calculus - i1 sequence and series of real numbers 1.1 sequence of real numbers suppose for each...

Calculus - I

M.Thamban Nair

Department of MathematicsIndian Institute of Technology Madras

June 2006/Aug-Dec. 2010/Aug-Dec.2011

Contents

Preface vi

1 Sequence and Series of Real Numbers M.T. Nair 1

1.1 Sequence of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Convergence and divergence . . . . . . . . . . . . . . . . . . . 2

1.1.2 Monotonic sequences . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.3 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.4 Further examples . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.5 Cauchy sequence . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.1.6 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . 19

1.2 Series of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.2.1 Convergence and divergence of series . . . . . . . . . . . . . . 21

1.2.2 Some tests for convergence . . . . . . . . . . . . . . . . . . . 23

1.2.3 Alternating series . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.2.4 Absolute convergence . . . . . . . . . . . . . . . . . . . . . . 27


2 Limit, Continuity and Differentiability of Functions M.T. Nair 32

2.1 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.1.1 Limit point of a set D ⊆ R . . . . . . . . . . . . . . . . . . . 32

2.1.2 Limit of a function f(x) as x approaches a . . . . . . . . . . 33

Limit of a function in terms of a sequences . . . . . . . . . . 35

2.1.3 Some properties . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.1.4 Left limit and right limit . . . . . . . . . . . . . . . . . . . . 39

ii

Contents iii

2.1.5 Limit at ∞ and at −∞ . . . . . . . . . . . . . . . . . . . . . 40

2.1.6 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 42

2.2 Continuity of a Function . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.2.1 Definition and some basic results . . . . . . . . . . . . . . . . 43

2.2.2 Exponential and logarithm functions . . . . . . . . . . . . . . 46

2.2.3 Some properties of continuous functions . . . . . . . . . . . . 51

Attaining max f and min f . . . . . . . . . . . . . . . . . . . 52

Intermediate value theorem . . . . . . . . . . . . . . . . . . . 52

2.2.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . 53

2.3 Differentiability of functions . . . . . . . . . . . . . . . . . . . . . . . 54

2.3.1 Some properties of differentiable functions . . . . . . . . . . . 55

2.3.2 Maxima and minima . . . . . . . . . . . . . . . . . . . . . . . 59

2.3.3 Some important theorems . . . . . . . . . . . . . . . . . . . . 61

Rolle’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Lagrange’s mean value theorem . . . . . . . . . . . . . . . . . 62

Cauchy’s generalized mean value theorem . . . . . . . . . . . 63

L’Hospital’s rules . . . . . . . . . . . . . . . . . . . . . . . . . 64

Taylor’s formula . . . . . . . . . . . . . . . . . . . . . . . . . 66

2.3.4 Increasing and decreasing functions . . . . . . . . . . . . . . . 70

2.3.5 More about local maxima and local minima . . . . . . . . . . 71


3 Definite Integral M.T. Nair 74

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.2 Lower and Upper Sums . . . . . . . . . . . . . . . . . . . . . . . . . 75

3.3 Integrability and Integral . . . . . . . . . . . . . . . . . . . . . . . . 77

3.3.1 Some necessary and sufficient conditions for integrability . . . 79

3.4 Integral of Continuous Functions . . . . . . . . . . . . . . . . . . . . 83

3.5 Some Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.6 Some Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

iv Contents

3.6.1 First fundamental theorem . . . . . . . . . . . . . . . . . . . 88

3.6.2 Second fundamental theorem . . . . . . . . . . . . . . . . . . 89

3.6.3 Applications of fundamental theorem . . . . . . . . . . . . . . 90

3.7 Appendix M.T. Nair . . . . . . . . . . . . . 92

4 Improper Integrals M.T. Nair 94

4.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

4.1.1 Typical examples . . . . . . . . . . . . . . . . . . . . . . . . . 96

4.2 Integrability by Comparison . . . . . . . . . . . . . . . . . . . . . . . 97

4.2.1 Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . 99

4.3 Integrability Using Limits . . . . . . . . . . . . . . . . . . . . . . . . 100

4.4 Additional Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5 Geometric and Mechanical Applications of Integrals M.T. Nair103

5.1 Computing Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

5.1.1 Using Cartesian Coordinates . . . . . . . . . . . . . . . . . . 103

5.1.2 Using Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 104

5.2 Computing Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.2.1 Using Cartesian Coordinates . . . . . . . . . . . . . . . . . . 105

5.2.2 Using Polar Coordinates . . . . . . . . . . . . . . . . . . . . . 106

5.3 Computing Volume of a Solid . . . . . . . . . . . . . . . . . . . . . . 108

5.4 Computing Volume of a Solid of Revolution . . . . . . . . . . . . . . 109

5.5 Computing Area of Surface of Revolution . . . . . . . . . . . . . . . 110

5.6 Centre of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

5.6.1 Centre of gravity of a material line in the plane . . . . . . . . 111

5.6.2 Centre of gravity of a material planar region . . . . . . . . . 112

5.7 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5.7.1 Moment of inertia of a material line in the plane . . . . . . . 113

5.7.2 Moment of inertia of a circular arc with respect to the centre 114

5.7.3 Moment of inertia of a material sector in the plane . . . . . . 115


Contents v

6 Sequence and Series of Functions M.T. Nair118

6.1 Sequence of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 118

6.1.1 Pointwise Convergence and Uniform Convergence . . . . . . . 118

6.1.2 Continuity and uniform convergence . . . . . . . . . . . . . . 123

6.1.3 Integration-Differentiation and uniform convergence . . . . . 124

6.2 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125


7 Power Series M.T. Nair131

7.1 Convergence and Absolute convergence . . . . . . . . . . . . . . . . . 131

7.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . 133

7.3 Differentiation and Integration . . . . . . . . . . . . . . . . . . . . . 134

7.3.1 Series that can be converted into a power series . . . . . . . . 135


8 Fourier Series M.T. Nair136

8.1 Fourier Series of 2π-Periodic functions . . . . . . . . . . . . . . . . . 136

8.1.1 Fourier series and Fourier coefficients . . . . . . . . . . . . . . 136

8.1.2 Even and odd expansions . . . . . . . . . . . . . . . . . . . . 138

8.2 Fourier Series of 2`-Periodic Functions . . . . . . . . . . . . . . . . . 141

8.2.1 Fourier series of functions on arbitrary intervals . . . . . . . . 141


References 144

Index 145

Preface

This is based on a course that the author gave to B.Tech. students of IIT Madrasmany times since 1996 for the use of students and teachers of IIT Madras.

Comments and suggestions from the readers are welcome.

June 2006 M. T. Nair

[email protected]

Revised: August-November 2010

Revised: August-November 2011

vi

1

Sequence and Series of Real Numbers

1.1 Sequence of Real Numbers

Suppose for each positive integer n, we are given a real number an. Then, the listof numbers,

a1, a2, . . . , an, . . . ,

is called a sequence, and this ordered list is usually written as

(a1, a2, . . . , . . .) or (an).

More precisely, we define a sequence as follows:

Definition 1.1 A sequence of real numbers is a function from the set N of naturalnumbers to the set R of real numbers. If f : N→ R is a sequence, and if an = f(n)for n ∈ N, then we write the sequence f as (an) or (a1, a2, . . .).

A sequence of real numbers is also called a real sequence.

Remark 1.1 (a) It is to be born in mind that a sequence (a1, a2, . . . , . . .) is differentfrom the set an : n ∈ N. For instance, a number may be repeated in a sequence(an), but it need not be written repeatedly in the set an : n ∈ N. As an example,(1, 1/2, 1, 1/3, . . . , 1, 1/n, . . .) is a sequence (an) with a2n−1 = 1 and a2n = 1/(n+ 1)for each n ∈ N, where as the set an : n ∈ N is same as the set 1/n : n ∈ N.

(b) Instead of sequence of real numbers, we can also talk about a sequence ofelements from any nonempty set S, such as sequence of sets, sequence of functionsand so on. Thus, given a nonempty set S, a sequence in S is a function f : N→ S.For example, for each n ∈ N, consider the set An = j ∈ N : j ≤ n. Then weobtain a sequence of subsets of N, namely, (A1, A2, . . .).

In this chapter, we shall consider only sequence of real numbers. In some of thelater chapters we shall consider sequences of functions as well.

EXAMPLE 1.1 (i) (an) with an = 1 for all n ∈ N – a constant sequence withvalue 1 throughout.

(ii) (an) with an = n for all n ∈ N.

(iii) (an) with an = 1/n for all n ∈ N.

1

2 Sequence and Series of Real Numbers M.T. Nair

(iv) (an) with an = n/(n+ 1) for all n ∈ N.

(v) (an) with an = (−1)n for all n ∈ N – the sequence takes values 1 and −1alternately.

Question: Consider a sequence (a1, a2, . . .). Is (a2, a3, . . .) also a sequence? Why?

1.1.1 Convergence and divergence

A fundamental concept in mathematics is that of convergence. We consider conver-gence of sequences.

Consider the sequences listed in Example 1.1 and observe the way an vary as nbecomes larger and larger:

(i) an = 1: every term of the sequence is same.

(ii) an = n: the terms becomes larger and larger.

(iii) an = 1/n: the terms come closer to 0 as n becomes larger and larger.

(iv) an = n/(n+ 1): the terms come closer to 1 as n becomes larger and larger.

(v) an = (−1)n: the terms of the sequence oscillates between −1 and 1, and doesnot come closer to any number as n becomes larger and larger.

Now, we make precise the statement “an comes closer to a number a” as nbecomes larger and larger.

Definition 1.2 A sequence (an) in R is said to converge to a real number a if forevery ε > 0, there exists a natural number N (in general depending on ε) such that

|an − a| < ε ∀ n ≥ N,

number a is called the limit of the sequence (an).

Remark 1.2 (a) Note that, different ε can result in different N , i.e., the numberN may vary as ε varies. We shall illustrate this in Example 1.2.

(b) In Definition 1.2, the relation |an − a| < ε can be replaced by |an − a| < c0εfor any c0 > 0 which is independent of n. In other words, a sequence (an) in Rconverges to a ∈ R if and only if for any c0 > 0 and for every ε > 0, there existsN ∈ N such that |an − a| < c0 ε for all n ≥ N .

If (an) converges to a, then we write

limn→∞

an = a or an → a as n→∞

or simply as

an → a.

Sequence of Real Numbers 3

Note that|an − a| < ε ∀n ≥ N

if and only ifa− ε < an < a+ ε ∀ n ≥ N.

Thus, limn→∞ an = a if and only if for every ε > 0, there exists N ∈ N such that

an ∈ (a− ε, a+ ε) ∀n ≥ N.

Thus, an → a if and only if for every ε > 0, an belongs to the open interval(a−ε, a+ε) for all n after some finite stage, and this finite stage may vary accordingas ε varies.

Remark 1.3 Suppose (an) is a sequence and a ∈ R. The to show that (an) doesnot converge to a, we should be able to find an ε > 0 such that infinitely may an’sare outside the interval (a− ε, a+ ε).

Exercise 1.1 Show that a sequence (an) converges to a if and only if for every openinterval I containing x, there exits N ∈ N such that an ∈ I for all n ≥ N . J

EXAMPLE 1.2 The sequences (1/n), ((−1)n/n), (1 − 1n) are convergent with

limit 0, 0, 1 respectively:

For the sake of illustrating how to use the definition to justify the above state-ment, let us provide the details of the proofs:

(i) Let an = 1/n for all n ∈ N, and let ε > 0 be given. We have to identify anN ∈ N such that 1/n < ε for all n ≥ N . Note that

1

n< ε ⇐⇒ n >

1

ε.

Thus, if we take N =

⌈1

ε

⌉+ 1, then we have

|an − 0| = 1

n< ε ∀ n ≥ N.

Hence, (1/n) converges to 0.

Here dxe denotes the integer part of x.

(ii) Next, let an = (−1)n/n for all n ∈ N. Since |an| = 1/n for all n ∈ N, in thiscase also, we see that

|an − 0| < ε ∀ n ≥ N :=

⌈1

ε

⌉+ 1.

Hence, ((−1)n/n) converges to 0.

(iii) Now, let an = 1 − 1n for all n ∈ N. Since, |an − 1| = 1/n for all n ∈ N, we

have

|an − 1| < ε ∀ n ≥ N :=

⌈1

ε

⌉+ 1.

Hence, (1− 1/n) converges to 0.

EXAMPLE 1.3 Every constant sequence is convergent to the constant term inthe sequence.

To see this, let an = a for all n ∈ N. Then, for every ε > 0, we have

|an − a| = 0 < ε ∀ n ≥ N := 1.

Thus, (an) converges to a.

EXAMPLE 1.4 For a given k ∈ N, let Let an = 1/n1/k for all n ∈ N. Then an → 0as n→∞.

To see this, first let ε > 0 be given. Note that

1

n1/k< ε ⇐⇒ n >

1

εk.

Hence,1

n1/k< ε ∀ n ≥ N :=

⌈1

εk

⌉+ 1.

Thus, 1/n1/k → 0.

Exercise 1.2 Corresponding to a sequence (an) and k ∈ N, let (bn) be defined bybn = ak+n for all n ∈ N. Show that, for a ∈ R, an → a if and only if bn → a. J

Definition 1.3 A sequence (an) is said to be eventually constant if there existsk ∈ N such that ak+n = ak for all n ≥ 1.

Exercise 1.3 Show that every eventually constant sequence converges. J

Theorem 1.1 Limit of a convergent sequence is unique. That, is if an → a andan → a′ as n→∞, then a = a′.

Proof. Suppose an → a and an → a′ as n→∞, and suppose that a′ 6= a. Now,for ε > 0, suppose N1, N2 ∈ N be such that

an ∈ (a− ε, a+ ε) ∀n ≥ N1, an ∈ (a′ − ε, a′ + ε) ∀n ≥ N2.

In particular,

an ∈ (a− ε, a+ ε), an ∈ (a′ − ε, a′ + ε) ∀n ≥ N := maxN1, N2.

If we take ε < |a − a′|/2, then we see that (a − ε, a + ε) and (a′ − ε, a′ + ε) aredisjoint intervals. Thus the above observation leads to a contradiction.

An alternate proof. Note that

|a− a′| = |(a− an) + (an − a′)| ≤ |a− an|+ |an − a′|.

Now, for ε > 0, let N1, N2 ∈ N be such that

|a− an| < ε/2 for all n ≥ N1, |a′ − an| < ε/2 ∀n ≥ N2.

Then it follows that

|a− a′| ≤ |a− an|+ |an − a′| < ε ∀n ≥ N := maxN1, N2.

Since this is true for all ε > 0, it follows that a′ = a.

Exercise 1.4 Prove the following.

1. Let b ≥ 0 such that b < ε for all ε > 0. Then b = 0.

2. Let an ≥ 0 for all n ∈ N such that an → a. Then a ≥ 0.

J

The following theorem can be easily proved.

Theorem 1.2 Suppose an → a, bn → b as n→∞. Then we have the following :

(a) an + bn → a+ b as n→∞,

(b) For every real number c, can → c a as n→∞

(c) If an ≤ bn for all n ∈ N, then a ≤ b.

(d) (Sandwitch theorem) If an ≤ cn ≤ bn for all n ∈ N, and if a = b, thencn → a as n→∞.

Exercise 1.5 Prove Theorem 1.2. J

Exercise 1.6 If an → a and there exists b ∈ R such that an ≥ b for all n ∈ N, thenshow that a ≥ b. J

Exercise 1.7 If an → a and a 6= 0, then show that there exists k ∈ N such thatan 6= 0 for all n ≥ k. J

Exercise 1.8 Consider the sequence (an) with an =

(1 +

1

n

)1/n

, n ∈ N. Then

show that limn→∞

an = 1.

[Hint: Observe that 1 ≤ an ≤ (1 + 1/n) for all n ∈ N.] J

Exercise 1.9 Consider the sequence (an) with an =1

nk, n ∈ N. Then show that

for any given k ∈ N, limn→∞

an = 0.

[Hint: Observe that 1 ≤ an ≤ 1/n for all n ∈ N.] J

Definition 1.4 A sequence which does not converge is called a divergentsequence.

Definition 1.5 (i) If (an) is such that for every M > 0, there exists N ∈ N suchthat

an > M ∀ n ≥ N,

then we say that (an) diverges to +∞.

(ii) If (an) is such that for every M > 0, there exists N ∈ N such that an < −Mfor all n ≥ N , then we say that (an) diverges to −∞.

Definition 1.6 If (an) is such that anan+1 < 0 for every n ∈ N, that is an changessign alternately, then we say that (an) is an alternating sequence .

An alternating sequence converge or diverge. For example, (Verify that) thesequence ((−1)n) diverges, whereas ((−1)n/n) converges to 0.

Definition 1.7 A sequence (an) is said to be bounded above if there exists areal number M such that an ≤M for all n ∈ N; and the sequence (an) is said to bebounded below if there exists a real number M ′ such that an ≥M ′ for all n ∈ N.A sequence which is bound above and bounded below is said to be a boundedsequence.

Exercise 1.10 Show that a sequence (an) is bounded if and only if there existsM > 0 such that |an| ≤M for all n ∈ N. J


1. If (an) is not bounded above, then there exists a strictly increasing sequence(kn) of natural numbers such that akn → +∞ as n→∞.

2. If (an) is not bounded below, then there exists a strictly increasing sequence(kn) of natural numbers such that akn → −∞ as n→∞.

J

Theorem 1.3 Every convergent sequence is bounded. The converse is not true.

Proof. Suppose (an) converges to x. Then there exists N ∈ N such that |an−x| ≤1 for all n ≥ N . Hence

|an| = |(an − a) + a| ≤ |an − a|+ |a| ≤ 1 + |a| ∀n ≥ N,

so that|an| ≤ max1 + |a|, |a1|, |a2|, . . . , |aN−1| ∀n ∈ N.

To see that the converse of the theorem is not true, consider the sequence ((−1)n).It is a bounded sequence, but not convergent.


The above theorem can be used to show that certain sequence is not convergent,as in the following example.

EXAMPLE 1.5 For n ∈ N, let

an = 1 +1

2+

1

3+ . . .+

1

n.

Then (an) diverges: To see this, observe that

a2n = 1 +1

2+

1

3+ . . .+

1

2n

= 1 +1

2+

(1

3+

1

4

)+

(1

5+

1

6+

1

7+

1

8

)+

. . .+

(1

2n − 1+ . . .+

1

2n

)≥ 1 +

n

2.

Hence, (an) is not a bounded sequence, so that it diverges.

Using Theorem 1.3, the following result can be deduced

Theorem 1.4 Suppose an → a and bn → b as n→∞. Then we have the following.

(i) anbn → ab as n→∞.

(ii) If bn 6= 0 for all n ∈ N and b 6= 0, then an/bn → a/b as n→∞.


Exercise 1.13 If (an) converges to a and a 6= 0, then show that there exists k ∈ Nsuch that |an| ≥ |a|/2 for all n ≥ k, and (1/an+k) converges to 1/a. J

1.1.2 Monotonic sequences

We can infer the convergence or divergence of a sequence in certain cases by observ-ing the way the terms of the sequence varies.

Definition 1.8 Consider a sequence (an).

(i) (an) is said to be monotonically increasing if an ≤ an+1 for all n ∈ N.

(ii) (an) is said to be monotonically decreasing if an ≥ an+1 for all n ∈ N.

If strict inequality occur in (i) and (ii), then we say that the sequence is strictlyincreasing and strictly decreasing, respectively.

Also, a monotonically increasing (respectively, monotonically decreasing) se-quence is also called an increasing (respectively, a decreasing) sequence.

We shall make use of some important properties of the set R of real numbers.

Definition 1.9 Let S be a subset of R. Then

(i) S is said to be bounded above if there exists b ∈ R such that x ≤ b for allx ∈ S, and in that case b is called an upper bound of S;

(ii)S is said to be bounded below if there exists a ∈ R such that a ≤ x for allx ∈ S, and in that case a is called a lower bound of S.

Definition 1.10 Let S be a subset of R.

(i) A number b0 ∈ R is called a least upper bound (lub) or supremum ofS ⊆ R if b0 is an upper bound of S and for any upper bound b of S, b0 ≤ b.

(ii) A number a0 ∈ R is called a greatest lower bound (glb) or infimum ofS ⊆ R if a0 is a lower bound of S and for any lower bound a of S, a ≤ a0.

Thus, we have the following:

• b0 ∈ R is supremum of S ⊆ R if and only if b0 is an upper bound of S and ifβ < b0, then β is not an upper bound of S, i.e., there exists x ∈ S such that β < x.

• a0 ∈ R is infimum of S ⊆ R if and only if a0 is a lower bound of S and ifα > a0, then α is not a lower bound of S, i.e., there exists x ∈ S such that x < α.

The above two statements can be rephrased as follows:

• b0 ∈ R is supremum of S ⊆ R if and only if b0 is an upper bound of S and forevery ε > 0, there exists x ∈ S such that b0 − ε < x ≤ b0.

• a0 ∈ R is infimum of S ⊆ R if and only if a0 is a lower bound of S and forevery ε > 0, there exists x ∈ S such that a0 ≤ x < a0 + ε.

Exercise 1.14 Show that supremum (respectively, infimum) of a set S ⊆ R, ifexists, is unique. J

Exercise 1.15 Prove the following:

(i) If b0 is the supremum of S, then there exists a sequence (xn) in S whichconverges to b0.

(ii) If a0 is the infimum of S, then there exists a squence (xn) in S whichconverges to a0. J

EXAMPLE 1.6 (i) If S is any of the intervals (0, 1), [0, 1), (0, 1], [0, 1], then 1 isthe supremum of S and 0 is the infimum of S.

(ii) If S = 1n : n ∈ N, then 1 is the supremum of S and 0 is the infimum of S.

(iii) For k ∈ N, if Sk = n ∈ N : n ≥ k, then k is the infimum of Sk, and Skhas no supremum.

(iv) For k ∈ N, if Sk = n ∈ N : n ≤ k, then k is the supremum of Sk, and Skhas no infimum.

The above examples show the following:

• The supremum and/or infimium of a set S, if exists, need not belong to S.

• If S is not bounded above, then S need not have supremum.

• If S is not bounded below, then S need not have infimum.

However, we have the following properties for R:

Least upper bound property: If S ⊆ R is bounded above, then S has a leastupper bound. We may write this least upper bound as lub(S) or sup(S).

Greatest lower bound property: If S ⊆ R is bounded below, then S has agreatest lower bound, and we write it as glb(S) or inf(S).

Theorem 1.5 (i) Every sequence which is monotonically increasing and boundedabove is convergent.

(ii) Eery sequence which is monotonically decreasing and bounded below is con-vergent.

Proof. Suppose (an) is a monotonically increasing sequence of real numberswhich is bounded above. Then the set S := an : n ∈ N is bounded above.Hence, by the least upper bound property of R, S has a least upper bound, say b.Now, let ε > 0 be give. Then, by the definition of the least upper bound, thereexists N ∈ N such that aN > b− ε. Since an ≥ aN for every n ≥ N , we get

b− ε < aN ≤ an ≤ b < b+ ε ∀ n ≥ N.

Thus we have proved that an → b as n→∞.

To see the last part, suppose that (bn) is a monotonically decreasing sequencewhich is bounded below. Then, it is seen that the sequence (an) defined by an = −bnfor all n ∈ N is monotonically increasing and bounded above. Hence, by the firstpart of the theorem, an → a for some a ∈ R. Then, bn → b := −a.

Note that a convergent sequence need not be monotonically increasing or mono-tonically decreasing. For example, look at the sequence ((−1)n/n).

1.1.3 Subsequences

Definition 1.11 A sequence (bn) is called a subsequence of a sequence (an) ifthere is a strictly increasing sequence (kn) of natural numbers such that bn = aknfor all n ∈ N.

Thus, subsequences of a real sequence (an) are of the form (akn), where (kn) isa strictly increasing sequence natural numbers.


For example, given a sequence (an), the sequences (a2n), (a2n+1), (an2), (a2n) aresome of its subsequences. As concrete examples, (1/2n), and (1/(2n + 1)), (1/2n)are subsequences of (1/n).

A sequence may not converge, but it can have convergent subsequences. Forexample, we know that the sequence ((−1)n) diverges, but the subsequences (an)and (bn) defined by an = 1, bn = −1 for all n ∈ N are convergent subsequences of((−1)n).

However, we have the following result.

Theorem 1.6 If a sequence (an) converges to x, then all its subsequences convergeto the same limit x.


What about the converse of the above theorem? Obviously, if all subsequencesof a sequence (an) converge to the same limit x, then (an) also has to converge tox, as (an) is a subsequence of itself.

Suppose every subsequence of (an) has at least one subsequence which convergesto x. Does the sequence (an) converges to x? The answer is affirmative, as thefollowing theorem shows.

Theorem 1.7 If every subsequence of (an) has at least one subsequence which con-verges to x, then (an) also converges to x.

Proof. Proof is left as an exercise.

We have seen in Theorem 1.3 that every convergent sequence is bounded, but abounded sequence need not be convergent.

Question: For every bounded sequence, can we have a convergent subsequence?

The answer is in affirmative:

Theorem 1.8 (Bolzano-Weirstrass theorem). Every bounded sequence of realnumbers has a convergent subsequence.

Proof: For a first reading this proof can be omitted. 1 Let (an) be a bounded se-quence in R. For each k ∈ N, consider the set

Ek := an : n ≥ k,

and let bk := supEk, k ∈ N. Clearly, b1 ≥ b2 ≥ . . ..1From the book: Mathematical Analysis: a straight forward approach by K.G. Binmore.

We consider the following two mutually exclusive cases:

(i) For every k ∈ N, bk ∈ Ek

(ii) There exists k ∈ N such that bk 6∈ Ek.

In case (i), we may write (bk) is a monotonically decreasing subsequence of (an),and since (an) is bounded, (bk) converges.

Now, suppose that case (ii) holds, and let k ∈ N such that bk 6∈ Ek. Then forevery n ≥ k, there exists n′ > n such that an′ > an. Take n1 = k and n2 = n′.Then we have n1 < n2 and an1 < an2 . Again, since n2 ∈ Ek, there exists n3 > n2such that an21 < an3 . Continuing this way, we obtain an increasing subsequence(anj )

∞j=1. Again, since (an) is bounded, (anj )

∞j=1 is bounded, so that it converges.

Remark 1.4 We may observe that the proof of Theorem 1.8 can be slightly mod-ified so that we, in fact, have the following: Every sequence in R has a monotonicsubsequence.

Exercise 1.17 Prove the statement in italics in Remark 1.4. J

1.1.4 Further examples

EXAMPLE 1.7 Let a sequence (an) be defined as follows :

a1 = 1, an+1 =2an + 3

4, n = 1, 2, . . . .

We show that (an) is monotonically increasing and bounded above.

Note that

an+1 =2an + 3

4=an2

+3

4≥ an ⇐⇒ an ≤

3

2.

Thus it is enough to show that an ≤ 3/2 for all n ∈ N.

Clearly, a1 ≤ 3/2. If an ≤ 3/2, then an+1 = an/2 + 3/4 < 3/4 + 3/4 = 3/2.Thus, we have proved that an ≤ 3/2 for all n ∈ N. Hence, by Theorem 1.5, (an)converges. Let its limit be a. Then taking limit on both sides of an+1 = 2an+3

4 wehave

a =2a+ 3

4i.e., 4a = 2a+ 3 so that a =

3

2.

EXAMPLE 1.8 Let a sequence (an) be defined as follows :

a1 = 2, an+1 =1

2

(an +

2

an

), n = 1, 2, . . . .

It is seen that, if the sequence converges, then its limit would be√

2.

Since a1 = 2, one may try to show that (an) is monotonically decreasing andbounded below.

Note that

an+1 :=1

2

(an +

2

an

)≤ an ⇐⇒ a2n ≥ 2, i.e., an ≥

√2.

Clearly a1 ≥√

2. Now,

an+1 :=1

2

(an +

2

an

)≥√

2 ⇐⇒ a2n − 2√

2an + 2 ≥ 0,

i.e., if and only if (an−√

2)2 ≥ 0. This is true for every n ∈ N. Thus, (an) is mono-tonically decreasing and bounded below, so that by Theorem 1.5, (an) converges.

EXAMPLE 1.9 Consider the sequence (an) with an = (1+1/n)n for all n ∈ N. Weshow that (an) is monotonically increasing and bounded above. Hence, by Theorem1.5, (an) converges.

Note that

an =

(1 +

1

n

)n= 1 + n.

1

n+n(n− 1)

2!

1

n2+ · · ·+ n(n− 1) . . . 2.1

n!

1

nn

= 1 + 1 +1

2!

(1− 1

n

)+

1

3!

(1− 1

n

)(1− 2

n

)+

· · ·+ 1

n!

(1− 1

n

)· · ·(

1− n− 1

n

)≤ an+1.

Also

an = 1 + 1 +1

2!

n(n− 1)

n2+ · · ·+ 1

n!

n(n− 1) . . . 2.1

nn

≤ 1 + 1 +1

2!+

1

3!+ · · ·+ 1

n!

≤ 1 + 1 +1

2+

1

22+ · · ·+ 1

2n−1

< 3.

Thus, (an) is monotonically increasing bounded above.

Exercise 1.18 Show that (n1/n)∞n=3 is a monotonically decreasing sequence. J

In the four examples that follow, we shall be making use of Theorem 1.2 withoutmentioning it explicitly.

EXAMPLE 1.10 Consider the sequence (bn) with

bn = 1 +1

1!+

1

2!+

1

3!+ · · ·+ 1

n!∀n ∈ N.

Clearly, (bn) is monotonically increasing, and we have noticed in the last examplethat it is bounded above by 3. Hence, by Theorem 1.5, it converges.

Let (an) and (bn) be as in Examples 1.9 and 1.10 respectively, and let a and btheir limits. We show that a = b.

We have observed in last example that 2 ≤ an ≤ bn ≤ 3. Hence, taking limits, itfollows that a ≤ b. Notice that

an = 1 + 1 +1

2!

(1− 1

n

)+

1

3!

(1− 1

n

)(1− 2

n

)+

. . .+1

n!

(1− 1

n

). . .

(1− n− 1

n

).

Hence, for m,n with m ≤ n, we have

an ≥ 1 + 1 +1

2!

(1− 1

n

)+

1

3!

(1− 1

n

)(1− 2

n

)+

· · ·+ 1

m!

(1− 1

n

). . .

(1− m− 1

n

).

Taking limit as n→∞, we get (cf. Theorem 1.2 (c))

x ≥ 1 +1

1!+

1

2!+

1

3!+ · · ·+ 1

m!= bm.

Now, taking limit as m→∞, we get a ≥ b. Thus we have proved a = b.

The common limit in the above two examples is denoted by the letter e.

EXAMPLE 1.11 Let a > 0. We show that, if 0 < a < 1, then the sequence (an)converges to 0, and if a > 1, then (an) diverges to infinity.

(i) Suppose 0 < a < 1. Then we can write a = 1/(1 + r), r > 0, and we have

an = 1/(1 + r)n = 1/ (1 + nr + · · ·+ rn) < 1/(1 + nr)→ 0 as n→∞.

Hence, an → 0 as n→∞.

An alternate way: Let xn = an. Then (xn) is monotonically decreasing and boundedbelow by 0. Hence (xn) converges, to say x. Then xn+1 = an+1 = axn → ax. Hence,x = ax. This shows that x = 0.

(ii) Suppose a > 1. Then, since 0 < 1/a < 1, the sequence (1/an) converges to0, so that (an) diverges to infinity. (Why ?)

EXAMPLE 1.12 The sequence (n1/n) converges and the limit is 1.

Note that n1/n = 1 + rn for some sequence (rn) of positive reals. Then we have

n = (1 + rn)n ≥ n(n− 1)

2r2n,

so that r2n ≤ 2/(n− 1) for all n ≥ 2. Since 2/(n− 1)→ 0, by Theorem 1.2(c), thatrn → 0, and hence by Theorem 1.2(c), n1/n = 1 + rn → 1.

Remark 1.5 The proof of the result in Example 1.12 does not require the knowledgethat n1/n → 1, but uses the facts that 2/(n − 1) → 0 (which can be shown easily)and Theorem 1.2. But, if one is asked to show that n1/n → 1, then we can giveanother proof of the same by using only the definition, as follows:

Another proof without using Theorem 1.2. Let ε > 0 be given. To find n0 ∈ N suchthat n1/n − 1 < ε for all n ≥ n0. Note that

n1/n − 1 < ε ⇐⇒ n1/n < 1 + ε

⇐⇒ n < (1 + ε)n = 1 + nε+n(n− 1)

2ε2 + . . .+ εn.

Hencen1/n − 1 < ε if n > 1 + 2/ε2.

So, we may take any n0 ∈ N which satisfies n0 ≥ 2(1 + /ε2).

EXAMPLE 1.13 For any a > 0, (a1/n) converges to 1.

If a > 1, then we can write a1/n = 1+rn for some sequence (rn) of positive reals.Then we have

a = (1 + rn)n ≥ nrn so that rn ≤ a/n.

Since a/n → 0, by Theorem 1.2(c), rn → 0, and hence by Theorem 1.2(c), a1/n =1 + rn → 1.

In case 0 < a < 1, then 1/a > 1. Hence, by the first part, 1/a1/n = (1/a)1/n → 1,so that an → 1.

Exercise 1.19 Give another proof for the result in Example 1.13 without usingTheorem 1.2. J

EXAMPLE 1.14 Let (an) be a bounded sequence of non-negative real numbers.Then (1 + an)1/n → 1 as n→∞:

This is seen as follows: Let M > 0 be such that 0 ≤ an ≤M for all n ∈ N. Then,

1 ≤ (1 + an)1/n ≤ (1 +M)1/n ∀n ∈ N.

By Example 1.13, (1 +M)1/n → 0. Hence the result follows by making use of part(d) of Theorem 1.2.

EXAMPLE 1.15 As an application of some of the results discussed above, considerthe sequence (an) with an = (1+1/n)1/n, n ∈ N. We already know that lim

n→∞an = 1.

Now, another proof for the same:

Note that 1 ≤ an ≤ 21/n, and 21/n → 1 as n→∞.

EXAMPLE 1.16 Consider the sequence (an) with an = (1 + n)1/n. Then an → 1as n→∞. We give two proofs for this result.

(i) Observe that an = n1/n (1 + 1/n)1/n. We already know that n1/n → 1, and

(1 + 1/n)1/n → 1 as n→∞.

(ii) Observe that n1/n ≤ (1 + n)1/n ≤ (2n)1/n = 21/nn1/n, where n1/n → 1 and21/n → 1 as n→∞.

