe. In the 17th century, there seemed to be little concern for the convergence of the, infinite series. But during the 18th century, two ~rench*mathematicians D' Alethbert and Cauchy devised

, remarkable tests for the convergence of infinite series under certain conditions which we shall discuss in this unit. Also, we shall discuss, in this unit, a few more tests for the

Sequences and Series , convergence of the infinite series when these two basic tests fail to help us in knowing the convergence of the Infinite Series.

' Objectives

Therefore, after studying this unit, you should be able to * define an infinite series as well as a positive term series . I.. define the associated sequence of partial sums associated with an infinite series and

hence its convergence

* use the Ratio and Root Tests to determine the convergence of infinite series

I.. apply Integral Test and few more tests to discuss convergence of positive term series. -

.- - -- 1

6.2 INFINITE SERIES

" 1 Consider the sequence (-). Its term as you know, are n

With the help of the terms of this sequence, form an expression

1 which is nothing but a summation of the infinite number of terms of ( - ). Such an n expresiion is known as an infinite series.

In general, we define an inf~nite series as follows:

DEFINITION 1: INFINITE SERIES

If ( un ) be r sequence of real numbers, then the expression

is called an infinite series of real numbers.

The series is generally written in the. form u, or sibply I: u, , where n.1

ul, u,, ......, u, ........ are respectively called the first term, the second ............ ........., . term, the nth. term of the series. We jlhall write just

"infinite serieszor "series*' in place of "infinite series of real numbers". - ..+. . ...... EXAMPLE 1: i) ?he series a t (atd) t (a+2d) + .....: is a 6 infinite' series.

You are familiar with it. It is an Arithmetic Series with 'a' as the first term and 'd' as the common difference.

i i ) The series atar+ar2 + ......... is an infinite series.

You know that it is a Geometric Series with 'a' as the first term and 'r' as the common ratio. I i i i ) The expression 1 -1 + 1 -1 ........ is an infinite series.

\ . .......... It has been formed by using the terms of sequence (1, -1, 1, -1

I

)

Consider the series I

We find the sum of first renn, first two terms, first three terins, ..........., first n terms of the series and denote them by,s,, 3, s,, ........ s, respectively. Then, we have

30 s1 = l (sum of the f"st term)

1 s 2 = 1 4 - z (sirn~ of the first two terms)

, ,

Positive Term Gx-i~@

1 1 1 sn = 1 + - + - + . . + suit^ of the first n ternls)

2 4

From the above series, we obtained a sequence (s,, s,, s,, ......... ) - (s,J.

This sequence (s,,) is called the sequence'of partial sums or the sequence associated with the series. In general, we define the associated sequence of ally series as follows.

DEFINITION 2: SEQUENCE OF PARTIAL SUMS (ASSOCIATED SEQUENCE)

Let C un be a given series. Define s,, s,, s,, ... as follows :

Then the sequence ( s n ) = (s,, s,, ...) is called the sequence of partial sum or (simply associated sequence) of ihc series C I I ~ , where s, is calIed the first partial sum, s, tile 2nd partial sum, s, the 3rd partial sum, ..., sn the nth partiaI sullr and so on.

f

.... For instance, in case of scries C (-1)"" = 1-l+l-I-t s, - 1, s, = 0 (sum of first two I I 02 l

terms), s, = I (sum of first three terms), and so on. 'That is, 1

I 1 .... However, if we consider the series C (-1)" - -I+l-1.1-!-I+ thed I nt 1

otherwise . .

Now, you try the following exercise.

EXERCISE 1 Find the sequence of partial sums of the following series:

1 1 i) l t T + T + . . . ; ii) 1 +2+3+.. . --,-

I

Having defined a sequence associated with a series, we are in a position to define convergence of a scries and then, give a meaning to the phrase 'surn of an infinite series'.

03 ... .Let XU,, be an infinite series and (sn) be its sequence of partial sum i.e., sn = rr, + u, t t n=l

u,, V n 2 I. I f the sequence ( s n ) converges to s, wc say that the series C usconverges Note t h ~ t when

03 ra m I to s and we write 2 s, = s. The11 Is' is called the sum of series C un. u, = converge, the symbol

k= 1 11=1 ' 11*1

1 m n j: u,, i s uvcd to k-1

In other words T: u, =lim Z a,, provided the limit on r.h.s. exists. n-l n-bV r"l

denote not only the inflnitc 1 series, hut also its sum.

1 3 1

Sequences cYr Scries However, if the associated sequence (sw) diverges, we say that the ,

CO

series C 11x1 u,, diverges. If the sequence ( sn ) diverges to +* or -*, we write

a

un = +m or Z un = - m, respectively. I I= I n-l

As'you have seen above, for the series x (-I)"", the associated sequence is (1, 0, 1, 0. n 2 1

.....), which does not converges. Thus the series C (-1)"" is divergent. However, it n 2 1

neither diverges to + a, nor to -03.

EXAMINE 2: Examine the convergence of the following infinite series:

" 1 I m

ii) n; . iii) 'x [l+(-l)"+'l. n=l n=l

k l i I 1 1 SOL TION ': (i) Consider the series C --, . Here sn = + --j- + ... + . n=l 2 2

Using, simple induction arguements, it is easy to see that

" O 1 Since ( s,, ) converges to I , the series C converges to I (by above definition) and

n=12

1' we can write C - = 1

n= 1 2"

.... ii) Consider the series 1 + 2 + 3 + In this case,

CO

The sequence ( s n ) is unbounded and, so, is divergent. Thus the series E n is divergent. n=I

m

iii) Consider the series. C[1 + (-1)"+']. As before, ( sn ) is 2 or 0 according as n is n= I

odd or even. Thus, ( s n ) = (2, 0, 2, 0, ...) is divergent. Therefore, the series is divergent.

Note that the following results follow immediately froin the definition of convergence of the series.

I . The addition, omission or change of a finite number of terms of a series does not affect its behaviour regarding its convergence or divergence.

11. Multiplying the terms of a series by a non-zero uumber does not affect its behaviour as regards its convergence or divergence.

EXERCISE 2 . m

i) Let a t (a+d) + (a+2d) + ......... = x (a+ (n-1)d) be an arithmetic series. Prove n= l

that the series diverges to +* o r -* according as d 5 0 or d < 0. What can you say if d = O?

......... ii) If a series u, + converges to s, then prove that u, + u, + u, + ........

m 1 iii) Prove that the series C converges to the sum 1.

I "=I n(n+l)

he following theorems are immediate consequence of the theorems on sequences which you have studied in Unit 5. ,

Positive Tcrln Series m m

THEOREM 1: If C un converges to s and C vn converges to t, then n=I n= I

- i) x ( u n + vn) converges to s+t; and

n=I

ro

ii) &un converges to cs, for c ER. n=l

CI) a3

PROOF : Let ( s " ) and ( t , , ) be the sequences of partial sums of 1 un and C vn , m l l ~ l u=l

respectively. Then the nIh partial sum of the series 1 (ull + v n ) is 11=1

- We know that, lim (sll 1- tll) = !im sll + li~n tll = si-t. I-Jence z (un + vll ) converges to s+t.

n + r n-,* n-P r 11-1

This proves (i). m

For (ii), observe that the n'll partial sum of the series cun is 11" 1

CUI + CUI -t- ... + Cull = C (u I + u2 -+ .,.. -1. un) = CSI1.

"0, Since 1i1-n (csn) = c lirn s,] = CS, therefore the series cun converges to cs.

n+m 11-m 11=1

In the following theorem, we have shown that if the series is convergent, then all the terms after some stage must beco~ne arbitrary small.

93

THEOREM 2: If z u,, is a convergent series, then lirn un = 0. n= l 1 1 3 '

m

PROOF: Suppose 2 un = s. l'hen lim sll = s, where ( s n ) is the sequence of partial sums 11- 1 n.+r

u3

of tlie series C un. Now. since l i~n sn = s, therefore lini s ~ - ~ = s as well. Also, n-l n-) r 11-P~C

ul, = sn - sn-, implies that l i ~ n un = lirn ( sn - sn-[ ) = lirn sn - l i~n so_, = s - s = 0. n-1 n-P c n- r~ j n- bm

m

Thus, linl un = 0 is a necessary co~ldition for un to converge. n-tx nz l

At times, this little result is very useful. We would suggest you to make use of this result as a tool while tesing the colivergence or divergence of a series. So, if for a series

un, you are able to show that l i~n un s 0, then it is easy to deduce that n->'c

00

I series 2 un does not converge. For example, using Theorem 2, you can quickly deduce I n= 1

00 n ~ n that the series - is not convergent becausc lint -- = I + 0 in.this case.

I n+= n f 1 n-w n+l

I However, the converse of tlie above heorem is not true i.e., the condition lirn un - 0 is n+x

I m

I not a sufficient condition for the convergence of x u n . In other words, tliere are divergent n;.l ,. '

m ' . series 1 un with un = 0. For cxnmple, lim .& = 0, but the sequence of partial sums

n=I n+x

1 .of the series Cx is known to be a divergent series. (see pp.16-17 of the block)

r In Unit 5, you have learnt the Caucliy criterion for the convergence of a sequence. 1

? Closely connected with this. is Cauchy's Pri~lciple of convergence of infinite series. We

I state and prove this theorem as follows.

I (1

Sequences & Scrics THEOREM 3: CAUCHY PRINCIPLE OF CONVERGENCE m

The series 1 un converges if and only if, given E > 0, there exists m E N n=l

such that C ur < G, whenever n, k 3 m (n > k). 1: ,+. 1 PROOF: Let (sn ) be the sequence of partial sums of the series C un. By Cauchy's principle of convergence of sequence, (s,) is convergent iff, given F, > 0, there exists

m E N such that 1s" - sk 1 < 6, for n, k > m, (n > k). But sn - sk = i: u,, and since r=k+ l

convergence of the sequence (sn ) implies the convergence of C u , it follows that C un

is convergent iff, given E > 0, tilere exists m EN such that

/ f: ur ( < a, for n, k > n(u > k). r k + l

Let us use Cauchy's principle of convergence to test the convergence of the harrnoriic series.

1 1 EXAMPLE 3: Test the convergence of the series 1 + - + - + ...

2 3

SOLUTION : Suppose the series converges. By Cauchy's Principle of convergence, given E > 0, 3 m, E N such that

I g U , 1 < c, for n, L > m, (n > k).

Take k = m and n = 2m, where m > m,. Then

1 1 But L.H.S. > - + -+ 1 1 1 4- - = m - = - ....... 2m 2m 2m 2m 2 '

1 This will lead to a contradiction, if we choose E < - . Hence, the series

2 converge.

EXERCISE 3

1 1 1 i) Find the sum of the series 1 - - + - - -+.,, .

4 16 64

ii) Does g, (-$-) converges?

00 OD 00

iii) Prove that if C un converges and C vn diverges, then C (un + vn) diverges. n=l n=l n=l

ca

iv) Give an example of a series C un such that riel

... ... (u, + u,) + (u, + u,) i- converges, but.u, + u, + u, + u, + does not converge.

The infinite series series have been divided into two major classes: The positive term series and the series with arbitrary terms both positive and negative t e n s , called the

-general series. We shall study the positive term series in this unit while the series with arbitrary terms (general series) will be discussed in Unit 7. .

4.3 SERIES OF POSITIVE TERMS ~ o s i t i v c ' ~ e ~ . m Series

A series u,,, where u,, , 0 for all n, is called a series of positive terms or a positive term series.

Recall that the behaviour of a series is defined in terms of it's associated sequence of partial sums. The sequence (s, , ) associated with a series C un of positive terms is a ~iionotonic increasing sequence, because s n l l - sn = \ i n+ , :. 0. We know that a monotonic sequence converges if it is bounded and diverges to + 03 if it is unbounded. Tl~us, as a silnple consequence of this fact, we have the following theorem.

THEOREM 4: Let C II,, be a series of positive terms with associated sequence (s,,). n=l

UJ m

Then C 11" converges if ( st,) is bounded, and C 11" diverges to +a if ( s n ) is n=l n= l

unbounded.

For example, consider the Hannonic Serics x I Let ( r ) bc the sequence of partial 11..

sums of this series. We claim that the sequcnce (sn ) is unbounded, Indeed

1 1 1 1 sl = 1; s,-s, =- ; s , - 5 =-. f -. > - ' 2 ' 2 3 4- 2 '

1 1 1 1 . 1 s - s = --+ - 4- - + -- > - ; and, in general, 8 6 7 8 2

I S21; - S2k-l =

1 1 f -- + ... + - 5. 7k-I

1 1 -- -. - - ,, , and so, 2L-1+1 2 -1- 2k - ' 2k .G

I

5'1 This shows that ( sn ) i s itself unbounded. Hence the series 1 -, divergers to + oo,

a,

Next, we consider another important series C xn, called the geometric series with n= I

00 1 EXAMPLE 4: (Geometric Series) Tlie scries x x" converges to - , if 0 < x < 1,

n=l 1-x and diverges to cc, if x 2 1.

SOLUTlON ; If x>l, tlleli the sequence ( sll ) associated with the series

1

I 5 rn ~f positiva tenns is unbounded, Indeed n=!

1 But (n) is unbounded and hellce divs@ent. Thercforc ( s n ) is divergent and hence the giver! series is divergent for x 2 I .

For the case 0 < x .< 1 , we have

Therefore, ( sn) is convergent and hence xn = = , f o r0 . : x<1 .

Seqqences and Serics EXERCISE 4

i) Show that if u, + u, + ......, converges to s, then' so dues m, + O + u2 + 0 + u, + 0 +....... More generally, show that any number of zero terms may be inserted anywhere ( or removed from anywhere) in a convergent series without affecting its convergence or its sum.

w

I i i ) Determine the convergence or divergence sf the series log (I + - ).

n = l n

From Exercise 4 (i), it follows that the behaviour of a series of non-negalive terms is determined by that of a series of positive terms. In other words, the convergence or divergence of the positive term series and the non-negative term series is same.

Now, let us study some tests of convergence of the positive term series. In Section 6.4, we discuss some general tests and in Section 6.5, we shall study some special tests of convergence.

6.4 GENERAL TESTS OF CONVERGENCE

So far, you have seen the convergence of a series. Jt is defined in terrrls of convergence of its associated sequence of partial sums. However, it is not always easy to find the sequence ( s,, j and its convergence. Then, how to tesl the convergence of such scries? For this, we state and prove some general tests for the convergence of series of positive terms.

COMPARISON TEST

The most common tests of convergence of the positive term series are the comparison tests. In these tests, we compare the series x u, with a series x v, with known behaviour. Accordingly, we decide whether the series C u,, is convergent or divergent. This is sometimes, reworded by saying that the behaviour of the series x u, in terftls of its convergence or divergence is dominated by h e behaviour of the series C v, . In other words, we say that if a posilive term series u, is dominated by a positive lerm serie:; C v, which is convergent, then u, is also convergent. Similarly, if a positive te~m serics C u,, dominates another positive term series bI, and C u, is divergent, then C v, is also divergent.

We discuss the comparison tests in the form of the following Lheorem:

THEOREM 5: Let x u, and x v, be any two series of positive terms. n=1 n = l

I ) Suppose there exists a positive real number k such that u, c k v, V n. Then if Cv, is convergent, Xu, is ctinvergent and if Xun is divergent, Cv, is divergent.

u n 11) Suppose !$ - = A, where A is a finite non-zero real number.

n Then Xun and Cv, converge or diverge together.

u n V 11 111) Suppose there exists a positive integer m such that- 2 - for

" n + 1 v n - i ~ n 2 m. Then if Cv, is convergent, xu,, is convergent and if Cu, i s . divergent, Cv, is diverge*>+

.e=-.

PNOOF: I. since' u, < kv, for all n, therefore

Suppose C vn is convergent.

Since X v, converges, there musoexis1 a positive number A such that

Consequently, u,+uZ + . . . . . . + u, < kA, V n.

Positive 're,-rn Series

--..- . .

monotonic. Hence

u,, is convergent.

Sinlilarly you can show that C u, is divergent implies that 2 v,is divergent.

l a Since C, u, and C v, are two series of positive terms, therefore

U" which implies that !2 -- 2 0. V"

In other words, A 10.

But by our assumption A $0, therefore A > 0.

Now let us choose an E > 0 (however small) such that A - F. > 0.

U Since --"- = A, therefore there exists a positive integer m " I, such that

Consider

Using (I), it follows that if 2 v, converges. u,, also converges. Further if 2 u, diverges, then vn also diverges.

Now consider the inequality

(A -" E)V*, < u,,

Then

1 v, <- (A*) un

Thus, aga~n it follows that if C u, converges, then Z, v , also converges and if C v , diverges, then C, u, diverges. Hence Z, u, and C v,, converge or diverge together.

You may note that in thc case when convergence of C u, follows from the convergence of v, , then A may or inay not be zero. But conversely when the coilvergence of 2 v,

follows from the convergence of C u, , then A niust not be zero ( A # 0).

. . . . ID) Putting n=m, m+l , m-e2, m+3, n-2, n-1 in the given inequality, we get

Sequences and Series Multiplying the correspop2ing sides of the above inequalities, we get L, L V n > m

U " V" U" V i.e. - < - V n > m um V r n

Since rn is a fixed integer, u, and v, both are positive, therefore

" is a positive fixed number. Let %= k, where k is a fixed positive number. Then "m vm obviously, by using I, it follows that if C v, converges, then C u, converges

and if 1 u, diverges, then v, diverges.

This completes the proof of the theorem.

EXAMPLE 5: Test the convergence of the following series:

SOLUTION: (i) Consider the series

Compare the series with the convergent geometric series

1 for each n. That is, each term of the first series is less than the Clearly 2"-'+ p corresponding term of the second series. Hence, by *the Comparison Test (I) the given series converges.

1 1 1 ii) Consider the series 1 + - + - + . . . . . . . . + - +- : . . .

4-5 43 & Let us compare this series with the Harmonic Series

You have seen earlier that the Harmonic Series diverges.

1 1 Now, for each n,.- 2 2 In other words, each term of the given Series is greater than the

& corresponding term of the harmonic series. Hence, by the Comparisoq Test (I) the serjeg diverges.

iii) In the series 1 1 1 1 . . . . . . . . . . . . I + - + - + - + . * + - + I! 2! 3! n! .

you know that n! is 'n factorial'. That ia nl = n(n-1) ... 3.2.1,

Does this converge? With which series shall we compare it? Let us examine it.

We know that deletion of a finite number of terms does rlot @$r the canvergeqce 07 divergence of a series. So, let us consider the terms of the above series f ~ ~ r p the third term

,

onwards. 1 1

Now, - = - 2! 2

Positive .Term Series

I I -- 1 1 < 1 - 1 -- ?,,-I , and so on. ... . . . n!-n(n-1) 3.2 2 x 2 ~ x 2

1 1 1 We also know that - + - + - + ... converges. Hence, by Comparison Test (I), the given 2 22 23 . -

1 1 1 1 ............ series converges. This series 1 + - + - + - + + 7 + ......... is a very l ! 2! 31 n.

important series. The number to which it converges, is denoted by e, which, as you know, is called the exponential number or transcendental number.

You would have !~oted that, in order to use the Colnparison Test I, you must liave a large number of known convergent and divergent series.

8 Let us now discuss important series which is frequently used for the Colnparisoi~ Tests. 1 1 1 1 ......... ............ This is known as the p-series namely 7 + - + - + + - + 1 2P 3P nP

I where p is a positive real number. Let us investigate the behaviour of the p-series for different values of p.

The p-series is one of most important series. Its behaviour changes from divergence to . convergence as we go from p=l to p'> 1. We state and prove the following theorem known as p-test for its convergence which depends upon the values of p.

THEOREM 5 : ( p-Test)

1 A-positive term series - p > 0 is convergent if p > 1 and is

n=l nP divergent if p 5 1.

PROOF: There are three cnses namely p = 1, p < 1 and p > 1. We discuss these cases as follows:

Case 1: Let p = 1. The series is just the Harmonic Series, which has already been shown to be divergent.

Case 2: Let p < 1. 1 1

, Since p < 1, np 5 n and hence - 2 -for each n. In other words, each term of the series is nP n 1 1 1 ... . . . . greater than the corresponding term of the divergent scries 1 + 5 + 5 + + - +

n Hence, in this case also, p-series diverges.

Case 3 : Let p r 1.

To'consider this case, we use the following series for comparison:

(The pattern should be clear).

It is clear that each term of the p-series is less than or equal to the corresponding term of this b series. We claim that this series converges. Indeed, it is clear that

2 which is a geometric series with common ratio -- < 1.

2p

Thus this series converges. Hence by Comparison Test I, the p-series also converges.

This completes the proof of the theorem.

EXERCISE 5 ,

In the proof of Theoren] 5, we have grouped some terms of the series of positive terms. Prove that such grouping does 180: affect the nature of thr. series. - To see how p-tests works, let us discuss a case in.Lhe following example.

n .EXAMPLE 6: Sllow that the series 2 - diverges.

,,=I n1 + 1 I1 tl 1

SOLUTION : For large values of n, - behaves like ': i.e. - So, for comparison, n 2+ I n- n

" 1 we can take the harmonic series , . As seen above, this is a divergent series.

n= l

n So, we take un = --- 1

n2+ I and vn = -, then

n

u n2

Iim = liln -- - - 1 , which is non-zero and finite. n->m Vn n - t r n2 +l

Hence by the Comparison Test (JI), it follows that the given series diverges.

Try the following exercises.

EXERCISE 6

Determine whether the foHowing series are convergent ?

EXERCISE 7 m

Show that if the series C u,, of positive terms converges, then the series 11-1

1 5 $ also converges. Hence, deduce that the series -- diverges. I I = ~ d n

6.5 SOME SPECIAL; TESTS OF CONVERGENCE

In section 6.4, w e discussed some general tests to know the convergence 01. divergence O K infinite series. These tests enable us to deal with a fairly large numbcr of positive term series. I-Iowevcr, the scope is really quite limited and we are forced to look for other tests to handle a few more series. In this section, we shall develop some special tests which can be used to test the convergence of a still a larger number of infinite series. We begin our discussion with the two basic tests which are more useful and frequently employed.

Thc first test, called the Ratio Test, is due to J.DIAlembert [1717-17831 and the other, called the Root Test is due to Cauchy, both eminent French Mathematicians.

In the Ratio Test, we discuss the convergence of a given series by studying the sequence of the ratios of the consecutive terms. Comparison Test needs another series with known behaviour for the purpose o f comparison but in ratio test we use only the t e ~ m of the given series. We now state and prove D'Alembert's Ratio Test.

I

THEOREM 6: D' ALEMBERT'S RATIO TEST

u I I + ~ Let if. urn be a series of positive terms such that lim - = L. Tllcn the following n-1 n-tm Un I

hold. I i) If L < 1, then bn converges.

n=l

ii) If L > 1, then un diverges. 11=1

i i i) If L = 1, the test fails to give any definite information about the convergence of the series.

L1n+ 1

PROOF: Case (i) ~ e t , / E - = L < 1.

I-et r be a real number Such that L < r < 1. Choose a number E> 0 such that L+E = r.

un+ 1 Since lim - = L, there exists a positive integer III such that . n- U,

But L + E < r, therefore

i.e, u,,, < r u, fur n 2 m.

Thus,

< r urn,, < r2 urn, for n = m+l

Positive Tcr~n Series

u,,, < r umt2 < I3 urn, for 11 = m+2.

In general, u,, < rk un, for k = 1, 2, 3 ,.......

Hence, by Colnparison Test, urn+, + u~, ,+~ ..... ... converges, since it is do~ninated by the . convergent geometric series

u,, r + u,, r2 . . . . .+ u,, rk . . . , L.,

Therefore, the series u,, also converges. n= l

Case (ii) Let - = L > 1 .

Choose a number E > 0 such thar La-& > 1 .

u"+l Since - = L, there exists a posi t i~e integer m such that

u " t l i.e. L-E < - < L + E for n 1 m

4 un - Thus

That is, u,,, > u, for all n 2 m. This means that fi5 u, # 0.

- m

Hence x u, caanot converge. Thus x u,, is divergent to m. P I n=l

Case (iii) L = 1 . The test fails because the series may converge or may diverge. The reason is I l ial there are convergent scries of positive terms with

I .- .-.

"n+~ and there are divergent series of positive terms with lim - = 1. n.+= Un

"0 I For instance - diverges. Here un = , so that

n=i n n

1 On tlie other hand, n=I conveizes by p-test. Here II,, =-- n2 SO that

112 l i n l VEL = lim -- = I . n-+x uII n ill,

Note tllal i t . 111 the st:iternent of the D'Alernbert's Test, had we taken !i!n -- = L then Clll I I

w

L. > I \r':l~dr.l ~mply conierpcnct. and L < 1 would imply divergetrce of u, . n= I

' n r ~ You may also note ~hnl if liyj - = then the cerieb u, is divergent. You may prove i t

U"

by ;ipply~np rhc pl.ocodurt of the case ( i i ) .

EXilMPI.E 7: Test the convergence of the series

211- 1 :!n+ l SOLUTION: Here u, = -- 2"

. SO lhat u ,+~ = 2"+1 .

",+I Since lim -- = < 1. the series converges.

n -ko 11, 2 - .. - - - - -- -

EXERCISE 8

1 1 1 i ) Show, using the Ratio Test, that the series e = l + -- I! + 2i + - 3! + ...........

converges.

i i ) For what positive values of x does the series .-l. converge? . 11=I n

i i i ) Find all positive values of x for which the series 1 + 2x + 3x2 + 4x3

+ ....... converges.

rn

5" i v ) Test for convergence the series -- ,,=I (2n+ l ) !

You have seen that D'Alembert's Ratio Test fails to give any definite information about thc convergence or divergence or the series in some situations. I n such cases, sometimes Cauchy's Root Test is helpful. But mostly Cauchy's Root Test is more suitable for those

. series whose nth term contains n, n2 etc. in the exponent. In the Root Test, the convergerlce of a given series is based on the behaviour ol: the sequence formed by taking the nth root of the terms of the given series. Let us state and prove this test as the following thcclrern:

THEBREM 7: CAUCHY'S ROOT TEST Let 7 un be a series of positive terms such that !;E (un)" = L. . nS , I

di) I f L < 1, then u,, converges. n = l

Positivc Term Series

rn

i i ) I f L > 1, then LI,, diverges. 11=1

i i i ) I f I , = 1. the test fails and the series may converge or diverge.

PROOF: Case ( i ) Let L < I

Cl~oose a real number r S L I C ~ that L. < 1. < I.

Let E > 0 be a number such ~ h n t L+E = r.

Since lim (un)li" = L, there exists m E N such tliat n-+a

I(un)" - LJ < E, for 11 2 m;

i.e. L - E < (u,)"" < L+E, for n 2 m;

for n 2 m;

for n 2 m.

Since c P is a gedmetric series wit11 common ratio r, wt~icli is less than I . so it is convergent. Thus by Co~nparison Test, it follows that un is Convergent.

Case (ii) Let L > 1. Choose a real number s such that L > s > 1:

Let E > 0 be a number such tliat L-E = S.

Since lirn (un)lA = L, there exists in E N such :iiaL, for n 2 m.

Since sn is a geometric series with colnmon ratio s, which is greater than I , so it is divergent. Hcnce by Cornparis011 Test, C un is-divergent.

Casc (iii) Let L, -- 1. In this case, the test fails to f~irnish any definite infornlation about the convergence or divergence of the series. For exahple, consider the convetLont series

1 1 ;;i and the divergent series 2 ;r . In both tlic cases lirn (u,)lh = 1 .

n=l n , I

, l<X,\ hI1'1.1: X : 'l'cst 1'01. Convergence the Scries

Sincc n occurs in tlic cxponcnt ol' LI,, , SO wc ~ ~ p p l y C ; I L I C ~ ~ ' \ Root TC\[. Hcrc i

I I l l , ~ / ~ ~ i n !1+' = + < 1 , 11 -. - 11 ) - 211-1 ,

Hcncc the series cc?nverpes.

111 tlic abovc cxample. i f YOLI wish LO apply the Kulio 'l'est. you will have to c\ta!i~i~lc.

I ~ I I I .!.!\?I i s C + 11 + " LI,,

43

I

Sequences Sr Series

which is certainly not easy. Now we take an example where the Ratio Test fails, but the Roor Test gives a definite answer.

DO

EXAMPLE 9: Consider the series un where n = l

( 2-""IT;if n i s o d d un = I

2-n36" i f n is e v e n

SOLUTION: .Let us try the Ratio Test.

, k n + l Hence ,lz - = 0.

%

U2n - On the other hand. - - Tzn- - 1.. fi+@ u~~~ , 2-2n+l -W) - 2

%+I In ,other words, the sequence- does not have a limit.

u,

Thus, the ratio test is not applicable in this case. However, the root test is applicable as is evident from the following:

- 1 I -I+- 1 For ( uln )a = 2 ' 6 , SO that lim ( uzn lrn = - n+- 2

-1-4- - 1 and (uzn+, = 2 &+I so that !E (u2n+l)zn+1 - - - 2

1 1 Thus, tz ( u n ) i -- < 1. - 2 Hence the series converges.

You may note that whatever the ratio test determines the nature of a series, so does the

root test. In other woids, if = L then it is true that lim (un)Ih = L. But n->m

converse may not be true as.is clear from the above example. Thus the root test is more powerful than the ratio test.

EXERCISE 9

Test the following series for convergence:

Sometimes to discuss the nature of a series, we associate an iategral to the series and discuss its convergence which is easier. This method is given by Cauchy's Integral Test . which we now discuss.

i

Before introducing the integral test, you may recall some prel; e improper Integrals.

Let f be a.real valued function with domain [a, .o [.

t

suppose that f(x) is such that if(x) dx has a meaning for every t 2 a. d

Then we.write

m

If lim 4 (t) exists, then we say that the integral [f(x) dx is convergent or that it exists. In 1 3 - B

that case, we write

jf(x)dx = lim jf(x) dx. a I-+-,

m

If lim $(t) does not exist, then it follows that jf(x) dx does not exist. 1 3 - a

m

If fi~ $ (t) = w, then the integral 1 I(x) dx is said to be divergent. a

' Positive Term Series

For example, let f(x) = -$ be a function defined on the interval J l . 00 [: Then, we have

c4

Since 4 (t) = m, therdore the integral f(x) dx is not convergent. I 1

Let f(x) = - be another function defined on the interval [ I , m[. x4

Then, we have

1 Hence ?$ @ (t) = - 3'

w 1 In this case, we say that I f(x) dx is convergent and that its value is j.

, a I

THEOREM 8: CGUCHY'S INTEGRAL TEST

Let f be a real valued function with domain [ 1, m [ such that

i ) f(x) 2 0, V x 2 1 (f is non-negative) 1: i i ) x c y + f(x) r f (y), (f is a monotonically decreasing function)

i i i ) f(x) be integrable for x > 1 such that f(n) = u, i.e.' f (n) is associated : with the series I: u,.

t 45 i E i

Sequences & Serifs m /-

Then Z f(n) is convergent if m d only if I /f(x)dxis convergent and % f(u) is

divergent if and only if . a

/f(x) dx is divergent. I

i.r r f ( x ) I d. and n-I 2 un converge or d iv~ igr together.

PROOF: Since f is a deerellsing function on [ l , m[, we have f(n) I f(x) 5 f(n-1) for n-1 2 x 5 n, n=2, 3 ............

n

i.e. f(n) I I f(x) dx 5 f(n--1) for n = 2. 3 ......... n-l

Thus,

But,

Therefore, for n 2 2,

n II n- 1 n

where (s, ) denotes the sequence of partial surns of the series C u,. Therefore,

n

If we write A, = s" - I I f(x) dx, we have

n+ l a = u,, - ( f (x )dxS 0

Therefore, An+, I An V n. Thus the sequence (A,) is ml ,notonically decreasing sequence. Also A, 2 u, 2 0 V n, therefore the sequence (A,) is bounded below. Consequently (A,) is convergent.

Now

I n

S, = A, + I f(x) dx 1

The convergence of ( A, ) implies that ( s, ) and f(x) dx ) converge or diverge together.

DD

i ! C < Hence X q, and f(x) dx converge or diverge together. 1

The idea underlying Liie Cauchy's Integral Test and its proof is self-explanatory frorn the Figure I .

P ~ ~ s i l i v e Term Serifs

Fie. 1

You may note that if t he co~iditions of Cauchy's Integral Test are satisfied for x 2 k ( a Od 00

positive inleger), then x u n and J f(x) dx converge or diverge together. This can be seen n=k k

from the following example: m 4

EXAMPLE 10: Discuss the .Convsrge~lee of the p-series Z $ 7 p z 0 by n t l

using the Integral Test. 1 SO1,UTION : Here u, = --

n P 1 Let f(x) = - xp

For p > 0, f is a decreasing, positive integrable func~ioi~. So by Cauchy's Integral Test,

and p ( x ) dr converge or diverge together. *I nQ'

log x if p = 1

x ' - ~ -1 = i-, i f p + 1

4 I 1

OQ i f O < p S 1 .

I;

as x-+a, I

Therefore j $,

( I ) 6 x 1 dx converges for p > 1 and diverges for 0 < p S' 1 and hence the series "7 1

f f l t e ;, *? 5 ' converges for p > I and diverges for 0 < p < 1. -,

,,=I nQ 47 t i

Sequences and Series oa j 1" EXAMPLE 11: Test the 'convergence of the seriesx y

n=2 n ( log n I p where p > 8.

SOLUTION: Let f(x) = x(log for x > 2.

If p > 0, then f is a positive, decreasing, integrable function on [2, -1. Hence by Cauchy's

0

and I Integral Test, x converge or diverge together. n=2 n (log nIp x (log x)P

log (log x) - log (log 2) if p =1

x (log x ) ~ log x)'-p - (log 2) 1-p

2 if p#l

1-P

This shows that f(x) dx converges if p > 1 and diverges if 0 < p I 1. Therefore the given I m

converges if p > 1 and diverges if 0 < p 5 1. Series x n (log n). n=2

EXERCISE 10

Discuss the convergence of the series

In general, it is difficult to determine whether an arbitrary positive term series is covergent or divergent. There is no single universal test or method that will deal with all possible cases. We have discussed several useful tests including the popular ones like the Ratio Test and the Root Test. Most of these tests have been derived in some way from one of the forms of the comparison test. We now discuss some more tests which may be applied when all the earliest tests fail. In particular, some of these new tests will be helpful when the Ratio Test and the Root Test fail. We have selected only two tests to be discussed in our course, These are Raabe's Test and Gauss's Test.

Raabe (1801-1859) was Professor at Zurich. He made lot of important contributions to Geometry and Analysis. He gave a test for the convergence of a series of positive terms, which is often decisive when the D'Alembert's Test fails. We state the test, without proof and discuss examples to illustrate its use.

THEOREM 9: RAABE'S TEST

Let a, be a series of positive numbers such that n = l

Then i ) u,, converges if L > 1 n=l

DO

i i ) u, diverges, if L < 1 n= l

i i i ) the test Is inconclusive, if L = 1. Positivc 'Perm Series

Let us look at an example.

EXAMPLE 12 : Test the conlvergence of the series

un 2n+5md Hence - = - un+, 2n+4

un 2n+5 lim - = lim - = 1. n+ un+, n-- 2n+4

Thus, the ratio test fails. But

Hence, by Raabe's Test u,, diverges. - %I

Nolc 1hi11 IIIC c:luias I. = + and I . = - 0 ~ ;ur illstr incil~dc(i in ( i j nrrd ( i i ) in Ihc test.

EXERCISE 11

Test the convergence of the series

We end this section by discussing Gauss's Test.

Gauss (1777-1 855) an eminent German mathematician, gave a very powerful test for convergence which is applicable if Raabe.'s Test fails. It is not essential that first we apply Ratio Test, then Raabe's Test (if Ratio Test fails) and finally Gauss's Test (if Raabe's Test also fails). We can straightaway apply Gauss Test. Both D'Alembert's Ratio Test and Raabe's Test are included in this test. We only state this test and then illustrate it by an example.

THEOREM 10: GAUSS'S TEST . . -

Let x un be a series of positive terms. Suppose

where a,. b, p E R, a > 0, p > 1 and (r,) is a bounded sequence,

Then:

i) u, converges, if a > 1 n=I

; ii) x u, diverges, if a < 1, *I

I (4

1 v ' u, diverges, if 'a = 1, b 5 1 1 . n.1 I 49

Grqwcnces i31 bcrics E%A4MPH,i3 K3: Test the eoraverl,renc~ of the series

1 L Flt.fr.{* I.,, .-- -. :' .I- A ) ~ \ S : ( ~ S of -. '1.1121 :f<:i..e. j !;; ! is i~ bcralnrfcd srrlui7:rlce. Sirucc ttne coe ltii:i~..nt c.f

,:j ?I

. Discuss the conalergencr of the sedle.ui

where x, a, j3, y are rtaasieiac inaganiaers. <---.- - -..- *~"-"",-."--------.L.~-z"-*.-.~-,,------..-"-~----- ---." ,.-.. "-"

In this unii, yt1i.i have bcco ir~krodured to th.? ~?cr i~m of an infinite scrie:; :tnd the concept of c.:~nvcrg":cc of u ! ~ irifir~itc scries. Series of positive tennr;. .iGc.rc. tiiki:~ lip for r:onsider.ation, itlrt! v:tr.ric:bus tesis of conve:.gi.ilcc 1.1f itcrirs!; of positive fer!ns were c2iscussed.

l11 Section 6.2 we giivt' rl.re definiiinri i>i":rr~ ir~l'il~itc serirs i u ~ i gave a ;:-itsa:~Ii~g of thi: infinite sum as its qonvengence. Alt1.1ou~h 2n infiriite sumnletion seeam to be :xt.iifici:ll. yct hy rtsiri,g the powerful totlls of the limit concept, we are able Lo give a very concrete r~leaning ta 8.n

infhite sum. The convergence of an infinite series is shown in temls ciF the convergcnci of '

the associated (correspondiiig) sequence of partial sums'of its ternis. The basic technique is to find an explicit fohula.for the nth .term of the sequence uf partial sums. The convergence of this seqdence in~plies the convergence of t,he corresponding series.

. ~nfinite series have bcen ,divided ints Lwo n~ajor tlanscs - the one with positive terms and ' . the other consisting of the arhitilry terms. 111 Section 6.3, we deal with the positive teon szricr;. The notion of convCt'gence of these scries ha,s becn introduced,.Ciener~il tests of efovergence such as the comparison tests under different conditions and Be p-test have been i discussed in-Section 6.4. 1 In 6.5, we lrave discussed some special tests. Motable among these slre'the two basic tests - i

D'Alcrnhcl-1's Ralio Test and Cauchy's Root Test. We also studicd anotlrer important test- ! , Cwchy'r, lntcgr'll Tcst. Fjnally, we haue cliscusscd two more useful tests namely Kaiibc's I

Test and Gauss's T a t to1 tile cncvcrgenc? ol'ttle prrsitive ttm series. f i

~ 2 ) i) For the series a+(a+d) + (a+2d) -t- .... , the nh partial sum is

n - [ 2 a + ( n - l ) d ] , n . = l , 2 , 3 ,... =I, - - 2

If d =i 0, then s, = na.

If a ;t: 0, then (s,) = (na) is divergent.

If a = 0, then(sn) = (0) and it converges to 0.

If d ='0. the series diverges if a + 0

and converges if a = 0.

Now suppose d > 0. Let k > 0 be given.

n s, > k if 5 [2a+(n-l.)d] > k

Hence s, > k if 2a + (n-1) d > k.

This happens whenever (n-1) d > k -2a,

that is, :.hencvrrr n :> k+d-2a

Lel m E h S L I C ~ I that tn > li+(!-2a.

Then sn > k whenever n 2 m.

Hence ( s, ) diverges to +m. 'F'heretore, by definitit,m, the seric.:

00

[a+(n-l)d] diverges to +m.

n = l

You should be able to lake care of the case d < 0.

i i ) L.el '(s, !: he the seqlience associated with u ,sku2 + ........

am1 (t,, ) the sequence associated with us+ 4. ...... ...,..

Clearly tn-, = s, -. u,. (n 2 2)

Hence . iim t,-, = lip1 S, - U, E s .- u,, n-m n-+**

'rherefore, u2 + u3 + ...,-..., converges to s-. u,.

I rim sn = lim ( 1 --- ) = 1 . n . tar n - . ~ n+l

" 1 Hence, the series n(n.r-1)

converges to I . 11% 1

Sequences B6 Series

1 4 Hence t h i series 1 - + - - + .......... converges to -.

4 16 64 5

[Recall from LJnit 5 that l i ~n x h 0 if -1 < x < I ] . n 3 -

P+ 1 n+ l ii) Here un = - n+2 and n-i - 1h-t un = lim - = 1 + 0,

n+- n+2

P)

Hence, Z] u,, does not converge. n = I

iii). If'(s,) and (t, ) are sequences of partial surns of Cu, and

Xv, respectively, then sequence of partial sums of C (u, + v,) is (s,,+t,,) which is &vergent, since (s,,) is convergent and (t,) is divergent.

1,

Hence (u, + v, ) is divergent. n = 1

iv) Consider the series 1-1+1-1+1 -1+ ............ 1 if n is odd

Here 11, = -1 if n is even.

You can see that the nth par'tial sum s, is given by

f l i f n i s o d d s,, = 1

0 if n i s even.

Hence (s,,) is not convergent. Thus the.series 1-!+I-l+ .......... is not convergent

Howcver the series (1-1) + (1-1) + (1-1) + .......... is the series O+O+O+ ............. which, clearly, coverges to 0. , . .

IE(s, ) and (t, ) are the sequences associated with the series u, -c u, + ........... and u , + o + u , + o + ..'...........,

s,if n = 2m you can see that t, =

s, if n = 2x11-1

That is, the sequence ( t l , t2 , 5 .............. ) is

(sl, s,, s,, s2, s3, s3 .......... ). It is easy to see that

. (s,, st, s2, 3, s3, s3,.. ......... ) also converges to s.

* 1 . 1 ii) Here, s, =log (l+l) + log (li-5) + .............. + log (I+- n ) .

3 4 = log (2) $ log ( ;r ) + log ( 5 ) + ........ log (59 . 3 . 4 ' ............ n + 1

= log [2, 5 5 - 1 = log (nt 1). n

Jim s,, = log (lim (n+l)) = m, n-rm n 4 m

cd

1 Hence the series log ( It;; ) diverges to w. IF1

E 5 ) In E 3 (iv) you saw that grouping terms of a series attered its behaviour.

FIowever, such a thing will not happen in case of a series of positive terms. I :I

make this point clear, is the purpose of the exercise. c d

Suppose u,, is a convergent series of positive terms. Hence Is, ! is an increasing n=l

sequence of positive terrns bounded above, where

If the terms of z u, are grouped, and if (t, ) is the sequence associated with the new I%= I

series, then, it is easy to see that (t, ) is a subsequence of (s,) a ~ ~ d hence converges to the same limit as that of (s, ) .

To sllow that a divergent series of positive terns remains divergent rlnder grouping is easier to prove. Try it.

E 6) i) Here ti, = - 1 Take v,, = 2

3n+l.' U I1 I

Then lim 2 = lirn - = - , r vn "-+f 3n+l 3 '

00 " 1 Also z v,, = ; diverges.

n = l n = 1

1 Hence, - 3n+ diverges.

n=l

1111 ' 1 ii) Here u,. = - . Take v,, = -p

n2-4 112 lim U"= l i r n = 1.

n->- V, n-i.. n2-4

1 A I S O ~ 5 is the p-series will p = 3/2 > 1, and hence converges.

'

,=I n-

d n Thusz converges. ,,=I n 4

m

E 7) Since 2 u, converges, biz u, = 0. n=l

Hen'ce, there is M EN such that u, < 1 for n 2 M. 2

SO, for nl M, un < u, . ca

Hence, by Comparison Test I, u2converge.s. n = ~

1 1 E 8 ) i ) Here u, =$ u,,, =- (ncl)!

1 Hence, - converges.

n=l n!

. ii) Let us first use the ratio test. Here

* '-'"+I Hence bm = . l i rn

- x" Thus, when x < 1, - converges, and n=l n

Po9itive Term Scrirn

Ssq~aencei; and Series '.2

when x > 1. ' diverges. I 6 1 n

It remains to consider the case x = 1 because then ratio test fails. When x =I , the

" I series kcornes - , the.haqnonic series, which diverges.

$,=I 11

ms xn

Thus, finally, - converges if 0 < x < 1 1-1

and diverges if x 2 1.

i i i) Here un = n xW-', un+, = (n+I) xn.

a

So, nxn-I converges when x < 1 and diverges when x > I by n=l

D'Alembert Ratio Test and test Fdils for x = 1.

However, when x = I, the series becomes

1 t2+3-I-4+ ............ . which obviously di\rarge:,

5"' (20+ 1 )! Hence lirn %. = lirn ------ -- n4 .h u n-r- (2nt3)! ' 5"

= 5 lirn --- I . = o < 1.

n 3 - (21143) (2n+2)

Hence the seiies converges.

1 E 9) i) u, =L ---- , 1 . So (u, )E = -

(log n)n log n I

lirn un n = lirn --- - n+-

- 0 < 1 . n+ log n

1 Hel~ce ---- (log n)n converges.

n = l

I 1 lirn . = l im - = 0. n+ 'n n-m 11

J Hence, - converges.

,i=l n"

1 lo) Let f (x)=x lag x [log log x)p

f(x) is a decreasing, positive, integrable function of x and

f (n) = 1 for n 2 3

n log n [log (log n)lp

Cr)

I 1 Hence n log n [log (log n) lP and

n=3

converge or diverge tugother. Positive Term Series

X ( 1% (log (Inp x) 1- log (log (log 3) )

dx .--- --

-- i i f p == 1

x log x [.log (log x)]P

1-P

Hence the integral converges when p > 1 atld diverges when pS1. Hence, also, the given series converges when p > 1 and diverges whe11 p 5 1.

By Ratio Test, the series converge!; if x .< ! and diverges if x > 1.

If x = 1, Ratio Test fails.

U,, 211.1-2 - When x = I,-- ---- so that u,,~ 2n+1

1 lim n (k -1) = and so the rerics diverges by Rsaba'r Test. n.+m

Hence the series is convergent for x € I and divergent for x 2 1.

. a ( a + l ) ...,. (a+,n- I ) B(P+l) ..,. (P-bn-1) E 12) Here = - xn

1.2 ,.... 6 .... n y (yt-1) .... (yt-d-1)

Un = (n-kl)(yt-n) 1 n2+(y+I )n+y 1 1 - - --- -- - --- -_ U"+l (a + n) (P + n) x n2 + (or + P) n-kaP

lirn .!h~- = x and SO by Katio Test, series converges for x < 1 arid diverges for n-+- U,

For ,,, =, , % + =a-PL!'ily-aB) ""+I n2 + ( at-P)n + alp

:= 1 -I- - - ) + 5 k r so~ne houndai s e q a e n c e s,,. n n

By Gauss' Test, the given series converges if

PI-a-P r 1 i,e. y z a -t- I3 and diverges when yc I --cc--p~l.

i,e. y S a+p .

Hence (i) if x < 1, series converges for all positive value of a, p, y

ii) if x > 1, series diverges for all positive values of a, P, y

i i i ) if x - 1 , series converges fill. y , tx . 1) and dl\ VILJC~ f i )~ . j . 1 + ( 5

GENERAL SERIES

Structure 7.1 Introduction

Objeclives

7.2 Alternating Series Leibnitz'? Test

7.3 Absolute and conditional Convergence 7.4 Rearrangement of Series 7.5 Summary 7.6 Answers/Hints/Solutions

7.1 INTRODUCTION . P

In Unit 6, we dealt with the positive term series. Accordingly, we developed the convergence tests which are applicable only to the positive term series. But, as you are aware that, an infinite series need not be always a positive term series. In fact, an infinite series, in general, can have both an infinite number of negative tems as well as an infinite number of positive terms. The series which have both negative and positive terms may be classified into two major categories. The first category consists of those infinite series whose terms are alternately positive and negative. Such series are called Alternating Series. The other category is one in which tems need not necessarily be alternately positive and negative that is to say, the infinite series whose terms are mixed and do not follow any pattern of being positive or negative. For example, the infinite series

has alternately positive and negative terms, whereas the infinite series

does not follow any specific pattem.

The question, therefore, arises: How to test the convergence of such infinite series? The convergence tests discussed in Unit 6 are not suitable enough for the purpose because these tests in their present form cannot be applied to these series. Hence, we have to either modify these tests or devise new tests of convergence for such series.

'

We shall discuss an important test applicable to the series with terms alternately positive and negative. This test is known as Leibnitz test.

To determine the convergence of other infinite series having no pattern of p~sitive and negative terms, we shall introduce the notions of Absolute and Conditional Convergence of the infinite series. Finally, +e shall have a brief discussion on the method of Rearrangement'of series. There are a few more categories of general series viz. Power-Series which we intend to discuss in some other advanced level course in Analysis at a later stage.

Objectives

Therefore, after studying this unit, you should be able to

* recognize an Alternating Series

m- apply the Leibnitz Test to know the convergence of an Alternating Series I

* identify an absolutely convergent series and a conditionally Convergent Series

* have an idea about the method of rearrangement of the terms of an infinite series to '

know its convergence or divergence.

- ---- --------- -- -- Gcnerol Series

7.2 ALTERNATING SERIES

In this section, we shall consider serieb whose terms are alternately positive and negative. Such series are called 'Alternating Series'. For example, the infinite series

and -1 "I- 2- 3 + 4 - 5 + ..............

are alternating series. Formally, we define an alternating series in the following way : m

DEFINITION 1 : An infinite ser iesz u, i s called an Alternating Series if n= l

any two consecutive terms of the series are of opposite sign. (Y

An Alternating Series may, thus, be writtcn as (- 1) " + I u, = u, -u, + u, - u4 + ............ 1Cl

where, each u, =. 0.

If the first term i s negative, then it can be written as

m

G (-1) l1 U, = -11, "I- u2- us -+ ii4+ .. .......,..... 1 ~ 1

The second series can be obtained from the first if you multiply each term of the first series by -1. There:. ;e, it is enough to discuss the convergence of the first series. Thcre is a vely simple test for the convergelice of itn Alternating Series provideti there is a

............... sequence (s,) of partial sums of u , , u ,. which is ~nonatonically decreirsing and converg

e

nt to 0. This test is known as Leibnitz Tes.t after the name of Leibnitz, the eminent Gernlan mati~cmatician.

THEOREM 1 : (Leibnitz Test). - Let (- 1) + ' u u , be an Alternating Series sucli that

n = l

...... (i) u i > O V i = 1, 2 , 3 ,

( i i ) u , 2 u, 2 u, 2 ............... i,e.(u,) is a monotonically decreasing sequerice

( i i i ) l i m u,, = 0. n-3-

m

Then the series (- 1) + I u, is convergent. n = l

PROOF : Note that we are assuming that the odd terms of the alternating series are positive and the even terms are negative. Let (s,,) be the sequence associated with the series

Let us first consider the partial sunis with odd index namely those ending in positive terms 1

i.e.

S I , S3, SS, ............... Then, we have s l = u , >O

s j = U , - U? + u3 = sI -(uZ -4 ) 5 S, since u2 2 u3. Similarly, we have Sg=Ui-U2-+U3-Uj+Ug

= S3 + Us - U4

sS - S 3 5 uS-u4<O i.e. ss 5 s3 5 s,.

In general, -

'2n.t.1 - '2"-1 - ( ~ 2 " - uzn+l) S Z ~ - I . 1

Scquenses & Series

This shows that (s,,-, ) is ,t mol~otonic decreasing sequence. Also

S?,-l = (u! -u?) + (u) -Uq ) +.....-.+ ( u~n -~ - ~ ? n - ? ) + U2n-I 2 0.

This shows that (s2%,) is bounded below with 0 as the lower bound.

Therefore (s2,,-]) is a monotonic decreasing sequence which is bounded below. Hence it is convergent.

Suppose converges to,a li~nit s.

Then :?,m_ s ~ ~ - ~ = s.

In the same way, you can compute for the sequence of even partial sums and show that

S2n+? 2 s2,,

i.e. the sequence (s2") OT even partial sums is monotonically increasing.

Also

But u,, I; u, . V i = 2, 3, ... Therefore u2 - u, 2 0

...................... Hence

s, = u, - some positive quantity (number)

which, in turrm, shows that s, < u, , V n.

i

I 1'11~s (st,) is bounded above.

Hence (sZn) is convergent. Suppose i t converges to a limit t. Then

l i rn "->- SZn = t.

Since u ~ , = s,,, - s2,-,, we have, therefore, by the condition ( i i)

Thus s = t.

Thus both (sZ,,) and ( s~ , -~ ) converge to the same limit s.

Hence

lim s, ,+- -" s*

Finally, we shall show that the sequence ( s,) converges to s. Let E > 0 be given. Sipcg the ,

sequence (s2,, )converges to s, therefore there exists a positive integer n l , such that

Similarly, given E > 0, there exists a positive integer m, such that

1 s s < a v (2n-1) > m,

Thus i t follows that

I s,, -s / < E V n > max. (m,. m2)

This implies that ( s, ) converges to s or

lirn S, = S. n-+-

Therefore it follows that the alternating series <General Series

11, - U., + u3 -uq + ...............a

converges to the limit s, which, in fact, is the sum of the series. -- - -

EXERCISE 1

Prove Theorem 1 for the Alternating Series of the form m

(-1)" u, [that is, where the odd terms are negative and even terms are n= l

posit ive].

From Unit 6, you know that the condition pya un = 0 is necessary for the convergence of c4

every infinite series u,. But according to Leibnitz Test, if the given infinite series is an n=l

alternating series decreasing in absolute values, then the conditioniFm 11" = 0 is also

suf'icient for the convergence. Let us now study some examples and exercises:

EXAMPLE 1: Test the convergence of the Alternating Series

SOLUTION: This series is known as the Alternating Harmonic Series,

In this series the conditions of the LRibnitz Test are satisfied. Here (i) each u i > O i = 1,2, .........

(ii) u l > U, >u3 > u4 ...........

Hence, the series is convergent. The limit or sum of this series is well-known and is equal to log 2.

(See Example 6 in this section).

EXERCISE 2

0 n Show that x. (-l)n+l - 3n +2 diverges.

n = l

EXAMPLE 2: Test the convergence of the Alternating Series

L L .. SO1,UTION: Here since 3 > 32 > 33> ..........., therefore first aud second conditions of the

Leibnitz test are satisfied. 1

However, :2m u, = !z 3. = 1 + 0. (R~call from Unit 6).

Since the third condition of the Leibnitz test is not satisified, therefore the given series is divergent.

EXERCISE 3

For what values of p does the series , 1 1 1 1 ............... + converge?

Now cansider, once again the Harmonic Series I

Nrqurnres and Serics You know that this series diverges. But the Alternating 'Harmonic Series (as discussed in Example 1) namely .

converges. Thus, we have a series that converges only because some of its terms are negative. If all the negative terms arc replaced by the corresponding positike terms, then the convergence is demolished. To study this phenomenon in a !nore general way, we introduce the notions of absolute convergence and conditio~~al convergence in the next section.

7.3 ABSBLIJTE AND CONDITIONAL CONVERGENCE

Consider the following two series:

and

By the Leibnitz Test, both series converge.

Again co~lsicler the two series

1 1 1 1 iii) I t r + - + - -c - .............. and

2 4 8 1 6

obtained from (i) and (ii) by replacing each teim by its positive value.

The series (iii) converges, while the series (iv) diverges.