EXAMPLE 1.17 Suppose an > 0 for al n ∈ N such that limn→∞

an+1

an= ` < 1. We

show that an → 0.

Since limn→∞

an+1

an= ` < 1, there exists q such that ` < q < 1 and N ∈ N such

thatan+1

an≤ q for all n ≥ N . Hence,

0 ≤ an ≤ qn−NaN ∀ n ≥ N.

Now, since qn−N → 0 as n→∞, it follows that an → 0 as n→∞.

Exercise 1.20 Suppose Let an > 0 for al n ∈ N such that limn→∞

an+1

an= ` > 1.

Show that (an) diverges to ∞. J

EXAMPLE 1.18 Let 0 < a < 1. Then nan → 0 as n→∞.

To see this let an := nan for n ∈ N. Then we have

an+1

an=

(n+ 1)an+1

nan=

(n+ 1)a

n∀ n ∈ N.

Hence, limn→∞

an+1

an= a < 1. Thus, the result follows from the last example.

Exercise 1.21 Obtain the the result in Example 1.18 by using the arguments inExample 1.11. J

Remark 1.6 Suppose for each k ∈ N, a(k)n → 0, b

(k)n → 1 as n → ∞, and also

a(n)n → 0, b

(n)n → 1 as n→∞. In view of Theorems 1.2 and 1.4, one may think that

a(1)n + a(2)n + · · ·+ a(n)n → 0 as n→∞

and

b(1)n b(2)n · · · b(n)n → 1 as n→∞.

Unfortunately, that is not the case. To see this consider

a(k)n =k

n2, b(k)n = k1/n k, n ∈ N.

Then, for each k ∈ N, a(k)n → 0, b

(k)n → 1 as n→∞, and also a

(n)n → 0, b

(n)n → 1 as

n→∞. But,

a(1)n + a(2)n + · · ·+ a(n)n =1

n2+

2

n2+ · · ·+ n

n2=n+ 1

2n→ 1

2as n→∞

andb(1)n b(2)n · · · b(n)n = 11/n21/n · · ·n1/n = (n!)1/n 6→ 1 as n→∞.

In fact, for any k ∈ N, if n ≥ k, then

(n!)1/n ≥ (k!)1/n(kn−k)1/n = (k!)1/nk1−k/n = k( k!

kk

)1/n.

Since for any given k ∈ N, k(k!kk

)1/n→ k as n→∞, it follows that (n!)1/n 6→ 1. In

fact, there exists n0 ∈ N such that

(n!)1/n ≥ k

2∀ n ≥ maxn0, k.

Thus, the sequence is((n!)1/n

)is unbounded.

EXAMPLE 1.19 Let an = (n!)1/n2, n ∈ N. Then an → 1 as n→∞. We give two

proofs for this.

(i) Note that, for every n ∈ N,

1 ≤ (n!)1/n2 ≤ (nn)1/n

2= n1/n.

Now, since n1/n → 1, we have (n!)1/n2 → 1.

(ii) By GM-AM inequality, for n ∈ N,

(n!)1/n = (1.2. . . . .n)1/n ≤ 1 + 2 + . . .+ n

n=n+ 1

2≤ n.

Thus,1 ≤ (n!)1/n

2 ≤ n1/n.

Since n1/n → 1 we have (n!)1/n2 → 1.

1.1.5 Cauchy sequence

Theorem 1.9 If a real sequence (an) converges, then for every ε > 0, there existsN ∈ N such that

|an − am| < ε ∀n,m ≥ N.

Proof. Suppose an → a as n → ∞, and let ε > 0 be given. Then we know thatthere exists N ∈ N such that |an − a| < ε/2 for all n ≥ N . Hence, we have

|an − am| ≤ |an − a|+ |a− am| < ε ∀n,m ≥ N.

This completes the proof.

Definition 1.12 A a sequence (an) is said to be a Cauchy sequence if for everyε > 0, there exists N ∈ N such that

|an − am| < ε ∀n,m ≥ N.

Theorem 1.9 show that every convergent sequence is a Cauchy sequence. Inparticular, if (an) is not a Cauchy sequence, then (an) does not converge to anya ∈ R. Thus, Theorem 1.9 may help us to show that certain sequence is notconvergent. For example, see the follwing exercise.

Exercise 1.22 Let sn = 1 + 12 + . . . + 1

n for n ∈ N. Then show that (sn) is not aCauchy sequence. [Hint: For any n ∈ N, note that s2n − sn ≥ 1

2 .] J

Let us observe certain properties of Cauchy sequences.

Theorem 1.10 Let (an) be a Cauchy sequence. The we have the following.

(i) (an) is a bounded sequence.

(ii) If (an) has a convergent subsequence with limit a, then the sequence (an)itself will converge to a.

Proof. Let (an) be Cauchy sequence and ε > 0 be given.

(i) Let N ∈ N be such that |an − am| < ε for all n,m ≥ N . In particular,

|an| ≤ |an − aN |+ |aN | ≤ ε+ |aN | ∀n ≥ N.

This proves (i).

(ii) Suppose (ank) is a convergent subsequence of (an), and let a = limk→∞

ank . Let

n0 ∈ N be such that |ank − a| < ε for all k ≥ n0. Hence, we have

|ak − a| ≤ |ak − ank |+ |ank − a| < 2ε ∀ k ≥ maxN,n0.

Thus, an → a as n→∞.

In fact, the converse of Theorem 1.9 also holds:

Theorem 1.11 Every Cauchy sequence of real numbers converges.

Proof of Theorem 1.11. Let (an) be a Cauchy sequence. By Theorem 1.10(i),(an) is bounded. Then, by Bolzano-Weierstrass theorem (Theorem 1.8), (an) has aconvergent subsequence (ank). Let a = lim

k→∞ank . Now, by Theorem 1.10(ii), an → a

as n→∞.

An alternate proof without using Bolzano-Weierstrass theorem (Theorem 1.8). Let(xn) be Cauchy sequence, and ε > 0 be given. Then there exists nε ∈ N such that|xn − xm| < ε for all n,m ≥ nε. In particular, |xn − xnε | < ε for all n ≥ nε. Hence,for any ε1, ε2 > 0,

xnε1 − ε1 < xn < xnε2 + ε2 ∀ n ≥ maxnε1 , nε2. (∗)

From this, we see that the set xnε − ε : ε > 0 is bounded above, and

x := supxnε − ε : ε > 0

satisfies

xnε − ε < x < xnε + ε ∀ n ≥ nε. (∗∗)

From (∗) and (∗∗), we obtain

|xn − x| < ε ∀ n ≥ nε.


Here is an example of a general nature.

EXAMPLE 1.20 Let (an) be a sequence of real numbers. Suppose there exists apositive real number ρ < 1 such that

|an+1 − an| ≤ ρ|an − an−1| ∀ n ∈ N, n ≥ 2.

Then (an) is a cauchy sequence. To see this first we observe that

|an+1 − an| ≤ ρn−1|a2 − a1| ∀ n ∈ N, n ≥ 2.

Hence, for n > m,

|an − am| ≤ |an − an−1|+ . . .+ |am+1 − am|≤ (ρn−2 + . . .+ ρm−1)|a2 − a1|≤ ρm−1(1 + ρ+ . . .+ ρn−m−3)|a2 − a1|

≤ ρm−1

1− ρ|a2 − a1|.

Since ρm−1 → 0 as m→∞, given ε > 0, there exists N ∈ N such that |an−am| < εfor all n,m ≥ N .

Exercise 1.23 Given a, b ∈ R and 0 < λ < 1, let (an) be a sequence of real numbersdefined by a1 = a, a2 = b and

an+1 = (1 + λ)an − λan−1 ∀ n ∈ N, n ≥ 2.

Show that (an) is a Cauchy sequence and its limit is (b+ λa)/(1− λ). J

Exercise 1.24 Suppose f is a function defined on an interval J . If there exists0 < ρ < 1 such that

|f(x)− f(y)| ≤ ρ|x− y| ∀ x, y ∈ J,

then the for any a ∈ J , the sequence (an) defined by

a1 = f(a), an+1 := f(an) ∀ n ∈ N,

is a Cauchy sequence. Show also that the limit of the sequence (an) is independentof the choice of a. J

1.1.6 Additional exercises

1. Prove that xn −→ l as n→∞ if and only if |xn − l| → 0 as n −→ ∞, and inthat case |xn| − |l| −→ 0 as n→∞.

2. Establish convergence or divergence of the sequence (an) each of the followingsequences, where an is:

(i)n

n+ 1(ii)

(−1)nn

n+ 1(iii)

2n

3n2 + 1, (iv)

2n2 + 3

3n2 + 1.

3. Suppose (an) is a real sequence such that an → 0 as n → ∞. Show thefollowing:

(a) The sequence (a2n) converges to 0.

(b) If an > 0 for all n, then the sequence (1/an) diverges to infinity.

4. Let xn be a sequence defined recursively by xn+2 = xn+1 + xn, n ≥ 1 withx1 = x2 = 1. Show that xn diverges to ∞.

5. Let xn =√n+ 1 −

√n for n ∈ N. Show that xn, nxn and

√nxn are

convergent. Find their limits.

6. Let xn =∑n

k=11

n+k , n ∈ N. Show that xn is convergent.

7. If (an) converges to x and an ≥ 0 for all n ∈ N, then show that x ≥ 0 and(√an) converges to

√x.

8. Prove the following:

(a) If xn is increasing and unbounded, then xn −→ +∞ as n −→∞.

(b) If xn is decreasing and unbounded, prove that xn −→ −∞ as n −→∞.

9. Let a1 = 1, an+1 =√

2 + an for all n ∈ N. Show that (an) converges. Also,find its limit.

10. Let a1 = 1 and an+1 = 14(2an + 3) for all n ∈ N. Show that an is monotoni-

cally increasing and bounded above. Find its limit.

11. Let a1 = 1 and an+1 =an

1 + anfor all n ∈ N. Show that an converges. Find

its limit.

12. Prove that if (an) is a Cauchy sequence having a subsequence which convergesto a, then (an) itself converges to a.

13. Suppose (an) is a sequence such that the subsequences (a2n−1) and (a2n) con-verge to the same limit, say a. Show that (an) also converges to a.

14. Let xn be a monotonically increasing sequence such that x3n is bounded.Is xn convergent? Justify your answer.

15. Show that, if a sequence (an) has the property that an+1− an → 0 as n→∞,then (an) need not converge.

16. Let a1 = 1 and an+1 = 14(2an + 3) for all n ∈ N. Show that (an) is monotoni-

cally increasing and bounded above. Find its limit.

17. Give an example in support of the statement: If a sequence an has theproperty that an+1 − an → 0 as n→∞, then an need not converge.

18. Let a > 0 and xn = an−a−nan+a−n , n ∈ N. Discuss the convergence of the sequence

(xn).

19. If 0 < a < b and an = (an+bn)1/n for all n ∈ N, then show that (an) convergesto b. [Hint: Note that (an + bn)1/n = b(1 +

(ab

)n)1/n.]

20. If 0 < a < b and an+1 = (anbn)1/2 and bn+1 = an+bn2 for all n ∈ N with a1 = a,

b1 = b, then show that (an) and (bn) converge to the same limit. [Hint: Firstobserve that an ≤ an+1 ≤ bn+1 ≤ bn for all n ∈ N.]

21. Let a1 = 1/2 and b = 1, and let an+1 = (anbn)1/2 and bn+1 = 2anbnan+bn

for alln ∈ N. Show that (an) and (bn) converge to the same limit. [Hint: Firstobserve that bn ≤ bn+1 ≤ an+1 ≤ an for all n ∈ N.]

22. Let a1 = 1/2, b = 1, an = (an−1bn−1)1/2 and 1

bn= 1

2( 1an

+ 1bn−1

). Prove that

an−1 < an < bn < bn−1 for all n ∈ N. Deduce that both the sequences anand bn converge to the same limit α, 1/2 < α < 1.

23. Prove that if an is a Cauchy sequence having a subsequence which convergesto a, then an itself converges to a.

Series of Real Numbers 21

1.2 Series of Real Numbers

Definition 1.13 A series of real numbers is an expression of the form

a1 + a2 + a3 + . . . ,

or more compactly as∑∞

n=1 an, where (an) is a sequence of real numbers.

The number an is called the n-th term of the series and the sequence sn :=n∑i=1

ai

is called the n-th partial sum of the series∑∞

n=1 an.

1.2.1 Convergence and divergence of series

Definition 1.14 A series∑∞

n=1 an is said to converge (to s ∈ R) if the sequencesn of partial sums of the series converge (to s ∈ R).

If∑∞

n=1 an converges to s, then we write∑∞

n=1 an = s.

A series which does not converge is called a divergent series.

A necessary condition

Theorem 1.12 If∑∞

n=1 an converges, then an → 0 as n→∞. Converse does nothold.

Proof. Clearly, if sn is the n-th partial sum of the convergent series∑∞

n=1 an,then

an = sn − sn−1 → 0 as n→∞.

To see that the converse does not hold it is enough to observe that the series∑∞n=1 an with an = 1

n for all n ∈ N diverges whereas an → 0.

The proof of the following corollary is immediate from the above theorem.

Corollary 1.13 Suppose (an) is a sequence of positive terms such that an+1 > anfor all n ∈ N. Then the series

∑∞n=1 an diverges.

The above theorem and corollary shows, for example, that the series∑∞

n=1nn+1

diverges.

EXAMPLE 1.21 We have seen that the sequence (sn) with sn =∑n

k=11k! con-

verges. Thus, the series∑∞

n=11n! converges. Also, we have seen that the sequence

(σn) with σn =∑n

k=11k diverges. Hence,

∑∞n=1

1n diverges.

EXAMPLE 1.22 Consider the geometric series series∑∞

n=1 aqn−1, where a, q ∈ R.

Note that sn = a+ aq + . . .+ aqn−1 for n ∈ N. Clearly, if a = 0, then sn = 0 for alln ∈ N. Hence, assume that a 6= 0. Then we have

sn =

na if q = 1,a(1−qn)

1−q if q 6= 1.

Thus, if q = 1, then (sn) is not bounded; hence not convergent. If q = −1, then wehave

sn =

a if n odd,0 if n even.

Thus, (sn) diverges for q = −1 as well. Now, assume that |q| 6= 1. In this case, wehave ∣∣∣∣sn − a

1− q

∣∣∣∣ =|a||1− q|

|q|n.

This shows that, if |q| < 1, then (sn) converges to a1−q , and if |q| > 1, then (sn) is

not bounded, hence diverges.

Theorem 1.14 Suppose (an) and (bn) are sequences such that for some k ∈ N,an = bn for all n ≥ k. Then

∑∞n=1 an converges if and only if

∑∞n=1 bn converges.

Proof. Suppose sn and σn be the n-th partial sums of the series∑∞

n=1 an and∑∞n=1 bn respectively. Let α =

∑ki=1 ai and β =

∑ki=1 bi. Then we have

sn − α =

n∑i=k+1

ai =

n∑i=k+1

bi = σn − β ∀ n ≥ k.

From this it follows that the sequence (sn) converges if and only if (σn) converges.

From the above theorem it follows if∑∞

n=1 bn is obtained from∑∞

n=1 an byomitting or adding a finite number of terms, then

∞∑n=1

an converges ⇐⇒∞∑n=1

bn converges.

In particular,∞∑n=1

an converges ⇐⇒∞∑n=1

an+k converges

for any k ∈ N.

The proof of the following theorem is left as an exercise.

Theorem 1.15 Suppose∑∞

n=1 an converges to s and∑∞

n=1 bn converges to σ. Thenfor every α, β ∈ R,

∑∞n=1(αan + β bn) converges to α s+ β σ.

1.2.2 Some tests for convergence

Theorem 1.16 (Comparison test) Suppose (an) and (bn) are sequences of non-negative terms, and an ≤ bn for all n ∈ N. Then,

(i)∑∞

n=1 bn converges =⇒∑∞

n=1 an converges,

(ii)∑∞

n=1 an diverges =⇒∑∞

n=1 bn diverges.

Proof. Suppose sn and σn be the n-th partial sums of the series∑∞

n=1 an and∑∞n=1 bn respectively. By the assumption, we get 0 ≤ sn ≤ σn for all n ∈ N, and

both (sn) and (σn) are monotonically increasing.

(i) Since (σn) converges, it is bounded. Let M > 0 be such that σn ≤M for alln ∈ N. Then we have sn ≤M for all n ∈ N. Since (sn) are monotonically increasing,it follows that (sn) converges.

(ii) Proof of this part follows from (i) (How?).

Corollary 1.17 Suppose (an) and (bn) are sequences of positive terms.

(a) Suppose ` := limn→∞anbn

exists. Then we have the following:

(i) If ` > 0, then∑∞

n=1 bn converges ⇐⇒∑∞

n=1 an converges.

(ii) If ` = 0, then∑∞

n=1 bn converges ⇒∑∞

n=1 an converges.

(b) Suppose limn→∞

anbn

=∞. Then∑∞

n=1 an converges ⇒∑∞

n=1 bn converges.

Proof. (a) Suppose limn→∞anbn

= `.

(i) Let ` > 0. Then for any ε > 0 there exists n ∈ N such that `− ε < anbn< `+ ε

for all n ≥ N . Equivalently, (`− ε)bn < an < (`+ ε)bn for all n ≥ N . Had we takenε = `/2, we would get `

2bn < an <3`2 bn for all n ≥ N . Hence, the result follows by

comparison test.

(ii) Suppose ` = 0. Then for ε > 0, there exists n ∈ N such that −ε < anbn< ε

for all n ≥ N . In particular, an < εbn for all n ≥ N . Hence, we get the result byusing comparison test.

(b) By assumption, there exists N ∈ N such that anbn≥ 1 for all n ≥ N . Hence the

result follows by comparison test.

EXAMPLE 1.23 We have already seen that the sequence (sn) with sn =∑n

k=11k!

converges. Here is another proof for the same fact: Note that 1n! ≤

12n−1 for all

n ∈ N. Since∑∞

n=11

2n−1 converges, it follows from the above theorem that∑∞

n=11n!

converges.

EXAMPLE 1.24 Since 1√n≥ 1

n for all n ∈ N, and since the series∑∞

n=11n diverges,

it follows from the above theorem that the series∑∞

n=11√n

also diverges.

Theorem 1.18 (de’Alembert’s ratio test) Suppose (an) is a sequence of positiveterms such that limn→∞

an+1

an= ` exists. Then we have the following:

(i) If ` < 1, then the series∑∞

n=1 an converges.

(ii) If ` > 1, then the series∑∞

n=1 an diverges.

Proof. (i) Suppose ` < q < 1. Then there exists N ∈ N such that

an+1

an< q ∀ n ≥ N.

In particular,an+1 < q an < q2an−1 < . . . < qn a1, ∀ n ≥ N.

Since∑∞

n=1 qn converges, by comparison test,

∑∞n=1 an also converges.

(ii) Let 1 p > 1n ≥ N.

From this it follows that (an) does not converge to 0. Hence∑∞

n=1 an diverges.

Theorem 1.19 (Cauchy’s root test) Suppose (an) is a sequence of positive termssuch that limn→∞ an

1/n = ` exists. Then we have the following:

(i) If ` < 1, then the series∑∞

n=1 an converges.

(ii) If ` > 1, then the series∑∞

n=1 an diverges.

Proof. (i) Suppose ` < q < 1. Then there exists N ∈ N such that

an1/n < q ∀ n ≥ N.

Hence, an < qn for all n ≥ N . Since the∑∞

n=1 qn converges, by comparison test,∑∞

n=1 an also converges.

(ii) Let 1 p > 1 ∀ n ≥ N.

Hence, an ≥ 1 for all n ≥ N . Thus, (an) does not converge to 0. Hence,∑∞

n=1 analso diverges.

Remark 1.7 We remark that both d’Alembert’s test and Cauchy test are silentfor the case ` = 1. But, for such case, we may be able to infer the convergence ordivergence by some other means.

EXAMPLE 1.25 For every x ∈ R, the series∑∞

n=1xn

n! converges:

Here, an = xn

n! . Hence

an+1

an=

x

n+ 1∀ n ∈ N.

Hence, it follows that limn→∞an+1

an= 0, so that by d’Alemberts test, the series

converges.

EXAMPLE 1.26 The series∑∞

n=1

(n

2n+1

)nconverges: Here

an1/n =

n

2n+ 1→ 1

2< 1.

Hence, by Cauchy’s test, the series converges.

EXAMPLE 1.27 Consider the series∑∞

n=11

n(n+1) . In this series, we see that

limn→∞an+1

an= 1 = limn→∞ an

1/n. However, the n-th partial sum sn is given by

sn =n∑k=1

1

k(k + 1)=

n∑k=1

(1

k− 1

k + 1

)= 1− 1

n+ 1.

Hence sn converges to 1.


n=11n2 . In this case, we see that

1

(n+ 1)2≤ 1

n(n+ 1)∀n ∈ N.

Since∑∞

n=11

n(n+1) converges, by comparison test, the given series also converges.

EXAMPLE 1.29 Since∑∞

n=11n2 converges and

∑∞n=1

1n diverges, by comparison

test, we see that the series∑∞

n=11np converges for p ≥ 2 and diverges for p ≤ 1.


n=11np for p > 1. To discuss this example,

consider the function f(x) := 1/xp, x ≥ 1. Then we see that for each k ∈ N,

k − 1 ≤ x ≤ k =⇒ 1

kp≤ 1

xp=⇒ 1

kp≤∫ k

k−1

dx

xp.

Hence,n∑k=2

1

kp≤

n∑k=2

∫ k

k−1

dx

xp=

∫ n

1

dx

xp=n1−p − 1

1− p≤ 1

p− 1.

Thus,

sn :=

n∑k=1

1

kp≤ 1

p− 1+ 1.

Hence, (sn) is monotonically increasing and bounded above. Therefore, (sn) con-verges.


Now, we have a more general result.

Theorem 1.20 (Integral test) Suppose f(x) is a continuous, non-negative anddecreasing function for x ∈ [1,∞). For each n ∈ N, let an :=

∫ n1 f(x)dx. Then

∞∑n=1

f(n) converges ⇐⇒ (an) converges.

Proof. First we observe that an =∫ n1 f(x) dx =

∑nk=2

∫ kk−1 f(x) dx. Now, since

f(x) is a decreasing function for x ∈ [1,∞), we have for each k ∈ N,

k − 1 ≤ x ≤ k =⇒ f(k) ≤ f(x) ≤ f(k − 1).

Hence, for k = 2, 3, . . .,

f(k) ≤∫ k

k−1f(x) dx ≤ f(k − 1)

so thatn∑k=2

f(k) ≤n∑k=2

∫ k

k−1f(x) dx ≤

n∑k=2

f(k − 1).

Thus,n∑k=2

f(k) ≤∫ n

1f(x) dx ≤

n∑k=2

f(k − 1).

Now, let sn :=

n∑k=1

f(k) for n ∈ N. Then from the above inequalities, together

with the fact that (sn) is a monotonically increasing sequence, it follows that (an)converges if and only if (sn) converges.

1.2.3 Alternating series

Definition 1.15 A series of the form∑∞

n=1(−1)n+1un where (un) is a sequence ofpositive terms is called an alternating series.

Theorem 1.21 (Leibniz’s theorem) Suppose (un) is a sequence of positive termssuch that un ≥ un+1 for all n ∈ N, and un → 0 as n → ∞. Then the alternatingseries

∑∞n=1(−1)n+1un converges.

Proof. Let sn be the n-th partial sum of the alternating series∑∞

n=1(−1)n+1un.We observe that

s2n+1 = s2n + u2n+1 ∀ n ∈ N.

Since un → 0 as n → ∞, it is enough to show that (s2n) converges (Why?). Notethat

s2n = (u1 − u2) + (u3 − u4) + . . .+ (u2n−1 − u2n),


s2n = u1 − (u2 − u3)− . . . (u2n−2 − u2n−1)− u2nfor all n ∈ N. Since ui − ui+1 ≥ 0 for each i ∈ N, (s2n) is monotonically increasingand bounded above. Therefore (s2n) converges. In fact, if s2n → s, then we haves2n+1 = s2n + u2n+1 → s, and hence sn → s as n→∞.

By the above theorem the series∑∞

n=1(−1)n+1

n converges. Likewise, the series∑∞n=1

(−1)n+1

2n−1 and∑∞

n=1(−1)n+1

2n also converge.

Remark 1.8 The series∑∞

n=1(−1)n+1

2n−1 appear in the work of a Kerala mathemati-cian Madhava presented in the year around 1550 by another Kerala mathematicianNilakantha. The discovery of the above series is normally attributed to Leibniz andJames Gregory after nearly 300 years of its discovery.

Suppose (un) is as in Leibniz’s theorem (Theorem 1.21), and let s ∈ R be such

that sn → s, where sn is the nth partial sum of∞∑n=1

(−1)n+1un.

How fast (sn) converges to s?

In the proof of Theorem 1.21, we have shown that s2n is a monotonicallyincreasing sequence. Similarly, it can be shown that s2n−1 is a monotonicallydecreasing sequence.

Since (s2n−1) is monotonically decreasing and (s2n) is monotonically increasing,we have

s2n−1 = s2n + u2n ≤ s+ u2n, s ≤ s2n+1 = s2n + u2n+1.

Thus,s2n−1 − s ≤ u2n, s− s2n ≤ u2n+1.

Consequently,|s− sn| ≤ un+1 ∀ n ∈ N.

1.2.4 Absolute convergence


n=1 an is said to converge absolutely, if∑∞

n=1 |an|converges.

Theorem 1.22 Every absolutely convergent series converges.

Proof. Suppose∑∞

n=1 an is an absolutely convergent series. Let sn and σn bethe n-th partial sums of the series

∑∞n=1 an and

∑∞n=1 |an| respectively. Then, for

n > m, we have

|sn − sm| =∣∣∣ n∑j=m+1

an

∣∣∣ ≤ n∑j=m+1

|an| = |σn − σm|.


Since, σn converges, it is a Cauchy sequence. Hence, form the above relation itfollows that sn is also a Cauchy sequence. Therefore, by the Cauchy criterion, itconverges.

Another proof without using Cauchy criterion. Suppose∑∞

n=1 an is an absolutelyconvergent series. Let sn and σn be the n-th partial sums of the series

∑∞n=1 an and∑∞

n=1 |an| respectively. Then it follows that

sn + σn = 2pn,

where pn is the sum of all positive terms from a1, . . . , an. Since σn converges,it is bounded, and since pn ≤ σn for all n ∈ N, the sequence pn is also bounded.Moreover, pn is monotonically increasing. Hence pn converge as well. Thus,both σn, pn converge. Now, since sn = 2pn − σn for all n ∈ N, the sequencesn also converges.


n=1 an is said to converge conditionally if∑∞

n=1 anconverges, but

∑∞n=1 |an|diverges.

EXAMPLE 1.31 We observe the following:

(i) The series∑∞

n=1(−1)n+1

n is conditionally convergent.

(ii) The series∑∞

n=1(−1)n+1

n2 is absolutely convergent.

(iii) The series∑∞

n=1(−1)n+1

n! is absolutely convergent.

EXAMPLE 1.32 For any α ∈ R, the series∑∞

n=1sin(nα)n2 is absolutely convergent:

Note that ∣∣∣∣sin(nα)

n2

∣∣∣∣ ≤ 1

n2∀ n ∈ N.

Since∑∞

n=11n2 converges, by comparison test,

∑∞n=1

∣∣∣ sin(nα)n2

∣∣∣ also converges.

EXAMPLE 1.33 The series

∞∑n=3

(−1)n log n

n log(log n)is conditionally convergent. To see

this, let un =log n

n log(log n). Since n ≥ log n ≥ log(log n) we have

1

n≤ log n

n log(log n)≤ 1

log(log n)(∗)

so that un → 0. It can be easily seen that un+1 ≤ un. Hence, by Leibnitz theorem,the given series converges. Inequality (∗) also shows that the series

∑∞n=3 un does

not converge.

Here are two more results whose proofs are based on some advanced topics inanalysis



n=1 an is an absolutely convergent series and (bn) is asequence obtained by rearranging the terms of (an). Then

∑∞n=1 bn is also absolutely

convergent, and∑∞

n=1 an =∑∞

n=1 bn.


n=1 an is a conditionally convergent series. The forevery α ∈ R, there exists a sequence (bn) whose terms are obtained by rearrangingthe terms of (an) such that

∑∞n=1 bn = α.

To illustrate the last theorem consider the conditionally convergent series∞∑n=1

(−1)n+1

n. Consider the following rearrangement of this series:

1− 1

2− 1

4+

1

3− 1

6− 1

8+ · · ·+ 1

2k − 1− 1

4k − 2− 1

4k+ · · · .

Thus, if an = (−1)n+1

n for all n ∈ N, the rearranged series is∑∞

n=1 bn, where

b3k−2 =1

2k − 1, b3k−1 =

1

4k − 2, b3k =

1

4k

for k = 1, 2, . . .. Let sn and σn be the n-th partial sums of the series∑∞

n=1 an and∑∞n=1 bn respectively. Then we see that

σ3k = 1− 1

2− 1

4+

1

3− 1

6− 1

8+ · · ·+ 1

2k − 1− 1

4k − 2− 1

4k

=

(1− 1

2− 1

4

)+

(1

3− 1

6− 1

8

)+ · · ·+

(1

2k − 1− 1

4k − 2− 1

4k

)=

(1

2− 1

4

)+

(1

6− 1

8

)+ · · ·+

(1

4k − 2− 1

4k

)=

1

2

[(1− 1

2

)+

(1

3− 1

4

)+ · · ·+

(1

2k − 1− 1

2k

)]=

1

2s2k.

Also, we have

σ3k+1 = σ3k +1

2k + 1, σ3k+2 = σ3k +

1

2k + 1− 1

4k + 2.

We know that sn converge. Let limn→∞ sn = s. Since, an → 0 as n→∞, it thenfollows that

limk→∞

σ3k =s

2, lim

k→∞σ3k+1 =

s

2, lim

k→∞σ3k+2 =

s

2.

Hence, we can infer that σn → s/2 as n→∞.



(These problems were prepared for the students of MA1010, Aug-Nov, 2010.)

1. Using partial fractions, prove that∞∑n=1

3n− 2

n(n+ 1)(n+ 2)= 1

2. Test the following series for convergence:

(a)∞∑n=1

(n!)2

(2n)!(b)

∞∑n=1

(n!)2

(2n)!5n (c)

∞∑n=1

(n

n+ 1

)n2

(d)

∞∑n=1

(n

n+ 1

)n2

4n (e)

∞∑n=1

1

n

(√n+ 1−

√n)

(f)

∞∑n=1

(−1)(n−1)√n

(g)∞∑n=1

1

n2 + 2n+ 3(h)

1

2+

2

22+

3

23+ ...+

n

2n+ ...

(i)1√10

+1√20

+1√30

+ ...+1√10n

+ ... (j) 2 +3

2+

4

3+ ...+

n+ 1

n+ ...

(k)13√

7+

13√

8+

13√

9+ ...+

13√n+ 6

+ ...

(l)1

2+

(2

3

)4

+

(3

4

)9

+...+

(n

n+ 1

)n2

+... (m)1

2+

2

5+

3

10+...+

n

n2 + 1+...

3. Is the Leibniz Theorem applicable to the series:

1√2− 1

− 1√2 + 1

+1√

3− 1− 1√

3 + 1+ ...+

1√n− 1

− 1√n+ 1

+ ...

Does the above series converge?

Justify your answer.

4. Find out whether (or not) the following series converge absolutely or condi-tionally:

(a) 1− 1

32+

1

52− 1

72+

1

92+ ...+ (−1)n+1 1

(2n− 1)2+ ...

(b)1

ln 2− 1

ln 3+

1

ln 4− 1

ln 5+ ...+ (−1)n

1

lnn+ ...

5. Find the sum of the series:1

1.2.3+

1

2.3.4+ ...+

1

n.(n+ 1).(n+ 2)+ ...

6. Test for the convergence of the following series:

(a)

∞∑n=1

1

(n+ 1)√n

(b)

∞∑n=1

1

n!(c)

∞∑n=1

(√n4 + 1−

√n4 − 1

)


(d)

∞∑n=1

1

(a+ n)p(b+ n)p, a, b, p, q > 0 (e)

∞∑n=1

tan

(1

n

)

(f)∞∑n=1

(3√n3 + 1− n

)(g)

∞∑n=1

√2n!

n!(h)

∞∑n=1

1

n(n+ 1)

(i)

∞∑n=1

(n+ 6)−1/3 (j)

∞∑n=1

(log n)−n (k)

∞∑n=1

(1 +

1

n

)n2

(l)

∞∑n=1

(nx

1 + n

)n(m)

∞∑n=1

(−1)n−1n

n+ 1(n)

∞∑n=1

(−1)n−1√n(n+ 1)(n+ 2)

(o)∞∑n=1

(−1)n+1(√n+ 1−

√n)

(p)∞∑n=1

(−2)n

n2

7. Examine the following series for absolute / conditional convergence:

(a)∞∑n=1

(−1)n+1

(2n− 1)2(b)

∞∑n=1

(−1)n+1n

3n−1(c)

∞∑n=1

(−1)n

n(log n)2

(d)∞∑n=1

(−1)n log n

n log log n(e)

∞∑n=1

(−1)n+1 1

n(f)

∞∑n=1

(−1)n+1 1

n2n

8. Let (an) be a sequence of non-negative numbers and (akn) be a subsequence(an). Show that, if

∑∞n=1 an converges, then

∑∞n=1 akn also converges. Is the

converse true? Why?

2

Limit, Continuity and Differentiabilityof Functions

In this chapter we shall study limit and continuity of real valued functions definedon certain sets.

2.1 Limit of a Function

Suppose f is a real valued function defined on a subset D of R. We are going todefine limit of f(x) as x ∈ D approaches a point a, not necessarily in D. First wehave to be clear about what “x ∈ D approaches a point a” means.

2.1.1 Limit point of a set D ⊆ R

Definition 2.1 Let D ⊆ R and a ∈ R. Then a is said to be a limit point of D ifthere exists a sequence (an) in D which is not eventually constant such that an → aas n→∞.

We may recall that a sequence (an) is said to be eventually constant if thereexists k ∈ N such that an = ak for all n ≥ k.

Theorem 2.1 A point a ∈ R is a limit point of D ⊂ R if and only if for everyδ > 0,

(a− δ, a+ δ) ∩ (D \ a) 6= ∅.

Proof. Suppose a ∈ R is a limit point of D. Then we know that there exists asequence (an) in D which is not eventually constant such that an → a. Hence, forevery δ > 0, there exists N ∈ N such that an ∈ (a − δ, a + δ) for all n ≥ N . Inparticular, there exists n ≥ N such that an ∈ (a− δ, a+ δ) ∩ (D \ a).

Conversely, suppose for every δ > 0, (a−δ, a+δ)∩(D\a) 6= ∅. Then, for eachn ∈ N, taking δ = 1/n, there exists an ∈ D \ a such that an ∈ (a− 1/n, a+ 1/n).Hence, 0 ≤ |an − a| < 1/n for all n ∈ N, showing that an → a. Clearly, (an) is noteventually constant.

32

Limit of a Function 33

EXAMPLE 2.1 (i) Every point in an interval is its limit point.

(ii) If I is an open interval of finite length, then both the end points of I arelimit points of I.

(iii) If D = x ∈ R : 0 < |x| < 1, then every point in the interval [−1, 1] is alimit point of D.

Remark 2.1 (i) For a ∈ R, an open interval of the form (a − δ, a + δ) for someδ > 0 is called a neighbourhood of a, or sometimes, a δ-neighbourhood of a.

(ii) By a deleted neighbourhood of a point a ∈ R we mean a set of the formDδ := x ∈ R : 0 < |x− a| < δ for some δ > 0, i.e., the set (a− δ, a+ δ) \ a.

With the terminologies in the above remark,

• a point a ∈ R is a limit point of D ⊆ R if and only if every deleted neighbour-hood of a contains at least one point of D.

Now, we define limit of f(x) as x approaches a.

2.1.2 Limit of a function f(x) as x approaches a

Definition 2.2 Let f be a real valued function defined on a set D ⊆ R, andlet a ∈ R be a limit point of D. We say that f(x) has the limit b ∈ R as xapproaches a if for every ε > 0, there exists δ > 0 such that

|f(x)− b| < ε whenever x ∈ D, 0 < |x− a| < δ,

and in that case we write

limx→a

f(x) = b or f(x)→ b as x→ a.

Thus, limx→a f(x) = b if and only if for every open interval Ib containing b thereexists an open interval Ia containing a such that

x ∈ Ia ∩ (D \ a) =⇒ f(x) ∈ Ib.

In the following, whenever we talk about limit of a function f as xapproaches a ∈ R, we assume that f is defined on a set D ⊆ R and a isa limit point of D.

Exercise 2.1 Show that, if limit of a function, if exists, is unique. J

Exercise 2.2 Show that, if limx→a

f(x) = b, then there exists a deleted neighbourhood

Dδ of a and M > 0 such that |f(x)| ≤M for all x ∈ Dδ. J

34 Limit, Continuity and Differentiability of Functions M.T. Nair

Remark 2.2 Suppose f is a real valued function defined on an interval I and a ∈ I.What do we mean by the statement that

“ limx→a

f(x) does not exist”?

If you see the definitiion of existence of a limit, then you see that, the above state-ment means the following:

For any b ∈ R, there exists ε > 0 such that for any δ > 0, there is atleastone xδ ∈ (a− δ, a+ δ) such that f(xδ) 6∈ (b− ε, b+ ε).

How do we show, by first principle, that is, from the definition, that a function fdoes not have a limit as x approaches to a particular point x0?

Assume that the limx→x0 f(x) exists and it is equal to β for some β ∈ R.Then we know that for every ε > 0, there exists δ := δε such that

0 < |x− x0| < δε =⇒ |f(x)− β| < ε.

Then, for a particular choice of ε, say ε = ε0, one must be able to findan x′ such that |x′ − x0| < δε0 but |f(x′)− β| ≥ ε0.

We illustrate the above remark by a simple example.

EXAMPLE 2.2 Let f : [−1, 1]→ R be defined by f(x) =

0, −1 ≤ x ≤ 0,1, 0 < x ≤ 1.

We

show that limx→0

f(x) does not exist. Suppose limx→0 f(x) = β for some β ∈ R and

let ε > 0 be given. Then, we know from the definition of the limit that there existsδ := δε such that

x 6= 0, x ∈ (−δ, δ) =⇒ f(x) ∈ (β − ε, β + ε).

We know that, f(x) takes the values only 0 and 1 and the length of the interval(β− ε, β+ ε) is 2ε. Hence, if we take ε0 such that 2ε0 < 1, then f(x) cannot belongto (β−ε0, β+ε0) for all x in a neighbourhood of 0. To see this consider the followingcases:

case (i): β = 0. In this case, f(x) 6∈ (β − ε0, β + ε0) for any x > 0.

case (ii): β = 1. In this case, f(x) 6∈ (β − ε0, β + ε0) for any x < 0.

case (iii): β 6= 0, β 6= 1. In this case, f(x) 6∈ (β − ε0, β + ε0) for any x0.

Thus, we arrive at a a contradiction to our assumption.

Limit of a function in terms of a sequences

Let a be a limit point of D ⊆ R and f : D → R.

Suppose limx→a

f(x) = b. Since a is a limit point of D, we know that there exists

a sequence (xn) in D which is not eventually constant such that xn → a. Doesf(xn)→ b? The answer is “yes”. In fact, we have more!

Theorem 2.2 If limx→a

f(x) = b, then for every sequence (xn) in D such that xn → a,

we have f(xn)→ b.

Proof. Suppose limx→a

f(x) = b. Let (xn) be a sequence in D such that xn → a. Let

ε > 0 be given. We have to show that there exists n0 ∈ N such that |f(xn)− b| < εfor all n ≥ n0.

Since limx→a

f(x) = b, we know that there exists δ > 0 such that

x ∈ D \ a, |x− a| < δ =⇒ |f(x)− b| < ε. (∗)

Also, since xn → a, there exists n0 ∈ N such that |xn−a| < δ for all n ≥ n0. Hence,from (∗), we have |f(xn)− b| < ε for all n ≥ n0.

What about the converse of the above theorem? The converse is also true, in aslight restricted sense.

Theorem 2.3 If for every sequence (xn) in D which is not eventually constant, thesequence (f(xn)) converges to b, then lim

x→af(x) = b.

Proof. Suppose for every sequence (xn) in D which is not eventually constant,the sequence (f(xn)) converges to b. Assume for a moment that f does not have thelimit b as x approaches a. Then, by the definition of the limit, there exists ε0 > 0such that for every δ > 0, there exists at least one xδ ∈ D such that

xδ 6= a and |xδ − a| < δ, but |f(xδ)− b| > ε0.

In particular, for every n ∈ N, there exists xn ∈ D \ a such that

xn 6= a and |xn − a| <1

n, but |f(xn)− b| > ε0.

Thus, (xn) is not eventually constant such that xn → a but f(xn) 6→ b. Thus wearrive at a contradiction to our hypothesis.

Remark 2.3 The advantage of Theorem 2.2 is that, if we can find a sequence (xn)in D such that xn → a, but (f(xn)) does not converge, then we can assert thatlimn→∞

f(x) does not exist. Similarly, by Theorem 2.3, we can determine limn→∞

f(x) if

we are able to show convergence of (f(xn)) to some b for any arbitrary (not for aspecific) non-eventually constant sequence (xn) in D which converges to a.

2.1.3 Some properties

The following two theorems can be proved using Theorems 2.2 and 2.3, and theresults on convergence of sequences of real numbers.

Theorem 2.4 We have the following.

(i) If limx→a

f(x) = b and limx→a

g(x) = c, then

limx→a

[f(x) + g(x)] = b+ c, limx→a

f(x)g(x) = bc.

(ii) If limx→a

f(x) = b and b 6= 0, then f(x) 6= 0 in a deleted neighbourhood of a

and

limx→a

1

f(x)=

1

b.

Theorem 2.5 If f and g have the same limit b as x approaches a, and if h is afunction such that f(x) ≤ h(x) ≤ g(x) for all x deleted neighbourhood of a, thenlimx→a

h(x) = b.

The following two corollaries are immediate from Theorem 2.4.

Corollary 2.6 If limx→a

f(x) = b, limx→a

g(x) = c, and c 6= 0, then g is nonzero in a

deleted neighbourhood of c and

limx→a

f(x)

g(x)=b

c.

Corollary 2.7 If limx→a

f(x) = b, limx→a

g(x) = c and f(x) ≥ g(x) for all x in a deleted

neighbourhood of a, then b ≥ c.

Theorem 2.8 Suppose limx→a

f(x) = b and limy→b

g(y) = c. Then limx→a

g(f(x)) = c.

Proof. Let ε > 0 be given. Then there exists δ1 > 0 such that

|y − b| < δ1 =⇒ |g(y)− c| < ε.

Also, let δ2 > 0 be such that

|x− a| < δ2 =⇒ |f(x)− b| < δ1.

Hence,|x− a| < δ2 =⇒ |f(x)− b| < δ1 =⇒ |g(f(x))− c| < ε.


Exercise 2.3 Write details of the proof of Theorem 2.4 and Corollary 2.6 andCorollary 2.7. J

Exercise 2.4 Suppose ϕ is a function defined in a neighbourhood of a point x0such that lim

x→x0ϕ(x) = x0. If f is also a function defined in a neighbourhood of x0

and limx→x0

f(x) exists, then prove that limx→x0

f(ϕ(x)) exists and

limx→x0

f(ϕ(x)) = limx→x0

f(x).

J

EXAMPLE 2.3 If f(x) is a polynomial, say f(x) = a0 + a1x + . . . + akxk, then

for any a ∈ R,

limx→a

f(x) = f(a).

We obtain this by using Theorem 2.4.

Let us show the same by using the definition, i.e., using ε − δ arguments: Letb = f(a) and let ε > 0 be given. We have to find δ > 0 such that |x − a| < δ =⇒|f(x)− b| < ε. Note that

f(x)− f(a) = a1(x− a) + a2(x2 − a2) + . . .+ ak(x

k − ak),

where

xn − an = (x− a)[xn−1 + xn−2a+ . . .+ xan−2 + an−1].

Now, suppose |x− a| < 1. Then we have |x| < 1 + |a| so that

|xn−jaj−1| < (1 + |a|)n−1

and hence,

|xn − an| < |x− a|n(1 + |a|)n−1.

Thus, |x− a| < 1 implies

|f(x)− f(a)| ≤ |x− a|(|a1|+ |a2|2(1 + |a|) + . . .+ |ak|k(1 + |a|)k−1

),

Therefore, taking α := |a1|+ |a2|2(1 + |a|) + . . .+ |ak|k(1 + |a|)k−1, we have

|f(x)− f(a)| < ε whenever |x− a| < δ := min1, ε/α.

EXAMPLE 2.4 Let D = R \ 2 and f(x) = x2−4x−2 . Then lim

x→2f(x) = 4.

Note that, for x 6= 2,

f(x) =(x+ 2)(x− 2)

x− 2= (x+ 2).

Hence, for ε > 0, |f(x)− 4| < ε whenever |x− 2| < δ := ε.

EXAMPLE 2.5 Let D = R\0 and f(x) = 1x . Then lim

x→0f(x) does not exist. To

see this consider the sequence (xn) with xn = 1/n for n ∈ N. Then we have xn → 0but (f(xn) diverges to infinity. Therefore, by Theorem 2.2, lim

x→0f(x) does not exist.

Alternatively, for any b ∈ R,

|f(x)− b| ≥ |f(x)| − |b| > 1 whenever |f(x)| > 1 + |b|.

But,

|f(x)| > 1 + |b| ⇐⇒ |x| < 1

1 + |b|.

Thus, for any b ∈ R,

|f(x)− b| > 1 whenever |x| < 1

1 + |b|.

Thus, we have proved that it is not possible to find a δ > 0 such that |f(x)− b| < 1for all x with |x| < δ.

EXAMPLE 2.6 We show that (i) limx→0

sin(x) = 0 and (ii) limx→0

cos(x) = 1.

From the graph f the function sinx, it is clear that

0 < x <π

2=⇒ 0 < sinx < x.

Also,

−π2< x < 0 =⇒ 0 < | sinx| < |x|.

Hence, from Theorem 2.5, we have limx→0| sinx| = 0. Thus, lim

x→0sin(x) = 0.

Also, since cosx = 1− 2 sin2(x/2) and limx→0

sin(x/2) = 0, Theorem 2.4(i) implies

limx→0

cosx = 1.

EXAMPLE 2.7 We show that limx→0

sinx

x= 1.

It can be seen, using the graph of sinx that

0 < x <π

2=⇒ sinx < x < tanx.

Hence,

0 < x <π

2=⇒ cosx <

sinx

x< 1.

Since sin(−x)−x = sinx

x and cos(−x) = cosx, it follows that

0 < |x| < π

2=⇒ cosx <

sinx

x< 1.

Therefore, by Theorem 2.4(iv) and Example 2.6(ii), we have limx→0

sinx

x= 1.

Now a general example.

Exercise 2.5 Let f : R→ R be such that f(x+y) = f(x)+f(y). Suppose limx→0

f(x)

exists. Prove that limx→0

f(x) = 0 and limx→c

f(x) = f(c) for every c ∈ R.

Hint: Use the facts that f(2x) = 2f(x), Theorem 2.8 and f(x)− f(c) = f(x− c). J

Exercise 2.6 Suppose ϕ is a function defined in a neighbourhood I0 of a point x0such that

x ∈ I0, |x− x0| < r =⇒ |ϕ(x)− x0| < r ∀ r > 0.

If f is also a function defined in a neighbourhood of x0 and limx→x0

f(x) exists, then

prove that limx→x0

f(ϕ(x)) exists and limx→x0 f(ϕ(x)) = limx→x0 f(x). J

2.1.4 Left limit and right limit

Definition 2.3 Let f be a real valued function defined on a set D ⊆ R, and leta ∈ R be a limit point of D.

(i) We say that f(x) has the left limit b ∈ R as x approaches a from leftif for every ε > 0, there exists δ > 0 such that

|f(x)− b| < ε whenever x ∈ D, a− δ < x < a.

(ii) We say that f(x) has the right limit b ∈ R as x approaches a fromright if for every ε > 0, there exists δ > 0 such that

|f(x)− b| < ε whenever x ∈ D, a < x < a+ δ.

If f(x) has the left limit b ∈ R as x approaches a from left we write

limx→a−

f(x) = b or f(x)→ b as x→ a−,

and if f(x) has the right limit b ∈ R as x approaches a from right we write

limx→a+

f(x) = b or f(x)→ b as x→ a+.

The following theorem can be proved easily.

Theorem 2.9 Let f be a real valued function defined on a set D ⊆ R, and let a ∈ Rbe a limit point of D. Then lim

x→af(x) exists if and only if lim

x→a−f(x) and lim

x→a+f(x)

exist and limx→a−

f(x) = limx→a+

f(x), and in that case

limx→a

f(x) = limx→a−

f(x) = limx→a+

f(x).

In view of the above theorem, if (i) limx→a−

f(x) does not exist or (ii) limx→a+

f(x)

does not exist or (iii) both limx→a−

f(x) and limx→a+

f(x) exist but limx→a−

f(x) 6= limx→a+

f(x),

then limx→a

f(x) does not exist.

Exercise 2.7 Give examples to illustrate the above statements. J

2.1.5 Limit at ∞ and at −∞

Definition 2.4 Suppose a function f is defined on an interval of the form (a,∞) forsome a > 0. Then we say that f has the limit b as x→∞ and write lim

x→∞f(x) = b,

if for every ε > 0, there exits M > a such that

|f(x)− b| < ε whenever x > M.

Definition 2.5 Suppose a function f is defined on an interval of the form (−∞, a)for some a < 0. Then we say that f has the limit b as x → −∞ and write

limx→−∞

f(x) = b, if for every ε > 0, there exits M < a such that

|f(x)− b| < ε whenever x < M.

Exercise 2.8 Prove the following:

1. If f is defined in (a,∞) with a > 0, then the function g(x) := f(1/x) is definedon a (0, 1/a), and

limx→∞

f(x) exists ⇐⇒ limx→0

g(x) exists

and in that case limx→∞

f(x) = limx→0

g(x).

2. If f is defined in (−∞, a) with a < 0, then the function g(x) := f(1/x) isdefined on a (1/a, 0), and

limx→−∞

f(x) exists ⇐⇒ limx→0

g(x) exists

and in that case limx→−∞

f(x) = limx→0

g(x).

J

Definition 2.6 1. limx→a

f(x) = +∞ if for every M > 0, there exists δ > 0 such

that0 < |x− a| < δ =⇒ f(x) > M.

2. limx→a

f(x) = −∞ if for every M > 0, there exists δ > 0 such that

0 < |x− a| < δ =⇒ f(x) < −M.

3. limx→+∞

f(x) = +∞ if for every M > 0, there exists α > 0 such that

x > α =⇒ f(x) > M.

4. limx→+∞

f(x) = −∞ if for every M > 0, there exists α > 0 such that

x > α =⇒ f(x) < −M.

5. limx→−∞

f(x) = +∞ if for every M > 0, there exists α > 0 such that

x < −α =⇒ f(x) > M.

6. limx→−∞

f(x) = −∞ if for every M > 0, there exists α > 0 such that

x < −α =⇒ f(x) < −M.

For the next example, we require the following definition.

Definition 2.7 For positive numbers a and b, and p, q ∈ N, we define

ap/q := (ap)1/q.

If b is a positive real number and (bn) is s sequence of positive rational numberssuch that bn → b, then we define

ab := limn→∞

abn .

(Of course, one has to show that this limit exists; which is true!).

EXAMPLE 2.8 Recall that limn→∞

(1 +

1

n

)nexists, and we denoted it by e. Now

we show that

limx→∞

(1 +

1

x

)x= e.

Let ε > 0 be given. We have to find an M > 0 ∈ N such that

e− ε <(

1 +1

x

)x< e+ ε whenever x > M. (∗)

Now, we can see that, for every n ∈ N, if x ∈ R is such that n ≤ x ≤ n+ 1, then

1 +1

n+ 1≤ 1 +

1

x≤ 1 +

1

n

so that (1 +

1

n+ 1

)n≤(

1 +1

x

)x≤(

1 +1

n

)n+1.

Thus is is same as

αn ≤(

1 +1

x

)x≤ βn,

where

αn :=(

1 +1

n+ 1

)−1(1 +

1

n+ 1

)n+1, βn :=

(1 +

1

n

)n(1 +

1

n

).

We know that αn → e and βn → e as n → ∞. Therefore, there exists n0 ∈ N suchthat for all n ≥ n0,

e− ε < αn < e+ ε, e− ε < βn < e+ ε.

Now, take M = n0 and let x > M . Take n ≥ n0 such that n+ 1 ≥ x ≥ n. For thisn and x, we have

e− ε < αn ≤(

1 +1

x

)x≤ βn < e+ ε.

Thus, we obtained an M > 0 such that

e− ε <(

1 +1

x

)x< e+ ε whenever x > M.

Thus, we have proved (∗).

Using the arguments used in the above example, we obtain a more general result.

Theorem 2.10 Suppose (αn) and (βn) are sequences of positive real numbers andf is a (real valued) function defined on (0,∞) having the following property: Forn ∈ N, x ∈ R,

n < x < n+ 1 =⇒ αn ≤ f(x) ≤ βn.

If (αn) and (βn) converge to the same limit, say b, then limx→∞

f(x) = b.

2.1.6 Additional Exercises

1. Using the definition of limit, show that limx→3

x

4x− 9= 1.

2. Show that the function f defined by f(x) =

x, if x < 1,1 + x, if x ≥ 1

does not

have the limit as x→ 1.

3. Let f be defined by f(x) =

3− x, if x > 1,1, if x = 1,2x, if x, 1.

Find limx→1

f(x). Is it f(1)?

4. Let f be defined on a deleted neighbourhood D0 of a point x0 and limx→x0

f(x) =

b. If b 6= 0, then show that there exists δ > 0 such that f(x) 6= 0 for everyx ∈ (x0 − δ, x0 + δ) ∩D0.

Continuity of a Function 43

5. Let f be defined by f(x) =

1, if x ∈ Q,0, if x 6∈ Q. Show that

(i) limx→0

f(x) does not exist, and

(ii) limx→0

xf(x) = 0.

6. Suppose limx→∞

f(x) =∞ and limx→∞

g(x) = b. Show that limx→∞

g(f(x)) = b.

7. Let f : (0,∞)→ R be such that limx→0

f(x) = b. Show that limx→∞

f(x−1) = b.

2.2 Continuity of a Function

2.2.1 Definition and some basic results

Definition 2.8 Let f be a real valued function defined on an interval I and a ∈ I.The f is said to be continuous at a if lim

x→af(x) = f(a).

Definition 2.9 Let f be a real valued function defined on an interval I. Then f issaid to be continuous on I if f is continuous at every a ∈ I.

Thus, f is continuous at a ∈ I if and only if for every ε > 0, there exists δ > 0such that

|f(x)− f(a)| < ε whenever x ∈ I, |x− a| < δ.

By Theorems 2.2 and 2.3, we have the following:

Theorem 2.11 A function f : I → R is continuous at a ∈ I if and only if for everysequence (xn) in I with xn → a, we have f(xn)→ f(a).

Also, using Theorem 2.4, we obtain the following.

Theorem 2.12 Suppose f and g are defined on an interval I and both f and g arecontinuous at a ∈ I. Then we have the following.

(i) f + g and fg are continuous at a.

(ii) If g(a) 6= 0, then there exists δ0 > 0 such that g(x) 6= 0 for every x ∈ I with|x− a| < δ0 and f/g is continuous at a.

From Theorem 2.8, we have the following.

Theorem 2.13 Suppose f is continuous at a point x0 and g is continuous at thepoint y0 := f(x0). Then g f is continuous at x0.

The following property of a continuous function is worth noticing.

Theorem 2.14 Suppose f is a continuous function defined on an interval I andx0 ∈ I is such that f(x0) > 0. Then there exists δ > 0 such that f(x) > 0 for allx ∈ I ∩ (x0 − δ, x0 + δ).

Proof. Suppose the the conclusion in the theorem does not hold. Then for everyδ > 0, there exists x ∈ I ∩ (x0 − δ, x0 + δ) such that f(x) ≤ 0. In particular,for each n ∈ N, taking δn = 1/n, there exists xn ∈ I ∩ (x0 − 1/n, x0 + 1/n) suchthat f(xn) ≤ 0. Thus, we have xn → x0 as n → ∞. Hence, by Theorem 2.11,f(xn)→ f(x0). Since f(xn) ≤ 0 for all n ∈ N, we have f(x0) ≤ 0, which contradictsthe assumption that f(x0) > 0.

EXAMPLE 2.9 If f(x) is a polynomial, say f(x) = a0 + a1x+ . . .+ akxk, then f

is continuous on R

EXAMPLE 2.10 For given a ∈ R, let f(x) = |x−a|, x ∈ R. Then f is continuousat every x0 ∈ R. To see this, note that

|f(x)− f(x0)| = ||x− a| − |x0 − a|| ≤ |x− x0.

Hence, for every ε > 0, we have

|x− x0| < ε =⇒ |f(x)− f(x0)| < ε.

EXAMPLE 2.11 Let f(x) = x2−4x−2 for x ∈ R \ 2 and f(2) = 4. Then f is

continuous on R.

EXAMPLE 2.12 The functions f and g defined by f(x) = sinx and g(x) = cosxfor x ∈ R are continuous at 0.

Note that for x, y ∈ R,

sinx− sin y = 2 sin(x− y

2

)cos(x+ y

2

)so that

| sinx− sin y| ≤ |x− y| ∀x, y ∈ R.

Hence, for every ε > 0 and for every x0 ∈ R,

|x− x0| < ε =⇒ | sinx− sinx0| < ε.

Thus, f(x) = sinx is continuous at every x ∈ R. Since cosx1− 2 sin2(x/2), x ∈ R,it also follows that g(x) = cosx is continuous at every x ∈ R.

EXAMPLE 2.13 Let f(x) = sinxx for x 6= 0 and f(0) = 1. Then f is continuous

at every point in R.

EXAMPLE 2.14 Let f(x) = 1x for x 6= 0. Then there does not exist a continuous

function g defined on R such that g(x) = f(x) for all x 6= 0.

EXAMPLE 2.15 The function f defined by f(x) = 1/x, x 6= 0 is continuous atevery x0 6= 0:

Note that for x 6= 0, x0 6= 0,∣∣∣∣1x − 1

x0

∣∣∣∣ =|x− x0||xx0|

≤ 2|x− x0||x20|

whenever |x− x0| ≤|x0|2

since |x| = |x0 − (x0 − x)| ≥ |x0| − |x0 − x|. Hence, for any ε > 0,∣∣∣∣1x − 1

x0

∣∣∣∣ < ε whenever |x− x0| < δ := minεx20

2,x02.

Thus, f is continuous at every x0 6= 0.

EXAMPLE 2.16 Let f be defined by f(x) = 1/x on (0, 1]. Then there does notexist a continuous function g on [0, 1] such that g(x) = f(x) for all x ∈ (0, 1].

Suppose g is any function defined on [0, 1] such that g(x) = f(x) for all x ∈ (0, 1].Then we have 1/n→ 0 but g(1/n) = f(1/n) = n→∞. Thus, g(1/n) 6→ g(0).

EXAMPLE 2.17 The function f defined by f(x) =√x, x ≥ 0 is continuous at

every x0 ≥ 0:

Let ε > 0 be given. First consider the point x0 = 0. Then we have

|f(x)− f(x0)| =√x < ε whenever |x| < ε2.

Thus, f is continuous at x0 = 0. Next assume that x0 > 0. Since |x − x0| =(√x+√x0)|√x−√x0|, we have

|√x−√x0| =

|x− x0|√x+√x0≤ |x− x0|√

x0.

Thus,|√x−√x0| < ε whenever |x− x0| < δ := ε

√x0.

More generally, we have the following example.

EXAMPLE 2.18 Let k ∈ N. Then the function f defined by f(x) = x1/k, x ≥ 0is continuous at every x0 ≥ 0:

Let ε > 0 be given. First consider the point x0 = 0. Then we have

|f(x)− f(x0)| = x1/k < ε whenever |x| < εk.

Thus, f is continuous at x0 = 0. Next assume that x0 > 0. Let y = x1/k and

y0 = x1/k0 . Since

yk − yk0 = (y − y0)(yk−1 + yk−2y0 + . . .+ yyk−2 + yk−10 ),

so that

x− x0 = (x1/k − x1/k0 )(yk−1 + yk−2y0 + . . .+ yyk−2 + yk−10 ).

Hence,

|x1/k − x1/k0 | =|x− x0|

yk−1 + yk−2y0 + . . .+ yyk−2 + yk−10

≤ |x− x0|yk−10

.

Thus,

|x1/k − x1/k0 | < ε whenever |x− x0| < δ := εyk−10 = εx1−1/k0 .

Thus, f is continuous at every x0 > 0.

EXAMPLE 2.19 For a rational number r, let f(x) = xr for x > 0. Then usingthe above example together with Theorem 2.13, we see that f is continuous at everyx0 > 0.

We know that given r ∈ R, there exists a sequence (rn) of rational numbers suchthat rn → r. For n ∈ N, let fn(x) = xrn , x > 0. Since each fn is continuous forx > 0, one may enquire whether the function f defined by f(x) = xr is continuousfor x > 0.

First of all how do we define the xr for x > 0?

We shall discuss this issue in the next subsection, where we shall introduce twoimportant classes of functions, namely, exponential and logarithm functions. In fact,our discussion will also include, as special cases, the Examples 2.17 - 2.19.

2.2.2 Exponential and logarithm functions

We have already come across expression such as ab for a > 0 and b ∈ R, though wehave not proved existence of such numbers, and also defined a number denoted by

e as the limit of the sequence

(1 +

1

n

)1/n

or the series

∞∑n=0

1

n!. Now, we formally

define the following functions:

• Exponential function: ex, x ∈ R.

• Natural logarithm function: lnx, x > 0.

• Exponential function: ax, x ∈ R for a given a > 0.

• Logarithm function with base a > 0: loga x, x > 0.

First, we observe that for every x ∈ R, the series∑∞

n=0xn

n! converges absolutely.This can be seen by using the ratio test. This series plays a very significant role inmathematics.

Definition 2.10 For x ∈ R, let

exp(x) :=

∞∑n=0

xn

n!.

The function exp(x), x ∈ R, is called the exponential function.

From the above definition, it is clear that

exp(0) = 1 and exp(1) = e.

In order to derives some of the important properties of the function exp(x), thestudent is urged to do the following exercise.

Exercise 2.9 Suppose that series∑∞

n=0 an and∑∞

n=0 bn are absolutely convergent.Then, the series

∞∑n=0

cn with cn :=n∑k=0

akbn−k (∗)

is absolutely convergent. Further, show that if α =∑∞

n=0 an and β =∑∞

n=0 bn, then∑∞n=0 cn = αβ.

The series∑∞

n=0 cn defined in (∗) is called the Cauchy product of the series∑∞n=0 cn and

∑∞n=0 cn. J

Using the conclusion in the above exercise, it can be proved that

exp(x+ y) = exp(x) exp(y) ∀x, y ∈ R. (∗∗)

Exercise 2.10 Prove (∗∗) above. J

We observe the following properties:

• exp(−x) =1

exp(x)∀x ∈ R. In particular, since exp(x) > 0 for x ≥ 0, we

haveexp(x) > 0 ∀x ∈ R.

[This follows from (∗∗)]

• exp(x) > 1 ⇐⇒ x > 0 and exp(x) = 1 ⇐⇒ x = 0.

[From the definition, x > 0 implies exp(x) > 1. Next, suppose x ≤ 0. If x = 0, thenexp(x) = exp(0) = 1. If x < 1, then taking y = −x, we have y > 1, and hence fromthe first part, exp(y) > 1, i.e., 1/ exp(x) = exp(−x) > 1 so that exp(x) < 1. Hence,exp(x) > 1 ⇐⇒ x > 0. From this, we get exp(x) = 1 ⇐⇒ x = 0.]

• x > y ⇐⇒ exp(x) > exp(y).

[x > y ⇐⇒ x− y > 0 ⇐⇒ exp(x− y) > 1. But, exp(x− y) = exp(x)/ exp(y).]

• exp(kx) = [exp(x]k ∀x ∈ R, k ∈ Z. In particular, taking x = 1 and x = −1,

exp(k) = ek ∀ k ∈ Z.

• Since e = exp(1) = exp(k/k) = [exp(1/k]k ∀ k ∈ N, we have

exp(1/k) = e1/k ∀ k ∈ N.

• exp(m/n) = em/n ∀m,n ∈ N. Hence,

exp(r) = er ∀ r ∈ Q.

We know that every real number is a limit of a sequence of rational numbers.Thus, if r ∈ R, there exits a sequence (rn) of rational numbers that rn → r. So, itis natural to define

er = limn→∞

ern

provided the above limit exists. Thus, our next attempt is to show that the functionexp(x), x ∈ R, is continuous.

In view of these observations, we use the following

Proposition 2.15 The following results hold.

(i) exp(x)→∞ as x→∞.

(ii) exp(x)→ 0 as x→ −∞.

Proof. For x ≥ 0, we have

exp(x) = 1 + x+

∞∑n=2

xn

n!≥ 1 + x.

From this (i) follow. To see (ii), let x < 0 and y = −x. Then y > 0 so that by (ii),

exp(x) = exp(−y) =1

exp(y)→ 0 as y →∞.

Thus, exp(x)→ 0 as x→ −∞.

NOTATION: For brevity of expression, we shall use the notation ex for exp(x).

Theorem 2.16 The function exp(x) is continuous on R

Proof. Let x, x0 ∈ R. Then we have

ex − ex0 = ex0(ex−x0 − 1) = ex0∞∑n=1

(x− x0)n

n!= ex0(x− x0)

∞∑n=1

(x− x0)n−1

n!.

Thus, if |x− x0| ≤ 1, then

|ex − ex0 | ≤ ex0 |x− x0|∞∑n=1

1

n!= ex0(e− 1)|x− x0|.

Hence, for every ε > 0,

|ex − ex0 | < ε whenever |x− x0| < min1, ε/[ex0(e− 1)]

so that ex is a continuous function for x ∈ R.

Theorem 2.17 The function ex is bijective from R to (0,∞).

Proof. First we observe that, for x1, x2 in R

ex2 − ex1 = ex1 [ex2−x1 − 1].

Thus,ex2 = ex1 ⇐⇒ ex2−x1 = 1 ⇐⇒ x1 = x2,

showing that the function x 7→ ex is one-one.

Next, to show the function is onto, let y ∈ (0,∞). Since, by Proposition 2.15,

ex → 0 as x→ −∞, ex →∞ as x→∞,

by intermediate value property, there exists x ∈ R such that ex = y.

Definition 2.11 For b > 0, the unique a ∈ R such that ea = b is called the naturallogarithm b, and it is denoted by ln b. The function

lnx, x > 0,

is called the natural logarithm function.

Definition 2.12 For a > 0 and b ∈ R, we define

ab := eb ln a.

Remark 2.4 We note that ln e = 1 so that if a = e, then the Definition 2.12matches with Definition 2.10.

Theorem 2.18 Let a > 0. Then the function ax is continuous and bijective fromR to (0,∞).

Proof. Note that for x ∈ R, ax := ex ln a. Hence, the result is a consequence ofTheorems 2.16 and 2.17, and the Definition 2.12, and using the fact that compositionof two continuous functions is continuous.

Definition 2.13 Let a > 0. For c > 0, the unique b ∈ R such that ab = c is calledthe logarithm of c to the base a, and it is denoted by loga c. The function

loga x, x > 0,

is called the logarithm function.

We observe that following.

• For y ∈ R, y = lnx ⇐⇒ ey = x.

• For a > 0 and y ∈ R, y = loga x ⇐⇒ ay = x.

• For a > 0 and x > 0, loga x =lnx

ln a.

Exercise 2.11 For a > 0, b > 0, show that logb a loga b = 1. J

Theorem 2.19 The functions lnx and loga x for a > 0 are continuous on (0,∞).

Proof. Let x, x0 belong to the interval (0,∞), and let y = lnx and y0 = lnx0.Then we have ey = x and ey0 = x0. Assume, without loss of generality that x > x0.Since ea > 1 if and only if a > 0, we have y > y0, and hence

x− x0 = ey − ey0 = ey0(ey−y0 − 1) = ey0∞∑n=1

(y − y0)n

n!≥ ey0(y − y0).

Hence,

|y − y0| ≤ e−y0 |x− x0|.

Thus, for ε > 0, we have |y − y0| < ε whenever x− x0| < ey0ε, lnx is continuous on(0,∞). Since loga x = lnx/ ln a, the function loga x is also continuous on (0,∞).

Theorem 2.20 For r ∈ R, the function f defined by f(x) = xr is continuous atevery point x ∈ (0,∞).

Proof. For r ∈ R and x > 0, we have xr = er lnx. Hence, the result follows fromTheorem 2.19 and Theorem 2.13.