This leads us to divide the coIlvergant series into two classes, namely, the abso11.1tely convergent series and the conditionally convergent series, which we define as follows:

00

DEFINITION 2::,et u, be a n infinite series of real numbers. n=l

i ) If lunl converges, then we say that the series u, converges n= l n = l

absolutely.

i i ) If Z u, converges b u t Z 1 unl diverges, we say t h l Z u, converges n=l n= l n= l

conditionally.

Thus the series (i) converges absolutely, wnile series (ii) converges condirin- '

Nore that in (ii) we have defined Z u, to be conditionally conve. .,, 1s convergent IFI d

m 4 w

but 1 u,, / is divergent In (i) we have defined u,, to be absolute& coqverpnt if re1 ml

Iun/ is convergent. but we have not said anything about the behaviour o r . - " ( - 1 )n t l

EXAMPLE 3: (i) the series is absolutely convergent. n = ~ n2

( i i ) The series

" ( ) n t l 1 (L- (Alternating Harmonic Series) n = l

and

are condi t io~~al ly convergent.

The following theorern provides thar we can produce exa~i~plcs of ;tbholutcly convel.geu[ series by changislg i3,lpebl';iic sig~;s ill- somc or all of the lenils of ;\ collvergimt ~t'l.ii~s of pi'sitive terms.

THEOREM 2: If an infinite series is absolutely cornPe~-gelt;, i l len it is convergent.

PROOF: Let 2 u,, be an absolutely convergcn, series i.e. ; LI,) is collcrrgent. 1-hcn MT

have to prove [hat C u,, is convtrge1;t.

Let ! s, ) be the sequence of partial s\ims of Z u,, . Then

It is enough to show rhal (s,,) is a Cnuchy sequence.

Let ( t , , ) be the aeq~lellce associated with the series 1 1 u,, 1 , Sincc Z LI,, 1 is c1111verge1ll.

therefore (r,,) is also convergent. Thu\ ( I ,,) is a Cauchy sequcncc. In olher wol.rls. for an E > 0, there exists H positive inteper 111 such \li;\~

/ t, - tk I < E f i r rl rn. k ;. nl.

Suppose n > k. Then

Which shows t h n ~ ( s,, ) is ;i Cauchy sequence. This colnplete!, [he piaoof of !he llicorem.

Thus every absolutely convergent series is convergent. Tlw corivcrsc. howcvcr. is ncll rue. That is to say that if a series is convergent, thrn i t lliay ti01 be absolulcly converrrent. Can you give an example ? Try i t .

EXAMPLE 4: Ttst the absolute and co~lditional convergence of the series

I SOLUTION: Here 1 u , , = 5----

n= 1 ,,=I -11+ 1

I Letv, = - for n = 1 , 2 , 3 ...............

n

I Then the series v,, =x ; is diverselit.

11-1 11-1

Also,

I l n 1 lim --= lim---. tI )LU vll 11-*.2n+ I - 2

I-Iencc by comparison test. il I'ollows that

1 Sequences and Series rn z 1 u, 1 is divergent. Thus n= 1

- (-1)11+l z 2,,+, is not absolutely convergent. 11= 1

L However, since -- > - ~ n . 2n+l 211-1

1 - (-I)"+' and lim- = 0, therefore by Leibnitz test, is convergent. In other words, it

n+.2n+l ,=, 2n+l is a case of conditional convergence.

Now try the following exercises.

EXERCISE 4

i) Test the convergence and' absolute convergence of the series

i i ) Determine the values of p for which the series

1 1 - - - 1 1 1 - 2'

+ - 3 I)

4P + ...,......,.... converges conditionally.

All the tests of convergence of infinite series discussed in Unit 6 for series of positive term can be used to decide absolute convergence of general series.

=a

c o s n x EXAMPLE 5: Test the series -F 9 x E R, for convergence.

n= 1

COS 11 X SOLUTION: Here u, =n,

1 Butx 2 converges, (Recall from Unit 6, why it so?)

ii=l

m

This means that the series u, is absolutely convergent. IF= I

m m

Hence, the series u, i.e. cos n x

is convergent. IF1 .*I n2

m cos n x

Therefore I: - converges absolutely for all x G R. rI=1 n2

EXERCISE 5 ca

i ) Let un be absolutely convergent and (5,) be a bounded sequence. n=l

DO

Prove that s, un is also absolutely convergent, n=l

i i ) Can a series of positive terms be conditionally convergent?

7.4 REARRANGEMENT OF SERIES

You know that to find the sum of a finite number of real numbers, the order in which they are added does not matter. But this is not the case when you have-to find the sum of infinite series of numbers. The order in which the terms occur in an infinite series may affect i ts nature and the sum i.e, the convergence of the series.

In this section, we shall discuss this aspect of the infinite series. This is, also, sonletirnes referred to as the Rearrangement Convergence, where every rearrangement of the series converges. We first have the following definition:

DEFINITION 3: Let x u, be an infinite series. Let K be a 0ii:*-t0*9ne n=l

w

function from N onto N. Then C uXb, is said to be s rearrangement of 0-1

1 1 1 For example I + - + - + - - - + .........,.. is a rearrangcrnent of the series 3 2 5 7 4

We state two results (without proof) which will indicate the effect on the convergence of i! series due to the order in which the ternls occur in the series.

m

I If C un is an absolutely convergent series converging to s, then n= l

m

every rearrangement of u, also converges to s. t l = I

Thus, the order in which the ternrs occur is immaterial in absolutely convergent series.

What of conditionally convergent series? To answer this question, we state the following result of Riemann.

m

I1 Let u, be a conditionally convergent series. Given any a E R . n=l

m

there is a rearrangement of the series x u, which converges to a. n= 1

Let us give an illustration of a Rearrangement and show how the sun1 or tlie convergence is altered:

EXAMPLE 6: Show that

Evaluate the sum of the rearranged series

SOLUTION: Set a number 5 as

Then this is a decreasing sequence of positive numbers. It is a well-known result that the sequence (rn) converges to a limit y called Euler's constant and is approximately equal to 0.577). Let us, therefore, assume that lirn r,, = r

n - m

Let (sn) denote the sequence of the partial sum of the series

and (tn) denote the sequence of the partial sum of the rearrangement of the series namely

Then, we have

General Serics

Scquenres and Serico\

= [r,,, -I. log 21-11 - ('r, + l o g n]

= [r2,, - r,,] + log 711 - log n

= [r:,, - r,,] + log 2

Since (r,,) is c ~ ~ ~ L ~ ~ , v E I I [ , therefore (r,,) is a Caucl~y Sequence. Consequently, therc exists I

rnE. N such that / r2,, - r,, < E for n 5 rn where E>O is any number.

This in~plies thai

%,, = log 2. 1,-*w,

Now. i t is casy to show [hat

Jim s,, = lug 2. I,->-,

For the heclijenc~> r ,,. we have a

Again since ir,,) is a Cauchy Sequence, therefore,

1 1 1 Since t,,,, = t3, + -- and t,,,, = t,, + -- + -. - .

.In +l 4 n + 1 4n+3

Therefore

3 lirn t , = - log 2. n+- 2

This shows that the arrangement of a conditional convergent series lnay change its sum.

--"-- - - -------

EXERCISE 6 m

Suppose un is a series of positive terms diverping to +w. Show that n= l

D1

every rearrangement of 2 u, also diverges lo , -. n=1

------.-- -pH -----

7.5 SUMMARY

This unit has been mili~ly concerned with series of al-bilrc~l-y real numbers. A very importatlt exatnple of such a series is an Alternating Series. To test convergence of an Allematihg Series. wc apply the most useful test known as Leibnitz 'Test, which we hnvc discussed in

Section 7.2. In Section 7.3, we dealt with another category, the series of arbitrary terms. the one which does not'follow any pattern of its terms. Such series may be Absolutely Convergent Series and conditionally convergent series. Absolutely Convergent Series are stable under any reamngement, in the sense, that no rearrangement can disturb the convergence or sum of an absolutely convergent series. On the other hand, you can make a conditionally convergent series behave as you wish by a suitable rearrangement which we defined and demonstrated in Sectioi~ 7.4.

Precisely speaking, in this unit we have studied three notions related to the infinite series of arbitrary terms namely

i) the convergence of the Alternating Series

ii) the absolute and conditional convergence of the series

. iii) the rearrangement convergence, where every rearrangement of the series converge.

7.6 ANS WERS/HINTS/SOLUTIONS

E 1) 'I'he alternating series is - -ul + u2 -u3 + uj ................ Wher:! u, . u2, fi3, ................ are positive ilurnhers satisfying

..................... u, 2 u2> u3 2 :.. ad

lim u, = 0. n--t-

Let ( t, ) be the sequence of the partial sums of this series. Also, let ( s, ) be the sequence of the partial sums of the series

It is obvious that t, = -s, for each n.

By Theorem 1, (sn ) converges. Hence ( t, ) = ( -s, ) also converges.

n E 2) Here u, = (-I)"+' -

3n+2

Since vyw u, is not equal 0, therefore the series is divergent.

E 3) Case (i) Let p > 0. Then

1 lim - = 0. n-w nP

Hence the series converges.

Case (ii) Let p = 0. Then the series is

But !?- un = - lim (-1In does not exist. n+=

Therefore, the series diverges.

Case (iii) Let p < 0. Again

lim un = lirn n-+m n+- nP

does not exist. Hence the sbries diverges.

(;enera1 Series

Sequences and Series 1 1 1 + ---, and E4) i) Since - + ;;5 > - n3 (II+I)~ (n+1)5 1 1

lim n-w (-+;;j)=O. n3

therefore by Leibnitz test,

converges. - " I = ' l The series Z (ij + $) =Z 7 + 7 - also converges.

n=l ,=I n ,SI nS 00

ThusZ ( - l )n t l ($ + L, is absolutely convergent. n=z n3

ii) In E 3) we have senn that

converges only when p > 0.

We also know that the p-series

converges only for p > 1.

is conditionally convergent for 0 p S 1.

E 5) i) Let Isn( < k for each n E N,

Let (a,) and (t,) be the sequences ~ssociated with the series

ca

(ud and 2 Isnu,, I, respectively. n- l n-1

Since lenj converges therefore (a> is an increasing sequence of positive

numbers bounded above, m

TO show that n-I 2 Isn u,I converges, we have to show that (t,) is also bounded

above,

Note that t, = Is, u, I + Is, u, I + ... + I s,, u,, I = I s I I I u I I + I s 2 1 l ~ 2 I +..,+ Isnl IunI

Since (sn) i s bounded above, therefore (ka,,) is also bounded above. Hence (tn) I

- . €.--- - - - - I

is bounded above. Thus x s, u, converges absolutely. I F 1

ii) No, Justify it. CQ DJ

E 6) Suppose a rearrmgament u, (,, of u, converges. IEl rF1

09 - No@ tha ts u, is itself a rearrangement of x uu,(;,.

n=l -1 m

Hence, u, would converge, contradicting the hypothesis, that -1

u, diverges. ~1

m

Hence, every rearrangement of u, also diyergec. I * *I

REVIEW General Series,

In this Block, you have studied the notion of a sequence and its convergence. Also, you have been introduced to the infinite series and their convergence. The infinite series have been classified into two types of series namely positives term series i.e. the series with positive terms only and the series with a mjx of both positive as well as negative2erhs. Accordingly, various tests for the convergence of the corresponding series have been discussed. You should now attempt the following self-test questions to ascertain whether or

, not you have achieved the main objectives of leanling the material in this block. You may compare your solutions/ answers with the ones given at the end.

1 Given below are the sequences whose nth term is given. Write the range of each of these sequences and determine which of these are hounded and unbounded.

(ii) (-IT I

(iii) nn

nn (iv) cos .-

3

1 (v) (1 ,k --In

2 For each of the following sequences determine whether it converges or not. If it converges, then give its limit.

n2 + 3 (ii) s, =----

n 2- 3

(iii) s, = 2 --I1

(iv) s, = 3 + (-1)"

(-4J (v) S" "

11

3 Determine which of the following sequences are monotonic '?

3n+2 ('1 2n-5

n 2 + 1 (ii) -- n 2n2- 1

..(iii) ~~m I I (iv) I + 7 I n

(v) n+(-1)" )r

4 Find the sum of the first n terms of the following series and hence decide whether each series converges or diverges. If the series converges, find its sum :

1 2 4 ; (i) - + - +- + ............'

3 9 27

(ii) 2 - 4 + 6 - 8 + ................ 1 1 i ) + + ...........

3 4 ........ (iv) log 2 + log 5 + log - + -. 3

. -

I UNIT 8 LIMIT OF A FUNCTION I STRUCTURE

i 8.1 ~ntroduction Objectives

8.2 Notion s f Limit Finite Limits Infinite Limits

' 8.3 Sequential Limits 8.4 Algebra of Limits / 8.5 Summary 8.6 Answers/Hints/Solutions

I

In Unit 5, we dealt with sequences and their limits. As you know, sequences are functions whose domain is the set of natural numbers. In this unit, we discuss the limiting process for the real functions with domains as subsets of the set K of real

I numbers and range also a subset of R. What is the precise meaning for the intuitive idea of the values f(x) of a function f tending to or approaching a numbar

I A as x approaches the number a? The search for an answer to this question shall enable you to understand the concept of the limit which you have used in calculus. We shall give a rigorous meaning to the intuitive idea of the limit of a function in 1 Section 8.2. The relation between the limit of a function and the limit of a

I sequence is established in Section 8.3. The effect of algebraic operations of 1 addition, subtraction, multiplication and division on the limits of functions is I examined in Section 8.4. It wilt then be extended to study the effect of these I

' / algebraic operations on the continuity of a f~~nctioii in Unit 9.

Objectives I

I

I Thus after studying this unit, you should be able to 1 I

i e dcfine limit of a function at a point and find its value

1 e know sequential approach to limit of a function e find the limit of sum, difference, product and quotient of functions.

- I 8.2 NOTION OF LIMIT I

-- The intuitive idea of limit was used both by Newton and Leibnitz in their independent invention of Differential Calculus around 1675. Later this notion of limit was also developed by D'Alembert. " When the successive values attributed Lo a variable approach indefinitely s fixed value so as to end by differing from it by as little as one wishes, this last is called the limit of all the others."

Consider a simple example in which a function f is defined as

f(x) = 2x + 3, V x E R , x # 1.

Give x the values which are near to 1 in the following way:

When x = 1.5, 1.4, 1.3, 1.2, 1.1, 1.01, 1.001

f(x) = 6, 5.8, 5.6, 5.4, 5.2, 5.02, 5.002

/ When x = .5, .6, .7, .8, .9, .99, 399

Limit and Contirluity You can form a table for these values as follows:

The limit of a function f a t a point a is meaningful only if a is a limit point of its domain. That is, the condition f (x ) ?, cr as n + a would make sense only when there does not exists a nbd. U of a for which the set U n Dorn(fl\la) is empty i.e..

You see that as the values of x approach 1, the values of f(x) approach 5.This is expressed by saying that limit of f(x) is 5 as x approaches 1. You may note that when we consider the limit of f(x) as x approaches 1, we do not consider the value of f(x) at x = 1. .

. . . . . - u ~(Dorn (f))'. Thus, in general, we can say as follows:

Let f be a real function defined in a neighbourhood of a point x = a except possibly at a. Suppose that as x approaches a, tjhe values taken by f approach more and more closely a value A. In other wirds, suppose that the numerical difference hetween A and the values taken by f can be made as small as we please hy taking values ~f x sufficiently close to a. Then we say that f tends to the limit A as x tends to a. We write

f(x) -b A as x -k a or t[lz fGa) = A.

This intuitive idea of the limit of a fimction can be expressed mathematically as formulated by the German mathematician Karl Veierstrass in the 18th Century. Thus, we have the following defmition:

DEFINITION 1 : Limit of a Function L e t a function f be defined in a neighbourhood of a point 'a' except possibly at 'a'. The function f is said to tend Pa or'approach a number A as x tends to or approaches a number 'a' if for any E > 0, there exists a number 6 > 0 such that

(f(x) - A1 < E for 0 i JX - a) < 6.

4 We write it as lim X-+R f(x) = A. You may note that

Iqx) - A1 < E for 0 < IX - a1 < 6.

can be equivalently written as (see Unit 3)

f(x) € ] A - E , A + E [ for x E] a - 6, a + 6 [ and x * a.

Geometrically, the above definition says that, for strip S, of any given width around the point A, if it is possible to find a strip Sn of some width around the point a such that the values that f(x) takes, for x in the strip Sa (x # a), lies in the shaded box formed by the intersection of strips S, and S n , then lii f(x) = A.

This is shown geometrically in Figure 1 below. The inequality 0 < Jx - a1 < 6 determines the interval ] a - 6, a + 6[ minus the point 'a' along the x-axis and the inequality If(x) - A( < E determines the interval ] A - E, A + E[ along the y-axis.

y-axis+

. .

A++-.- . . A--

A-c---

,

0

............. ... ................................................. ........... .............................. ............................................................. " _._ - .......... ........................................ .... .............. ............................................... ........................ ....-... " " ....-....... .. .-...

. . , . Fig. 1 . . .

. ,

EXAMPLE 1 : Let a function f: R-+ R be defined as . ,

6 f(x) = x2, Q x ER. ' 1

. . -

Find its limit when x -b 2.

SOLUTION : By intuition, it follows that

Let us verify this with the help of E-6 definition. In other words, we have to show that for a given E > 0, there exists a 6 > 0 such that

Suppose that an E > 0 is fixed. Then consider the quantity /qx) - 41, which we can write as

If(x) - 41 = (xZ- 41 = I (X - 2)(x + 2)l.

Note that the term (x - 2) is exactly the same that appears in the 6-inequality in the definition. Therefore this term shoilld be less than 6. In other words,

We rest~ict 6 to a value 2 so that x lies in the interval ] 2 - 6, 2 + S[ c ] 0, 4[. Accordingly, then (x + 21 < 6. Thus, if 6 ;s 2, then

I x - 2 1 < 2 3 O < J x + 2 1 < 6 ,

and further that

I X - 21 < 6 4 2 3 I X + 21 (X - 21 < 6 I X - 21 < 66.

If 6 is small then so is 66. In fact it can be made less than E by choosing 6 suitably.

Let us, therefore, select 6 such that 8 = min.(2, d6). Then

0 < Ix - 21 < 6 a If(x) - 41 < 6 Ix - 21 < 6. 6 .S 6. ~ / 6 = E.

This completes the 'solution.

Note that the first step is to manipulate the term Jf(x) - A( by using algebra. The second setp is to use a suitable strategy to manipulate If(x) - A1 into the form

Ix - a\ (trash)

where the 'trash' is some expression which has the property that: it is bounded provided that 6 is sufficiently small. Why we use the term 'trash' for the expression as a multiple of Jx - a[? The reason is that once we know that it is bounded, we can replace it by a number and forget about it.

In Example 1, the number 6 arose by virtue of this 'trash'. If you take 6 5 3 (instead of 6 S 2), you,can still show that 6 will be replaced by 7. In that case you can set 6 as

and the proof will be complete. Thus, there is nothing special about 6. The only thing is that such a number (whether 6 or 7) has to be evaluated by the restriction placed on 6.

Finally, note that in general, 6 will depend upon E.

Now you should be able to solve the following exercises:

EXERCISE 1

For a function I: R -s R defined by f(x) = x2, find its limit when x tends to 1 by the E - 6 approach.

EXERCISE 2

x t - x + 18 Show that Lim = 4, using the E - 6 definition.

x*z 3x - 1

Limit of o Ponrtiw

Ei'mit and Continuity In Unit 5, we proved that a convergent sequence cannot have more than one limit. I n the same way, a function cannot have more than one limit at a single point of its domain. We prove it in the following theorem:

THEOREM 1 : If Em. f(x) = A, lim f(x) = B, then A = B. x - a x - a

PROOF : In short, we have to show that i f lirn f(x) has two values say A and B, x - a

then A = B. Since lim f(x) = A, lirn f(x) = B, given a number E >' 0, there x - a x - a

exists numbers S 1 , A2 > 0 such that I f(x) - A1 < '&/2 whenever 0 < I x - a 1 <

and (f(x) - B ( < 8/2 whenever 0 < Ix - a ( < 62.

If we take 6 equal to minimum of 6, and B2, then we have I f ( x ) - A / . < &/2and ( f(x)- BI < E/2 whenever0 < ( x -- a ( < 6.

Choose an xo such that 0 < 1% - a1 < 8- Then

[ A - B I = ( A - f(xo) + f(xo) - B ( 5 ( A ;- f(xo) ( + I f'(x0) B (

6 &/2 + E/2 = 8.

E is arbitrary while A and B are fixed. Hence I A - BI is less than every positive number E which implies that [ A - BI = 0 and hence A - R. (Fcrr otherwise, if A # B then A - B = C # 0 (say). We can choose I <: (CI which will be a contradiction to the fact that 1 A - BI < & for every & > 0.)

In the example considered before defining limit of a function, we allowed x to assume values both greater and smaller than 1. If we consider values of x greater than 1 that is on the right of 1, we see that values of f(x) approaches 5. We say that f(x) tends to 5 as x tends to 1 from the right. Similarly you see that values of f(x) approach 5 as x tends to 1 from the left i.e. through va1uc.s smaller than 1. This leads us to define right hand and left hand limits as under : DEFINITION 2 : Right hand limits and Left hand limits Let a function f be defined in a neighbourhood of a point 'a' except poisibly at 'a'. I t is said to tend to a number A as x'tends to a number ';s' from the right or through values greater than 'a' if given a number E > 0, there exists a number 6 > 0 sucli that

If(x)- A1 < & f o r a < x < a + 6.

We write. it as lirn f(x) = A or lirn f(x) = A or f(a +) = A.

X- a+ x-n+O See figure 2(a) . The function f is said to tend to a number A as x tends to 'a' from the left or through values smaller than 'a' if given a number E > 0, there exists a number 6 > 0 such that

(f(x) - AJ < E for a - 6 < x < a

We write it as

l im f(x) = A or lim f(x) = A or f(n -) = A, x-a- x-a - 0

>'See figure 2(b). Since the definition of limit of a function employs only values of x differeat from

a 'a' it is totally immaterial wbat the value of the function is at x = a or whether f X

is defined at x = a at all. Also it is obvious that lim f(x) = A if and only if Fig. Z(b) f (a+) = A, f(a-) = A. x - a

This we prove in the next theorem. First we consider the following example to illustrate it. . EXAMPLE 2 : Find the limit of the function f defined by

xZ 4- 3x f(x) = for x # 0

2x when x - 0

0 SOLUTION : The given function is not defined at x = 0 since f(0) = - which is meaningless. 0

x + 3 If x $ 0, then f(x) = --- . Therefore

2

Right Hand Limit = lirn f(x) x-d+o

.= lim (0 + h) + 3

. . h-0 2 (k > 0)

Left Hand Limit = lirn f(x) x-0-0

= lim f(x) = (0 - h) + 3

2 Ol > 0)

h-0

Since both the right hand and left hand limits exist and are equal,

You can similarly solve the following exercise.

EXERCISE 3 Find the limit of the functioa P defined as

2x2 + x f(x) = , x # 0 when x tends to 0.

3x

We, now, discuss the theorem concerning the existence'of limit and that of right and the left hand limits.

THEOREM 2 : Let.f be a real function. Then

lim f(x) = A if and only. if Iim f(x) and lirn f(x) x - I x-n+ x-a-

both exist and are equal to A.

PROOF : If lim f(x) = A, then corresponding to any E > 0, thirc'exists a x- a+ ._ _ _ . - I

I'

6 > 0 such that

If(x) - A1 < E whenever 0 < Ix - a1 < 6

i.e. I&))-A1 <$whenever a - 6 < x < a + 6 , x f a

This implies that 1 f(x) - A ( < E whenever a - 6 < x < a

and If(x) - A1 < 6 whenever a < x < a + 6,

Hehce both the leh hand and right hand iimits exist and are equal to A. Conversely, if f(a+) and f(a-) exist and are equal to A say, then corresponding to E > 0, there exist positive numbers and A2 such that

~f(x)-A1 < E whenevera < x < a + 1

and If(x) - Aj < & whenever a - b2 < x < a.

Let 6 be the Jtiinimum of 61 and 62. Then

(f(x) - A[ < E whenever a - 6 < x < a + 6, x # a

: i.e. (f(x) - AJ < 4 if 0 < Ix - a1 < 6 . 'which proves that

, lim f(x) exists and lim f(x) = A. x-a x-a . 9'

Idslfi surd Colbtlndtg E LE 3 : Consider the function f defined by

Flnd its limit as x -- 1.

- SOLUTION : Note that f(x) is not defined at x = i . (Why ?).

x2 - 1 For any x $ 1, f(x) = - = x + l .

x - l

lim f(x) = lim (x + 1) = 2 x- 1 + x-I+

lirn f(x) = lim (x -t- 1) = 2 x-I- x-l-

Since lirn f(x) = lim f(x), by Theorem 2, lim f(x) = 2, x - I + x-1- x- 1

lirn f(x) = 2 can be seen by 8-6 definition as follows : x- 1

Corresponding to any number E > 0, we can choose 6 = E itself. Then, it is clear that

From Theorem 2, it follows that f ( l+ ) and f(1-) also exist and are both equal to 2.

EXAMPLE 4 : Let f: R -- R be defined as

Find its limit when x - 0.

SOLUTION : You are familiar with the graph of f as given in Unit 4. It is easy to see that lirn f(x) = 0 = f(0 + ) = f(0-). The fact that f(0) = 3 has neither any

x-0

bearing on the existence of the limit of f(x) as x tends to 0 nor on the value of the lim f (~) . x- 0

Now try the following exercise:

EXERCISE 4

Find, if possible, the limit of the following functions.

when x tends to 2.

- 1 (ii) f(x) = - x # 0

e l lX + 1

when x tends to O. I

I

EXAMPLE 5 : Define f on the whole of the real Line as follows:

1 i f x > O ,

7 0 -1 if x < 0. ,

Find its lirnit when x tends to 0.

SOLUTION : Since f(x) = 1 for all x > 0,

Limit of a Function

Similarly f(0- ) = -1. Since lim f(x) f lim f(x), !$ f(x) does not exist. x+l+ X-+I-

Now, we give another proof using E - S definition.

I For, if lim f(x) = A, then for a given E > 0, there must exist some 6 > 0, such that x - 4

1 If(x) - A1 < E. Let us choose x, > 0, x, < 0 such that Ix,] < S and Ix,l < 6. Then

1 2 = If(x,) - f(x,)/

I I If(x,) - AJ + ]A - f(x,)l I

< 2 ~ , (because ( x,- 0 1 < S and (x, - 0 I < 6)

for every E which is clearly impossible if E < I . Non-existence of lim f(x) also follows from Theorem 2, since f(O+ ) # f(0- ).

x 3 0

I The above example shows clearly that the existence of both f(a + ) and f(a- ) alone is I not sufficient for the existence of linl f(x). In fact, Cor lirn f(x) to exist, they both should

~4 X 3 1

be equal.

1 Now consider, the function f defined by f(x) = y for x + 0.

r The graph of f looks as shown in the Figure 3; You know that it is a rectangular l

hyperbola. Here none of the lirn f(x) and lirn f(x) exists. Hence lim f(x) does not exist x 4 + x 4 + x+O

This can be easily seen from the fact that 1/x becomes very large numerically as x approaches 0 either from the left or from the right. If x is positive and takes up larger and larger values, then values of l /x i.e. f(x) is positive and becomes smaller and smaller. This is expressed by saying that f(x) approaches 0 as x tends to ao. Similarly if K is negative and numerically takes up larger and larger values, the values of f(x) is negative and numerically becomes smaller and smaller and we say that f(x) approaches 0 as x tends to -m. These two observations are related to the notion of the limit of a function at infinity.

Let us now discuss the behaviour of a function f when x tends to. co.

Let a function f be defined for all values of x greater than a fixed number c. That is to say that f is defined for all sufficiently large values of x. Suppose that as x increases indefinitly, f(x) takes a succession of values which approach more and more closely a value A. Further suppose that the numerical difference between A and the values f(x) taken by the function can be made as small as we please by taking values of x sufficiently large. Then we say f tends to'the limit A as x tends to infinity. More precisely, we have tlie following definition:

DEFINITION 3 : A function f tends to a limit A, as x tends to infinity if having chosen a positive number E , there exists a positive number k such that

The number E can be made as small as we like. Indeed, however small & we may take, we can always find a number k for which the above inequality holds. We rewrite this definition in the following way:

A function f(x) -- A as x - w if for every E > 0, there exists k > 0 such that

If(x) -.A[ < E for al lx s k.

We write it as , Lim f(x) = A. x - m

This notion of the limit of a function needs a slight modification when x tends to -a. This is as follows:

We say that Lirn f(x) = A, if for a given & > 0, there exists a number k < 0 such that - -"

(f(x) - A1 < E whenever x 5 k.

We write it as lirn f(x) = A. x - -m

Instead of f(x) approaching a real number A as x tends to + oo or -m, we may also have f(x) approaching t 03 or ~ o o as x tends to a real number 'a'. For example, if f(x) = l/x Z, x # 0 and x takes values near 0, the values of f(x) becomes larger and larger. Then we say that f(x) is tending to + OJ as x tends to 0. We can also have f(x) tending to + m or -m as x tends to + oo or -03. For example f(x) = x tends to -t cx, or -a, as x tends to + m or -m. Again: the function f(x) = -x tends to + m or -m as x tends to -m or -t a. We formulate the following definition to cover all such cases of infinite limits.

DEFINITION 4 : Infinite Limits of a Function Suppose a is a real number. We say that a function f tends to + m when x tends to a, if for a given positive real number M there exists a positive number 6 such that

f(x) > M whenever 0 < Ix - a1 < 6.

We write it as

lirn f(x) = + m. x - a

In this case we say that the function becomes unbounded and tends to +w as x tends to a.

In the same way, f is said to -oo as x tends to a if for every real number -M, there is a positive number 6 such that

f(x) < - M whenever 0 < Ix - a1 < 6.

We write it as

lim f(x) = -CQ. X - a

In this case also f(x) is unbounded and tends to -00 as x tends to a. You can give similar definitions for f(a+) = + a, f(a-) = -t a, f(a+) = -w, f(a-) = -w.

12 /-

- d

Now we define lim f(x) = oo. x - -

f is said to tend to w as x tends to 03 if given a number > 0, .there exists a number k > 0 such that

f(x) > M for x r k.

We may similarly define lim f(x) = + a , lirn f(x) = -w, lim f(x) = -w.

X--m x - + a x--m

In all such cases we say that the function f becomes unbounded as x tends to + ca or -w as the case may be. ,

It is easy to see from the definition of limit of a function that the limit of a constant function at any point in its domain is the constant itself. Similarly if lim f(x) = A, then lirn cf(x) = cA for any constant c where c is a real number.

x-a x-a

EXAMPLE 6 : Justify that

lim 1

= 03.

x-2 (x - 2)2

SOLUTION : You have to verify that corresponding to a given positive number M, there exists a positive number 6, such that

1 > M whenever 0 < Ix - 21 < 6. (x - 212

Indeed for x # 2,

Take 6 = I. Then you can see that &I

1 > M whenever 0 < Ix - 21 < 6. (X - 2)2

Hence

lirn 1

= a. x - 2 (x-2)2

Now try the following exercise.

EXERCISE 5 (i) Consider f(x) = 1x1, x € R. Show that lim f(x) = + m.

and lim f(x) = + w x-f m

x--m

and f(O+) = f(0-) = 0 = f(0)

(ii) Let f(x) = - 1x1, x E R. Prove that lim f(x) = -m

and lim f(x) = +a, x- + m

x--0,

and f(0) = f (O+) = f(0-) = 0.

We have already stated that if a function f Is definea by r(x) = l/x, x # 0, then the limits f(O+) and f(0-) and lirn f(x) do not exist, It simply means that these

x-0 limits do not exist as real numbers. In other words, there is no (finite) real number A such that f (O+) = A f(0-) = A, or 'lim f(x) = A.

x-0

You can easily solve the following exercise:

EXERCISE 5 1

(i) Let f(x) = - , x # 0. Show that lim f(x) = + m y lim f(x) - 03

1x1 x-O+ x -0-

and lim f(x) = +a. x-0

1 (5) Let f(x) = - - , x # 0. Show that lim f(x) = -a, ]im f(x) = - o ~

1x1 x-O+ x-0 - and lim f(x) = -a.

x-0

1 (iii) Let f(x) = --, x f 0. Prove that lim f(x) = + a, liln f(x) = --oo

x x-O+ x-o-

1 (iv) Let f(x) = - -, x $ 0. Prove that lira f(x) = -m, lim f(x) = oo.

X x-a+ r;-&.

8.3 SEQUENTIAL LIMITS

In Unit 5, you studied the notion of the limit of a sequence. You also know that a sequence is also a function but a special type of function. What is special about a sequence? Do you remember it? Recall it from Unit 5. Naturally, you would like to know the relationship of a sequence and an arbitrary real function in terms of their limit concepts. Both require us to find a fixed number A as a first step. Both assume a small positive number E as a test for closeness. For functions we need a positive number 6 corresponding to the given positive number E and for sequences we need a positive integer m which depends on E . So, then what is the difference between the two notions? The only difference is in their domains in the sense that the domain of a sequence is the set of natural numbers whereas the domain of an arbitrary function is any subset of the set of reak numbers. In the case of a sequence, there are natural numbers only which exceed any choice of m. But for a function with a domain as an arbitrary set of real numbers, this .js not necessary the case. Thus in a way, the notion of the limit of a function at infinity is a generalization of that of limit of a sequence.

Let us now, therefore, examine the connection between the limit of a function and the limit of a sequence called the sequential limit. We state and prove the following theorem for this purpose:

THEOREM 3 : Let a'function f be defined in a neighbourhood of a point 'a' except possibly at 'a'. Then f(x) tends to a limit A as x tends to 'a' if and only if for every sequence (x,), x, # a for any natural number n, converging to 'a', f(x,) converges to A.

PROOF : Let, lim f(x) = A. Then for a number E > 0, there exists a 6 > 0 such X-a

that for 0 < Jx - a1 < 6 we have

Let (x,) be a sequence (x, # a for any n € N) such that (x,) converges to a i.e. ' x , - a .

Then corresponding to 6 > 0, there exists a natural number m such that fo r all n 2 m I

(x, - a( < 6. t

I

Consequently, we have

lf(x,J - A[ < E ; V n r m. I

This implies that f(x,,) converges to A. I

Conversely, let f(x,) converge to A for every sequence .x, which converges to a, x, # a for any n.

I

14 1 I

L

Suppose lim f(x) 2 A. x-a

hen there exists at least one E , say E = Eo such that for any 6 > 0 we hive an x6 such that

0 < Ix6 - a1 < 6 and

If(x6) - A1 2 Eo.

We get a a sequence (x,) such that x, = xs where 6 = l/n and

1 0 < Jx, - a1 < - for n = 1, 2 ,.....

n and

I f(x,) .- A 1 2 Eo.

0 < lx, - a1 * x, $ a for any n.

1 1 Since - - 0 and Ix, - a1 < -, it follows that x, + a.

n n I.. I. ~ i . l t I f(x,jW- A 1 r Eo * f(x,) -f. A i.e, f(xJ does not tend to A.

Therefore x, # a V a and x, tends to a as n tends to oo whereas fcx,,) does not converge to A, contradicting our hypothesl. This completes the proof of the theorem.

You may note that the above theorem is true even when either a or A is infinite or both a a i d A are infinite (i.e. +a, or -w).

By applying this theorem, wc can decide about the existence or non-existence of limit of a function at a point. Consider the following examples:

0 if x is rational EXAMPLE 7 : Let f(x) =

1 if x is irrational

§how thrt at no point a in the real line R lim f(x) exists. 4

x-a

SOLUTION : Consider any point 'a' of the real line. Let (p,) be a sequence of rational numbers converging to the point 'a'. Since p, is a rational number, f(p,) = 0 for all n and consequently lirn f ( p ~ = 0. Now consider a sequence (q,) of irrational numbers converging to 'a'. Since q, is an irrational number, f(qn) = 1 for all n and consequently lirn f(q,) = 1. So for two sequences @J and (q,) converging to 'a'; sequences (f(p,,)) and (f(q,)) do not converge to the same limit. Therefore lim f(x) cannot exist for if it exists and is equal to A, then

x-a both (f(p,)) and (f(q,J) would have converged to the same limit A.

i EXAMPLE 8 : Show thnt for the function f: R - R defined by

f(x) = x Q x € R, lim f(x) exists for every a E R. X - P

u' SOLUTION : Consider any point a E R. Let (x,) be a sequence of points of R I converging to 'a'. Then f(x,) = x, and consequently lirn f(xJ = lim (xJ = a, So

for every sequence <x,> converging to 'a' (f(xJ) converges to 'a'. So by Theorem 3, lirn f(x) = a. Consequently lim f(x) exists for every a € R.

x - a X - a

Now try, the following exercises.

EXERCISE 7

Show that for the function f: R - R defined by f(x) = x2,

llim f(x) exists for every a € R. x-P

15

EXERCISE 8

Show that hn 2X = 2 by proving that for any sequence (x3, xn # 1, cornverging x-1

. to 1, 2% converges to 2.

8.4 ALGElBlEaA OF LIMITS

We discussed the algebra of limits of sequences in Unit 5. In this section we apply the same algebraic operations to limits of functions. This will enable us to solve the problem of finding limits of functions. In other words we discuss limits of sum, difference, product and quotient of functions. Before we do this, let us first recall the meanings of the sum, difference, product, quotient of two functions which you have studied in Unit 4.

DEFINITION 5 : Algebraic Operations on Functions Let f and g be two functions with domain D C R. Then the sum, difference, product, quotient of f and g denoted by f + g, f -'g, fg, f/g an: functions with domain D defined by

, provided in the last case g(x) # 0 for all x in ID.

Now we prove the theorem.

THEOREM 4 If X-a Mm f(x) = A and X- P Urn g(x) = B, where A and B are real numbers,

(i) $2 (f + g) (x) = A + B = Um x- a f(x)+ lim X- a g(x),

(ii) X-a lim (f - g) (x) = A - B = x- a Um f(x) - X-a Iim g(x),

(iii) lim (f g) (x) = A . B = x-a

(iv) If further lirn g(x) # 0, then Ilm f/g (x) exists and x- a X-a

- f Urn f(x) x-.

lim - (x) = A A = , X-9 g pJ g(x)

PROOF : since lim f(x) = A and lirn g(x) = B, corresponding to a number x- a x-a

E > 0. There exist numbers 6, > 0 and J2 > 0 such that 0 < Ix - a1 < 6, * \f(x) - A1 < &/2 (1) 0 < Ix - a1 < 62 * (g(x) - BI < &/2 (2)

. Let 6 = minimum (dl, d2). Then from (1) and (2) we have that 0 < Ix - a( < 6 (f(x) + g(x) - (A + B)( s (f(x) - A ( + Ig(x) - BI

< &/2 + E / 2 = E. I

Which shows that lim (f + g) (x) = lim f(x) + g(x) = A + B X-B X-8

This proves part (i). The proof of (ii) is exactly similar. Try it yourself.

(iii) I f(x) g(x) - AB I = I (f(x) - A) g(x) + A (g(x) -I B) I 5 lf(x) - Al Ig(x)l + IAI. I(g(x) - B)l . (3)

I 6 Since lim g(x) = B corresponding to 1, there exists a number % > 0 x-8

such that 0 < I X - "1 < aa 3 IgCx) .-- I31 < 1.

Limft a l a Function

which implies that Ig(x) 5 Ig(x) - BI + IBI _< 1 + (BI = K (say) (4) Shce ?I f(x) = A, corresponding to E < 0, there exists a number 6, > 0 such that

nuniber 6,< 0 such that

0 .: Ix - a1 < 6 , 3 (f(x) - A1 < e/2K I

I Since lim g(x) = B, corresponding to a n u d e r E > 0, there exists a number 6, > 0 such %-+a

that 1 E 0 < Ix - a1 ' 6, = If(x) - BI <

I 2(IAl + 1) (6)

I Let 6 = min (a,, S,, G,).Then using (4), (5) aird (6) in (3),

I we have for 0 < Ix - a/ < 6,

Therefore, lim g(x) = AB i.e. lim (fg) (x) = AB = lirn f(x), lirn g(x), which proves x+a X+D x+a x+a

part (iii) of the theorem.

(iv) First we show that g does not vanish in a neighbourhood of a.

lim g(x) = B and B ;c 0. Therefore IBI > 0. Then corresponding to x-Pa

I B I -;;- we have a number p > 0 such that for 0 < Ix - a/ < p,

Now by triangle inequality, we have

IBI In other words, 0 < (x - a1 < p a Ig(x)l >- 2 . (7)

Again since lim g(x) = B, for a given number E > 0, we have a number p' > 0 such x+a

that 0 < Ix - a1 < p' implies that

I

I.

Let 6 = min (p, p'). Then if 0 < Jx - a1 < 6 , from (7) and (8)

1' we have

1 1 This proves that ?. - = -

g(x) B '

Now by part (iii)'of this theorem, we get that

f(x) lim - - 1 1 - $2 f(x). - = !% Rx). !& - x+a g(x) g(x)

l l l l l &\A) X+L xsT (2) (X) )= A/kl = lim g(x) ' x+a

This completes the proof of the theorem. You may note the theorem is true even when a = f co. You may also see that while proving (iv), we have proved that if

Limit and Contlnulty 1 1 lirn g(x) = B # 0, then lirn - = -. x-a x-a g(x) %

Before we solve some examples, we prove two more theorems.

5 : Let f and g be defined in the domain D and let f(x) 5 g(x) for all x in D. Then if lim' t(x) and Urn g(x) exist,

x-8 x-n

Urn f(x) s lim g(x). x- n I- a

PROOF : Let lirn f(x) = A, lirn g(x) = B. If possible, let A > B. x-a x-a

A - B for & = - , there exist 6,, S2 > 0 such that

2 0 < ( x - a / < 6, * ( f (x ) -A( < E

and 0 < ( x - a1 < a2 =$ (g(x) - I31 < E. If 6 = min. S2), then for 0 < ( x - a( < 6, g(x) E ]B - E, B + E[

A + B andf(x) E ] A - & , A + E [ . B u t B -I- E = A - E = - . Therefore

2 g(x) < f(x) for 0 < ( x - a ( < 6 which contradicts the given hypothesis. Thus A 5 B.

TWEOlREM 6 : Let S and T be non-empty subsets of the real set R, and let f: S 4 T be a function of S onto T. Let g: U - R be a function whose domain U C R contains T. Let us assume that lim f(x) exists amdi is equal to b

x - a and Jim g(y) exists and is equal to e. Then Ilim g(f(x)) exists and is equal to c.

Y - b x - n

PROOF : Since lim g(y) = c, given a number E > 0, there exists a number Y - b

cue > 0 such that

0 < (Y - b ( < cro * (g(y) - C ( < 6 .

Since lirn f(x) = b, corresponding to cq > 0, there exists 6 > 0 such that x - a

Hence taking y = f(x) and combining the two we get that for '

0 < Ix - a l < 6, (g(f(x)) -4 = Ig(y) - c l < (since (f(x) - b ( < a,).

This co~npletes the proof of the theorem. Finally we give one more result without proof.

RESULT : If X - a Jim i(x) = A, A > 0 and X - a lim g(x) = 1B where A and B are f nite .,

real numbers theu

lirn f(x)g4) = A ~ . x -- n

Now we discuss some examples. You will see how the above results help us in reducing the problem of finding h i t of complicated functions to that of finding limits of simple functions.

EXAMPLE 9

Find lim (2x + 7) (3x - 11) (4x + 5)

x - 0 4x3 t X - 1

SOLUTION

lirn (2x + 7) (3x - 11) (4x + 5)

x - 0 4x3 + x - 1

We divide the numerator and denominator by x3 since x3 is neither zero nor w..

(2x + 7) (3x - 11) (4x + 5) = lim . . x - w 4x3 + X - 1

= lim - - 2 x 3 ~ 4 = -6; x - O l 1 1 4

x2 - 9 EXAMPLE 10 : Find Urn3 - x 2 - 4 x + 3

x2 - 9 SOLUTION : lim = lim (x - 3) (x + 3

- 3 x2-4x + 3 x - 3 ( x - 3 ) ( x - 1)

x2 - 9 x + 3 Hence lim = lim -

" ' 3 x 2 - 4 x + 3 x - 3 x - 1

x2 - 9 The function f{x) = is not defined at x = 3. But we arc

x2 -4x + 3 considering only the values of the function at those points x in a neighbourhood of 3 for which x # 3 and hence we can cancel x - 3 factor

(1 + x)l12 - 1 EXAMPLE 11 : Evaluate Jim - 0 (1 + x)l13 - 1

SOLUTION : To make the problem easier, we make a substitution which enables us to get rid of fractional powers 1/2 and 1/3. L.C.M. of 2 and 3 is 6. So, we put 1 + x = y 6

Then we have

(1 + x)ln - 1 y3- 1 lim = lim - ' ( Y - ~ ) ( Y ~ + Y + ~ )

=i lim x + o (1 t X y n - 1 y - r l yZ- 1 Y + l ( y- 1 ) ( y f 1)

Y 2 + y + 1 3 = lirn - - -.

Y - 1 y + l 2 Try the following exercises:

I EXERCISE 9 I

1 Find

r

(i) lirn (2x + 313 (3x - 212

X - w x5 + 5

(3) lirn (x3 + 1)II3

x - w x + 1

EXERCISE 10

.

2x for 0 s x < 1 If g(x) = 4 for x = 1

5-3x for 1 < x s 2.

find =el g(x) I

19

Fhd

3x2 - x - 10 (ii) Find lim -

x - 2 x 2 + 5 x - I 4 P - ctas r

Oii) lim - x - 0 x 2 X - 4 1 xz

(vii) li)im (-) sin x - sin a x - - 2 x + I

(iv) llim x - 8 x - 8

In this unit, you have been introduced to the concept of a limit of a function. In Section 8.2, we started with the intuitive idea of a limit of a function. Then we derived the rigorous definition of the limit of a function, popularly called G - 6 definition of a libit. Further, we gave the notion of right and left hand limits of a function. It has been proved that lim f(x) = A if and only if both right hand

X - 8

and left hand limits are equal to A i.e. lim f(x) = Xllm f(x) = A. In the same X - a + - 8-

section we discussed the limit of a function as x tends to 3. oo or -oo. Also we . discussed the infinite limit of a function. In Section 8.3, we studied the idea of

sequential limit of a function by connecting the idea of limit of an arbitrary function with the limit of a sequence. It has been shown how this relationship helps in finding tlne limits of functions. In Section 8.4, we defined the algebraic operations of sum, difference, product, quotient of two functions. We proved that the limit of the sum, difference, product and quotient of two functions at a point is equal to the sum, difference, product and quotient of the limits of the functions at the point provided in the case of quotiexit, the limit of the function in the denominator is non-zero. Finally in the same section, the usefulness of the algebra of limits in finding the limits of complicated functions has been illustrated.

El) We claim. that lim f(x) = 1. To verify this, let E > 0 be a fixed real number. Then x - 1

If(x) - 11 = Jx" 11 = J x - 1 ) ( X + 11.

2 - 6 < x < 2 -1- 6,:A @ 2I.e. I e x < 3, x # 2. Then -10 < x.- 11 < -8and2 < 3x- 1 < $so that Ix- 111 < IOand 13x- 1.1 3 2

x - 11 Thus I - 1 < 5. Now if E > 0 is given and if simultaneously

3x - 1

1 1 < 5 then I x2 - X + 18 ( x - 21 e E / 5 and - 4 <@.

3x - 1

Hence we can choose S = min ( 8 4 1) I

Then for 0 < Ix - 21 < S we have

In fact, in ihis problem, f(2) is defined md takes the value 4.

, 2x + 1 E3) When x Z 0, f(x) = -- . Therefore

3 . Right hand limit =: lim f(x)

x - o+

Left hand limit = limo f(x) x -.

= lim 2(0 - h) + 1

h - 0 fh > 0)

3

Since both the right hand and left hand limits kist and are equat, therefire 1

lim f(x) = -. x - 0 3

12 + h - 21 E4) (i) Right hand limit = f(2+ ) = lirn ---

h - 0 2 - I - h - 2 (h ' 0) = 1.

Similarly left hand limit = -1

,, lirn f(x) does not exi%t. x - 2

Since f(O 9 ) 7t f(0-), lim f(x) does not exist. X -

E5) (i) When x is positive or zero f(x) = x, and when x is iiigative, f(x) = -. A.

f(0 + ) = lim x = 0 and f(O-) = lim - x = 0. Also f(0) = 0 X -. 04 - :*.

lirn f(x) = lim 1x1 = lirn x =-w.Pnfa@tforanyM>O, X - m X - C O x - f a

f ( x ) > M i f x ~ kwl i thk= M + 1.

S i d k l y lim f(x) = lh I x 1 = fim -.x = @. X -C -00 x - -m X - -m

(ii) It is s~~ to ti).

Limit and Continuity E6) (i) Let h > 0

lirn f(x) = lirn 1 1 = lim - - - - 00.

x - O + h - 0 10 + hJ 11 -0 h

1 1 lim f(x) = ,m - = lim - = 03. x - 0- - 0 ( 0 - h l h - 0 h

Hence it follows that lirn f(x) = oo. It can also be proved as follows: x - 0

1 If M > 0 is any number, f(x) = - 1 > M if 0 < 1x1 < - that is

1x1 M 1

f(x) > M if 0 < 1x1 < 6 where 6 = -. I M

Hence lirn f(x) = a. x - 0

(ii) is similar to (i). 1 1

- - (iii) lim f(x) = lirn - = a~ and lim f(x) = lim - - 00 X-O+ h-0 h X-0- h-0 -h

(iv) is similar to (iii).

E7) If (x,) be a sequence converging to 'a', f(x,) = x:--a2 and so by Theorem 3,

lim f(x) = aZ. x-a

E8) Let (x,) be any sequence belonging 'to the domain of definition of f converging to 1 and such that x, # 1 for any n. Given E > 0 we want to find an M such that for all n r M

Choose E l = logz (1 + &/2). It is clear that E l > 0. Since lirn xn= 1, therefore corresponding to E, > 0, there exists a positive integer M such that n-tm

Thus for n 2 M, we have

2". < 21% = 2.2e1 = 2.2 '0~2(It&) = 2(1 + €12) = 2 + &.

:. 2 - E < 2% C 2 + E, for n 2 M i.e., (2Xn - 21 < E, for n 2 M.

This proves that 2'. tends to 2. From theorem 3, it follows that lirn 2X = 2. x- l

(2x + 313 (3x - 212 E9) (i) lirn

x-'= xS -+ 5

( 2 + 3 / ~ ) ~ ( 3 - 2 / x ) ' 23.32 = lirn - - - = 72.

x-Ol 1 + 5/x5 1

(ji\ lirn 3m - - l i rn

3diTm - = I. 1 + l/x

2x for 0 r x < 1

5 - 3x for 1 < x s 2.

lirn g(x) = lirn (5 - 3x) = 5 - lirn 3x = 5 - 3 = 2 x-l+ x - l + x - l +

and lim g(x) = lim 2x = 2 x- 1- x-l-

Hence lirn g(x) = 2. x- 1

f i - 2 El l ) ( i ) lirn - = lina

(6 - 2) (v5 + 2) x-4 x - 4 "-4 (x - 4) (fi + 2)

3x2 - X - 10 (ii) lirn -

x-2 x2 + 5x - 14

(3x + 5) (x - 2) 3x + 5 = Iim = lim -

x-2 (x + 7) ( x- 2) x-2 X + 7

1 - cos x 2 sin2 x/2 1 sinZ x12 (iii) lim = lim = lim - -

x-0 x2 X-+O 4.x2/4 x-tO 2 (Xm2

, a I

1 sin2 x/2 sin x/2 - - - since lim = lim ----- = 1 2 ' x-0 (x/212 x-0 ( x/2 )

x + a x - a 2 cos - sin - sin x - sin a

(iv) lim = lim 2 2 x- a X - a x-a x - a

sin - x + a = lim cos - lim 2

x-a 2 X-a X - a

. x - 1 1 - l /x lim - = lim - = 1. x-= x + 1 x-= 1 + l /x

x - 1 lim (*). = lim [I + - - 1]I x-rn X + 1 x-w x + l

x + l - 2 -- since lim x- op [I + (=)I = e.

or lirn (1 - l/xIx x- = 1.I (">' =

X-m x + l lim (1 + l/x)' x-

Jim [(I - x-m

- - lirn (1 + l / ~ ) ~ x- m

sin 2x (v0 $3 (y-)

Limit of a Function

sin 2x $3 (?) = 2 and -1im X-o (1 + 3 = 1

Eimit and Ontlnplty

9.1 introduction Objectives

9.2 Continuous Functions 9.3 Algebra of Continuous Functions 9.4 Non-continuous Functions 9.5 Summary 9.6 Answers/Hints/Solutions

Suppose that you have functions which are defined ona an intewal, eilher open; or clrjsed. If you draw the graph of these functions, you will observe that some of these can be sketched down in one smooth 'continuaus' sweep of your pen, wllile others have many breaks or jumps. For example, draw the graph of the fo!lowing two functions :

(a) f ( x ) = x 2 , x ~ [ - 2 , 2 ] ;

These are as shown in figures l(a) and I@).

Pig. I(a) Fig. l(b)

You can see that while the gra!)h of the first fimction can be drawn in the 'continuous' motion withnut lifting the pen from the paper while the eraph o f the other functi~n cannot %tie drawn in this manner. This is an interesting property of the first finction which is nut possessed by the second function. It is, therefore, natural to wonder if it can be given some mathematical nicaning. In fact, mathematicians of the past several centuries did confront this question, namely:

"Is there a way to specify those curves which can be drawn with a single stroke of one's pen?"

UNIT 9

9.1 Introduction Objectives

9.2 Continuous Functions 9.3 Algebra of Continuous Functions 9.4 Non-continuous Functions 9.5 Summary 9.6 Answers/Hints/Solutions

9.1 INTRODUCTION

Suppose that you have functions which are defined on an interval, either opet; or closed. If you draw the graph of these functions, you will observe that some of these can be sketched down in one smooth 'continuous' sweep of your pen, v~hile others have many breaks or jumps. For example, draw the graph of the following two functions :

(a) f(x) = xZ, x e[- 2, 21;

These are as shown in figures l(a) and l(b).

Pig. f(a) Fig. l(b)

You can see that while the graph of the first fi~nction can be drawn in the 'continuous' motion without lifting the pen from the paper while the graph of the other function cannot Ije &awn in this manner. This is an interesting propertj of the first finction which is nut possessed by the second function. It is, therefore, natural to wonder if it can be given some mathematical meaning. In fact, mathen~aticims of thc past several centuries did confront this question, namely:

"1s there a way to specify those curves which clan be drawn with a single stroke of one's pen?"

Llrrplt and Co~flnd6' The answer is yes and the functions representing such curves are given the names

- as Continuous Fuactlomrs. What is, then, the mothem~&a1 menacing, of en continuous bunstion? What rare the Punctia~ns slush as t b ope Ima flgaaae l(b)T We shall try to answer these questions and a few more in this unit.

In Unit 8, we made clear our intuitive idea of tlte vdues of a function f(x) approaching a number A as the variable x approaches a given point a. In continuous graphs of ft~nctions, as you have seen in the figures l(a) and (b) that as 'x approaches a, the functional values approach f(a). When there is break (or jump) in the graph, then this property fails at that point. This idea of continuity is, therefore, connected with the value of lim f(x) and the value of the function f

x-a at the point a. We define in this unit the continuity of a function at a given point a in precise mathematical language. Therefore extend it to the continuity of a function on a non-empty subset of the domain of f q ~ ~ h i ~ h could be the whole of the domain of f also. We study the effect of thc algebraic operations s f addition, subtraction, multiplication and division on coretltnlousl furkctisns.