Remark 2.5 Often, the notation log x is used for the natural logarithm functioninstead of lnx.


2.2.3 Some properties of continuous functions

Recall that a subset S of R is said to be bounded if there exists M > 0 such that|s| ≤M for all s ∈ S, and set which is not bounded is called an unbounded set.

Recall that if S is a bounded subset of R, then S has infimum and supremum.

Exercise 2.12 Let S ⊆ R. Prove the following:

(i) Suppose S is bounded, and say α := inf S and β := supS. Then there existsequences (sn) and (tn) in S such that sn → α and tn → β.

(ii) S is unbounded if and only if there exists a sequence (sn) in S which isunbounded.

(iii) S is unbounded if and only if there exists a sequence (sn) in S such that|sn| → ∞ as n→∞.

(iv) If (sn) is a sequence in S which is unbounded, then there exists a subsequence(skn) of (sn) such that |skn | → ∞ as n→∞.

(v) If (sn) is a sequence in S such that |sn| → ∞ as n → ∞, and if (skn) issubsequence of (sn), then |skn | → ∞ as n→∞. J

Definition 2.14 A real valued function defined on a set D ⊆ R is said to be abounded function if the set f(x) : x ∈ D is is bounded. A function is said tobe an unbounded function if it is not bounded.

The following can be easily deduced from the definition:

• A function f : D → R is bounded if and only if there exists M > 0 such that|f(x)| ≤M for all x ∈ D.

• A function f : D → R is not unbounded if and only if there exists a sequence(xn) ∈ D such that the |f(xn)| → ∞ as n→∞.

Theorem 2.21 Suppose f is a real valued function defined on a closed and boundedinterval [a, b]. Then f is a bounded function.

Proof. Assume for the time being that f : [a, b]→ R is not a bounded function.Then for every n ∈ N, there exists a sequence (xn) in [a, b] such that |f(xn)| → ∞as n → ∞. Since (xn) is a bounded sequence, by Bolzano-Weierstrass property ofR, there exists a subsequence (xkn) of (xn) such that xkn → x for some x ∈ [a, b].Therefore, by continuity of f , f(xkn) → f(x). In particular, (f(xkn) is a boundedsequence. This is a contradiction to the fact that |f(xn)| → ∞ as n→∞.

Thus, we have proved that continuous function cannot be unbounded.


Attaining max f and min f

Theorem 2.22 Suppose f is a continuous real valued function defined on a closedand bounded interval [a, b]. Then there exists x0, y0 in [a, b] such that

f(x0) = α := inff(x) : x ∈ [a, b], f(y0) = β := supf(x) : x ∈ [a, b].

Proof. By the definition of infimum and supremum, there exist sequences (xn)and (yn) in [a, b] such that f(xn) → α and f(yn) → β as n → ∞. Since (xn)and (yn) are bounded sequences, there exist subsequences (xkn) and (ykn) of (xn)and (yn), respectively, such that xkn → x and ykn → y for some x, y in [a, b]. Bycontinuity of f , f(xkn)→ f(x) and f(ykn)→ f(y) as n→∞. But, we already havef(xkn)→ α and f(ykn)→ β. Hence, α = f(x) and β = f(y).

Remark 2.6 By Theorem 2.22, we say that the infimum and supremum of a con-tinuous real valued function f defined on a closed and bounded interval [a, b] areattained at some points in [a, b], and in that case, we write

inff(x) : x ∈ [a, b] = mina≤x≤b

f(x), supf(x) : x ∈ [a, b] = maxa≤x≤b

f(x).

The conclusion in the above theorem need hold if the domain of the function is notof the form [a, b] or if f is not continuous. For example, f : (0, 1] → R defined byf(x) = 1/x for x ∈ (0, 1] is continuous, but does not attain supremum. Same is thecase if g : [0, 1]→ R is defined by

g(x) =

1x , x ∈ (0, 1],1, x = 0.

Thus, neither continuity nor the fact that the domain is a closed and boundedinterval cannot be dropped. This does not mean that the conclusion in the theoremdoes not hold for all such functions! For example f : [0, 1)→ R defined by

f(x) =

0, x ∈ [0, 1/2),1, x ∈ [1/2, 1).

Then we see that neither f is continuous, nor its domain of the form [a, b]. But, fattains both its maximum and minimum.

Intermediate value theorem

Suppose f is a continuous real valued function f defined on a closed and boundedinterval [a, b], and

α := mina≤x≤b

f(x), β := maxa≤x≤b

f(x).

Clearly,

α ≤ f(x) ≤ β ∀x ∈ [a, b].

Now, the question is whether every value between α and β is attained by the func-tion. The answer is in affirmative. In fact we have the following general theorem,known as Intermediate value theorem1

Theorem 2.23 (Intermediate value theorem) Suppose f is a continuous realvalued function defined on an interval I. Suppose x1 and x2 are in I such thatf(x1) < f(x2), and c is such that f(x1) < c < f(x2). Then there exists x0 lyingbetween x1 and x2 such that f(x0) = c.

Proof. Without loss of generality assume that x1 < x2. Let

S = x ∈ [x1, x2] : f(x) < c.

Then S is non-empty (since x1 ∈ S) and bounded above (since x ≤ x2 for all x ∈ S).Let

α := supS.

Then there exists a sequence (an) in S such that an → α. Note that α ∈ [x1, x2].Hence, by continuity of f , f(an) → f(α). Since f(an) < c for all n ∈ N, we havef(α) ≤ c. Note that α 6= b (since f(α) ≤ c < f(x2)).

Now, let (bn) be a sequence in (α, x2) such that bn → α. Then, again bycontinuity of f , f(bn)→ f(α). Since bn > α, bn 6∈ S and hence f(bn) ≥ c. Therefore,f(α) ≥ c. Thus, we have prove that there exists x0 := α such that f(x0) ≤ c ≤ f(x0)so that f(x0) = c.

The following two corollaries are immediate consequences of immediate from theabove theorem.

Theorem 2.24 Let f be a continuous function defined on an interval. Then rangeof f is a an interval.

Corollary 2.25 Suppose f is a continuous real valued function defined on an in-terval I. If a, b ∈ I satisfy a 0, andif 0 < β < f(c), then show that there exists δ > 0 such that f(x) > β for allx ∈ (c− δ, c+ δ) ∩ [a, b].

2. Let f : R → R satisfy the relation f(x + y) = f(x) + f(y) for every x, y ∈ R.If f is continuous at 0, then show that f is continuous at every x ∈ R, and inthat case f(x) = xf(1) for every x ∈ R.

1Proof taken from the book: A Course in Calculus and Real Analysus by S.R. Ghorpade andB.V. Limaye (IIT Bombay), Springer, 2006.


3. There does not exist a continuous function f from [0, 1] onto R – Why?

4. Find a continuous function f from (0, 1) onto R.

5. Suppose f : [a, b]→ [a, b] is continuous. Show that there exists c ∈ [a, b] suchthat f(c) = c.

6. There exists x ∈ R such that 17x19 − 19x17 − 1 = 0 – Why?

7. If p(x) is a polynomial of odd degree, then there exists at least one ξ ∈ R suchthat p(ξ) = 0.

8. Suppose f : R→ R is continuous such that f(x)→ 0 as |x| → ∞. Prove thatf attains either a maximum or a minimum.

9. Suppose f : [a, b]→ R is continuous such that for every x ∈ [a, b], there exists

a y ∈ [a, b] such that |f(y)| ≤ |f(x)|2

. Show that there exists ξ ∈ [a, b] such

that f(ξ) = 0.

10. Suppose f : [a, b]→ [a, b] is continuous such that there |f(x)−f(y)| ≤ 1

2|x−y|

for all x, y ∈ [a, b]. Show that there exists ξ ∈ [a, b] such that f(ξ) = ξ.

2.3 Differentiability of functions

Definition 2.15 Suppose f is a (real valued) function defined on an open intervalI and x0 ∈ I. Then f is said to be differentiable at x0 if

limx→x0

f(x)− f(x0)

x− x0

exists, and in that case the value of the limit is called the derivative of f at x0.

The derivative of f at x0, if exists, is denoted by

f ′(x0) ordf

dx(x0).

Exercise 2.13 Show that f : I → R is differentiable at x0 ∈ I if and only iflimh→0

f(x0+h)−f(x0)h exists. J

Exercise 2.14 Suppose f is defined on an open interval I and x0 ∈ I. Show thatf is differentiable at x0 ∈ I if and only if there exists a continuous function Φ(x)such that

f(x) = f(x0) + Φ(x)(x− x0),

and in that case Φ(x0) = f ′(x0). J

Differentiability of functions 55

Exercise 2.15 Let Φ be as in Exercise 2.14. Then f is differentiable at x0, if andonly if for every sequence (xn) in I\x0 which converges to x0, the sequence Φf (xn)converges, and in that case f ′(x0) = limn→∞Φf (xn). J

Exercise 2.16 Suppose f and g defined on I are differentiable at x0 ∈ I andα ∈ R. Show that the functions ϕ(x) := f(x) + g(x) and ψ(x) := αf(x), x ∈ I aredifferentiable at x0, and

ϕ′(x0) = f ′(x0) + g′(x0), ψ′(x0) = αϕ′(x0).

J

2.3.1 Some properties of differentiable functions

Theorem 2.26 (Differentiability implies continuity) Suppose f defined on Iis differentiable at x0 ∈ I. Then f is continuous at x0.

Proof. Note that

f(x0 + h)− f(x0) =f(x0 + h)− f(x0)

hh→ f ′(x0).0 = 0 as h→ 0.

Thus, f is continuous at x0.

For the following theorem, we may recall that if ϕ is continuous at a point x0 andϕ(x0) 6= 0, then there exists an open interval I0 containing x0 such that ϕ(x) 6= 0for all x ∈ I0.

Theorem 2.27 (Products and quotient rules) Suppose f and g defined on Iare differentiable at x0 ∈ I. Then the function ϕ(x) := f(x)g(x) is differentiable atx0, and

ϕ′(x0) = f ′(x0)g(x0) + f(x0)g′(x0). (∗)

If g(x0) 6= 0, then the function ψ(x) := f(x)/g(x) is differentiable at x0, and

ψ′(x0) =g(x0)f

′(x0)− f(x0)g′(x0)

[g(x0)]2. (∗∗)

Proof. Note that

ϕ(x0 + h)− ϕ(x0) = f(x0 + h)g(x0 + h)− f(x0)g(x0)

= [f(x0 + h)− f(x0)]g(x0 + h) + f(x0)[g(x0 + h)− g(x0)]

so that

ϕ(x0 + h)− ϕ(x0)

h=

f(x0 + h)− f(x0)

hg(x0 + h) + f(x0)

g(x0 + h)− g(x0)

h→ f ′(x0)g(x0) + f(x0)g

′(x0) as h→ 0.


Hence, ϕ is differentiable at x0, and

ϕ′(x0) = f ′(x0)g(x0) + f(x0)g′(x0).

Also, since

ψ(x0 + h)− ψ(x0) =f(x0 + h)g(x0)− f(x0)g(x0 + h)

g(x0 + h)g(x0)

=[f(x0 + h)− f(x0)]g(x0)− f(x0)[g(x0 + h)− g(x0)]

g(x0 + h)g(x0),

we have

ψ(x0 + h)− ψ(x0)

h=

f(x0+h)−f(x0)h g(x0)− f(x0)

g(x0+h)−g(x0)h

g(x0 + h)g(x0)

→ f ′(x0)g(x0)− f(x0)g′(x0)

[g(x0)]2as h→ 0.

Thus, ψ is differentiable at x0, and ψ′(x0) = g(x0)f ′(x0)−f(x0)g′(x0)[g(x0)]2

.

Theorem 2.28 (Composition rule) Suppose f is defined on an open interval Iand x0 ∈ I, and g is defined in an open interval containing y0 := f(x0). Let

ϕ = g f.

Then we have the following.

(i) Suppose f is differentiable at x0 and g is differentiable at y0. Then ϕ isdifferentiable at x0 and

ϕ′(x0) = g′(y0)f′(x0).

(ii) Suppose ϕ is differentiable at x0, g is differentiable at y0 and g′(y0) 6= 0. Thenf is differentiable at x0 and

f ′(x0) =ϕ′(x0)

g′(y0).

(iii) Suppose ϕ is differentiable at x0, f is differentiable at x0 and f ′(x0) 6= 0. Theng is differentiable at y0 and

g′(y0) =ϕ′(x0)

f ′(x0).

Proof. For x 6= x0, let

Φ(x) :=(g f)(x)− (g f)(x0)

x− x0.


Let (xn) be a sequence in I \x0 which converges to x0. Then, taking yn := f(xn),n ∈ N, and y0 = f(x0), we have

Φ(xn) =(g f)(xn)− g f)(x0)

xn − x0

=g(yn)− g(y0)

xn − x0

=g(yn)− g(y0)

yn − y0× f(xn)− f(x0)

xn − x0.

(i) Suppose f is differentiable at x0. Now, since xn → x0 we have, by continuityof f at x0, yn → y0. Therefore,

f(xn)− f(x0)

xn − x0→ f ′(x0),

g(yn)− g(y0)

yn − y0→ g′(y0).

Thus, Φ(xn)→ g′(y0)f′(x0) showing that g f is differentiable at x0 and

(g f)′(x0) = g′(y0)f′(x0).

(ii) Suppose g f is is differentiable at x0 and g′(y0) 6= 0. Then we haveΦ(xn)→ (g f)′(x0) and

f(xn)− f(x0)

xn − x0=

Φ(xn)g(yn)−g(y0)yn−y0

→ (g f)′(x0)

g′(y0).

Hence, f is differentiable at x0 and f ′(x0) =(g f)′(x0)

g′(y0).

(iii) Proof of this part is analogous to the proof of (ii). Hence, we omit thedetails.

Remark 2.7 The part (ii) and (iii) of Theorem 2.28 is not available in standardbooks on Calculus. I found it useful while discussing derivative of logarithm functionin next section (Section ??).

Exercise 2.17 Prove part (iii) of Theorem 2.28. J

We shall assume that the students are familiar with the following:

• For c ∈ R, if f(x) = c, x ∈ R, then f ′(x) = 0 ∀x ∈ R.

• If f(x) = x, x ∈ R, then f ′(x) = 1 ∀x ∈ R.

• If f(x) = sinx, x ∈ R, then f ′(x) = cosx ∀x ∈ R.

• If f(x) = cosx =⇒ f ′(x) = − sinx.


From these, using theorems in the last subsection, we obtain the following:

• For n ∈ N, if f(x) = xn, x ∈ R, then f ′(x) = nxn−1 ∀x ∈ R.

• If f(x) = cosx = 1− 2 sin2(x/2), x ∈ R, then f ′(x) = − sinx ∀x ∈ R.

• If f(x) = tanx for x ∈ D := x ∈ R : cosx 6= 0, then f ′(x) = sec2x ∀x ∈ D.

EXAMPLE 2.20 The function ex is differentiable for every x ∈ R and

(ex)′ = ex ∀x ∈ R.

We note that for h 6= 0,

ex+h − ex

h− ex =

ex

h(eh − 1− h) =

ex

h

∞∑n=2

hn

n!.

Now, if |h| ≤ 1, then |h|n ≤ |h|2 for all n ∈ 2, 3, . . .. Thus,

|h| ≤ 1 =⇒∣∣∣∣ex+h − exh

− ex∣∣∣∣ ≤ ex|h| ∞∑

n=2

1

n!= ex|h|(e− 2).

From this we obtain that ex is differentiable at x and its derivative is ex.

EXAMPLE 2.21 For a > 0, the function ax is differentiable for every x ∈ R and

(ax)′ = ax ln a ∀x ∈ R.

By the composition rule in Theorem 2.28,

(ax)′ = (ex ln a)′ = ex ln a ln a = ax ln a.

EXAMPLE 2.22 The function lnx is differentiable for every x > 0, and

(lnx)′ =1

x, x > 0.

To see this, let f(x) = lnx and g(x) = ex. Then we have g(f(x) = x for every x > 0.Since g f is differentiable, g is differentiable, and g′(y) = ey 6= 0 for every y ∈ R,by Theorem 2.28, f is differentiable for every x > 0 and we have g′(f(x))f ′(x) = 1.Thus,

1 = elnx(lnx)′ = x(lnx)′

so that (lnx)′ = 1/x.


EXAMPLE 2.23 For a > 0, the function loga x is differentiable for every x > 0,and

(loga x)′ =1

x ln a, x > 0.

We know that

loga x =lnx

ln a.

Hence, (loga x)′ =1

x ln afor every x > 0.

EXAMPLE 2.24 For r ∈ R, let f(x) = xr for x > 0. Then f is differentiable forevery x > 0 and

f ′(x) = rxr−1, x > 0.

By the composition rule in Theorem 2.28,

f ′(x) = (er lnx)′ = er lnxr

x=xrr

x= rxr−1.


(i) The function ln |x| is differentiable for every x ∈ R with x 6= 0, and

(ln |x|)′ = 1

x, x 6= 0.

(ii) For a > 0, the function loga |x| is differentiable for every x ∈ R with x 6= 0,and

(loga |x|)′ =1

x ln a, x 6= 0.

J

2.3.2 Maxima and minima

Recall from Theorem 2.22 that if f : [a, b]→ R is a continuous function, then thereexists x0, y0 in [a, b] such that

f(x0) ≤ f(x) ≤ f(y0) ∀x ∈ [a, b].

In this case, we write

f(x0) = mina≤x≤b

f(x) and f(y0) = maxa≤x≤b

f(x).

Definition 2.16 A (real valued) function f defined on an interval I (of finite orinfinite length) is said to attain

(a) global maximum at a point x1 ∈ I if f(x1) ≥ f(x) for all x ∈ I, and

(b) global minimum at a point x2 ∈ I if f(x2) ≤ f(x) for all x ∈ I.

The function f is said to attain global extremum at a point x0 ∈ I if f attainseither global maximum or global minimum at x0.

Thus, a continuous fiunction f defined on a closed and bounded interval I attainglobal maximum and global minimum at some points in I.

In Remark 2.6 we have seen that a function f defined on an interval I need notattain maximum or minimum if either I is not closed and bounded or if f is notcontinuous. However, maximum or minimum can attain in a subinterval. To takecare of these cases, we introduce the following definition.

Definition 2.17 A (real valued) function f defined on an interval I (of finite orinfinite length) is said to attain

(a) local maximum at a point x1 ∈ I if there exists δ > 0 such that

f(x1) ≥ f(x) ∀x ∈ I ∩ (x1 − δ, x1 + δ),

(b) local minimum at a point x2 ∈ if if there exists δ > 0 such that

f(x2) ≤ f(x) ∀x ∈ I ∩ (x2 − δ, x2 + δ).

The function f is said to attain local extremum at a point x0 ∈ I if f attainseither local maximum or local minimum at x0.

Remark 2.8 It is conventional to omit the adjective local in local maximum, localminimum and local extremum. Thus when we say a function has maximum at apoint x0, we generally mean a local maximum at x0. Similar comments apply tominimum and extremum.

Theorem 2.29 (Necessary condition) Suppose f is a continuous function de-fined on an interval I having local extremum at a point x0 ∈ I. If x0 is an interiorpoint of I (i.e., x0 is not an end point of I) and f is differentiable at x0, thenf ′(x0) = 0.

Proof. Suppose f attains local maximum at x0 which is an interior point of I.Then there exists δ > 0 such that (x0 − δ, x0 + h) ⊆ I and f(x0) ≥ f(x0 + h) for allh with |h| < δ. Hence, for all h with |h| < δ,

f(x0 + h)− f(x0)

h≥ 0 if h < 0,

f(x0 + h)− f(x0)

h≤ 0 if h > 0.

Taking limit as h→ 0, we get f ′(x0) ≥ 0 and f ′(x0) ≤ 0 so that f ′(x0) = 0.

By analogous arguments, it can be shown that if f attains minimum at a pointy0 ∈ (a, b), then f ′(y0) = 0.

Remark 2.9 A function can have more than one maximum and minimum. Forexample, consider

f(x) = sin(4x), [0, π].

We see that f has maximum value 1 at π/8 and 5π/8, and has minimum value −1at 3π/8 and 7π/8.

Remark 2.10 (a) In view of Theorem 2.29, if a function f is differentiable at aninterior point x0 of an interval I and f ′(x0) 6= 0, then f can not have local maximumor local minimum at x0.

(b) It is to be observed that in order to have a maximum or minimum at a pointx0, the function need not be differentiable at x0. For example

f(x) = 1− |x|, |x| ≤ 1,

has a maximum at 0 and

g(x) = |x|, |x| ≤ 1,

has a minimum at 0. Both f and g are not differentiable at 0.

(c) Also, if a function is differentiable at a point x0 and f ′(x0) = 0, then it is notnecessary that it has loal maximum or local minimum at x0. For example, consider

f(x) = x3, |x| < 1.

In this example, we have f ′(0) = 0. Note that f has neither local maximum norlocal minimum at 0.

Definition 2.18 Suppose f is defined n an interval I and x0 is an interior pointof I. If f ′(x0) exists and f ′(x0) = 0 or if f ′(x0) does not exist, then x0 is called acritical point of f .

In Section 2.3.5 we shall give some sufficient conditions for existence of localexrema of functions. Now, let us derive some important consequences of Theorem2.29.

2.3.3 Some important theorems

Rolle’s theorem

Theorem 2.30 (Rolle’s theorem) Suppose f is a continuous function defined ona closed and bounded interval [a, b] such that it is differentiable at every x ∈ (a, b).If f(a) = f(b), then there exists c ∈ (a, b) such that f ′(c) = 0.

Proof. Let g(x) = f(x)−f(a). Then we have g(a) = 0 = g(b), and g′(x) = f ′(x)for every x ∈ (a, b).

We know that g attain maximum and minimum at some points x1 and x2,respectively, in [a, b], i.e., there exists x1, x2 in [a, b] such that

g(x2) ≤ g(x) ≤ g(x1) ∀x ∈ [a, b].

If g(x1) = g(x2), then g is a constant function and hence g′(x) = 0 for all x ∈ [a, b].Hence, assume that g(x2) < g(x1). Then, either g(x1) 6= 0 or g(x2) 6= 0. Assumethat g(x2) 6= 0, so that x2 6= a and x2 6= b. Hence, by Theorem 2.29, g′(x2) = 0.Thus, f ′(x2) = 0.

Similarly, if g(x1) 6= 0, then we shall arrive at f ′(x1) = 0.

Exercise 2.19 Show that between any two roots of the equation ex cosx − 1 = 0,there is at least one root of the equation ex sinx− 1 = 0. J

Lagrange’s mean value theorem

As a corollary to Rolle’s theorem we obtain the following.

Theorem 2.31 (Lagrange’s mean value theorem) Suppose f is a continuousfunction defined on a closed and bounded interval [a, b] such that it is differentiableat every x ∈ (a, b). Then there exists c ∈ (a, b) such that

f(b)− f(a) = f ′(c)(b− a).

Proof. Let

ϕ(x) := f(x)− f(a)− f(b)− f(a)

b− a(x− a), x ∈ [a, b].

Note that ϕ is continuous on [a, b], differentiable in (a, b), ϕ(a) = 0 = ϕ(b), and

ϕ′(x) := f ′(x)− f(b)− f(a)

b− a, x ∈ (a, b).

By Rolle’s theorem (Theorem 2.30), there exists c ∈ (a, b) such that ϕ′(c) = 0.Thus, f(b)− f(a) = f ′(c)(b− a).

EXAMPLE 2.25 Let f be continuous on [a, b] and differentiable at every point in(a, b). Suppose there exists c ∈ R such that

f ′(x) = c x ∈ (a, b).

Then there exists b ∈ R such that

f(x) = c x+ b ∀x ∈ [a, b].

In particular, f ′(x) = 0 for all x ∈ (a, b), then f is a constant function.

To see this consider x0 ∈ (a, b). Then for any x ∈ [a, b], there exists ξx betweenx0 and x such that

f(x)− f(x0) = f ′(ξx)(x− x0) = c(x− x0).

Hence, f(x) = f(x0) + c(x− x0). Thus, f(x) = c x+ b with b = f(x0)− c x0.

Cauchy’s generalized mean value theorem

Suppose f and g are continuous functions on [a, b] which are differentiable on (a, b).Suppose further that g′(x) 6= 0 for all x ∈ (a, b). Then, by Lagrange’s mean valuetheorem, there exist c1, c2 in (a, b) such that

f(b)− f(a)

g(b)− g(a)=f ′(c1)

g′(c2).

Question is whether we can assert the existence of a single point c ∈ (a, b) suchthat

f(b)− f(a)

g(b)− g(a)=f ′(c)

g′(c).

Answer is in affirmative as the following theorem shows.

Theorem 2.32 (Cauchy’s generalized mean value theorem) Suppose f andg are continuous functions on [a, b] which are are differentiable at every point in(a, b). Suppose further that g′(x) 6= 0 for all x ∈ (a, b). Then, there exists c ∈ (a, b)such that

f(b)− f(a)

g(b)− g(a)=f ′(c)

g′(c).

Proof. First note that from the assumption on g, using Mean value theorem,g(b(6= g(a). Now, let

ϕ(x) := f(x)− f(a)− f(b)− f(a)

g(b)− g(a)[g(x)− g(a)], x ∈ [a, b].

Note that ϕ is continuous on [a, b], differentiable in (a, b), ϕ(a) = 0 = ϕ(b), and

ϕ′(x) := f ′(x)− f(b)− f(a)

g(b)− g(a)g′(x), x ∈ (a, b).

By Rolle’s theorem (Theorem 2.30), there exists c ∈ (a, b) such that ϕ′(c) = 0. Thiscompletes the proof.

Exercise 2.20 Let 0 < a < b. Show that for every n ∈ N, a <n[bn+1 − an+1]

(n+ 1)[bn − an]< b.

[Hint: take f(x) = xn+1 and g(x) = xn.] J

If f is defined in a closed interval [a, b] and x0 = a or x0 = b, then by limx→x0f(x)we mean limx→x+0

f(x) if x0 = a and limx→x−0f(x) if x0 = b.

L’Hospital’s rules

Theorem 2.33 (L’Hospital’s rule)2 Suppose f and g are continuous functionson [a, b] which are differentiable at every point in (a, b), except possibly at x0 ∈ [a, b].

Suppose f(x0) = 0 = g(x0) and limx→x0

f ′(x)

g′(x)exists. Then lim

x→x0

f(x)

g(x)exists and

limx→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x).

Proof. Since limx→x0

f ′(x)

g′(x)exists, there exists a deleted neighbourhood D0 of x0

such that g′(x) 6= 0 for x ∈ D0∩ [a, b]. By Cauchy’s generalized mean value theorem(Theorem 2.32), for every x ∈ D0, there exists ξx between x and x0 such that

f(x)

g(x)=f(x)− f(x0)

g(x)− g(x0)=f ′(ξx)

g′(ξx).

Since |ξx−x0| < |x−x0| and limx→x0

f ′(x)

g′(x)exists, by using the limits of composition of

functions, limx→x0

f ′(ξx)

g′(ξx)exists and it is equal to lim

x→x0

f ′(x)

g′(x)(See Exercise 2.4). Thus,

limx→x0

f(x)

g(x)exists and lim

x→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x). This completes the proof.

The following theorem is proved by modifying the arguments in the proof ofTheorem 2.36 .

Theorem 2.34 (L’Hospital’s rule) Suppose f and g are continuous functions on[a, b] which are differentiable at every point in (a, b), except possibly at x0 ∈ [a, b].

Suppose limx→x0

f(x) = 0 = limx→x0

g(x) and limx→x0

f ′(x)


x→x0

f(x)

g(x)exists

and

limx→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x).

Proof. Let f(x) =

f(x) ifx 6= x00 ifx = x0

and g(x) =

g(x) ifx 6= x00 ifx = x0

. Then,

the result is obtained from Theorem 2.36 by taking f and g in place of f and g,respectively.

The following theorem is proved using the above theorem by using the changeof variable x 7→ y := 1/x.

2L’Hospital is pronounced as Lopital. The rule is named after the 17th-century French mathe-matician Guillaume de l’Hospital, who published the rule in his book Analyse des Infiniment Petitspour l’Intelligence des Lignes Courbes (i.e., Analysis of the Infinitely Small to Understand CurvedLines) (1696), the first textbook on differential calculus.

Theorem 2.35 (L’Hospital’s rule) Suppose f and g are differentiable at every

point in (a,∞) for some a > 0. Suppose limx→∞

f(x) = 0 = limx→∞

g(x) and limx→∞

f ′(x)

g′(x)

exists. Then limx→∞

f(x)

g(x)exists and

limx→∞

f(x)

g(x)= lim

x→∞

f ′(x)

g′(x).

Proof. Let f(y) = f(1/y) and g(y) = g(1/y) for 0 < y < 1/a. We note that

limx→∞

f(x) = 0 = limx→∞

g(x) ⇐⇒ limy→0

f(y) = 0 = limy→0

g(y).

Also, since

f ′(y) = [f(1/y)]′ = f ′(1/y)(−1/y2), g′(y) = [g(1/y)]′ = g′(1/y)(−1/y2),

we have

limx→∞

f ′(x)

g′(x)exists ⇐⇒ lim

y→0

f ′(y)

g′(y)exists.

Hence, applying Theorem 2.36 to f , g instead of f, g, we obtain the result.

The following theorem also holds.

Theorem 2.36 (L’Hospital’s rule) Suppose f and g are continuous functions on[a, b] which are differentiable at every point in (a, b), except possibly at x0 ∈ [a, b].

Suppose limx→x0

f(x) = ∞ = limx→x0

g(x) and limx→x0

f ′(x)


x→x0

f(x)

g(x)exists

and limx→x0

f(x)

g(x)= lim

x→x0

f ′(x)

g′(x).

Proof. First consider the case of β := limx→x0

f ′(x)

g′(x)6= 0. In this case, since

limx→x0

f(x) =∞ = limx→∞

g(x) ⇐⇒ limx→x0

(1/f(x)) = 0 = limx→x0

(1/g(x)),

the result follows from Theorem 2.35 by interchanging the roles of f and g.

To consider the general case where β is not necessarily zero, let α, x ∈ R suchthat |x− x0| < |α − x0| and α sufficiently close to x0 such that γ(x) 6 g(α). This isguaranteed because, g′(x) 6= 0 for x sufficiently close to x0. Then there exists ξx,αlying between α and x such that have

f ′(ξx,α)

g′(ξx,α)=f(x)− f(α)

g(x)− g(α)=f(x)

g(x)

[1− f(α)

f(x)

][1− g(α)

g(x)

]

so that

f(x)

g(x)=f ′(ξx,α)

g′(ξx,α)

[1− f(α)

f(x)

][1− g(α)

g(x)

] .Now, for each such α we have

limx→x0

[1− f(α)

f(x)

]= 1 = lim

x→x0

[1− g(α)

g(x)

].

Also, since |ξx,α − x0| < |α − x0|, by using the limits of composition of functions(See Exercise 2.4) we have

limα→x0

f ′(ξx,α)

g′(ξx,α)= lim

x→x0

f ′(x)

g′(x).

Hence, using (ε, δ) arguments, it can be shown that limα→x0

f ′(ξx,α)

g′(ξx,α)

[1− f(α)

f(x)

][1− g(α)

g(x)

] exists

so that limx→x0

f(x)

g(x)exists and

limx→x0

f(x)

g(x)= lim

α→x0

f ′(ξx,α)

g′(ξx,α)

[1− f(α)

f(x)

][1− g(α)

g(x)

] = limx→x0

f ′(x)

g′(x).


Exercise 2.21 Fill gaps in the proof of the above theorem. J

Remark 2.11 The cases

(i) limx→−∞

f(x) = 0 = limx→−∞

g(x),

(ii) limx→x0

f(x) = −∞ = limx→x0

g(x)

can be treated analogously to the cases already discussed in the above theorems.

Taylor’s formula

Now, we state another important formula in calculus.

Theorem 2.37 (Taylor’s formula) Suppose f is defined and has derivativesf (1)(x), f (2)(x), . . . , f (n+1)(x) for x in an open interval I and x0 ∈ I. Then, forevery x ∈ I, there exists ξx between x and x0 such that

f(x) = f(x0) +

n∑j=1

f (j)(x0)

j!(x− x0)j +

f (n+1)(ξx)

(n+ 1)!(x− x0)n+1.


Equivalently, for h with x0 + h ∈ I, there exists ξx between x and x0 such that

f(x0 + h) = f(x0) +

n∑j=1

f (j)(x0)

j!hj +

f (n+1)(ξx)

(n+ 1)!hn+1.

Proof. Let x ∈ I with x 6= x0, and let

Pn(t) = f(x0) +n∑j=1

f (j)(x0)

j!(t− x0)j , t ∈ I.

Then Pn is a polynomial of degree n, Pn(x0) = f(x0) and

P (j)n (x0) = f (j)(x0), j ∈ 1, . . . , n.

Now, letg(t) = f(t)− Pn(t)− ϕ(x)(t− x0)n+1, t ∈ I,

where

ϕ(x) :=f(x)− Pn(x)

(x− x0)n+1.

Note that, by this choice of ϕ(x), we have g(x0) = 0 and g(x) = 0. Also, we have

g(1)(x0) = 0, g(2)(x0) = 0, . . . , g(n)(x0) = 0.

Hence, by Rolle’s theorem, there exists x1 between x0 and x such that g′(x1) = 0.Again, by Rolle’s theorem, there exists x2 between x0 and x1 such that g′′(x2) = 0.Continuing this, there exists ξx := xn+1 between x0 and xn such that g(n+1)(ξx) = 0.But,

g(n+1)(t) = f (n+1)(t)− P (n+1)n (t)− ϕ(x)(n+ 1)! = f (n+1)(t)− ϕ(x)(n+ 1)!.

Thus, we have

ϕ(x) =f (n+1)(ξx)

(n+ 1)!,

so that

f(x) = f(x0) +

n∑j=1

f (j)(x)

j!(x− x0)j +

f (n+1)(ξx)

(n+ 1)!(x− x0)n+1.

Thus the proof is complete.3

Definition 2.19 In the Taylor’s formula (Theorem 2.37, the term

Rn(x) :=f (n+1)(ξx)

(n+ 1)!(x− x0)n+1

is called the remainder term in the formula. 3This proof adapted from S. Ghorpade & B.V. Limaye: A Course in Calculus and Analysis,

Springer, 2006


We observe that if f is infinitely differentiable and if

|Rn(x)| → 0 as n→∞

for every x ∈ I, then

f(x) = f(x0) +∞∑n=1

f (n)(x0)

j!(x− x0)n, x ∈ I. (∗)

Definition 2.20 If f can be represented as a series as in (∗), for all x in a neigh-bourhood of x0, then such a series is called the Taylor’s series of f around thepoint x0.