We shall use these results in Unit 10 to discuss the properties of continuous functions and the concept of uniform continuity.

Objectives After studying this unit, you should be able to:

s define the continuity of a function at a point of its domain, @ determine whether a given function is continuous or not, @ construct new continuous functions from a given class of continuous functions.

9.2 CONTINUOUS FUNCTIONS

We have seen that the limit of a function f as the variable x approaches a given point a in the domain of 'a function f does not depend at all on the value of the function at that point a but it depends only on the values of the function at tlie points neq a. In fact, even if the function f is not defined at a then lim f(x)

x-a may exist. For example lim f(x) exists when

x-1

x2 - -1 f(x) = - though f is not defined at x = 1.

x - 1

We have also seen that lim f(x) may exist, still it need not be the same as f(a) x-a

when it exists (see example 2, Unit 8). Naturally, we would like to examine the special case when both lim f(x) and f(a) exist and are equal. If a function

x- a has these properties, then it is caIled a continuous function at the point a. We give the precise definition as follows:

DEFINITION 1 : Continuity of a Function st a Point A function f defined on a subkt S of the set R is said to be continuous at a point a E S, if

i) lim f(x) exists and is finite x-9 . . li) lim f(x) = f(a). x-a

Note that in this definition, we assume that S. contains some open interval containing the point a. If we assume that there exists a half open (semi-open) interval [a, c[ contained in S for some c E R, then in the above definition , we can replace lim f(x) by Iim f(x) and say that the function is continuous from

x-a X-a+ the right of a or f is right continuous at a.

Similarly, you can define left continuity at a, replacing the rob of lim f(x) 1

x-a by lim f(x). Thus, f is contjnuous from the right at a if and only if

26 x-a-

f (a+) = f(a)

It is contipuous from the left at a if and only if

f(a-) = f(zz)-

From the definition of continuity of a function f d a point a and properties of limits it follows that f (a+) = f(a-) = f(a) if and only if, f is continuous at a. If a function is both continuous from the right and co~ltinuous from the left at a point a, then it is continuous at a and conversely.

The definition 1 is popularly known as the Limit-Definition of ont ti nu it^. Since lim f(x) is also defined in terms of E and 6, we have an equivalent

x-a formulation of the definition 1 in terms of E and 6. Note that whenever we talk of continuity of a function f at a in S, we always assume that S coiitains a neighbourhood containing a. Also remember that if there is orie such neighbaurhood there are infinitely many such neighbourhoods. An equivalent definition of continuity in tenils of t: and 6 is given as follows:

DEFINITION 2 : ( E , 8)-Definitiom of Cuntiwuity A filnction f is continuous nt x = a if f is defined in a neighbsurbood of a and caprtesponding to ra given n~arnber E > 0, there exists some number cS > 0 i+sh that ( x - a1 < 6 implies (fdx) -- f (a)( .= E. Note that unlilce in the definition of limit, we should have

(f(x) - f(a)( 4 E for ( x - a ( < 6.

The two definitions are equivalent. Though this fact is alnlost obvious, i f will be appropriate to prove it.

THEOREM 1 : The limit definition of continuity and the (E, 6)-definitioir of conlti~~~aity are equivalennl.

PROOF : Suppose f is continuous at a point a in the sense of the limit definition. Then given E > 0, we have a 6 :. 0 such that 0 <: Ix - a1 < 8 implies Jf(x) - f(a)l

E. When x = a, we trivially have

Hence, ( x - a ( 6 .6 3 (f(x) - f(a)J < E which is the (E, 6)-definition.

Conversely we now assume that f is continuous in the sense of (6, 6)-definition. Then for every E > 0 there exists a 6 > 0 such that

Leaving the point 'a', we can write it as

0 6 ( x - a ( <: 6 3 ( f (x)- f (a)J < E.

This implies the existence, of lim f(x) and that lim f(x) = f(a), %-*a X - R

Note that 6 in the definition 2, in general, depends on the given function f, E and the point a. Also J x - a ( < 6 if and only if a - 6 6 x < a + 6 and ]a - 6, a + 6[ is an open interval containing a. Similarly (f(x) - f(a)J < & if and only if

f(a) - & < f(x) < f(a) + E .

We see that f is continuous at a point a, if corresponding to a given (open) E-neighbourhood U of f(a) there exists a (open) 6-neighbourhood V of a such that f(V) C U. Observe that this is the same as x E V =+ f(x) E U. This formulation of the continuity at a is more useful to generalise this definition to more general situations in Higher Mathematics.

A function f is said to be continuous on a set S if It is continuous at every point of the set S. It is clear that a constant function defined on S is continuous on S.

Let us, now, study some examples and exercises:

EXAMPLE 1 : Examine the continuity of the following functions: i) The absolute value (Modulus) function, ii) The signum function.

SOLUTION i) You know from Unit 4, that the absolute value function

f: R - R is defied a f(x9 = 1x1, V x E W. The function is continuous at every point x E R. For given G > 0, we can choose 6 = E itself. If a f R be any point then1 ] x - aj < 6 = G implies that

ii) The signum function, as you know fram Unit 4, is a function f : R -. R defined as

f(x) = 1 i f x > 0

= 0 i f x = O

- - - 1 i f x < O

Tliis function is aot continrnous ~t the point x = (i. We have already seen In 3rd 8 that f(O + ) = 1, f(0-) = - 1. Since f(0 -t ) ;.f f(O-), lim f(x) docs not exist and

x - l ? consequently the function is not contln~ious at x = C. For every poifit x f 0 tl:c function f is continuous. This is easily seen from the graph of the fi-instion f ;is described in Unit 4. The~z is a jump at the point x = 3 in the vaiurs of f j r r) defined in a neighbourhood of 0.

Note that if f : R - R is defined as,

f(x9 = 1 i Q x r 0.

= - 1 i f x < O . then, it is easy to see that this function is corrtinuous from th-:: r.ilg.ht at x ;- 11. ." _i

not fiom the left. It is continuous at every point x f 0.

Similarly, if f is defined by f(x) = 1 if x > 0 = - 1 i f x s 0

then f is continuous from the left at x = 0 but not from the right.

EXERCISE 1 Examine the continuity of the following functions:

i) The function f: R - (0) -. R defined as

at the point x = 0

ii) The function f: R - [ I ) - B defined as

ill) The function f: R - {O ] - R defined as

EXAMPLE 2 : Discuss the continuity of the function sin x on the red Pse R.

SOLUTION : Let f(x9 = Sin x V x E R. We show by the (E, &definition that f is continuous at every point of R.

Consider an arbitrary point a E R. We have

x - a x + a If(x) - f(a)l = lsin x - sin a1 = cos - I

2

x + a

2 8

X - t d eas --;-- 1 5 I )

L

From Trigonometry, yon know that 1 SZrk 8 ) .r: 181.

Consequently ( f(x) - f(a) 1 s 1 % - a 1 < 6 if jx - a / < 6 where 6 =. E.

So f is continuous at the point a. But a is any saint of R. H e ~ c e Sin x is continuous on the real line W.

EXERCISE 2 Discuss the continuity of cos r srn the acal h e B,

In Unit 8, we have coniiected the limit of a ?.unction wit11 the limit of a scqr.rence sf rcal numbers. In the same yay, we can discuss the continuity of :I fundism in the !angnage sf' the sequence of red. nurnbers in the ~Jomain of the fuuciiosn. 'Chis is explained in. the following the ore!:^.

THEdPmM 2 : A franrcaon f : $3 -. W Is continuous ef point an in $ If arn8 aalg 31 far every sequence (x,), (xg2 E 8) eornwrrgimg do a, f(x,) converges 90 f(&.

PROOF : Let us SUppOsr? thrtt f is cohatinuous at a. Then linn f(x) = f(a). x - j

Given E > 0, there exists a 8 :, 10 such that ( X - a1 < 6 * If(x) - f(a)( < E.

If x, is a sequence converging to %I,, then corresponding to 6 > 0, there exists a positive integer M such that

Ix, - a1 < 6 for n I M.

Thus, for n 2 M, we have Ix, - a1 < 6 which, in turn, implies that

If(x,) - f(a)l < 8, proving thereby f(x,) converges to f(a).

Conversely, let us suppose that whenever x,, converges to a, f(xd) converges to f(a). Then we have to prove that f is continuous at a. For this, we have to show that corresponding to an E > 0, there exists some 6 > 0 such that

(f(x) - f(a)l e E, whenever J x - a1 < 6.

If not, i.e., if f is not continuous at a, then there exists an E > 0 such that whatever 6 > 0 we take there exists an xg such that

Ixg - a1 < 6 but If(%) - f(a)l 2 E.

By taking 6 = 1, 1/2, 1/3, ... . in succession we get a sequence (x, ) , where x, = xg for S = I/n, such that Jf(x,) - f(a)( z E. The sequence [x,] converges to a. For, if m > 0, these exists M such that l /n < m for n r M and therefore (x, - a1 < m for n r M. But f(xJ does not converge to f(a), a contradiction to our hypothesis. This completes the proof of the theorem.

Theorem 2 is sometimes used as a definition of the continuity of a function in terms of the convergent sequences. This is popularly known as the Sequential Definition of Continuity which we state as follows:

DEFINITION 3 : Sequential Continuity of a Function bet f be a real-valued function whose domdn is a subset of the set B. The function f Is said to be continuous st a point a if, for every sequence (xJ in the . domain of f converging to a, we have,

lim f(xJ = f(a) I l - O r r

The next example illustrates this definition.

EXAMPLE 3 : Let f : R -. R be defined as Umit and Continuity

f(x) = 2x2 + 1, V x E R

Prove that f is continaous on R by using the sequential definition of the continuity of a function.

SOLUTION : Suppose (x,) is a sequence which converges to a point 'a' of R. Then, we have

lirn f(xn) = lirn (2x: + 1) = 2 (lim x,,)' + 1 = 2a2 + 1 = f(a) fi-m n - 0 n-m

This shows that f is continuous at a point a E R. Since a is an arbitrary element . of R, therefore, f is continuous everywhere on R.

EXERCISE 3 Prove by sequential definition of continuity that the function f: W -+ R defined by f(x) = 4% is continuous at x = 0.

9.3 ALGEBRA OF CONTINUOUS FUNCTIONS

As, in Unit 8, we proved limit theorems for sum, difference, product etc. of two functions, we have similar results for continuous functions also. These alge'r. operations on the class of continuous functions can be deduced froni the corresponding theorems on limits of functions in Unit 8, using the limit de,anition of continuity. We leave this deduction as an exercise for you. However, we give a formal proof of these algebraic operations by another method which illustrates the use of Theorem 2. We prove the following theorem:

TmOREM 3 : Let f and g be any real functions both csntinuo~ls at a point ---. a E R. Then,

i) af defined by (af) (x) = af(x), is continuous for any real number a,

ii) f + g.defined by (f + g) (x) = f(x) + g(x) is continuous at a,

%) f - g defined by (f - g) (x) = f(x) - g(x) is continuous at a,

iv) fg defined by (fg) (x) = f(x) g(x) is continuous at a,

v) f/g defined by (f/g) (x) = - f(x) , is continuous at a provided g(a) # 0. g(x)

PROOF : Let x, be an arbitrary sequence converging to a. Then the continuity of f and g imply that the sequences f(xn) and g(xJ converge to f(a) and g(a) respectively. In other words, lim f(xn) = f(a), lirn g(xn) = g(a).

Using the algebra of sequences discussed ih Unit 5, we can conclude that lim af(x,,) = af(a1,

lirn (f + g) (x3 = lim f(xnl + lirn g(xn) = f(a) + g(a), lim (f - g) (xn) = lim f(xn) - lim g(x,) = f(a) - g(a),

lirn (f - g) (xJ = lim f(xJ lim g(xn) = f(a) g(a),

such that g(%) = 0, then g(s) - g(a) implies that

This proves the parts (i), (ii), (iii) and (iv). To prove the part (v) we proceed as follows: Since g(a) # 0, we can find a > 0 such that the interval ]g(a) - a, g(a) + a[ is either entirely t o the right or to the left of zero depending on whether g(a) > 0 or g(a) < 0. Corresponding to a > 0, there exists a 6, 2 0 such that (x - a1 < implies Jg(x) - g(a)l < a, i.e., g(a) - a! < g(x) < g(a) + at. Thus, for x such that Ix - a 1 < til, g(x) # 0. If (xJ converges t o a, omitting a finite number of terms of the sequence if necessary, then we can assume that g(x,) # 0, for all n. Hence, f(x3 f(a) converges to - E

and so - is,continuous at a. This completes the proof of k3(x,) g(a) ' g

the theorem.

In part (v) if we define f by f(x) = 1, then it follows that if g is continuous at 'a' and g(a) # 0, then its reciprocal function l/g is continuous at 'a'.

Continuity

Now, we prove another theorem, which shows that a continuous function of a continuous function is continuous.

THEOREM 4 : Let f and g be two real functions such that the range of g is contained in, the domain of f. If g is continuous at x = a, f is continuous at b =

g(a) and h(x) = f(g(x)), for x in the domain of g, then h is continuous a t a.

PROOF : Given E > 0, the continuity of f a t b = g(a) implies the existence of an q > 0 such that for

Corresponding to q > 0, from the continuity of g at x = a, we get a 6 > 0 such that

Ix - a/ < S implies Ig(x) - g(a)l < q ....( 2)

Combining (1) and (2) we get that

Ix - a1 < S implies that

where we have taken y = g(x). Hence h is continuous at a which proves the theorem.

Let us now study the following example:

EXAMPLE 4 : Examine for continuity the following functions:

(i) The polynomial function (Refer to Unit 4) f: R + R defined by

f(x) = a, + a,x + a,xz + .... + axxn.

ii) The rational function (Refer to Unit 4) f: R + R defined as

f(x) , V x for which q(x) + 0. f(x) = - q(x)

SOLUTION : i) It is obvious that the function f(x) = x, x E R, is continuous on the whole of the

real line. It follows from theorem 3(iv) that the functions xZ, x3, ...., are all continuous. Again from theorern 3(i) and 3(ii) and the fact that constant functions are continuous, we get that any polynomial f(x) in x, i.e., the function f defined by f(x) = a, +.a,x + a,xZ + ... + a,, xn,

is continuous on R.

(ii) It follows from theorem 3(v) that a rational function f, defmed by,

is continuous at every point a E R for which q(a) # 0.

Try the following exercise.

EXERCISE 4

Examine the continuity of the function f: R -+ R defined as,

i) f(x) = x3 a t a point a E R;

(ii) f(x) - 1 , i f x + 2

1 , i f x Z 2

You have seen that a function may or may Irot be contitluwle at a pcirlk of. thc domain of the function. Let us now examine why a filnci.ion 5 1 s to ki~e continuous.

A function E: S -. R fails to be cc~atinuous on in doinain Y iS is laat coctirrucatr$; at a particular point of S. This means that there e:iists a poitri u E S such ih~ir. either

i) lim f(x) docs irot exist X - a

or ii) lim f(s) exists but 3s noi equal to i"q'ci). x - a

But you know that a fp~.ncticsn f is contiauous at a pninr. c if as2 41-fi93r i s f(a + > = f(a-) r= 4a).

Thus, if f is not continuous st a, then sne of the fallotving will he;;izlrn:

i) cirlier f(a t- ) or f(a-) does not exist (this includes the case IW hela both .++ ; .:lad f(a-) do not cxlst).

ii) both f(a 4 ) and f(a-) exist but f(a s- ) f f(a-).

iii) both f (a4) and f{a-') exist and f(a+) = f(a-) but tfiey are not equl;l t::.: >.;,:

If a function f: S - R is discontinuous f ~ r each b E S, then we say that " totally discontinuous an S. Functions which are elstally discontinuous ale : + o:') :,.

encountered but by no means rare. We give an ex~rnple.

E LE 5 : Zxamirne whether or not the function f: W -. R defined as,

I, If x is irrational f(x) =

0, if x is raiional

is totally discontinuous.

SOLUTION : Let b be an arbitrary but fixed real number. Choose & .= I/2. . Let 6 > 0 be fixed. Then the interval defined by

is (x: b - 6 < x < b + 6)

or ] b - 6 , b + 6 [

This interval contains both rational as well as irrational numbers. Why? (Refer to Unit 2 for the answer.) .

If b is rational, then choose x in the interval to be irrational, If b is irrational then ' choose x in the interval to be rational. In either case,

0 < I x - b J < 6

I and If(x) - f(b)l = I > E .

Thus, f is not continuous at b. Since b is an arbitrary element of S, f is not I continuous at any point of S and hence is totally discontinuous.

Now you should be able to try the following exercise:

EXERCISE 5 Show that the function f: R -, R defined by

1, if x is rational f(x) =

0, if x t imtional is totally diseopainuous. Does f(a+) and f(a-) exist at any point a E R4

Titere are ceaiairl discontinuities whish can be removed. 'rhese me known as removable discontinuities, A di~o~&$n.as.3@ a of a give8 PUBC~BDQ f : S Ha gS sdd 26: Bs P & ~ Q V B ~ ~ C if f h ~ H d O of f ( ~ b x t~%ds to ts edsh om$ that

lim f(x1 8 %(a] x - 8

PI) other words, f has removable diss;o~t%m3ity zt x - a if f(rr + 3 - f(a-). but none 15 equal to f(a).

The removnble discontinuities of a fu~icsion can be removed simply by changing the value of tfne function at the point m of dj.scoatinuity. For this a function with removable dlscontinuil.ies caa be thought of as being dmose cgntinilous. We discuss the following cxmnple to illnstrate a few cwcs of removable discsntinuitks.

EXAMPLE 6 : Discuss the ssrture sf the dlsnazatielsGtles of She P~89swiag furaeticens:

i) This fuiiction is discontinuous at x = 2. This is a removable discontinuity, for if we redefine f(x) = 4, then we can restore the continuity of f at x = 2.

ii) This is again a case of removable discontinuity at 3. Therefore, if f is defined by f(x) = 3 V x E R, then it is continuous at x = 3.

iii) This function is discontinuous at x = 0. Why? This is a case of discontinuity which is removable. To remove the discontinuity, set f(0) = 0. In other words, define f as

f(x) = x2, X E ] - 2, 0 [U] 0, 2 [ = 0, x = o

This is continuous at x = 0. Verify it.

E%AM!PLE 7 : Let a function f : R -- R he defined as,

Test the continuity of the function. Determine the type of discontinuity if' it exists.

SQLUTIOM

i) Here f(0 +) and f(O-) both do not exist (as fidte real numbers) and so function is discontinuous. This is not' a case of removable discohtinuity.

I

ii) In this case, f(0) does not exist whereas f<0+) exists and f(0-) = f(0) = 1. This is not a case of removable discontinuity, 3 3

t ? n ~ i t ; I I I ~ c o r ~ t i ~ ~ u i t ! Now try the following exercises:

EXERCISE 6

Prove that the function f defined by f(x) = x sin l /x if x # 0 and f(0) = 1 has a removable discontinuity at x = 0.

EXERCISE 7 Prove that the function If 1 defined by I f 1 (x) = (f(x)l for every real x is continuous on R whenever f is continuous on R.

EXERCISE 8

i) Find the type of discontinuity at x = 0 of the function f defined by f(x) = x + 1 if x > 0, f(x) = - (X + 1) if x < 0 amd f(0) = 0.

fi) The function f is defined by

1 f(x) = sin -, x # O

X

= 0, x = o 1s f continuous at 01

In this unit you have been introduced to the concept of the continuity of a function at a point of its domain and on a subset of its domain. The limit definition and (8, - @-definition of continuity have been given in Section 9.2. It has been proved that both the definitions are equivalent. In the same section, sequential definition of continuity has been discussed and illustrations regarding its use for solving problems have been given. In Section 9.3, the algebra of co~ltinuous functions is considered and it has been proved that the sum, difference, product and quotient of two continuous functions at a point is also continuous at the point provided in the case of quotient, the function occurring in the denominator is not zero at the point. In the same section, we have proved that a continuous function of a continuous function is continuous. Finally in Section 9.4, discontinuous and totally discontinuous functions are discussed. Also in this section, one kind of discontinuity that is removable discontinuity has been studied.

E l ) i) The function f is not defined at x = 0. Therefore, f.is not continuous at x = * O .

ii) This function is continubus at every point x # 1, since f(a) is defined at every point a # 1 and lim f(x) = f(a). It is not defined at x = 1 and

x - a .so not continuous at x = 1. If we define f(1) = 2, then the function is continuous at x = 1.

iii) In this case again, f is continuous at all x # 0. See the graph i~ the - figure l@). Justify it by (8, 6)-definition. f is not defined at 0 and so not continuous at 0.

x + a E2) ]cos x - cos a1 = sin -

2 a - x I x + a x - a

= 2 Isin --;- sin - I m

5 2 1 sin- 2 Y - a l

Continuity

5 I X - al.

Then proceeded as in Example 2. Hence cosx is continuous on R.

]E3) Suppose (xn) is a sequence which converges to 0. Then,

' lirn f(xn) = lirn (fin) = d G n = @ = 0 = f(O),

which shows that f is continuous at x = 0.

E4) (i) The function f(x) = x3 is continuous at x = a, for if (xn) is a sequence which converges to a, then lirn f(x,) = lirn x: = (lim xnJ3 = a3 = f(a), which shows that f is continuous at a.

(ii) This function is not continuous at x = 2 because lirn f(x) = 4 whereas x+2

f(2) = 1.

E5) Let a be rational in R. Then f(a) = 1. We have irrationals x as close to a as we want, i.e., there exist points x in every neighbourhood of a such that If(x) - f(a)( = 1 and so if E < 1, we cannot find a 6 > 0 such that for (x - a1 < 6, Jf(x) - f(a)l < E, i.e., the function is not continuous at a. Similar argument holds good when a is irrational. Hence, the function f is discontinuous

' everywhere. It is clear from the above argument that f(a+ ) and f(a- ) also do not exist at any point a.

E6) Fix) = x sin llx, if x s 0 and f(0) = 1. Then

If(x)( = Ix sin 11x1 I 1x1,

since sine function is a bounded function with absolute value bounded by 1.

For E > 0, if 6 = E, O < 1x1 < S 3 If(x)\ < E, i.e., lirn f(x) = 0. x-10

Hence, f(O+ ) = f(0- ) = Ff; f(x) = 0.

If we redefine f(0) = 0 instead of 1, we see that f is continuous at 0. Hence 0 is removable discontinuity.

E7) Since f is given to be a continuous function on R, f is continuous at any point a in R. Hence, given E > 0, there exists a 6 > 0 such that, I

(x - a( < 6 implies (f(x) - f(a)l < E. I

Now by triangle inequality for I I we get, i

which proves that / f J : x -+ (f(x)l is continuous at a. Now a E R being arbitrary, (q is continuous on R.

E8) i) lim f(x) = lim (x + 1) = 1 i.e., f(O+ ) = 1. . . x+o+ x+o

lirn f(x) = lim (- (X + 1)) = - 1 i.e., @- ) = - 1, and so, X-PO- X 4

f(O+ ) + f(0- ).

Hence, 0 is a discontinuity which is not a removable discontinuity.

ii) The function f(x) = sin llx for x + 0, f(0) = 0 has an irremovable * discontinuity at x = 0 since neither f(O+ ) nor @- ) exists.

3 5

19. B Introduction Objectives

10.2 Continuity on Bounded Closed Hnakervals 11 0.3 Poin%wise Contiw~ity rand Uniform Continuity 10.4 Sumnlary 18.5 Sollrstions/Himtj:/Answers

-- -- -...---- -.-

10, B INTRODUCTION ---- -

%%wing studied in the last two 11Pliis limit and continuity of a function at a point, dgcbra of limits and continuous functions, the connection between limits and carltinuity etc., we now take up tlie study of the Gehaviour of continuous functions a n bounded closed intervals on the real line. In Section 10.2 you will learn that continuous functions on such intervals are bounded and attain their bounds; they take all values in between any two values taken at p~ in ts sf such intervals. In Section 10.3 you will also be introduced to the concept of uniform continuity and further you will see that a continuous function on a bounded closed interval is uniformly continuous. This means that continuous functions are well-behaved on bounded closed intervals. Thus, we will see that bounded cbsed intervals form an important subclass of the class of subsets ~f the real line which are known as compact subsets of the real line. You will study more about this in higher mathematics at a later stage. We will henceforth call bounded closed intervals of R as compact intervals.

The results of this unit play an important and crucial role in Real Analysis and so for further study in analysis, you must understand clearly the various theorems given in this unit. Some of the deep theorems of Block 3 are contained in this unit.

It may be noted that an interval of R will not be a compact interval if it is not a bounded or closed interval.

Objectives /.

After the completion of the study of this unit, you should be able to @distinguish between the properties of continuous functions on bounded closed

intervals and those on intervals which are not closed or bounded. e understand the important role played by bounded closed intervals in Real

Analysis. @know the concept of uniform continuity and its relationship with continuity.

CONTINUITY ON BOUNDED CLOSED INTERVALS

We now consider functions continuous on bounded closed intervals. They have properties which fail to be true when the intervals are not bounded or closed. Firstly, we prove the properties and then with the help of examples we will show the failures of these properties. To prove these properties, we need an important property of the real line that was discussed in Unit 1. This property called the completeness property of R states as follows:

Any nonempty subset of the Real Hne R wbkh is boa* above has the least oppcr bound. Or equivakrtly, any inonempty subset of R which is bounded below hns the patest lowet bound.

In the following theorems we prove the properties of functions continuous on bounded closed intervals. In the first two theorems we show that a continuous function 0n.a bounded closed interval is bounded and attains its bounds in the interval. Recall that f is bounded on a set S, if there exists a constant M > 0 such that If(x)l 5 M for all x E S. Note also that a real function f defined on a domain D (whether bounded or not) is bounded if and only if its range f(D) is a bounded subset of R.

THEOREM 1 : A function f continuous on a bounded and closed interval [a, b] is necessarily a bounded function.

PROOF : Let S be the collection of all real numbers c in the interval [a, b] such that f is bounded on the interval [a, c]. That is, a real number c in [a, b] belongs to S if and only if there exists a constant M, such that (f(x)l 5 M, for all x in [a, c]. Clearly, S f 4 since a E S and b is an upper bound for S. Hence, by completeness property of R, there exists a least upper bound for S. Let it be k (say). Clearly, k 5 b. W prove that k E S and k = b which will complete the proof of the theorem.

Corresponding to E = 1, by the continuity of f at k (5 b) there exists a d > 0 such that

Jf(x) - f(k)l c E = 1 whenever Jx - kl < d, x E [a, b].

By the triangle inequality we have

I 1 f(x1 l - I f(k) l I l f(x) - f(k) I < 1

Hence, for all x in [a. b] for which (x - k( <d, we have that

I f (d l < lf(k)l + 1 . . . (1)

Since k is the least upper bound of S, k-d is not an upper bound of S. Therefore, there is a number c E S such that

k - d < c 5 k

Consider any t such that k 5 t < k + d. If x belongs to the interval [c, t] then Ix -kl < d. For,

x E [ c , t ] * c s x s t * k - d < c ~ x ~ t < k + d . . . (2)

Now c E S implies that there exists M, > 0 such that.for all

x E [a, cl, If(x)I 1 Mc . . . (3)

x E [a, t] = [a, c] U [c, t] 3 either x 6 [a, c] or x E [c, t].

If x E [a, c], by (3) we have

If(x)l 5 Mc < Mc + If@)( + 1.

If, however, x E [c, t] then by (1) and (2) we have

If(x)l <If(&)l + 1 < M, + lf(k)l + 1

In any case we get that x E [a, t] implies that

If(x)l Mc + lf(k)l + 1

This shows that f is bounded in the interval [a, t] thus proving that t E S whenever k s t < k + d. In particular k E S. In such a case k=b. For otherwise we can choose a 't' such that k < t < k + d and t E S which will contradict the fact that k is an upper bound. This completes the proof of the theorem.

Having proved the boundedness of the :function continuous on a bounded closed interval, we now prove that the function attains its bounds that is it has the greatest and the smallest values.

THEOREM 2 : If f is a ~ontinuous function on the bounded c l o d Lntemal [a, bl then there exists points xl and x2 in [I, b] auch that f(xa S f(x) 5 f(x3 for all x E [a, bl-(i.e. f attalm its bun*).

Properties of continuous^

fi~nctlam

3 !wit und Continuity PRQQ)F : From Theorem 1, we know that f is bounded oa Ba,b]. Therefore there exists M such that I f(x)l s M V x € [a, bj, Hence, the collection (f(x) : a a x s b] has an upper bound, since f(x) s (f(x)l 5 M V x € [a, b]. So by the completeness property of 8, the set (f(x) : a s x I b] has a least upper bound.

Let us denote by K the least upper bound of (f(x) : a s, x 5 b]. Then f(x) 5 K for all x such that a s x s b. We claim that there exists xz in [a, b] such that f(xz) = K. If there is no such x2, then K - f(x) > 0 for all a r x s b. Hence, the function g given by,

is defined for all x in [a, b] and g is contiiiuous since f is continuous (Refer Unit 9). Therefore by Theorem 1, there exists a constant M r > O such that

lg(x)l 5 M r bf x E [a, bl

Thus, we get -\

i.e., 1

f(x) 5 K - 7 V x E [a, bl. M

But this comradicts the choice of K as the least upper bound of the set [f(x) : a I x 1: b 1 . This contradiction, therefore, proves the existince of air 5,. in [a,b] such that f(xz) = K r f(x) for a I x I; b. The existence of xl in [a,b] such that f(xl) 5 f(x) for a G x I b can be proved on exactly similar lines by taking the g.1.b. of ( f(x) : a s x 5 b ] instead of the 1.u.b. or else by considering -f instead of f. (Try it).

Theorems 1 and 2 are usually proved using what is called the Heine-Borel property on the real line or other equivalent properties. The proofs given in this unit straightaway appeal to the completeness property of the red line (Unit 2) namely that any subset of the real line bounded above has least upper bound. These proofs may be slightly longer than the conventional ones but it does not make use of any other theorem except the property of the real line stated above.

As remarked earlier, the properties of continuous functions fail if the intervals are not blinded or closed, that is, the intervals of the type

1% b[, la, bl, [a, bt, [a,' m[, la, 4, I -m, a], ] -a, a[ or ] -m,oo[. We illustrate them with the help of the followirrg examples and exercises.

EXAMPLE 1 : Show that the function f defined by f(x) = 3 V x E [O, a[ is continuous but not bounded.

SOLUTION : The function f being a polynomial function is continuous in [O, a [ . The domain of the function is an unbounded closed interval. The function is not bounded since the set of values of the function that is the range of the function is (x2 : x € [0, oo[] = [0, a [ which is not bounded.

1 EXAMPLE 2 : Show that the function f defined by f ( ~ ) = - V x E 10, 1[ is continuous but not bounded. X

SOLUTION : The function f is continuous being the quotient of continuous functions F(x) = 1 and G(x) = x with

G(x) z 0, x E 10, I[ (Refer Unit 9).

Domain of f is bounded but not a closed interval. The function is nqt bounded since its range is ( l/x : x E 10, I [ ] = 11, m[ which is not a bdunded set.

EXERCISE 1 Show that the function f defined by f(x) = x V x E I--, a [ is contin~oas but mot bounded.

Properlies of Continuous

EXERCISE 2 : Show that the.functlo~ f given by f(x) =. 1

V x E ] 2,3 [ is contilauous but mot bouaded. (X - 212

E LE 3 : Show that the function f smcb that f(x) = x V x E 10, PI is continuous but does not attain its boaands.

SOLUTION : As mentioned in Example 2, the identity function f is continuous in 10, I[. Here the domain of f is bounded but is not a dosed interval. The function f is bounded with least upper bound 0.u.b) = 1 and greatest lower bound (g.1.b) = 0 and both the bounds are not attained by the function, since range of f = 10, I[.

1 EXAMPLE 4 : Show that tbe function f such that

I is continuous but does not attain its g.1.b.

SOILIJTJON : The function G given by G(x),= x 2 V x E 10, 1[ is continuous and G(x) z 0 V x E 10, 11 therefore its reciprocal function f(x) = 1/x2 is continuous in 10, I[ (Refer Unit 9). Here the: domain f is bounded but is not a closed interval. Further 1.u.b. of f does not exist whereas its g.1.b. is 1 which is not attained by f;

EXERCISE 3 Show that the function f given by f(x) = sin x, x E 10, r/2[ is continuous but does not attain any of its bounds. EXERCISE 4 Prove that the function f given by f(x) = x 2 V x E ]-OD, O[ is continuous but does not attain its g.1.b.

I We next prove another important property known as the intermediate value

I property of a continuous function on an interval I. We do not need the assumption that I is bounded and closed. This property justifies our intuitive idea of a continuous function namely as a function f which cannot jump from one value to another since it takes on between any two values f(a) and f(b) all values lying between f(a) and f(b).

THEOREM 3 : (Intermediate Value Theorem). Let f be a continuous function on an interval containing a and b:If K Is any number between f(n) and f(b) then there is a number c, a s c s b such that f(c) = K.

PROOF : Either f(a) = f(b) or f(a) < f(b) or f(b).< f(a). If f(a) = f(b) then K = f(a) = f(b) and so c can be taken to be either a or b. We will assume that f(a) < f(b). (The other case can be dealt with similarly.) We can, therefore, assume that f(a) < K < f@).

Let S denote the collection of all real numbers x in [a, b] such that f(x) < K. Clearly S contains a, so S # and b is an upper bound for S. Hence, by completeness property of R, S has least upper bound and let us denote this least upper bound by c. Then a s c s b. We want to show that f(c) = K.

Since f is continuous on [a, b], f is continuous at c. Therefore, given E > 0, there exists a 6 > 0 such that whenever x is in [a, b] and 1 % - c ( < 6, If(x) - f(c)( < G,

i.e., f(c) - & < f(x) < f(c) + E. . . . (4)

If c # b, we can clearly assume that c + 6 < b. Now c ier the least upper bound of S. So c - 6 is not 'an upper bound of S, Hence, there exists a y in S sucb'that c - 6 < y s c. Clearly 1 y - cl < 6 and so by (4) above, we have

f(c) - E < f(y) < f(c) + G. Since y is in S, therefore f(y) < K. Thus, we get , 3 9

f(c) - & < K Y

Limit and Continuity If now c = b then K - & < K < f(b) = f(c), i.e., K < f(c) + E. If c # b, then c < .b; then there exists an x such that c < x < c + 6, c + 6, x E [a, b] and for this x, f(x) < f(c) + E by (4) above. Since x > c, K r f(x), for otherwise xG would be in S which will imply that c is not an upper bound of S. Thus, agdn we have K 5 f(x) < f(c) + E. In any case,

K < f(c) + & . . . (6) Combining (5) and (ti), we get for every E > 0

f(c) - E < K < f(c) + E which proves that K = f(c), since G is arbitrary while K, f(c) are fixed. In fact, when f(a) < K < f(b) and f(c) = K, then a < c < b.

L

COROLLARY 1 : If f is a continuous funcqion on the closed interval [a, b] and If f(a) and f(b) have opposite signs (i.e., f(a) f(b) < 01, them thew is a point q in ]a, b[ at which f vanishes. (i.e., f(q) = 0).

Corollary follows by taking K = 0 in the theorem.

COROLLARY 2 : Let f be a continuous function defined on a bounaled closed interval [a, b] with values in [a, b]. Then there exists a point c in [a, b] such that f(c) = c. (i.e., there exists a fixed point c for the function f on [a, b]).

PROOF : If f(a) = a or f(b) = b then there is nothing to prove. Hence, we assume that f(a) # a and f(b) # b.

.)

Consider the function g defined by g(x) = f(x) - x, x E [a, b]. The function being the difference of two continuous functions, is continuous on [a, b]. Fcrqitkr, since f(a), f(b) are in [a, b], f(a) > a (since f(a) # a, f(a) E [a, b]) and f(b) < b. (Since f(b) # b, f(b) E [a, b]). So, g(a) > O and g(b) < 0. Hence, by Corollary 1, there exists a c in ]a, b[ such that g(c) = 0, i.e., f(c) = c: Hence, there exists a c in [a, b] such that f(c) = c.

The above Corollary 1 helps us sometimes to locate some of the roots of polynomials. We illustrate this with the following example.

EXAMPLE 5 : The equation x4 + 2x - 11 = 0 has a real root lying between 1 and 2.

SOLUTION : The function f(x) = x4 + 2x - 11 is a continuous function on the closed interval [l, 21, f(1) = - 8 and f(2) = 9. Hence, by Corollary I, there exists an x, E 11, 2[ such that f(x,) = 0, i.e., x, is a real root of the equation x4 + 2x - 11 = 0 lying in the interval ] 1, 2[.

Try the following exercises:

EXERCISE 5 Show that the equation 16x 4 + 64x3 - 32x 2 - 117 = 0 has a real root > 1. EXERCISE 6 Prove that the equation cos x - x = 0 possesses a root lying L the Interval 10, r 1. EXERCISE 7 Prove that any polynomial of odd power with real coefficients has at least one real mot.

EXERCISE 8 Show that the equation 4x3 - 9 x 2 - 6x + 2 = 0 has a real root in each of the intervals ]'-I, 0 1, ] 0, 1 [ and ] 2, 3 [.

io.3 POI NTWI SE CONTI NUI TY AND UNIF ORM CONTINUITY

In this section, you will be introduced with the concept of uniform continuity of a function. The-concept of uniform continuity is given in the whole domain of the

Pdnction whereas the concept of continuity is pointwise that is it is given at a point of the domain of the function. If a function f is continuous at a point a in a set A, then corresponding to a number E > 0, there exists a positive number 6(a) (we are denoting 6 as 6(a) to stress that 6 in general depends an the point a chosen) such that ( x - a 1 € 6(a) implies that If(x) - f(a)l c 6. The number 6 (a) also depends on E. When the point a varies &(a) also varies. We may or may not have a 6 which serves for all goints a in A. If we have such a 6 common to all points a in A, then we say that f is uniformly continuous on A. Thus, we have the following definition of uniform continuity.

DEFINITION : Uniform Continuity of a Funsetion Let f be a function defined on ;a subset A contained in the set R of all reals. If corresponding to any number G > 0, there exists a number 6 > 0 (depending only on G) such that

then we say that f is uniformly continuous on the subset A.

An immediate consequence of the definition of uniform continuity is that uniform continuity in a set A implies pointwise continuity in A. This is proved in the following theorem.

THEOREM 4 : If a function f is uniformly continuous in a set A, then it is continuous in A.

PROOF : Since f is uniformly continuous in A, given a positive number E, there corresponds a positive number 6 such that

lx - , Y I < 6; x, Y €i A * If(x) - f(y)l < . . . (7)

Let a be any point of A. In the above result (I), take y = a. Then we get,

Ix- a1 < 6;x € A * Jf(x)- f(a)J < E

which shows that f is continuous at 'a'. Since 'a' is any point of A, it follows that f is continuous in A.

Now we consider some examples.

EXAMPLE 6 : Show that the functibn f: R - R given by

f(x) = x V x E R . is uniformly continu,ous on R

SOLUTION : For a given & > 0, 6 can be chosen to be 6 itself so that

Ix - y J < 6 = G * If(x) - f(y)l = J x - Y I C 6 .

EXAMPLE 7 : Show that the f ,untt io f: R - R given by

f(x) = x2 v x E R

is not uniformly continuous on R.

SOLUTION : Let 6 be any positive number. Let 6 > 0 be any arbitrary positive number. Choose x > E/6 and y = x + 6/2. Then

(fix) -.f(y)l = JxZ - y21 = Ix + Y I Ix - Y I

That is whatever 6 > 0 we choose, there exist real numbers x, y such that lx - y( < 6 but (f(x) - f(y)( > G which proves that 6 is not unifirmly continuous. But we know that f is a continuous function on R.

Properties of Continuous Fu%ttions

EXAMPLE 8 : In the abase example if we restrict the domain of f to be the clawd interval [-I, 11, theam show that f Is uniformly c o n ~ u o u s on [-I, I].

8 SOLUTION : Given E > 0, choose 6 < -. If Ix - y 1 < 6 and x, y E [-I, 11,

2 then using the triangle inequality for I Iwe get,

lf(x) - f(y)l = 1x2 - y21 = tx + Y I Ix - Y I 6 (1x1 + IYI)

5 26 (since 1x1 s 1, lyl 5 1)

< E .

You should be able to solve the following exercises:

EXERCISE 9 Show that f(x) = xu, n > 1 is not uniformly continuous on K even tho~ngb for each n > 1, it is a continuous function on R.

EXERCISE 10 1 Show that the function f(x) = - for 6) c x < 1 is continuous for every x but not uniformly o n 10, I[. X

EXERCISE 11 1 Show that the function f(x) = sin - is not uniformly continuous on the interval

X 10, I[ even though it is colntinuous in that interval.

EXERCISE 12 ' Show that f(x) = cx where c is a fixed non-zero real number is a uniformly

cbntinuous function on R.

We have seen in Exercise 10 that the function defined by f(x) = l /x on the open interval 10, 1 [ is not uniformly continuous on 10, 1 [ even though it is a continuous function on 10, 1 [. Similarly, in Example 7, the function f defined as f(x) = x2 is continuoos on the entire real line R but is not uniformly continuous on R. However, if we restrict the domain of this function to the bounded closed interval [-I, 11, then it is uniformly continuous. This property is not a special property of the function f, where f(x) = x2 but is common to all continuous functions defined on bounded closed intervals of the real line. We prove it in the following theorem.

THEOREM 5 : If f is a continuous function on. a bounded and closed interval [a, b] then f is uniformly continuous on [a, b].

PROOF : Let f be a continuous function defined on the bounded closed interval [a, b]. L a S be the set of all real numbers c in the interval [a, b] such that for a given E > 0, there exists positive number d, such that for points xl, xz belonging to closed interval [a, c],

I f(xl) - f(xz) 1 < E whenever Jxl - xz 1 < d,.

(In other words f is uniformly continuous on the interval [a, c]. Clearly a E S so that S is non-empty. Also b is an upper bound of S. Prom completeness property of the real line S has least upper bound which we denote by k. k r b. f is continuous at k. Hence given E > 0, there exists positive real number dk such that

(f(x) - f(k) 1 < El2 whenever Ix - k 1 < dk . . . (8) 1

Since k is the least upper bound of S, k - - dk is not an upper bound of S. 2

Therefore there exists a point c E S such that

k - 1/2 dk < C S k. . . . (9) since c E S, from the definition of S we see that there exists d, such that

42 ( f(xl) - E(x2) ( < E whenever lxl - x2( < dc, xr, xz E [a, cl. ...( 10)

Let d = min ((112) dk, d,) and b' - min. (k + (1/2) dk, b).

NOW let XI, x2 E Ia, b'l and 1x1 - ql < d. Then if x,, x2 E [a, c ] , Ix, - x,] < d s d, by the choice of and 4, then If(x11 - f(xz)l < E by (10). If one of xl, xz is not in fa, cl, then both XI, x2 belong to the interval ]k - dk, k + dk!. For xl g [a, c], implies b' 2 XI > c > k - (112) dk > k - dk'by (9) above. This means xl 5 b' implies xl S k + (112) dk < k + dk by the choice of b'. i.e.

k - dk < k - (112) dk < X l < k 4- (1/2) dk < k + dk ...( 11)

, Ixl - x21 < d implies that XI - (112) dk < x2 < x1 + (1/2) dk since d r (1/2) dk by the choice of d. Thus we get from (11) above that

k - d k < XI -(1/2)dk < ~2 < x1+(1/2)dk < k+

'& Then (11) and (12) show that xl, x2 E ]k - dk, k + dk[.

Thus we get that 1 . ~ 1 -kJ < dk and Ix2 -. kl < dk, which in turn implies, by (8) above, that If(xl) - f(k)l < El2 and If(xz) - f(k)l .= E12. . '

Thus I f(xl) -- f(xz)l 5 I f(x1) - f(k)l + If&) - f(xp)/ < €/2+ E/2 = E. In other words, if Ixl - xzl < d and XI, xz are in [a,V] then /f(xl) - f(xz)( < E which proves that b' E S i.e. b' 5 k. But k s; b' by the choice of b' si11ce.k is k+(1/2) dk and k a b. Thus we get that k = b'. This can happen only when k = b. For if k < b. i.e. k = b ' = min (k+(1/2) dk, b) < b, then it implies that inin (k + (1/2) dk, b) = (k -t (1/2) elk = b', where b ' E S i.e. k+ (1/2) dk is in S and is greater than k which is a contradiction to the fact that k is the 1.u.b of S. Thus we have shown that k = b E S. In other words there exists a positive number db (correspoilding to b) such that Jx l - xzl < db, XI, x2 E [a, b] implies (f(xl) - f(xz)( < E . Therefore f is uniformly continuous in [a, b].

You may note that uniform continuity always implies continuity but not conversely (see Exercise 10). Converse is true when continuity is in the bounded closed interval.

Before we end this unit, we state a theorem without proof regarding the continuity of the inverse function of a continuous function. "

THEOREM 6 : Inverse Function Theorem Let f: I -- J be a functioe which is both one-one and onto. If f is continuous on I, then f-' : J -. I is cbntinuous on J.

For example the function

1 f: 1 ?, -- [-I, I ] defined by

f(x) = sin x, is both one-one and bnto. Besides f is continuous on [ $, f ] . Therefore, by Theorem 6, the function

f : - 1 1 - Li, $1 defined by

I fm'(x) = sin-'x

is continuous on [-I, I.].

10.4 SUMMARY

In this unit you have been introduced to the properties of continuous functions on bounded closed intervals and you have seen the failure of these properties if the intervals are not bounded and closed. In Section 10.2, these properties have been studied. It has been proved that if a function f is continuous on a bounded and closed interval, then it is bounded and it also attains its bounds. In the same section we @roved the Intermediate Vdue Tbeorem that is if f is continuous on an intervd containing two points a and by then f takes every value between f(a) and f(b). In Section 10.3, the notion of u n i f o ~ continuity is discussed. .We have proved that if a function f is .uniformly continuous in a set ,A, then it is

Limit and Continuity continuous in A. But converse is not w e . It has been proved that if a function is i continous on a bounded and closed interval, then it is uniformly continuous in that

interval. These properties fail if the intervals are not bounded and closed. This has been illustrated with a few examples.

El) Continuity at any point c of 1-cw, m[ follow easily because (f(x) - f(c)J = Ix - cl < E if JX - CI < 6 where 6 = E.

Range of f = ]-a, m[ which is not bounded and so f is not bounded.

E2) The function F and G given by F(x) = 1 and G(x) = ( ~ - 2 ) ~ , V x €12, 3[, are continuous and G(x) s 0 in 12, 3[ and,. so,

*

F(x) 1 = - 1 i.e., f(x) = - is continuous in 12, 3[. Its range 11, m[

G(x) (x-2)2 (x-2)l

is not bounded.

E3) Continuity of f is proved in Unit 9.

Range of f = 10, I[. g.1.b. f = 0 & 1 .u.b. f = 1 and they are not attained.

E4) Continuity o f f can be proved easily, (Refer Unit 9) g.1.b. o f f is 0 which is not attained by f,

E5) f(x) = 16x4 + 64x4 - 32x2 - 1 17 being a polynomial is a continuous function on the interval

Hence by Corollary I of Theorem 3 there exists an xo in 11, 2[ such that f(xd = 0, i.e., there exists a root xo, 1 < xo < 2, of the equation

E6) Let f(x) = cos x -x. Then f is a continuous knction on [0, n]. f(0) = 1 > 0 and fin) = - (1 + n) < 0. Hence there exists an x in ] 0, x[ such that f(x) = 0. i.e. there exists a real root for cos x-x = 0 between 0 and x .

E7). Let f(x) = azntl xht' + ...... + a, be a polynomial of odd degree, a,,,, * 0. It is a continuous function on the whole of the real line R. We will suppose without loss of generality that > 0.

' f(x) Then, lim - = > 0. X-MO xZntl

~ e n c e we can find a real number b large enough so that f(b) > 0.

(Justify this) lim = + > 0. Hence we can find a real X-PJ

number a such that f(a) < 0 (when x is negative, x2"*' is negative). f is continuous on the interval [a, b], f(a) < 0 and @) > 0. Hence, by Corollary 1

, of Theorem 3, there exists a real number xo in ]a, b[ such that f(xJ = 0. So, xo is a real root of the polynomial f

then f is a continuous fhnction on the whole real line R and hence on the intervals [-I, 01, [O, 11 and [2, 31 also.

f(-1) = -4-9 + 6+2 < 0 and f(0) = 2 s 0.

Hence there exists a root x, in the interval 1-1. Cl[ s.t. f(xJ = 0. Again,

Hence there exists a root xi in the interval ]0,1[ s.t. f(x,) = 0. f(2) = 32 - 36 - 12 + 2 < 0 f(3) = 108 - 81 - 18 -t- 2 0.

Therefore, again, there exists a root x, in the interval ] 2,3 [.

~ g ) f(x) = xn, n > 1. Already we have proved in Example 7, that f(x) = xZ is continuous on R but not uniformly continuous. The proof for a general n > 1 is very much similar. Let now f(x) = xn, 6 > 0 be arbitrarily chosen and kept futed. Let E be any positive number. Choose x > 1, where

2& i.e. n > - . Take y = x + 612, then (x-y( = E/2 < S and x, y > 1.

6

2& > (612) n > jN2) - = e.

S Therefore f is not uniformly continuous.

1 E10) f(x) = -, for 0 -= x < 1.

X

Let a be fixed such that 0 .< a < 1. Then xn (0 < xn < 1) converges to a

1 1 implies that - converges to - , hence f is continuous at a, So, a ~ ] 0 , 1 [

x; . a

being arbitrary, f is continous on 10, I[. Let 6' > 0 be an arbitrarily chosen positive number and kept fixed.

X Take x < 6 and y = - , Clearly 0 < y < x, so (x - yl = x - y < x < S, and

1 3-x

1 1 - - - = 1. Thus, there does not exists any 6 > 0 fulfilling the job description Y x

when E = 1. SO, f is not unifonnly continuous.

E l l ) f(x) = sin l/x, 0 < x < 1.

1 The function g(x) = -; is a coitinuous function on ]0,1[ and also

h(x) = sinx is a continuous function on 10, 11. Hence f(x) = h(g(x)) is a continuous function on 10, l[. We will now prove that it is not uniformly continuous on 10, l[. Let 0 < E < 1 and 6 > 0 be any positive number.

Properties oTContinuoua Fuoctions

1 Take 1 x = - 1

= (nn -t- ~ 1 2 ) , Then

n7c

If(x) - f(y)l = 1.

So, for 6 > 0, choose n so that x < 6. Then, x, y E 10, 6[, implies that

(x - y( < 6 but (f(x) - f(y)J = 1 > E.

Hence f is.not unifonnly continuous.

E12) Let E > 0. Choose 6 < s l JcJ. Then, whenever (x - yl < 6, we have If(x) - f(y)( = Icx - cyl = I c 1 Jx - yl < I c 1 6 .< E and, so, f is uniformly continuous.

In this block, you have been introduced to the concept of the limit of a functic~n f(x) as x tends to a point 'a'. You were also acquainted with the notion of the sequential limit. Subsequently, the notion of continuity ail9 uniform continuity of a function has been discussed. Further the properties of functions continuous on bounded closed intervals have been proved. You have also seen the failure of' these properties if the functions are continuous on intervals which are not bounded or closed. You should now attempt the following self-test questions to ascertain whether or not you have achieved the main objectives of learning the material hl this block. You may compare your solutions/mwers with those given at the end.

1. Find the h i t s of the fouowing 'fuGctions:

1 (i) f (x) = xcos -, x # 0, a s x - 0.

X

1x1 (ii) f (x) = - , x # 0, a s x +m X

sin x (iii) f (x) = - , x # 0, asx - 00

2. For the following functions, find the limit, if it exists:

fi-6 ' (i) f (x) = -- for x # b where b > 0, as x -- b

X - b

(ii) f (x) = 1

for x # 0, as x -. O f

1 + e- 'IX

1 - x when x 4 1 (iii) f (x) = , a s x - 1 .

2xwhenx > 1.

3. Test whether or not the limit exists for the following: 3 - x whenx > 1

(i) f(x) = lwhenx = 1 , a s x - 1. 2x when x < 1.

x2 - 4 (ii) f(x) = - , x E R, asx - 1.

x2 4- 4

(iii) f(x) = I'2T-F- 2

,.x # Oasx - O X

4. Discuss the continuity of the followirlg functions at the points noted against each.

x2 for x z 1. (i) f(x) =

0 for x = 1.

a s x - 1.

x2 - 4 (iii) f(x) = - when x # 1.

x - 1

(iv) f(x) = (1 ; x)'" if x # 0

i f x = O asx - 0.

1 x2 sin - if x z 0

if x = 0

5. Show that the function f : R - R defined as

does not attain its infimum.

6. Show that the function f : R -. R such that f(x) = x is not bounded but is continuous in [I, oo[.

7. Which of the following functions are uniformly continuous,in the interval noted against each? Give reasons.

(i) f(x) = tan x, x E [0, ?r/4]

1 (ii) f(x) = - on [ I , 41.

x2 - 3

8. Which of the following functions have removable discontinuity at x = O?

sin x for x f 0

for x Z 0

C 1 for x = o

9. Give an example of the fqllowing;

(i) A function which is nowhere continuous but its absolute value is everywhere continuous.

(ii) A function which is continuous at one point only.

(iii) A linear function which is continuous and satisf~s the equation f(x + y) = f(x) + f(y).

(iv) Two uniform continuous functions whose product is not uniformly continuous.

10. State whether or not the following are true or false.

(i) A polynomial function is continuous at every point of its domain.

(ii) A rational function is continuous at every point at which it is defined.

(iii) If a function is continuous, then it is always uniformly continu~us,

(iv) The functions ex and log x are inverse functions for x > 0 and both are continuous for each x > 0.

' (v) The functions cos x and cos-'x are continuous for all real x.

(vi) Every continuous function is bounded.

(vz) A continuoas function Is always monotonic.

(viii) The function dn x is monotonic as we!l as csntilalnous for x E 0, - . [ 3 (ix) The function cos x is continuous as weil as monstonic for every

x E R.

(x) The function 1x1, x E R is continuous.

1. (i) Limit is zero, since x cos - 5 ! X I I :I and limit of 1x1 as x tends to 0 is zero.

A

(ii) lim f(x) = Iim - = 1. x-Q, x - m X

sin x 1 ' (iii) 1 1 zs -

1 for x # 0 and Iim - = 0.

1x1 x-- 1x1

So lim f(x) = 0. x- or)

1 2. ji) -

2 6 '

(ii) 1.

(ii) lirn f(x) = 0 and lim f(x) = 2. So lim, f(x) does not exist. x-1- x-I+ x- 1

3. (i) lim f(x) = 2 and lim f(x) = 2. x-I- x- l t

:. lim f(x) exists and is 2. x- 1

3 (ii) - -

5 '

1 (iii) -.

4 1

(iv) -. 32

4. (i) Lim f(x) = 1 but f(1) = 0. So f is discontinuous at 1. x- I

(ii) Lim f(x) = 1 and Lirn f(x) = 0. So Lim f(x) does not exist. x-It x-1- x-1

So f is discontinuoys at x = 1.

(Ti) Lim f(x) does not exist. So f is discontinuous, x- 1

(iv) Lim f(x) = e but f(0) = 1. So f is discontinuous at 0. x- 0

(v) Lim f(x) = 0 but f(0) = 1. So f is discontinuous at 0. x- 0

5. Inf. f = 0 which is not attained by f.

6. Range of f = [l, oo[ which is not bounded.

7. Both the functions are uniformly continuous since they are continuous in bounded closed intervals.

8. (i) and (iii).

9. [ f(x) = 1 if x is rational

= - I if x is irrational

(ii) [ f(x) = x if x is rational

= - X if x is hratiational the only point of continuity is 0.

(iii) f(X) = CX, V x E R where C is a fixed constant. (iv) f(x) = x, g(x) = sin x, V x E R

Both f(x) and g(x) are unifowy continuous but their product f(x) g(x) = x sin x is not uniformly wntinuous on R.

10. (i) True (ii) True (iii) Fallse (iv) True (v) True (vi) False (vii) False (viii) True (ix) False (x) True

UNIT 11 IDEWIVAIFIVES

Structure

1 I. 1 Introduction Objectives

11.2 Derivative of a Function Geometrical [nterpretation

1 1.3 Differtntiability and Continuity 1 1.4 Algebra of Derivatives 1 1.5 Sign of a Derivative 11.6 Summary 1 1.7 Answers/Hints/Solutions

-- I1 .I INTRODUCTION

You have been introduced to the limiting process in various ways. 117 Block I, this process was discussed in terms of the limit point of a set. The limit concept as applied to sequences was studied in Block 2. In Block 3, the limit concept was formalized for any function in general. It was used to define the continuity of a function. We now consider another important aspect of the limiting process. This is in relation to the development of the derivativc of a function.

You may think for a while that perhaps therc is some cl~ro~~ologicnl order in the historical development of the limiting process. However, this is, perhaps not the case. As a matter of fact. historically Differential Calculus was created by Newton and Leibnitz long before the structure of real members was put on the firm foundation.

Moreover, the concept of limit as discussed in Unit 8 was framed much later by Cauchy in 1321. How, then, is the limit concept used in the development of the definition of the

' derivative of a ftinction? This is the first and thc forcmost question, we have to tickle in this '

unit. Besides, we have to answer a few more related questions viz. What is the geometrical. I meaning of the derivative of a fi~nction? The answer to this questi~n will help you in

appreciating the geometrical significance of some impcit.siit theorems to be discusscd in Unit 12,

The limit concept is common to both continuity and differentiability of a fimction. Does it indicate some connection between the notions of contin~rity and tlifferenliability? If so, then what is the relationship between the two notions? We shall find suitable answers to these questions. Also, we shall discuss the characterization of lhe monotonic functions (refer to Unit 4) with the help of the derivative of the function

I

i OBJECTIVES I I Therefore, after studying this unit, you should be able to I define the derivative of a fi~nction at a point and give its geometrical meaning

@ apply the algebraic operations.of addition, subtraction, multiplication and division on the derivatives of functions

, @ obtain a relationship between the continuity and differentiability of a function characterise the monotonic functions with the help of their derivatives.

1 . 2 DERIVATIVE OF A FUNCTION

The well-known British Mathematician Issac Newton (1642-1727) and the eminent German ;mathematician G.W. Leibnitz (1646-1716) share tlie credit of initiating Calculus towards the ' end of seventeenth century. To some extent, it was an atternlit to answer proble~ns already

'

tackled by ancient Greeks but primarily Calcullls was created t9 trcat some major problcms viz.

, i) To find the velocity and acceleration at any instant of a moving object, givcn a funclion describing the position of the object with respect to time.

Differentiability '. ii) To find the tangent to a curve at a given point.

i To frnd the maximum or minimum value of a function.

These were some of the problems among others which led to the development of the derivative of a function at a point. We define it in the following way :

DE#IN~TION 1: DERIVATIVE AT A POINT Let f be a real function defined on an open interval'la, b[. Let c be a point of this interval so that a c c 2: b. The function f is said to be differentiable at the point x = c if

lim f(x) - f(c) x+c X - c

.-• . . exists and is finite. . . We denote it by f l(c) and say that T is derivable at x = c r or 'f has derivative at x = c' or simply that f'(c) exists. Further, f'(c) is called the derivative or the differential co-efficient of the function f at the point c.

Note that in the definition of the derivative, to evaluate the limit of the quotient '

3x1 - f(c) X - C

at the point c, the quotient must be defined in a NBD of the point c. In other words, the function f must be defined in a NBD of the point c. It is because of this reason why we define the derivative of a function at a point c in an open interval ]a, b[.

However, at the end points a and b of the interval [a,bJ, we can define one sided derivatives as follow. For, let

exists and is fmite, then we say that f is derivable from the right at c. It is denoted by f r (c +) or Wr(c). Also it is called the right hand derivative of f at c. ,Similarly, if

exists and is finite, then we say that f is derivable from the left at c. It is denoted by I fr(c-) or Lf r(c). It is also called the left hand derivative of f a t c,

From the definition of limits in Unit 8, if follows that fr(c) exists if and only if Lf'(c) and Rfl(c) exist and Lfl(c) = Rfl(c)

i.e., f'(c) exists ~f'(c), Rf'(c) exists and Lf'(c) = Rfl(c). For example, consider the function f defined on ]a, b[ as

f(x) = x2, Vx €]a, b[,

Let c be an interior point of ]a, b[ i,e., a < c < b. Then

Lf'(c) = lim ax) - flc) X - C

I - \

Similarly, you can calculate Rf'(c) and obtain Rfr(c) = 2c. This shows that Lf'(c) = Rf'(c) = 2c. Hence f'(c) exists and is equal to 2c.

Now, let us conside: the question : What happens if f is defined in a closed interval [a, b] and either c = a or c : b or c takes any value in the interval? To answer these

I

questions, we give the following defmitions:

DEFJN~TIO~ 2: DERIVATIVE IN AN INTERVAL Let the function f bydefined on the closed interval [a, b]. Then

6 (i) f is said to be derivable a t the end point a i.e. f'(a) exists, if -- - --

l ib f(x) - f(a) exists. In other words, I+.+ x - a

ff(a) = lim f(x) - f(a) a+@+ x - 11

ii) Likewise we say f is derivable at the end point b, if

lim 0 ) - f(b) a+b- X - 11 exists and

f(x) - f(b) I ft(b) = lirn r+b- x - b

i 1 iii) If the function f i s derivable at each point of the open interval ]a, b[, then it is said to be derivable in the interval ]a, bl.

(iv) I f f is derivable at each point of the open interval ]a, b[ and also at the end points a and b, thkn f is said to be derivable in thc closed interval [a, b].

We can similarly define the derivability in [a, b[ or ]a, b] or ] - a, a[ or ] - a, a[ or ]a, m[ or [a, m[ or R = ] - w, m[.

Note that for finding lirn '(*) - ") , generally we write x = c + h, so that x c is x-c+ X -> C

equivalent to h 4 0. AEcordingly, then we'have

and f'(c) = lirn f(c + h) - f(c)

h h-+O

Now let us discuss the following example :

EXAMPLE 1 : Let f : R + R be a function defined as +

I

(i) f(x) = x", V x ER, where n is a Bxcd positive integer, and

I

ii) f(x) = k, V x ER, where k is any fixed real number.

i Discuss the differentiability of 'f at'any point x ER.

SOL'IJTION : (i) Let c be any point of R. Then C

I L f(x) - xn - cn I = lim - I X-SC X - C X - M X - C

Since 'c is arbitrary point of R, therefore fl(x) exists for all x ER. It is.given by , f (x) =' nx"', V x c R, . 8

EXAMPLE 2 : Let a function f :[0,5] 4 R be defined as

2x + 1, when 0 s x < 3 Rx) =

x2 - 2, when 3 s x s 5

. Der ivat ives

Is f derivable at ic = 3 1

Differentiability SOLUTION ; f y 3 -) = lim f(x) - f(3)

; 3 X - 3

2(x - 3) = lim = 2,

x+3 X - 3

and f (3 +) = lim f(x) - f(3) x+3. X - 3

(x2 - 2) - 7 = lirn

3 X - 3

= lim (x + 3) = 6, and, so, x+3

f'(3 -) # f '(3 +) a ft(3) does not exist i.e. f is not derivable at x = 3.

Now, try the following exercises.

EXERCISE 1 Let f: R + R be defined

show that f'(0 +) * f'(0 -).

EXERCISE 2 (i) Find the paints at which the function f: R + R defined by

f(x) = I X - 11 + I X - 21, V x ER, is not derivable.

ii) Prove that fi R + R defined by f(x) = x 1x1, Q x ER, is derivable at the origin.

1

EXAMPLE 3 : Let f: R + R be defined as f(x) = x2 cos (11~) if x # 0 and f(0) = 0. Find the derivative of f a t x = 0, if it exists.

f(x) - floe) = lim x2cos(l/x) SOLUTION : lim

x+o x - 0 x+O X

1 = lirn x cos (-)

X x 4

1 1 Also cos -takes values between - 1 and 1 and thus is bounded i.e. Jcos - 1 5 1. Hence

X X

lirn 1 - '(O) = limx cos - = 0,

1-0 X - 0 x - 0 X

. So that P(0) exists and is equal to 0.

EXERCISE 3 Let f:R-.R bedefinedas

= 0, i f x = O Is f derivable at x = 07

EXAMPLE 4 : For the function, f defined by

f(x) = (log x I (x ) 01, determine f' (1 +) and f ' (1 -).

Derivatives

SOLUTION : ff(l +) = lim 4x1 - (1)

x+l+ X - 1

f(l + h) - f(1) = lim'

h+Ot h

= lim Ilog(l+ h) I - (log 1 I 'h+O h

= lim log (1 + h)"" h+O

1 = log e = 1.

log(1 - h) Also f'(1 -) = lim = -1.

h-,O h

I EXERCISE 4 !

-e-I/' (i) Given : f(x) = x. + e-l" ' if x + 0 and f(0) = 0. Determine ff(O +) and ff(O -),

(ii) Let f be a function defined by X

f(x) = - V X E R . ,

1 + 1x1

Show that f is defferentiabIe everywhere.

(iii) If the function given by

axZ + b, x 2 0 f(x) =

x2 Iog, x 0

possesses derivative at x = 0, then find a and b.

(iv) Let f be an even function defined on R. If fl(0) exists, then fihd its value.

11.2.1 Geometrical Interpretation of the Derivative

One of the important problems of Geometry is that of finding or drawing the tangent at any point on a given curve, The tangent describes the direction of the curve at the point

i and to define it, we have to use the notion of limit. A convenient measure of the direction of the curve is provided by the gradient or the slope of the tangent. This slope varies from point to point on the curve. You will see that the problem of finding the tangent and its gradient (slope) at any point on the curve is equivalent to the problem of finding the derivative of: the function y = f(x), which represents the curve. Thus, the tangent to the curve y = f(x) at the point with abscissa x exists if the function has a derivative at the point x and the tangent slope = fl(x). This is what is called the geometrical interpretation of the derivative of a function at a point of the domain of the function. We explain it as follows.

i

Fig. 1 7 .

Differentiabil ity Let f be a differentiable function on an interval I. Tlie graph of f is the set

{(x, y)) : y = f(x), x E: I ) .

Let c, c -F h EI, SO that P(c, f(c)) and Q(c + h, f(c + h)) are two points on the graph of i.:

Therefore, the slope of the line PQ is the nurnber

f(c + h) - -- f(c) . !.e., - f(c + h) - fjc) h

- = tan LQ, (C -t 11) - h

Also as h +O, Q -+ P. By definition, the derivative o f f at c is

f(c 4- h) -- f(c) f'(c) = lim -

II+O h

= lim (slope of PQ). Q+I'

In the limit, when Q + P, the line PQ becomes the tangent at P. Therefore,

f'(c) = lirn (slope of PQ) = slope of the tangent line ro the curve y = f(x) at P. Q-tP

Thus, when f'(c) exists, it gives the slope of the tangent line to the graph of f at the poirit (c, f(c)). That is fyc) is the tacgent of the angle which this tangent line at (c, f(c)) makes with the positive direction of the axis of x.

If fl(c) = 0 the tangent line to the graph o f f at x = c is parallel to'the axis of x and if f'(c) exists and does not have finite value, then the tangent line is parallei to the axis of y.

11.3 DIIFFEWENTXDILITY AND CONTINUITY

You have seen that the notion of limit is essential and common for both the continuity and the dii'ferentiability of a function at a point. Obviously then there should be some relation.between the continuity of a function and its derivative. This relation is sane as the one between curve, the graph of the function and existance o f a tangent to the curve. A curve may have taiagant st all points on it. It inay have no tangent at some points o ~ i it. For instance in the Figure 2(a), the curve has tangents at all points on it while the curve in Figure 2@) has a point P, s sharp point P, where no tangent exists. In fact, differentiability of function at a point implies smooth turning of the corresponding curve along that point. And, so; a curve can't have a tangent at sharp points.

The fact that a curve is continuous does not neccssarily imply that a tangent exists at all . points on the curve. I-Iowever, intuitively it follows that if a curve has a tangent at a point, then

, the curve must be continuous at that point. Thus, it follows that the existence of a derivative (tnngent to a curve) of a function at a poult implies that the fcnction is continuous at that

point. Hence differentiability of a function implies the continuity of the function. However, a continuous function may not be always differentiable. For example, the absolute value function f: R -, R defined as f(x) = 1x1, 'd x ER, is continuous at every point of its domain but it is not differentiable at the point x = 0 because, at x = 0, there is a sharp bending in its graph. This is evident from the graph of this function which you can easily see in Unit 4.

Now we prove it in the form of the following theorem.

THEOREM 1 : Let a function f be defined on. an interval I. I f f is derivable s t a point c €1, then it is continuous at c.

PROOF : Since f is derivable at x = c, therefore, lirn f(x) - f(c) exists and is equal to x+c x - C

f f(c). Now, f(x) - f(c) = f(X) - f(C) . (X -,c), for x + C. X - C

This implies that

lim fix) = f(c). X+C

That is, f is continuous at x = c.

We have given the proof for the case when c is not an end point of the interval 1 If G is an end point of the interval, then lirn is to be replaced by lirn or lirn according as c is left

X 3 C x--)E+ X+c-

end point or the right end point of the interval.

Thus, it follows that continuity is a necessary condition for derivability at a poi;^!. However it is not sufficient. Many functions are readily available which are continliolls at a point but not derivable there at. We give below example of two such functions.

EXAMPLE 5 : Left f : R -, R be the function given by

Then f is continubu~ at x = 0 but it is not derivable there at.

SOLUTION : Recall from Unit 4 that f(x) is of the form

We claim that f is continuous at x = 0, for

lirn f(x) = lirn f(x) = 0 = f(0). (See the graph in Unit 4). h+O+ h+O-

Now, f(x) - f(0) - x - 0

f'(0 +) = lim - lim - - h+O+ x - 0 h+O x - 1 ,

- x - 0 and,. ff(O -) = lirn - = - 1 .

x+O X - 0

Thus f is not derivable at x = 0.

.I . EXERCISE 5 Justify that f : R + R defined as ,

(i) f(x) = 1x1 -t- Ix - 11 is continuous but not derivable at x = 0 and x = 1.

(ii) f(x) = 1x1 -t- Ix - 11 + IX - 21 is continuo;s but not derivable at x * 0, 1, 2.

EXAMPLE 6 : Let f: R + R be defined as

x, for 0 2 x < I f(x) =

1, for x 2 1.

Derivatives

Differentiability Then f is not derivable at x = 1 but it is continuous at x = 1.

SOLUTION : Clearly, lirn f(x) = lirn f(x) = 1 = f(1). X+l+ x+l-

This shows that f is continuous at x, = 1. Now

f(cj'L f(l) f'(1 +) = lirn

x+l+ X - 1

x - 1 and f'(1 -) = lim -- = 1 i.e., fr(1 +) # lim ff(l -1.

x+l x - 1

This shows that f is not derivable at x = 1.

From the above examples, it is clear that derivability is a more restrictive prope1-Q than continuity. One might visualise that if a function is continuous on an interval, then it might fail to be derivable at finitely many points at the most in the said interval. This, however, is not true; there exists functions which are continuous on R but which are not derivable at any point whatsoever. In 1872, German Mathematician, K. Weierstrass, first gave an example of such a function. Here we mention an example due to Van der Waerden. The function is defined as

where a = 112 or -112 according as x 2 0 or x < 0. This function is known to be co~rtinuous everywhere but derivable nowhere. , -

Now try the following exercise.

EXERCISE 6

Prove that a function f : R + R defined as

1 f(x) = x sin -, x # 0; and f(O) = 0, is continuous but not derivable a t the origin.

X

11.4 ALGEBRA OF DERIVATIVES

You have seen that whenever we have a new limit -definition a natural question arises. How does it behave with respect to the algebraic operations of addition, subtraction, multiplication and division? We discussed the algebra of convergent sequences in Unit 5. We also discussed algebra of limits and continuous function in Block 3.

In this section, we shall discuss some theorems regarding the derivability of the sum, product, quotient and composite of a pair of derivable functions.

I. SUM OF TWO DERIVGBLE FUNCTIONS.

Let f and g be two functions both defined on an interval I. If these are derivable at c €1 then f + g is also derivable at x = c and

(f + g)' (c) = f '(c) + gl(c).

PROOF : By definition, we have

Then

lim (f +.g) - (f + g)(c) - -Ilm , - f(x) + g(x) - f(c) -g(c) K-bC X - C X - C

= lim f(x) - KC) + lim g(x) - ,g(c)' x-c x- C x-c X - C

= f '(c) + g'(c). 5 .

=. ( f + g)'(c) = f'(c) + g'(c). Thus f + g is derivable at x = c.

In the same way you can also prove that f - g is also derivable at x = c and . (f - g)'(c) = f'(c) - gl(c).

I1 PRODUCT OF TWO DERIVABLE FUNC'TIONS

Let f and g be two functions both defined on an interval I. If these are derivable at c E I, then f.g, is also derivable at x = c and

PROOF : By definition, you ]lave

f(x) - f(c) = t!!! x - c

and lini g(x) - g(c) = g.(c) x--c+ X - C

By using the above two definitions of f'(c) and g'(c) as well as the algebra of limits (refer to unit 8$, we have

I

I l i n~ (fg) (') - "') ('? exists and is equal to a-.L x - c

f'(c) . g(c) + f(c) . g'(c). =-- (fg)' (c) = f'(c) . g(c) + f(c) . g'(c) ijevce fg i s derivable at x = c. ,

If a function f is derivable at a point c, thcn for each real number k, the funclion kf is also derivable at c and

I (Itf)' (c) = k . f'(c).

1 For the prooc take f = k, g = f in Result I1 and use the fact that derivative of a consta~~t function is zero everywhere.

III QUOTIENT OF TWO DERI VABL E FUNCTIONS

I Let f and g be two hnctions both defined on an interval I. Iff and g are derivable at a point '

c E 1 and g(c) # Q, then the function f/g is glso derivable at c and

/ (f[,); (c) = ycc, "(0) - !(c) . s'Cc) i I g(c) l2 I , PROOF : By definitions, we havc

f(x) - f(c) = f.((,) lim 1 \ ' L x - C

and

. g(x) - g(c) Iim = g'(c) - X - c .

Now '

Differentiability

Proceeding to limits as x + c, keeping in mind that f and g are derivable at x = c and g(c) z 0, we get

which proves the result. 1 In particular, let f be derivable at c and let f(c) # 0, then 7 is derivable at c and

(1 19 ' (c) = - f '(C)/{~(C))~.

This is known as the Reciprocal Rule for Derivatives. For its proof, take f(x) = 1, and g = f in result I11 and use the fact that derivative of a constant function is zero everywhere.

Iv. CHAJN RULE. Let S and T be subsets of R and f : S + T, g: T -h R be two functions. I f f is derivable a t c ES and g is derivable at f(c) ET, then g.f is derivable at c and

(go0 (c + h) - (gof)(c) PROOF : Let ~ ( h ) =

h , h#O.

Now, we have to show that lim ~ ( h ) exists and is equal to gr(f(c)).f'(c). Let us define h-to

a new function 4 : T + R as

Observe that

if h + 0. Then lim ~ ( h ) = lim +(h).fl(c), by derivability of the function f at c, provided h-to h 4

lim &h) exists. h-DO

Thus, the proof of the theorem will be complete if we can show that l i i $ (h) exists and is equal to g'(f(c)).

Now, to show that liy ~ ( h ) = gr(f(c)), observe that, by derivability of g,

which implies that given E > 0, 3 6 > 0, such that

And f is derivable at c a f is continuous at c 3 for 6 > 0, 3 5' > 0 silch that Ih( 6' =, 1 f(c + h) - f(c)l< 6.

Derivatives

Ler us consider a number h such that Jhl < 6'. We have to consider ihe two cases :

(i) f(c + h) = f(c), and (ii) f(c + h) # f(c).

In case (i), by definition of 4 (h),

I (C (h) - gl(f(c)) 1 = 0 4 E. (4)

Zn case (il), if we write f(c + h) - f(c) = k + 0, then, by the dkfinition o i ~ ( I I ) ,

Now, by (3!,

JliJ .: 6' 3 Jf(c -1.. h) - !'{c)) 1 4 6 =. O < IkJ < 6, by the definition of k,

by (5). By (4) and (6), we get

lhl < 6 ' s I $(h) - gf(f(c)) I c c z lirn 4 (h) = gJ(f(c)), as was to be shown. This conlpletes the proof.

h-,O

I dx I Alternately, we can say that if y = f(x) and z = g(y), wherc both - and 3 exist, then I dy dx I I dz

- exists and is given by i dx

Recall that this form of chain rule is generally used in problems of Calculus. I I

For sxa~nple, to find the derivative of the function

f(x) = (x2 f x2 + 2)25, Iet :ft = h(u) = 1 1 ~ ~ ~ where u - x J + xZ + 2. Then

We now show how to differentiate the inverse of ,a difi'erentiablc function. Let f be a o?e-one differentiable function on an open interval I. Thcn f is strictly increasing or decreasing arid the range f(J) o f f is an interval J, say. Then the inverse function g = f-' has, the domain J and

fog = i,, gbf = i,,

where i, and i, are the identidjr function on I and J, respectively. The11 you know that ~ ( x ) = Y g(y)=x, V X E I , ' ~ E J .

Differentiability Consider any point c of I. We have as'sumed that f is derivable at c. A natural question arises: Is it-possible for g to be derivable at f(c)? If it is so, then under what conditions? We discuss this question as follows :

Now f is derivable at c. If g is derivable at f(c), then, by the chain rule for derivatives, g o f is derivable at c and (g o f)' (c) = g ' (f(c)) ff(c).

But ( g o f l (x) = g(f(x)) = X, V x €1. Therefore,

(go f)' (x)= 1, trx €1.

In particular for x = c, we get

(g 0 f) '(c) = 1 3 g '(f(c)). f'(c) = 1 3 f'(c) * 0.

n u s for g to be derivable it is necessary that ff(c) #,O i.e., the condition for the inverse o f f to be derivable at a point f(c) is that its derivative f' must not be zero at point c i.e., ff(c) $ 0. In other, words, we can say that, if ff(c) = 0, then the inverse of f is not' derivable at c. Thus we find that a necessary condition for the derivability of the inverse function of f at c 'is that f'(c) # 0. Is this condition sufficient also? To answer this question, we state and prove the following important theorem :

THEOREM 2 : INVERSE m C I 2 O N THEOREM Suppose f is one-one continuous function on an open interval I and let J = f(I). I f f is differentiable at x, €1 and if f'(x,,) $0, then f1 is differentiable at yo = f(x,) EJ and

1 Crl>' 6,) = 7

f (xd PROOF : Note that J is also an open interval, by Intermediate Value Theorem. Since f is differentiable at x, €1, therefore

Since ff(x,) s 0 and f being one-one, f(x) $ f(xo), for x # x,, we have

1 - 1 X - X, lim . - -- i.e., iim

1 =- -x. fCx) - fix,,) f'(x0) *-UO f ( ~ ) - f(XO) ~'(xO)

a

X - X,

So, given E > 0, there exists 6 > 0 such that

Let g = PI. Since f is one-one continuous function on I, therefore, by inverse function ' theorem for continuous finctions, the inverse hnction g is continuous on J. In particular, g is continuous at yo. Also, g is one-one. Hence, there exists q > 0 such that

0 < l'g(y) - ~ ( Y J I < 6 , for 0 c I Y - Y,I < 71

i.e., 0 < (g(y) -xoJ <ti, for 0 < ( y -yo( <q . (8)

From (7).and (a), we get

< E , for 0 < ( y - y,l < 9.

1 lt follows that lim g(y) - '0 = - Y% f(g(y)) - f(xo> fl(xO) '

Now yo = f(xo) x, = g&) and f(g(y)) = y. Therefore, lim 1 g(y) - ~ [ Y J -

y 3 . 3 Y - Y o f'(xo) 1

Hence g is differentiable at yo and gf(y0) = - , Replacing g by f I, we can say that f' (xo)

f1 is differentiable at yo and

To illustrate the above theorem, consider the following example,

EXAMPLE 7 : Find the derivative at a point yo of the domain of the inverse function of the function f, where f(x) = sin x, x E:] - 7~12, 71/2[.

SOLUTION : You know that the inverse function g o f f is denoted by sin-'. Domain of g is ] -1, I[. Since f is one-one continuous function on ] - d 2 , x/2[ and it is differentiable at all points of ] - 7~12, 71/2[, using the above theorem, you can see that g is differentiable in ] - 1, I [ and if yo = sinx, is any point of ] - 1, 1[, where x, EJ -7112, n/2[, we have

1 1 gf(yo) =r = - f(xo) cosx,'

And, since cos x, = -o = q. 1 1

Hence, gf(yJ =- i.e., (sin-')' (yo) = - 4q? fi; ' Try the following exercise.

EXERCISE 7 Find the derivative at a point yo of the domain of the inverse function of the function f, where f(x) = log x, x €10, [.

11.5 SIGN OF A DERIVATIVE

In this section, we shall discuss the meaning of the derivative of a function at a point I being positive or negative. For this, we have to recall the idea of monotonic or monotone

functions which we have already discussed in Unit 4. But here we require the concept of

, increasing or decreasing function at a point of the domain of the function. So we give all I these concepts in the following definition.

DEFINITION 3 : MONOTONIC FUNCTIONS

! Let f be a function with domain as interval I and let c €1. Then f is said to be an i i increasing (or a decreasing) function in the interval I if, for x,, x, EI, I

x, < x, f(x,) 5 f(xJ (or f(x,) 2 f(x2), respectively). I

I Also, f is said to be strictly increasing (or decreasing) in I if, for x,, x, €1,

I xl < x2 3 f(x,) l f(x,) (or f(x,) 2 f(xJ, respectively).

I Using these concepts, we say that f is an increasing function at a point x = c if, there exists a 6 > 0 such that f is increasing in the interval ]c -6, c + 61.

I 1 Again, we say that f is a decreasing function at a point x = c if, there exists a S > 0

j such that f is decreasing in the interval ]c - 6, c + 61.

Finally, f is said to be monotone or monotonic in I if either it is increasing in I or it is I decreasing in I. We can similarly define strictly monotone (or monotonic) functions. 1

Obviously the function f defined by f (x) = xZ in [0, I ] is an increasing function. And, the function f defrned by

. f(x) = llx in [ l , 21 is a decreasing function.

Now we give the significance of the sign of the derivative of a function at a point. I

Derivatives

It is often possible to obtain valuable information a'noul a function fro111 he knowledge of the sign of the derivative of a hnction.

We discuss the two according as the derivative is positive or negative i.e.,

for some x in the domain off .

Case ( i ) Let c be any interior point of the domain [a, bj sf a function f.

L,el fl(c) exist. Suppnsc f' jc) > 0.

f(x) - f(c) This means lim = f (c) :: 0.

x-bc x - c

Thus, for a giver! E (0 < E 5: f'(c)), theie exists a 6 v 0 such that

f(x) - qcj . i.e., x E]C - 8, c + 61, x + c a fl(c) - E <

X - C f'(c) + E

3 f'x)-f(c) > 0, by the choice of E which is less than P(c). X - C

Therefore, for all x E]C, c -t- 61, f(x) f(c) and, for all x E]C- 6, c[, YX) <f(c). Thus f increadng at x = c. Now, let P(c) 0. Define a function 4 as

$ (x) = - f(x), \d x E [a, b]. So +Yc) = -f'(c) 0. Therafore, using the above proved result, there exists F > 0 such that

\dx E]C, C + 6[, #(x) > 4 ( ~ ) 3 f(X) < f ( ~ ) .

and, x E]C - S ,c[, &x) < S(C) r s ~ f(x)> qc).

Thus f is decreasing at x = c.

We now consider the end points of the interval [a, b].

Case (ii) Consider the end point '8 ' . You can show as in case (I), if f'(a) exists, ihere exists 6 > 0 such that

f'(a) > 0 3 f(x) > f(a), for x €]a, a 3- 5[,

and fya) < 0 3 f(x) < f(a), far x la, a + 6[ .

Case (iii) Consider the end point @bl. You can show that there exist 6 r 0 such that f'(a) > 0 3 f(x) < f(b), for x ~ ] b - 6, bl,

and fl(h) < 0 2 f(x) > f(b), for x ~ ] b - 6, b[.

Consider the following examples to make the idea clear.

EXAMPLE 8 ; Slrow that the function f, defined on R by

f(r) = x3 - 3x2 + 3x - 5, V x ER.

is increasing in every interval,

SOLUTION : Now f(x) = x3 - 3x1 i- 3x - 5. Therefore, f'(x) = 3x2- 6x + 3 = 3(x -

fl(x) > 0, when x # 1.

I I Let c be any real number less than 1. Then f is continuous in [c, I] and f'(x) > 0 in ]c, I[. , 1 This implies that f is increasing in [c, 1 [.

I

18 Sjinilmiy, f is increasing in every interval ] 1, dl, where; d is any real number greater than 1. We find that f is increasing in every interval.

I ,

EXAMPLE 9 : Seperate the intervals in which the function f defined on R by f(x) = 2x3 - 15xz + 36x + 5, V x ER, is increasing or decreasing.

SOLUTION : Here f(x) = 2x" 15 x2 + 36x + 5, therefore

ff(x) = 6x2 - 3 0 ~ + 36 = 6(x2 - 5x + 6)

= 6 ( ~ - 2 ) ( ~ - 3)

so that ff(x) > 0, whenever x > 3 or x < 2. Thus f is increasing in the intervals ] -03, 21 and [3, a[. Also ff(x) < 0, for 2 < x < 3. Therefore f is decreasing in the interval [2, 31.

Now try the following exercises.

I

i EXERCISE 8 Separate the intervals in which the function, f, defined on R by

1 f(x) = x3 - 6x 2 + 9x + 4, V x ER, I I is increasing or decreasing.,

EXERCISE 9 I Show that the function f, defined on R by

f(x) = 9 - 12x + 6x 2- x2,V XER, 1

is decreasing in every interval.

Let f be a fbnction with domain as an interval I c R.

I Let I, = {x, GI: f f(x,) exists ). If I, # 0, we get a function f'with domain I,. We call flthe I derivative or the first derivative of f . We also denote the first derivative o f f by fly or Df.

If we write y = qx), x €1, then the fmt derivative of f(x)= y is also written as * or y, or DY. dx

Again, let I, = {t EI,J ff (t) exists ). If I, + 0, we get a function (f3' with domain I,, which we call second order derivative of f and denote it by f" or 5. We can define

I

I higher order derivative o f f in the same way. We shall discuss higher order derivatives in Unit 13. In the meantime, let us study the following example.

EXAMPLE 10 : Let f : R -- R be defined as

Show that f"(0) exists. Find its value.

SOLUTION : For x # 0, clearly 1 1

ff(x) = 4x3 sin (-1 - x2 cos (T) X

while

ff(0) = lim f(x) - f(0) x-0 X ' 0

1 = limx3 sin (-) = 0. ,

x-0 X

Thus we get 1 1

f!(x) = 4x3*sin (;-) - x2 cca (-;), if x + 0

Derivatives

f '(x) - f '(0) Now f"(0) = lim -

x-0

4x3 sin (1 /x) - x2 cos (1 /x) = lim

x -0 X

= l i i[4x2 sin (1 /x) - x cos (1 /I)] = 0.

Now try the following exercise.

EXERCISE 10 I f f :R--Risdehedas

f(x) = sin (sin x) +f x E R, then show that

P'(x) + tan x f'(x) + cos2x f(x) = 0.

In this unit, we have discussed the differentiability of a function. In Section 11.2, we defined the derivative of a function fat a point c of its domain, an open interval ]a, b[. If

lim f(x) - exists, then the limit b called the derivative off at 'cl and is denoted by fl(c). If x-c X - C

we consider the right hand limit, lim f(x) - f(c) and it exists, then it is called the right hand x - c t X - C

derivative of f a t 'c' and is denoted by RfP(c). Likewise lirn . f(x) - f(c) , if it exists, is called x-c- x - ' C

the left hand derivative off at c and is denoted by L fg'(c). From the definition of limit i t follows that f'(c) exists Lff(c) and Rfg'(c) bolh exist and Lfr(c) = Rff(c). If f is derivable at each point af the open interval ]ti, b[, then iJ is said to be derivable in ]a, b[. If Lhe function f is defined in the closed interval [a, b], then f is said to be derivable at the left end

point 'a' if lim ~ - b - f(x) x - - f(a' a exists and the limit is called derivative off at 'a' and denoted by

f '(a).

Similarly, if lirn - f(b) exists, that f is said to be derivable at 'b' and the limit is denoted x-b- x - b

by f'(b) and is called the derivative off at 'b'. The function f is said to be derivable in [a, b] if it is derivable in the open interval ]a, b[ and also at the end points 'a' and 'b'. In the same section, geometrical interpretation of the derivative is discussed and you have seen that the derivative f'(c) of a function fat a point 'c' of it domain represents the slope of the tangent at the point (c, f(c)) on the graph of the function f. In Section 11.3, the relationship between the differentiability and continuity is discussed. We have proved that a function which is derivable at a'point is continuous these at and illustrated that the converse is not true always. In Section 11.4, the algebra of derivatives was considered. It has been proved that iff and g are derivable at a point c, then f + g, fg are derivable at 'c' and (f f g)' (c) = f '(c) k g

g'(c),

(fg)' (c) = f(c) g'(c) + fg'(c) g(c). Further if g(c) # 0, then f/g is also derivable a t c and

Also in this section the chain rule for differentiation is proved, that is, if f and g are two functions such that the range off is contained in the domain of g and f, g, are derivable respectively at c, f(c) then g o f derivable at c and (gof)'(c) = g'(f(c)), f'(c). Result concerning the differentiatioh of inverse function is discussed in the same section. If f is one-one continuous function on an open interval I and f(1) = J and iff is differentiable at

1 xo E I, f'(x0) # 0, then f- ' is differentiable at yo = f(xo) f J and ( f I ) ' (yo) - . Finally

f 7x0) in Section 1 1.5, you have seen that a function f is increasing or decreasing at a point 'c' of its domain if its derivative f t(c) at the point is positive or negative.

Derivatives

X ~ 1 ) fyo-)=lim f(x)-f(0) = l i m - = l i m l = 1 .

x-to- X - 0 x-to X x-to

0 ff(O+) =lim f(x)-f(0)= lim - = lim 0 e 0.

x-tot X - 0 x-to X x-to

So, ff(O -) # ff(O +).

E 2) (i) The given function f, is

The critical points are 1 and 2. At other points, f is derivable. Show that ft(l -) = - 2, fT1 +) = 0, f'(2 -) = 0, ft(2 +) = 2.

The only points, where f is not derivable, are x = 1 and x = 2.

(ii) The given function is

f(x) = x2 if x 2 0.

x2 x2

f'(0 +) = lim - =: lim -= lim (x) = 0 x-0 X x-0 X x-0

- X L

f'(0-) =l im-=l im(-x)=O x.-0 X x-0

Since f.'(O +) = f'(0 -), therefore f is derivable at 0.

f(x) - f(0) E 3) ff(0) = lirn - x-0

I 1 I = limsin -

x-0 X

Limit does not exist. Hence f is not derivable at x = 0.

E 4 ) (i) f '(O+)= 1, f f (O-)2z- 1.

(ii) The given function is

I X X Since - I + x (X > 0) and - (x < 0) have non-zero polynomials in their

1 - x denominators, they are derivable in their respective domains. Check directly at x = 0 and find that

ff(O +) = f'(0 -) = 1. So f is derivabley x E R.

(iii) Given that f has dcrivativ* at x = 0 Therefore ff(O +) = f'(0 -)

x210g x - b i.e, lim = lim (ax2 + b) - b

x-0 X k-0 X

i 1 = lim (ax) = a . 0 = 0.

x-to

1 1 Since f is derivable at 0, therefore, f is continuous at x = 0.

Then, f(0 +j = f(0 -) 3 lim x2 log x = b. x-Ml

Differentiability It is known that lim x210g x = 0" x - 0 4

So b = 0. Then from ( l ) , a . O = limx log x

x-0

= 0, which holds tt a E R. Hence a is arbitrary real number and b = 0.

(iv) Since f is an even funaion, f(- x) = f(x) U x E R. Also f'(0) exists implies f '(0 t) = f'(0 -) = f '(0)

Therefore f'(0) = lirn f(x) - f(0) x-0 x - 0

f(- x) - f(0) = lim (f is an even function f(X) = f(- x))

x-0 X - 0

f(- x) - f(0) = - lim = - f'(0)

- - 0 - X - 0

$hich gives 2f'(0) = 0 =3 f'(0) = 0.

E 5 ) 0) I x i f x 1 O Now Ix I =

- x i f x < O

x - 1 i f x 1 1 and Ix - 1 I =

1 - x i f x < I

f(x) can be written as

lim f(x) = lhn 1 = 1 and lirn f(x) = lirn (1 - 2x) = 1. x--rO+ x+O+ x-ro- x-ro

Thus, lim f(x) = lirn f(x). x+o+ x-io

Therefore, f is continuous at 0.

Now lim f(x) = lim (2x - 1) = 1 and lim f(x) = lim 1 = 1. x+l+ x-i l x-11- X-r l

Thus, lim f(x) = lirn fix) = f(1). x-r l- X+l+

Therefore, f is continuous at 1.

f(x) - f(0) = lh 1 -2x- 1

ff(O -) = lim = - 2, and so, x - 1 0 - X - 0 x 4 X

1 fr(O +) z fl(0), which implies that f is not derivable at 0.

I f(x) - f(1) = lim 2 x- 1- 1 f'(1 +) = lim = lim 2 = 2;

x+l+ x - 1 X+l X - 1 x-rl

1 - 1 f f( l -)= lirn f ( ~ ) - f ( l ) = , ~ ~ - - 0.

x+l- x - 1 x - b l X - 1

f f( l+) r f'(1-), which implies that f is not derivable at 1. (ii) Proceed as in (i).

1 E 6) limf(x) = limx sin - = o

x-0 x-0 X

1 (sin. -is bounded and limx = 0)

X x-0

22 So F is continuous at x = 0

f(x) - f(0) f '(0) :- lim

1 = lirnsin - which does not exist.

r-O X - 0 h-0 X

E 7) 'The inverse functic~n e, o f f is given by g(y) =- eY. Domain of p is R, the sc! of real nurnbers. Since i is one-oile con:inuous function 10n 10, ,=[ and i t is dnfferentiahie at all poinis of 10, mi, SO g is differentiable in R and ~f yo r= leg xo be any point of R where xoE 10. "[, we l~eve

Nov; yo = lo& XI, 3 XO = eY0

Hence g'(y~l) I= eSo.

E S ) Iiere f(x) = xJ - 6x2 'r Yx + 4V x E l i

:. f'(x) = 3(x7 4x + 3) = 3(x - l ) ( x -- 2 )

So that f ' i x ) > 0, whenever x > 3 or x c: 1

Therefwe f is increasifig in the intervals, ] - x, I ] and 13, on[.

Also Y(x) :: 0 for 1 < x < 3.

Therefore f is decreasing in the interval 11, 31.

E 9) f(x) = 9 - 12x I- 6x' -- x' Therefore fJ(x) = -- 12 -I- 12.u -- 3 x 2 - -- 3 (x? --- 4x + 4)

= -- 3 (X - 2j2 .< 0, fur J: + 2. Lxt c be arip member Ic.js tha!i 2. Then ff(xi < 0 in ]c, 21. 'Therefore f is decreasing in [c, 2j. Similarly f is decrensing in i2, dl, whcre t l is any real number greater than 2

Hence we get that f is decrensing in every interval.

E 10) f'(x) -- cos (:sin x) ctis x, f"'(x) = -- sin (sjn x) cos2x cos (sin x) . sin:;

f '(x) = - f(x) cus'x - --- sinx

COS X

UNIT 312 MEAN -VALUE THEOREMS

Structure 12.1 Introduction

Objectives

12.2 Rolle's Theorem 12.3 Mean Value'Theol.em

Lagrmge's Meal] Value Theorem Cauchy's Mean Value Theorem Generalised Mean Value Theorem

12.4 Intermediate Value Theorem for Derivatives Darboux Theorem

12.5 Summary 12.6 A~swers/Hints/Solutions

12.1 INTRODUCTION

In Unit 11, you were introduced to the notion of derivable functions. Some interesting and very useful properties are associated with the functions that are continuous on a closed interval and derivable in the interval except possibly at the end points. These properties are formulated in the form of some theorems, called Mean Value Theorems which we propose to discuss in this unit. Mean value theorems are very imporjant in Analysis because many useful and significant results are deducible ftom them. First we shall discuss the well-known Rolle's theorem. This theorem is one of the simplest, yet the most fundarncntal theorem of real analysis. It is used to establish the mean-value theorems. Finally, we shall illustrate the use of these theorems in solving certain problems of Analysis.

Objectives After studying this unit, you should be able to

know Rolle's theorem and its geometrical meaning 9 deduce the mean value theorems of differentiability by using Rolle's the or en^ + give the geometrical interpretation of the mean value theorems 9 apply Mean Value Theorems to various problems of Analysis @ understand the Intermediate Value Theorem for derivatives and the related Darboux

Theorem.

12.2 ROLLE'S THEOREM

The first theorem which you are going to study in this unit is Rolle's theorem given by Michael Rolle (1652-1719), a French mathematician. This theorem is the foundation stone for all the mean value theorems. First we discuss this theorem and give its gemetrical interpretation. In the subsequent sections you will see its application to various types of problems. We state and prove the theorem as €allows

THEOREM 1 : (ROLLYS THEOREM) If a function f : [a, b] -. R is

(i) continuous on [a, b] (ii) derivable on ]a, b [, and (i) f(a) = f(b), then there exists at least one real number c €]a, b[ such that f'(c) = 0.

PROOF : Since the function f is continuous on the closed interval [a, b], it is bounded and attains its bounds (refer to Uei: 19). Let sup. f = M and inf. f = m. Then there are points c, d E [a, b ] such that f(c) = M and f(d) = m. Only two possibilities arise : Either M = m or M Z m. Case (i) When M = m.

Then M = m 3 f is constant over [a, b] .- f(x) = k V x E [a, b], for some fixed real number k. 3 f'(x) = OV x G[a, b].

Case (ii) : When M Z m. Then we proceed as follows :

Since f(a) = f(b), therefore at least one of the numbers M and m, is different from f(a) (and also different from f(b)).

Suppose that M is different from f(a) i.e. M # f(a). Then it follows that f(c) f f(a) which implies that c # a.

Also M # f(b). This implies that f(c) # f(b) which means c f b. Since c # a and c # b, therefore c E ]a, b[.

Again, f(c) is the supremum off on [a, b]. Therefore

f(x) 5 f(c) V x E [a, b] * f(c - h) 5 f(c), =

I

I for any positive real numbers h such that c - h E [a, b]. Thus

for a positive real number h such that c - h E [a, b].

Taking limit as h - 0 and observing that ff(x) exists at each point x of ]a, b[, in particular at x = C, we havr f'(c -) 2 0 Again f(x) I f(c) also implies that f(c + h) - f(c)'

5 0 h

for a positive real number h such that c 4- h E ra, b]. Again on taking limits as 11 - 0, we get ff(c +) 1 0.

But f'(c -) = f'(c +) -- f'(c). Therefore f'(c -) 2 0 and f'(c +) 5 0 imply tint

f'(c) 5 0 and f'(c) 2 0

which gives f'(c) = 0, where c E]a, b[. You can discuss the case, m # f(a) and m # ((b) in n similar manner.

Note that under the conditions stated, Rolle'r theorem guarantees the existence of at least one c in ]a, b[ such that' f'(c) = 0. It does not say *mything about the existence or otherwise of a

more than one such number. , I s we shall see in pral~lems, for a given f, therc may exist several numbers c such that f'(c! = 0.

Next we give the geometrical significance of the theorem.

Geometrical Interpretation of Rolle's Theorem

, Mean-Value Theorems

Fig. 1

Differentiability You know that f'(c) is the slope of the tangent to the graph of f at x =.c. Thus the theorem simply states that between two end points with. equal ordinates on the graph off, there exists at least one point where the tangent is parallel to the axis of X, as shown in the Figures I.

After the geometrical interpretation, we now give you the algebraic interpretation of the theorem.

Algebraic Jnterpt-etation of Rolle's Theorem You have seen that the third condition of the hypothesis of Rolle's theorem is that f(a) = f(b). If for a function f, both f(a) and f(b) are zero that is a and b are the roots of the equation

' f(x) = 0, then by the theorem there is a point c of ]a, b[, where ff(c) = 0 which means that c is a root of the equation f'(x) =: 0.

Thus Rolle's ttleorem implies that between two roots a and b of fix) ='a, there always exists at least one root c of f'(x) = 0 where a < c < b. This is the algebraic interpretation of the theorem.

Before we take up problems to illustrate the use of Rolle's theorem you may note that the hypothesis of Rolle's theorem cannot be weakened. To see this, we consider the following three cases :

Case (i) Rolle's theorem does ncjt hold i f f is not continuous in [a, b].

For example, consider f where

I x i f O 5 x < 1 f(x) =

O i f x = 1. Thus f 1s continuous everywhere between 0 and 1 except at x = I . So f is not continuous in [O, 11. Also it is derivative in 10, I[ and f(0) = f(1) = 0. But f'(x) = 1 Y x E]O, I[ i.e. f'(x) + OU x x 10, 11.

Case (ii) The theorem no more iemains true i f f ' does not exist even at one point in ]a, b[. Consider f where

f(x) = I X 1 v- x E I- I , ir.

Here /is continuous in [- 1, I], f(- 1) = f(l), but f is derivable'ff x E 1- 1, 1 [ except at x = 0:

- 1,- 1 < x < O Also fl(x) =

Hence there is not point c G 1- 1, I[ such that ff(c) I= 0.

Case (iii) The theorem does not hold if f(a) # f(b). For example iff is tbe function such that f(x) = x in [l,2], then I

f(1)' = 1 z 1 + 2 = f(2).

Also ff(x) = 1 V x G ] I , 2[ i.e, there is no point c E ] 1, 2[ such that E'(c) =: 0.

Now we consider one example which illustrates the theorem :

EXAMPLE 1 : Verify Rolle's theorem for the fi~nction f defined by

(i) f(x) = x3 - 6x2 f: 11x - 6 V x E [I, 31.

(ii) f(x) = (x - a)"' (x - b)" x E [a, b] where m and n are positive integers.

SOLUTION : (i) Being a polynomial function, f is continuous on [ I , 31 and derivable in 11, 3[. Also €'(I) = f(3) = 0. Now ff(x) = 3x2 - 12x + 1 1 = O

4 1 I - x=2+- - - , 2 - - - 6 n

Clearly bolh of them lie in ]1,3[.

(4) f(x) = (X - a)" (X - b)"

Obviously f is continuous in [a, b] and derivable in ]a, b[.

Also f(a) " f(b) = 0.

NOW f'(x) = m(x - a)"-' ( X - b)" +(x - a)" ( X - b)"-' = 0 implies that ( X - a)"-' (x - b)"-' [rn(x.- b.) t n(x - a)] = 0 i.e, m(x - b) + n(x - a) = 0.

(As x f a or b : we wan1 those points which are in ]a, b[). na + rnb

Thus x = + , This is point c and i t clearly lies in ]a, b[. YOU may note from Example I(i) that point c is not unique. Now you should be able to try the following exercises :

EXERCISE 1 Verify Rolle's Theorem for the function f where f(x) = slin x, x E [- 2rr, 2 4 .

EXERCISE 2 Examine the validity of the hypothesis and the conclusion of Rolle's theorem for the function f defined by

(a) f(x) = CQS x'o' x E [- ~ / 2 , ~ / 2 [

(b) f ( ~ ) = 1 + (X E LO, 21.

Next we give an example which shows application of Rolle's Theorems to the theory of equations.

EXAMPLE 2 : Show that there is no real number A for which the equation x3 - 27x + A = 0 has two distinct roots in [O, 21.

SOLUTION : Let f(x) =. x ' - 27x + A.

Suppose for some value of A, f(x) = 0 has two distinct root (Y and P that is f has two zeros' n and p, a # p in [O, 21. Without any loss of generality, we can suppose, cu < P.

Therefore [a, P] C [0, 21.

Now f is clearly continuous on [a, PI, derivable in ]a, ,!3[ and (a) = f(P) = 0.

Therefore by Rolle's theorem, 3 c E ](Y, P[ such that f'(c) = 0 - 3c2 - 27 = 0 - c2 -9=0==3c=&3 .

' Clearly none of 3 or - 3 lies in 10, 2[, whcnce - 3 or 3 $ ]a , p[.

Thus we arrive at a contradiction. Hence the result.

EXERCISE 3 Prove that between any two real roots of ex sin x = 1, there is at least one real root of excosx+ 1 = O .

, EXERCISE 4 i Prove that if ao, a], ..., a, E R be such that

a0 ;a1 - +-+ ... +a,-l n t l n n + a, = 0, then there exists at least one real number x between 0

and 1 such that

a0xn + alxW-' + ... + a, = 0.

Next examples show how Rolle's Theorem helps in solving some difficult problcrns.

Mean-Value Theorems

EXAMPLE 3 : I f f and g are continuous in [a, b3 and derivhble in ]a, b[ with g'(x) Z OY x E ]a, b[; prove that thete Cxisb c C ]a, b[ such that

SOLUTION : The result to be proved can be written as .' ' f(c) g'(c) 4 fl(c) g(c) - f(a) g'lc) - g(b) f'(c) = 0 the left hand side of which is the derivative of the function f(x) g(x) - f(a) g(x) - g(b) f(k) at x = c. This suggests that we should aljply Rolle's Theorem to the function d where

44x1 = f(x) g(x) - f(a) g(x) - g(b) f (x) ,v x E [a, bl.

Since f a;d g are continuous in [a, b] and derivab!e in ]a, 8, therefore #J is continuous in [a, b] and derivable in ]a, b[. Also $I (a) = - g(b) f(a) = +fb);! So 4 satisfies all the conditions of Rolle's Theorem. Thus there is a point c in ]a, b[ such that G'(c) = 0 that is

f(c) gf(c) f ff(c) g(c) - f(a) gf(c) - g(b)f'(c) = 0

which proves the result.

EXAMPLE 4 ; If a function f is such that its derivative, f' is continuous on [a, b] and derivable on ]a, b[, then show that there exists a number c E ]a, b[ such that

1 I@) = f(a) + @ - a) f'(a) 4- y (b - a)' Ff(c).

SOLUTION : Clearly the functions f and f ' are continuous and ddrivable on [a, b].

Consider the function where 4lx) = f(b) - f(x) - (b - X) fO(x) - (b - x)' A , V x E [a, b] wherc A is a constant to be determined such that

4 (a) = 4 (b). .'. f(b) - f(a) - (b - a) ff(a) - (b - a ) ' ~ = 0 ( 1 ) Now +, being the sum of continuous and derivable functions, is itself continuous 6n

, [a, b] and derivable on ]a, b[ and also +(a) = +(b), for the value of A given by (1).

Thus #J satisfies all the conditions of Rolle's theorem. Thereforc there exists c E ]a, b[ such that #Jf(c) = 0.

Now $(x) = - f l ( x ) + f f (x) - (b - X ) ff'(x) 4 2(b - x)A This gives 0 = 4'(c) = - (b - c) ff'(c) + 2(b - c)A

I which means A = - f"(c) since b # c. 2

,Putting the value of A in ( I ) , you will get 1

f(b) = f(a) + (b - a)' f '(a) 4 - (b - a)' f "(c). 2

EXERCISE 5 Assuming f" to be continuous on [a, b], show that

b- c c- a 1 f(c) - f(a) . - - f(b) * = - (c - a) (c - b) f"(d)

b - ,a b a 2 where both c and d lie in [a, b].

Note that the key to our proof of the above examples 3 and 4 and Exercise 5 and many more such situations, is the judicious choice of the function, 4, and many students compare it with the magician's trick of pulling a rabbit from a hat. If -one can hit at a proper choice of 4. the problems are more than half done.

12.3 MEAN VALUE THEOREM

In this section, we discuss some of the most useful results in Differential Calculus known as the mean-value theorems given again by the two famous French mathematicians Cauchy and

LagrangL'Lagrange pr0vi?d a result only by using the first two conditions of Rolle's theorem. Hence i t is called Lslgrango's Mean-Value Theorem. Cauchy gave another mean value theorem in which lie used two functions inslcad of one function as in the case of Rolle's theorem and Lagrange's Meqn-Value Theorem, You will see later that Lagrange's theorem is a particular of ~ $ ~ c h ~ ' s Mean Value Thcorem. Finally, we discuss the generalized form of these two theorems. We begin with Mean-Value Theorem given by J.L. Lagrange 11736-181 31

THEOREM 2 : LAGRANGE'S MEAN VALUE THEOREM

If a function f : [a, b] - P is (i] continuous on [a, b] and (ii) derivable on ]a, b[, theq there exists at least one point c E ]a, b[ such that

f'(c) = f(b) - f(a)

b - a

Mean-Value Theorems

(Tllis is also known as the First Mean Value Theorem of Differential Calculus.) J.L. L.ugrnsgu

PROOF : We define a new function $ as follows : cp(x) = f(x) -I- AxV x E [a, b] where A is a conetaut to be chosen such that 4(a) = $(b). $(a) = f(a) + Aa and $(b) f(b) + Ab. $(a) = &(b) gives

Now the function @, being the sum of two continuous and derivable functions is itself

(i) continuous on [a, b] (ii) derivable on ]a, b[, and (iii) @(a) = $(b).

Therefore by,Rolle's theorem 3 'a real number c E ]a, b[ such that +'(c) = 0. But cpf(x) = f'(x) + A So Q = $'(c) - fl(c) + A

f(b) - f(a) which means that f'(c) = - A = - a In the statement of the above theorem, sometimes b is replaced L;' i i + h, so that the number c between a and b a n be taken as n + Oh where 0 < 0 < 1. Accordingly then, the theorem can bc restated as follows : '

Let f be defined and continuous on [a, a 4- h] and derivable on ]a, a + h[, then there exists 4, 0 < 6 .f: 1 such that ffa + h) = f(a) + bf' (a + Oh).

Certain important and useful results can be deduced from Lagrange Mean-Value Theorem. We state and prove these results as follows :

You already know that derivative of a constant function is zero. Conversely if the derivative of a function is zero, then it is a constant function. This can be formalized in the following way :

I, If a function f I s continuous on [a, b], derivable on ]a, b[ and f'(x) = O V x E ]a, q, then f(x) = k T x E [a, b], where k is some fixed real number.

.To prove it, let A be any point of [a, b].

.Then [a, A] C [$, bl.

Thus f is

i) continuous on [a, A]

ii) derivable on ]a, A[

Therefore, by Lagmnge's mean value theorem, 3 c E ]a, A[ such that

Differentinbility Now f'(x) = 0 V x E ]a, b[ * f'jx) = OV x E ]a, h[ i. f'(c) = 0

f(h) = f(a) V h €[a, b]

But X is any arbitrary point of [a, b]. Therefore

f(x) = f(a) = k (say) V x E [a, b].