A natural question that one may ask is:

Doe every infinitely differentiable function in a neighbourhood of x0 hasa Taylor’s series expansion?

Unfortunately, the answer is negative. For example, if we define

f(x) =

e−1/x

2, x 6= 0,

0, x = 0,

then it can be seen that f(0) = 0 and fk)(0) = 0 for all k ∈ N. Thus, f does nothave the Taylor’s series expansion around the point 0,

EXAMPLE 2.26 Using Taylor’s formula, we shall show that

sinx =∞∑n=0

(−1)nx2n+1

(2n+ 1)!∀x ∈ R.

For this, let f(x) = sinx and x0 = 0. Since f is infinitely differentiable, and

f2j(0) = 0, f2j−1(0) = (−1)j ∀ j ∈ N,

we have

f(x) = f(x0) +

2n+1∑j=1

f (j)(0)

j!xj +

f (2n+2)(ξx)

(2n+ 2)!x2n+2

= f(x0) +

n∑j=0

f (2j+1)(0)

(2j + 1)!x2j+1 +

f (2n+2)(ξx)

(2n+ 2)!x2n+2

= f(x0) +n∑j=0

(−1)j

(2j + 1)!x2j+1 +

f (2n+2)(ξx)

(2n+ 2)!x2n+2

Also, since | sinx| ≤ 1, we have

∣∣∣f (2n+2)(ξx)x2n+2

(2n+ 2)!

∣∣∣ ≤ |x|2n+2

(2n+ 2)!→ 0 as n→∞.

Therefore,

∣∣∣f(x)−[f(x0) +

n∑j=0

(−1)j

(2j + 1)!x2j+1

]∣∣∣→ 0 as n→∞

and hence, sinx =∞∑n=0

(−1)nx2n+1

(2n+ 1)!∀x ∈ R.

Exercise 2.22 Suppose f is infinitely differentiable in an open interval I and x0 ∈ I.Further, suppose that there exists M > 0 such that

|f (k)(x)| ≤M ∀x ∈ I, ∀k ∈ N ∪ 0.

Then show that

f(x) = f(x0) +∞∑n=1

f (n)(x0)

j!(x− x0)n, x ∈ I.

J

Exercise 2.23 Using Taylor’s formula, prove the following:

(i) cosx =∞∑n=0

(−1)nx2n

(2n)!for all x ∈ R.

(ii) ex =

∞∑n=0

xn

n!for all x ∈ R.

(iii)1

1− x=

∞∑n=0

xn for all x with |x| < 1.

(iv) tan−1 x =

∞∑n=0

(−1)nx2n+1

2n+ 1for all x ∈ R.

Deduce Madhava-Gregory series:π

4=

∞∑n=0

(−1)n

2n+ 1.

J

2.3.4 Increasing and decreasing functions

Definition 2.21 Let f be a function on a set D ⊆ R. Then f is said to be

(i) monotonically increasing or increasing on D if

x, y ∈ D, x ≤ y =⇒ f(x) ≤ f(y),

(ii) strictly increasing on D if

x, y ∈ D, x < y =⇒ f(x) < f(y),

(iii) monotonically decreasing or decreasing on D if

x, y ∈ D, x ≤ y =⇒ f(x) ≥ f(y).

(iv) strictly decreasing on D if

x, y ∈ D, x < y =⇒ f(x) > f(y).

Theorem 2.38 Let f be continuous on [a, b] and differentiable on (a, b). Then

(i) f is increasing iff f ′(x) ≥ 0 for all x ∈ (a, b).

(ii) f is decreasing iff f ′(x) ≤ 0 for all x ∈ (a, b).

(iii) f is strictly increasing if f ′(x) > 0 for all x ∈ (a, b).

(iv) f is strictly decreasing if f ′(x) < 0 for all x ∈ (a, b).

Proof. (i) Suppose f is increasing and x ∈ (a, b). Note that

f ′(x) = limh→0

f(x+ h)− f(x)

h= lim

h→0+

f(x+ h)− f(x)

h.

Since, f(x+h)−f(x)h ≥ 0 for h > 0, from the above equality we obtain f ′(x) ≥ 0.

To see the converse and (iii), let x1, x2 ∈ [a, b] with x1 < x2. Then, by meanvalue theorem, theorem there exists ξ ∈ (x1, x2) such that

f(x2)− f(x1) = f ′(ξ)(x2 − x1).

Hence, if f ′(x) ≥ 0 (respectively, f ′(x) > 0) for every x ∈ (a, b), then f(x1) ≤ f(x2)(respectively, f(x1) < f(x2)). Thus, (i) and (iii) are proved. Similar arguments willlead to the proof of (ii) and (iv).

EXAMPLE 2.27 Consider the function f(x) = x4 for x ∈ R. Then we havef ′(x) = 4x3 for all x ∈ R. Note that

f ′(x) > 0 ∀x > 0 and f ′(x) < 0 ∀x < 0.

Hence,

f is strictly increasing on (0,∞), and

f is strictly decreasing on (0,∞).

2.3.5 More about local maxima and local minima

Theorem 2.39 (A sufficient condition) Suppose f is continuous on an intervalI and x0 is an interior point of I. Further suppose that f is differentiable in adeleted nbd of x0.

(i) If there exists an open interval I0 ⊆ I containing x0 such that

f ′(x) > 0 ∀x ∈ I0, x < x0 and f ′(x) < 0 ∀x ∈ I0, x > x0,

then f has loal maximum at x0.

(ii) If there exists an open interval I0 ⊆ I containing x0 such that

f ′(x) < 0 ∀x ∈ I0, x < x0 and f ′(x) > 0 ∀x ∈ I0, x > x0,

then f has local minimum at x0.

Proof. (i) Let x ∈ I0. Then, by mean value theorem, there exists ξx between x0and x such that

f(x)− f(x0) = f ′(ξx)(x− x0).

By assumption,

x < x0 =⇒ f ′(ξx) > 0 and x > x0 =⇒ f ′(ξx) < 0.

Hence, in both the cases, we have f(x) < f(x0) so that has maximum at x0. Thus,(i) is proved.

Similar arguments will lead to the proof of (ii).

EXAMPLE 2.28 Consider

f(x) = x4, g(x) = 1− x4, |x| < 1.

Then f ′(x) = 4x3 is negative for x < 0 and positive for x > 0. Hence, by Theorem2.39, f has local minimum at 0. Also, g′(x) = −4x3 is positive for x < 0 andnegative for x > 0. Hence, by Theorem 2.39, g has local maximum at 0.

Remark 2.12 The conditions given in Theorem 2.39 are cannot be dropped. Forexample, consider f(x) = x3, x ∈ R. Then we see that f ′(x) = 3x2 > 0 for positivex and negative x. Note that f does not have extremum at 0.

Theorem 2.40 (Another sufficient condition) Suppose f is defined on an in-terval I and x0 is an interior point of I. Further, suppose that f continuously twicedifferentiable in a neighbourhood of x0 and f ′(x0) = 0. Then we have the following:

(i) If f ′′(x0) < 0, then f has local maximum at x0.

(ii) If f ′′(x0) > 0, then f has local minimum at x0.

Proof. By Taylor’s theorem, there exists an open interval I0 containing x0 suchthat for every x ∈ I0, there exists ξx between x0 and x such that

f(x)− f(x0) = f ′(x0)(x− x0) +f ′′(ξx)

2(x− x0)2 =

f ′′(ξx)

2(x− x0)2. (∗)

(i) Suppose f ′′(x0) < 0. Since f ′′ is continuous in a nbd of x0, there exists anopen interval I1 containing x0 such that for all x ∈ I1,

f ′′(x) ≤ f ′′(x0)

2.

In particular, from (∗), we obtain

f(x)− f(x0) =f ′′(ξx)

2(x− x0)2 < 0 ∀x ∈ I1.

Thus, f has a maximum at x0.

(ii) Suppose f ′′(x0) > 0. Then, we obtain reverse of the inequalities in the proofof (i), and arrive the conclusion that f has a minimum at x0.

Remark 2.13 The conditions given in Theorem 2.40 are only sufficient conditions.There are functions f for which non of the conditions (i) and (ii) are satisfied at apoint x0, still f can have local extremum at x0. For example, consider

f(x) = x4, g(x) = 1− x4, |x| < 1.

Then f ′(0) = 0 = g′(0), f has local minimum at 0 and g has local maximum at 0.But, f ′′(0) = 0 = g′′(0).

Remark 2.14 How to identify critical points and extreme points of a function?

1. Suppose f is defined on an open interval I.

(a) Find those points at which either f is not differentiable or f ′ vanish.These points are the critical points of f .

(b) Suppose f ′(x0) = 0.

i. If f ′(x) has the same sign for x on both side of x0, then f does nothave an extremum at x0. Otherwise,

ii. use the test for maximum or minimum as given in Theorem 2.39.

2. Suppose f is continuous on [a, b] and differentiable on (a, b).

(a) f can have maximum or minimum only the at the end points of [a, b] orat those points in (a, b) at which f ′ vanishes.

(b) Use the tests as in Theorem 2.39 or Theorem 2.40.


1. Prove that the function f(x) = |x|, x ∈ R is not differentiable at 0.

2. Consider a polynomial p(x) = a0 + a1x2 + . . . + anx

n with real coefficients

a0, a1, . . . , an such that a0 +a12

+a23

+ . . .+ann+ 1

= 0. Show that there exists

x0 ∈ R such that p(x0) = 0.

[Note that the conclusion need not hold if the condition imposed on the coef-ficients is dropped. To see this, consider p(x) = 1 + x2.]

3. Let I and J be open intervals and f : I → J be bijective and differentiable atevery x0 ∈ I. If f ′(x0) 6= 0, then show that the inverse function f−1 : J → Iis also differentiable at x0 and and (f−1)′(x0) = 1/f ′(x0).

4. Using Taylor’s theorem, show that

(1 + x)n = 1 + nx+n(n− 1)

2!x2 +

n(n− 1)(n− 2)

3!x3 + . . .+ xn.

5. Show that there exists no differentiable function f : [0, 1]→ R which is differ-

entiable on (0, 1) such that f ′(x) =

0, if 0 < x < 1/2,1, if 1/2 ≤ x < 1.

[Hint: Use Example 2.25 in the interval [0, 1/2] and [1/2, 1] taking x0 = 1/2,and show that the resulting function f is not differentiable at x0 = 1/2.]

6. Suppose f is differentiable on (0,∞) and limx→∞

f ′(x) = 0.

Prove that limx→∞

[f(x+ 1)− f(x)] = 0.

3

Definite Integral

3.1 Introduction

In school you must have come across the definition of integral of a function f :[a, b]→ R between the limits a and b, i.e.,∫ b

af(x)dx

as the number g(b)− g(a), where g : [a, b]→ R is such that its derivative is f .

One immediate question one raises would be the following:

Given any function f : [a, b] → R, does there exist a differentiable functiong : [a, b]→ R such that g′(x) = f(x)?

Obviously, it is not necessary to have such a function g (See the exercise below).

Exercise 3.1 Consider the function f(x) =

−1, −1 ≤ x ≤ 0,

1, 0 ≤ x ≤ 1. Show that there

does not exist a differentiable function g : [−1, 1]→ R such that g′(x) = f(x) for allx ∈ (0, 1).

Another point one recalls from school is that if g : [a, b] → R is a differentiablefunction, then g′(x) has a geometric meaning, namely, it represents the slope of thetangent to the graph of g at the point x.

Do we have a geometric meaning to the integral∫ ba f(t)dt?

We answer both the above questions affirmatively for a certain class of functionsby giving a geometric definition of the concept of integral.

Suppose f : [a, b] → R is a bounded function. Our attempt is to associate anumber γ to such a function such that, in case f(x) ≥ 0 for x ∈ [a, b], then γ is thearea of the region, say Rf , bounded by the graph of f , x-axis, and the ordinates ata and b. We may not succeed to do this for all bounded functions f .

Suppose, for a moment, f(x) ≥ 0 for all x ∈ [a, b]. Let us agree that we havesome idea about what the area under the graph of f . Then it is clear that

m(b− a) ≤ area(Rf ) ≤M(b− a), (3.1.1)

74

Lower and Upper Sums 75

where m = infx∈[a,b] f(x) and M = supx∈[a,b] f(x). Thus, we get estimates forarea(Rf ). To get better estimates, let us consider a point c such that a < c < b.Then we have

m1 ≤ f(x) ≤M1 ∀x ∈ [a, c]; m2 ≤ f(x) ≤M2 ∀x ∈ [c, b],

where

m1 = infx∈[a,c]

f(x), m2 = infx∈[c,b]

f(x), M1 = supx∈[a,c]

f(x), M2 = supx∈[c,b]

f(x).

Then it is obvious that

m1(c− a) +m2(b− c) ≤ area(Rf ) ≤M1(c− a) +M2(b− c). (3.1.2)

Sincem(b− a) = m(c− a) +m(b− c) ≤ m1(c− a) +m2(b− c),

M(b− a) = M(c− a) +M(b− c) ≥M1(c− a) +M2(b− c),

we can infer that the estimates in (3.1.2) are better than those in (3.1.1). We canimprove these bounds by taking more and more points in [a, b]. This is the basicidea of Riemann integration.

3.2 Upper and Lower Sums

Let f : [a, b] → R be a bounded function, and let P be a partition of [a, b], i.e., afinite set P = x0, x1, x2, . . . , xk of points in [a, b] such that

a = x0 < x1 < x2 < . . . < xk = b.

We shall denote such partitions also by P = xiki=0.

Corresponding to the partition P = xiki=0 and the function f , we associatetwo numbers:

L(P, f) :=k∑i=1

mi∆xi, U(P, f) :=k∑i=1

Mi∆xi,

where, for i = 1, . . . , k, ∆xi = xi − xi−1, and

mi = inff(x) : xi−1 ≤ x ≤ xi, Mi = supf(x) : xi−1 ≤ x ≤ xi.

Remark 3.1 If f is continuous, then L(P, f) abd U(P, f) can be represented as

L(P, f) :=k∑i=1

f(ci)∆xi, U(P, f) :=k∑i=1

f(di)∆xi,

respectively, for some ci, di, i ∈ 1, . . . , k.

76 Definite Integral M.T. Nair

Definition 3.1 The quantities L(P, f) and U(P, f) are called lower sum andupper sum, respectively, of the function f associated with the partition P .

Note that if f(x) ≥ 0 for all x ∈ [a, b], then L(P, f) is the total area of therectangles with lengths mi and widths ∆xi, and U(P, f) is the total area of therectangle with lengths Mi and widths ∆xi for i = 1, . . . , k. Thus, it is intuitivelyclear that the required area, say γ, under the graph of f must satisfy the relation:

L(P, f) ≤ γ ≤ U(P, f)

for all partitions P of [a, b].

Throughout this chapter, functions defined on [a, b] are considered to be boundedfunctions, and P denotes the set of all partitions of [a, b]. Thus we have

L(P, f) ≤ U(P, f) ∀P ∈ P.

NOTATION: If P and Q are partitions of [a, b], then we denote by P ∪ Q thepartition obtained by taking all points in P and Q with the condition that commonpoints would be taken only once.

Exercise 3.2 For P,Q ∈ P, let P = P ∪Q. Prove that

L(P, f) ≤ L(P , f), U(P , f) ≤ U(Q, f) and L(P, f) ≤ U(Q, f).

J

We observe that

m(b− a) ≤ L(P, f) ≤ U(P, f) ≤M(b− a) ∀P ∈ P,

where m = inff(x) : a ≤ x ≤ b and M = supf(x) : a ≤ x ≤ b. In viewof this, the set L(P, f) : P ∈ P is bounded above by M(b − a) and the setU(P, f) : P ∈ P is bounded below by m(b− a). Hence,

αf := supL(P, f) : P ∈ P,βf := infU(P, f) : P ∈ P

exist.

Exercise 3.3 Show that

L(P, f) ≤ αf ≤ βf ≤ U(Q, f)

for all P,Q ∈ P. (Hint: Exercise 3.2.) J

Integrability and Integral 77

3.3 Integrability and Integral

Definition 3.2 If there exists a unique γ such that

L(P, f) ≤ γ ≤ U(P, f) ∀P ∈ P,

then we say that f is Riemann integrable on [a, b], and this γ is called theRiemann integral of f , and it is is denoted by∫ b

af(x) dx.

Remark 3.2 In the due course, Riemann integral will be simply referred to asintegral

The proof of the following theorem is left as an exercise.

Theorem 3.1 A bounded function f : [a, b]→ R is integrable if and only if αf = βf .

Definition 3.3 The quantities αf and βf are known as lower integral and upperintegral, respectively, and they are usually denoted by∫ b

af(x)dx and

∫ b

af(x)dx, respectively.

Remark 3.3 Not all functions are integrable! For example, consider f : [a, b]→ Rdefined by

f(x) =

0, x ∈ Q,1, x 6∈ Q.

For this function we have L(P, f) = 0 and U(P, f) = b − a for any partition P of[a, b]. Thus, in this case αf = 0, βf = b− a, and hence, f is not integrable.

Theorem 3.2 Suppose f is integrable on [a, b], and m, M are such that

m ≤ f(x) ≤M ∀x ∈ [a, b].

Then

m(b− a) ≤∫ b

af(x) dx ≤M(b− a).

In particular, if M0 > 0 is such that |f(x)| ≤M0 for all x ∈ [a, b], then∣∣∣∣∫ b

af(x) dx

∣∣∣∣ ≤M0(b− a).


Proof. We know that for any partition P on [a, b],

m(b− a) ≤ L(P, f) ≤∫ b

af(x) dx ≤ U(P, f) ≤M(b− a).

Hence the result.

Definition 3.4 Suppose f : [a, b]→ R is integrable. Then we define∫ a

bf(x) dx := −

∫ b

af(x) dx.

Also, for any function function f : [a, b]→ R, we define∫ τ

τf(x) dx := 0 ∀ τ ∈ [a, b].

EXAMPLE 3.1 Let f(x) = c for all x ∈ [a, b], for some c ∈ R. Then we have

L(P, f) = c(b− a), U(P, f) = c(b− a)

Hencec(b− a) = L(P, f) ≤ αf ≤ βf ≤ U(P, f) = c(b− a).

Thus, f is integrable and∫ ba f(x)dx = c(b− b).

EXAMPLE 3.2 Let f(x) = x for all x ∈ [a, b]. Let

xi = a+ i(b− a)

n, i = 0, 1, . . . , n.

Then Pn = xini=0 is a partition of [a, b]. In this case we have

mi = xi−1, Mi = xi, ∆xi =b− an

for i = 1, . . . , n.

Hence

L(Pn, f) =

n∑i=1

xi−1∆xi =

n∑i=1

[a+ (i− 1)

(b− a)

n

](b− a)

n,

U(Pn, f) =

n∑i=1

xi∆xi =

n∑i=1

[a+ i

(b− a)

n

](b− a)

n.

It is see that

L(Pn, f) = a(b− a)− (b− a)2

n+ (b− a)2

n(n+ 1)

2n2,

U(Pn, f) = a(b− a) +(b− a)2

n2n(n+ 1)

2

as n → ∞. Thus, both L(Pn, f) and U(Pn, f) converge to the same limit(b2 − a2)/2. Now, since

L(Pn, f) ≤ αf ≤ βf ≤ U(Pn, f) ∀n ∈ N,

the function f , is integrable and∫ ba x dx = (b2 − a2)/2.

3.3.1 Some necessary and sufficient conditions for integrability

In Example 3.2, what we have showed is that for a sequence (Pn) of partitions,the sequences U(Pn, f) and L(Pn, f) converge to the same point. We shall see(Corollary 3.4 below) that this is true in general.

Theorem 3.3 A bounded function f : [a, b]→ R is Riemann integrable if and onlyif for every ε > 0, there exists a partition P of [a, b] such that

U(P, f)− L(P, f) < ε.

Proof. Suppose f is integrable and let γ be its integral. Then we have αf = βf .Let ε > 0 be given. By the definition of αf and βf , there exists partitions P ′ andP ′′ of [a, b] such that

αf − ε < L(P ′, f) < αf + ε, βf − ε < U(P ′′, f) < βf + ε.

Let P = P ′ ∪ P ′′ . Then we have

αf − ε < L(P ′, f) ≤ L(P, f) ≤ U(P, f) ≤ U(P ′′, f) < βf + ε.

Since αf = γ = βf , we have

γ − ε < L(P, f) ≤ U(P, f) < γ + ε.

This proves that U(P, f)− L(P, f) < ε.

Conversely, suppose for every ε > 0, there exists a partition Pε of [a, b] suchthat U(Pε, f) − L(Pε, f) < ε. Since L(Pε, f) ≤ αf ≤ βf ≤ U(Pε, f) we obtain thatβf − αf < ε for every ε > 0. This shows that αf = βf . Thus, f is integrable.

Corollary 3.4 A bounded function f : [a, b]→ R is Riemann integrable if and onlyif there exists a sequence (Pn) of partitions of [a, b] such that

U(Pn, f)− L(Pn, f)→ 0 as n→∞

and in that case

L(Pn, f)→∫ b

af(x) dx and U(Pn, f)→

∫ b

af(x) dx

as n→∞.

Proof. The proof follows from Theorem 3.3. However, we give details:

Suppose f is integrable. Then, by Theorem 3.3, for each n ∈ N, there exists apartition Pn of [a, b] such that

U(Pn, f)− L(Pn, f) <1

n.

Thus, U(Pn, f)− L(Pn, f)→ 0 as n→∞.

Conversely, suppose there exists a sequence (Pn) of partitions of [a, b] such thatU(Pn, f)− L(Pn, f)→ 0 as n→∞. Since

L(Pn, f) ≤ αf ≤ βf ≤ U(Pn, f) ∀n ∈ N,

we obtain that αf = βf . Thus, f is integrable.

We use Corollary 3.4 to prove the following theorem.

Theorem 3.5 Let f and g be integrable over [a, b]. Then f + g is integrable and∫ b

a[f(x) + g(x)]dx =

∫ b

af(x) dx+

∫ b

af(x) dx.

Proof. By Corollary 3.4, there exists sequences of partitions (P ′n) and (P ′′n ) of[a, b] such that

U(P ′n, f)− L(P ′n, f)→ 0, U(P ′′n , g)− L(P ′′n , g)→ 0

as n→∞. For each n ∈ N, let Pn = P ′n ∪ P ′′n . Then we have

L(P ′n, f) ≤ L(Pn, f) ≤∫ b

af(x)dx ≤ U(Pn, f) ≤ U(P ′n, f),

L(P ′′n , f) ≤ L(Pn, f) ≤∫ b

ag(x)dx ≤ U(Pn, f) ≤ U(P ′′n , g).

Hence,U(Pn, f)− L(Pn, f)→ 0, U(Pn, g)− L(Pn, g)→ 0 (∗)

as n→∞. We note that for every y ∈ [c, d] ⊆ [a, b],

infc≤x≤d

f(x) + infc≤x≤d

g(x) ≤ f(y) + g(y) ≤ supc≤x≤d

f(x) + supc≤x≤d

g(x).

Hence,

L(Pn, f) + L(Pn, g) ≤ L(Pn, f + g) ≤ U(Pn, f + g) ≤ U(Pn, f) + U(Pn, g). (∗∗)

Hence, from (∗),

U(Pn, f + g)− L(Pn, f + g)→ 0 as n→∞.

Thus, by Corollary 3.4, f + g is integrable on [a, b]. Writing

an := L(Pn, f) + L(Pn, g), bn := U(Pn, f) + U(Pn, g),

(∗∗) implies

an ≤∫ b

af(x)dx+

∫ b

ag(x)dx ≤ bn,

an ≤∫ b

a[f(x) + g(x)]dx ≤ bn.

Note that bn − an → 0 as n→∞. Thus, we obtain∫ b

a[f(x) + g(x)]dx =

∫ b

af(x) dx+

∫ b

af(x) dx.


Theorem 3.6 Suppose f is integrable on [a, c] and [c, b]. Then f is integrable on[a, b], and ∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.

Proof. Let f1 = f |[a,c], f2 = f |[c,b]. Let ε > 0 be given. Since f1 and f2 areintegrable, there exist partitions P1 and P2 of [a, c] and [c, b] respectively such that

U(P1, f1)− L(P1, f1) <ε

2, U(P2, f2)− L(P2, f2) <

ε

2.

Suppose P = P1 ∪ P2. Then, it can be seen that

L(P, f) = L(P1, f1) + L(P2, f2), U(P, f) = U(P1, f1) + U(P2, f2).

Hence,

U(P, f)− L(P, f) = [U(P1, f1)− L(P1, f1)] + [U(P2, f2)− L(P2, f2)] < ε.

Thus f is integrable. Since

L(P, f) ≤∫ b

af(x)dx ≤ U(P, f),

L(P1, f1) + L(P2, f2) ≤∫ c

af(x)dx+

∫ b

cf(x)dx ≤ U(P1, f1) + U(P2, f2),

it follows that ∣∣∣∣∫ c

af(x) dx+

∫ b

cf(x) dx−

∫ b

af(x) dx

∣∣∣∣ < ε.

This is true for all ε > 0. Hence the final result.


Theorem 3.7 If f is integrable on [a, b] such that f(x) ≥ 0 for all x ∈ [a, b], then∫ ba f(x) dx ≥ 0. More generally, suppose f and g are integrable on [a, b] such thatf(x) ≤ g(x) for all x ∈ [a, b]. Then∫ b

af(x) dx ≤

∫ b

ag(x) dx.

Proof. Since L(P, f) ≤∫ ba f(x)dx for every partition P on [a, b], and since

L(P, f) ≥ 0 by the the assumption on f , we obtain∫ ba f(x)dx ≥ 0. The general

case follows from the above by applying the above result for the function ϕ definedby ϕ(x) = g(x)− f(x), x ∈ [a, b].

Corollary 3.8 Suppose f is integrable on [a, b]. Then∣∣∣∣∫ b

af(x)dx

∣∣∣∣ ≤ ∫ b

a|f(x)|dx.

Exercise 3.4 Prove Corollary 3.8.

Definition 3.5 Let f : [a, b] → R be a function, P = xi : i = 0, 1, . . . , k be apartition of [a, b], and T = ti : i = 1, . . . , k be such that xi−1 ≤ ti ≤ xi for alli = 1, . . . , k. Such a set T is called a set of tags on P or simply a tag. The number

S(P, f, T ) :=

k∑i=1

f(ti)∆xi

is called the Riemann sum for f corresponding to the partition P and the set Tof tags on P .

We may observe that for any partition P and for any set T of tags on P ,

L(P, f) ≤ S(P, f, T ) ≤ U(P, f).

This observation together with Corollary 3.4 gives the following result.

Theorem 3.9 If (Pn) is a sequence of partitions of [a, b] such that

U(Pn, f)− L(Pn, f)→ 0 as n→∞,

then f is integrable, and

S(Pn, f, Tn)→∫ b

af(x) dx,

where Tn is any set of tags on Pn, n ∈ N.

In the appendix (Section 3.7), we shall give a characterization of Riemann inte-grability in terms of Riemann sums (see Theorem 3.20).

Integral of Continuous Functions 83

3.4 Integral of Continuous Functions

Definition 3.6 Let P = xi : i = 0, 1, . . . , k be a partition of [a, b]. Then thequantity maxxi − xi−1 : i = 1, . . . , n is called the mesh of the partition P and itis denoted by µ(P ), i,e.,

µ(P ) := maxxi − xi−1 : i = 1, . . . , k.

Theorem 3.10 Suppose f : [a, b] → R is a continuous function. Then f is inte-grable. In fact, we have the following:

(i) For every ε > 0 there exists a δ > 0 such that for every partition P of [a, b]with µ(P ) < δ, we have

U(P, f)− L(P, f) < ε.

(ii) Suppose (Pn) is a sequence of partitions of [a, b] such that µ(Pn) → 0 asn→∞, and for each n ∈ N, let Tn be a set of tags on Pn. Then the sequencesU(Pn, f), L(Pn, f) S(Pn, f, Tn) converge to the same limit

∫ ba f(x) dx.

The main property of a continuous function f : [a, b] → R that is required toprove the first part is its uniform continuity.

Definition 3.7 A real valued function f defined on an interval I is said to beuniformly continuous on I if for every ε > 0, there exists δ > 0 such that

x, y ∈ I, |x− y| < δ =⇒ |f(x)− f(y)| < ε.

Clearly, every uniformly continuous function is continuous. But, the converse isnot true. To see this, consider the function

f(x) =1

x, 0 < x < 1.

Clearly, f is continuous on I := (0, 1). Now, let ε > 0 be given. For x, y ∈ (0, 1),

|f(x)− f(y)| < ε ⇐⇒ |x− y|xy

< ε ⇐⇒ |x− y| < εxy.

Thus, for f to be uniformly continuous, there must exist a δ > 0 such that δ ≤ε|xy|, which is not possible. In particular, we are not in a position to choose a δindependent of the points x, y.

To see this fact, more clearly, consider xn = 1/n and yn = 1/(n + 1). Then weknow that |xn − yn| → 0 as n → ∞ and |f(xn) − f(yn)| = 1 for all n ∈ N. Hence,the condition in the definition of uniform continuity is not satisfied if we take ε < 1.

The above kind of situation will not arise if the interval I is closed and bounded.More specifically, we have the following theorem.

Theorem 3.11 Every real valued continuous function defined on a closed andbounded interval is uniformly continuous.

Proof. Suppose f : [a, b] → R is continuous. We have to show that for everyε > 0, there exists δ > 0 such that

|x− y| < δ =⇒ |f(x)− f(y)| < ε.

Suppose this is not true. Then there exists ε0 > 0 such that for every δ > 0, thereare x, y such that

|x− y| < δ but |f(x)− f(y)| ≥ ε0. (∗)

Taking δ = 1/n, there exists xn, yn such that

|xn − yn| < 1/n but |f(xn)− f(yn)| ≥ ε0.

Since (xn) is a bounded sequence, it has a convergent subsequence, say xkn → cfor some c ∈ R. Since |xkn − ykn | → 0, we also have the convergence ykn → c.Now, by the continuity of f , we have f(xkn) → f(c) and f(ykn) → f(c). Thus,|f(xkn)− f(ykn)| → 0. This is a contradiction to (∗).

Proof of Theorem 3.10. Let f : [a, b] → R be a continuous function, and letε > 0 be given. Let P : a = x0 < x1 < x2 . . . < xk = b be a partition of [a, b]. Then

U(P, f)− L(P, f) =

k∑i=1

(Mi −mi)(xi − xi−1).

Since f is continuous on the closed interval [a, b], there exists ξi, ηi in [xi−1, xi] suchthat Mi = f(ξi), mi = f(ηi) for i = 1, . . . , k. Hence,

U(P, f)− L(P, f) =

k∑i=1

[f(ξi)− f(ηi)](xi − xi−1).

Again, since f is uniformly continuous on [a, b], there exists δ > 0 such that

|f(t)− f(s)| < ε/(b− a) whenever |t− s| < δ.

Hence, if we take P such that µ(P ) < δ, then we have

U(P, f)− L(P, f) =k∑i=1

[f(ξi)− f(ηi)](xi − xi−1) < ε.

Therefore, by Theorem 3.3, f is integrable, and the proof of (i) is also over.

Next, suppose (Pn) is a sequence of partitions such that µ(Pn) → 0 as n → ∞.Let N ∈ N be such that µ(Pn) < δ for all n ≥ N . Then, it follows by (i) above that

Integral of Continuous Functions 85

U(Pn, f) − L(Pn, f) < ε for all n ≥ N , showing that U(Pn, f) − L(Pn, f) → 0 asn→∞. Now, the concludions in (ii) follows from the observations

L(Pn, f) ≤∫ b

af(x)dx ≤ U(Pn, f),

L(Pn, f) ≤ S(Pn, f, Tn) ≤ U(Pn, f)

for all n ∈ N.

Exercise 3.5 Suppose f : [a, b]→ R is a continuous function. Show that for everyε > 0 there exists a δ > 0 such that for every partition P of [a, b] with µ(P ) < δ, wehave

(a)∫ ba f(x)dx− L(P, f) < ε

(b) U(P, f)−∫ ba f(x)dx < ε

(c) |S(P, f, T )−∫ ba f(x)dx| < ε for any tag T on P .

Remark 3.4 The property (c) in the above exercise is written as

limµ(P )→0

S(P, f, T )→∫ b

af(x)dx.

EXAMPLE 3.3 Let f(x) = ex for all x ∈ [a, b]. Then, f is continuous. Let

hn =(b− a)

n, xi = a+ ihn, ti = xi−1, i = 1, . . . , n.

Then with Pn = xini=1 and T = tini=1, we have µ(Pn)→ 0, and

S(Pn, f, Tn) = hn

n∑i=1

ea+(i−1)hn = hnea

n∑i=1

α(i−1)n = hne

aαnn − 1

αn − 1,

where αn = ehn . Since αnn = eb−a, we have

S(Pn, f, Tn) = hneaα

nn − 1

αn − 1= ea[eb−a − 1]

hnehn − 1

= [eb − ea] hnehn − 1

.

Since limn→∞ehn−1hn

= 1, we have S(Pn, f, Tn) =→ ea[eb−a − 1] = eb − ea.

Remark 3.5 It can be also shown that if a bounded function f : [a, b] → R ispiecewise continuous, i.e., there are at most a finite number of points in [a, b] atwhich f is discontinuous, then f is integrable. For a proof of this, see Theorem 3.23in the appendix (Section ??).


3.5 Some Properties

For continuous functions f and g, we can give a simpler proof for Theorem 3.5, asin the following.

Theorem 3.12 Suppose f and g are continuous functions on [a, b]. Then∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx∫ b

ac f(x) dx = c

∫ b

af(x) dx ∀ c ∈ R.

Proof. Suppose (Pn) is a sequence of partitions on [a, b] such that µ(Pn)→ 0 asn → ∞. For each n ∈ N, let Tn be a set of tags on Pn. Since f , g are continuous,the functions f + g and c f are also continuous for every c ∈ R, and


af(x) dx, S(Pn, g, Tn)→

∫ b

ag(x) dx,

S(Pn, f + g, Tn)→∫ b

a[f(x) + g(x)] dx.

Now, since

S(Pn, f + g, Tn) = S(Pn, f, Tn) + S(Pn, g, Tn),

S(Pn, c f, Tn) = c S(Pn, f, Tn) ∀ c ∈ R,

it follows that ∫ b

a[f(x) + g(x)] dx =

∫ b

af(x) dx+

∫ b

ag(x) dx,

∫ b

ac f(x) dx = c

∫ b

af(x) dx ∀ c ∈ R.