Note that if the derivatives of two functions are equal, lhen they diffzr by a constant. We have the following formal result :

11. If two functions f and g are ( i ) continuous in [a, b], (ii) derivable in la, b[ and (iii) f'(x) = g'(x) V x € ]a, b[, then f - g is a constant function.

PROOF : Define a function & as

$(x) = f(x) - g(x)Yx E [a, b]

Therefore &'(x) = 0 Y x t- ]a, b[ because it is given that

f'(x) = gf(x) for each x in ]a, b[

Also 4 is continuous in [a, b], therefore,

+(x) = k, Y x € [a, b],

where k is some fixed real number. This means that

i(x) - g(x) = k V x € [a, b]

i.e. (f - g) (x) = k Y x E [a, b]. Thus f - g is a constant function in [a, b] The next two results give us method to test whether the given function is increasing or decreasing.

111. If a function f is (i) contiriuous on [a, b] (ii) derivable on ]a, b[ and (iii) f'(x) > 0 V x E ]a, b[, then f is strictly increasing 011 [a, b].

For the proof, let X I , x2 ( X I < x?) be any two points of [a, b]. Then f is continuous in [X I , x:] and derivable in 1x1, XZ[, SO by Lagrange's mean value theorem, f(x2) - f(x1)

= ff(c) > 0, for x, < c < xz X? - X I

which implies that

f(x2) - XI) > 0 a f(xz) > ~ ( X I ) for xz > x,

Thus f(x2) > f(x1) for xz > X I .

Therefore f is strictly increasing on [a, b]. If the condition (iii) is replaced by ff(x) 2 0 'd- x E [a, b], then f is increasing in [a, b] since you will get f(xz) > f(x1) for xz > X I .

IV. If a function f is (i) continuous on [a, b] (ii) derivable on la, b[ and (iii) f'(x) < 0 V x E la, b[ tlien f is strictly decreasing on [a, b].

Proof is similar to that of 111. Prove it yourself. If condition (iii) in IV is replaced by ff(x) 5 0 V x E ]a, b[, then f is decreasing in [a, b].

The result 111 and IV remain true if instead ol[a, b] we have the intervals [a, a[, ']- m, b], I- *, a[, la, a [ , I- b[, etc. Note that the conditions of Lagrange's mean value theorem cannot be weakened. To see this, consider the following examples :

(1) Let f be the function defined on [I , 21 as follows :

! 1 i f x = 1 f(x) = x' if l < x < 2

2 i f x = 2

Clearly f is continuous on [ I , 2[ and derivable on 11, 2[, it is not continuous only at x = 2 i.e. the first condition of Lagrange's 'Mean Value Theorem is violaled.

Also f(2) - f(1) = 2 - 1 = 1 .

2- 1

and f'(x) = 2x for 1 < x < 2.

If this theorem is to be true then f'(x) = 1 i.e. 2x = I i.e. x = 1/2 must lie in 11, 2[, which is clearly false.

(2) L,et f be the function defined on [- 1, 21 as

' f(x) = 1 x 1 .

Here f is continuous 011 [- 1, 31 and derivable at all point of 1- 1, 2[ except at x = 0, so that the second condition of Lagrange's Mean Value Theorem is violated.

( 2 - f - 1 - 2 - + 1 ) - 1 Also - - - - -

2 - (- 1) 2 3

f(2) - f(- 1) so that # f'(x) for any x in 1- 1,2[.

2 -- (- 1) We may remark that the conditions of Lagrange's mean value theorem are only sufficient. They are not necessary for the conclusion. This can be secn by considering the function on [0, 21 defined as :

I 1 1 For-<x <-yf'(x) = 1.

4 2 In particular, f'(3/8) = 1.

I f(2) - f(0) - 2 - 0 Also - 1 = f ' (3/5)

2- 0 2- 0 even though f is neither continuous in the interval [O, 21 nor it is derivable on 10, 2[, since f is neither continuous nor derivable at 1/4 and 1 /2. Now you will see the geometrical significance of Lagrange's Mean Value Theorem. Geometrical Interpretation of Lagrange's Mean Value theorem

Fig. 2

Mean-Value Theorems

Differentiability Draw the graph of the function f between the two points A(a, and B(b, f(b)). The

number f(b) - f(a) gives the !lope of the chord AB. Also fl(c) gives the slope of the tangent b - a

to the graph, at the point P(c, f(c)). Thus the geometrical meaning of Lagrange's Mean Value theorem is stated as above :

If the graph o f f is continuous between two points A and B and possesses a unique tangent at each point of the curve between A and B, then there is at least one point on the graph lying between A and B, where the tangent is parallel to the chord AB.

Before considering example, we have another interpretation of th'e theorem.

We know that f(b) - f(a) is the change in the function f as x changes from a to b so that

{f(b) - f(a))/(b - a) is the average rate of change of the function over the interval [a, b]. Also f'(c) is the actual rate of change of the function for x = c. Thus, the Lagrange's mean value theorem states that the average rate of change o fa function over an interval is also the actual rate of change of the function at some point of the interval.

This interpretation of the theorem justifies the name 'Mean Value' for the theorem.

Now we consider an example which verifies Lagrange's Mean. Value Theorem.

EXAMPLE 5 : Verify the hypothesis and conclusion of Lagrange's mean value theorem for the functions defined as :

1 ii) f ( x )= logxVxE [1,1 te].

SOLUTICJN : (i) Here f(x) = l/x; x E [ l , 41.

Clearly f is continuous in [I, 41 and derivable in 11, 4[. So f satisfies the hypothesis of Lagrange's mean value theorem. Hence there exists a point c E ]1,4[ satisfying

Putting the values off and f', you get

which gives c2 = 4 i.c. c = t 2. Of these two values of c, c = 2 lies in ]I, 4[.

(ii) Here f(x) = log x; x E [ l , 1 t e-'1. Clearly f is continuous in [ l , 1 + e-'1 and derivable in 11, 1 t e-I[.

Therefore the hypothesis of Lagrange's mean value theorem is satisfied by f. ~herefdre there exists a point

c E 11, 1 t eFl[ such that

Putting the values off and f', you get 1 log (1 + e-') - log 1 -= c e- I

which gives c = [ e log (1 + e-I) 1".

I You can use the inequality

Now try the following exercises.

EXERCISE 6 Verify Lagrange's Mean Value theorem for the function f defined in [0, n/2] wlwre f(x) - ms x V x E [0, n/2].

EXERCISE 7 Find 'c' of the Iagange's M a n Value Theorem for the function f defined as

s f(x) = x(x - 4 ) (X - 2)V x E [O, 31.

Nc\w you wi!l be givcn cxaniplcs ihowing the use of Lagrange's Mean Vi~lue Theorcm in solving different types of roblems.

EXAMPLE 6 : Piove that for any quadratic functioli, 1x2 + mx f - n, the value of 8 in 1

Lagrange's Mean Value theorem 'is always -, whatever I, m, n, a and h may be. 2 SOL,UTION : l,ct f(,x) =- lx? -1- rnx t n; x E [a, a -I- h]. f being a polynoniial funcliol~ IS conlinuous in [a, a + h] and desivablc in ]a, a + h[. Thus f satisrics the condi~io~is of I,agrangels Mcan Value thcorem.

Therefore thcrc exists 0 ( O c: (I .I I j such 1?1a~ f(a -I- h) - 1'(;1) I - hf'(a -1- (!h)

.Putting the values of C and f' you will gel

lin -I 11)' + ni(a t- 'h) -i. II - la' + 1113 4:. n -1- Ii [2 1 ( i ~ + 0 11) t m] i . ~ . z.2.. lh' - 21 ()h2 which gives O .= 1 /"), ulh:itt:v~~. ill 11, I, In, 11 II~!LJ/ be.

EXAMPLE 7 : if a and b {a < b) arc ~.e!il ~~urnbers, ifiel~ there exists a real number c between a and \, sucli that

SOLJU7'1(9N : Consicier the function, f, dd'ined by i'(i) - x' V x t? [a, bJ.

Clearly f satisfies thc hypothesis of Lagrange's mean value theorcm. Therefore thcre exists c t: ]a, h[ such t1i;lt

[(b) f(a) = -- h - a

which gives

EXERCISE 8 Show that an the curve, y = ax2 + bx + c, (a, b, c E R a # O), the chord joining the points whose abscissae are x = m and x = n, is parallel to the tangent at the point whose abscissa is given by x = (m + n)/2.

EXERCISE 9 Let f be defined and continuous on [a - h, a + h] and derivable on ]a - h, a + h[. Prove that there exists a real number 8 (0 < 6 < 1) for which

f(a + h) + f(a - h) - 2f(a) = h[fl(a + Oh) - f'(a --- Oh)].

With the help of Lagrange's Mean Value Theorcm we,can prove some inequalities in Analysis. We consider the followilig example.

EXAMPLE 8 : Prove that sin x < x for O < x 5 n/2.

SOLLITION : Let f(x) = x - sin x; O 5 x 5 7r/2. f is continuous in [O, n/2] and derivable i l l 10, n/2[.

Differentiability Also, ffx) = 1 - cosx > 0, for 0 < x < xl2.

Therefore f is strictly increasing in [O, ~121, which means that f(x) > f(0) for 0 < x I xI2. (Using corollary I11 of Lagrange's Mean Value Theorem) i.e., x - sin x > 0, for 0 < x 5 xI2. i.e., sin x < x for 0 < x S 7112.

We can also start with the function g(x) = sin x - X, for 0 I x I nl2. Then we have to use corollary IV of Lagrange's Mean Value Theorem to arrive at the desired result.

EXAMPLE 9 : Prove that tan x > x, whenever 0 -= x 7112.

SOLUTION : Let c be any real number such that 0 < c < nl2. Consider the function f, defined by f(x) - tan x - x, V E [0, c].

The function f is continuous as well as derivable on [0, c]: Also, fi(x) = sec2 x - 1 = tan2 x > 0, b' x E]O, C[

Thus f is strictly increasing in [0, c]. Consequently, f(0) < f(c) 3 0 < f(c), which shows that 0 < tan x - x, when x = c.

, ,

This implies, tan x > x, when x ='c'. Since c -is any real number such that 0 < c < 7112, therefore, tanx > x, whenever 0 < x < ~ 1 2 .

X EXAMPLE 10 : Show that - c ]log (1 + x) -= x, b' x > 0.

l + x

SOLUTION :Let f(x) = x - log (1 + x), x 2 0.

1 - X Therefore, f'(x) = 1 - - - - l + x l + x ' '

Clearly, fi(x) > 0, for x > 0. Therefork, f is strictly increasing in [o, m[. Therefore,

f(x) > f(0) = 0, b' x > 0

i.e., x > log (1 + x), V x > 0

i.e., log (1 + x) < x, V x > 0. X

Again, let g(x) = log (1 + x) - -- , x 2 0. Then l + x

Clearly, gt(x) > 0, b' x > 0.

X i.e., log (1 k x) > - V x > O

1 + x ' . x i.e., - < log (1 + x), b' x > 0.

1 + x

Now try the following exercise.

EXERCISE 10

Prove that

' i) x - x' < tan-' x, if x > 0; and

ii) e-' > 1 - x, if x > 0.

Cauchy generalized Lagrange's Mean Value Theorem by using two function as follows.

THEOREM 3 : CAUCHY'S MEAN VALUE THEOREM

Let f and g be two function defined on [a, b] such that

i) f and g are continuous on- La, b]; -.

ii) f and g are derivable on ]a, b[, and

iii) g'(x) # 0 tf x € ]a, b[, '

then there exists at least one.rea1 number c € ]a, b[ such that

I

{This is also known as Second Mean Value Theorem of Differential Calculus.)

ROOF : Let us first observe that the hypothesis implies g(a) # g(b) . i

$Since g(a) = g(b), combined with the other two conditions g has, means g satisfies the 'hypothesis of Rolle's Theorem. Thus there exists c E ]a, b[ such that g'(c) = 0, which violates 3.cohrlition (iii)).

L,et a function $ be defined by

c$(;.) = f(x) + A g(x)tf x € [a, b], wh~xe A is a constant to be chosen such that

.which gives

A = - (f(b) - f(a)j (g(b) - g(a)). _I

(A,, proved above, g(b) - g(a) # 0). Nc w (1) $ is continuous on [a, b], since f and g are so,

(2) $ is derived on ]a, b[, since f and g are so, ant1 (3) $(a) = 4(b).

Thus 4 satisfies the conditions of Rolle's Theorem. Therefore there is a point c E ]a, b[ such thai $'(c) = 0 which means that f'(c) + Ag'(c) = 0

I Alternative statement of Cauchy's Mean Value Theorem If in the statement of above theorem, b is replaced by a + h, then the number c E ]a, b[ can be written as a + Oh where 0 < 8 < 1. The above theorem then can b$ restated as :

Let'f and g be defined and continuous on [a, a + h], derivable on ]a, a + h[ and g'(x) # 0 t fx E ]a, a + h[, then there exists a real number 8(0 < 0 < 1) such that f'(a + Oh) - f(a +h) - f(a) - g'(a + Oh) g(a + h) - g(a) . .

, .As remarked earlier, Lagrange's Mean Value Theorem can be deduced from Cauchy's 1 Mean Value Theorem in the following way 1 In Cauchy's mean value theorem, take g(x) = x. Then g'(x) = 1 and have g'(c) = 1. ~ i s o I g(a) = a, g(b) = b. Result of Cauchy's mean value theorem becomes

I This holds if (i) f is continuous in [a, b] and (ii) f is derivable in ]a, b[ which is nothing but I &agrange's mean value theorem.

Dote that you might be tempted to prove Cauchy's mean value theorem by applying Lagrange's mean value theorem to the two functions f and g separately and then dividing. The desired result caeo t be obtained in this manner. In fact, we will obtain

1 :'(CI) _'f(b)-f(a) --- i t

(~2) g(b) - g(a) ' where CI E ]a, b[ and cz E ]a, b[. Note that here cl is not necessarily equal to cz.

Mean-Value Theorems

1 'AS in the case of Rolle's Theorem and Lagrange's Mean Value Theorem we give geometrical significance of Cauchy's Mean Theorem

1 Geometrical Interpretation of Cauchy's Mean Value Theorem

The conclusion of Cauchy's mean value theorem may be written sr .

Differentiability

This nleans

slope of rh:: chord joining (a, f(a!) and (h, f(bi) -- slope of tht. chord joining (a, g(aj) and (b, g(b))

- slope of the tangent to y = f(x) at (c, f(c)) - - --- --A-

slope of r k ~ iar~gent to y : g(x) at (c, g(c)!

Suppose that two curves y = f(x) and y := g(x) arc continuously d a w n between the two ordinates x = a hnd x = b ds shown in ihe 1:;gul.e 3. Suppose furthc.1 that the Langent can be drawn to each of thc curves at each point lying between Lhrse abscissae ant1 no where does tlie tangent to the curve, y = g(x), between these abscissae become parallel to the X-axis. Then there exists a point c between a and b such that the ratio of the slopcs of thc chords joining the end points of the curves is equal to ratio of the slopes of ihe tangents to the curves at the points obtained by the intersection of the curves and the ordinate at c.

I

Fig. 3

As in the case of Rolle's Theorem and Lagrange's Mean Value Theorem, are now give examples concerhing the verification and application of Cauchy's Mean Valut: Theorem.

EXAMPLE 11 : VerQ Cauchy's Mean ~ k u e Theorem for the functions f and g defined as f(x) = x2, g(x) = x4'd- x E [2, 41.

SOLUTION : The function f and g, being polynomial functions, are conti~luous in [2, 41 and derivable in 12, 4[. Also g'(x) = 4x3 $ O'f x E ]2,4[. All the conditions of Cauchy's Mean Value Theorem are satisfied. Therefore there exists a point c E 12, 4[ such that f(4) - f(2) ff(c) - g(4) - g(2) g'(4

i t . c k l./E c - fi lies in ]2,4[

So Cauchy's Mean Value Theorem is verified.

EXAMPLE 12 : Apply Cauchy's Mean Value Theorem to the functions f aqd g defined as

f(x) = x2, g(x) = x.V x € [a, b], and show tiiat 'c' is the arithmtrtic mean of 'a' and 'b'.

1Mea1a-Vnlue Theorems SOLUTION : Clearly Lhe filnciion f and g satisfy the hypothesis of C a u ~ h ~ ' ~ Mcnn Vallle Theorem. Therefort 3 c E ]a, b[ such that ft(c) f(b) - f(a) - - --- g'(c) 6th) - g(4 Puttirlg the valul::; or' f, g, f'. g' we get 2~ t>2 - ii? _- - _I- _ - -- h -k a 1 b - a -4..

1 - c = -

2 (;\ + 13j. which shows that c is the arithmetic mean of 'a' and 'h'.

sin a! - sin ,!j EXAMPLE 83 : Show that --------- - cot 0.

cos 63 - CCIS IY

I SOP,IJTION : Let f(x) = sin x and gjx) = cos s.

7r where x E 10, 01 C ]O,, [. - Now f'(x) - cos x and g ' (x ) = - sin x Fil~ictions f , ~ n d g are both continuol!~ on [a, /3]. derivable on ]a, Pi', and

g'(x) + O i/- x E ]a, @[.

By Cnuchy's filean vnlue theorem, there. exists O E ]a, f i [ such that sin /3 -- sin a cos - - -- cos p - cos a. - sin O

sir1 ru -- sir) p -= COl O.

cos p - cos eu

Try the following exercise.

, .% a- L -0 4 - d , . - . - - W u ~ --.-- EXERCISE 1 l '

Verify the Cauchy's mean value theorem for the fuactions, f(x) = sin x, g(x) = cosx in the interval 1- x/2 ,0 ] .

I I

' 1 EXTRCSE 12

1 I Let the functions f and g be dsfined us : f(r) = ex and g(x) = c", V x E [a, b]. I

Show that 'r' obtained from Ceuehy's mean value theorem is $ 1 ~ arithmetic mean or's and b.

EXERCISE 13 I Lei f(r) = 41 and g(x) = I/ &, V x a [a, bl given that 0 < a < b. VC& Cnuehy's *earl

1 value theorem and slliow thai c obtained thus is ill@ geometric mearm of a nnd b.

EXERCISE 114 Two functiorls f and g are defined as : f@) = x-' and g(x) =- X-2, V x E (a, b], given tlant 0 e [a, b].

i ' Apply Cauchyrs mean value thecarem and show that c thus obtained is the harmonic I mean of a and b.

The following theorern generalises bath Lngrar~ge's and Couchy's mcun value theorems. In this theorem, threc functions f, g, h we involved. Both Lagrange's and Cauchy's mean value theorems are its spec,ial cases.

THEOmM4 : GENEWJSED MEAN 'bTAl,UETHEOR%M If three functions, f, g and h we continuo~s in [a, b] and derivable in ]a, bh, illen there exists a real number c ~ ] a , b[ such that

Differentiability PROOF : Define the'function, $I , as

for all x in [a, b].

Similarly 4(b) = 0.

Since each of the functions f, g and-h is continuous on [a, b] and derivable on ]a, b[, therefore 4 is also continuous on [a, b] and derivable on ]a, b[. . .

f(a) g(a) h(a)

Thus 4(a) = 4(b)

4(a) =

Therefore 4 satisfies all the conditions of Rolle's theorem.

So there exists c E ]a, b[ such that

f(a) g(a) h(a) f(b) g(b) h(b)

which proves the theorem. .

= 0, since two rows of the determinant are identical.

Now we show that hgrange's and Cauchy's mean value theorems are deducible from this theorem by choosing the functions f and g specially.

i) First we deduce Lagrange's Mean Value Theorem from the Generalized Meal) Value Theorem.

Take g(x) = x and h(x) = 1 V x E [a, b], so that

#(x) = 1

Now 4'(c) = 0 gives fl(c) = f(b) - f(a) which is Lagrange's mean value theorem. b - a

ii) Next we deduce Cauchy's mean value theorem from the Generalized Mean-Value Theorem Take h(x) = 1 V x E [a, b].

f(x) g(x) 1 So that d(x) = f(a) g(a) 1

fl(x) g'(x) . 0 --. 4'(x)= f(a) ~ ( a ) 1 = f'(x) [g(a) - g(b)] - g'(x) [f(a) - f(b)l

Now 4'(c) = 0 * f'(c) [g(a) - g(b)] - gl(c) [f(a) - f(b)] = 0

a-- ff(C) - - f(a) provided gf(x) # 0 for x E b, b[. g'(c) g(b) - g(a)

which is the Cauchy's mean value theorem.

12.4 INTERMEDIATE VALUE THEOREM Mean-Value Theorems

We end this unit by discussing Intermediate Value Theorem for derivatives. Just as you Intermediate Value Theorem for continuous functions in Unit 10, there is an

Intermediate Value Theorem for derivable functions, which we now state and prove.

THEOREMS : (DARBOUX) LNTEEMEDIATE VALUE THEOREM FOR DEWlVATlVES If a function f is derivable on [a, b] and ff(a) # fl(b), then for each k lying between fl(a) and fl(b), there 'exists a point c €]a, b[ such that f'(c) = k.

In case, gr(a) < 0 and gl(b) > 0, then

-gt(a) > 0 and -gf(b) 0.

Therefore, at some point c ~ ] a , b[,

--g'(c) = 0 or - f'(c) + k = 0 i.e., ff(c) = k.

Theorem 5 is due to French mathematician, J.G. Darboux [1842-!917'], which is usefill in determining the rnaxirnuni or minimum values of a function. This is popularly known as Darboux Theorem. Another important result, which is a particular case of Darboux's Intermediate Value Theorem, is as given below.

THEOREM 6 : Let f be derivable in [a, hj. If P(a) aeld ff(b) are of opposite signs, then there exists a poirei c ~ j a , b[ such that fr(c) =: 0.

PROOF .: Since ft(a) and fyb) are of opposite signs, thereforz one of ft(aj or f t(b) is positive and other is negative. Take k = 0 in the Darboux Tt~eoaern. You get a point c ~ ] a , b[ such that f'(c) = 0.

An immediate deduction froan above theorem is that if the der~vative of a hnctioil does not vanish for any point x in ]a, b[, then the derivative has the same sign for all x in ]a, b[. This is proved in the following example.

EXAMPLE 14 : I f f is derivalblle in ]a, b[ and f'(x) + 0, 'd x la, b[, then ff(%!, retains t11c same sigrl, positive or negative, for ali x inn la, b[.

SOLUTION : If possible, suppose x,, x, ~ ] a , b[, x, <: x,, are such that t"(x,), ft(x,) have opposite signs. By Theorem 6 , there exists a point c ~ ] x , , x,[ c]a, b[ such that f'(c) = 0, which is a contradiction. I-Ience f r(x) retains the same sign, for all x in ]a, b[.

---- 82.6 SUMMARY - In this unit mean value theorems of differentiability have been proved. In Section 12.2, Rolle's theorem, the fundamental tlleorem of Real Analysis is proved. According to this theorem: if f : [a, b] -P R is a function, continuous in [a, bj, derivable in ]a, b[ m d f(a) =

f(b), then there is at least one point c ~ ] a , b[ such that f'(c) = 0. 'The geometric significance of the theorem is also given. Geometrically, on the graph of the fu~lction f, there is at least one point between the end points, where the tangent is parallei to the x-axis. Using Rolle's theorem, Lagrange's Mean Value Theorem is proved in Section 12.3. It states that if a function f : [a, b] --+ R is continuousin [a, h] and derivable in ]a, b[, there

f(b) - f(a) is at least one point c in ]a, b[ such that -- = f'(c). An irnportarit consequence of b - a

the theorem is that i f f is continuous on [a, b] and derivable on ]a, b[ with ft(x) = 0 on ]a, b[, then f is a constant function on [a, b]. Another important deduction from the ,,-

theorem is that i f f is continuous in [a, b] and derivable in ]a, b[ then (i) f is increasing or decreasing on [a, b] according a s fl(.x) 2 0, 'd x ~ ] a , b[ or f(x) 5 0, V x ~ ] a , b[ (ii) f is strictly increasing or strictly decreasing in [a, b] according as f'(x) > 0, 'd x €la, b[ or f'(x) .: 0, V x ~ ] a , b[. Applying these results, some inequalities in real analysis are established, With the help of Rolle's theorem, Cauchy's theorem is proved in Section 12.4. It states that if f and g be two functions tiom [a, b] to R such that they are continuous in [a, b], derivable in ]a, b[ and g'(x) + 0, V x ~ ] a , b[, then there exists at least

one point c in ]a, b[ such that f(b) - f(a) = - fl(c) . Lagrange's Mean Value Theorem is

g(b) - g(a) g'(c)

particular case of Cauchy's mean value theorem if we choose the fhnction g as g(x) = x 'd x ~ [ a , b]. A more general theorem, known as generalised mean value theorem is given in Section 12.5. You have seen that it is also established with the help of Rolle's , Theorem. According to this theorem, i f f , g, 11 be three functions from [a, h] to R such that they are continuous in 1% b] derivable in ]a, b[, then there exists at leakone point c e]a, b[ such that

v .

Both Lagrange's and Cauchy's theorem are particular cases of ihis iheorem. [f you tske g(x) = x and h(x) -- I +if s E [a, b]. then you get Lagrange's theorem lroni it. Cauchy's mean value theorem l'olluws from this general iheorern if you take only h(x) - I -V. x E [a, h]. Finally, in this section, lnterlnediaie Value Theorcm for derivatives is given according to which iff is derivable in [a, b], f'(a) ?' f(b) snd k is any !lumber lying beiween fl(a) and fP(b), then there exists a point c E ]a, b[ such that f l (c) - k From this ibI!ows 1)arboux Theorem naineiy iff :s$erivable in [a, I.)] and f'ja). f'(b) < 0, then there is a point c'in ]a, b[ such that f'(c) = 0.

Mean-Value Theorems

E I ) f is ~oiltinuous in [- 27i., 2sr] and derivabie in 1- 2rr, 2rr[ f(- 2 7 ~ j -- f(2.rr) = 0. All conditions of Kolle's Tlieoren~ arc satis!ied. Therefore there exists z point c in I - - 2n, 31d such that

f'(q I7 0 --., -, (.US C = 0 2. -- ,. c: I Tr/:!

Both the points k IT/?. E I-- 2x, 27r[.

E 2) (a) Clearly f(x) - - cos x :;atisfie> the hypothcsis of Roliiz's ~hccbren~ over [-- sr/2, rr i?]. So conclusion of the theorean will alsv be truc.

Thus there i s point c E 1- r/2, 7r/2] SUCII that fl(c) = 0 Here f ' (x> = sin x and f'(x) -- 0 irnpiies x - 0. SO c = 0 E 1- 71./2, n./2[.

2 (b) Contlitlo~~s of Rolle's The~rem are not satisfied, as t ' ix) == (u - I ) ' ' 3

f c ~ x f 1 and f is no1 dcnvable at x =- 1 a point of 10, 21. So f is not llerivablc in 10, 2[. So hypothcsis of Rolle's theoren1 is not valid. As f'(u) f O For ally x in 10, 2[, so conclusi'on of the theorem is not tme.

E 3) Let a and b, ;I # b, be ally two roots of e'sin x = 1 ':<- sin x -;- e '

.&%<> e-' -. :;in x = 0. . I .. c " -- sin a O and e"' - sin b = O

Let f(x) - e "- sin x -V x E [a, h]. Clearly I' is continuous in [a, b'j and derivable in ]a, b[. Also ffa) = f(b) -= Q

l'hcrefbrc thc hypothesis of Rolle's Theorem is satisfird by f ovcr [ti, b].

Therefore there exists c E ]a, b[sucli that fr(c) z: O which implies - e-' - cos c -- 0 * eC cos c .t I = 0.

So e" cos x -1 I = I) has a root c tbr some c i-- ]a, b[.

f E 4) Consider t!ie function, 1: defined as : sox"+' alx"

f(x) = f -- -t. . . . . . -1 , E [Q, I]. n f l n 1

Here f is continuous on [0, I], derivable on 10, 1[ and f(0) = 0 al I a I

f(1) = - + - - I - . . . . . f a,, = 0 (given condition) n f 1 n

.'. f(O) = f(1)

So f satisfies the hypotllesis of Rolle's theorem ancl therefore there is a point c E 10, 1[ such that f'(c) - 0 that is there is x E 10, 1[ such that f'(x) = 0 that is

E 5 ) We have to show that 1 (b - a) f(c) - (b -- c) [(a) - (C - a) [(I.)) = - (13 -- a) jc - a) fP'(d) for'sorne

2 c, d, E [a, b].

Consider the functioli 4, defined by

$(x) = (b - a) f(x) - (b - x) f(a) - (x -- a) f(b) - (b - a) (x T a) ( x - b) A where the constant A is to be determined such that +(c) = 0.- This implies

(b - a) f(c) - (lj - c) f(a) - (c - a) f(b) - (b - a) (c - a) (c - b) A = 0. (2) It is given that f" is continuous on [a, b] which implies that f ~ , f', f" are continuous on

[a, bl. +(a) = +(b) = 0 and 4 is differentiable in [a, b]. So C#J sat~sfies all the conditions of Rolle's theorem on each of the intervals [a, c] and [c, b] Thus there exists two numbers CI , c2 respectively in ]a, c[ and ]c, b[ such that

#(cr) = 0 and c#~'(cz) = 0.

Again @(x) = (b - a) fP(x) f(a) - f(b) - (b - a) [2x - (a + b)] J$ which i s continuous and derivable in [a, b] and in particular on [c,, c2]. Also ~ ' ( c I ) = (b'(c2) = 0. By Rolle's theorem, L3 d E ]c ,, c2[ such that

r#)"(d) = 0

Now @'(x) = (b - a) fJ'(x) - (b - a). 2A 1

f (d) = (b - a) f"(d) - (b - a) 2A = 0 ==s- A = --.f"(d) 2 (3)

where a < cl < d < cz < b and the result follows from (2) and (3).

E 6) Here f(x) = cos X, x € [O, n/2]. f is continuous in [O, n/2] and derivable in 10, n/2[.

By Lagrange's Mean Value Theorem, there exists a pt, c in 10, .rr/2[ such that

2 i.e. sin c = -

n-

2 i.e, c = sin-' - E 10, n/2[.

7T

E 7). Here f(x) = x' - 3x2 + 2x Therefore f'(x) = 3x2 - 6x + 2

Let qsolve the equation ff(c) = f(b) - f(n) - f(3) - f(O) - b- a 3 - 0

-,> ; Since 0 does not lie in 10, 3[, this value of c is rejected. So the required value of c which lies in 10, 3[ is c = 2.

E 8) Apply Lagrange's Mean Value theorem to the function f, given by

f(x) = ax2 + bx + d3' E [m, n].

You will get a c E In, n[ satisfying

f(n) - f(m) (Assume : n > rn) f'(c) = n- m

(an2 i- bn + d) - (an2 + bm + d) + 2ac i- b = n- m

= a ( n + m ) S b m + n

* C = ~ and c E ]m, n[

nl + n which implies that at x = - 2 , the tangent to the given curve is panllel to the

chord joining the points whose abscissae are x = m and x n.

E 9) Define a function. 4, by setting .42 $(x) = f(a i- hx) + f(a - hx)Y x E [0, I].

' As x varies over [0, 11, a - hx varies over [a - h, a] and a t hx varies over . [a, a t h].

I ' Therefore Q, is continuous in [0, 11 and derivable in 10, I[. By Lagrange's mean value theorem 3 8(0 < 8 < 1) such that

E 10) i) Consider F(x) = tan-' x - (x - x3/3), x 1 0. 1 x4

F '(x) = --- - (1 - x2) = - I + x2 1 t x2 > 0 for x > 0.

+ Thus F is svictly increasing in [0, m[.

:. F(x) > F(0) for x > 0 x3

i.e. tan-' x - (x - - j > 0 for x > 0 3 .

i.e. tan-' x > x - x3/3 for x > 0.

ii) Consider F(x) = e-" (1 - x) for x 1 0 and proceed as in (i).

E 11) The given'functions satisfy the hypothesis of Cauchy's mean value theorbm. -

:. '3 . . 8 E ] - .rr/2,O[ such that

which clearly lies in 1- 7r/2,0[.

E 12) ~2 find c from

. Thus 2c = a -I b G = (a t b)/2. which means that c is the arithmetic mean of a & b.

'E 13) We find c from

which gives c = 6 6 = a q-" Thus c is the geometric mean of a and b, -

E 14) We find c from

r(d - f(b) - f(a) -- g'(c) g(b) - g(a)

Thus c is the harmonic mean of a and b.

Mean-Value Theorems

UNIT 13 HIGHER ORDER DERIVATIVES

Structure

13'. 1 introduction Objectives

13.2 Taylor's Theorem M:tclaurin's Expansion

13.3 Indeterminate Forms 13.4 Extreme Values 13.5 Summary 13.6 Answers/Hints/Solutions

In ll~iit 13, you have lea i~~t Rolle's theorem and have seen how to apply this theorem in proving mean value theorems. In these theorems only the first derivative of the functions are involved. In thii unit, you will study tile application of Rolle's 'Theorem in proving theorems involving th6 higher order derivativa of functions.

Given a real function f(x), can we find an infinite seriesof real-numbers say of the form

ao + a , x +a2 x 2 C a3 x 3 + . . . .: . . . . . whose sun1 is precisely the given function? To answer this question we have to approximate a functioil with an infinite'serics of the above form which is also known as thc infinite polynomial or power series. 'This appioach of approximating a function was known io Newton around 1676 but it was developed late by the two British mathematicians Brook Taylor [1685-17311 and S.C. Maclaurin [1698-17461. The functions which can be represented a5 infinite series of the above form are some of the very special functions.

Such a representation of a fi~nction requires a number of dcrivatives of the function i.e. the derivatives of higher orders particularly at x = 0 which we intend to discuss in this Unit. Some work done by Taylor in this direction has found recent applications in the mathematical treatment of Photogranrmetry-lhr sciencr ofsur~ie~ing I!v mews of pho1ograplr.s ttrkaz from at2 naroplanc.

Besides, we sha!l also demonstrate the use of derivatives for finding the limits of indeterminate forms and the maximum and minimum values of functions in this unit.

Objectives After studying this unit, you should, thcrefore, be able to @ know theorems involving higher order dcrivatives viz. Taylor's Theorem @ expand functions in a power series v i ~ . Maclaurin's'series

evaluate the limits of indeterminate forms find the maximum and minimum values of functions.

13.2 TAYLOR'S THEOREM

In this seclion, we shall discuss the use of Rolle's theorem in proving theorems involving higher order derivatives of functions, Beforc proving these theorems, you will bc introduced to the idea of higher derivatives through the following definition :

DEFINITION 1 : HIGHER DERIVATIVES Let f be a function with domain D as a subset of R, Let Dl Z C$ be the set of p in ts of D at which f is derivable. We get another function with domain D l such that i ts -.slue at any point c of D, is f'(c). We call this function the derivative of f or first der i~ duve of f and denote it by f'. If the derivative of f' at any point c of its domain Dl exists, then it is called second derivative off at c and is denoted by f" (c). If D2 # C$ be the set of all those p in ts of D l at which f' is derivable, we get a function with domain D2 such that its value at any p i n t c of

Differentiability Therefore, fia) = #(b) = 0 a

A = f(b) - f(a) - (b - a)f'(a) - - 2!

Now, r) q5 is continuous in [a, b], since f, fl, ...., P-I) and (b - x)P, for all positive integers

p, are all continuous in [a, b];

ii) 4 is derivable in ]a, b[, since f, fr, ....., P I ) and (b - x)Q, for all positive integers p, are all derivable in ]a, b[; and

Therefore by Rolle's theorem, 3 c E ]a, b[ such that q5' (c) = 0.

7 (b - c)"-' (b - c)'-' r$' (c) = - (n - l)! f'" (c) + Ap = 0

(b - a)'

(b - c)"' (b - a)' ,,, which gives A = p . (n - I ) ! f (c).

Substituting this value of A in (2), we'obtain, . (b - a)"-'

(b - a)2 ftf (a) + . , . + f(b) = f(a) + (b - a) f ' (a) + - (n - I)!

f'"-"(a) 2!

This completes the proof of the theorem,

The expression, (b - a)" (b - eln-' f'nl (c)

R n = @ . ( , , - I ) !

which occurs in (3), after n terms, is called Taylor's remainder after n terms. he form (4),is called Schlomilch and Roche form of remainder.

From this we deduce two special forms of remainder after n terms.

i) Take p = n in (4), (b - a)*

R, = - f'" (c). n!

This is called Lagrange's form of remainder.

ii) Take p = 1 in (4), (b - a) (b - c)"'

R, = f'" (c). (n - I)!

This is called Cauchy's form of remainder.

The Taylor's theorem with Lagrange's form of remainder states :

If a function f defied on [a, b] be such that f'"" is continuous on [a, b] and derivable on b8 b[ then 3 a real number c E ]a, satisfying

f(b) = f(a) t (b - a) f'(a) + ?!-I-!$ Y (a) + . . . . . 21

Alternative form of Taylor's theorem with Lagrange's form of remainder is obtained if instead 1

of interval [a, b], we have the interval [a, a + h]. .

If we put b = a t h then we can write c = a + 0 h for some 8 between 0 and 1 and the theorem can be restated as :

46 If f'"" is continuo^ on [a, a + h] imd derivable on la, a t h[, then

h2 hn-r f(a + h) = f(a) + hft(a) + - fd'(a) + ... + - hn

2! P-') (a) + - f "(a + 0 h), (n - I)! n!

Higher. Order Derivatives

(0

for some real 8 satysfying 0 < 8 < 1. I

Now, let x be any point of [a, b]. I f f satisfies the condition of Taylor's theorcm on [a, b], then it also satisfies the condition of Taylor's theorem on [a, x], where x > a. Tkcrefore,

, froin (9, we have

(x - 5 ~ ) ~ (x - a)"-I (x - a)" f(x) = f(a) + (x - a)fl(a) + - fl(a) + ... + P-I) (a) + -

2! (n - l)! n! f'(c), (7)

where c is some real number in ]a, x[.

Note that (7) also holds when x = a because, then (7) reduces to the identity f(a) = f(a), as the remaining term on the right hand side of (7) vanish.

You may note that we can have forms similar to ( 5 ) , (6) and (7) for Taylor's theorem with Cauchy's form of remainder.

If in Taylor's theorem, we take a = 0, then we get a theorem known as Maclaurin's theorem. We give below Maclanrin's theorem with Lagrange's and Cauchy's form of . remainders. You can also write Schlomilch and Roche form of remainders.

THEOREM 2 : (MACLAURIN'S TIbOREM WITH LAGRANGE'S FORM OF -ER)

I f f be a function defined on [OZ b] such that fcn-') is continuous on [0, b] and derivable .on 10, b[, then for each x in [0, b], there exists a real number c (0 < c < x) such that

. PROOF : Take a = 0 in (7) above and proof is complete.

We can similarly write Maclaurin's theorem with Cauchy's form of remainder as: '. '

3 Xn-l x(x - c)"-I flx) =flO) + xf'(0) + - f"(0) + ... + - E"-I) (0) +

2! Pn) (c).

(n - I)! (n - I)!

You may note that

xn 9 K, (x) = - f")(c) = - f r (Ox) (0 < 0 < l),

n! n !

in case of Lagrange's form of remainder and

= xn( 1 - €))"-I

(n - l)! f(")(ex) (O < e < 11, I

. in case of Cauchy's form of remainder.

By applying Taylorls theorem or Maclaurin's theorem, also we can prove some inequalities of real analysis. Earlier, in the last unit, you were given a method of proving the inequality of examining the sign of derivative of some function. Consider the following example now.

EXAMPLE 2 : Using Maclaurin's theorem, prove that

SOLUTION : Foi x = 0, result is obvious. Now, let x > 0 and consider f(t) = cost. Then f has derivatives of all orders, for all t in R. By Maclaurin's theorem with

Di f f c re~~ t i a t r i l i t y remdinder at'tcr Lwo terms applied to f i n [@, X I ,

x2 f (x j - f (O) 3- x f'(0) + - 2! f " (Ox) where 0 c: d C: I .

Pt~lting thc values oft; f', f " we have x -

cos x x ] - -- cos f 0 x )

X ? 1

x- . X- Now cos trx 5 1 and ao 1 - - cos Ox 2 I -- - 1.e. cos x 2 1 - -

2 ? 2

I t x < 0. Lhcn - \ > 0 and therefore cos ( - x) 2: I - (!- x)' 1 .I '.

that is cos x 1 1 -- , . Hence cos x 2 1 - -"--t' x E R.

You should be able to solvc the following exercise. - - EXERCISE 2 Using h4aclaurin1s tlleorem, show that

Now you will see how to find the Maclaurin's expansiorl of certain elementary functions

of the type, €7, sin x, cos s, (1 -I- x)"' and log (1 + x ) in terms of an infinite series (power series) 3s a, t a,x + a2x? + ..., with the help of Tay lo l * ' s .~~~d Maclaurin's tlleorems.

We have sezn before that

(x - a)"-1 + -- trw-" (a) + Rn(x), (n - l)!

where R,(x) is the Taylor's renlainder after n tenns. Put

( X - (X - a)"-' Sn(xj = f(a) + (X - a)fl(a) + -- 2! f " (~ ) + . .. . + ------ fln-1) (a).

(n - I)! Then, f(x) = Sn(x) + Rn(x).

A natural qucstion arises iis to whether we can express f(x) in the form of the infinite series (X - a)".'

f(a) + (X - a) f'(a) + . . . . . + --- f in - " (a) + . . . . . (11 - I)!

and if so under what conditions? This question can be split up in the following situations:

i) Under what conditions on f is each term of the series defined?

ii) U,nder what conditions does the series (9) converge?

iii) Under what conditions is the sum of the series (9), f(x)? We examine these now one by one. i) Each term of the series (2) is defined iff f n (a) exists for all positive integers n. ii) Assuming fU(a) exists .tf n, we have from (8),

S, = f(x) - Rn(x) (assuming the conditions for Taylor's theorem are satisfied in some interval [a, a + h])

From this, it follows that < S, > converges iff lim R,(x) exists and the series (9) converges n-03

' . iff lim R,(x) exists. I)--

iii) Assuming the series (9) converges, we find from above that its sum is f(x) - lim R,(x). "-.m

48 Now f(xj - l i~n R,~(x) = f(x) iff lim' R,(x) = 0, n-= n-=

showing that the series (9) converges to f(x) iff lim R,(x) = 0. n-=

Higher Order Derivatives

Summing up the above discussion, we have the following results.

THEOREM 3 : I f f : [a, a 4- h] -- R be a function such that

i) f("' (x) exists for each positive integer n, for all x E [a, a + h].

ii) lim kn(x) = 0 V x E [a, a + h], 0--

then

for every x E [a, a + h]. This is called Taylor's infinite series expansion of f(x). We also sometimes call i l the expression for f(x) as a power series in (x - a).

. We give an example to illustrate Taylor's series for a function.

~: EXAMPLE 3 : Assuming the validity of expansion, show that 7,- (x - 7,-/4) Ir(x - .rr/412

tan-' x = tan-' - + - 4 1 + 2 / 1 6 4(l + ~ ' / 1 6 ) ~

+ . . . . . V x E R

SOLUTION : Let f(x) = tan-'x = tan-' (7r/4 + x - n/4)

Here a = 7r:/4, h = x - 7r/4. fyx) exisb "T x and Y n.

1 2x . Now ff(x) =- , f" (x) = -

I + xZ (1 + X2)2 '""'

By Taylors series, h2

f(a + h) = f(a) + h fl(a) + - f"(a) + ..... 2!

Putting the values o f f , f', f", ..., we obtain

EXERCISE 3 Assuming the validity of expansion, expand cos x in powers of (x-x14).

If you put a=O in the Taylor's series you will get the following result.

THEORJ3M 4 : Let f : [0, h] + R be a function such that

i) fin)(x) exists for every positive integer n and for each x E [0, h]; t

ii) lim Rn (x) = 0, for each x E [0, h]. n 9 m

xz xn ' Then, f(x) = f(0) i- xft(0) + fV(O) + ,.. + - V(0) + ,..., for every x 6 10, h].

n !

This series i; called the Maclaurin's infinite series expansion of qx).

Note that Taylor's series remains valid in the interval [a - h, a 4- h] and Madaurin's series remains valid in the interval [- h, h] also provided the requirements of the eqpansion are satisfied in the intervals.

You may also note that one h a y consider any form of remainder R,(x) in the above discussion. We shall now ~ns ider Maclaurin's series expansions of the functions ex, cos x and log (1 + x).

EXAMPLE 4 : Find the Maclawin series expansion of e: m x and log (1 -k x).

Differentiabil ity SOLUTION : Expansion o: ex. Let f(x) = ex, Q x ER. Then I?") (x) = ex, Vx E [-h, h], h > 0 and for all positive integers n. In otlier words, fcn)(x) exists for each n and for all x in W.

Let us now consider the limit of the remainder, Rn(x). Taking Lagrange's form of remainder, we have

X" So, &Rn(x) = lim e".

n+m

xn But, & 3 = 0 as shown below

Let un = , then n!

So, by Ratio test, C lunl is convergent and, therefore, Xun is convergent and consequently

x!' lim un = lim - = 0, if x # 0. n+m n+m n!

xn

If x = 0, then also i i ~ ,! = 0. Therefore J,izRn(x) = 0.

Thus the conditions of Maclaurin's infinite expansion are satisfied ...

Also, f(0) = 1 and f " ) (O) = 1, n = 1.2, .... Hence

Expansion of cosx. Let f(x) = cos x, Q x ER.

n r Then fc")'(x) = cos (x +-), for n = 1, 2, ...

2

Therefore, f(0) = 1 and P)(O) = cos (nd2), Q n.

Clearly f and all its derivatives exists for all real x.

Taking Lagranges form of the remainder,

Therefore, IRn(x) ( = I z !~ - . I cos OX+?)^ 5 7 + 0 as n + w, V x ER (Proved in the expansion of ex.), 1::l which implies lirn R,(x) = 0, Q x E W.

n+m

Thus the conditions of Maclaurin's infinite expansion are satisfied,

n a From f ("'(0) = c6s (-i-), we get

f(zm+qO) = cos ((2m + I ) ~ ) = 0 and f(2m)(0) = cos ((2rn)rd2) = cos m n = (-1". 2

substituting these values in the expansion, we have

xz x4 x6 x2n cosx = 1 - - + - - - + ...... + (-1)" - - ..., Q x ER.

50 2' 4! 6 ! - Qn)!

3. EXPANSION OF log (1 + x)

Let f(x) = log (1 + x) for - 1 < x 5 1. (- 1ln-'(n - I)!

Then f'"(x) = (1 + x)" x > - 1.

We shall consider the following cases:

i) 0.5 x 5 1. Taking Lagrange's form of remainder after n terms, we have

n Since 0 I x 5 1, 0 < 8 < 1, therefore

1 Also - - 0 as n -

n Hence lim R, ix) = 0.

n--

So the conditions of Maclaurin's infinite expansion are satisfied for 0 5 x 9 1.

ii) - 1 < x < 0,

In this case, x may or may no1 be numerically less than 1 + ex; so that nothing can be said

about the limit of ( - ; ex )'I as n - m. Thus Lagrange's form of remainder does not help to

draw any definite conclusion. We now take the help of Cauchy's form of remainder, which is

X" (1 - e)"--l Rn(X)= (n - 111 f'" (ex)

Therefore - - 0 as n - a. (:;BBx)n-l

Also x"- 0 as n -+ a, r

1 and - < 1

and'it is independent of n. 1 C O X i -'ix 1

? Thus lim Rn(x) = 0.

n--=

Hence the conditions of Maclaurin's series expansion are satisfied also when - 1 < x < 0.

Thus substituting f(0) = 0, f'") (0) = (- 1)"' (n - l)! in the expansion, we get

EXERCISE 4

Prove that sin x = x - ;;+f - +... . .+ (- 1ln-l P I . , . . V t x E R . (2n - I)!

EXERCISE 5 m(m -

Prove that (1 + x )m = 1 f mx f ') x Z+ . . . . . 21

Higher Order Derivatives

for 'all integers m and L iien ( x 1 < 1. /

EXERCISE 6 Assuming the validity of expansion, expand log (1 4- sin x) in powers of x, upto. 4th power of X.

13.3 INDETERMINATE FORMS

We have proved in Unit 8 (Block 3) that

lim f(x) f(x) - x -a

lim - - --- x - a g(x) lim g(x)

x- a

provided lim f(x) and lirn g(x) both exist and lirn g(x) f 0. It may sometimes happen that x- a x- a X - 8 '

lirn { f(x)/g(x) ] exists even though lim g(x) = 0. One can easily see that if lim g(x) = 0, then X- a x-a x-8

f(x) a necessary condition for lirn -to exist and be finite is that lim f(x) = 0. X-a g(x) 11-01

In fact, if lirn { f(x)/g(x) ] = k, x- n

then lim f(x) = lirn [ f(x)/g(x) . g(x) ] x - a x- n

= lirn { f(x)/g(x) ] . lim g(x) X-8 X-8

= k-.o = 0.

In this section we propose to discuss the method of evaluatingh { f(x)/g(x) ] when both lirn f(x) X-8 X - 8

f(x) and lirn g(x) are zero or infinite. In these cases -are said to assume indeterminak forms x-a g(x)

0/0 or m/m respectively as x -- a.

0 DEFINITION 2 : INDETERMINATE FORM

f(x) 0 If lirn f(x) = 0, lim g(x) = 0 then -issaid to assume the indeterminate form - w x tends to 'a' x- a X-8 g(x) 0

DEFINITION 3 : INDETERMINATE FORM '

00

f(x) 00 If lim f(x) = 00, lirn g(x) = 00, then i s said to assume the indeterminate forq ; as ;r X- a x- 8 ~ ( x )

tends to 'a'. Other indeterminate forms are 0 x m, .=a - W, oO, 1" and m0 which can be similarly defined. Ordinary methods of evaluhting the limits are of little help. 60me speqial methods are required to evaluate these peculiar limits. This Special method, geporglly called, L, Hopital's Rule is due to the French mathematician, L' Hopital(1661-1704). Iq fact, this method is due to J, Bernoulli, who happened to be a teacher of L' Hopital and hb (Bernoulli's) lectures were published by the latter in the book form in 1696, but sukgequently

0 Bernoulli's name almost disappeared. Let us consider the indeterminate from - and discuss a some related. theorems. Note the differences in the hypothesis of these theoftms B P ~ me line, of proof should be very carefully noted.

THEOREM 5 : Let f and g be two functions such that

i) lim f(x) = 0, lim g(x) = 0, a- 1 x - l .-

iii f' (a) and g' (a) exist, and (iii) g'(a) 2 0. Then h

PROOF : By hypothesis, f and g are derivable at x = a * they are continuous at x = a

and lim g(x) = g(a). x - a

Therefore by condition (i), f(a) = 0 = g(a).

Higher Order Derivatives

Also f' (a) = lirn f(x) - f(a) f(x) = lim -

x- a x - a X- a x - a

g(x) - g(a) = lim g(x) and g' (a) = lirn x- a x - a X- a x - a

. f ' (a) - lim f(x) / (x - a) . . f(x) = lim - . g' (a) x - a g(x) / (x - a) x - a g(x) .

We may remark that condition (i) in the above theorem can be replaced by f(a) = g(a) = 0.

0 THEOREM 6 : (L' Hopital's rule for form)

I f f and g are two functions such that

lirn f(x) = lim g(x) = 0, i, x- a

ii) f' (x) and g' (x) exist and g' (x) # 0 for all x in ]a - 6, a + 6[, 6 > 0, except possibly at a, and

f' exists, iii) lim - x - a g' (x)

f(x) f ' (XI then lirn - = lim - x - 0 g(x) I-.. g' (x)

PROOF : Define two functions 4 and (v such that

g ( x ) , b ' x ~ ] a - 6 , a + 6 [ a n d x * a lim g(x), if x = a, X 3 P . 1 .I

S~nce f ' (x) and g' (x) exlst %' x E ] a - 6, a + S [ except possibly at a, 4 and J, are continuous and derivable on ] a - 6, a + S [ except.possibly at a.

Also since lirn '(x) = lirn f(x) = 4 (a) x- a X - n

and lirn $(x) = lim g(x) = t,!~. (a), X d l l X-8

therefore gb and 9 are continuous at x = a, as well.

Let x be a point of ] n - 6, a 4- 6 [ such that x # a. For x > a, 4 and satisfy the conditions of Cauchy's mean value theorem on [a, x] so that

-- '(x) - '(a) - '1 (

el for some c e I a, x [.

$ 1 - $ 8 * ( 4 But +(a) = lirn f(x) = 0 & 9 (a) = lim, g(x) = 0

x- a x -n

Proceeding to limits

4 (x) - 4' (c) 4 lim - lim -=

4' (XI lim 7

x - a t * (x) ,-a+ *' (c) x-,+ * ( 4

f(x) ==$' lim -= f' (x) lirn -'

.--a+ g(x) ,-a+ g (x) For x < a, we can similarly prove that

f(x) - f' (XI lirn -- lim -. g(x) x-a- g' (XI

f' (x) f ' (XI f' (XI But lirn -= lim -=lim- x-a+ g' (x) x-n- g' (x) x -a g (x)

f(x) f ' (XI .. . Hence lirn - = lirn - x-n g(x) g: ( 4 ,'

0 You may note that if the expression lirn represents the indeterminate form - and the

X - a g ( 4 0 functions f'(x) and g'(x) satisfy the conditions of the above,theorem, then

f(x) f ' (XI f" (x) lirn - = lirn = lim - x-ag(x) x-ug (x) x-ag (x) In fact the above rule can be generalised as follows :

Iff and g are two functions such that

i) f'" ' (x), g'"' (x) exist and g(r' (x) # 0 (r = 1, 2, . . . . . , n) for any x in ]a - 6, a 4- 6 [ except possibly at x = a, lim f(x) = lim f' (x) = . . . . . - lim f (n-11 - (x) = 0

- - lim g'"-" as x - a, ii) lim g(x) = lim g' (x) = . . . . . (XI = o f'"' (x)

iii) lirn 7 exists, then x-8 "I (x)

f'"' (x) lim fo = lim -. X-"(x) x -a g'"' (x) This is known as Generalised L' Hopital's Rure for form.

0

Note that L; Hopitai's Rule is valid even if x -- m.

In fact, we have

1 .

f(x) f h-- - l h - 1 , where x = 7 ,

x+m g(x) Z-+(K 1 i2

1 1 f' ( T I ( - 2 )

= lim z+O+ I (by L' Hopital's Rule)

1 g f (y ) (- 2)

= lirn --- z-0t I

9' ( y)

f' (x) = Iim -. 2--+ g' (x)

Now we give examples to illustrate the use of L' Hopitals's rule in evaluating the limits of 0

indeterminate form - . 0

EXAMPLE 5 : Evaluate each of the following limits :

SOLUTION : JZ - 2 cos ( r / 4 + x) - f(x)

i) Let us write -- X ~ ( X I '

where f(x) = @ - 2 cos ( ~ / 4 4- x) and g(x) = x.

lirn f(x) = fi - 2 cos 7r/4 = 0 and lirn g(x) = 0. x-0 x - 0

f(x)/g(x) is, therefore, of the form 0/0 as x -- 0.

Applying L' Hopital's rule,

& - 2 cos ( ~ / 4 + x) 2 sin [ ~ / 4 + x) lim. = lim = 2sin z= fi

54 x - 0 X x - 0 1 4

ii) Writing tanx-x . -Lmx-x x - .- x2 sin x x3 sill x

we find that tan x - x tan x - x x

lirn = lirn lim - A-o x2 sip x x3 x-Q sin x

tan x - x 0 = lirn

x-0 x3 ( ,f om) 2 sec x - , I = lim ----

x- 0 3x2 (By L' Hopital's Rule)

1

x'cos x - log (1 + x) iii) lim

x- 0 ( + form )

x2

1 cos x - x sin x - -

~ ----... 1 -b x

= lirn 0 X-0 (again -form) 2x 0

1 - - . - sin x - (sin x 4- x cos x) + 1/(1 -t x ) ~ lim 2 x- 0 1

EXAMPLE 6 : Determine the values of a and b for which

exists and equals 1 /6.