Theorem 3.13 (Mean-value theorem) Suppose f is continuous on [a, b]. Thenthere exists ξ ∈ [a, b] such that

1

b− a

∫ b

af(x) dx = f(ξ).

Proof. Since f is continuous, we know that there exist u, v ∈ [a, b] such thatf(u) = m := min f(x) and f(v) = M := max f(x). Hence, by Theorem 3.2,

f(u) ≤ 1

b− a

∫ b

af(x) dx ≤ f(v).

Some Properties 87

Hence, by intermediate value theorem, there exists ξ ∈ [a, b] such that

1

b− a

∫ b

af(x) dx = f(ξ).

Hence the result.

Theorem 3.14 (Generalized mean-value theorem) Suppose f and g are con-tinuous on [a, b] where g(x0) 6= 0 for some x0 ∈ [a, b] and g(x) ≥ 0 for all x ∈ [a, b].Then there exists ξ ∈ [a, b] such that∫ b

af(x)g(x) dx = f(ξ)

∫ b

ag(x)dx.

Proof. Let m := inf f(x) and M = sup f(x). Then we have

m

∫ b

ag(x)dx ≤

∫ b

af(x)g(x) dx ≤M

∫ b

ag(x)dx.

Since g(x0) 6= 0 for some x0 ∈ [a, b] and g(x) ≥ 0 for all x ∈ [a, b], it follows that∫ ba g(x)dx > 0. Hence,

m ≤∫ ba f(x)g(x) dx∫ b

a g(x)dx≤M.

Therefore, by the intermediate value theorem, there exists ξ ∈ [a, b] such that

f(ξ) =

∫ ba f(x)g(x) dx∫ b

a g(x)dx.

Hence the result.

Let us illustrate the above two mean value theorems by an example.

EXAMPLE 3.4 Let f be continuous in [a, b] where a > 1. We show that thereexist ξ, η ∈ [a, b] such that∫ b

a

f(x)

xdx =

(b− a)

bf(ξ) +

(1

a− 1

b

)(∫ η

af(t)dt

).

To see this, first we apply the product rule for integration:∫ b

a

f(x)

xdx =

[1

x

∫ x

af(t)dt

]ba−∫ b

a

(− 1

x2

)(∫ x

af(t)dt

)dx. (∗)

Now, by Mean Value Theorem 3.13, there exists ξ ∈ [a, b] such that∫ x

af(t)dt = (b− a)f(ξ),


and by Generalized Mean Value Theorem 3.14, there exists η ∈ [a, b] such that∫ b

a

(− 1

x2

)(∫ x

af(t)dt

)dx =

(∫ η

af(t)dt

)∫ b

a

(− 1

x2

)dx.

Hence, from (∗), using the fact that

∫ b

a

(− 1

x2

)dx =

(1

b− 1

a

), we get

∫ b

a

f(x)

xdx =

(b− a)

bf(ξ) +

(1

a− 1

b

)(∫ η

af(t)dt

).

In particular, if a = 1, b = 2, then∫ 2

1

f(x)

xdx =

1

2f(ξ) +

1

2

(∫ η

af(t)dt

).

3.6 Some Results

3.6.1 First fundamental theorem

Theorem 3.15 Suppose f is continuous on [a, b], and for x ∈ [a, b], let

ϕ(x) =

∫ x

af(t)dt.

Then ϕ is differentiable and ϕ′(x) = f(x) ∀x ∈ (a, b).

Proof. Let x ∈ (a, b) and h > 0 be such that x+ h ∈ [a, b]. Then we have

ϕ(x+ h)− ϕ(x) =

∫ x+h

af(t)dt−

∫ x

af(t)dt =

∫ x+h

xf(t)dt.

By mean-value theorem, there exists ξh in the interval with endpoints x, x+h suchthat

1

h

∫ x+h

xf(t)dt = f(ξh).

Since f is continuous at x, we have f(ξh)→ f(x) as h→ 0. Hence

limh→0+

ϕ(x+ h)− ϕ(x)

h= f(x).

Similarly, we have

limh→0−

ϕ(x+ h)− ϕ(x)

h= f(x).

Thus, ϕ′(x) exists and ϕ′(x) = f(x).

Definition 3.8 The function ϕ in Theorem 3.15 is called the indefinite integral.

Some Results 89

3.6.2 Second fundamental theorem

Definition 3.9 A function g is called an anti-derivative or primitive of f if gis differentiable and g′(x) = f(x) for all x ∈ [a, b].

By the Theorem 3.15, if f is a continuous function, then the indefinite integralϕ(x) =

∫ xa f(t)dt is an anti-derivative of f .

Exercise 3.6 If g1 and g2 are anti-derivatives of f , then show that g1 − g2 is aconstant function. J

Theorem 3.16 (Second fundamental theorem) Suppose f is continuous on[a, b] and suppose that g is an anti-derivative of f . Then∫ b

af(t)dt = g(b)− g(a).

Proof. Let ϕ be the indefinite integral of f , i.e., ϕ(x) =∫ xa f(t)dt, x ∈ [a, b].

Then we have g′(x) = f(x) = ϕ′(x) for all x ∈ [a, b], i.e., g′(x) − ϕ′(x) = 0 for allx ∈ [a, b]. Hence, g − ϕ is a constant function. Hence, in view of the Theorem 3.15,we have

g(b)− g(a) = ϕ(b)− ϕ(a) =

∫ b

af(t)dt.


Remark 3.6 The conclusion of the above theorem is also known as Newton-Leibnizformula. The difference g(b)− g(a) is usually written as [g(x)]ba.

Here is another proof for the above theorem.

An alternate Proof. Let g be an antiderivative of f , i.e., g is differentiable andg′ = f . Let P : a = x0 < x1 < . . . < xn = b be any partition of [a, b]. Then byLagrange’s mean value theorem, there exists ξi ∈ [xi−1, xi] such that

g(xi)− g(xi−1) = g′(ξi)(xi − xi−1) = f(ξi)(xi − xi−1).

Hence,

g(b)− g(a) =

n∑j=1

[g(xi)− g(xi−1)] =

n∑j=1

f(ξi)(xi − xi−1).

Since the Riemann sum∑n

j=1 f(ξi)(xi − xi−1) is constant for any partition P , it

follows that∑n

j=1 f(ξi)(xi − xi−1) =∫ ba f(x)dx. Thus,∫ b

af(x)dx = g(b)− g(a).


Remark 3.7 For giving the alternate proof above, we have not used the continuityof f , but the fact that f is an anti-derivative, i.e., derivative of a function g. It isknown that, a function can be an anti-derivative without being continuous.

The following theorem shows that integral of a continuous function is a Riemannsum.

Theorem 3.17 Suppose f is a continuous function.Then for every partition P of[a, b], there exists a set T of tags on P such that

S(P, f, T ) =

∫ b

af(x) dx.

Proof. Let P = xi : i = 1, . . . , k be a partition of [a, b]. Since f is continuous,by mean value theorem (Theorem 3.13), there exists ξi ∈ [xi−1, xi] such that∫ xi

xi−1

f(x) dx = f(ξi)(xi − xi−1), i = 1, . . . , k.

Hence, taking T = ξi : i = 1, . . . , k,

S(P, f, T ) =

k∑i=1

f(ξi)(xi − xi−1) =

k∑i=1

∫ xi

xi−1

f(x) dx =

∫ b

af(x) dx.


3.6.3 Applications of fundamental theorem

Theorem 3.18 (Product formula) Suppose f and g are continuous functions on[a, b], and let G be an anti-derivative of g. If f is differentiable on [a, b], then∫ b

af(x)g(x) dx = [f(x)G(x)]ba −

∫ b

af ′(x)G(x) dx.

Proof. Recall that if u and v are differentiable, then

(uv)′ = u′v + uv′.

Hence, ∫ b

a[u(x)v(x)]′ dx =

∫ b

au′(x)v(x) dx+

∫ b

au(x)v′(x) dx.

Using fundamental theorem,

[u(x)v(x)]ba =

∫ b

au′(x)v(x) dx+

∫ b

au(x)v′(x) dx.

Thus, ∫ b

au(x)v′(x) dx = [u(x)v(x)]ba −

∫ b

au′(x)v(x) dx.

Now, taking f(x) = u(x) and v(x) = G(x), we obtain the required formula.

Some Results 91

Theorem 3.19 (Change of variable formula) Suppose ψ : [α, β] → R is adifferentiable function such that ψ(α) = a and ψ(β) = b. Then, for any continuousfunction f : [a, b]→ R, ∫ b

af(x) dx =

∫ β

αf(ψ(t))ψ′(t)dt.

Proof. Let F be an anti-derivative of f , i.e., such that F ′(x)f(x). Then takingG(t) = F (ψ(t)) for t ∈ [α, β], we have

G′(t) = F ′(ψ(t))ψ′(t) = f(ψ(t))ψ′(t), t ∈ [α, β].

Hence, by fundamental theorem,∫ β

αf(ψ(t))ψ′(t)dt =

∫ β

αG′(t)dt = G(β)−G(α) = F (ψ(β))− F (ψ(α)).

Hence, ∫ β

αf(ψ(t))ψ′(t)dt = F (b)− F (a) =

∫ b

af(x) dx.


The following examples have been worked out by knowing the antiderivatives ofcertain functions.

EXAMPLE 3.5 For k ≥ 0,∫ b

axk dx =

[xk+1

k + 1

]ba

=bk+1 − ak+1

k + 1.

EXAMPLE 3.6 For α 6= 0,∫ b

aeαx dx =

[eαx

α

]ba

=eαb − eαa

k + 1.

3.7 Appendix

Theorem 3.20 A bounded function f : [a, b] → R is Riemann integrable if andonly if there exists γ such that for every ε > 0, there exists a partition P of [a, b]satisfying

|S(P, f, T )− γ| < ε for every tag T on P

and in that case γ =∫ ba f(x)dx.

Proof. Let ε > 0 be given.

Suppose f is Riemann integrable. Then, by Theorem 3.3, there exists a partitionP of [a, b] such that U(P, f)−L(P, f) < ε. Since L(P, f) ≤ S(P, f, T ) ≤ U(P, f) forevery tag T on P , we obtain

|S(P, f, T )− γ| < ε for every tag T on P.

Conversely, let P be a partition of [a, b] be such that

γ − ε < S(P, f, T ) < γ + ε for every tag T on P. (∗)

Let

T ′n = t′i,n : i = 1, . . . , k, T ′′n = t′′i,n : i = 1, . . . , k

be tags on P : a = x0 < x1 < . . . < xk = b such that

t′i,n → mi, t′′i,n →Mi as n→∞.

Then

S(P, f, T ′n)→ L(P, f), S(P, f, T ′′n )→ U(P, f) as n→∞. (∗∗)

Also, from (∗), we have

γ − ε < S(P, f, T ′n) < γ + ε, γ − ε < S(P, f, T ′′n ) < γ + ε ∀n ∈ N.

Now, taking limit and using (∗∗), we have

γ − ε < L(P, f) < γ + ε, γ − ε < U(P, f) < γ + ε.

Therefore, U(P, f)− L(P, f) < ε. By Theorem 3.3, f is integrable.

As a consequence of the above theorem we have the following.

Corollary 3.21 Suppose f : [a, b]→ R is a bounded function. If (Pn) is a sequenceof partitions on [a, b] such that S(Pn, f, Tn) converges for every tag Tn on Pn foreach n ∈ N, then f is Riemann integrable, and∫ b

af(x)dx = lim

n→∞S(Pn, f, Tn).

Appendix M.T. Nair 93

Unlike Theorem 3.9, the above theorem and corollary does not seem to be usefulfor evaluation of integrals.

Knowing a sequence (Pn) of partitions, how to assert the convergence ofS(Pn, f, Tn) for every tag Tn Pn for each n ∈ N?

In this regard, analogous to Theorem 3.10(ii), we have the following result (See [3]1

for its proof).

Theorem 3.22 Suppose f : [a, b]→ R is a Riemann integrable (bounded) function.If (Pn) is a sequence of partitions on [a, b] such that µ(Pn) → 0 as n → ∞, thenS(Pn, f, Tn)


af(x)dx as n→∞

for every tag Tn on Pn for each n ∈ N.

Now, we specify a large class if functions of practical importance which areRiemann integrable.

Theorem 3.23 Suppose f : [a, b] → R is bounded and piecewise continuous,i.e.,there are at most a finite number of points in [a, b] at which f is discontinuous.Then f is integrable.

Proof. Let ε > 0 be given. We have to show that there exists a partition P suchthat U(P, f)− L(P, f) < ε.

Suppose that c ∈ (a, b) such that f is continuous on [a, c) and (c, b]. Let δ > 0 besuch that c+ δ a. Let f1, f2, f3 be restrictions of f to the intervals[a, c− δ], [c+ δ, b] and [c− δ, c+ δ], respectively. Since f is continuous on [a, c− δ]and [c+ δ, b], f is integrable on these intervals, so that there exist partitions P1 on[a, c− δ] and P2 on [c+ δ, b] such that

U(P1, f1)− L(P1, f1) <ε

3, U(P2, f2)− L(P2, f2) <

ε

3.

Assume that δ > 0 is so small that (M −m)δ < ε/6. Then for any partition P3 on[c− δ, c+ δ], we have

U(P3, f3)− L(P3, f3) <ε

3.

Now, for the partition P = P1 ∪ P2 ∪ P3 on [a, b], we have

U(P, f) = U(P1, f1) + U(P2, f2) + U(P3, f3),

L(P, f) = L(P1, f1) + L(P2, f2) + L(P3, f3).

Hence,U(P, f)− L(P, f) < ε.

Thus, f is integrable. The case of more than one (but finite number of) points ofdiscontinuity can be handled analogously.

1S.R. Ghorage and B.V. Limaye, A Course in Calculus and Analysis, Springer, 2006.

4

Improper Integrals

Recall that we defined definite integral of a function f for the case when f is abounded function defined on a closed interval [a, b]. In case the assumptions on fare not satisfied, then can we still have a notion of integral? We discuss a few suchcases.

4.1 Definitions

Definition 4.1 Suppose f is defined on [a,∞). If ϕ(t) :=∫ ta f(x) dx exists for

every t > a, and if limt→∞ ϕ(t) exists, then we define the improper integral of fover [a,∞) as ∫ ∞

af(x) dx = lim

t→∞

∫ t

af(x) dx.

Definition 4.2. Suppose f is defined on (−∞, b]. If ψ(t) :=∫ bt f(x) dx exists for

every t < b, and if limt→−∞ ψ(t) exists, then we define the improper integral off over (−∞, b] as ∫ b

−∞f(x) dx = lim

t→−∞

∫ b

tf(x) dx.

Definition 4.3. Suppose f is defined on R := (−∞,∞). If∫ c−∞ f(x) dx and∫∞

c f(x) dx exists for some c ∈ R, then we define the improper integral of fover (−∞,∞) as ∫ ∞

−∞f(x) dx =

∫ c

−∞f(x) dx+

∫ ∞c

f(x) dx.

We may observe that existence of limt→∞∫ t−t f(x) dx does not, in general, imply

existence of∫∞−∞ f(x) dx. To see this, consider the following example:

Let f(x) = x for every x ∈ R. Then we have∫ t−t f(x) dx = 0 for every t ∈ R,

but the integrals∫ c−∞ f(x) dx and

∫∞c f(x) dx do not exist.

94

Definitions 95

Next we consider the case when f is defined on an interval J of finite length, butlimx→x0 |f(x)| =∞, where x0 either belongs to J or it is an end point of J .

Definition 4.4. Suppose f is defined on (a, b]. If∫ bt f(x) dx exists for every

t ∈ (a, b), and if limδ→0

∫ ba+δ f(x) dx exists, then we define the improper integral

of f over (a, b] as ∫ b

af(x) dx = lim

δ→0

∫ b

a+δf(x) dx.

Definition 4.5. Suppose f is defined on [a, b). If∫ ta f(x) dx exists for every

t ∈ (a, b), and if limδ→0

∫ b−δa f(x) dx exists, then we define the improper integral

of f over [a, b) as ∫ b

af(x) dx = lim

δ→0

∫ b−δ

af(x) dx.

Definition 4.6. Suppose f is defined on [a, c) and (c, b]. If∫ ca f(x) dx and∫ b

c f(x) dx exist, then we define the improper integral of f over [a, b] as∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.

Now we combine the situations in Definitions 4.1, 4.2 with Definitions 4.4, 4.5,to consider improper integrals over intervals of the form (a,∞) and (−∞, b).

Definition 4.7. Suppose f is defined on (a,∞). If the integrals∫ ta f(x) dx and∫∞

t f(x) dx exist as improper integrals for every t > a, then we define the improperintegral of f over (a,∞) as∫ ∞

af(x) dx =

∫ t

af(x) dx+

∫ ∞t

f(x) dx.

Definition. Suppose f is defined on (−∞, b). If the integrals∫ t−∞ f(x) dx and∫ b

t f(x) dx exist as improper integrals for every t < b, then we define the improperintegral of f over (−∞, b) as∫ b

−∞f(x) dx =

∫ t

−∞f(x) dx+

∫ b

tf(x) dx.

In case an improper integral exists (resp. does not exists), then we also say thatthe improper integral converges (resp. diverges).

96 Improper Integrals M.T. Nair

4.1.1 Typical examples

EXAMPLE 4.1 Consider the improper integral

∫ ∞1

1

xdx. Note that

∫ t

1

1

xdx = [lnx]t1 = ln t→∞ as t→∞.

Hence,∫∞1

1x dx diverges.

EXAMPLE 4.2 Consider the improper integral

∫ ∞1

1

x2dx Note that

∫ t

1

1

x2dx =

[−1

x

]t1

= 1− 1

t→ 1 as t→∞.

Hence,∫∞1

1x2dx converges.

EXAMPLE 4.3 For p 6= 1, consider the improper integral

∫ ∞1

1

xpdx. In this

case, we have ∫ t

1

1

xpdx =

[x−p+1

−p+ 1

]t1

=t−p+1 − 1

−p+ 1.

Note that,

p > 1 =⇒ t−p+1 − 1

−p+ 1→ 1

p− 1as t→∞,

and

p < 1 =⇒ t−p+1 − 1

−p+ 1→∞ as t→∞,

Hence, ∫ ∞1

1

xpdx

converges for p > 1,diverges for p ≤ 1.

EXAMPLE 4.4 For p 6= 1, consider the improper integral

∫ 1

0

1

xpdx. In this case,

we have ∫ 1

δ

1

xpdx =

[x−p+1

−p+ 1

]1δ

=1− δ−p+1

−p+ 1.

Note that,

p > 1 =⇒ δ−p+1 − 1

−p+ 1→∞ as δ → 0,

and

p < 1 =⇒ δ−p+1 − 1

−p+ 1→ 1

1− pas δ → 0,

Integrability by Comparison 97

Hence, ∫ 1

0

1

xpdx

converges for p < 1,diverges for p ≥ 1.

EXAMPLE 4.5 Let a < b and α < 1. Then∫ ba

dx(b−x)α converges:

We observe that for a < t < b,∫ t

a

dx

(b− x)α=

∫ b−a

b−t

du

uα.

Now,

limt→b

∫ t

a

dx

(b− x)αexists ⇐⇒ lim

t→b

∫ b−a

b−t

du

uαexists

⇐⇒ limε→0

∫ b−a

ε

du

uαexists

⇐⇒ α < 1.

Exercise 4.1 Suppose f ≥ 0 and the integral∫ ba f(x)dx exists for every x ∈ [a, b)

and limx→b(b− x)αf(x) exists for some α < 1. Then∫ ba f(x)dx exists.

[Hint: Observe that for any ε > 0, there exists x0 ∈ [a, b) such that the numberβ := limx→b(b− x)αf(x) satisfies 0 ≤ f(x) ≤ β+ε

(b−x)α for all x ∈ [x0, b).]

4.2 Integrability by Comparison

We state a result which will be useful in asserting the existence of certain improperintegral by comparing it with certain other improper integral.

Suppose J is either an interval of finite or infinite length. Suppose f defined onJ , except possibly at one point (It can be generalized to the case when f is not bedefined at some finite number of point in J). We denote the improper integral of fover J by ∫

Jf(x) dx,

and say the the improper integral over J converges whenever it exists, and otherwise,we say that the improper integral

∫J f(x) dx diverges.

For example, if J = [a, b], then f may not be defined at a or at b or at somepoint c ∈ (a, b), and the corresponding improper integrals, by definitions are

limt→a

∫ b

tf(x) dx, lim

t→b

∫ t

af(x) dx, lim

t→c−

∫ t

af(x) dx+ lim

t→c+

∫ b

tf(x) dx.

Theorem 4.1 Suppose J is above, and f and g are defined on J .

(i) If 0 ≤ f(x) ≤ g(x) for all x ∈ J , and∫J g(x) dx exists, then

∫J f(x) dx

exists.

(ii) If∫J |f(x)| dx exists, then

∫J f(x) dx exists.

EXAMPLE 4.6 Since ∣∣∣∣sinxxp∣∣∣∣ ≤ 1

xp,

∣∣∣cosx

xp

∣∣∣ ≤ 1

xp

it follows from Example 4.3 and Theorem 4.1(ii) that the improper integrals∫ ∞1

sinx

xpdx and

∫ ∞1

cosx

xpdx

converge for all p > 1.

In fact∫∞1

sinxxp dx and

∫∞1

cosxxp dx converge for all p > 0 as we see in the next

example.

EXAMPLE 4.7 Let p > 0. Then for t > 0,∫ t

1

sinx

xpdx =

[1

xp(− cosx)

]t1

− p∫ t

1

1

xp+1cosx dx

=

[cos 1− cos t

tp

]− p

∫ t

1

cosx

xp+1dx.

By the result in Example 4.6,

∫ ∞1

cosx

xp+1dx converges for all p > 0. Also, cos t

tp →0 as t→∞. Hence, ∫ ∞

1

sinx

xpdx converges for all p > 0.

Similarly, we see that∫ ∞1

cosx

xpdx converges for all p > 0.

EXAMPLE 4.8 Since∣∣∣∣sinxxp∣∣∣∣ =

∣∣∣∣sinxx∣∣∣∣ 1

xp−1≤ 1

xp−1,

∣∣∣cosx

xp

∣∣∣ ≤ 1

xp

it follows from Example 4.4 above and Theorem 4.1(ii)that∫ 1

0

sinx

xpdx converges for all p < 2,

Integrability by Comparison 99

∫ 1

0

cosx

xpdx converges for all p < 1.

EXAMPLE 4.9 Observe that

sinx

xp=

sinx

x

1

xp−1≥ sin 1

xp−1∀x ∈ (0, 1].

Since∫ 10

1xp−1 dx diverges for p− 1 ≥ 1, i.e., for p ≥ 2, it follows that∫ 1

0

sinx

xpdx diverges for all p ≥ 2,

EXAMPLE 4.10 From Examples 4.8, 4.9, 4.7,∫ ∞0

sinx

xpdx converges for 0 0, the improper integral

Γ(x) :=

∫ ∞0

tx−1e−t dt

converges. The function Γ(x), x > 0, is called the gamma function.

Note that for tx−1e−t ≤ tx−1 for all t > 0, and∫ 10 t

x−1 dt converges for x > 0.Hence, by Theorem 4.1,∫ 1

0tx−1e−t dt converges for x > 0.

Also, we observe thattx−1e−t

t−2→ 0 as t → ∞, and

∫∞1 t−2dt converges. Hence, by

Theorem 4.2,∫∞1 tx−1e−t dt converges. Thus,

Γ(x) :=

∫ ∞0

tx−1e−t dt =

∫ 1

0tx−1e−t dt+

∫ ∞1

tx−1e−t dt

converges for every x > 0.

Beta function

We show that for x > 0, y > 0, the improper integral

β(x, y) :=

∫ 1

0tx−1(1− t)y−1 dt

converges. The function β(x, y) for x > 0, y > 0 is called the beta function.

Clearly, the above integral is proper for x ≥ 1, y ≥ 1. Hence it is enough toconsider the case of 0 < x < 1, 0 < y < 1. In this case both the points t = 0 andt = 1 are problematic. hence, we consider the integrals∫ 1/2

0tx−1(1− t)y−1 dt,

∫ 1

1/2tx−1(1− t)y−1 dt.

We note that if 0 < t ≤ 1/2, then (1−t)y−1 ≤ 21−y so that tx−1(1−t)y−1 ≤ 21−ytx−1.

Since∫ 1/20 tx−1 dt converges it follows that

∫ 1/20 tx−1(1− t)y−1 dt converges. To deal

with the second integral, consider the change of variable u = 1− t. Then∫ 1

1/2tx−1(1− t)y−1 dt =

∫ 1/2

0uy−1(1− u)x−1 du

which converges by the above argument. Hence,

β(x, y) :=

∫ 1

0tx−1(1− t)1−y dt, x > 0, y > 0

converges for every x > 0, y > 0.

4.3 Integrability Using Limits

Now some more results which facilitate the assertion of convergence/divergence ofimproper integrals, whose proofs follow from the definition of limits.

Theorem 4.2 Suppose f(x) ≥ 0, g(x) ≥ 0 for all x ∈ [a,∞),∫ ba f(x)dx and∫ b

a g(x)dx exists for every b > a. Suppose further thatf(x)

g(x)→ ` as x→∞.

(i) If ` 6= 0, then∫∞a f(x)dx converges ⇐⇒

∫∞a g(x)dx converges.

(ii) If ` = 0, then∫∞a g(x)dx converges ⇒

∫∞a f(x)dx converges.

Proof. (i) Suppose ` 6= 0. Then ` > 0, and for ε > 0 with `− ε > 0, there existsx0 ≥ a such that

`− ε < f(x)

g(x)< `+ ε ∀x ≥ x0.

Hence(`− ε)g(x) < f(x) < (`+ ε)g(x) ∀x ≥ x0.

Additional Exercises 101

Consequently,∫∞x0f(x)dx converges iff

∫∞x0g(x)dx converges. As

∫ x0a f(x)dx and∫ x0

a g(x)dx exist, the result in (i) follows.

(ii) Suppose ` = 0. Then for ε > 0, there exists x0 ≥ a such that

f(x)

g(x)< ε ∀x ≥ x0.

Thus, f(x) < εg(x) for all x ≥ x0. Hence, convergence of∫∞x0g(x)dx implies the

convergence of∫∞x0f(x)dx. From this the result in (ii) follows.

Exercise 4.2 Suppose f and g are non-negative continuous functions on J . Then∫ b

af(x)dx exists ⇐⇒

∫ b

ag(x)dx exists

in the following cases:

1. J = (a, b] and limx→a

f(x)

g(x)= ` and ` > 0.

2. J = [a, b) and limx→b

f(x)

g(x)= ` and ` > 0.

3. J = [a,∞) and limx→∞

f(x)

g(x)= ` and ` > 0.

4. J = (−∞, b] and limx→−∞

f(x)

g(x)= ` and ` > 0.

In 1-4 above, if ` = 0, then

∫ b

ag(x)dx exists =⇒

∫ b

ag(x)dx exists . J

4.4 Additional Exercises

1. Does

∫ ∞1

sin

(1

x2

)dx converge?

[ Hint: Note that∣∣sin ( 1

x2

)∣∣ ≤ 1x2

.]

2. Does

∫ ∞2

cosx

x(log x)2dx converge?

[Hint: Observe∣∣∣ cosxx(log x)2

∣∣∣ ≤ 1x(log x)2

and use the change of variable t = log x.]

3. Does

∫ ∞0

sin2 x

x2dx converge?

[Hint: Observe sin2 xx2≤ 1

x2for x ≥ 1 and sin2 x

x2, 0 < x ≤ 1 has a continuous

extension on [0, 1].]


4. Does

∫ 1

0

sinx

x2dx converge?

[Hint: Observe sinxx2

=(sinxx

)1x ≥

(sin 11

).]

5. Does

∫ ∞a0

f(x)dx exists implies

∫ b

af(x)dx→ 0 as a, b→∞.

[Hint: Note that∫ ba f(x)dx =

∫ ba0f(x)dx−

∫ aa0f(x)dx→ 0 as a, b→∞.]

6. Does

∫ ∞0

e−x2dx converge?

[Hint: Note that e−x2

is continuous on [0, 1], and e−x2 ≤ 1

x2for 1 ≤ x ≤ ∞.]

7. Does

∫ ∞2

sin(log x)

xdx converge?

[Hint: Use the change of variable t = log x, and the fact that∫∞log 2 sin t dt

diverges.]

8. Does

∫ 1

0lnxdx converge?

[Hint: Use the change of variable t = log x.]

5

Geometric and MechanicalApplications of Integrals

5.1 Computing Area

5.1.1 Using Cartesian Coordinates

Suppose a curve is given by an equation

y = f(x), a ≤ x ≤ b,

where f : [a, b] → R is a continuous function such that f(x) ≥ 0 for all x ∈ [a, b].Then, the “area under the curve”, i.e., the area of the region bounded by the graphof f , the x-axis, and the ordinates at x = a and x = b, is

limµ(P )→0

k∑j=1

f(ξi)∆xi =

∫ b

ay dx.

Suppose the curve is given in parametric form:

x = ϕ(t), y = ψ(t), , α ≤ t ≤ β,

such that a = ϕ(α), b = ψ(β). Then the area under the curve takes the form∫ β

αψ(t)ϕ′(t)dt.

If f takes both positive and negative values, but changes sign only at a finitenumber of points, then the area bounded by the curve, the x-axis, and the ordinatesat x = a and x = b, is given by ∫ b

a|f(x)| dx.

Suppose f : [a, b] → R and g : [a, b] → R are continuous functions such thatf(x) ≤ g(x) for all x ∈ [a, b]. Then the area of the region bounded by the graphs off and g, and the ordinates at x = a and x = b is given by

limµ(P )→0

k∑j=1

[g(ξi)− f(ξi)]∆xi =

∫ b

a[g(x)− f(x)] dx.

103

104 Geometric and Mechanical Applications of Integrals M.T. Nair

5.1.2 Using Polar Coordinates

Suppose a curve is given in polar coordinates as

ρ = ϕ(θ), α ≤ θ ≤ β,

where ϕ : [α, β] → R is a continuous function. Then the the area of the regionbounded by the graph of ϕ and the rays θ = α and θ = β is is given by

limµ(P )→0

k∑j=1

1

2[ϕ(ξi)∆θi]ϕ(ξi) = lim

µ(P )→0

k∑j=1

1

2[ϕ(ξi)]

2∆θi =1

2

∫ β

αρ2 dθ.

Thus,

Area :=1

2

∫ β

αρ2 dθ. (5.1.1)

EXAMPLE 5.1 We find the area bounded by the cures defined by

y =√x, y = x2, x ≥ 0 :

Note that the points of intersection of the curves are at x = 0 and x = 1. Also,√x ≥ x2 for 0 ≤ x ≤ 1. Hence, the required area is∫ 1

0

(√x− x2

)dx =

[x3/2

3/2− x3

3

]10

=1

3.

EXAMPLE 5.2 We find the area bounded by the ellipsex2

a2+y2

b2= 1. Let us

use the parametrization

x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.

Then the required area is

4

∫ a

0y dx = 4

∫ 0

π/2(b sin t)(−asint) dt = 2ab

∫ π/2

0(1− cos 2t dt = πab.

Next, let us consider the polar form of the ellipse. For this, consider the polarform of points (x, y) on the ellipse, i.e.,

x = ρ cos θ, y = ρ sin θ,

where (ρ, θ) satisfiesx2

a2+y2

b2= 1, i.e.

ρ2 cos2 θ

a2+ρ2 sin2 θ

b2= 1, i.e., ρ2 =

(cos2 θ

a2+

sin2 θ

b2

)−1.

Thus, the area can also be computed (Exercise) using the formula (5.1.1).

Computing Arc Length 105

EXAMPLE 5.3 We find the area bounded by one arch of the cycloid

x = a(t− sin t), y = a(1− cos t).

One arch of the cycloid is obtained by varying t over the interval [0, 2π]. Thus, therequired area is∫ 2πa

0y dx =

∫ 2π

0y(t)x′(t) dt =

∫ 2π

0a2(1− cos t)2 dt = 3πa2.

EXAMPLE 5.4 We find the area bounded by a circle of radius a. Without lossof generality assume that the centre of the circle is the origin. Then, the circle canbe represented in polar coordinates as

ρ = a.

Hence the required area is1

2

∫ 2π

0ρ2 dθ = πa2.

EXAMPLE 5.5 We find the area bounded by the lemniscate

ρ = a√

cos 2θ.

The required area is

2

[1

2

∫ π/4

−π/4ρ2 dθ

]= a2

∫ π/4

−π/4cos 2θdθ = a2.

5.2 Computing Arc Length

5.2.1 Using Cartesian Coordinates

Suppose a curve is given by and equation

y = f(x), a ≤ x ≤ b,

where f : [a, b]→ R is a continuous function which is differentiable except possiblyat a finite number of points. Then the length of the curve is given by

A := limµ(P )→0

k∑j=1

√(xi − xi−1)2 + (yi − yi−1)2,

where

yi − yi−1 = f(xi)− f(xi−1) = f ′(ξi)∆xi,


for some ξi ∈ [xi−1, xi], i = 1, . . . , k. Hence,

A := limµ(P )→0

k∑j=1

√(xi − xi−1)2 + (yi − yi−1)2

= limµ(P )→0

k∑j=1

√(∆xi)2 + [f ′(ξi)∆xi]2

= limµ(P )→0

k∑j=1

√1 + [f ′(ξi)]2∆xi

=

∫ b

a

√1 +

(dy

dx

)2

dx.

• If the curve y = f(x), a ≤ x ≤ b, is given in parametric form:

x = φ(t), y = ψ(t), α ≤ t ≤ β,

then1

A =

∫ b

a

√1 +

(dy

dx

)2

dx =

∫ β

α

√(dφ

dt

)2

+

(dψ

dt

)2

dt.

Remark 5.1 Curves in parametric form are assumed to be piecewise smooth, i.e.,having unique tangents except possibly at a finite number of points. Note thatif a curve is given in parametric form as x = φ(t), y = ψ(t) with α ≤ t ≤ b,then it has unique tangent at (x0, y0) if φ′(t0), ψ

′(t0) exists, where t0 is such that(x0 = φ(t0), y0 = ψ(t0), and |φ′(t0)|2 + |ψ′(t0)|2 6= 0.

5.2.2 Using Polar Coordinates

Suppose a curve is given in polar coordinates as

ρ = ϕ(θ), α ≤ θ ≤ β,

where ϕ : [α, β]→ R is a continuous function. Since

x = ρ cos θ, y = ρ sin θ, α ≤ θ ≤ β,

we have

A =

∫ β

α

√(dx

dθ

)2

+

(dy

dθ

)2

dθ.