SOLU'rIQN : The given function is of the form (0/0) for all values of a and b when x -- 0. x (a - cos x) + b sin x .'. limy

x-0 x 3

(a - cos X) + x sin x + b popx = lirn x- 0 3x2

The denominator tends to 0 as x tends to 0, the fraction will tend to a finite limit only if the numerator also tends to zero as x -- 0. This requires a - l + b = O

A Supposing (10) is satisfied, we have

a + ( b - l ) ~cosx+xs inx lirn x-0

f 3x2

- (b - I ) sin x + x cos x + sin x = lim -

X-o 6~

x cos x + (2 - b) sin x = lirn X-0 6x ( + form )

- x sin x + cos x + (2 - b) cos x = lirn x-[I 6

3-==.b 1 -. --=- 6 (given)

* b = 2 From (lo), a = - 1.

Now you should be able to solve the following exercises.

Higher Order Derivatives

EXERCISE 7 Evaluate the following limits :

sin 3x2

i) lim x -0 log cos (2x2 - x)

ii) litn sin h x - sin x

x-0 xtan2x

EXERCISE 8 sin 4x f a sin 2x

If the limit x3 as x -- 0 is finite, find the value of 'a' and the limit.

q EXERCISE 9

What is wrong wit4 the following application of L' Hopital's rule :

Find also the correct limit.

00 00

Next we consider the indeterminate form - , L' Hopital's rule for - form is similar to that OC, 00

03

for 0/0 form. We only state the result for form without proof. m

THEOREM 7 : (L' Hopital's rule for form) .

: Iff and g be two functions such that

I

1 ii) f'(x) and g'(x) wrist, g8(~)+ 0, %' x E ] a - 4 a f S [, S > 0 except possibly at a, and

iii) lim - ' exkts; then l3' (x)

f(x) - f' (x) . lim - - x-a g(x) x-k g (x)

The above theorem tells us that lim 3, when f(x) and g(x) bath tend to infinity as x - a, g(x)

can be dealt with in the same way as (% 1 form. In fact forms ( $ ) and ( $ ) can be

interchanged and care should be taken to select the foim which would enable us to evaluate the limit quickIy.

I The above theorem also holds in the case of infinite limits.

Now we consider examples to illustrate the application of L' Hopital's rule for finding the limit 00

of indeterminate form 2

EXAMPLE 7 Evaluate the following limits :

log tan 2x %* l M m r

locr (a > 0) ii) lim - x-- xo

SOLUTION : (i) Writing log'tan 2x - f(x) --,

56 1% tan x g(x) ,

I

where f(x) = log tan 2x and g(x) = log tan x, we find that the given expression is of the form

log tan 2x = lim 2 cot 2x sec2 2x :. lirn x - o t logtanx "O+ cotxsec2x

2 sinx cosx = lim 1 = lim - - - 1.

x-O+ sin 2x cos 2x x-O+ cos 2x

log x m ii) lim 7 (d > 0) is , fonn.

x+x

1 /x Therefore its value is equal to lirn -

x - - @.Xa-'

- EXERCISE 10 Evaluate the following limits :

7T log (x - - )

2 i) lirn

X - T ~ Z + tanX

log sin x ii) lh --

1-0- C2t X

Now we consider the indeterminate forms 0. m and m - a. These can be converted to 0 /0 or w/w forms as shown below :

i) lim f(x) = 0 and lirn g(x) = m, then x - P x-n

lim f(x) . g(x) is 0 . w form. x-n

We tan write

f(x) f(x) . g(x) = - g(x) or -3

l /g (x) l/f (x)

0 m which are respectively - or po forms and hence can be evaluated by L' Hopital's rule.

0

ii) If lirn f(x)l = lim g(x)=. m, then X- '0 I-R

lirn { f(x) - g(x) ) is - m form. x-0

0 This can be reduced to - form by writing

0

Higher Order Derivatives

and then we can apply L'Hopital's rule. The following example will clarify the procedure. First we consider 0 , ~ form

EXAMPLE 8 Evaluate:

i) lim x log x 1-0s.

7l-X ii) lim sec - log (l/x).

1- 1 2 SOLUTION : i) Take f(x) = x and g(x) = log x.

Then lirn f(x) = 0 and lirn g(x) = - *, x-O+ x-01

so that the given form is 0 X =. We can write it as

Lug x 09 Iim x fog x = lim --- ( - form) r-O+ .--ot l / x '

< -- lim x = 0. : lim -- -

x -o+ - ] /xZ x - 0

ii) Taking fjx) = log (l/x) and g(x) = sec (in;i2), We get that the given form is 0 X as x - 1 .

.'. lim sec (,lrx/2) lug (1 / x ) x - l

0 = lim -9 (l/x) ( - form) *-. I cos (7rx/2) 0

-- I/x = lirn

x- 1 - sin (nx/2). 7r/2

= 2:n.

EXERCISE 11 Evaluate the following limits :

i) F3 sin x log x2

ii) lirn (1 - x) tan (7rx/2). x-1

Now we considcr example for w - = form.

EXAMPLE 9 : Evaluate

SOLUTION 1

i) Let f(x) = 1

and g(x) = - . log (x - 3) x - 4

Both these tend to = as x -. 4. Thus the given limit is 03 - form. We can write it as

(x -- 4) - log (x - 3) 0 lim (- form) x - 4 (X -4)10g(x- 3) 0

I x - 3 = lirn -

x - 4 x - 4 log (x - 3) + -

x- 3 x - 4 = Iim 0

( - form) ~-4 (x -3 ) log (x - 3 ) + ( x - 4 ) 0

' = lim 1 - 1 - - .

I - x -41+ log (x -3 )+1 2

1 1- sin x - cos x ii) lim (sec x ---- ) = lim (it is rn - co form)

X + X ~ I- sin x .-..a cos x (1 - sin x)

- 1 - cos x + sin x i = lirn - sin x (1- sin x) - cos2x - 1 - 1 I

i - 1 - - - t . 2 i

-1pIII

i EXERCISE 12 !

i Evaluate the following limits :

I 1 1

1 58 ' 1 j '

Higher Order Derivatives

Finally we consider the Indeterminate forms I", mO, o0 For all these forms we have to evaluate

lim [,f(x) I""', x- a

where lim f(x) = 1, oc or 0 and lim g(x) = m or 0,0 (respectively). X- a x- a

We can write y = [ f ( x ) ~ ~ ( ~ ' Therefore log y = g(x) log f(x)

lim log y = lirn [ g(x) log f(x) 1. x -a , x- a

In each of these three cases, right hand side is 0 . form which can be evaluated.

Let lim [ g(x) log f(x) ,] = .I. x-.n

Therefore linl log y = 1 x - 8

which implies log [ lim y ] = I

x - a

=+ lirn y = e' x - n

=+ lim [ f(x' 'dX) = I A e I

X - 8

The following example discusses these indeterminate forms.

EXAMPLE 10 : Evaluate

ii) lim (sec x)'Ot x - n/2-

iii) lim (1 - X2)2/ l~~ (1-x) x-1-

SOLUTION :

i) It is of the form 1"

tan x 1/x2 Let y = (T) Therefore log y =

tan x

lirn log y = Vrn x - 0 - 0 ' x2

tanx x = lirn

. x-0 . 2x

x secL x - tan x 0 = lim - ( - form)

1-0 2x2tanx 0 2x sei2x tan x

= lirn x-o 2 [ 2x tan x + x2 secZ x ]

sec2 x tan x = lim

x-0 2 tanx + xsec2x

tanx 0 = lim ( - form)

a-0 sin 2x4- x 0

sec2 X = lirn

1 =-. ~ - 0 2 c o s 2 x + 1 3

which gives lirn y = elt3 x-0

ii) It is of the form mO. Let y = (sec x)"' " So log y = cot x log sec x.

log sec x O0 Therefore lim log Y = lim - ( - form)

x - r / 2 - x - T / ~ - tanx M

1 -. sec x tail x - lirn - set x

x-.ir/z- sec2 x = lirn (sin x cos x) = 0.

x - r / Z -

which implies log = lim y = 0 3 lirn y = eo = 1, . x - r / 2 - x - d l -

iii) It is of the form 0'. Let = (1 - X2) Z/log ( I - * )

EXERCISE 13 Evaluate the following limits :

ii) lirn (cot x ) " ~ x+w

iii) lim (cos x)'"". 14iT/Z-

So log y = 2

log (1 - x2) log (1 - x)

log (1 - x2) = 2.

log (1 - x)

log (1 - jc2) w lim log y = 2. lim ( - form) 1-1- x - I - log(1-xl

- 2x41 - x2) = 2. lim

x-1- - l / ( l - x \ (By L' Hopital's Rule)

X = 2*lim -= 2 x - I 1 + x

which gives lirn y = e2. : - I

13.4 EXTREME VALUES

In this section, we shall be concerned with the applications of derivatives and Mean Value theorems to the determination of the values of a function which are greatest or least in their immediate neighbourlioods; generally known as local or relative maximum and minimum values. The interest in finding the maximum or the minimum values of a function, arose from many diverse directions. During the war period, the cannon operators wanted to know if they could somehow maximize (and if so, to what extent) the distance travelled horizontally i.e. the range, when a cannon-ball is shot from the cannon. The position of the angle at which the cannon was inclinded to the ground mattered the most in such cases. Another direction was the study of motion of planets. It involved maxima and minima problems such as finding the greatest and the least distances of the planets from the sun at a particular time and so on.

We shall find below the necessary and sufficient conditions for the existence o maxima or minima. First we define extreme values of a function.

DEFINITION 4 : EXTREME VALUE OF A FL~CTION

60 Let f be a function defined on an interval I and let c be any interior point of I.

I) f is said to have a local or relative.maximum value (a local or relative maximum) at ' x = c if 3 a number 6 > 0 such that

i.e. f(c) is the greatest value of the function in me interval ]c - 6, c + 6 [ i.e. f(c) is a local maximum value of the function f if 3 6 > 0 such that f(c) > f(c 4- h)- f(c + h) - f(c) < 0 for 0 < (hl < 6.

2) f is said to have a local or relative minimum value (a local or relative miil imw) at x = c if 3 a number 6 > 0 such that $ + x E ] c - 6 , c + 6 [ , x # c =f(x)>f(c) or equivalently f(c + h) - f(c) > 0 for 0 < (hl < 6. or f(c) is the least value of the function f in the interval ] c - 6, c -t 6 [.

3) f is said to have an extreme value (an extremum or a turning value) at x = c, if it has either a local maximum or a local minimum at x = c.

The following simple examples will clarify your ideas about maximum and minimum values.

EXAMPLE 11 : Let f be a function defined on R as f(x) = x 2 T x E R, then f has a local minimum at x = 0. From the graph (Fig. l), the values in the neiglibourhood of the value at x = 0 is greater than 0.

EXAMPLE 12 : Let f be a function defined on R as f(x) = sin x V x € E; then f has a local milli~nuln at x = - n/2 and a local maximum at x = n/2, In fact, f has a minimum at x = 2n7-r - =/2 and a maximum at x = 2nn + ~ / 2 ; n being ally mteger as is evident from the following Figure 2:

Higher Order Derivatives

X )x

Y

Fig. 2

EXAMPLE 13 : Let f be a function f defined as : f(x) = x3tf x E R; then f has neither a maximum nor a minimum at x = 0. At x = 0 f(0) = 0. If we take any interval ] - d, d[ about the point 0, then it contains points XI, x2 such that xl > 0 and xz < 0. Now f(x1) > f(0) = 0 and f(x2) < f(0) = 0. Note that while ascertaining whether a value f(c) is an extreme value of f or not, we compare f(c) with the values off in any small neighbourhood of c, so that the values of the function outside the neighbourhood do not come in question.

Y

Fig. 3

Fig. 4

Thus a local maximum (minimum) value of a function may not be the greatest (least) of all the values of the function in a finite interval. In fact, a function can have several local maximum and minimum values and a local minimum value may even be greater than a

, maximum value. A glance at the above Figure 4 shows that the ordinates of the points PI, P3, Ps are the local maximum and the ordinates of the points Pz, P4 are the local minimum values

'of the corresponding function and that the ordinate of P4 which is a local minimum is greater than the ordinate of PI, which is a local maximum.

Further you must have noticed that the tangents at the points PI, P2, Pj, P4 PS in the above figure are paralIel to the axis of x, so that if CI, c2, c3, c4, cs are the abscissae of these points, then each of f'(cl), ~'(cz), f'(c3), fP(c4), ~'(cs) is zero.

We proceed to establish the truth of this result below :

THEOREM r 8 : A necessary condition for f(c) to be arl extreme value of a function f is that f'(c) = 0, in case it exists.

FROOF : Hex, we assume. f is derivable at c. Lei, further, f(c) be a local maximum value o! f. ~ h u s there exists a real number 6 > 0 such that

f(c t - 11) - f(c) NOW for 11 :> 0, we have ( P

h -

f(c -k h) - i(c) and for h q: I) we h ~ v e 7---- 3 0

h From ( :3) and (13j, we have

flc .+- h) -- fjc) f(c -I- 11) - f(c) iitr~ ---- .: 0 and lian ---

h 2 0, x. >a F l~-+i!-. h

which gives fyc) < 0 and f l(c) 2 0. Therefore, f'cc) = 0.

It call bc sin~ilarly shown that f'(c) = O, if f(c) is a local minimurn value off:

'Tlie vanishing of f'(c) is only a necessary but mot a sufficienl condition for f(c) to be a,, cstl.crne value as we now show with the help of the following example.

Consider a function, f, defirled by f(x) = x' U x E K Then T'(x) - 3xL, f'(0) =. 0. Also f(0) = 0. Clearly for x > 0, f(x) > 0 - f(0) and for x < 0, fix) < 0 = f(0) thus [(O) is not a local cxlrema value even though f'(0) - 0.

Further you can note that a fu~lction may hnve a local maxi~nuin or a local minimum value at a point without being derivable at that poiilt. For example, if f(x) - I x 1 "d- x EJ, then f is not derivable at x .=: 0, but has local minimum at x - 0.

Y

! We may remark that in view of the above theorem, we find that if a fnnclion f has a local exlrerne value at a point x = c, then either* f 1s not derivable at x = c or f'(c) = 0. Thus in order to investigate the local ~naxitna and minima of a function f, wc have to first find out the values of x for which f'(x) does not exist or if f'(x) exists, then it vanishes. (These values are generally called the critical values of f.) We then examine for which of these values, does the function ;ictually ]lave n local mfixirnu~n or a local minimum, The points where firit derivative of a Suuactic~n vanish nre callecl stationary poink.

DEFINPlrION 5 ; STATIONARY VALUE OF A FUNCTION x = c is called a stationary point for the function f if f'(c) = 0. Also f(c) is then c: e stationary value.

I .

You have seen that if a function f is derivable at an interior point c of its domain and I f'(c) = 0, then f muy not h ~ v c an extreme value nt C. To decide whether f has an extreme

Higher Order Derivatives

I

Differentiability value ornot at such a point, we need some method. By knowing the sign of the derivative on the left and right of the point we can decide whether f has a local maximum or local minimum, at the point. This is the purpose of the next theorem.

THEOREM 9 (FIRST DERIVATIVE TEST) Let a function f be derivable on an interval ] c - 6, c 4- 6 [, 6 > 0, and let f'(c) = 0. If

i) i ' ( x )>OVxE]c -& ,c [and f ' ( x )<OYxE]c lc+6 [ , then fhasa loca l maximum at x = c.

ii) f ' ( x ) < O V x E ] c - & , c [ a n d f ' ( x ) > O Y ~ E ] ~ , c - k 6 [ ~ t h e n f b a s a l o c a l minimum at x = c.

PROOF : i) Let b be an arbitrary point of ] c - 6, c [. Then f satisfies the conditions of Lagrange's

mean value theorem in [b, c], so that . ,. . f(c) - f(b) = (c - b) f'(a) for some a E ] b, c [.

Since f'(x) > 0, 'd x E]C- 6 , c[, therefore ff(a) > 0, and so f(c) - f(b) > 0. N o w b is any point of ]c - 8 , c[, .', f(c) - f(x) > 0, X E]C - 8, c[.

Let now d be an arbitrary point of ] c, c + 6 [. Then f satisties the conditions of Lagrange's mean value theorem in [c, dl, so that

f(d) - f(c) = (d - c) f' (P ) for some P E ] c, d [. f ' (x)<d-tt xE ] c , c f 6 [ .'. f'(P) < 0.

So, f(d)- f(c) < 0. Now d is any point ]c, c + 8[, therefore f(x) - f(c) < 0, 'v' x x] c, c + 6 [ . From (14) and (15), we find that

, V x E ]c - 6, c + 6 [, x # c -7L f(x) < f(c) =3 f has a local maximum at x = c

ii) You can similarly prove it.' If 3 6 > 0 such that

x E ] c - - 6 , c [ * f ' ( x ) > O and x E ] c, c + 6 [ f'(x) < 0, then we say that f'(x) changes sign from positive to negative as x passes through c. Simila~ly, if 3 .6 > 0 such that

x E ] c - 6 , c [ -? . f ' ( x )<O

and x f ] c, c 4- 6 [ f'(x) > 0, then we say that f'(x) changes sign from negative to positive as x passes through c.

In view of this terminology, the above theorem can be stated as follows :

Let f be derivable OQ an open interval I and let f'(c) = 0 at some point c E I. If ff(x) changes sign from positive to negative (negative to positive) as x passes through c, then f has a local maximum (minimum) at x = c. You may note that the conditions of the above theorem are sufticient but not necessary. For example, consider the function f, defined by

1 f(x) = x4 (2 + sin -) when x # 0,

X

and f(0) = 0.

This function f is derivable everywhere, f'(x) does not change sign from negative to positive as , x passes through 0 and yet f has a local minimum at x = 0.

You may further note that, if fl(x) does not change sign?.e., it has the same sign throughout the interval ]c - 6, c + 6 [ , for some 6 > 0, then f is either strictly increasing or strictly decreasing throughout this neighbourhood and, so, f(c) is not an extreme value off.

Geometrically interpreted, the above theorem states that the tangent to a curve at every point in a certain left handed neighbourhood of the point P whose ordinate is a local maximum (minimum) makes an acute (obtuse) angle and the tangent at any point in a certain right handed ncighbourhood of P makes an obtuse (acute) angle with the axis of X. In case the tangent on either side of P makes an acute angle (or abtuse angle, the ordinate of P is neither a local maximum nor a local minimum.

The following example shows the application of the above theorem for finding extreme values of a function.

EXAMPLE 14 : Examine the Function f given by

for extreme values.

SOLUTION : Here f(x) = (x - 2)4 (x + 1)' T ~ U S fyx) = 4 ( ~ - 2)' (X + 1)' i- 5 (X - 2)4 (X + 114

= (x - 2)'(x + 114 (9x - 6) So f'(x) = 0 for x = - 1, 2/3, 2.

Thus we expect the function to have extreme values for these values of x.

Now ff(x) >Ofor x < - 1, and f'(x) > 0 when x is slightly greater than L- 1.

'

Therefore f has neither maximum nor minimum at x = - 1.

Next f'(x) ch,lnges sign from positive to negative at x = 2/3, therefore f has a local maximum at x = 2/3.

Higher Order Derivatives

Also f'(x) changes sign from negative to pos~tive at x = 2 and therefore it has a local minimum thereat. '

EXERCISE 14 : Examine the polynomial function given by lox6 ---24x5 + 15x4 - 40x3 + 108 'V' x E R for local maximum and minimum values.

We can also decide about the maximum and minimum values of a function at a point c from the sign of second derivative at c, This, you will see, In the next theorem, called the second derivative test.

'THEOREM 10 : (SECOND DERIVATIVE TEST) Let f be derivable on an 'interval ] c - 6, c + 6 [ and f'(c) = 0. i) If f"(c) < 0, then f has a local maximum at x = c. ii) If fN(c) > 0, then f has a local minimum at x = c. .

PROOF : The existence of f"(c) implies that f and f' exist ~ n d are conl~nuous at x = c. Continuity at c implies the existence off and f'in a certain neighbourhood, ] c - 61, c + 61 [, 0 < 61 < 6.

(i) Let f"(c) < 0.

This implies that f' is a strictly decreasing function at x = c.

Thus there c ~ s t s 62 (0 < 62 < 61) such that:

ff(x) < f'(c) 440 'V' x E ] C, c i- 6 2 [ (16) and f f (x) > ff(c) = O ' f x E ] C - 6 2 , ~ [ (17)

Now (16) gives f f (x) < 0 'f tf- € ] c, c + 62 [ which implies that f is 8 decreasing function in -

[ c, c i- ] and (2).gives f'(x) > 0 3' x € ] c - 8 2 , c [ which implies that f is an increasing function in [ c - 62, c 1, so that at x = c f has a local maximum.

(ii) Ycu can similarly work out the proof.

We may remark that the above theorem ceases to be helpful if for some c, both f'(c) and f"(c) ' are zero. To provide for this deficiency, we need to consider higher order deiivatives. We make use of the Higher Mean Value theorem i.e. Taylor's theorem to obtain gencralisntion of this result after the following remark.

Differentiability l t n not poss~ble to draw ally conclus~on regarding extreme values of a function at a polnL x = c if fP'(c),= 0.

i) Let lhe function, be defined by f(x) = x3, tf x E R '

Here f'(0) = 0 = fM/O) and the functisa f has neither a local maxinlum nor a lccal minimuni at x = 0. '

ii) Let the function be defined by

f( i) = x", '.c x E R.

Here f'(0) = 0 '= f " (0) and f has a local minimum at x = 0. Similarly f(x) = - x4 , 'v' x E R has a local maximum at x = 0.

Now we give general criteria For fillding extreme values and the second derivative test is also special case of this.

THEOREM 11 : (GENERAL CMTERIA) Let f be a function defined on an interval I and let c be an interior point of I. Let

(i) f'(c) = f" (c) = . . . . fn-I (c) = 0

and (ii) fn(c) exkt and be different from zero, then if n is even, f(c) is a local minimum or a local maximum value o f f according as fn(c) > 0 or fn(c) < 0; if n is odd, f(c) is not an extreme value off.

PROOF : Since fn(c) exists, we have that

f, f' f ", . . . . , fn-.l all exlst and are continuous at x = c.

Also continuity at x = c implies the existence off, f', f", . . . fn-I in a certain ncighbau~*hood ] c - 61, c t 6~ [ of'c (61 > 0).

As f n ( ~ ) ' # 0, 3 a neighbourhood ] c 6, c t 6 [ (0 < 6 < 61) such that for f "(c) > 0,

fn-'(x) < fn-'(c) = O ' v ' x E]c - 6, c[ and fW1(x)> fn-'(c) = OY x E ] c , c 4- 6 [ (18) and for f"(c) < 0, fn-'(x)> fn-'(c) = 0-V x E l c - 6, C [ (19) and f"-'(x) < fn-'(c) = 0 +f x € ]c, c + 6 [ Again for any real number h, where I h 4 < 6, we have by Taylor's theorem with Lagraqge's form of remainder after (n - 1) terms, '

h2 hn- I f(c+ h)= f(c) + hfl(c) + - f"(c) -1 ....+ -

24 fen-')(c+Bh) (O<O.: l ) .

(n - I)! From which we get

hn-I f(c + h) - f(c) = ---- fn-'(c + Oh)

(n - I)! (21)) i

where c + 0 h E ] c - 6, c -t 6 [. (Putting ft(c), f"(c), . . . . . fn-' (c) equal to zero).

f t i Let n be odd : ; 1 Cleariy hn-I > 0 for any real number h and further, when f"((c) > 0, we deduce f rop (18) that I

i t ' for h negative c + Oh E ] c - 6, c [ and fW1(c -I- Oh) < 0 and for h positive,

! i fn-'(c + Oh) > 0.

So from (20), f(c + h) < f(c) T c f h E ] c - 6, c [ and f(c + h) > f(c) V c + h E ] c, c + 6 [ which shows that f(c) is not an.extrerne value.

When fn(c) < 0, it may similarly be shown that f(c) is not an extreme value.

Let n be even : In this case, h"' is positive or negative according as h is positive or negnt: ,, . ~ e deduce from (18) and (20) as before that if fyc) > 0, then for every point c + h E ] c - 6, c f 6 [, f(c + h) > f(c) which means that P has a local minimum at .x ---- c.

. It may similarly be shown from (19) and (201 that f has a local maximum at = if fnlc) < 0. Higher Order Derivatives

The second derivative test can be deduced from this general criteria by taking n = 2. From this theorem, we see that extreme values exist only if the first non-vanishing derivative is of even order. In the following example, you will see the application of this general criteria.

EXAMPLE 15 : Examine the function (x - 3)' (x + 1)4 for extreme values. I SOLUTION : Let f(x) = (x - 315 (x + 114 Then f'(x) = (x - 314 (x f (9x - 7),

f"(x) = 8(x - 313 (X + (9x2 - 14x + l), f"'(x) = 24 (x - 312 (X + 1) (21x3 - 49x2 + 7x + 13), fW(x) = 24 (x - 3) ( 3 ~ - 1) (21xL 49x2 7x + 13)

+ 168 (x - 312 (x + 1) (9x2 - 14x + l), and f v (x) = 48(3x ' 5) (21x3 - 49x2 f 7x + 13)

+ 336 (x - 3) (3x - 1) (9x2 - 14x + 1) + 336 (x - 3)' (X + 1) (9x - 7),

7 Now f ' vanishes for x = - 1, - 7 3.

9 Let us now test these for extreme values.

At x = - 1, fW,is the first non-vanishing derivative and fW(- 1) = - 24.4.4.64 < 0. Therefore x = - 1 is a point of local maxima.

7 At x = - , f " is the first non-vanishing derivative

9

3 16 40,0. and fff 1;) = 8. ( $) -9 j-

7 $0 x = - is a point of local minima.

9 At x = 3, the first non-vanishing derivative is f v, and i! is of odd order.

Thus x = 3 is neither a point of local maxima nor a point of local minima for the function.

EXAMPLE 16 : Show that the function sin x ( I + cos x) has a local maxim4 at x = n/3, (0 s x527.r).

SOLUTION : Lct f(x) = sin x (1 i- cos x) Y x E [ 0 , 2 n 1. Then ff(x) = cos x (1 + cos x) - sin2 x = cos x 4- cos 2x and f"(x) = - sin x - 2 sin 2x.

7

'I'r Therefore f has a local maxima at x = - 4 3 Try the following exercises.

EXERCISE 15 Find the local maximum and minimum values of the function f defined by

i) f(x) = 4x-' - (P - I)-',v lc€R - {O, 1). 1 1

ii) f(r)= sinr + - - s i r 1 2 ~ + ~ ~ i n 3 x T xE[O,rr] 2 EXERCISE 14 Show that the function f denned by f(x) = xm (1 - x)" x X R, where m and n are positive integers has a local maximum value at some point of its domain, whatever the values of m and n may be.

EXERCISE 17 Show that the local maximum value of ( $) '

. --. Differentiabilitv We end this section by gwing a method of finding greatest and least values of a function in an

interval provided the function is derivable at all interior points of the interval.

The greatest and the least values of a function are also its extreme values in case they are attained at points within the interval so that the derivatives must be zero at the corresponding points.

The greatest value of a function is also called global or absolute maximum. Similarly the least value of a function is also known as global or absolute minimum.

If cl, c2, . . . . , ck be the roots of the equation, f'(x) = 0 which belong to ]a, b[, then the greatest and the least values of the function fin [a, b] are the greatest and the least members ,

respectively of the finite set

Ma), ~ (c I ) , f(c21, . . . . , f(ck), f(b)l. Consider the following example.

EXAMPLE 17 : Find the greatest and the least values of the function f defined by f(x) = 3x4 - 2x3 - 6%' + 6x + 1 in the interval [O,2].

SOLUTION : We have

f(x) = 3xL 2x3 - 6x2 t 6x + 1 Therefore f1(x) = 12x3 - 6x2 - 1 2 ~ f 6

= 6(x - 1) (x t 1) (2x - 1)

==3 f'(x) = Oforx = I , - 1, 4- 1/2.

The number - 1 does not belong to the interval [O, 21 and is not to be considered. Now

f(1) = 2, f - = - t f(0) = 1 and f(2) = 21. (;) :: Thus the greatest value o f f in [O, 21 is 21 and the least value is 1.

Try the following exercise.

EXERCISE 18 Find the least and the greatest value of the function f defined by :

f(x) = x4 - 4x3 - 2x2 t .12x + 1

in the interval [ -. 2, 5 1.

13.5 SUMMARY

In this unit, some theorems involving higher order derivatives of a function have been proved and also the applicatio~ of derivatives for finding the limits of indeterminate forms and finding the exteme values of a function has been discussed.

In Section 13,2, Taylor's Theorem has been proved with the help of Rolle's Thgorern. According to this.theorem, iff : [a, b] - R is a function such that its (n - l)[h derivative f"' is continuous in [a, b] and derivable on ]a, b[, then thece is at least one real npmber c E la, b[ such that .

2 (b - a)"-' f "'(a) f(b) = f(a) t (b - a) f'(a) + 9 f"(a), + .,,.. +

(n - 1)l

+ (b - alp '(b - G)"-~ p(n - I)! f n ( 4

p being any positive integer.

(b - a)P (b.- c)"-~ The term - f "(c) is called Taylor's Remainder after n tcrms and deooted p(n - l)! by Rn and this form of remainder in due to Schlomitch and Roche.,By putting p = n and .

68 p,= 1, we get respectively Lagrange's and Cauchy's form of remainder. If we put a = 0 is

Taylor's theorem, we obtain Maclaurin's theorem, In the same section, you have seen how to obtain Maclaurin's series expansion of a function. If f: [a, b] - R is a function such that 'fn(x) exists for any positive integer n and for each x E [0, h] and lirn Rn(x) = 0 for each x E [O, h],

n-== then for all x in [0, h] ,

* which is Maclaurin's series for f(x). Using this result, Maclaurin's series expansions of e" sin x, cos x, log (I + x), (1 + x)ln have been obtained as :

x3 xS X 2 n t l

sin x - x - - + - ..... + (- 1)" 3! 5 ! (2n + 1)1

+ .....%' x E R,

x2 x4 x2S

cosx = 1 - - + - -t- .,... + (- 1)11 - 2! 4! (2n)! + ..... +t x G R

(1 + x)" = 1 + rnx + m(m - x2 + ....., 1x1 < I . 2!

0 In Section 13.3, methods for finding limits of Indeterminate forms O, z, 0, rn, oo - m, 1; .

0 mo, 0' have been given. All these an based on L'Hopital's Rule for g form. If lim f(x) = 0,

1-8

\

f(x) 0 lirn g(x) = 0, then - is said to assume the indeterminate forms - as x tends to 'a'. 1 -n g(x) 0

0 L'Hopital's Rule for - form states that if lim f(x) = lim g(x) =. 0, f'(x), g'(x) exist and 0 x-0 1-0

f(x1 - ft(x) gt(,x) # 0 for all x in ]a - 8, a + S[ (6 > 0) and lima exists, then lirn- - lim-- TYI

X-B gJ(x) %-a g(x) x-n g'(x) 7

L'Hopital's rule for form is similar.

In Section 13.4, application of derivatives for finding extreme valuts of a function is given. Iff is a function defined on an open interval I, and c is any interior point of I, then f is said to have a local or relative maximum at c if there exists a number 6 > 0 such that x E ]c - 6, c + 6[, ,

x # c --J f(x) < f(c). Likewise, f is said to have a local or relative minimum at c if there exists a number 6 > 0 such that x E jc - 6, c + 6[, x # c * f(x) > f(c). f is said to have an extreme value at c if it is either a local maximum or a local minimum at c. You have seen that the' necessary condition for f to have an extreme value at c is that f'(c) = 0 provided it exists. The condilion f'(c) = 0 is not sufficient for f to have an extreme value at c. For example the function f defined by f(x) = I x ( Y x E R it has a local minimum at x = 0 but f'(0) does not exist. For deciding whether a function f has an extieme value at a point c, we have the following general test.

Suppose that f is a function defined on an interval I and c is an interior point of I such that

* f'(c) = f"(c) = ..... fT1-'(c) = 0 and fn(c) # 0. Then if n is odd, then f does not have an . , extreme value at c and if n is even, then f has a local maximum 01. local minimum at c

' accordiog as f n(c) < 0 or f n(c) > 0.

E 1) f is derivable for ail x in R and ft(x) = cos x'tf x ER. So the first derirjative f ' has domain R. The function cos x is also derivable at all points of R. So the second derivative f " has also the domain R and

f"(x) = - hi11 \ . x L R. In general the nth derivative f has also @main R. We can write ff(x) = sin (x + n/2), fU(x) = sin (x + 2 . 7r/2) and in general.

z- fn(x) = sin (x + n 1 ), V x E R.

Higher Order Derivatives

E 2) (i) If x = 0, the statement is trivially true, consider x > 0. By applying Taylor's Theorem to the function f defined by f(t) .= sin t in [0, x] and writing the remainder after 3 terms, we get -

x2 x3

f(x) = f(0) 4- x fl(0) + f"(0) + f"(6 x) where 0 < 6 < 1)

x3

i.e. sin x = x - - cos (Ox) (putting the values off, f', f", f"'). Since cos Ox 5 I, 3!

whatever Ox may be and x > 0,

Again, applyiig Taylor's theorem to the function fin [O, x] and writing the remainder after 5 terms,, we get

x3 x3 sin x = x - x + j l c o s ( 6 1 x), where0 < dl < 1.

Since cos (81 x) I 1, whatever 81 x may be and since x > 0, therefore

From (21) and (22) we have x3 x3 xS

x--<s inx I x - -+ -wheneverx> 0 3! - 3! 3! Let now x < 0. Set y = - x, then y > 0.

From (23), we have

Put y = -x in it and simplify. You will get

E 3) (i) Let f(x) = cos x

Here a = n/4, h = x - n/4. n T

Now f'"(x) = cos (x + 2' 'r T n7r

Therefore f'"(-) = cos (- + 4 4

Put n = 1,2, 3, ..... we get 7r 7r 7r T

ff(7r/4) = - sin - , f "(-) = - cos 7 fU'(7r/4) = sin - , ..... 4 4 4

Assuming the possibility of expansion, we have by Taylor's Series,

h2

f(a .+ h) = f(a) -I-. hf '(a) + f "(a) + ..... I I n ' lr IT 1 = 2 7r i.e. na x = cos - + (x - ( - sin - ) + - (x - T) { - cos - ) 4 4 21 4

1 7r lr + -(x - T) sin - 4 ""' 3!

sion of cos x.

E 4) Prmed.as in the expansion of cos x, in Example 4 by takirig n n

f(x) = sin x, fn(x) = sin (x 4- 2).

E 5 ) Two case3 arise :

(i) m is a positive integer. ~ e t f(x) = (1 + a x E R. We find that V n E N, f'"(x)exists-V x ~ R R I Y ~ ~ ' ~ ( x ) = OV n > n ~ a n d Y x E E .

Therefore Rn(x) = 0 for all n > m.

Which implies lim R,(x) = 0 and, we have I!--

xm f(x) = f(O) 4- xfl(0) ..... + x f ' " " ( ~ ) , ~ x E R,

since the.other terms all vanish. Substituting the values of f(x), f(O), f'(O), .... fim'(0), we have

(ii) m is not a positive integer. In this case, we find that if we write

f(x) = (1 + x)", whenever m # - I, then f'"(x) = m (m - 1) ,.... (m - n + 1) (1 + x)'~-", Thus for each positive integer n, 'f'"' is defined in [- h, h] for each h E 10, l[.

If R,(x) denotes Cauchy's form of Taylor remainder after n terms we have

We know that for ( x I < 1,

1 - 8 Also 0 < ---- I + ex

< 1 f o rO<8< 1 andfor- 1 < x < 1.

Therefore

Next (1 + ex)"-' < (1 + Ix I)'-' if m > 1 as 0 < 8 < 1,

< 1 and (1 + ex):-' = when m < 1.

(1 + ~ x ) ' - ~ (1 - Jx I)'-"

Hence the conditions of Maclaurin's infinite expansion are satisfied. Making the substitutions,

f(0) = 1, f'(0) = rn, f "(0) = m(m - l), ..... ' fn(0) = m(m - 1) ..... (m - n 4- 11, we get

m(m - 1) x2 +. m(m - l)(m - 2) (1 f x)' = 1 + mx + , 2! 3! x 3 + .,...

for Jx 1 < 1.

Note that when m is not a positive integer, the expansion is not pbssible if 1x1 > 1,forthenas n --*,

m(m - 1) ..... (m - n 4- 1) xn and so R,(x), does not tend to z&o, (n - l)!

Higher Order Derivatives

Differentiability E 6) Let f(x) = log (1 + sin x) COS X

Then f'(x) = -$,;,,

1 fP'(x) = - 1 + sin x

COS X f"'(x) = (1 + sin x)'

sin x + sin2x 4- 2 COS'X fW(x) = -

. (1 + sin x13

Putting x = 0, we get

....... f(0) = 0, f'(0) = 1, f"(0) = - 1, f"'(0) = 1, fN(0) = - 2

Substituting these values in x2

f(x) = f(0) + xf'(0) t 7 f"(0) + ...... we get

sin 3x2 0 (- form) ') !i5 ,lag tos (2x2 - x) 0

6x cos 3x2 lim X - a -. tan (2xZ - x) (4x - 1)

x(2x - 1) 1 = - 6 lim cos 3x2. lim lim

x 4 .+ tanx (2x - 1) X+O (2x - 1)(4x - 1)

sinh x - sin x 0 iii lim

x-to x t d x (iyfom)

sinh x - sin x = lim

x-bo x3

sinh x - sin x 0 x = liln (- form) (- -+ 1 as x -+ 0)

x-bO -- x3 0 tan x

cos h x - cos x 0 = lim (- form)

r - o 3x2 . 0

sin h x + sin x 0 = lim (- form)

X- o 6~ 0

' 1 (1 + x)"" e + - ex 2

iii) lim ' is (0/0) form as x-0 xZ

Let y = (1 t x)""

1 Therefore, log y = - log (1 + x).

X

Higher Order Derivatives

where o (x3) stands for those terms of x containing x3 or higher powers. Thus the given limit becomes

Now sin 4x + a s in .2~ 0

lim ( - form) x -- 0 x3 0

4 cos 4x -I- 2a cos 2x = lim I

x - (1 3x2 its denominator tends to zero for x - 0, f i e fraction will tend to a finite limit only if the numerator also tends to zero as x - 0. This requires 4 + 2 a = O * a = - 2 .

0 When this is satisfied, we have - form and the given limit

0

- ;116sin4~-4asin '2x 0 = lim ( - form)

x -0 6~ 0

- 64 cos 4x - 8a cos 2x = l i y

x - 6

- - - 64 .- 8a 6

= - B ( a = - 2).

' 3x1 - 4 The expression -

2 x + 1 is not of the form 0/0 as x -- 1.

3x2 - 4 Therefore it is not correct to apply L' Hopital's rule to evaluate lim ------ .

x - 1 2x + 1

lirn (3x2 - 4) 3 x 2 - 4 x - I - 1 In fact lirn - = =-.

x - 1 2 x + I l im(2x+1) x - l

3

7r log (x - - )

2 09 i) lim ( form)

x - = I ~ + tanx

Differentiability

cos2 x 0 = lim - (, form)

x - t a t

2 cos x (- sin x) = lirn = 0.

, x - r /Z+ I 00

ii) It is form. Apply L' Hopital's Rule. The limit is 0.

E 11) i) Given form is 0.a. lirn sin x log x2

x-0 log x2

= lirn - ( form) x - 0 COSeC X

- 2 = lirn

x - o X C O S ~ C ~ ~ ~ ~ X

sin x tan x -1 iq- I - 2 . x

Q ' ii) The given form is 0.m. Convert it to - form and then fmd the limit, The limit is 0

2

1 E 12) i) The limit is - -

2

1 1 ii) l i m ( T - - ~ ) ( m - " f o r m )

x-0 x t a n x

tan2 x - x2 0 =.lim ( - form)

K-o x2tan2x 0

-tan2 x - x2 x 2 = lirn

K O x4, (G? tan2 x - x2 x 2

= lim Urn(-) '-0 X4 1-0 tan x

t an2x -x2 0 = lim

r-o X4 ( 7 form)

tan x. sec2 x - x = lirn

x-to w (By L'Hopital's Rule)

SecZ x + 3 tanZ x secZ x - 1 tan2x + 3 t d x sec2x = lirn w = lim

XUJ K-M 6x2

ianx 2 1+3sec2x 2 =lim(-) =-

x 4 x 3

is 1" form. a

R sec2 (-) Let y ,= [sin' (-i-- )] 2 - bx

- ax Then

?r log y = secv- log s i n z ( L )

2-bx 2-sx

-

'R log sin2 (- Higher Order Derivatives

2 - $ 0 lim log y = lim X+O x~ (-ij- form)

X i c0s2 (- 1

2 - b x - T r a

2cot (- )- 2 - ax,, (2 - ax12

>z.lim x -0 2 n nb

- sin ( - ) 2 - bx (2 - bx)'

cot ( - - 2a

) 2 - ax - - - lim . lim (2 - bx12

b x - o 2 ~ . )

,-o (2 - ax)2 sin ( -

2 - bx n-

cot ( - - 2a 2- a x ) 0 - - - lim ( - form)

b a - 0 277 0 sin ( -

2 - bx )

n: 1

.na - cosec2 ( -

- 2a 2 - ax (2 - ax)2 - - - lirn b x - o 27r )

27rb ms(-

2 - bx (2 - b ~ ) ~

- 2a a - - - . - b 2b

- - a2 - -

b2

=+ lim y = e- n'/b'. 1-0

ii) 1. iii) 1.

E 14) f f(x) = 60 x2(x + I ) ~ (x - 2). Apply first derivative test and show that at x = 0, f has neither local maxima nor local minima and at x = 2, f has local minima.

4 E15) i) f(x)=---- V x E R - ( 0 . 1 )

x x - 1

- 4 1 =+ ff(x) = 7 + (X--l)i

Now f'(x) = 0 * x = 2, 2/3.

8 2 Also f"(x) = - - -

X~ ( ~ - 1 ) ~ Clearly then f has local maximum at x = 2 and local minimum at x = 2/3.

ii) Maximum at x = 7r/4 and 3 ~ / 4 , minimum at x = 27r/3.

E 16) Here f(x) = xm( l - x ) " ~ x E R Therefore f'(x) = mxm-' (1 - x)" - nxm (1 - x)"-I

- Xm-l - (1 - x)"-' (m - mx - nx) If both m = n = 1,

1 ' thenf'(x)= 1 - 2 x = 0 -J x = - 9

I 'I 2 - ' and f"(x) = - 2 < 0 i.e. f has a local maximum at x = 1/2 in this case.

Therefore we can assume either m > 1 or n > 1. In any case, f'(x) = 0 yields x = m/(m + n) and i'(x) changes sign from positive to negative at this point. Thus f has a local maximum at x = m/(m n).

For extreme valuer, dy '= 0, which means dx

log x f 1 = 0 * x = e-'.

Now (24)

- = - dy y (log r + l j dx

dZy Y =+ - = - dy (log x + I ) - -

dx 2 5 x

I d2y At x = e- , we have 7 = - e'" . e < 0 dx

Therefore at x = e-I, y has a local maximum and the maximum value of y is el"

In this block, you have been introduced to the rigorous notion of the derivative of a function. !

'Also, the relationship between the continuity and differentiability have been explained. Further, the algebra of the derivatives has been discussed. All these have been deslt with in Unit 11. In Unit 12, certain mean-value theorems have been discussed while in Urrit 13, the results of these theorems have been extended to higher derivatives in the form of Taylor's

q theorem and Maclaurin's series. Further, these theorems have been used to give the power : series expansions of some algebraic and transcendental functions m well as evaluating the

limits of indeterminate forms and extreme values of some functions. Now try the following questions so as to test for yourself your conceptual understandingof the material.

1) Give an example of each of the following : a) A functio~l which is not differentiable at ode point of its domain. b) A function which is not differentiable at two points of its domain. c) A function which is not differentiable at three points of its domain.

2) State whether the following statements are true or false. a) A continuous function is always differentiable. b) A differentiable function is always continuous. c) Every monotonic function is differentiable. d) Every differentiable function is monotonic.

3) Give an example of each of the following : a) Two functions f and g such that f -t g is derivable but f and g may not be derivable b) Two functions f and g such that f . g Is differentiable but f and g may not be

differentiable. c) Two functions f and g such that f - g is differentiable but f and g may not be

differentiable. d) Two functions f and g such that f/g is defined and differentiable but f and g may not

be differentiable. I

4) Give an example OF a function f such that it is not derivable but I f I is derivable at every point of the domain.

1 5) Give an example of each of the following : , a) A function f to which Rolle's theorem is applicable.

b) A function f tb which Rolle's theorem is not applicabie.

6) Ilsing the Maclaurin series expansions of sin x and cos x, find the power serics expansion of co: 2x and sin x cos 2x.

7) Is stationary value of a function necessarily an extreme value? Justify your answer.

8) Prove that i) if f-is continuous in [a, w[ and f'(x) > 0 Y x E ]a, -[, then f is strictly increasing in

[a, =-[.

ii) if f is continuous in [a, w[ and f'(x) < 0 Y x € ] a, m [, then F is strictly decreasing in [a, cx, [.

I 9). What is wrong with the following use of L'Hopitd's Rule x3. - 4x2 3x2 - 8x;- G x - 8 - 4

lim = lim - lim - - - - a--0 9x2 - 2x x-0 18x - 2,. x - 0 18 9

1 1 10) Which indeterminate form is lim [ (4 4 - ) 7 - 1. Find the limit.

x - ~ x sin x

1) a) F(x) = I x 1 V x E R. f is not differentiable at one point '0' of its domain. b) f(x) = 1 x 1 4 1 x 4 1 ( 'v'x E R. Differentiable except at the points - 1 and 0 of

thc domain. c) f(x) L- ( x I + ( x + 1 I + I x - 1 ( VX E R. Differentiable except at the

points - 1, 0 and 1 of the domain

2) a) False. The function f givcn,by f(x) = 1 x - 1 ( Y x E R is continuous but not differentiable at I.

Differentiability b) True.

c) False. The function f : R -- R defined by

I 1 when x > 0 f(x) = - 1 when x'< 0

0 when x = 0 is monotonic but not differentiable at 0.

d) False. The function f : R -- R defined by x3

f(x) =-- 4xVx E R 3

is differentiable in R but it is monotonic because f'(x) = x2 - 4 = (X - 2) (x + 2) which is positive if x > 2 or x < - 2 and is negative for - 2 < x < 2 and consequently f is mdilotonically increasiog for x I - 2 and x 1 2 and f is monotonically decreasing; in [ - 2 ,2 1.

3) a) Take f(x) = I x I and g(xj = - 1 x I Y x E R.

b) Take f(x) = g(x) = ( x I V x E,R.

I 1 when x is rational 4) consid& f(x) =

- 1 when x is irrational So f is not derivable at any point of R but I f I is derivable and I f I '(x) = 0 V x E R.

5) a) Takef(x) = x4, a = - 1, b = 1

b) Take f(x) = sin x; a = - n/2, b = r/2.. 1 1

6 ) cos22x = ( + 4X ) = 7 [ 3/2 + 2 cos 4x + - 2 cos 8x ] 1.

1 sin x cos 2x = - (sin 3x - sin x)

2 Now expand cos 4x, cos 8x, sin 3x, sin x in power series.

7) Consider the function f(x),= (x V x <R. f'(1) = 0 and so f(1) is a stationary value. But f(x) > f(1) for lx >>.I and f(x) < f(1) for x < 1 and consequently f(1) is not an extreme value.

8) Proceed exactly in the same way as that for the finite interval [a, b]. 0. . . 3x2 - 8x

9) lim is not of - form. L Hopital's ruk cannot be applied. , x-0 18x - 2 0

10) It is .o - 00 form. The given limit can be written as 4x sin x 4- sin x - x

lim 0

a Now it is - form. Apply L'Hopital's rule and you will get h-0 , x sin x 0

the limit as 4.

NOTES

IJNIT 14 THE RIEMANN INTEGRATION

Structure 14.1 Introduction

Objectives

14.2 Riemann Integrability 14.3 Riema~in Integrable Functions ' 14.4 Algebra of Integrable Functions 14.5 Conlputing an Integral 14.6 Summary 14.7 Answers/ Hints/ Solutions

14.1 INTRODUCTION

You are quite fanliliar with the words 'differentiation' and 'integrpti~n'. You know that in ordinary language, differe~itiatio'n refers to separating a r distinguislliIlg things while integration means putting things together. In Mathematics, particulnrly in Calculus and Analysis, differentiation and integration are considered as some kind of operations on functions. You have used these operations ip your study of Calculus. You have also studied differentiation in a rigarous way in Unit 11. 114 this unit, you will be introduced to the operation of integratipq in a rigaraps nianner.

There are essentially two ways of describing the operation of integration. One way i s to view it 3s the inverse operation of differentiation. The other way i s to treat it as some sort of limit of a sum.

The first view gives rise to an integral which is the resu!t of reversing the prooess of differentiation. This is the view which was gengrally cansidered during the eighteenth century.

Accordingly, the method is to obtain, from a given function, another function whicll has the first function as its derivative. This secorld function, if it b# obtained, is called the indefinite integral of the first function. This is also called the 'primitive' or anti-derivative of the first funclion. Thus, the intcgral of a function f(x) is obtained by findipg an anti-derivative o r primitive function F(x) swpb that F'(x) = f ( ~ ) . The indefinite integral of f(x), is symbolized by thp nota t i~ l l l f (x ) dx.

The second view is relatcd to the limiting process. I t giyes rise Lo an integral which is the limit of all the values of a function in an interval. This i ~ j the integral of tl function f(x) over an interval [a,b,]. It is called the definite ip#eq#l and is donoted by *

b

1 f(x) dx. d

The definite integral is a number since geometrically it ~or respor~ds to an area of a region enclosed by the graph of a function.

I

Although both the notions of integration are closely related, yet, you will see Lter, the definite integral turns out to be a mare fuqdawental concept. In fact, i t is the starting Ijoint for some importan1 genera\iza\inr+s like the double iniegrals, triple integrals, lirle inlegrals etc, which you may study in the Cayrse MTE pn Advanced Calculus,

F

The integral in the anti-derivative sense was given by Neyrtan. This notion was found Lo be adequate so long as the functions to be integrated were pmtinuous. Bul in the carly 19th ceiltury, Fourier brought to light the need for makipg integration meaningful far the f~!nct~ons that are not c~n l ipuous. He csme across such functioqs in applied problems. Cauchy forn~ulated -rigorous definition a f the integral of a function. He essentially provided a general theory of integration but only for continuous functions. C~*uchy's theory of Inlegration for continuous functions is sufficient for piece-wise continuous functions as well as for the functions having isolated

discontinuities. However, it was G.B.F. Riemann [1826-18661 a German mathematician who extended Cauchy's integral to the discontinuous functions also. Riernann answered the questiorl " what is the meaning of l f (x) dx??'

The concept ot definite integral was given by Riemann in the middle of the nineteenth century. That is why, it is called Riemann Integral. Towards the end of 19th Century, T.J. Stieltjes [1856-18941 of Holland, introduced a broader concept of integration replacing certain linear functions used in Riemann Integral by functions of more general forms, In the beginning of this century, the notion of the measure of a set of real numbers paved the way to the foundation of modern theory of Lebesgue Integral by an eminent French Mathematician H. Lebesgue [1875-19411, a beautiful generalisation of Riemann Integral which you may study in some advanced courses of Mathematics. In this unit, the Riemann Integral will be defined without bringing in the idea of differentiation. In unit 12, you will see the usual connection between the Integration and Differentiation. Just by applying the definition, it is not always easy to test the integrability of a function. Therefore, condition of integrability will be derived with the help of which it becomes easier to discuss the integrability of functions. Then just as in the case of continuity and derivability, we will also consider algebra of integrable functions. Finally, in this unit, second definition of integral as the limit of a sum will be given to you and you will be shown the equivalence of the two definitions.

Objectives

After the study of this unit, you should, therefore, be able to -- define the Riemann Integral of a function - derive the c~ndit ions of Integrability and determine the class of functions which

are always integrable -- discuss the algebra of integrable functions -- compute the integral as a limit of a sum.

14.2 RIEMANN INTEGRATION

The study of the integral began with the geometrical consideration of calculating areas of plane figyres. You know that the well-known formula for computing the area of a rectangle i s equal to the product of the length and breadth of the rectangle. The questiotl that arises from this formula is that of finding the correct modification of this formula which we can apply to other plane figures. To do so, consider a function defined on a closed interval [a,b] of the real line, which assumes a constant value K 2 0 throughout the interval. The graph of such a function gives rise to a rectsngular region bounded by the X-axis and the ordinates x = a, x = b as shown in tho Figure I.

k (b-a) k I

I I I I . 0, a b

)% '

Obviously, the area enclosed is k (b-a). NOW, suppose that ]a, b[ is further divided into smaller intervals by inserting points of division, say

a = x, S x , I x , I ' x , I x , = b,

and the function f is defined so as to take a constant value at each of the resulting sub- intervals i.e.,

k,, if x E [xI , x2[ f(x) =

Further, suppose that di = length of the ith interval ]xi, xi_, [ i.e.,

Then, we get four rectangular regions and the area of each region is A, = kldl, A, = k2dz, A, = k,d,, and A, = k,d,, as shown in Fig. 2.

Fig. 2

The total area enclosed by the graph of the function, X-axis and tlie o'rdin:ates x=a, x=b is equal to the sum of these areas i.e.

Area = A1 + A2 + A3 + A4

= kid I + k2d2 -k k3d3 + k4d4.

Note that in the last equation, we have generalized the notion of area. In other words, we are able to compute the area of a region which is not of rectangular shape. How did we get it? By breaking up the region into a series of non-overlapping rectangles which include the totality of the figure and summing up their respective areas. This is simply a slight obstraction of the same process which is used in Geometry.

Since the graph of the function in figure 2 consists of 4 different steps, such a function, as you know f rom unit 4, is called a step function. What we have obtained is the area of a region bounded by

I i) a non-negative step function ii) the vertical lines defined by x=a and x=b iii) the X-axis.

This area is just the sum of the areas of a finite number cff non-overlapping rectangles resulting from the graph of the given function. The area is nothing but kreal number.

Now suppose that the graph of a given function is as shown in the figure 3.

Uoes it make any sense to obtain the area of the region under the graph o f f ? If so, how can we compute its villue'? T o answer this question, we introduce the notion of the integral of a function as given by Riemann.

Fig. 3

T o introduce the notion of an integral of a function, we will require such a real number which results frola applying the function and which represents the area of the region bounded by the graph o f f , the vertical lines x=a, x=b and the X-axis. This can be achieved by approximating the given function by suitable step functions. The area of the region will, then, be approximated by the areas enclosed by these step functions, which in turn are obtained as sum of the areas of non-overlapping rectangles as we have computed for the figure 2. This is precisely the idea behind the formal treatment of the integral which we discuss in this section. First, we intioduce some terminology and basic notions'which will be used throughout the discussion.