Note thatdx

dθ= ρ′ cos θ + ρ(− sin θ),

dy

dθ= ρ′ sin θ + ρ cos θ.

1We used the fact that dydx

= (dy/dt)(dx/dt)

, which follows from the rule: dydt

= dydx

dxdt

.

Computing Arc Length 107

Hence, it follows that

A =

∫ β

α

√(dx

dθ

)2

+

(dy

dθ

)2

dθ

=

∫ β

α

√ρ2 +

(dρ

dθ

)2

dθ.

EXAMPLE 5.6 We find the length of the circumference of a circle of radius a.

Without loss of generality assume that the centre of the circle is the origin,i.e.,the circle is given by x2 + y2 = a2. The required length is

L := 4

∫ a

0

√1 +

(dy

dx

)2

dx, y =√a2 − x2.

Thus,

L := 4a

∫ a

0

dx√a2 − x2

= 2πa.

EXAMPLE 5.7 Now we find the length of the circle when it is represented by theequations

x = a cos θ, y = a sin θ, 0 ≤ θ ≤ 2π.

The required length is

L := 4

∫ π/2

0

√(dx

dθ

)2

+

(dy

dθ

)2

dθ

= 4

∫ π/2

0

√a2 sin2 θ + a2 cos2 θ dθ = 2πa.

EXAMPLE 5.8 Let us find the length of the ellipse

x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π.


L := 4

∫ π/2

0

√(dx

dθ

)2

+

(dy

dθ

)2

dθ

= 4

∫ π/2

0

√a2 sin2 θ + b2 cos2 θ dθ

= 4

∫ π/2

0

√a2(1− cos2 θ) + b2 cos2 θ dθ

= 4

∫ π/2

0

√a2 − (a2 − b2) cos2 θ dθ

= 4a

∫ π/2

0

√1− β2 cos2 θ dθ,

where β =√a2−b2a . The above integral is not expressible in standard form unless

β = 1, i.e., unles b = a in which case the ellipse is the circle. But, the integral canbe approximately computed numerically.

EXAMPLE 5.9 We find the length of the astroid: x = a cos3 t, y = a sin3 t.


L := 4

∫ π/2

0

√(dx

dθ

)2

+

(dy

dθ

)2

dθ

= 4

∫ π/2

0

√9a2 cos4 t sin2 t+ 9a2 sin4 t cos3 t dθ

= 12a

∫ π/2

0

√cos2 t sin2 dt = 6a.

EXAMPLE 5.10 We find the length of the cardioid ρ = a(1 + cos θ).

The required length is L :=∫ 2π0

√ρ2 + ρ′2 dθ. Since

ρ2 = a2(1 + cos θ)2, ρ′2 = a2 sin2 θ,

we have

L =√

2a

∫ 2π

0

√1 + cos θ dθ = 4a

∫ 2π

0

∣∣∣∣cosθ

2

∣∣∣∣ dθ = 8a.

5.3 Computing Volume of a Solid

Suppose that a three dimensional object, a solid, lies between two parallel planesx = a and x = b. Let α(x) be the area of the cross section of the solid at the point x,with cross section being parallel to the yz-plane. We assume that the function α(x),x ∈ [a, b] is continuous. Now, consider a partition P : a = x0 < x1 < . . . < xk = bof the interval [a, b]. Then the volume of the solid is given by

limµ(P )→0

k∑j=1

α(ξi)∆xi =

∫ b

aα(x) dx.

EXAMPLE 5.11 Let us compute the volume of the solid enclosed by the ellipsoid

x2

a2+y2

b2+z2

c2= 1.

For a fixed x ∈ [−a, a], the boundary of the cross section at x is given by the equation

y2

b2+z2

c2= 1− x2

a2,

Computing Volume of a Solid of Revolution 109

i.e.,

y2

φ(x)2+

z2

ψ(x)2= 1, where φ(x) = b

√1− x2

a2, ψ(x) = c

√1− x2

a2.

Hence,

α(x) = πφ(x)ψ(x) = πbc

(1− x2

a2

),

and the required volume is

V :=

∫ a

−aα(x) dx = πbc

∫ a

−a

(1− x2

a2

)dx =

4

3πabc.

In particular, volume of the solid bounded by the sphere x2 + y2 + z2 = a2 is 43πa

3.

5.4 Computing Volume of a Solid of Revolution

Suppose a solid is obtained by revolving a curve y = f(x), a ≤ x ≤ b, with x-axisas axis of revolution. We would like to find the volume of the solid.

In this case the area of cross section at x is given by

α(x) = πy2 = π[f(x)]2, a ≤ x ≤ b.

Hence, the volume of the solid of revolution is

V :=

∫ b

aα(x) dx = π

∫ b

ay2 dx.

EXAMPLE 5.12 Let us compute the volume of the solid of revolution of the curvey = x2 about x-axis for −a ≤ x ≤ a. The required volume is

V := π

∫ a

−ay2 dx = π

∫ a

−ax4 dx =

2

5a5.

EXAMPLE 5.13 We compute the volume of the solid of revolution of the catenary

y =a

2

(ex/a + e−x/a

)about x-axis for 0 ≤ x ≤ b. The required volume is

V := π

∫ b

0y2 dx = π

∫ b

0

a2

4

(ex/a + e−x/a

)2dx =

πa2

4

∫ b

0

(e2x/a + e−2x/a + 2

).

We see that

V :=πa3

8

(e2b/a − e−2b/a

)+πb3

8.

5.5 Computing Area of Surface of Revolution

Suppose a solid is obtained by revolving a curve y = f(x), a ≤ x ≤ b, with x-axisas axis of revolution. We would like to find the area of the surface of the solid.

The required area is

A := limµ(P )→0

k∑j=1

2πf(ξi)∆si,

where P : a = x0 < x1 < . . . < xk = b is a partition of the interval [a, b], and

∆si :=√

1 + [f ′(ξi)]2∆xi, i = 1, . . . , k.

Thus

A = limµ(P )→0

k∑j=1

2πf(ξi)√

1 + [f ′(ξi)]2∆xi = 2π

∫ b

ay

√1 +

(dy

dx

)2

dx.

EXAMPLE 5.14 We find the surface of revolution of the parabola y2 = 2px,0 ≤ x ≤ a for p > 0. The required area is

A = 2π

∫ a

0y

√1 +

(dy

dx

)2

dx

= 2π

∫ a

0

√2px

√1 +

p

2xdx = 2π

√p

∫ a

0

√p+ 2x dx

= 2π√p

2

3

[(2x+ p)3/2

1

2

]a0

=2π√p

3

[(2a+ p)3/2 − p3/2

].

5.6 Centre of Gravity

Suppose A1, A2, . . . , An are material particles on the plane at coordinates

(x1, y1), (x2, y2), . . . , (xn, yn)

and masses m1,m2, . . .mn respectively. Then the centre of gravity of the systemof these particles is at the point A = (xC , yC), where

xC :=

∑ni=1 ximi∑ni=1mi

, yC :=

∑ni=1 yimi∑ni=1mi

.

Now we attempt to define the centre of gravity of a material line and materialplanar region enclosed by certain curves.

Centre of Gravity 111

5.6.1 Centre of gravity of a material line in the plane

Suppose a curve L is given by the equation y = f(x), a ≤ x ≤ b. We assumethat this curve is a material line. Suppose the density of the material at the pointX = (x, y) is γ(X). This density is defined as follows: Suppose M(X, r) is the massof an arc of the line containing the point X with length r. Then the density of thematerial at the point x is defined by

γ(X) := limr→0

M(X, r)

r.

Now, in order to find the centre of gravity of L, we first consider a partitionP : a = x0 < x1 < . . . < xk, and take points ξi = [xi−1, xi], i = 1, . . . , n. Then wetake the the centre of gravity of the system of material points

(ξ1, f(ξ1), (ξ2, f(ξ2), . . . , (ξk, f(ξk)

as

xC(P ) =

∑ni=1 ξiγi∆si∑ni=1 γi∆si

, yC(P ) :=

∑ni=1 f(ξi)γi∆si∑n

i=1 γi∆si.

Here, ∆si is the length of the arcs joining (xi−1, yi−1) to (xi, yi), and γi is the densityat the point (ξi, f(ξi). Here yi = f(xi). Note that γi∆si is the approximate massof the arc joining (xi−1, yi−1) to (xi, yxi). Now, the centre of gravity of L is at(xC , yC), where

xC = limµ(P )→0

∑ni=1 ξiγi∆si∑ni=1 γi∆si

, yC := limµ(P )→0

∑ni=1 f(ξi)γi∆si∑n

i=1 γi∆si.

Assuming that the function γ(X) := γ(x, f(x)) is continuous on [a, b], we see that

xC =

∫ ba xγ(x, y)

√1 +

(dydx

)2dx

∫ ba γ(x, y)

√1 +

(dydx

)2dx

, yC =

∫ ba yγ(x, y)

√1 +

(dydx

)2dx

∫ ba γ(x, y)

√1 +

(dydx

)2dx

.

Example. We find the centre of gravity of the semi-circlular arc x2 + y2 = a2,y ≥ 0, assuming that the density of the material is constant. In this case,

y = f(x) :=√a2 − x2,

so that it follows that √1 +

(dy

dx

)2

=a√

a2 − x2.

Hence, since γ(x, y) is constant,

xC = 0 yC =

∫ a−a y

√1 +

(dydx

)2dx

∫ a−a

√1 +

(dydx

)2dx

=2a

π.


5.6.2 Centre of gravity of a material planar region

Next we consider the centre of gravity of a material planar region Ω bounded bytwo curves

y = f(x), y = g(x), with f(x) ≤ g(x) a ≤ x ≤ b.

Suppose that the density of the material at the point X is γ(X). This density isdefined as follows: Suppose M(X, r) is the mass of the circular region S(X, r) ⊆ Ωwith centre at x and radius r > 0, and α(X, r) is the area of the same circularregion. Then the density of the material at the point x is defined by

γ(X) := limr→0

M(X, r)

α(X, r).

Now, in order to find the centre of gravity of Ω, we first look at the following specialcase: Suppose Ω is a rectangle given by a1 ≤ x ≤ b1, a2 ≤ y ≤ b2. Then we caninfer that the centre of gravity of such rectangle is located at the point(

a1 + b12

,a2 + b2

2

).

Taking the above obervation into account, we consider a partition P = xiki=0

of the interval [a, b], and consider the rectangular strips:

Ri : xi−1 ≤ x ≤ xi, f(ξi) ≤ y ≤ g(ξi), i = 1, . . . , k,

where ξi = xi−1+xi2 , i = 1, . . . , k. If γ is the (constant) density of the material, then

the mass of the rectangular strip Ri is

mi = γ[g(ξi)− f(ξi)]∆xi, i = 1, . . . , k.

Assuming that the mass of the rectangular strip Ri is concentrated at its mid-point:

Xi :

(ξi,f(ξi) + g(ξi)

2

),

we consider the centre of gravity of the system of material points at Xi as

xC,P :=

∑ni=1 ξimi∑ni=1mi

, yC,P :=

∑ni=1

f(ξi)+g(ξi)2 mi∑n

i=1mi.

Now the centre of gravity of Ω is defined as

xC = limµ(P )→0

xC,P , yC = limµ(P )→0

yC,P ,

i.e.,

xC = limµ(P )→0

∑ni=1 ξiγ[g(ξi)− f(ξi)]∆xi∑ni=1 γ[g(ξi)− f(ξi)]∆xi

=

∫ ba x[g(x)− f(x)] dx∫ ba [g(x)− f(x)] dx

Moment of Inertia 113

yC = limµ(P )→0

∑ni=1

12 [f(ξi + g(ξi)]γ[g(ξi)− f(ξi)]∆xi∑n

i=1 γ[g(ξi)− f(ξi)]∆xi

=1

2

∫ ba [f(x+ g(x)][g(x)− f(x)] dx∫ b

a [g(x)− f(x)] dx

EXAMPLE 5.15 We find the coordinates of the centre of gravity of a segment ofa parabola y2 = a x cut off by the straight line x = a.

In this case

f(x) = −√a x, g(x) =

√a x, 0 ≤ x ≤ a.

Hence the coordinates of the centre of gravity are

xC =

∫ ba x[g(x)− f(x)] dx∫ ba [g(x)− f(x)] dx

=2∫ ba x√a x dx∫ b

a 2√a x dx

=3

5a.

yC =1

2

∫ ba [f(x+ g(x)][g(x)− f(x)] dx∫ b

a [g(x)− f(x)] dx= 0.

5.7 Moment of Inertia

Suppose there are n material points in the plane. Let their masses be m1,m2, . . .mn

respectively. Suppose that these points are at distances d1, . . . , dn from a fixed pointO. Then the moment of inertia of the system of these points with respect to thepoint O is defined by the quantity:

IO :=n∑i=1

d2imi.

If O is the origin, and (x1, y1), (x2, y2), . . . , (xn, yn) are the points, then

IO :=

n∑i=1

(x2i + y2i )mi.

5.7.1 Moment of inertia of a material line in the plane

Suppose a curve L is given by the equation y = f(x), a ≤ x ≤ b. We assumethat this curve is a material line. Suppose the density of the material at the pointX = (x, y) is γ(X).

Now, in order to find the moment of inertia of L, we first consider a partitionP : a = x0 < x1 < . . . < xk, and take points ξi = [xi−1, xi], i = 1, . . . , n. Then


we consider the moment of inertia of the system of material points at (ξ1, ηi), i =1, . . . , n. Here, ηi = f(ξi), i = 1, . . . , n.

IO,P :=n∑i=1

(ξ2i + η2i )mi.

Thus,

IO,P :=n∑i=1

(ξ2i + η2i )γi∆si.

Here, ∆si is the length of the arcs joining (xi−1, yi−1) to (xi, yi), and γi is thedensity at the point (ξi, ηi). Note that γi∆si is the approximate mass of the arcjoining (xi−1, f(xi−1) to (xi, f(xi). Now, assuming that the functions f(x) andγ(x) := γ(x, f(x)) are continuous on [a, b], the moment of inertial of L with respectto O is

IO = limµ(P )→0

IO,P

= limµ(P )→0

n∑i=1

(ξ2i + η2i )γi∆si

=

∫ b

a(x2 + y2)γ(x, y)

√1 +

(dy

dx

)2

dx.

5.7.2 Moment of inertia of a circular arc with respect to the centre

Suppose the given curve is a circular arc: ρ = a, α ≤ θ ≤ β. Following the argumentsin the above paragraph, we compute the moment of inertia using polar coordinates:

The moment of inertia, in this, case is given by

IO := limµ(P )→0

n∑i=1

d2imi,

where di = a, mi = γia∆θi, for i = 1, . . . , n, so that

IO = limµ(P )→0

n∑i=1

a2γi[a∆θi] = a3∫ β

αγ(θ)dθ.

Here, γ(θ) is the point density. If γ(θ) = γ, a constant, then

IO = a3∫ β

αγ(θ)dθ = (β − α)γa3.

In particular, M.I of the circle ρ = a, 0 ≤ θ ≤ 2π, is

IO = 2πγa3.

5.7.3 Moment of inertia of a material sector in the plane

The region is R : 0 ≤ ρ ≤ a, α ≤ θ ≤ β with constant density γ. To find the M.I. ofR, we partition it by rays and circular arcs:

P : α = θ0 < θ1 < θ2 < . . . < θn = β,

Q : 0 = ρ0 < ρ1 < ρ2 < . . . < ρm = a.

Consider the elementary region obtained by the above partition:

Rij : ρj−1 ≤ ρj ≤ a, θi−1 ≤ θi ≤ θi.

Assume that the the mass of this region Rij is concentrated at the point (ρj , θi),

where ρj ∈ [ρj−1, ρj ], θi ∈ [θi−1, θi]. Then the MI of the material point at (ρj , θi)is mijd

2ij where mij is the mass of the region Rij which is approximately equal to

[ρj∆θi∆ρj ]γ, and dij = ρj . Thus the MI of the sub-sector θi−1 ≤ θ ≤ θi is definedby

limµ(Q)→0

n∑j=1

mijd2ij = lim

µ(Q)→0

n∑j=1

[ρj∆θi∆ρj ]γρ2j

= limµ(Q)→0

n∑j=1

(γρ3j∆ρj

)∆θi

= γ

(∫ a

0ρ3 dρ

)∆θi

=γa4

4∆θi.

From this, it follows that, the moment of inertia of the sector α ≤ θ ≤ β is

limµ(P )→0

m∑i=1

γa4

4∆θi =

(β − α)γa4

4.

In particular, moment of inertia of a circular disc is

πγa4

2=Ma2

2,

where M = πa2γ is the mass of the disc.

Exercise 5.1 If M is the mass of a right circular homogeneous cylinder with baseradius a, then show that its moment of inertia is Ma2

2 .


1. Find the area of the portion of the circle x2 + y2 = 1 which lies inside theparabola y2 = 1− x.


[Hind: Area enclosed by the circle in the second and third quadrant andthe area enclosed by the parabola in the first and fourth quadrant. The therequired area is π

2 + 2∫ 10

√1− x dx.

Ans: π2 + 4

3 . ]

2. Find the area common to the cardioid ρ = a(1 + cos θ) and the circle ρ = 3a2 .

[Hind: The points of intersections of the given curves are given by 1+cos θ = 32 ,

i.e., for θ = ±π3 . Hence the required area is

2

[1

2

∫ π/3

0

(3a

2

)2

dθ +1

2

∫ π

π/3a2(1 + cos θ)2 dθ

].

Ans: 74π −

9√3

8 . ]

3. For a, b > 0, find the area included betwee the parabolas y2 = 4a(x + a) andy2 = 4b(b− x).

[Hind: Points of intersection of the curves is given by a(x+ a) = b(b− x), i.e.,

x = b2−a2a+b = b− a; y = 2

√ab. The required area is

2×[∫ b−a

−a

√4a(x+ a) +

∫ b

b−a

√4b(b− x) dx

].

Ans:83√ab(a+ b).]

4. Find the area of the loop of the curve r2 cos θ = a2 sin 3θ

[Hint:r = 0 for θ = 0 and θ = π/3, and r is maximum for θ = π/6. The area

is∫ π/30

r2

2 dθ. ]

5. Find the area of the region bounded by the curves x− y3 = 0 and x− y = 0.

[Hint: Points of intersections of the curves are at x = 0, 1,−1. The area is2∫ 10 (x1/3 − x)dx. Ans: 1/2 ]

6. find the area of the region that lies inside the circle r = a cos θ and outsidethe cardioid r = a(1− cos θ).

[Hint: Note that the circle is the one with centre at (0, a/2) and radius a/2.

The curves intersect at θ = ±π/3. The required area is∫ π/3−π/3(r

21 − r22)dθ,

where r1 = a cos θ, r2 = a(1− cos θ). Ans: a3

3 (3√

3− π) ]

7. Find the area of the loop of the curve x = a(1 − t2), y = at(1 − t2) for−1 ≤ t ≤ 1.

[Hint: y = 0 for t ∈ −1, 0, 1, and y negative for −1 ≤ t ≤ 0 and positive for0 ≤ t ≤ 1. Also, y2 = x2(a − x)/a so that the curve is symmetric w.r.t. thex-axis. Area is 2

∫ a0 ydx = 2

∫ 01 y(t)x′(t)dt. Ans: 8a2/15 ]


8. Find the length of an arch of the cycloid x = a(t− sin t), y = a(1− cos t).

[Hint: The curve cuts the x-axis at x = a and x = 2πa for t = 0 and t = 2πrespectively. Thus the length is

∫ 2π0

√[x′(t)]2 + [y′(t)]2dt. Ans: 8a. ]

9. For a > 0, find the length of the loop of the curve 3a y2 = x(x− a)2.

[Hint: The curve cuts the x-axis at x = a, and the curve is symmetric w.r.t.

the x-axis. Thus the required area is 2∫ a0

√1 +

(dydx

)2dx. Note that 6ayy′ =

(x− a)(3x− a), so that 1 + y′2 = (3x+a)2

12ax . Ans: 4a√3. ]

10. Find the length of the curve r = 21+cos θ , 0 ≤ θ ≤ π/2.

[Hind: ` :=∫ π/20

√r2 + [r′]2dθ = 2

∫ π/40 sec3 θdθ. Ans:

√2 + ln(

√2 + 1). ]

11. Find the volume of the solid obtained by revolving the curve y = 4 sin 2x,0 ≤ x ≤ π/2, about y-axis.

[Hint: writing y = 4 sin 2x, 0 ≤ x ≤ π/4 and y = 4 sin 2u, π/4 ≤ u ≤ π/2, the

required volume is∫ 40 (u2−x2)dy = π

∫ π/2π/4 u

2(8 cos 2u)du−π∫ π/40 x2(8 cos 2x)dx.

Also, note that the curve is symmetric w.r.t. the line x = π/4. Hence, the

required volume is given by π∫ π/40 [(π4 − x)2 − x2]dy. Ans: 2π2.]

12. Find the area of the surface obtained by revolving a loop of the curve 9ax2 =y(3a− y)2 about y-axis.

[Hind: x = 0 iff y = 0 or y = 3a. The required area is 2π∫ 3a0 x

√1 +

(dxdy

)2dx.

Ans: 3πa2. ]

13. Find the area of the surface obtained by revolving about x-axis, an arc of thecatenary y = c cosh(x/c) between x = −a and x = a for a > 0.

[Hind: The area is 2π∫ a−a y

√1 + y′2 dx = 2π c

∫ a−a cosh2 x

c dx. Ans: πc [2a +

c sinh 2ac ]. ]

14. The lemniscate ρ2 = a2 cos 2θ revolves about the line θ = π4 . Find the area of

the surface of the solid generated.

[Hind: The required surface is 2×2π∫ π/4−π/4 h

√ρ2 + ρ′2 dθ, where h := ρ sin

(π4 − θ

),

ρ = a√

cos 3θ so that ρ2 + ρ′2 = a2

cos 2θ . Ans: 4πa2. ]

15. Find the volume of the solid generated by the cardioid ρ = a(1 + cos θ) aboutthe initial line. [Ans: 8

3 . ]

6

Sequence and Series of Functions

6.1 Sequence of Functions

6.1.1 Pointwise Convergence and Uniform Convergence

Let J be an interval in R.

Definition 6.1 By a sequence of functions on J we mean an ordered set

f1, f2, . . .,

where fn is a (real valued) function defined on J for each n ∈ N, and in that case,we denote such a sequence by (fn).

More precisely, a sequence of functions on J is a map F : N → F(J), whereF(J) is the set of all real valued functions defined on F . If fn := F (n) for n ∈ N,then we denote F by (fn), and call (fn) as a sequence of functions.

Definition 6.2 Let (fn) be a sequence of functions on an interval J .

(a) We say that (fn) converges at a point x0 ∈ J if the sequence (fn(x0)) ofreal numbers converges.

(b) We say that (fn) converges pointwise on J if (fn) converges at every pointin J , i.e., for each x ∈ J , the sequence (fn(x)) of real numbers converges.

Definition 6.3 Let (fn) be a sequence of functions on an interval J . If (fn)converges pointwise on J , and if f : J → R is defined by f(x) = limn→∞ fn(x), x ∈J , then we say that (fn) converges pointwise to f on J , and f is the pointwiselimit of (fn), and in that case we write

fn → f pointwise on J.

Thus, (fn) converges to f pointwise on J if and only if for every ε > 0 and foreach x ∈ J , there exists N ∈ N (depending, in general, on both ε and x) such that|fn(x)− f(x)| < ε for all n ≥ N .

118

Sequence of Functions 119

Exercise 6.1 Pointwise limit of a sequence of functions is unique.

EXAMPLE 6.1 Consider fn : R→ R defined by

fn(x) =sin(nx)

n, x ∈ R

and for n ∈ N. Then we see that for each x ∈ R,

|fn(x)| ≤ 1

n∀n ∈ N.

Thus, (fn) converges pointwise to f on R, where f is the zero function on R, i.e.,f(x) = 0 foe very x ∈ R.

Suppose (fn) converges to f pontwise on J . As we have mentioned, it can happenthat for ε > 0, and for each x ∈ J , the number N ∈ N satisfying |fn(x)− f(x)| < ε∀n ≥ N depends not only on ε but also on the point x. For instance, consider thefollowing example.

EXAMPLE 6.2 Let fn(x) = xn for x ∈ [0, 1] and n ∈ N. Then we see that for0 ≤ x < 1, fn(x)→ 0, and fn(1)→ 1 as n→∞. Thus, (fn) converges pointwise toa function f defined by

f(x) =

0, x 6= 1,1, x = 1.

In particular, (fn) converges pointwise to the zero function on [0, 1).

Note that if there exists N ∈ N such that |xn| < ε for all n ≥ N and for allx ∈ [0, 1), then, letting x → 1, we would get 1 < ε, which is not possible, had wechosen ε < 1.

For ε > 0, if we are able to find an N ∈ N which does not vary as x variesover J such that |fn(x)− f(x)| < ε for all n ≥ N , then we say that (fn) convergesuniformly to f on J . Following is the precise definition of uniform convergence of(fn) to f on J .

Definition 6.4 Suppose (fn) is a sequence of functions defined on an interval J .We say that (fn) converges to a function f uniformly on J if for every ε > 0there exists N ∈ N (depending only on ε) such that

|fn(x)− f(x)| < ε ∀n ≥ N and ∀x ∈ J,

and in that case we write

fn → f uniformly on J.

120 Sequence and Series of Functions M.T. Nair

We observe the following:

• If (fn) converges uniformly to f , then it converges to f pointwise as well.Thus, if a sequence does not converge pointwise to any function, then it can notconverge uniformly.

• If (fn) converges uniformly to f on J , then (fn) converges uniformly to f onevery subinterval J0 ⊆ J .

In Example 6.2 we obtained a sequence of functions which converges pointwisebut not uniformly. Here is another example of a sequence of functions which con-verges pointwise but not uniformly.

EXAMPLE 6.3 For each n ∈ N, let

fn(x) =nx

1 + n2x2, x ∈ [0, 1].

Note that fn(0) = 0, and for x 6= 0, fn(x) → 0 as n → ∞. Hence, (fn) convergespoitwise to the zero function. We do not have uniform convergence, as fn(1/n) = 1/2for all n. Indeed, if (fn) converges uniformly, then there exists N ∈ N such that

|fN (x)| < ε ∀x ∈ [0, 1].

In particular, we must have

1

2= |fN (1/N)| < ε ∀x ∈ [0, 1].

This is not possible if we had chosen ε < 1/2.

EXAMPLE 6.4 Consider the sequence (fn) defined by

fn(x) = tan−1(nx), x ∈ R.

Note that fn(0) = 0, and for x 6= 0, fn(x) → π/2 as n → ∞. Hence, the givensequence (fn) converges pointwise to the function f defined by

f(x) =

0, x = 0,π/2, x 6= 0.

However, it does not converge uniformly to f on any interval containing 0. Tosee this, let J be an interval containing 0 and ε > 0. Let N ∈ N be such that|fn(x)− f(x)| < ε for all n ≥ N and for all x ∈ J . In particular, we have

|fN (x)− π/2| < ε ∀x ∈ J \ 0.

Letting x → 0, we have π/2 = |fN (0) − π/2| < ε which is not possible if we hadchooses ε < π/2.

Now, we give a theorem which would help us to show non-uniform convergenceof certain sequence of functions.

Theorem 6.1 Suppose fn and f are functions defined on an interval J . If thereexists a sequence (xn) in J and c 6= 0 such that an := fn(xn)−f(xn)→ c as n→∞,then (fn) does not converge uniformly to f on J .

Proof. Suppose (fn) converges uniformly to f on J . Then taking ε = |c|/2, thereexists N ∈ N such that

|fn(x)− f(x)| < |c|2∀n ≥ N, ∀x ∈ J.

In particular,

|fn(xn)− f(xn)| < |c|2∀n ≥ N.

Now, taking limit as n→∞, it follows that

|c| = limn→∞

|fn(xn)− f(xn)| < |c|2

which is a contradiction. Hence our assumption that (fn) converges uniformly to fon J is wrong.

In he case of Example 6.2, taking xn = n/(n+ 1), we see that

fn(xn) =

(n

n+ 1

)n→ 1

e.

Hence, (fn) does not converge to f ≡ 0 uniformly on [0, 1).

In Example 6.3, we may take xn = 1/n, and in the case of Example 6.4, we maytake xn = π/n, and apply Theorem 6.1.

Exercise 6.2 Suppose fn and f are functions defined on an interval J . If thereexists a sequence (xn) in J such that [fn(xn) − f(xn)] 6→ 0, then (fn) does notconverge uniformly to f on J . Why?

[Suppose an := [fn(xn)− f(xn)] 6→ 0. Then there exists δ > 0 such that |an| ≥ δfor infinitely many n. Now, if fn → f uniformly, there exists N ∈ N such that|fn(x) − f(x)| < δ/2 for all n ≥ N . In particular, |an| < δ/2 for all n ≥ N . Thus,we arive at a contradiction.] J

Here is a sufficient condition for uniform convergence. Its proof is left as anexercise.

Theorem 6.2 Suppose fn for n ∈ N and f are functions on J . If there exists asequence (αn) of positive reals satisfying αn → 0 as n→∞ and

|fn(x)− f(x)| ≤ αn ∀n ∈ N,

then (fn) converges uniformly to f .


Exercise 6.3 Supply detailed proof for Theorem 6.2. J

Here are a few examples to illustrate the above theorem.


fn(x) =2nx

1 + n4x2, x ∈ [0, 1].

Since 1 + n4x2 ≥ 2n2x (using the relation a2 + b2 ≥ 2ab), we have

0 ≤ fn(x) ≤ 2nx

2n2x=

1

n.

Thus, by Theorem 6.2, (fn) converges uniformly to f = 0.


fn(x) =1

n3log(1 + n4x2), x ∈ [0, 1].

Then we have

0 ≤ fn(x) ≤ 1

n3log(1 + n4) =: αn ∀n ∈ N.

Taking g(t) := 1t3

log(1 + t4) for t > 0, we see, using L’Hospital’s rule that

limt→∞

g(t) = limt→∞

4t3

3t2(1 + t4)= 0.

In particular,

limn→∞

1

n3log(1 + n4) = 0.

Thus, by Theorem 6.2, (fn) converges uniformly to f = 0.

We may observe that in Examples 6.2 and 6.4, the limit function f is not con-tinuous, although every fn is continuous. This makes us to ask the following:

Suppose each fn is a continuous function on J and (fn) converges to f pointwise.

• If f is Riemann integrable, then do we have∫ b

af(x)dx = lim

n→∞

∫ b

afn(x)dx

for every [a, b] ⊆ J?

• Suppose each fn is continuously differentiable on J . Then, is the function fdifferentiable on J? If f is differentiable on J , then do we have the relation

d

dxf(x) = lim

n→∞

d

dxfn(x)dx ?

The answers to the above questions need not be affirmative as the followingexamples show.


fn(x) = nx(1− x2)n, 0 ≤ x ≤ 1.

Then we see that

limn→∞

fn(x) = 0 ∀x ∈ [0, 1].

Indeed, for each x ∈ (0, 1),

fn+1(x)

fn(x)= x(1− x2)

(n+ 1

n

)→ x(1− x2) as n→∞.

Since x(1−x2) < 1 for x ∈ (0, 1), we obtain limn→∞

fn(x) = 0 for every x ∈ [0, 1]. But,

∫ 1

0fn(x)dx =

n

2n+ 2→ 1

2as n→∞.

Thus, limit of the integrals is not the integral of the limit.


fn(x) =sin(nx)√

n, x ∈ R.

Then we see that

limn→∞

fn(x) = 0 ∀x ∈ [0, 1].

But, f ′n(x) =√n cos(nx) for all n ∈ N, so that

f ′n(0) =√n→∞ as n→∞.

Thus, limit of the derivatives is not the derivative of the limit.

6.1.2 Continuity and uniform convergence

Theorem 6.3 Suppose (fn) is a sequence of continuous functions defined on aninterval J which converges uniformly to a function f . Then f is continuous on J .

Proof. Suppose x0 ∈ J . Then for any x ∈ J and for any n ∈ N,

|f(x)− f(x0)| ≤ |f(x)− fn(x)|+ |fn(x)− fn(x0)|+ |fn(x0)− f(x0)|. (∗)

Let ε > 0 be given. Since (fn) converges to f uniformly, there exists N ∈ N suchthat

|fn(x)− fn(x)| < ε/3 ∀n ≥ N, ∀x ∈ J.

Since fN is continuous, there exists δ > 0 such that

|fN (x)− fN (x0)| < ε/3 whenever |x− x0| < δ.

Hence from (∗), we have

|f(x)− f(x0)| ≤ |f(x)− fN (x)|+ |fN (x)− fN (x0)|+ |fN (x0)− f(x0)| < ε

whenever |x − x0| < δ. Thus, f is continuous at x0. This is true for all x0 ∈ J .Hence, f is a continuous function on J .

6.1.3 Integration-Differentiation and uniform convergence

Theorem 6.4 Suppose (fn) is a sequence of continuous functions defined on an in-terval [a, b] which converges uniformly to a function f on [a, b]. Then f is continuousand

limn→∞

∫ b

afn(x)dx =

∫ b

af(x)dx.

Proof. We already know by Theorem 6.3 that f is a continuous function. Nextwe note that ∣∣∣∣∫ b

afn(x)dx−

∫ b

af(x)dx

∣∣∣∣ ≤ ∫ b

a|fn(x)− f(x)|dx.

Let ε > 0 be given. By uniform convergence of (fn) to f , there exists N ∈ N suchthat

|fn(x)− f(x)| < ε/(b− a) ∀n ≥ N, ∀x ∈ [a, b].

Hence, for all n ≥ N ,∣∣∣∣∫ b

afn(x)dx−

∫ b

af(x)dx

∣∣∣∣ ≤ ∫ b

a|fn(x)− f(x)|dx < ε.


Theorem 6.5 Suppose (fn) is a sequence of continuously differentiable functionsdefined on an interval J such that

(i) (f ′n) converges uniformly to a function, and

(ii) (fn(a)) converges for some a ∈ J .

Then (fn) converges to a continuously differentiable function f and

limn→∞

f ′n(x) = f ′(x) ∀x ∈ J.

Series of Functions 125

Proof. Let g(x) := limn→∞

f ′n(x) for x ∈ J , and α := limn→∞

fn(a). Since the conver-

gence of (f ′n) to g is uniform, by Theorem 6.4, the function g is continuous and

limn→∞

∫ x

af ′n(t)dt =

∫ x

ag(t)dt.