Let f be a real function defined and bounded on a closed interval [a,b]. I Recall that a real function f is said to be bounded if the rangc o f f is a bounded subset of R, that is, if tl~ei-e exist numbers rn and M such that m 5 f(x) 5 M for each x E [a,b]. M i h an upper bound and m i s a lower bound o f f in [a,b]. You also know that when f is bounded, its supremum and infimum exist. We introduce the concept of a partition of [a,b] and other related definitions:

IDEFINITION 1 : PARTWION Let la,b] be a given interval. By a partition P of [a,b] we mean a finite set of points {x0, xl, .... ,., x,}, where

a = x , < x , i .... < x , - , < x , , = b .

We write Ax, = xi - xi-,, (i=l, 2, ..., n). So Axi is the length of the ith sub-interval given by the partition P.

DEFINlTION 2: NORM OF A PARTlTION Norm of a partition P, denoted by \PI, is defined by (PI= max Ax,. Namely, the norm of P

I S I S n

i s the length of largest subinterval of [a, b] induced by P. Norm of P is also denoted by AP).

There is a one-to-one correspondance between the partitions of [a,b] and finite subsets of ]a, b[. This induces a partial ordering on the set of partitions of [a,b]. So, we have the following definition.

1 DEFtNlTlON 3: REFINEMENT OF A PARTITION Let P, and P, be two partitions of (a,b]. We say that P, is finer than P, or P, refines P I 1 or P, is a refinement of PI if PI c P,, that is, every point of P, is a point of P,. I

You rnay note that, if P, and P, are any two partitions of [a,b], then P, u P, is a common

1 1 1 1 1 1 1 refinement of P, and P,. For example, if P, = ( 0 ,;i, 7 ,i , 1) and P,= {0,%, 5 ,q, i, 1 I

1 1 1 1 I -. I ) are partitions of [O, I], then P, is a zefinement of P, and PI v P, = { 0, 2 5, T , i

2 I 8

I ) is their common refinement. - j - ,T3 f I

We now introduce the notions of upper sums and lower sums of a bounded function f on an interval [a,b], as given by Darboux. These are sometimes referred to as Darboux Sums.

DEFINITION 4: UPPER AND LOWER SUMS Let f: [a,b] - K be a bounded function, and let P = {xo, X I ... x,) be a partition of [a,b]. For i = 1, 2 ....., n, let M i and r n ~ be defined by

Mi = lub {f(x)': xi-! I x I xi)

mi = glb (f(x) : xi-1 5 x 5 xi)

i.e. Mi and m i be the supremum and infimum of f i n the sub-interval [xi-,, xi].

Then, the upper (Riemann) sum of f corresponding to the partition P, denoted by U (P,f), is defined by

n U (Ptf) = Mi Axi

i= 1

The lower (Riemann) sum o f f corresponding to the partition P, denoted by L(P, f), is defined by

Before we pass on to the definition of upper and lower integrals, it is good for you to have the geometrical meaning of thc upper and lower sums and to visualize the above definitions pictorially. You would, then, have a feeling for what is going on, and why such definitions are made. Refer to Ggures 4(i), 4(ii), 4(iii).

Fig. 4

In figuce 4(i), the graph of f : [a,bj -. R is drawn. The partition P = {XO, X I , .... xn) divides the interval [a,b] into sub-intervals 1x0, XI], [XI, x.1, .... [x,-I, x,]. Consider the area S under the graph o f f . In thefirst sub-interval [XO, XI], ml is the g.1.b. of the set of values f(x) for x in [xo, XI]. Thus ml Ax, is the arca of the small rectangle with sides m l and Ax, as shown in the figure 4(ii). Similarly mr A x2 ... rn, A xn are areas of such

n small rectangles and 2 m, Ax, i.e. lower sum L (P,O is the area SI which is the sum of

1 = 1

In tegrab i l i t y , areas of such small rectangles. The area SI is less than the area S under the graph 01'1'. In the same way M I A X I is the area of the Large rectangle with sides MI and -1 X I

11 and 2: MI -1 x, i.e.the upper sum U (P,f) is thc area Sz which is the sun1 of areas or

16 1

such large ~.ectangles as shown in Figure 4(iii). The area S, is more than the area S under the graph o f f . It is intuitively clear that if the points in the pal-tition P are increased, the areas S, and S, approach the area S.

. , We claim that the sets oS upper and lower sums corresponding to different parti'tions of [a,b] are bounded. Indeed, let m and M be the infimum and supremum of f i n [a.b]. Then m 5 ml I MI 5 M and so

m h x 1 _ ( r n i A x , ~ M ~ 4 ~ ~ 5 M A x , Putting i = 1, 2, ........ n and adding, we get

n n m 2: 1 x, I L(P,f) 5 U(P,fl< M r; A x,.

1-1 , = I

I" I 1- I

Thus m (b-a) 5 L(P,f) 5 U(P,f) 5 M (b-a)

For every partition P, there is a lower sum and there is an upper sum. The above inequalities show that the set of lower sums and the set of upper sums are bounded,.so that their supremum and infimum exist. In particular, the set of upper sums have a n infimum and the set of lower sums have a supremum. This leads us to concepts of upper and lower i;tegrals as given by Riemanrl and popularly known as Upper and Lower Riemann Integrals.

DEFINITION 5: UPPER AND LOWER RIEMANN INTEGRAL Let f: [a,b] --> R be a bounded function. The infimum or the greatest lower hound o f , the set of ail upper sums is called the upper (Riemann) integral o f f o n [a,b] and is denoted by ,

- b

f(x) dx.

- b

$ f(x) dx. = g.1.b. (U(P,f): P is a partition of [a,b]]. a

The supremum o r the least upper bound of the set of all lower sums is called the lower (Riernann) integral o f f on [a,b] and is denoted by

,f f(x) dx = 1.u.b {L(P,f): P is a partition of [a,b]]. 4 - I - _ _ _- Now we consider some examples where we calculate upper and lower integrals.

EXAMPLE 1: Calculate the upper and lower integrals of the function f defined in [a.b] as follows:

I I 1 when x is rational f(x) = U when x is irratiorial

SOLUTION: Let P = {x~ , ,x i ...... x,] be any partition of [a,b]. Let MI and m l be respectively the sup. f and il.rf..f in [x,-1, x,]. You know that every inferval contains infinitely many rational as well as irrational numbers. Therefore, m, = 0 and M, = I for i = 1, 2 ... n. Let us find U(P,f) and L(P,f).

Therefore U(P,f) = b-a and L(P,f) = 0 for every, partition P df [a,b]. Hence ' r l~e Riemnnn Integration

b

I f(x) dx = g.1.b. {U(P,f): P is a partition of [a,b]] a -

= g.1.b. {b-a] = b-a. b

I f(x) dx = 1.u.b. {L(P,f): P is a partition of [a,b]] P

= 1.u.b. {O] = 0.

EXAMPLE 2: Let f be a constant function defined in [a,b]. Let f(x) = k y x E [a$]. Find the upper and lower integrals of f . .

SOLUTION: With the same notation as in example 1, Mi = k and mi ='k V i. n n

S o U(P,f) = r; Mi A xi = k z A xi = k (b-a) i= l i = l

n n a n d L (P,T) = z mi A xi = k c A xi = k (b-a)

i = l i= I

Therefore U(P,T) = k(b-a) and L(P,f) = k (b-a) for every partition P of [a,b]. - b b

Consequently I f(x) dx = k (b-a) and / f(x) d x = k (b-a) a n

Now try the following exercise.

EXERCISE 1

Find the upper and lower Riemariri integrals of the function f defined in [a,b] as follows

- 1 when x is rational -1 when x is irrational

You have seen that sometimes the upper and lower integrals are cqual (as in Example 2) and sonletinles they are not equal (as in Esample 1). Whcnever they are equal, the function is said to be integrable. SO integrability is defined as follows:

DEFINITION 6: RIEMANN INTEGRAI, I.CI f: [a,b] -- R be a bounded function. The function f is said to.be Riemann

b b integrable or simply integrable or R-integrable over [a,b] if / f(x) dx = / f(x) dx and

'a - i f f is Riemann integrable, we denote the common value by * f(x) dx.This is called the

J . il Hiemarin integral a r simply the integral o f f on [a,b].

EXAMPLE 3: Sl~ow that the function f corisidered in Exanlple 1 is not Rie~nanrl integrable.

b

SOLUTION : As shown in Example I , I f(x) dx.= b-a and b h a b I F(x) d x = 0 and s o l Sin) dx # f(x) dx and consequently f is not

a

Riernann integrable.

EXAMPLE 4: Show that a constant function is Riemann integrable h

over [a,b] and f ind1 f(x) dx. I -

b b SOLUTION: As proved i;l Example 2 , / f(x) d x = k(b-a) =I f(x) dx

a h 1

Therefore, f is Riemann integrable on [a,b] a n d I f(x) dx = k (b-a). I

EXERCISE 2. Show that the function f defined in Exercise 1 is no1 integrable.

I I

I t is n o [ ulw;~!s easy to find [hc upper and I(i\\cr Rii'lli;~lili ilitegrals o!' ;I gi\~,r l boundcd I'unu~ic~n t 'and therch! Jccldc \I hcrhcs the I'~1nctio11 is inrcgl.;~hlc o\cl. I llc gi\sen intcr\;ll o r not. FOI; th is. wc discu\s home conditions of in~egr-abilit!. \ + i t 1 1 1\11. help ol'which wc can dccidc inrcp.;thilit! ol' a function wirliout I'inding uppcl. i 1 1 i ~ I

lower integrals. For pro\:in.g tlicsc cclndi~icini ol' inrcgrability \vc. rccl~1i1.c \ omc r-c\~rl~h which we give below i n the I'orm of theorem,. Svmc :Ire pro\cd LA liilc otlicl.\ i11.c _ ' :I\ CII

without proor.

THEOREM I ; I f the partition P, is a refinement of the partition P, of [a,b], then L(P,,f) 2 L(P,, f) and U(P,,f) 5 UP,,?).

PROOF: Suppose P, contains one point more thall P,. Let this extra point be c. Let PI = (x,, x,, .... , xn} and x ,_ , (- c < x,. Let M, and 111, be respectively the sup. f and inf. f in [xi-, , xl]. Suppose sup. f and inf, f in [x,-, , c] are p, and q l and those in [c, x,] are p, and q,, respectively. Then,

L(P,,f) - L(P,>f) = q , (c-x,_,) + cl, (x,-c)-mlAxl

(since A x, = (xi-c) + (c-x,-,).) Silnilarly U (P?,f) - U(P1,f) = ( P I - MI) (c-x~ I ) + (PZ-M,) ( ~ I - C )

Now m, 5 q 1 MI m, I q l 5 p? 5 MI

Therefore L(P?, f) - L (pl,t] L 0 and U (P?, f) - U (P I , f) 5 0 Therefore L ( P I , ~ ) 5 UP?, 0 and U (Pz, f ) 2 U (P, , I).

I S P2 contains p po~nts more than PI', then adding these extra points one by one lo P I and using the above results, the theorem is proved. We can also write thc theorem ns

L(P1.f) 5 L(P2,17 5 U(P2,f) I U(P,,I) fl.orn which it follows that U(P2,f) - L(P2,f) 5 U(PI.I) - L(P ,,t). As nn illustration ol' thcorcm.1, we consider the following example.

EXAMP1,E 5: Verify Theorem 1 for the function f(x) = x + 1 defir~ed over [0,1 j and

I I I 3 SOI.LITION: For partition P I , n = 5, xcl = 0, X I = - , x? = - , X I z - , x 4 = i , x. = I 4 3 2 .

I I 1 I :111dsoAx1 = - . 1 ' ; 2 = ~ . A x 4 = - , A x 4 = - , 1 x 5 = - . 4 12 6 4 4

5 4 I-u~.~licr M, = I'(x,) bi ni, = I'(x, I ) lor. i = 1; 2, 3, 4, 5 and therefore MI = - , M2 = - , 4 3

5 29 and' U(P,,f) = C Mi Ax, = -

i= l 18 '

17 I

Similarly, L(P,,f) =- , and 12

19 U(P,,f) =-. Hence L(P,,f) 5 L(P,,f) and U(P,, f ) l U(P,,f).

12

Do the following exercise yourself.

EXERCISE 3

Verify Theorem I for the function f(x) = sinx defined over the interval 10, $1 and I

7r 7r the partitions PI = (0, - ,-] and P, = (0,

4 2

By applying Theorem 1, it is easily proved that lower integral of a R~nction is less than or equal to upper integral of the function. It is proved in the next theorem. From Examples 3 and 4, you can see the truth of this result.

'11

THEOREM 2: / f(x) dx L /f(x) dx. 9 P

PROOF: If PI & P, are two partitions of [a,b] and P = PI u P, is their common refinement, then using Theorem 1, we have L(P,,f) I L(P,f) I U(P,f) < U(P,,f) and

L(P,, f ) < L(P, f ) I U(P, f ) l U(P,, f) . Therefore, L(Pl, f ) I U(P,, f 1. Keeping P2 fixed and taking 1.u.b. over all P I, we get

b

j' f(x) dx 5 U (PI, f) a

Now taking g.1.b. over all Pz, - we obtain

This proves the result.

In Theorem 1, we have compared the lower and upper sums for a partition P I with those for a finer partition P2. Next theorem, which we state without proof, gives the estimate of the difference of these sums.

THEOREM 3: If a refinement P, of P I contains p more points and If(x)l < k, for all x E [a, b], then

and U(P,,f) 2 U(P,,f) I U(P,,f) -2p k 6, where 6 is the norm of‘^,. This theorem helps us in proving Darboux's theorem which will enable us to derive conditions of integrability. Firstly, we give Darboux's Theorem.

T.HEOREM 4: (DARBOUX'S THEOREM) I f f : [a,b] -> R is a bounded function, then to every 6 > 0, there corresponds 6 > 0 such that

- b

(i) U(P,f) <$ f(x) dx + E n

b (ii) L(P,f) > / f(x) dx - E

P

for every partition P of [a,b] with JPI < 6.

PROOF: We consider (i). As f is bounded, there exists a positive number k such that - b

If(x)I 5 k Y x E [a,b]. As / f(x) dx is the infimum of the set of upper sums, therefore B

to each 6 > 0, there is g partition PI of [a,b] such that b E

U(P,, f ) < /f(x) dx +- n 2 '

(1)

Let PI = {x,, x ,,...., xp} and 6 be a positive number such that 2 k (p-1) 6 = €12. Let P be a partition of [a,b] wlth lPl < 6. Consider the common refinement P, = P u PI of P and PI. Each partition has the same end points 'a' and 'b'. So P, is a refinement of P having at the most (p -1) more points than P. Consequently, by Theorem 3,

r Thus

6 =/f(x) dx + 6, with IPI < 6.

' l ' l~e Riemann Integration

EXERCISE 4 Write down the proof of part (ii) of Darboux's Theorem.

As mentioned earlier, Darboux's Theorem immediatley leads us to the conditions of integrability. We discuss this in the form of the following theorem:

'THEOREM 5: (CONDITION OF INTEGRABILITY) FIRST FORM : The necessary and sufficient condition for a bounded function f to be integrable over la,b] is that to every number c> 0 there corresponds 6 > 0 such that

U(P,f) - L(P,f) < E,V P with JPI < 6.

PROOF ; We firstly prove the necessity of the condition. Since the bounded function f is integrable on [a,b], we have

Let E > 0 be any number. By - Darboux Theorem, there is a number 8 > 0 such that b

U(P,f) <I f(x)dx + ~ / 2 b "

=I a f ( x )dx+ € 1 2 Y P with lPl < 6

Also, b b

L(P,f) > 1 f(x) dx - t / 2 =I f(x) dx - €12 b 5 B

i.e. - L(P,f) < - a 1 f(x) dx f el2 Y P with JPI < 6

Agding (2) and (3), we get U(P,f) - L(P,f) < E Y P with IPI < 6.

Next, we prove that condition is sufficient.

It is given that, for each number E> 0, there is a number 6 > 0 such that

U(P,f) - L(P,f) < E, V P with (PI < 6. Let P be a fixed partition with JPI < 6. Then

b b

Therefore, /f(x) a dx - /f(x) a dx 5 U(P,O - L(P,f) < E.

Since E is arbitrary, therefore the non-negative number

is less than every positive number. Hence it must be equal to zero that is1 f(x) dx b a

=I f(x) dx and consequently f is integrable over [a,b]. a -

Second Form : The necessary and sufficient condition for a bounded function f to be integrable over [a,b] is that to every number > 0, there corresponds a partition P of [a,b] such that

u (P,f) - L(P,f) < E .

The proof is left as an exercise.

EXERCISE 5 Establish the condition of integrability in the second form.

14.3 RIEMANN INTEGRABLE P'CTNCTIONS

Having derived the necessary and sufficient conditions for the integrability of a function, we can now decide whether a function is Riemann integrable without finding the upper'and lower integrals of the function. By using the sufficient part of

i n 1

the conditior!s, we test the integrability of the functions. In this sectlon we d~scuss f'unctions which are always integrable. We will show that a continuous functjqn is always Riemann integrable. The integrability is not affected even'when there are finites 11unlber of points of discontinuity or Lhe set of points of discontinui~y of the function has a finite number of limit points. It will also be shown that a monotonic function i s also always Riemann integrable.

We shall denote by R(a,b), the family of all Riemann integrable functions on [a,b]. First we discuss results pertaining to continuous functions in the form of the followillg theorems.

THEOREM 6: I f f : [a, bJ + R is a continuous function, then f is integrable over [a$], that is f ~R(a ,b) .

PROOF: Recall from unit 10 that i f f is a continuous function on [a,b] then f is bounded and is also uniformly continuous.

'To s11ow thaL f E R [a,b] you have to show that to each number E > 0, there is a partition P for which

U(P,f) - L(P,f) < €

L.eL t > 0 be given. Since f is uniformly continuous on [a,b], there is a number 6 > 0

E such that If(x)-f(y)(< - whenever Ix - yl <6. Let P be any partition of [a,b] with [PI <a.

b-a

We show that, for such a partition P, U (P, f) - L(P, f) < E.

where Axl = x, - xl-,, and MI = sup {f(x) Jx,-, 2 x S x,) = f(5,) (say), for same {, r [x ,-,, x,]. Such a E,! exists because a continuous function f attains its bounds on [q,-,, x,]. Similarly, m, = inf (f(x) Ixl-, I x 5 xl) = f(ql) (sqy), for some q l €[xl-,, x,]. Hence

MI - ml = f ( l l ) - f(ql) I lf(5,) - f(qJ < ~/b-a , for all i,

since - q , I < A xl < 6. Substituting in (4) we obtain

E = - (b-a) = E.

b-a

This completes the proof1 of the theorem.

Thus, every continuous function is Riemann integrqble,

BuL aa remarked earlier. even when there are dihcontinuitieb ot'the function, i t i s integrable. This is given in tho next two thcuterps which we state without proof,

THEOREM 7: 1-et the bounded function f: [a,b] -> Q have a finite number of discontinuities. Then f E R (a,b).

THEOREM 8: Let the secof points of discpntinuity of a, bQunded function f: [a,b] -> R has a finite number of limit points, Then f E R (a,b),

Tllc Rienialil~ Integration,

We illustrate these theorems with the help of examples.

EXAMPLE 6 : Show that the function f where f(x) = x2 is integrable in every interval

Ca,bl.

SQLUTION: You know that the function f(x) = x2 is continuous. Therefore it is integrable in every interval [a,b].

EXAMPLE 7: Show that the function f where f(x) = [x] is integrable in [0,2] where [x] denotes the greatest integer not greater than x.

I 0 i f O I x < I SOLUTION: [XI = 1 if 1 5 x < 2

2 i f x = 2

The points of discontinuity of f in [0,2] are 1 and 2 which are finite in number and so it is integrable in [0,2].

EXAMPLE 8: Show that the function F defined on the interval [0,1] by

1 1 2rx, when - < x <-- , where r is a positive integer

F(x) = ]+I r 0, elsewhere,

is Riemann integrable.

SOLUTION: The function I: is discontinuous at the points 0, I, , .... . The set of 2 ' 3

points of dlscantinuity has 0 as the only limit point. So, the limit points are finite in number and hence tho hnction F is integrable in [0,1], by Theorem 8.

You should now try the following exercise. - - -

' EXERCISE 6 Show that the function f where f(x) = x[x] is integrable in [0,2].

EXERCISE 7 Show that the functiori f defined in LO, 21 such that f(x) = 0, when

n X" - n+l 06 - (n = 1, 2, 3 ,.... ), and f(x) = 1 , elsewhere, is integrable.

n+l n

EXERCISE 8 Prove that the function f defined in [0, I ] by the condition that if r is a positive integer,

1 1 f(x) = (-I)'-' when 4 x I - r , and f(x) = 0, elsewhere, is integrable.

There is one more class af integrable functions and this cl~tsh is t l i :~t ol' monotonic function>, This wc prove i n the following theorem.

THEOREM 9: Every monotonic function i s integrable.

PROOF: We shall prove the theorem for thc case whcrc I': Ly.b] -> R is a monqtonicnlly incrcasipg function. The function ih bounded. l'(n) and [(b) bcing g.1.b. and 1.~1,b. Let t > 0 be given number, L,ct P bc o po5itiie intcpcr ~ L I C I I 1l:at

n > (b-a) u'(b) - f ( i~)] t

Dividc !he interval [a,b] into n equal suhinten nl, by lhc parlition 1' I-- (s,,, x l ..., s,,] 01'

The Riernrnn Integration - b-a - - c [f(x1) - f(xl-l)l

" 1=1

This proves that f is integrable. Discuss the case of monotonically decreasing function as an exercise.

-- -- EXERCISE 9 Show that a monotonically decreasing function is integrable. . Now we give example to illustrate the theorem.

1 EXAMPLE 9: Show that the function f defined by the condition f(x) = - 7"

f(0) = 0, is integrable in [O, 11

SOLUTION: Here we have f (0) = 0,

1 f(x)= 1 when - < x 5 1 2

I 1 1 f (x )= - when ( - ) < x 5 -

2 2 2

I Clearly f is monotonically increasing in [O, 11. Hence it is integrable.

1 1 1 Show that the functio~l f defined in [0,1], for integer a > 2, by f(x) = 7 , when- <x.:-

a' l

I 1 (r = 1,2,3, . and f(0) = 0, is integrable.

14.4 ALGEBRA OF INTEGRABLE FUNCTIONS

In unit 11, we discussed the algebra of the derivable functions. Likewise, we shall now study the algebra of the integrable functions. In the previous section, you have seen that there are integrable as well as non-integrable functions. In this section you will see that the set of all integrable functions on [a,b] i s closed under addition and multiplication by real numbers, and that the integral of a sum equals the sum of the integrals. You will also see that difference, product and quotient of two integrable functions is also integrable. All these results are given in the following theorems.

I i THEOREM 10: I f f ~ R ( a , b ) , and h is any real number, then h f ER (a,b) and

b b

fh f(x) dx = h /f(x) dx. B 0

PROOF: Let P = {x,, x ,,...., xn) be a partition of [a,bl. Let Mi and m; be the respective 1.u.b. and g.1.b. of the function f in [x ,-,, xl]. Then h M i and hm, are the respective 1.u.b. and g.1.b. of the functiorl h f in [xl-,, xl], if A 2 0, and h m, and h Mi are the respective 1.u.b. andg,l.b, of h f in [xi-,, xi], if h < 0.

When 1 2 0, then U(P,hI) = A Ml Axi = h 2 Ml Ax, = )iU(P,f). 1=1 1 ;I

6 li a / hf(x) dx = h f f(x) dx.

a

Integrability Similarly L(P,kf) = A L(P,O.

Similarly L(P,hf) = A U L(Pf . h b

=> L J Af(x) dx = A 1 f(x) dx B

Since f is integrable in [a,b], therefore 6 b b

/f(x) dx = /f(x) dx = /f(x) dx. a a a

6 b b

Hence /A f(x) dx = d /A f(x) dx = h a /f(x) dx,

whether h 2 0 or i 0.

b b

Hence h f E R [a,b] and h f(x) dx = A$ f(x) dx. n a

Now suppose that X = -1. In this case the theorem says that iff E R [a,b], then b b

(-0 E R [a,b] and a I [-f(x)] dx = - a $ f(x) dx.

THEOREM 11 : I f f E R [a,b], g E R [a,b],. then f + g E R [a,b] and Ir b

j (f+g) (XI dx =J ftx) dx + j g(x) dx. B a

PROOF: We first show that f+g € R [a,b]. Let E > 0 be a given number. Since f E R [a,b], g E R la,b], there exist partitions P and Q of [a,b] such that

U(P,f) - L(P7f) < €12 and U (Q,g) - L (Q,g) < €12 If T is a partition of [a,b] which refines both P and Q, then U(T,f) - L(T,f) < 4 2 [U(T,f) - L(T,f) 5 U(P,f) - L(P,fll. Similarly, 'J(T,g) - L(T,g) < €1 2 ( 5 ) i";iso note that, if Mi = sup {f(x) : xi-I 5 x 5 x,] and

Ni = SUP {g(x): xi-I 6 x 5 xi}

then, SUP (f(x)+ g(x): xi-I 5 x 5 xi) 5 M I + Ni. Using this, it readily follows that U(T, f+g) -( W , f ) + U(T,g) . for every partition T of [a,b]. Similarly YT,f+g) 2 L(TJ-14- L(T,g)

for every partition T of [a,bj. Thus U (T,f+g) - L (T;f+g) 5 [U(T,f) t U(T,g) - L [(T,f) + L(T,g)]

= [U(T,f) - L(T,f)] + [U(T,g) - L(T,g)] < 2 1 5 = E for 2 2 I

T occurring in (5). This shows that f + g E R(a,b) b b b

It remains to sl~ow that 1 [f(x) + g (x)] dx =I [f(x) dx +$ g(x) a .J rl

Now , - b b

, J (f + g) (XI d x = j (f + g) (x) dx s U(P, f + g) 5 U(P,Q + u ( P , ~ ) I

... (0) a

for any partition P of [a,b]. ,Given any t > 0 we can find a partition P of [a,b] 18 such that

I

h

b 6-5

U(P,f) </ f(x) (x) dx + E /2

, Substituting (7) in (6), we obtain b b b

I Since (8) holds for arbitrary r: > 0, we obtain I

b b b

i / (f + g) (x) dx 5s f(x) dx +/ g(xJ dx a a a

1 Replacing f and g by -f and -g in (9) we obtain b b b

/ (- f - g) (x) dx I/ {- f (x)} dx +/ {- g (x)} dx

b b b

/ (f + g) (x) d x 5 -/ f(x) dx -/ g(x) dx a n a

I This is equivalent to 1 b b b , / (f + g) (x) dx 2s f(x) dx +/ g(x) dx

a a n

1 combining (9) and (lo), we get '

1 which proves the theorem. ~ 1 THEOREM 12: I f f eR(n;b) and g ~ R ( a , b ) , then f - g ER(B, b) and

PROOF: Since g E R [a,b], therefore -g E R [a,b] and b b

J - rg (x)l d x = -/ g(x) dx 1 1 Now f E R [a,b] and -g E R [a,b] implies that f + (-g) E R [a,b] and I

1 iherefore,

t that is (f-g) E R [a,b] and b b b

I / (f - g) (x) d x = / f(x) dx -/ g (x) dx. n n or the product and quotient of two functions, we state the theorems without proof.

ITHEOREM 13: I f f sR(a,b) and g ~ R ( a , b ) , then Pg eR(a,b).

(THEOREM 14 : If IsR(a,b), g sR(a,b) and there exists a numbrr t > 0 such that fIg(x)( Z t .Vr ~ [ a , b ] , then flg sR(a,b). i 1 Now we give some examples. I

I EXAMPLE 10: Show that the functioll f, where f(x) = x + [x] is integrable is [0,2]. I 1 SOLUTION: The functioll F(x) = x, being continuous is integrable in [0,2] and the function G(x) = [x] is integrable as it has only two points namely, 1 and 2 as points of discontinuity. So their sum is., f(x) is integrable in [0,2],

EXAMPLE 11: Give an example of function f and g such that f -I- g is integrable but f and g are not integrable in [a,b].

SOLUTION : Let f and g be defined in [a,b] such that

0, when x is rational

I , when x is irrational,

Integrability lewhen x is rational 0 when x is irrational

f and g are not integrable but (f $ g) = 1 Y x E [a,b], being a constant function, is integrable.

EXERCISE 11 Give example of functions f and g such that f-g, fg, f / g are integrable but f and g may not be integrable over [a,b].

Example 11 and Exercise 11 show that converse of each of Theorems 11 to 14 may not be true.

14.5 COMPUTING AN INTEGRAL

So far, we have discussed several theorems for testing whether a given function is integrable on a closed interval [a,b]. For example, we can see that a function f(x) = x L Y x E [0,2] is continuous as well as monotonic on the given interval and hence it is integrable over [O,q. But ihis information does not give us a method for finding the value of the integral of this function. In practice, this is not so easy as we might think of. The reason is that there are some functions which are integrable by conditions of integrability but it is difficult to find th: values of their integrals. For example, suppose a function is given by f(x) = ex . This is continuous over every closed interval and hence it is integrable. But we cannot find its integral by our usual method of antiderivative since there is no fuhction for which eX2 is the derivative. If possible, try to find the antiderivative for this function!

In such situations, to find the integral of a given function, we use the basic definition of the integral to evaluate its integral. Indeed, the definition of integral as a limit of sum helps us in such situations.

In this section, we demonstrate this method by means of certain examples. We have h

found the integrall f(x) dx via the sums II(P,f) and L(P,f). The numbers Mi and mi a

which appear in these sums are not necessarily the values of f(x), i f f is not continuous. In fact, we shall now show that / f(x) d x can be considered as limit ofsums in which Mi and m, are replaced by values of f . This approach gives us a lot of latitudein

b

evaluatingl f(x) dx, as we shall see in several examples. L

Let f: [a,b] -> R be a bounded function. Let la = xo < X, < ...... xn b]

be a partition P of [a,b]. Let us choose points tl, .... t,, such that

xi-I 5 t, Xi (i = 1, ... n). Consider the sum

Notice that, instead of Mi in U(P,f) and mi in L(P,f), we have f(tJ in S(P,f). Since ti's are arbitrary points in [xi-,, xi], S(P,f) is not quite well-defined. However, this will not ' cause any trouble in case of integrable functions. I

S(P,f) is called Riemann Sum corresponding to the partition P. We say that lirn S(P,$) = A

IPI - 0

or S(P,f) -> A as 1 PI -> 0 if for every number c < 0 3 S > 0 such that

JS(P,f) -AJ < a for P with JPJ < 6.

We give a theorem which expresses the integral as the limit of S(P,r),

t b

THEOREM 15: I f lim S(P,f) exists, then f ~ R ( a , b ) and lim S(P,f) = /f(x) dx. IPI-tO IPI -t 0

0

PROOF: Let lim S(P,f) = A. Then, given a number E > 0, there exists a number 6 > 0 iPI-'O

such that ,

IS(P,f) -A1 < €14, for P will1 IPI < 6.-

i.e., A - ~ / 4 < S(P,f) < A + €14, fol. P with (PI < 6.

Let P = {x,, x,, ....., xn). Suppose the points t , , ..., t,, vary in the intervals [x,, x,], ..., [xn-,, xn], respectively. Then, the 1.u.b. of the numbers S(P,f) are given by

I

1 Silriilarly, g.1.b. S(P,f) = L(P,f). Then, from (I I), we get

A - €14 < L(P,f) 5 U(P,f) 5 A + €14

Therefore:

U(P,f) - L(P,f) 5 (A + €14) - (A - €14)

I = E /2 < E.

In other words, f E R(a,b). Thus

b h 1 Since L(P,f) < /f(x) dx 2 /f(x) dx < U(P.f), therefore a

I

I From (12) and (13), we gel j b I I A - €14 5 /f(x) dx 5 A + €14. i 0 , That is,

b b

Since E is arbitrary, therefore /f(x) dx - A = 0, that is, /f(x) dx = A = l i~n S(P,f). This 1 1 IPI-tO

i completes the proof of the theorem.

I To illustrate this theorem, we give two examples.

SOLUTION : Here, the fiinction fi [a,b] + R is the constant function f(x) = 1. Clearly, for any partition P = (xo, x,, ...., xn) of [a,b], we have

/ sm,o = (x, - x0)f(t1) + (x2- x l ) f(tJ + ....... + (xn - xn-,)f(tn)

t / Since S(P,f) = b - a, for all partitions, 4 1 dx = lim S(P,f) = b - a.

IPI -+ 0 !

The Riemann integration

b b2 - at EXAMPLE 13: Show t h a t b dx = 7.

Integrabil ity SOLUTION: The function f:[a,b] ;, R in this e,xample is the identity fbnctbn f(x) = x,

Let P = (a = x,, x,, ...., xn = b) be any partition of [a,b]. Then

StP,f) = (x, - x,)f(tl) + (3- x,) f(t,) + .... + (xn- xn-,) Ktn)' where t, €b0 , x,], t2 E[x,, x2], ,,.

tn E[x,-,, xn] are arbitrary. Let us choose

Then, S(P,f) = (x, - x,) - 2 2 2

I Here again, S(P,f) =- (b2 - a'), no matter what the partition P we may take, Hence

2

h b 1 /f(x) dx = / xdx = lirn S(P,f) =? (b2 - a').

IP I - rO

The converse of Theorem 15 is also true which we state without proof as the next theorem.

THEOREM 16: If a functiou f is Riemann integrable on a closed interval [a,b], then

b

lim S(P,f) exists and lirn S(P,f) = /f(x) dx. IPI+O lPl-+O a

One of the important application of Theorem 16 is in computing the sum of certain power series. For, let us consider a partition P of [a,b] having n sub-intervals, each of length h so that nh = b - a. Then P can be written as P = (a, a + h, a + 2h, ..., a + nh = b).

Lett, = a + ih, i = 1,2 ,...., n. Then n

S(P,f) = C f(ti) Axi = h[f(a + h) 4 f(a + 2h) + ... + f(a + nh)]. i= l

When liln S(P,f) exists, then IPI+O

h

lirn h [f(a + h) + f(a + 2h) + ... + f(a + nh)] = /f(x) dx. 6 Z0" a

In the above formulae, we can change the limits of integration from a, b to 0, a, where b- a

a E N. For, by changing h to -- , it is easy to deduce from above formula that an

@ - a) liml " I: *[a + +)+I = j q x ) dx. a n+r, n (14; 1

I

b

But, /f(x) dx = 7 l

a

I Therefore, from (14), we get I

1 I , lim - ' f [ a +- (b"a'"'~]=6jF[a+Ox]dx, n+m n ,=I a n a

(If)

I

.L i

In (15), put a = 0, b = a. We get the following result: I f f is integrable in [O,cr ] , then

Tllc Ricmann Integration

This gives us the following method for finding the limit of sum of n terms of a series:

1. Write the general rth term of the series.

1 r 1 r 11. Express it as - f(-), the product of - and a function of - . n n n n

r 1 111. Change- to x a n d n to dx and integrate between the limits 0 and a. The value of n

the resulting integral gives the limit of the sum of n terms of the series.

Since each term of a convergent series tends to 0, the addition or deletion of a finite number of terms of.the series does not affect the value of the limit. Similarly, you can verify that

3n 1 3

lim . X [ 7 ) (+I ] = /+(x) dx, and so on. n+= ~1

As an illustration of these results, consider the following examples.

EXAMPLE 14: Find the limit, when n tends to infinity, of the series

n I n l 1 SOLUTION : General (rth) term of the series is gl

= ; (-). rill Hence, !,ma C - ( - ) = /-& dx = log 2. r=l n 1+- n

EXAMPLE 15: Find the limit, when n tends to infinity, of the series

1 - 1 + - 1 1

47 &Ti + K F + ..... + Gqizy- I

n 1

1 SOLUTION : Here the rth term = C r=l J n z - ( r - ~ ) ~

1 Since it contains (r-1), we consider its (r+l)th term i.e., /

" 1 the term Z - - n 1 1 - x -

r W r=O n 2- n

Therefore, lim C 1 =I - dx. because yi 1 = 0.

n-r=lnJ- 0 4 7 .

n The value of this integral, on the r.h.s. of last equality, is - 2 '

3n nl EXAMPLE 16: Find lim try.

n-)x

I: We have

Integrrbilily Since the number of terms in the summation is 3n, the resulting definite integral will have the limits from 0 to 3.

Therefore, lim - (3 '

This integral you can evaluate easily.

Now try tlie following exercises yourself.

EXERCISE 12 Find the limit, when n tends 10 infinity, of the series

EXERCISE 13 Find the limit, when 11 tends to infinity, of the series

114.6 SUMMARY ,

In this unit, you have been introduced to the concept of integration without bringing in the idea of differentiation. In section 14.2, upper and lower sums and integrals of a bounded function f over closed interval [a,b] have been defined. You have seen that upper and lower Riemann integrals of a bounded function always exist. Only when the upper and lower Riemann integrals are equal, the function f is said to be Riemann integrable or simply integrable over [a,b] and we write it as f E R [a,b] and the value

h

of the integral of f over [a,b] is denoted by f(x) dx. The definition of Riemann a

integral is given in this section. Also in this section, it has been shown th.at in passing from a partition PI to a finer partition 'P2, the upper sum does not increase and the lower sum does not decrease. Further you have seen that the lower integrable of a function is less than or equal to t h e upper integral. Further condition of integrability has been derived with tlie help of which the integrability of a function can be decided without finding the upper and lower integrals. Using the condition of integrability, it has been shown in section 14.3 that a function f is integrable on [a,b] if it is continuous or it has a finite number of points of discontinuities or the set of points of discontinuities have finite number of limit points. Also in this section you have seen that a monotonic function is integrable. As in the case of continuous and derivable functions, the sum, difference, product and quotient of integrable functions is integrable. This has been discussed in section 14,4. Finally in section 14.5, Riemann sum S(P,f) of a function f for a partition P has been defined and you have been shown

I, I that lirn S(P,fi exists if and only i f f E R [a,b] and j f(x) dx = lirn S(P,f). Using this

I P -- 0 a IPI - 0

idea a number of problems can be solved.

E 1) If P = {XO, x l .... x , ) be any partition of [a,b] in [x, I , x,], then M, = I, 1171 = - 1 Y. = I , 2, .... n

and Mi, in, be 1.u.b. and g.1.b. o f f

b

I! (P,f) = b-a and L(P,f) = aLb. heref for el f(x) dx = b-a and I

/b f(x) d x = a-b. a - -

b h

E 2) In Exercise I , you see that] f(x) d x # / f(x) dx and therefore f is not il a

integrable in [a,b].

The Riemnnn Integration

77- 6 ~ ( p ~ , g = - ( 1 -I- - 1, SO L(PI , f) 5 Id(!?., 0 and LJ(Pz,0 5 U(P1, f)

24 4 E 4) As f is bounded, there exists a positive number k such that

b

(f(x)I 5 k y x E [a,b]. AS/ f(x) dx is the supremum of the set of' lower sums, B

for each E > 0, there exists a partition PI of [a,b] such that

b

L(P1,f) >l f(x) dx - e l2 (16) a

Taking P, = {x,, x,, ...., xp} and a positive number 6 defined by 2 k(p-1)6= €12 and proceeding as in part (i) of Darboux Theorem with P, = PI u P, and (PI < 6. Then L(P,f) 2 L(P,.f) -2 (p-l)k6 (Using Theorem 3).

L L(P,f) - 2 (p-1) k 6 h

>/ f(x) dx - €12 - €12 (Using (16) 5 I>

=/ f(x) dx - E Y P with IPJ < 6. 0 -

E 5) Condition is necessary: Proceeding as in First Form, you will get U(P,f) - L (P,f) < E,V P with I PI < 6. Now fix a partition P having I PI < 6. So far this partition P,

U(P,f) - L(P,f) < €.

Condition is sufficient: For E > 0, 3 a partition P of [a,bJ sucli that U(P,f) - L(P,f) < E. Then proceeding as in first form, you will get that f is integrable in [a,b,].

E h) O i f 0 1 x < 1 f(x) = x i f l l x < 2

4 i f x = 2

This function has 1,2, as the only points of discontinuity and so it is integrable in [0,2].

E 7) The function f has the following p'oints of discbntinuity:

1 2 2 3 3 4 1, -, -, - , - , -, -, ..... and this set of points of discontinuity has 1 as the only

" 2 1 3 2 4 3

limit point and so the function f is integrable by Theorem 8. 2 5

In tegrabi l i ty 1 1 1 ' E 8) The function f has the following points of discontinuity: 0, 1, - , - , - ..,.....

This set h a s 0 as the only limit point and hence f is integrable on [O,l].

E9) In this case f(a) is the 1.u.b. and f(b) is the g.1.b. o f f . If E > 0 is any number,

you have to choose a positive integer n > @-a) [f(a)-f(b)l . Proceed E

as in Theorem 9 by taking Mi = f(xi-,) and mi = f(x,).

1 1 I 1 E 10) Here f(0) = 0, f(x) = 1, when, x 5 1, f(x) =- a , when - a2 < x I - a , ...., and

SO on.

Then, f is monotonically increasing in [0,1] and, so, by Theorem 9, it is integrable in [0,1].

E 11) Consider the functions f and g, defined in [a,b] as follows:

1, when x is rational f(x) = g (x) =

- 1, when x is irrational

Then both f and g are not integrable (Proved in Ex. I). Also, qx) - g(x) = 0, tJ x ~ [ a , b ] and so it is a constant function and hence integrable in [a,b]. Note that

f(x)g(x) = I , V x E [a,b], and also

Thus, both fg and f/g are constant functions and, so, they are also integrable, -

n Jn n E 12) Here (r+l )th term

1

So, the required limit = 1

dx, which is easy to evaluate.

n 1 n 1 I E 13) Here (r+ 1)th tenn C - = C - -

r=o n + r ~0 n I l + k I s = 1 Since the number of t e n s in the given series is 2n, the required limit is [- dx,

, 0 I + x which you can easily evaluate.

. UNIT 15 INTEGRABILITY AND DIFFERENTIABILITY -

Structure 15.1 Introduction

Objectives

15.2 Properties of Riemann Integral 15.3 Fundamental Theorem of Calculus 15.4 Mean Value Theorems

First Mean Value Theorem Second Mcan Value Theorcm

15.5 Summary 15.6 Answers/ Hints/Solutions

15.1 INTRODUCTION

In unit 14, the notion of the integral of a function was developed as a limit of sums of the series. Nowhere, the concept of differentiation was used. Apparently, you may conclude as if there is no relation between the integration and differentiation. But this is not true in all cases. No doubt, the notion of integral as a limit of sums allows us to compute the integrals in some simple cases which you have seen in section 14.5.

' Nevertheless, it is not convenient for a large number of cases. We do require the process of differentiation to compute the integrals for a certain class of functions. This

' means there must be some relationship between differentiability and integrability of a function. What is that relationship between the two nqtions? We shall bring forth this intimate connection between the notions of differentiation and integration for a certain class of functions. In the case of continuous functions, this relationship is expressed in the form of an important theorem called the Fundamental Theorem of Calculus, which we discuss in section 15.3. Prior to this, we need a few important properties of the definite integral which you have studied in your previous course on Calculus. We &all review these properties in section 15.2. In section 15.3, we shall study two additional theorems of integrability which use the process of differentiation. These theorems are known as theMean-value Theorems of integrability analogous to the mean-value theorems of differentiability.

Objectives After the study of this unit you should, therefore, be able to -+ know some important properties of the Riemann Integral -- establish the inverse relationship between integration and differentiation -- apply Fundamental theorems of calculus to evaluate large number of integrals - learn the mean value theorems of integrability and their applications.

15.2 PROPERTIES OF RIEMANN INTEGRAL

In section 14.4, you were introduced to some methods which enabled you to associate with each integrable function f defined on [a,b], a unique real number called the

. b

integrall f(x) d x in the sense of Riemann. In section 14.5, a method of computing B

this integral as a limit of a sum was explained. All this leads us to consider some nice properties which are presented as follows:

PROPERTY I: I f f and g are integrable on [a,b] and if

f(x) 5 g(x) V x E [a&], then

j f ( ~ ) dx 5 j g(x) dx 27

, UNIT 15 INTEGRABILITY AND DIFFERENTIABILITY

Structure 15.1 Introduction

Objectives

15.2 Properties of Rie~nann Integral 15.3 Fundamental Theorem of Calculus 15.4 Mean Value Theorems

First Mean Value Theorem Second Mean Value Theorem

15.5 Summary 15.6 Answers/Hints/ Solutions

15.1 INTRODUCTION

In unit 14, the notion of the integral of a function was developed as a limit of sums of the series. Nowhere, the concept of differentiation was used. Apparently, you may collclude as if there is n o relation between the integration and differentiation. But this is not true in all cases. N o doubt, the notion of integral as a limit of sums allows us to compute the integrals in some simple cases which you have seen in section 14.5.

' Nevertheless, it is not collvenient for a large number of cases. We do require the , process of differentiation to compute the integrals for a certain class of functions. This ' means there must be some relationship between differentiability and integrability of a f~~nct ion , What is that relationship between the two notions? We shall bring forth this intimate connection between the notions of differentiation and integration for a certain class of functions. In the case of continuous functions, this relationship is expressed in the form of an important theorem called the Fundamental Theorem of Calculus, which we discuss in section 15.3. Prior to this, we need a few important properties of tke definite integral which you have studied in your previous course on Calculus. We shall review these properties in section 15.2. In section 15.3, we shall study two additional theorems of integrability which use the process of differentiation. These theorems are known as theMean-value Theorems of integrability analogous to the mean-value theorcms of differentiability.

Objectives After the study of this unit you should, therefore, be able to -- know some important properties of the Riemann Integral -- establish the inverse relationship between integration and differentiation + apply Fundamental theorems of calculus t o evaluate large number of integrals -- learn the mean value theorems of integrability and their applications.

15.2 PROPERTIES OF RIEMANN INTEGRAL

In section 14.4, you were introduced to some methods which enabled you to associate with each integrable function f defined on [a,b], a unique real number called the

b

integralJ f(x) dx in the sense of Riemann. In section 14.5, a method of computing a

this integral as a limit of a sum was explained, All this leads us to consider some nice properties which are presented as follows:

PROPERTY I: I f f and g are integrable on [a,b] and if

f(x) 5 g(x) V x E [a$], then

b f f(x) d x 5 g(x) dx 27 I

lntegra bil i ty PROOF: Define a function h: [a,b] -+ R as h - g - f .

Since f and g are integrable on [a,h], therefore, the difference h is also integrable on [a,b]. Since

f(x) 5 g(x) => g(x) - f(x) 2- 0, therefore h(x) 2 O for all x E [a,b].

Consequently and m, be the

If P = {XU, xi, ..... x,) be any partition of [a,b] inf. of I1 in [x, 1 , x,], then

m , 2 I ) Y i = 1 , 2 ,.... n

Thus for every pr:tition P, the lower sum L(P,h) 2 0. In other words, Sup. (1 (P,h): P is a partition of [a,b]) 2 0

0 I- h

1 f(x) dx LO a

Since h is integrable In [a,b], therefoie b b b

1 h(x) dx = 1 h(x) dx = / h(x) dx. d 4 8

Thus b

1 h(x) d x 2 O a

0 r b

1 (g-f) (x) dx 1 0 a

b b

which proves the properly

PROPERTY 11: If f,is integrable on [a,b] then If1 is also integrable on ' b b

[a,b] and (1 f(x) dxl 51 (f(x)l dx B 8

PROOF : The inequality follows at once from Property I provided it is known that I fl is integrable on [a,b]. Indeed, you know that - I f 1 I f 5 Ifl. Therefore,

b b b

-1 il Jf(x)J I dx 51 u f(x)dx 51 a If(x)J dx

which proves the required result. Thus, i\ remains to show that If( is integrable.

Let e > 0 be any number. There exists a partition P of [a,b] such that

U(P,f) - L(P,f) l c

Let P = {XO,.XI, xz, . ..., x"). Let M'i and mi denote the supremum and infimum of If/ sad Mi and mi denote the supremum and infimum off in [x,-1, x,). You can easily check that

n n This implies that .Z, (MI - mi) A xi 2 aC (M(- m;) dxi,

1= 1 I= 1

i.e., U(P,(f)-L(P,IfJ) 5 U(P,f)-L(P,f) < E .

This shows that /fl is integrable on [arb]. 2 8 1

t

Note that the inequality established in Property 11 may be thought of as a Integrability and Differentiability I , generalization of the well-known triangle inequality I

( a + b l I l a l + lbl discussed in Unit 3. In other words. the absolute value of the limit of a sum never exceeds the limit of the sum of the absolute values.

I

b

You know that in the integral1 f(x) dx, the lower limit a is less than the upper limit I

a

b. It is not always necessary. In fact the next property deals with the integral in which the lower limit a may be less than or equal to or greater than the upper limit b.

For that, we have the following definition: b

DEFINITION 1: Let f be integrable on [a,b], that is, 1 f(x) dx exists when b > a. Then

a

i

a

= -1 f(x) dx, if a > b. h

I Now have the following property. I '

PROPERTY 111: If a function f is integrable in [a,b]and (f(x)I 5 k Y x E [a,b], then h

PROOF: There arc only three possibilities namely either a < b or a > b o r a = b. We discuss the cases as follows:

Case (i) a < b: Since If(x)l I k Y x E [a,b], therefore - k < f ( x ) S k Y x E [ a , b ]

h b b

=>1 -kdx 51 f(x) d x I k dx (why?) a a

b

=> -k (b-a) I 1 f(x) dx I k(b-a) a

I which completes the proof of the theorem. I I Case (ii) a > b: I I

In this case, interchanging a and b in the Case (i), you will get a

( 1 f(x) d x 1 5 k (a-b) h

b

i.e. 1 -1 f(x) dxl 5 k (a-b) r b

i.e. ( 1 f(x) d x ( 5 k (a-b) = k 1 b-al. u

J Case (iii) a = b :

In this case also, the result holds,

b

since 1 f(x) dx = 0 for a = b and kl bya l = 0 for a = b. a

Let [a,b] be af ixed interval. Let R [a,b] denote the set of all Riemann integrable functions on this interval. W e have shown in Unit 14 that'if f,g C R [a,b], then f + g f.g and Af for A E R belong to R [a,b]. Combining these ;with Property 11, we can say that the set R [a,b] of Riemann integrable functions is closed under addition, multiplication, scalar multiplication and the formatian of the absolute value.

\ I

If we consider the integral as a function Int: R[a,b] -+ R defined by b

Int (f) =1 f(x) d x n

with domain R [a,b] and range contained in R, then this function has the following properties:

lnt (f+g) = Int (f) + Int (g), lnt (Xf) = Int ( f ) '

In other words, the function lnt preserves 'Vector sums' and the scalar products. In the language of Linear Algebra, the function lnt acts as a linear transformation. ~ h i i function also has an additional interesting property such as

a lnt ( f ) 5 lnt(g) whenever

f l g . We state yet another interesting property (without proof) which shows that the' '~ iemann Integral is additive on an interval.

PROPERTY IV: I f f is integrable on [a,b] and c E [a,b], then f is integrable on [a,c] and [c,b] and conversely. Further in either case

b C b

/ f(x) dx = / f(x) dx -I-/ f(x) dx. a 1 c

According to this property, if we split the interval over which we are integrating into two ?arts, the value of the integral over the whole will be the sum of the two integrals over the subintervals. This amounts to dividing the region whose area must be found into two separate parts while the to'tal area is the sum of the areas of the separate portions.

h

We now state a few more properties of the definite integral( f(x) dx which you ought d

to h a v studied in the course on calculus (MTE-01).

(i) ( f (x )dx=( f (a -x )dx . o

2n B rl

(ii) f(x) dx = / f(x) dx. $1 f(2a-x) dx. . 0 0 0

a

2/ f(x) dx i f f is an even function (iii) j f(x) dx =

- a > 0 i f f is an odd function. lld d

(iv) / f(x) dx = n( f(x) dx i f f is periodic with period 'a' and n is a positive integer 0 0

provided the integrals exist.

15.3 FUNDAMENTlfL THEOREM OF .CALCULUS

In section 15.1, we raised a question, "What is the relationship between the two notions of differentiation and integration? Now we shall try to find an answer to this questibn. 1n fact, we shall show that differentiation and Integration are intimately related in the sense that they are inverse operations of each other.

Let us begin by asking ourselves th.e following question : "when is a function f: [a,b] r> R, the derivative of some function I?: [a,b] -> R?" For example consider the function f: [-I, 11 --> R defined by

This function is not the derivative of any function F: [-I, 11 -> R. Indeed i f f is the derivative of a function F: [-1, 11 -> R then (Refer to Unit 12 for the intermediate ' value property of derivatives) f must have the intermediate value property. But clearly, the function f given above does not have the intermediate value property. Heuce f cannot be the derivative of any function F: 1-1, 11 -> R.

However i f f : [-1,1] -> R is continuous, then f is the derivative of a function F: [-I, 11 _> R. This leads us to the following general theorem:

THEOREM 1 : Let f be integrable on [a,b]. Define a function P on [a,b] as

I

i

Y

F(X) = /f(t) dt, V x ~ [ a , b ] . a

Then F is continuous on [a,bl. Furthermore, if f is continuous at a point x, of [a,b], then F is differentiable at x, and F1(x,) = f(x,).

PROOF: Since f is integrable on [a,b], it is bounded. In other words, there exists a positive number M such that If(x)l I M, V x ~ [ a , b ] .

e Let E > 0 be any number. Choose x,y E [a,b], x < y, such that (x - yl < - . Then

M

integrability a n d ~

D i f f e r c n t i a t ~ i l ~ t y

Y

5/ Mdt = M(y-x) < X

Similarly you can discuss the case when y < x. This shows that F is continuous on I

[a,b]. In fact this proves the uniform cantinuity of F. I

Now, suppose f is continuous at a point xo of [a,b] i

We can choose some suitable h # 0 such that X ~ I f h E [a,b]. Then

Thus I

Now

F(x, +h) - F(x,) I h - f(xdl = ~ $ r f ( t ) d t . - l h x0 & , ) d t l

1 xo*s

= - I i [f(t) - f(x,)ldt 1. lhl .. I

Since f is continuous at x,, given a number E > 0, 3 a number 6 > 0 such that

If(x) - f(x,)l < E 12, whenever Ix-x,l < 6 and x E [a,b]. So, if Ihl 6, then If(t) - f(x,)J < E 12,

' for t e Ex,, x, +h], and consequently I

x +h s .

I I

/ [f(q - f(xo)]dt 5 - 1 hl. heref fore I ,

F(xO+h)-F(xO) - f'(x,) I 1 j

i

Therefore, lim F(x, + h) - F(x )

h f(x0), i.e., F f ( x 0 ) = f(xo).

h-0

3 1

Which shows that F is differentiable at xo and F'(xo) = f(xo) From Theorem 1 , you can easily deduce the following theorem:

THEOREM 2: Let f: [a,b] -- R be a continuous function. Let F: [a,b] -- R be a function defined by

Y

F(x) =J a f(t) dt, x E [a,b].

Then F'(x) = f(x), a 5 x 5 b.

This is Lhe first result which links the concepts of integral and derivative. I t says tha.t, i f f is continuous on [a,b] then there is a function F on [a,b] such that F' (x) = f(x), ff x E [a,b].

You have seen that iff: [a,b] -> R is continuous, then there is a function F: [a,b] -. R such that F' (x) = f(x) on [a,b]. Is such a function F unique? Clearly the answer is 'no'. For, if you add a constant to the function F, the derivative is not altered. In

X

other words, if G(x) = c -b / f(t) dt for a 5 X 5 b then also G' (x) = f(x) on [a,b]. 1

Such a function F or G is called primitive of f . We have the formal definition as follows:

DEFINITION 2: PRIMITIVE OF A FUNCTION Iff and F are functions defined on [a,b] such that Ff(x) = f(x) for x E [a,b] then F is called a 'primitive' or an 'antiderivative' o f f on [a,b].

Thus from Theorem 1, you can see that every continuous function on [a,b] has a primitive, Also there are infinitely many primitives, in the sense that adding a constant to a primitive gives another primitive.

"Is it true that any two primitives differ by a constant?"

The answer to this question is yes. Indeed if F and ti are two primitives o f f in [a,b], then F'(x) = G'(x) = f(x) V x E [a,b] and therefore [F(x) - G(x)lf = 0. Thus F(x)-G(x) 5 k (constant), for x E [a,b].

Isaac Barrow Let us consider an example.

EXAMPLE 1 What is the primitive of f(x) = log x in [1,2]

d SO1,UTION: Since - (x log x -x) = log x y x E[1,2], therefore F (x) = x log x-x dx

is a primitive o f f in [ I ,2]. Also G(x) = x lag x -x -I- k, k being a constant, is a primitive o f f . Try the following exercise yourself.

EXERCISE 1 Find the primitive of the function f defined in [0,2] by

, According to this theorem, differehtiation and integration ark inverse operations.

We now discuss a theorem which establishes the required relationship between differentiation and integration. This is called the Fundamental Theroem of Calculus.

I It states that the integral of the derivative of a function is given by the function itself. ' !

The Fundamental Theorem of Calculus was given by an English mathematician Isaac I

Barrow [1630-16771, the teacher of great Isaac Newton.

, THEOREM 3: (FUNDAMENTAL THEOREM OF CALCULUS) I /

I b

I f f is integrable on [a,b] and F is a primitive o f f on [a,b], then f(x) dx = F(b)- 32 F(a). a ' 1

I I

A

1

I b I 1

1 PROOF: Since f E R [a,b], therefore lim S(P,f) = J f(x) dx Integrabil i ty and IPI - 0 a , Dif ferent iabi l i ty

, where P = {xo, X I , x2 ,...., x,] is a partition of [a,b]. The Riemann sum S(P,f) is given by

!=I I = I 1

/ Since F is the primitive o f f o n [a,b], therefore F' (x) 5 f(x), x E [a,b]. I I f l , Hence S(P,f) = C F' (6) (x1-xl-I). We choose the points t, as follows: I I= I

1 By Lagrange's Mean Value theorem of Differentiability (Unit 12), there is a point t, in , ]XI-,, x, [such that I

I F(x,) - F(x,-I) = F' (t,) (x, - x,-I)

n Therefore, S(P,f) =.C [F(xi)- F (xi-,>] = F(x,) - F(x0) = F(b) - F(a).

I= 1 I.

Take the limit as IPJ -+ 0. Then /f(x) dx = F(b) - F(a). This completes tlSe proof. a

Alternatively, the Fundamental Theorem of Calculus is also interpreted by stating that the derivative of the integral of a crlr,tinuous function is the function itself.

b

If the derivative f of a function f is integrable on [a, b], then (fl(x) dx = f(b) - f(a)

Applying this theorem, we.can find the integral of various functions very easily. Consider the following example:

I

EXAMPLE 2: Show that j sin x dx = 1 - cos t. 0

SOLUTION: Since g(x) = - cos x is the primitive of f(x) = sin x in the interval [O,t], therefore

( 1

j Sin x dx = g(t) - g(o) = I - cost. 0

Try the following exercises.

EXERCISE 2 . 2

~ i n d j f(x) dx whcre f is the function given in Exercise 1. 0

EXERCISE 3 ! h

\ ~ v a l u a t e j xn dx where n is,a positive integer. I

b

We have, thus, reduced the problem of evaluatingl f(x) d x to that of finding 0 b

primitive o f f on [a,b]. Once the primitive is known, the value o f j f(x) dx is'easily 1

given by the Futldamental Theorem of Calculus.

You may note that ally suitable primitive will serve the purpose because when thc primitive is known, then the process described by the Fundamental Theorem is much simpler than other methods. However, it is just possible that the primitive may not exist hile you may keep on trying to find it. It is, therefore, essential to formulate some. lnditions which can ensure the existence of a primitive. Thus now the next stcp is to find the conditions on the integrand (function to be integrated) which will ensure the existence of a primitive. One such condition is provided by the theorem?.

According to thebrtm-2 i f f is continuous in [a,b], then the function F given by 1

I I I

33 jc I I

! I

6

Intcgrahility 7

F (x) =/' f(t) dt, x F [a,b] is differentiable in [a,b] and Ff(x) = f(x) Y x € [a,b] d

i.e. F is the primitive of f in [a,b]

The following observations are obvious from the theorems 1 and 2.

(i) I f f is integrable on [a,b], then there is a function F which is associated with f through the process of integration and the domain of F is the same as the interval [a,b] over which f is intEgrated.

(ii) F is continuous. In other words, the process of integration generates continuous function.

(iii) If the function f is continuous on [a,b] Lhen F is differentiable on [a,b]. Thus, tile process of integration generates differentiable functions.

(iv) At any point of continuity of f, we will have f(c) = f(c) for c E [a,b[. This means that i f f is continuous on the whole of [a,b], then F will be a member of the family of primitives o f f on [a,b].

In the case of continuous functions, this leads us to the notion

for the family of primitives o f f . Such an integral, as you.know, is called an Indefinite integral of f . I t does not simply denote one function, but it denotes a family of functions. Thus, a member of the indefinite integral o f f will always l$e an anti- derivative for f.

Theorem 3 gives US a condition on the function to be integrated which ensures Ihc existence of a primitive. But how to obtain the primitives once this condition is satisfied. In the next section, we look for the two most important techniclucs for finding the primitives. Before we do so, we need to study two important mean-values theorems of integrability.

15.4 MEAN VALUE THEOREMS '

In Unit 12, we discussed some mean-value theorems concerning the differentiability ol' a Cunction. Quite analogous, we have'two mean value theorems of integrability which we intend to discuss in this section. You are quitt familiar with the two wcll-known techniques of integration narnely the integration by parts and idegration by substitution which you must have studied in your study of Calculus. How thcsc methods were devised? We shall discuss this question also in this section.

THEOREM 4: FIRST MEAN VALUE THEOREM Let f:[a,b] -+ R be a continuous function. Then there exists c €[a&] such that

7 f(x) dx = (b-a) f(c).

PROOF : We know that h

m(b-a) 5 /f(x) dx 5 M (b-a), thus a

[f(x) dx m~ I M, where

(b-a)

m =glb {f(x): x ~ [ a , b ] ) , and

M I lub {f(x) : x ~ [a ,b ] ) .

Since f is continuous in [a,bl, it attains its bounds and it also attains every value bemeen the bounds. Consequently, there is a point c €[a$] such that

b ,

/ f(x) dx = f(c j (b-a), a

which, equivalently, can be written as 1

f(c) = - / f(x) dx. b-a a

This theorem is usually referred to as the Mean Value theorem for integrals. The geometrical interpretation of the theorem is that for a non-negative continuous hnction f, the area between f, the lines x = a, x = b and the x-axis can be taken as the area of a rectangle having one side of length (b-a) and the other f(c) for some c ~ [ a , b ] as shown in the Fig. 1.

.--

Y

I Fig. 1

We now discuss the generalised form of the first mean value theorem.

THEOREM 5: THE GENERAZIlSED F1[IZST MEAN VALUE THEOREM

Let f and g be any two functions integrable in [a,b]. Suppose g(x)'keeps the same sign for all x E [a,b]. Then there exists a number ,u lying between the bounds o f f such that

$PROOF: Let us assume that g(x) is positive over [a,b]. Since f and g are both integrable in [a,b], therefore both are bounded. Suppose that rn and M are the g.1.b. and 1.u.b. o f f in [a,b]. Then I

Consequently,

mg(x) 5 qx) g(x) s Mg(x), V x ~[a,bl . Therefore,

m )g(x) dx 5)4x) g(x) dx 5 M /g(x) dx.

It then follows that there is a number p e[m,M] such that

Corollary: Let f, g be continuous functions on [a,b] and let g(x) 2 0 on [a,b]. Then, there exists a c E [a,b] such that

Integrability and Differentiability

PROOF: since f is continuous on [a,b], so, there exists a point c s [a,b] such that b

(4x) g(x) dx = f(c) fg(x) dx, ;here p = qc) is as in Theorem 5.

Integrabil i ty We use the first Fundamental Theorem of Calculus for integration by parts. We discuss it in the form of the following theorem:

THEOREM 6: I f f and g are differentiable functions Qn [a,b] such that the derivatives f'and g'are both integrable on [a,b], then

PROOF: Since f and g are given to be differentiable on [a,b], therefore both f and g are continuous on [a,b]. Consequently both f and g are Riemann integrable on [a,b]. Hence both fg' as well as f' g are integrable.

fg' + f'g = (fg)'. Therefore (fg)' is also integrable and consequently, we have

'

b b b

J (fg)' = / fg' +/ f'g. d a d

By FundamentalTheorem of Calculus, we can write

I

Hence, we have h h

/ fg' = f(b) g(b) - f(a) g(a) -1 f'g.

h b

/ f(x) g' (x) d x = [ f(x) g(x) 1: -/ a f'(x) g(x) dx.

This theorem is a formula for writing the integral of the product of two functions. What we need to know is that the primitive of one of the two functions should be expressible in a simple form and that the derivative of the other should also be simple so that the product of these two is easily integrable. You may note here that the source of the theorem is the well-known product rule for differentiation.

'The Fundamental Theorem of Calculus gives yet another useful technique of integration. This is known as method by Substitution also named as the change of variable method. In fact this is the reverse of the well-known chain Rule for differcntiation. In this method, we use the law of composition of functions which you have studied in Unit 1 . In other words, we compose the given function f with another function g so that the co~nposite fog admits an easy integral. We deduce this method

L in the form of the following theorem: \ I THEOREM 7: Let f be a fi~nction defined and continuous on the range of a function g. If

g ' is continuously diffrentiable on Ic,dj, then d

dx =/(fog) (x) gt(x) dx,

where a = g(c) and b = g(d).

b

PROOF : Lct F(x) = /f(t) dt be a primitive of the function 1: a

Note that the function F is defined on the range of g.

I Since f is continuous, therefore, by Theorem 2, it follows that F is differentiable and Fr(t) = f(t), for ally t. Denote G(x) = (Fog) (x).

Then, clearly G is defined on [c,d] and it is differentiable there because both F and g are so. By the Chain Rule of differentiation, it follows that

I G'(x) - (Fog)' Cx) g'(x) = (fog) (x) g1(x),

Also fig is continuous since both f and g are continuous. Therefore, fog is integrable. Since g' is integrable, therefore (fog) g' is also integrable. Hence

d > .

d

I /(fog)(x) gf(x) dx = /G'(x) dx

C

3 6 f

= G(d) - G(c) = F(g (dl) - F(g (c)) = F(b) - F(a)

b

=I f(x) dx. 1

Integrabilitj and Differe~~tisbilitl

(Why'?)

you hu\;c seen that the proof of the theorem is based on tlie Chain Rule for d;l.l.cl.cntiaiion. In fact, this theorem is sometimes treated as a Chain Rule for Illtegration cxcept that i t is used exactly the opposite way from the Chain Rule for dil'fcrentiation. The Chain Rule for differentiation tells us how to differentiate a conipobitc function while the Chain Rule for Integration or.tlie change of variable method tells us how to simplify an integral by rewriting it as a composite function. Thus, we are using the equalities in the opposite directions.

We conclude this section by a theorem (without Proof) known as the Second Mcan Valuc Thoorem for Integrals. Only tlie outlines of the proof are given.

THEOREM 8: SECOND MEAN VALUE THEOREM Let f alid g be any two furlctions integrable in la,l)j and g be monotonic in la,bJ, then there exists c E [a,bj such that

PROOF : Tlle proof is based on the following result knowrl as Bonnet's Mean Value Theorem, given by a French mathematician 0. Bonnet [ I 8 19-1 8921.

Let f and g be integrable fu~ictiotis in [a$]. If 4 is any monotonically decreasing function and positive in [a,b], tlicn lliere exists a point c E la,b] such that

Lct g bc monotonically clecreasing so that 4 where 4(x) = g(x) - g(b), is non-negative and monotonically decreasing in [a,b]. Then there exists a humber c E [a,b] such that

i.e. h c b

J f(x) g(x) dx = g(a)J f(x) dx + g(b)J f(x) dx. I a C

Now let g be monotonically increasing so thit -g is mono~onically decreasing. Then thcre exists a nunlbcr c E [a,b] such that

h c b

I' 0 ) [-g(x)l dx = -g(a)I' f(x) dx - g(b)I' f(x) dx

This completes the proof of the theorem.

There are several applications of the Second Mean Value Theorem. It is sometimes used to develop the trigonorlletric functions and their inverses which you may find in higher Mathematics. Here, we consider a few examples concerning the verification and application of the two Mean. Value theorems.

EXAMPLE 3: Verify the two Mean Value Theorems for the functions

f(x) = x, g(x) = ex in the interval [-I, 11.

SOLUTION : VERIFICATION OF FIRST MEAN VALUE THEOREM:

Since f and g hire continuous in [-1, I], so they are integrable in [-1,1]. Also g(x) is positive in [-I, 11. Uy first Mean yalue Theorem, there is a number p between the '

bounds of f such that

g.l.b.{f(x)l-'l 5 x 5 1 ) = - 1 and I.u.b.(f(x)J-1 5 x I 1) = 1 and, so, p E [-I, 11. First I

Mean Value Theorem is verified. - I

VERIFICATION OF SECOND MEAN VALUE THEOREM:

As shown above, f and g are integrable in [-1,1]. Also g is monotonically increasing in [-1,1]. By second mean value theorem there is a points c E [-],I] such that ,

I C I

J f ( ~ ) g(x) dx = g ( - ~ j J f(x) dx + g(l) J f(x) dx - 1 I

'I ' => - i l x e Q x = - e x d x i - e l x d x Y

2 - 1 c2 1 =) - -- 1 cZ

(- --) t e(- ---I. e e 2 2 2 2

2 - 5 - 2.29 i . e . c=* E L - I , 11 Therefore c2 = er - -

e - 1 6.29

Thus second mean value theorem is verified.

EXERCISE 4 I

Show that the second mean value theorem does not hold good in the interval [- 1, I] for f(x) = g(x) = x2.

What do you say about the validity of the first mean value thcarem.

Now we show the use of mean value theorems to prove some inequalities.

EXAMPLE 4: By applying the first mean value theorein of Integral calculus, prove that

SOLUTION: In the first mean value theorem, take f(x) = 1

Jm' I

1 1 g(x) = - , x E [0, -1. Being continuous functions, f and g are integrable in

J - i 7 2 ' 1

[ 0, -1. I

2

By the first mean value theorem, there is a number / L ~ [ n i , M ] such that I !h

1 ' dx ( d[(l-x2) (1- k2x2)]

dx = p/-= pn/6. 0

1 1 where in = g.l.b.(f(x)( 0 5, <-) and M = I.u.b.{l'(x) 1 0 5, -1. Now m = 1 2 2

1 and M = Therefore, . I

4- q i

4 1

1 . n p n I I > I

15/15 - i.e. - - < - I - - I ' 6 6 n' i :

4 4 r % 1 7I 1

i and; so, - < / XI-=.

6 -o d[(l -x2) (1- k2x2)] 3 8 6 41-E 4 I

\

q ' sin x EXAMPLE 5: Prove that I / - 2

d ~ 1 5 - , i f q > ~ > o . P X P

1 SOLUTION: Let f(x) = sin x, f ix) = - , x ~ [ p , q ] . Being continuous, these functions are

X

integrable in [p,q]. BY Bonnet form of second mean value theorern, there is a point C E [p,q] such that

9 . 1 5

1 i.e., / E dx =- / sin x dx =- (cos p - cos 5). P x Q P P

1

Hence 1 / P

sin x - 1 2 dx I < - [Icos pl + Jcos 511 _< - P P

EXERCJSE 5

Show that I Is in x2 (Ix I 5 , if b > a > 0. ll

15.5 SUMMARY

The main thrust of this unit has been to establish the relationship between differentiation and integration with the help of the Fundamental Theorem of Calculus.

I11 section 15.2. we have discussed some important properties of (be Ricmann Integral. We havc shown that the inequality between any two fu~ictions is preserved by their corresponding Rielnann integrals; the modulus of the limit of a sum nevcr exceeds thc limit of thc sum of their modulie and if we split the interval over which we are integrating a function into two parts, then the value of the integral over the wholc will be the sum of the two integrals over the subintervals;

In sectibn 15.3, primitive of a runction has been defined. It has been proved that a continuous function has a primitive. Using the idea of a primitive, F~~ndamenta l Theorem o r Calculus has been proved which shows that differentiation and ' integration are inverse process.

In section 15.4, indefinite integral also called the integral function of an integrable function is defined and you have seen that this function is continuous. This function 1s differentiable whcnevel- the integrable function is continuous. F.inally in this section the First and Second Mean Value theorem have been discussed, The First Mcan Valuc theorem states that i f f is :1 C O I ~ ~ ~ ~ U O L I S I'l~nction in [a,b], thcn the value of the

b

integral1 S(x) d x is (b-a) times f(c) wherc c E [a,b],According to Generalised First .I

Mean Value Theorem, i f f and g are integl-able in [a,b] and g(x) keeps the same sign, h 11 I1

then the value o f J ~ ( x ) g(x1.d~ is J f(x) g (x) dx = J g(x) dx wllcrc p lies n B I

between thc bounds o f f . But i n the second mean value theorem, if out of the integrable functions f ~und g, g is monotonic in

h L b

[ah], then the value oi'J f(x) g(x) dx is g(a) f(x) dx + g(b) f(x) dx where c is J i point 01 [a,b].

C

E I ) If we consider the function F defined i n [O, 21 by

if x E [O. 1 [ f(x) = + , i f, ,[,,I

then Ff(x) =f(x), 'd x E [0,2]. It is obviously true for all x s l . At x = 1, calculate Rff( l ) and Lf' (1) and show that both are equal to 1.

7

E 2 ) By Fundamental Theorem of Calculus, /f(x) dx = F(2) - F(O), where F is given 0

in E I . Put the values of F(2) and F(0) and you will get the values of the integral.

(j Xl!+l

E 3) Since - [---I = x" and, therefore, - is a primitive o f f in [a,b] and dx n + l n + l

by the Fundan~ental 'Theorem of Calculus,

E 4 ) Functions f and g being continuous are integrable in [-I, I ] . The function g is not monotonic in [-1, I], since in [-1,0], it is lnonotonically decreasing and in [0,1] it is monotonically increasing. So second mean value theorem does ro t hold good.

Now the function g is -1-ve in [-I, 11. So, first mean tlieoren~ holds good and by

I I

this theorem, there is a number jr ~ [ r n , M ] such that /f(x) g(x) dx = p jg(x) dx, -I - I

I I

where n i = g.l.b.{f(x))-1 S x I I } and M = I.u.b.{f(x)/-1 S x I I ) i.e., /x"x=,u /x2 dx -I - I

3 and, so, p = - .

5

Now ~u = 0, M = I. So, jr E[O,I]] and first mean valw theoreni is verified.

1 1 E 5 ) Sin x2 = 7 (2x sill x2) for x + 0. Take f(x) = 2x sin ,x2 and $(x) = - and - X 2x

apply Bonnet Form of Second Mean Value Theorem as in Exa~nple 6 .

SEQUENCES AND SERIES OF .

Structure 16.1 lntroduction

I

Objectives

16.2 Sequences of Functions Pointwise Convergence

16.3 Uniform Convergence Cauchy's Criterion .

164 Series of Functions 16.5 Surnnlary 16.6 Answers/ Hints/Solutions

16.1 ' INTRODUCTION

In unit 5, you were introduced to the notion of sequences of real numbers and their convergence. In units 6 and 7, convergence of the infinite series of real numbers was considered, in this unit, we ivant to discuss sequences and series whose members are functions defined on a subset of the set of real numbers. Such sequences or series are known as sequences or series of real functions. You will be introduced t o the concepts of pointwise and uniform convergence of sequences and series of functions. Whenever they are convergent, their limit is a function called limit function. The question arises whether the properties of continuity, differentiability, integrability of the members of a sequence or series of functions are preserved by the limit function. We shall discuss this question also in this unit and show that these properties are preserved by the Uniform convergence and not by the pointwise convergence.

Objectives After the study of this unit, you should be able to - define sequence and series of functions -+ distinguish between the pointwise and uniform convergence of sequences and

series of functions - know the relationship of uniform convergence with the notions of continuity, differentiability, and integrability.

16.2 SEQUENCES OF FUNCTIONS

In unit 5, you have stadied that a sequence is a function from the set N of natural numbers to a set B. In that unit, scquences of real numbers have been considered in detail. You may recall that for sequences of real numbers, the set B is a sub-set of real numbers. If the set B is the set of real functions defined on a sub-set A of R, wc get a sequence called sequence of functions. WC define it in the following way:

d

DEFINITION 1: SEQUENCE OF FUNCTIONS Let A be a non-empty sub-set of R and let B be the set of all rcal functions each defined on A. A mapping from thc set N of natural numbers to the set B of real functions is called a sequence of functions., The sequences of functions are denoted by (f,,), (g,) etc. i t (fn) is a sequence of functions defined on A, then its members f ~ , fi, f3, .,.... are real functions with domain

. as the set A. These are also called the terms of the sequence (f,,). For example, let

f, (x) = x", n = 1, 2, 3, ..., where x EA = {x: 0 5 x S I }. Then (f,) is a sequence of I

functions defined on the closed interval [0,1].

Similarly consider (f,), where f (x) = sin nx, n = I, 2, 3, .... x ER. Then {f"), is a sequence of functions defined on the set R of real numbers.

42

Suppose (fn) is a sequence oPfunctions defined on'a set A and we fix a point x of A, I then the sequence (fn(x)), formed by the values of the members of (f,), is a sequence of

real numbers. This sequence of real numbers may be convergent or divergent. For

1 example suppose that fn (x) = xn, x E [-I, 11. If we consider the point x = - , then the 2

1 " sequence (fn (x)) is ((.-) ) which converges to 0. If we take the point x = 1, the

2

sequence (fn (x)) is the constant sequence (1,1,1 .....) which converges to I . If x = -1, the sequence (f, (x)) is (-1, I, -1, 1, ....) which is divergent.

Thus, you have seen that the sequence (fn (x)) may or may not be convergent. If for a sequence (f,) of functions defined on a set A, the sequence of numbers (fn (x)) converges for each x in A, we get a function f with domain A whose value f(x) at any point x of A is lirn fn (x). In this case (f,) is said to be pointwise convergent to f. We

n - m

define it in the following way: w

DEFINITION 2: POINTWISE CONVERGENCE A sequence of functions (f,) defined on a set A is said to be convergent pointwise to f if

-. for each x in A, we have lirn fn (x) = f(x).Generally, we write f;, f (pointwise) on A

n - m

or lirn t (x) = f(x) pointwise on A. Also f is called pointwise limit or limit function of n - w

Equivalently, we say that a sequence { f " ) converges to f pointwise on the set A if, for each E > 0 and each x EA, there exists a positive integer 111 depending both 011 E and x such that

Ifn (x) - f(x)( < E, whenever n 2 m.

Now we consider some examples.

, EXAMPLE 1: Show that the sequence (f,) where fn (x) = x", x € [0,1] is pointwise I convergent. Also find the limit.

SOLUTION: If 0 I x < 1, then lim fn (x) = lirn xn = 0. (Recall unit 5). n - - n - w

If x = 1, then lim,f(x) = lim 1 = 1. n-00 n - m

Thus (f,) is pointwise convergent to the limit function f where f(x) = 0 for 0 5 x < 1.and f(x)= 1 for x = 1.

EXERCISE 1 Show that the sequence of functions (f,,) where f,, (x) = xn, for x E [-I, I] is not pointwise convergent.

"

EXAMPLE 2: Define the function f,, , n = 1, 2 ,....., foilows

1 0 , if n < x < l

Show that the sequence ( t ,) is pointwise convergent.

SOLUTION: The graph of function fn looks as shown in the figure 1.

When x = 0, fn (x) = 0 for n = 1,2, ..... Therefore, the sequence (fn(0)) tends to 0. 1

If x is fixed such that 0 < x I 1; then choose m large enough so that - < x or m

1 m > - .Then fm (x) = fm+, (x) = .... = 0. Consequently the sequence (fn (x)) -3 0 as X

n- 3 a.

Fig. 1

Thus we see that fn (x) tends to 0 f ~ r every x in 0 I x 5 1 and consequently (f,) tends pointwise to f where f(x) = 0 'SF x E [0,1].

EXAMPLE 3: Consider the sequence of functions f,, defined by f, (x) = cos nx for -- < x < i.e.x E R. Show that the sequence is not convergent pointwise for every real x.

SOLUTION: If x = n/4 then (fn (x)) is the sequence

(11 J2, 0, -1 / J2, -1, - I / J2, 0, ......I which is not convergent.

You should be able to solve the following exercises.

EXERCISE 2 sin n x Show that the sequence (f,) where fn (x) = , - , x E R, is pointwise

J n convergent. Also find the pointwise limit.

EXERCISE 3 Examine which of the following sequences of functions converge pointwise

(i) f, (x) = sin nx, -- < x < + w.

nx (ii) f, (x) =r - - ~ < x < + ~ '

l+n2x2

If the sequence of functions (f,) converges pointwise to a function f on a subset A of R, then the following question arises? "If each member of (f,) is continuous, differentiable or integrable, is the limit function f also continuous; differentiable or integrable?" . The answer is no if the convergence is only pointwise. For instance in example 1, each of the fu'nciions f, is continuous (in fact uniformly continuous) but the sequence of these functions converges to a limit function f(x)

f(x) = 0, for 0 5 x < 1 1, for x = 1

which is not continuous. Thus, the pointwise convergence does not preserve the property of continuity. To ensure the passage of the properties of continuity, differentiability or integrability to the limit function, we need the notion of uniform

44 convergence which we introduce in the next section. I

16.3 UNIFORM CONVERGENCE

From the definition of the convergence of the sequence or real numbers, i t follows that the sequences (f,) of functions converges pointwise.to the function f on A if and only if for each x E A and for every number c > 0, there exists a positive integer In

such that

I f,, (x) - f(x) ( < E whenever n 2 m.

Clearly for a given sequence (f,,) of functions, this m will, in general, depend on the given $ and the point x under consideration. Therefore it is, sometimes, written as m (c,~). . The foilowing example illustrates this point.

EXAMPLE 5: Define f,, (x) = "or -a < x < a. n

For each fixed x the sequence (f,(x)) clearly. converges to zero. For a given t > 0, we must show the existence of an m, such that for all n 2 m,

31xl 1x1 This can be achieved by choosing m = [- ] + 1 where [- ] denotes the t t

1x1 1x1 integral part of (i.e. the integer m is next to1 in the real line). Clearly this choice of m depends both on 6 and x.

1 1 For example, let E = - . If x =- then !?!- = I and, so, m can be chosen I 0' 10.' E

1x1 to be 2. If x = 1, then, = lo3 and, therefore, m should be larger than lo3. If x = lo',

1x1 then 7 = lo6 and, so, 111 should be larger than lob, Note that it is impossible to find

1x1 an 111 that serve for all x. For, if 'it were, then , < E, for all x, I11

Consequently 1x1 is slnaller than ~ m , which is not possible. Geometrically, the fn fs can be described as shown in the Figure 2.

1 I By putting y = f;, (x), we see that y = - x is the line with slope - . f , is the \ine ? n

S c q u c ~ ~ c c s and Sc~. ies of FI I I IC titrl~s

1 y = x with slope 1 , fz is the line with slope - and so on, As n tends to a, the lines

2

approach the X-axis. But if we take any strip of breadth 2 c around X- axis, parallel t o the X -axis as shown in the.figuse 2, it is impossible to find a stage m such that all the lines after the stage m, i.e, f,, f,+, ...... lie entirely in this strip.'

If it is possible to find m which depends only on c but is independent of the point x under consideration, we say that (f,) is uniformly convergent to f. We definc uniform convergence as follows:

DEFlNlTION 3: UNIFORM CONVERGENCE

A sequence of functions (f,) defined on a set A is said to be uniformly convergent to a function f on A if given a number t > 0, there exists a positive integer m depending only on c such that

fn(x)-f(x)l < ~ f o r n L m a n d Y x €A . We write it as f, -- f uniformly on A or lim fn (x) = f(x) uniformly on A. Also f is called the uniform limit o f f o n A.

Note that if f, -- f uniformly on the set A, for a given E > 0, there exists III such that

f(x) - E < fn (x) < f(x) f E

for all x t A and n 2 m. In other words, for n L m, the graph of fn lies in the strip between the graphs o f f - E and f + . As shown in the figure 3, the graphs of f,, Tor n 2 m will all lie between the dotted lines.

'T'

Fig. 3

From, the definition of uniform convergence, it follows that uniform convergence of a sequence of functions implies its pointwise convergence and uniform limit i s equal to the pointwise limit. We will show below by suitable examples that the converse is not , true.

X EXAMPLE 6: Show that the sequence if,,) where fn (x) = - , x E R is pointwise but n

not uniformly convergent in R.

SOLUTION: In example 5, you have seen that (f,) is pointwise convergent to f where f(x) = 0 Y x E R. In the same example, at the end, it is remarked that given 6 > 0, it

1x1 is not possible to find a positive integer m such that - < e for n l rn & Y x E R n

i.e. (f,(x) - f(x)/ < 6 for n l rn & Y x € R. Consequently (f,,) is not uniformly convergent in R.

EXAMPLE 7: Show that the sequence (f,) where fn (x) = xn is convergent pointwise but not uniformly on [0,1].

SOLUTION: 'In 'example 2, you have been shown that (f,) is pointwise convergent to , f on [Q,1] where

f(x) = 0 ~ x E [ O , ~ [ & f(l) = 1 Sequences and Serics of Function$

Let e > 0 be any number. For x = 0 or x = I, If,,(x) - f(x)l < 6 for n l 1.

log E For 0 <.x < 1, Jfn(x) - f(x) I < E if x"< e i.e. n log x < log E i.e n > ---- .

log x

since log x is negative for 0 < x < 1. If we choose rn = [* ] + I . log x

then If, (x) - f(x)J < E for n 2 m. Clearly m depends upon E and x.

We will now prove that the c.onvergence is not uniform by showing that i t is not possible to find an m independent of x.

Let us suppose that 0 < e < .I. If there exists m independent of x in [O,L] so that

If, (x) --f(x)I < e for all n 2 m, then x" < e for all n 2 m, whatever may be x in 0 < x < 1 . If the same m serves for all x for a given E > 0 then x'" < E for all x, 0 < x < t . This

log € implies that m > --- (since log x is negative). This is 110t possible since log x

log x

decreases to zero as x tends to 1 and so log ellog x is unbounded.

Thus we have shown that the sequence (f,) does not converge to the functi0n.f uniformly in [0, I ] even though it converges pointwise.

X EXAMPLE 8: Show that the sequence (g,) where g,(x) = - , x E [O, a[ is 1 + nx

uniformly convergent in [O, 00 [.

SOLUTION: lim g, (x) = 0 for all x in the interval [0, [. Thus (g,) is pointwise n - w

converged f.0 f where f(x) = 0 Y x E [0, w[.

X Now Ig, (x) - f(x)l =- 1 <, - for all x in [O, m'[, I f n x n

1 Since l i~n fi = 0, therefore given E > 0, ther,e exists a positive integer m such that

n - m

Thus m depends only on e. Therefore,

Ig,(x) -f(x)I < e for n L m &Y x E [O,W[. Therefore (g,) -> f uniformly in [0, qo [.

EXERCISE 4 . Test the uniform convergence of the following sequence of functions in the specified

domains

xn (iii) f,, (x) = - 0 5 x 5 1

1 + x"

Just ns you have studied Cauchy's Criterion for convergence of sequence of real nun-lber's, H e have Cauchy's Criterion for uniform conLcrgence of sequence of . I

I

functions which we now state and prove. 4 3

THEOREM 1: CAUCHY'S PRINCIPILE OF UNIFORM CONVERGENC~ The necessary and sufficient condition for a sequence of functions (f,) defined on A to converge uniformly on A is that fur every c > 0, there exists a positive integer m such that

1fn (x) - f~ (x)I Q E for n > k L rn & 'bL x E A

PROOF: Condition is Necessary. I t is given that (f,) is uniformly convergent on A. Let fn - f unifnr~nly on A. Then given E > 0, there exists a positive integer m such that

) f , , ( x ) - f(x)l < €12 for n > n & Tf x E A. a'. I fm(x) - h(x) ( = 1 {,(XI - f(x) 1 + f(x) - fk(x) 1

5 I fn(x) - f(x) 1 + If(x) - 6(x) 1 (By triangular inequality)

This proves the hecessary part. Now we prove the sufficient part.

Condition is sufficient : It is given that for every c > 0, there exists a positive integer m such that (fn(x) - f~ (x)( < e for n > k 1 m and for all x in A. But by Cauchy's principle of convergence of sequence of real numbers, for each fixed point x bf A, the sequence of numbers (f,, (x)) converges. III other words, (f,) is pointwise cbnvergent say to f on A. Now for eacn 6 > 0, there exists a positive inteyt.~. m such that

Fix k and let n -+ oo. Then fn (x) --+ f(x) and we get

This is true for k 2 m and for a11 x in A. This shows that (fn) is unifor~nly convergent to f on A, which proves the suficient part. I

As remarked in the introduction, uniform convergence is the form of convergence of \ the sequence of function (f,) which preserves the continuity, differentiability and

xi integrability of each term f, of the sequence when passing to the limit function f. In other words if each member of the sequence of functions (f,,) defined on a set A is

.: continuous on A, then the limit function f is also continuous provided the convergence is uniform. The result may not be true if the convergence is only pointwise. Similar results hold for the differentiability and integrability of the limit function f. Before giving the theorems in which these results are proved, we discuss some examples to illustrate the results.

EXAMPLE 9: Disalss for continuity tile convergence of a sequence of f~nctions (q), where fl,(x.) = I- 11 -x2J", x E ( X ) )I-x2J I I } = [-&, 61.

1, when 11-x21 < I SOLUTION: Here lim fn (x) =

n+ r 0, wlieri 11- x2( = 1 i.e.. .u = 0, 2 6.

Therefore the sequence (f,) is pointwise convergent to f where

f(x) = 1 when 1 I-x" < 1 0 when 1 1-x2 1 = 1

Now each member of the sequence (f,) is continuous at 0 but f is discontinuous at 0. Here (f,) is not uniformly convergent in [-fi, 41 as shown below.

I Suppose (f,) is uniformly convergent in [-A a], so that f is its uniform limit.

I Taking t = - , there exists an integer m such that 2

I f . (n)-f(x)l < i f o r n ? m & ~ x€[-& &?I. 2

4 8

Now I f ' , , , (x) - f{x)I = when ( I -x2( < I ~ 1 1 ~ 1 1 ( I - x ' ( = l

Sequences a n d 'Series of F u n c t i o ~ l s

Since lim 1 I - x ' I ~ ~ ' - A 1 , 3 a t v no. 6 huch that \ 0

I I - x ' / ' ~ - I 1 < I 1 4 T o r O < I x l < 6 i.e. 3;4< I I-x21"'<5,4Tor 1x1 < 6 .

1 So I 1-x"'" > - for 1x1 < 6, which is a contradiction.

2 Conhequentlq (f,,) is not uniSormly convergent i n [--a, 41. EXAMPLE 10: Discuss, for continuity, the convcrgencc of the aequcnce ( f , , ) where

X f, (x) = - x E [O, -[.

I + n x

SOLUTION: In example 8, you have seen that (f,,) - f uniformly in [0, whe1.c f(x) = 0, x E [O,M[.

Hcre each f,, is continuous in [O, m[ and also the uniform limit is C O I I ~ ~ I I U O U S ill [O , a[.

EXAMPLE 11: Discuss for differentiability tlie scquence (I;,) where

sin n x (fn) (x) =

6 1

, V x E R.

SOLUTION: Here (f,) -- f uniformly where ((x) = 0 Y x E R. You can see that each fn and f a r e differentiable in R and

f,' (x) = 6 cos nx & f' (x) = O V x E R. f;: (0) = 6 - 03 wl~eieas f' (0) = O l in1 f',, (0) # f (0) I1 -C

i.c. limit of the derivatives is not equal to the derivative of the limit.

As you will see in the theorem for the differentiability of rand the equality of the limit of thc derivatives and the derivative of the limit, we require the uniform convcrgencc of tlie sequcnce (f;).

EXAMPLE 12: Discuss for integrabitity the sequence (f,,) where

t = n x e-llx2, x E [o. I].

SOLUTION: I f x = 0, then T, (0) = 0 n x cz and lim C, (0) = 0. I f x # 0, lim f,, (x) = lim --, which is of the rorm - .

n-- - el"- DC

Applying L'HopitaPs Rule, we have

IIX X tim en, = lim - - - 0. n-tx "->= 2 n ~ e " ~ '

So (fil)+ f, pointwise, where f(x) = 0,V x 'E[o,I].

1 You may find that fn(x) dx = - (I-e-") and

2

1 I

Therefore, lim hl (x) dx = - r /f(x) dx = flim ( f (x)) dx. That is, the integral of the limit n-,x o 2 o 0 n - + r

is not equal to be limit of the sequence of integtals. In fact, (f,) i s not uniformly convergent to f in [0,1]. This we prove by the contradiction method. If possible, let the

I sequence be uniformly convergent in [O, I]. Then, for E = -;i- , there exists a positive

1 integer m such that Ifn (x) - f(x)l < - , for n 2 111 ancl V x E[O,I].

4 , nx I I.e., --p < -, for n 2 m and V x ~ [0 .1 ] .

4

Integrabi l i ty 1 Choose apositive integer M 2 m such that - E [0,1].

M I

Take n = M and x - --- . We get 6

which is a contradiction. Hence (f,) is not uniformly convergent in [0.1].

Now try the following exercises. - --

EXERCISE 5 k Show that the limit function of the sequence (f,) where (f,) (x) =-, x E. K, is

colltinuous in R while (f,,) is not uniformly convergent. n

EXERCISE 6. Show that fo r the sequence (f,) where (f,) (x) = n x (1-x2)", x x [0,1], the iuregral of the limit is not equal t o the limit of the sequence of integrals.

Now we,give the theorems without proof which relate'uniform convergence with continuity, differentiability and integrability of the limit function of a sequence of functions.

THEOREM 2: (UNIFORM CONVERGENCE AND CONTINUITY) If (fn) be a sequence of continuous functions defined on [a,b] and (fn) * f uniformly on [a,b], then f is continuous on [a,b].

THEOREM 3: (UNIFORM CONVERGENCE AND DIFFERENTIATION) '

Let (fn) be a sequence of functions, each differentiable on [a,b] such that (fn(x0)) converges for some point x, of [a,b]. If (f,') converges uniformly on [a,b] then (fn) converges uniformly on [a,b] to a function f such that

f'(x) = lirn f,' (x); x E [a,b]. n-00

THEOREM 4: (UNIFORM CONVERGENCE AND INTEGRATION) If a sequence (fn) converges uniformly to f on [a,b] and each function f, is integrable on [a,b], then f is integrable on [a,b] and b b

J f(x) dx = ~irn J fn (x) dx a n-k

16.4 SERIES OF FUNCTIONS

Just as we havestudied series of real numbers, we can study series formed by a , sequence of functions defined on a given ser A. The ideas of pointwise convergence

and unifprm convergence of sequence of functions can be extended to series of functions.

DEFINITION 4: (SERIES O F FUNCTIONS)

A series of the forth f~ + f2 -k f3 + ...... + fn -k ,... where the C s are real functions defined on a given set ACR is called a series of functions and is denoted by f,.

I The function fn is called nth term of the series. 1:

n= I

Foreach x in A, f r (x) + fi (x) -t f3 (x) +......+ is a series of real numbers. We put

1 1 S, (x) = f i (x). Then we get a sequence (S,) of real functions defined on A. We

L= 1

say that the given series f~ -t fj -t ...... t f n 3- ..... of functions converges to a function ! pointwise if the sequence (Sn) associated to the given series of functions converges '

pointwise to the function f. i.e. (S,, (x)) converges to f(x) for every x in A. 50

We also say that f is the pointwise sum of the series 2 f;, on A. Seqoerrces r ~ n d Scries of

If the sequence (S,,) of functions converges unifornlly to the fi~nction f. then we say that Functicl l~s

the given series f , + f, + ...... + fn+ ......, of functions converges uniformly to

the function f on A and f is called uniform sum of? f on A. The function Sn is i= l

called the sum of n terms of the given series or the nIh partial sum of the series and

tbe sequence (S") is called the sequence of partial sums of the series 9 \. To make the I= I

ideas clear, we consider some examples.

EXAMPLE 13: Let fn (x) = xn-' where x0 = 1 and - r I x I r where 0 < r < 1. Then the associated series is 1 + x + x" ......

1 - xn In this case, S, (x) = I , + x + x2 + ........ +.xn-I. I t is clear that S,, (x) = -

1 - x

This sequence (S, (x)) of functions is easily seen to convkrge pointwise to the function

I , since xL> 0 as n -> w , since I x I < r < I but the f (x) = --

I - x

convergence is not uniform as shown below:

Let t > 0 be given.

Ixln - r I S, (x) - f(x) ( = - c - < E if1.I' < 6 ( I -- r)

I I-xl 1-1.

log ( € ( I - r)) i.e. n >

log r

l o g ( € ( l - rl)] + ,, then I f m = [

log r

I s, (x) - f(x) I < 6 if n> m and for - r F x I r. Therefore (S,) converges uniformly in [-r, r]. Thus the geometric series 1 + x t x2 +

I ..... converges uniformly in [-r, r] t o the sum'function f(x) = - I - x

EXAMPLE 14 : Let f, (x) = n x e-"" - (n - 1) x e- '11-1'x2 , xfE [(I,]]. 00

Consider the series Z f,, (x). n= l

In this case S . (x) = ; (k x-*"~ - (k - 1 ) x i 1k- "x2 7

) = n x e-nx- k= l

In example 12, you have seen that this sequence (S,) is pointwise but not uniformly convergent to the function f where f(x) = O.,x E (0, 1). Thus the series C f, (x) is pointwise convergent but not uniformly to the function f whcre f(x) = 0, x E [0, 11. Try the following exercises,

EXERCISE 7 Show that the series I

f

X - + X + X + ....... is unifornlly x + l (X + 1) (2x + 1) (2x + 1) (3x + 1)

i convergent in ]k, a[ where k is a positive number.

EXERCISE 8

X . Show that the series x- is uniformly convergent in [O,k] where k is any

n ( n + 1)

positive number but it does not converge uniformly in [0,-1.

'There is a very useful method to test the uniform col~vergence of a series of functions. Integrability In this method, we relate the terms of the series with those of a series with constant

terms. This method is popularly called Weierstrass's M-test given by the German mathematician K.W.T. Weierstrass (1815-1 897). We state this test in the form of the following theorem (without proof) and illustrate the method by an example.

t

THEOREM 5: WEIERSTRASS M- TEST Let C fn be a series of functions defined on a subset A of R and let (M,) be a sequence of real numbers such that Mn is convergent-and If,, (x)l 5 Mn, 'v' n and V x EA. Then C fn is uniformly and absolutely convergent on A.

Consider the following example and tlie exercise.

X EXAMPLE 14: Test the uniform convergence of the series 2

,,=I n2 (11 + 1)'

X Solution: Since Ifn (x) I = --

k 5 -j- ,Vn and \J x ~[o,k] .

n2 (n + 1) n

1 Now the series C Mn = k C - is known to be convergent, by p-test.

n3

Therefore, by Weierstrass M-test, the given series is uniformly convergent in the set [O , kl.

-- - -

EXERCISE 9

1 Show that the series 9 - converges uniformly, b' x E R.

,,-I n4 + x2

In this unit you have learnt how to discuss the pointwise and uniform convergence of sequences and series of functions. In section 16.2, sequence of functions is defined and pointwise convergence of the sequence of functions has been discussed. We say that a sequence of functions (f,) is pointwise convergent to f on a set A if given a number e > 0, there is a positive integer m such that

]f,,(x) -f(x)( < 6 for n > m , x E A. m in general depends on E and the point x under consideration. If it is possible to find rn which depends only o n E and not the point x under consideration, then (f,,) is said to be uniformly convergent to f on A. Uniform convergence has been defined in section 16.3. Further in this section, Cauchy's criteria for uniform convergence is discussed. Also in this section you have seen that if the sequence of functions (f,) is uniformly convergent to a function f on [a,b] and each f, is continuous or integrable, then f is also continuous or integrable on [a,b]. Further it has been discussed that if (f,) is a sequence of functions, differentiable on [a,b] such that (fn (x,)) converges for some point xo of [a,b] and if (f;) converges uniformly on [a,b], then (f,) converges uniformly to a differentiable function f such that P(x) = lim C (x); x E [a,b].

n - m

Finally in section 16.4, pointwise and uniform convergence of series of functions is given. The series of functions is said to be pointwise or uniformly convergent on a set A according as the sequence of partial sums (sn) of the series is pointwise or uniformly convergent on A,

16.6 ANSWERS/HINTS/SOLUTIONS TO EXERCISES

52 E 1) When x = - 1, lim xn does not exist. When x= 1, lim xn = 1.

When 1 x ( < 1, lirn xn = 0. heref fore (fn(x)) converges at all points of 1-1. 11 cxcept x = -1. Since the sequence (fn(x)) does not converges at each point of [- I,I], i t is not pointwise convergent - - i_n[-l,l]. However it is pointwise convergent in 1-l,l].

Sequcnccs rtnd Scr ics al' Funct ions

1 sin nx E 2) Since sin n x is bounded, and - converges to 0. ;herefore d n (

converges to 0 for every fixed value of x in R and so lim fn(x) = 0, x E R. n-m

Therefore (fn) is pointwsie convergent to the limit functionf where f(x) = 0 Y x E R. I

E 3) ( i ) For any value of x Z 0, the sequence of numbers (sin n x) is not convergent. 1

n r For example for x = r l 2 , the sequence is (sin- ) i.e (1,0, -1,O, ....) which is 9

not convergent. Thus sequence (f,) is not pointwise convergent. ( i i ) For x = 0, (fn(x)) = (0, 0, 0, ....) which is convergent. For x # 0,

= 0 as n -> m. So (fn) is paintwise convergent t0.f where fn (X) = +

f(x) = 0 for - 00 < x < m.

E 4) (i) lim fn (x) = 0, 0 < x < m. SO (fn) is pointwise convergent to f where f (x) = 0, n-m

0 < x (m. Let E > 0 be given. I

Ifn(x)-f(x)l = - ( O < X < ~ ) nx

1 If m=[-1; + I , then for n 2 m , I fn(x) -f(x)( < E . 'EX

1 m depends upon E & x. - is unbounded for 0 < x < m, so it is not possible to EX

find 111 which serves for all x. If such a m exists. then we have

1 - < E, for n 2 m and V x in 10, a[. I1 X

'1 1 :. < E 1.e.. m > - , which is not possible,

m x E X

I , since - + m as x + 0". E X

So (f,) is not uniformly convergent in lop[ .

(ii) (Cl) pointwise converges to f, where f(x) = 0, for -a < x < a. It is not

1 uniformly convergent. If it is, then for E =- , there is a positive integer m such 4

1 . nx 1 that ]fil(x) - f(x)( < - ~.e., I - I < - , Torn 2111 and for-m<x<co. Take

4 1i-n2x' 4 .

:. 1 nx 1 I I =-<- I + n2x2 2 4

which is a contradiction. S o (f,) is not uniformly convergent in - < x < m.

-. xn

-' i i ) For 0 5 x < 1 - -- 0 as n -- m, since xL 0. 1 + xn

1 .'. (f,,) - f pointwise where ffx) = 0 for 0 5 x < 1 and f(x) = -for x = 1. 2

let E > 0 be given. F o r x z O o r 1 , Ifn(x)-f(x)l = O < 6 f o r n 2 1 .

E i.e. n log x < log -

I - € E

i.e. n < logs' -- - a (log x is negative) ,log xe -

Take rn = [ l O ? l - - a ]+ 1. Then ' log x

(fn(x) -f(x)I < E for;> m. m depends upon e and x. Since log x -- 0 as x tends to I , it is not possible to find same rn for all x.

Therefore, the convergence is not uniform.

1 -nx - (iv) (f,) -> f pointwise where f(x) = 0 for 0 5 x < m, since lim - e - 0. n-m n

Let 6 > 0 be given.

1 ' 1 Since - -> 0, so there exists a positive integer m such that - < E for n L m.

n n

Therefore I f , (x) - f(x)J < E for n L m and for 0 I x < m and consequently f, -- f uniformly for 0 I x <

X E 5) lim - = 0, so (fn\ -> f pointwise in R where f(x) = 0 Yx E R. Obviously each

n - m n

f,, being a polynomial function, is continuous and also f being a constant function is continuous in R. (f,) is not uniformly convergent, since if it were, for a given e > 0, there is a positive integer m'such that I fn(x) - f (~) l < ~ f o r n L m & Y x E R .

X i.e. - < E for n 2 m and Y x E R.

1 Take n = m & x = m, then 1 <-which is a contradiction. So (fn) is not 2

uniformly convergent in. K.

E 6) fn(0) = 0 = f, (1).

n M For0 < x < I , lim fn ( x ) = x lim 'n -cr ; n-w (1 -x2)-"

= x lim 1, (L' Hopital Rule) II-- (1-x2)-" log (I-x2)

= 0

54 Therefore, (f,) -> f pointwise where f(x) = 0 'bT x E [O, 11.

I 1 n I f(x)dx = 0 &I fn (x) d x = - - 1 - as n- -- w.

o 2 ( n f 1) 2

Thus iim j f, (x) dx tj f(x) dx =/ lim f. (x) dx. n - = G o o n - m

E 7) I f S,(x) denotes the sum of n terms of the series, then

. X X Sn(x) = - + + ..,, upto n terms

x + 1 (x+l) (2x+1)

- 1 - I-----. and, so, nx+1'

(S,)) is pointwise convergent to f, where f(x) = I , V x ~ ] k , m[. NOW

1 - I - 0 an so for given 6 > 0, there is an integer m such that - < c for n 2 m. nk nk

Consequently I Sn(x) - f(x)i < e for n Z rn & y x Elk,-=[.

Therefore (S,) is uniformly convergent and so is the given series in ]k, m[.

E 8) If S, (x) is the sum of n terms of the series, then

I f - + x - - - +.. . . . .+ x - - - = ~ ( t ;) (; ? ) (I, " + I ' 1

S,,(s) -- x as n -- -. ' l 'hc~~cl'ore (S,) -> f pointwise where

f(x) = x Tor 0 I x < w

k In [O,k]. IS,,(x) - f(x)l 5 -

k and - - O a s n- m.

n + I n t l

For given E >'.O, there is a positive integer rn such that JS,(x) - S(x) I < E Tor n 2 m & Y x € [O, k].

'l'hcr-elf I-c (S,,) is uniforn~ly convergent in [0, k]. I ! ' ( h , , ) is ullirorrnly convergent in [0, a[, then for E = 112 therc is a positive intcgcr- rn s~rch that

X I S,,(x) - f(x1I = -

I <- for n 2 m & x E [0, w[ n f l 2

rake n = m & x = m f I and t l~cn tlicre is a contradiction. S o (S,,) is not uniformly convergent in [ O , m[.

. t lcncc thc berieh is uniformly convergent in [O, k] but is no[ uniformly con1 crgent irl 10, m[.

Sequences and Scrics of Func t io l~s

Integrabil ity,

1 < - f o reve ryn21 and V X E R .

n4 '

1 But C - is a convergent series. Therefore, by Weierstrass M-test, the series

1i4

1 C f, (x) = - , is uniformly convergent in R. n4 + , x 2

REVIEW This is the fifth and the last block in the course on Real Analysis.

It consists of three units namely the Units 14, 15 and 16. InUnit 14, you have been ? introduced to the fundamental notion of the integral of a function, popularly known

as the Riemann Integral. This is introduced through the supremum principle and is defined as the limit of a set of suitable sums thus ruling out the common misconception that integration is merely the reverse of differentiation. Also, the criteria for the integrability of a function have been given.

Some important properties of integrable functions have been discussed in Unit 15. The Fundamental Theorem of Calculus is, then, established which brings forth the relationship between differential and Integral Calculus. I t turns out that for a certain class of functions, integration is indeed a reverse process of differentiation. Two mean-value theorems and the well-known basic techniques of integration namely integration by parts and integration by substitution (change of variables) have been deduced with the help of the Fundamental Theorem of Calculus. Finally, Unit 16 dcals with sequences and series of functions which is a generalization of the notions of the sequences and series of real numbers discussed in block 2. The notion of pointwise and uniform convergence ha$ been introduced as well as their relevance t o the notions of continuity, differentiability and integrability of functions has been discussed.

r

You will do well if you try to attempt the following questions as a self-test to know whether or not you have actually grasped the material give^: in this block. You may check your answers with the ones given at the end.

, 1. If f(y,x) = 1 + 2x for y rational and F(y,x) = 0 for y irrational. Calculate

F(y) =/ f(y,x) dx. 0

2. By applying the generalised first mean theorem of integral calculus, show that

Show by means of an example that the product of two non-integrable functions I

may be integrable.

If f(x) = 0 y x E [a,b], then show'that

Give an example to show that f is integrable in [a,b] but has (i) finite number of points of discontinuity (ii) infinite number of points of discontinuity.

4

Show that 1 [x] dx = 6 0

Does the First Mean Value Theorem of Integral Calculus apply to t he function (i) f (x)= 1x1 i f O 5 x 5 2 (ii) f(x) = 2 if 2 I x 53.

Let (fn) be a sequence such that

Show that the limit function f is discontinuous in [ 0, 11

Using Weierstrass M-Test, show that the following series converge uniformly: m 1 1 (i) 2 n2 xn, X E [ -- ,- I n= I 2 2

(ii) 2 en" cos n x, x E [T , 2n] n= l

10. Test the convergence of the following series:. -

(i) 2 rn copverges and T, r"sin n (3 v Y 0 E R and r E [0, I ] 2

- ( i i ) 2 n-" xc [ l + LY QO [ ( a > 0)

ANSWERS/WINTS 1. F(y) = 2 if y is rational and F(y) = 0 if y is irrational

2. Apply generalised first mean value theorem by taking

I 1 f(x) = - and g(x) -

f i 42x2 Jr=-x2

3. Take f(x) = g(x) = 1 when x is rational - 1 when x is irrational, x E [a,b].

Then (fg) (x) = I Y x E[a,b]. I1

4. A constant function f(x) = k Y x E [a,b] is integrable in [a,b] a n d J f(x) dx = a

k (b-a) (Recall unit 14) Take k =,O.

5 . (i) Take f(x) = 1 when I 5 x < 2 3 when 2 I x 5 3.

1 1 (ii) f(x) = - when - 1

4n+l <XI - , n = O , 1,2 .... 4" 4"

f(0) = 0 a = O , b = l

4 1 2 3 4

6. o 1 [XI d x =I o [XI d x tl I Ex] d x l 2. [x] dx t 3 J [x] d x

I 2 3 4

= I d x + / ! d x t J 2 d x - t - 1 3 d x 0 I 2 3

7. Yes, since the functions are continuous

It is discontinuous at 1.

n' C - is convergent. 2"

(ii) 1 e-nx cos nx 1 5 e-"% ee-"""x E [T, 2n] C e-"" is a G.P. with common ration less than unity and so it is convergent.

10. (i) Ir"sin n 81 I rn. Z rn is a G.P. with common ratio r which is less than 1 and so it is convergent. '

1 (ii) Ix-'l =- 5 - Y x E [ I + u , ~ [ a n d Z is convergent. nX n ~ + ~

N O T E S