Let ϕ(x) :=∫ xa g(t)dt, x ∈ J . Then ϕ is differentiable and ϕ′(x) = g(x) for x ∈ J .

But,∫ xa f′n(t)dt = fn(x)− fn(a). Hence, we have

limn→∞

[fn(x)− fn(a)] = ϕ(x).

Thus, (fn) converges pointwise to a differentiable function f defined by f(x) =ϕ(x) + α, x ∈ J , and (f ′n) converges to f ′.

6.2 Series of Functions

Definition 6.5 By a series of functions on a interval J , we mean an expressionof the form

∞∑n=1

fn or

∞∑n=1

fn(x),

where (fn) is a sequence of functions defined on J .

Definition 6.6 Given a series∑∞

n=1 fn(x) of functions on an interval J , let

sn(x) :=n∑i=1

fi(x), x ∈ J.

Then sn is called the n-th partial sum of the series∑∞

n=1 fn.

Definition 6.7 Consider a series∑∞

n=1 fn(x) of functions on an interval J , and letsn(x) be its n-th partial sum. Then we say that the series

∑∞n=1 fn(x)

(a) converges at a point x0 ∈ J if (sn) converges at x0,

(b) converges pointwise on J if (sn) converges pointwise on J , and

(c) converges uniformly on J if (sn) converges uniformly on J .

The proof of the following two theorems are obvious from the statements ofTheorems 6.4 and 6.5 respectively.

Theorem 6.6 Suppose (fn) is a sequence of continuous functions on J . If∑∞

n=1 fn(x)converges uniformly on J , say to f(x), then f is continuous on J , and for [a, b] ⊆ J ,∫ b

af(x)dx =

∞∑n=1

∫ b

afn(x)dx.

Theorem 6.7 Suppose (fn) is a sequence of continuously differentiable functions onJ . If

∑∞n=1 f

′n(x) converges uniformly on J , and if

∑∞n=1 fn(x) converges at some

point x0 ∈ J , then∑∞

n=1 fn(x) converges to a differentiable function on J , and

d

dx

( ∞∑n=1

fn(x)

)=

∞∑n=1

f ′n(x).

Next we consider a useful sufficient condition to check uniform convergence. Firsta definition.

Definition 6.8 We say that∑∞

n=1 fn is a dominated series if there exists asequence (αn) of positive real numbers such that |fn(x)| ≤ αn for all x ∈ J and forall n ∈ N, and the series

∑∞n=1 αn converges.

Theorem 6.8 A dominated series converges uniformly.

Proof. Let∑∞

n=1 fn be a dominated series defined on an interval J , and let (αn)be a sequence of positive reals such that

(i) |fn(x)| ≤ αn for all n ∈ N and for all x ∈ J , and

(ii)∑∞

n=1 αn converges.

Let sn(x) =∑n

i=1 fi(x), n ∈ N. Then for n > m,

|sn(x)− sm(x)| =

∣∣∣∣∣n∑

i=m+1

fi(x)

∣∣∣∣∣ ≤n∑

i=m+1

|fi(x)| ≤n∑

i=m+1

αi = σn − σm,

where σn =∑n

k=1 αk. Since∑∞

n=1 αn converges, the sequence (σn) is a Cauchysequence. Now, let ε > 0 be given, and let N ∈ N be such that

|σn − σm| < ε ∀n,m ≥ N.

Hence, from the relation: |sn(x)− sm(x)| ≤ σn − σm, we have

|sn(x)− sm(x)| < ε ∀n,m ≥ N, ∀x ∈ J.

This, in particular implies that sn(x) is also a Cauchy sequence at each x ∈ J .Hence, sn(x) converges for each x ∈ J . Let f(x) = limn→∞ sn(x), x ∈ J . Then,we have

|f(x)− sm(x)| = limn→∞

|sn(x)− sm(x)| < ε ∀m ≥ N, ∀x ∈ J.

Thus, the series∑∞

n=1 fn converges uniformly to f on J .

Series of Functions 127


n=1cosnxn2 and

∑∞n=1

sinnxn2 are dominated series, since

∣∣∣cosnx

n2

∣∣∣ ≤ 1

n2,

∣∣∣∣sinnxn2

∣∣∣∣ ≤ 1

n2∀n ∈ N

and∑∞

n=11n2 is convergent.


n=0 xn is a dominated series on [−ρ, ρ] for 0 < ρ < 1,

since |xn| ≤ ρn for all n ∈ N and∑∞

n=0 ρn is convergent. Thus, the given series is a

dominated series, and hence, it is uniformly convergent.


n=1x

n(1+nx2)on R. Note that

x

n(1 + nx2)≤ 1

n

(1

2√n

),

and∞∑n=1

1

n3/2converges. Thus, the given series is dominated series, and hence it

converges uniformly on R.


n=1x

1+n2x2for x ∈ [c,∞), c > 0. Note

thatx

1 + n2x2≤ x

n2x2=≤ 1

n2x≤ 1

n2c

and

∞∑n=1

1

n2converges. Thus, the given series is dominated series, and hence it

converges uniformly on [c,∞).


n=1

(xe−x

)nis dominated on [0,∞): To see this,

note that (xe−x

)n=

xn

enx≤ xn

(nx)n/n!=n!

nn

and the series∑∞

n=1n!nn converges.

It can also be seen that |xe−x| ≤ 1/2 for all x ∈ [0,∞).


n=1 xn−1 is not uniformly convergent on (0, 1); in

particular, not dominated on (0, 1). This is seen as follows: Note that

sn(x) :=

n∑k=1

xk−1 =1− xn

1− x→ f(x) :=

1

1− xas n→∞.

Hence, for ε > 0,

|f(x)− sn(x)| < ε ⇐⇒∣∣∣∣ xn

1− x

∣∣∣∣ < ε.

Hence, if there exists N ∈ N such that |f(x) − sn(x)| < ε for all n ≥ N for allx ∈ (0, 1), then we would get

|x|N

|1− x|< ε ∀x ∈ (0, 1).

This is not possible, as |x|N/|1− x| → ∞ as x→ 1.

However, we have seen that the above series is dominated on [−a, a] for 0 < a < 1.

EXAMPLE 6.15 The series

∑∞n=1(1−x)xn−1 is not uniformly convergent on [0, 1];

in particular, not dominated on [0, 1]. This is seen as follows: Note that

sn(x) :=n∑k=1

(1− x)xk−1 =

1− xn if x 6= 10 if x = 1.

In particular, sn(x) = 1−xn for all x ∈ [0, 1) and n ∈ N. By Example 6.2, we knowthat (sn(x)) converges to f(x) ≡ 1 pointwise, but not uniformly.

Remark 6.1 Note that if a series∑∞

n=1 fn converges uniformly to a function f onan interval J , then we must have

βn := supx∈J|sn(x)− f(x)| → 0 as n→∞.

Here, sn is the n-th partial sum of the series. Conversely, if βn → 0, then the seriesis uniformly convergent. Thus, if

∑∞n=1 fn converges to a function f on J , and if

supx∈J |sn(x)− f(x)| 6→ 0 as n→∞, then we can infer that the convergence is notuniform.

As an illustration, consider the Example 6.15. There we have

|sn(x)− f(x)| =xn if x 6= 10 if x = 1.

Hence, sup|x|≤1 |sn(x)−f(x)| = 1. Moreover, the limit function f is not continuous.Hence, the non-uniform convergence also follows from Theorem 6.6.

Exercise 6.4 Consider a series∑∞

n=1 fn and an := supx∈J |fn(x)|. Show that thisseries is dominated series if and only if

∑∞n=1 an converges. J

Next example shows that in Theorem 6.7, the condition that the derived seriesconverges uniformly is not a necessary condition for the the conclusion.


n=0 xn. We know that it converges to

1/(1− x) for |x| < 1. It can be seen that the derived series∑∞

n=1 nxn−1 converges

uniformly for |x| ≤ ρ for any ρ ∈ (0, 1). This follows since∑∞

n=1 nρn−1 converges.

Hence,

1

(1− x)2=

d

dx

1

1− x=

∞∑n=1

nxn−1 for |x| ≤ ρ.

The above relation is true for x in any open interval J ⊆ (−1, 1); because we canchoose ρ sufficiently close to 1 such that J ⊆ [−ρ, ρ]. Hence, we have

1

(1− x)2=∞∑n=1

nxn−1 for |x| < 1.

We know that the given series is not uniformly convergent (see, Example 6.14).


1. Let fn(x) =x2

(1 + x2)nfor x ≥ 0. Show that the series

∞∑n=1

fn(x) does not

converge uniformly.

2. Let fn(x) =x

1 + nx2, x ∈ R. Show that (fn) converge uniformly, whereas (f ′n)

does not converge uniformly. Is the relation limn→∞

f ′n(x) =(

limn→∞

fn(x))′

true

for all x ∈ R?

3. Let fn(x) =log(1 + n3x2)

n2, and gn(x) =

2nx

1 + n3x2for x ∈ [0, 1]. Show that

the sequence (gn) converges uniformly to g where g(x) = 0 for all x ∈ [0, 1].Using this fact, show that (fn) also converges uniformly to the zero functionon [0, 1].

4. Let fn(x) =

n2x, 0 ≤ x ≤ 1/n,−n2x+ 2n, 1/n ≤ x ≤ 2/n,0, 2/n ≤ x ≤ 1.

Show that (fn) does not converge uniformly of [0, 1].

[Hint: Use termwise integration.]

5. Suppose (an) is such that∑∞

n=1 an is absolutely convergent. Show that

∞∑n=1

anx2n

1 + x2n

is a dominated series on R.

6. Show that for each p > 1, the series

∞∑n=1

xn

npis convergent on [−1, 1] and the

limit function is continuous.

7. Show that the series

∞∑n=1

(n+1)2xn+1−n2xn(1−x) converges to a continuous

function on [0, 1], but it is not dominated.



∞∑n=1

[ 1

1 + (k + 1)x− 1

1 + kx

]is convergent on [0, 1], but

not dominated, and∫ 1

0

∞∑n=1

[ 1

1 + (k + 1)x− 1

1 + kx

]dx =

∞∑n=1

∫ 1

0

[ 1

1 + (k + 1)x− 1

1 + kx

]dx.

9. Show that

∫ 1

0

∞∑n=1

x

(n+ x2)2dx =

1

2.

7

Power Series

7.1 Convergence and Absolute convergence

Power series is a particular case of series of functions.

Definition 7.1 Given a sequence (an) of real numbers, a series of the form∑∞

n=0 anxn

is called a power series.

Note that a power series∑∞

n=0 anxn converges at the point x = 0. What can we

say about its domain of convergence?

Theorem 7.1 (Abel’s theorem) Consider the series∑∞

n=0 anxn.

(i) If the above series converges at a point x0, then it converges absolutely atevery x with |x| < |x0|.

(ii) If the above series diverges at a point y0, then it diverges at every x with|x| > |y0|.

Proof. (i) Suppose∑∞

n=0 anxn converges at a point x0 6= 0. Let x be such that

|x| < |x0|. Then, since we have

|anxn| = |anxn0 |∣∣∣∣ xx0∣∣∣∣n ∀n.

Since |anxn0 | → 0 as n→∞, there exists M > 0 such that |anxn0 | ≤M for all n, sothat we have

|anxn| ≤M∣∣∣∣ xx0∣∣∣∣n ∀n.

Now, since∣∣∣ xx0 ∣∣∣ < 1, it follows, by comparison test that

∑∞n=0 |anxn| converges.

(ii) Suppose∑∞

n=0 anxn diverges at a point y0 6= 0. If

∑∞n=0 anx

n converges at apoint x with |x| > |y0|, then by part (i), it converges at y0 as well, which contradictsour assumption. Hence, the conclusion holds.

Theorem 7.1 shows that if the series∑∞

n=0 anxn converges at some nonzero point

x0, then it converges for all x in an interval of the form (−a, a), where a ≥ |x0|.

131

132 Power Series M.T. Nair

Definition 7.2 The domain (or interval) of convergence of a power series∑∞n=0 anx

n is the set

D := x ∈ R :∞∑n=0

anxn converges at x

and

R := sup|x| : x ∈ D

is called the radius of convergence of∑∞

n=0 anxn.

Thus, a number R with 0 < R <∞ is the radius of convergence of∑∞

n=0 anxn if

and only if∑∞

n=0 anxn converges for all x with |x| < R, and diverges at all x with

|x| > R.

If the power series∑∞

n=0 anxn converges only at the point 0, then the radius of

convergence is 0, and if it converges at all points in R, then sup|x| : x ∈ D doesnot exists, and in that case we say that the radius of convergence is∞, i.e., we writeR =∞.

EXAMPLE 7.1 Consider the power series∑∞

n=0 xn. In this case, we know that

the series converges for x with |x| < 1, and diverges for x with |x| > 1. Also, theseries diverges at x ∈ 1,−1. Hence, its radius of convergence is 1, and its domain(interval) of convergence is D = x : −1 < x < 1 = (−1, 1).


n=0xn

n . In this case, we know that theseries converges at x = −1 and diverges at x = 1. Hence, its radius of convergenceis 1, and its domain (interval) of convergence is [−1, 1).


n=0xn

n2 . We know that this seriesconverges at x = 1 and x = −1. Since

|xn+1/(n+ 1)2||xn/n2|

= |x| n

n+ 1→ |x| as n→∞,

by ratio test the series∑∞

n=0

∣∣xnn2

∣∣ converges for x with |x| < 1 and diverges for xwith |x| > 1. Therefore, the radius of convergence is 1, and the domain (interval)of convergence is [−1, 1].


n=0xn

n! . Since

|xn+1/(n+ 1)!||xn/n!|

= |x| 1

n+ 1→ 0 as n→∞,

by ratio test the series converges at every x ∈ R. Hence, the radius of convergenceis ∞, and the domain (interval) of convergence is R.

For finding the radius of convergence and domain of convergence, the followingtheorem will be useful.

Uniform Convergence 133

CONVENTION: In the following, we use the following convention:

(i) If an ≥ 0 for all n ∈ N and an →∞ as n→∞, then we write limn→∞ an =∞.

(ii) If c = 0, then we write 1/c =∞, and if c =∞, then we write 1/c = 0.

Theorem 7.2 Consider the power series∑∞

n=0 anxn, and let R be its radius of

convergence.

(a) If limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = L ∈ [0,∞], then R = 1/L.

(b) If limn→∞

|an|1/n = ` ∈ [0,∞], then R = 1/`.

Proof. Taking un(x) = anxn, n ∈ N, the proofs follow from Abel’s Theorem 7.1

and applying ration test and root test.

Exercise 7.1 Find the domain of convergence of the following power series.

(i)∑∞

n=1xn

2n−1 . (ii)∑∞

n=1xn

n×4n , (iii)∑∞

n=0xn

n! , (iv)∑∞

n=0 n!xn.

[Answers: (i) [−1, 1), (ii) [−4, 4), (iii)R, (iv) 0.] J

7.2 Uniform Convergence

Theorem 7.3 Suppose R > 0 is the radius of convergence of a power series∑∞n=0 anx

n. Then

(i)∑∞

n=0 anxn converges uniformly on [−ρ, ρ] for any ρ with 0 < ρ < R, and

(ii) the function f defined by f(x) :=∑∞

n=0 anxn, x ∈ (−R,R), is continuous

on (−R,R).

Proof. Let 0 < ρ < R, and r such that ρ < r < R. Then for every x with |x| ≤ ρ,we have

|anxn| ≤ |anrn|(xr

)n| ≤ |anrn|

(ρr

)n.

Since the series∑∞

n=0 anrn is convergent, the sequence (anr

n) is bounded, say|anrn| ≤ M for all n ∈ N, for some M > 0. Also, since ρ

r < 1,∑∞

n=0 anxn is

a dominated on [−ρ, ρ]. Hence the series∑∞

n=0 anxn is uniformly convergent on

[−ρ, ρ] and the function f defined by

f(x) :=

∞∑n=0

anxn

is continuous on [−ρ, ρ]. Since, this is true for any ρ < R, the function f is continuousat every x ∈ (−R,R). Indeed, for any x0 ∈ (−R,R), we may take ρ such that|x0| < ρ < R.]

134 Power Series M.T. Nair

7.3 Differentiation and Integration


n. Then the radius of convergence of∑∞

n=1 nanxn−1 is R. In particu-

lar, for any ρ with 0 < ρ < R, the series∑∞

n=1 nanxn−1 converges uniformly on

[−ρ, ρ] and it represents a continuous function on (−R,R).

Proof. Let x0 ∈ (−R,R), and let ρ such that |x0| < ρ < R. Then, we have

|nanxn−10 | = n|anρn−1|(|x0|ρ

)n−1∀n ∈ N.

Since∑∞

n=0 anρn converges, |anρn−1| → 0 so that it is bounded, say |anρn−1| ≤ M

for all n ∈ N. Thus,

|nanxn−10 | ≤Mn

(|x0|ρ

)n−1∀n ∈ N.

Since∑∞

n=0 n(|x0|ρ

)n−1converges, the series

∑∞n=0 |nanx

n−10 | also converges. Thus,∑∞

n=1 anxn−1 for every x ∈ (−R,R).

It remains to show that R is the radius of convergence of∑∞

n=1 nanxn−1. Suppose∑∞

n=1 nanxn−1 converges at some point y0 with |y0| > R. Then taking r with

|y0| > r > R, we see that∑∞

n=1 nanrn−1 converges (by Abel’s theorem applied

to this series). But, |nanrn−1| ≥ |anrn|/r so that by comparison test∑∞

n=1 anrn

converges. This is not possible since r > R. Thus, the radius of convergence of∑∞n=1 nanx

n−1 is R.


n, and let f(x) :=∑∞

n=0 anxn for |x| < R. Then the following hold.

(i) f(x) is a continuous function for |x| < R, and for [a, b] ⊆ (−R,R),∫ b

af(x)dx =

∞∑n=0

ann+ 1

[bn+1 − an+1].

(ii) f(x) is differentiable for every x ∈ (−R,R), and

d

dxf(x) =

∞∑n=1

nanxn−1.

Proof. The result in (i) follows from Theorems 7.3 and 6.6. Also, by Theorems7.4 and 6.7 it follows that

d

dxf(x) =

∞∑n=1

nanxn−1

for every x ∈ [−ρ, ρ], where 0 < ρ < R. Now, if x ∈ (−R,R), then we may take ρsuch that x ∈ [−ρ, ρ], and the result is valid for such x as well.

7.3.1 Series that can be converted into a power series

Some of the series may not be in the standard form∑∞

n=0 anxn, but can be converted

into this form after some change of variable. For example, consider the series

(i)

∞∑n=0

an(x− x0)n, (ii)

∞∑n=0

anx3n, (iii)

∞∑n=0

an1

xn, (iv)

∞∑n=0

an sinn x.

In each of these cases, we may take a new variable as follows: In (i) y = x− x0, in(ii) y = x3, in (iii) y = 1

x , and in (iv) y = sinx.


1. Find the interval of convergence of the following power series.

(i)∞∑n=1

(−1)n

nxn, (ii)

∞∑n=1

(−1)n3n

(4n− 1)xn.

(iii)

∞∑n=1

n(x+ 5)n

(2n+ 1)3, (iv)

∞∑n=1

2n sinn x

n2.

[Answers: (i): (−∞,−1) ∪ [1,∞); (ii): (−∞,−3) ∪ [3,∞);(iii): [−6,−4]; (iv): [−π

6 + kπ, π6 + kπ], k ∈ Z.]

2. Find the radius of convergence of∞∑n=0

(n− 1)!

nnxn. Ans: e

3. Find the radius of convergence and interval of convergence of the followingseries:

(i)

∞∑n=0

αnxn for α > 0; (ii)

∞∑n=0

αn2xn for α > 0; (iii)

∞∑n=0

xn!.

(iv)∞∑n=0

(−1)n

nxn(n+1); (v)

∞∑n=0

sin(nπ/6)

2n(x− 1)n; (vi)

∞∑n=0

(−i)n

4nnαx2n.

4. Show that

(i) log(1 + x) =∞∑n=1

(−1)n+1xn

nfor −1 < x ≤ 1,

(ii) tan−1 x =

∞∑n=0

(−1)nx2n+1

2n+ 1for −1 < x ≤ 1,


∞∑n=1

xn

n!is dominated on any closed interval [a, b]. Is it

dominated on R?

8

Fourier Series

While studying the heat conduction problem in the year 1804, Fourier found itnecessary to use a special type of function series associated with certain functions f ,later known as Fourier series of f . In this chapter we study such series of functions.

8.1 Fourier Series of 2π-Periodic functions

8.1.1 Fourier series and Fourier coefficients

Definition 8.1 Let (an) and (bn) be sequences of real numbers. Then a series ofthe form

c0 +

∞∑n=1

(an cosnx+ bn sinnx)

is called a trigonometric series.

We observe that the functions cosnx and sinnx are 2π-periodic. Hence, if c0 +∑∞n=1 (an cosnx+ bn sinnx) converges to a function f(x), the f(x) also has to be

2π-periodic. Thus, only a 2π-periodic function is expected to have a trigonometricseries expansion.

Definition 8.2 A function f : R → R is said to be T -periodic for some T > 0 iff(x+ T ) = f(x for all x ∈ R.

Now, suppose that f is a 2π-periodic function. We would like to know whetherf can be represented as a Fourier series. Suppose, for a moment, that we can write

f(x) = c0 +

∞∑n=1

(an cosnx+ bn sinnx) ∀x ∈ R.

Then what should be an, bn? To answer this question, let us further assume thatthe series can be termwise integrated. For instance if the above series is uniformlyconvergent to f , then termwise integration is possible. Recall that if (an) and (bn)

are such that

∞∑n=0

(|an|+ |bn|) converges, then the above series is dominated and

136

Fourier Series of 2π-Periodic functions 137

hence is uniformly convergent. Observe that∫ π

−πcosnx cosmxdx =

0, if n = mπ, if n 6= m,∫ π

−πsinnx sinmxdx =

0, if n = mπ, if n 6= m,∫ π

−πcosnx sinmxdx = 0.

Therefore, under the assumption that the series can be integrated termwise, we get∫ π

−πf(x)dx = 2πc0,∫ π

−πf(x) cosnxdx = anπ ∀n ∈ N,∫ π

−πf(x) sinnxdx = bnπ ∀n ∈ N.

Thus, c0 = 12π

∫ π−π f(x)dx and for all n ∈ N,

an =1

π

∫ π

−πf(x) cosnxdx, bn =

1

π

∫ π

−πf(x) sinnxdx.

Definition 8.3 The Fourier series of a 2π-periodic function f is the trigonometricseries

a02

+∞∑n=1

(an cosnx+ bn sinnx) ,

where

an =1

π

∫ π

−πf(x) cosnxdx, bn =

1

π

∫ π

−πf(x) sinnxdx.

We may write this fact as

f(x) ∼ a02

+

∞∑n=1

(an cosnx+ bn sinnx) .

The numbers an and bn are called the Fourier coefficients of f .

The following two theorems give sufficient conditions for the convergence of theFourier series of a function f to the function f at certain points x ∈ R.

Theorem 8.1 Suppose f is a bounded monotonic function on [−π, π). Then theFourier series of f converges to a function s(x), where

s(x) =

f(x) if f continuous at x,12 [f(x−) + f(x+)] if f not continuous at x.

138 Fourier Series M.T. Nair

Theorem 8.2 (Dirichlet’s theorem) Suppose f : R→ R is a 2π-periodic functionwhich is piecewise differentiable on (−π, π). Then the Fourier series of f convergesto the s(x) defined by

s(x) =

f(x) if f continuous at x,12 [f(x−) + f(x+)] if f not continuous at x.

Remark 8.1 It is known that there are continuous functions f defined on [−π, π]whose Fourier series does not converge point wise to f . Its proof relies on conceptsfrom advanced mathematics (cf. M.T.Nair, Functional Analysis: A First Course,Prentice-Hall of India, new delhi, 2002).

We may observe the following:

• Suppose f is an even function. Then f(x) cosnx is an even function andf(x) sinnx is an odd function. Hence bn = 0 for all n ∈ N, so that in this case theFourier series of f is

s(x) =a02

+∞∑n=1

an cosnx

with

an =2

π

∫ π

0f(x) cosnxdx.

In particular,

s(0) =∞∑n=0

an, s(π) = a0 +∞∑n=1

(−1)nan.

• Suppose f is an odd function. Then f(x) cosnx is an odd function andf(x) sinnx is an even function. Hence an = 0 for all n ∈ N ∪ 0, so that inthis case the Fourier series of f is

∞∑n=1

bn sinnx with bn =2

π

∫ π

0f(x) sinnxdx.

In particular,

s(π/2) =

∞∑n=0

(−1)nb2n+1.

8.1.2 Even and odd expansions

Suppose a function is defined on [0, π). Then we may extend it to [−π, π) in anymanner, and then extend to all of R periodically, that is by defining f(x±2π) = f(x).In this fashion we can get many series expansions of f all of which coincide on [0, π].

For example, by defining

fodd(x) =

f(x) if 0 ≤ x < π,−f(−x) if − π ≤ x < 0,

,

Fourier Series of 2π-Periodic functions 139

feven(x) =

f(x) if 0 ≤ x < π,f(−x) if − π ≤ x < 0,

then we may observe that

fodd(−x) = −fodd(x), feven(−x) = feven(x) ∀x ∈ [−π, π],

so that fodd and feven are the odd extension and even extension of f respectively.Therefore,

f(x) ∼ a02

+∞∑n=1

an cosnx, x ∈ [0, π), (∗)

and

f(x) ∼∞∑n=1

bn sinnx, x ∈ [0, π), (∗∗)

with

an =2

π

∫ π

0f(x) cosnxdx, bn =

2

π

∫ π

0f(x) sinnxdx.

The expansions (∗) and (∗∗) are called, respectively, the even and odd expansionsof f on [0, π).

EXAMPLE 8.1 Consider the 2π-periodic function f with f(x) = x for x ∈ [−π, π].Note that the f is an odd function. Hence, an = 0 for n = 0, 1, 2, . . ., and the Fourierseries is

∞∑n=1

bn sinnx, x ∈ [0, π]

with

bn =2

π

∫ π

0x sinnxdx =

2

π

[−xcosnx

n

]π0

+

∫ π

0

cosnx

ndx

=

2

π

−π cosnπ

n

=

(−1)n+12

n.

Thus the Fourier series is

2∞∑n=1

(−1)n+1

nsinnx.

In particular (using Dirichlet’s theorem), with x = π/2 we have

π

4=∞∑n=1

(−1)n+1

nsin

nπ

2=∞∑n=0

(−1)n+1

2n+ 1.

EXAMPLE 8.2 Consider the 2π-periodic function f with f(x) = |x| for x ∈[−π, π]. Note that the f is an even function. Hence, bn = 0 for n = 1, 2, . . ., andthe Fourier series is

a02

+

∞∑n=1

an cosnx, x ∈ [0, π], an =2

π

∫ π

0x cosnxdx.

It can be see that a0 = π, and

a2n = 0, a2n+1 =−4

π(2n+ 1)2, n = 0, 1, 2, . . . .

Thus,

|x| ∼ π

2− 4

π

∞∑n=0

cos(2n+ 1)x

(2n+ 1)2, x ∈ [0, π].

Taking x = 0 (Using Dirichlet’s theorem), we have

π2

8=∞∑n=0

1

(2n+ 1)2.

EXAMPLE 8.3 Consider the 2π-periodic function f with

f(x) =

−1, −π ≤ x < 0,

1, 0 ≤ x ≤ π.

Note that the f is an odd function. Hence, an = 0 for n = 0, 1, 2, . . ., and the Fourierseries is

∞∑n=1

bn sinnx, x ∈ [0, π]

with

bn =2

π

∫ π

0f(x) sinnxdx =

2

π

∫ π

0sinnxdx =

2

π(1− cosnπ).

Thus

f(x) ∼ 4

π

∞∑n=0

sin(2n+ 1)x

2n+ 1.

Taking x = π/2, we have

π

4=

∞∑n=0

(−1)n

2n+ 1.

EXAMPLE 8.4 Consider the 2π-periodic function f with f(x) = x2, x ∈ [−π, π].Note that the f is an even function. Hence, bn = 0 for n = 1, 2, . . ., and the Fourierseries is

a02

+∞∑n=1

an cosnx, x ∈ [0, π], an =2

π

∫ π

0x cosnxdx.

It can be see that a0 = 2π2/3, and an = (−1)n4/n2. Thus

x2 ∼ π2

3+ 4

∞∑n=1

(−1)n cosnx

n2, x ∈ [−π, π].

Fourier Series of 2`-Periodic Functions 141

Taking x = 0 and x = π (Using Dirichlet’s theorem), we have

π2

12=

∞∑n=1

(−1)n+1

n2,

π2

6=

∞∑n=1

1

n2

respectively.

8.2 Fourier Series of 2`-Periodic Functions

Suppose f is a T -periodic function. We may write T = 2`. Then we may considerthe change of variable t = πx/` so that the function f(x) = f(`t/π), as a functionof t is 2π-periodic. Hence, its Fourier series is

a02

+∞∑n=1

(an cosnt+ bn sinnt)

where

an =1

π

∫ π

−πf

(`t

π

)cosntdt =

1

`

∫ `

−`f(x) cos

nπx

`dx,

bn =1

π

∫ π

−πf

(`t

π

)sinntdt =

1

`

∫ `

−`f(x) sin

nπx

`dx.

In particular,

• if f is even, then bn = 0 for all n and

an =2

`

∫ `

0f(x) cos

nπx

`dx,

• if f is odd, then an = 0 for all n and

bn =2

`

∫ `

0f(x) sin

nπx

`dx,

8.2.1 Fourier series of functions on arbitrary intervals

Suppose a function f is defined in an interval [a, b). We can obtain Fourier expansionof it on [a, b) as follows:

Method 1: Let us consider a change of variable as y = x− a+b2 . Let

ϕ(y) := f(x) = f(y +a+ b

2), where − ` ≤ y ≤ `

with ` = (b−a)/2. We can extend ϕ as a 2`-periodic function and obtain its Fourierseries as

ϕ(y) ∼ a02

+

∞∑n=1

(an cos

nπ

`y + bn sin

nπ

`y)

where

an =1

`

∫ `

−`ϕ(y) cos

nπx

`ydy,

bn =1

`

∫ `

−`ϕ(y) sin

nπx

`ydy.

Method 2: Considering the change of variable as y = x − a and ` := b − a, wedefine ϕ(y) := f(x) = f(y + a) where 0 ≤ y < `. We can extend ϕ as a 2`-periodicfunction in any manner and obtain its Fourier series. Here are two specific cases:

(a) For y ∈ [−`, 0], define fe(y) = f(−y). Thus fe on [−`, `] is an even function.In this case,

ϕ(y) ∼ a02

+∞∑n=1

an cosnπ

`y

where ` = (b− a)/2 and

an =1

`

∫ `

−`ϕ(y) cos

nπx

`ydy.

(b) For y ∈ [−`, 0], define fo(y) = −f(−y). Thus fo on [−`, `] is an odd function.In this case,

ϕ(y) ∼∞∑n=1

bn sinnπ

`y

where

bn =1

`

∫ `

−`ϕ(y) sin

nπx

`ydy.

From the series of f we can recover the corresponding series of f on [a, b] bywriting y = x− a.


1. Find the Fourier series of the 2π- period function f such that:

(a) f(x) =

1, −π

2 ≤ x <π2

0, π2 < x < 3π

2 .

(b) f(x) =

x, −π

2 ≤ x <π2

π − x, π2 < x < 3π

2 .

(c) f(x) =

1 + 2x

π , −π ≤ x ≤ 01− 2x

π , 0 ≤ x ≤ π.

(d) f(x) = x2

4 , −π ≤ x ≤ π.

2. Using the Fourier series in Exercise 1, find the sum of the following series:

(a) 1− 1

3+

1

5− 1

7+ . . ., (b) 1 +

1

4+

1

9+

1

16+ . . ..

(c) 1− 1

4+

1

9− 1

16+ . . ., (d) 1 +

1

32+

1

52+

1

72+ . . ..

3. If f(x) =

sinx, 0 ≤ x ≤ π

4cosx, π

4 ≤ x <π2

, then show that

f(x) ∼ 8

πcos

π

4

[sinx

1.3+

sin 3x

5.7+

sin 10x

9.11+ . . .

].

4. Show that for 0 < x < 1,

x− x2 =8

π2

[sinxπ

13+

sin 3πx

33+

sin 5πx

53+ . . .

].

5. Show that for 0 < x < π,

sinx+sin 3x

3+

sin 5x

5+ . . . =

π

4.

6. Show that for −π < x < π,

x sinx = 1− 1

2cosx− 2

1.3cos 2x+

2

2.4cos 3x− 2

3.5cos 4x+ . . . ,

and find the sum of the series

1

1.3− 1

3.5cos 4x+

1

5.7− 1

7.9+ . . . .

7. Show that for 0 ≤ x ≤ π,

x(π − x) =π2

6−[

cos 2x

12+

cos 4x

22+

cos 6x

32+ . . .

],

x(π − x) =8

π

[sinx

13+

sin 3x

33+

sin 5x

53+ . . .

].

8. Assuming that the Fourier series of f converges uniformly on [−π, π), showthat

1

π

∫ π

−π[f(x)]2dx =

a202

+∞∑n=1

(a2n + b2n).

9. Using Exercises 7 and 8 show that

(a)

∞∑n=1

1

n4=π4

90, (b)

∞∑n=1

(−1)n−1

n2=π2

12

(c)∞∑n=1

1

n6=

π6

945(d)

∞∑n=1

(−1)n−1

(2n− 1)3=π3

32

10. Write down the Fourier series of f(x) = x for x ∈ [1, 2) so that it converges to1/2 at x = 1.

References

[1] R.G.Bartle and D.R.Sherbert, Introduction to Real Analysis, Third Edition,John Wiley & Sons, Inc. 2000

[2] K.G. Binmore, Mathematical Analysis: a straight forward approach, CambridgeUniversity Press, 1991, Third Print 1999.

[3] S.R. Ghorpade and B.V. Limaye, A Course in Calculus and Analysis, Springer,2006

[4] N. Piskunov, Differnetial Integral Calculus, Vol.I, Mir Publishers, Moscow,1974.

[5] N. Piskunov, Differnetial Integral Calculus, Vol.II, Mir Publishers, Moscow,1974.

144

Index

Abel’s theorem, 120Absolute convergence, 27Alternating series, 26anti-derivative, 79Arc length, 95Area, 93

beta function, 91bounded

above, 6below, 6

bounded function, 49bounded set, 48

Cartesian coordinates, 93, 95Cauchy product, 45Cauchy sequence, 16, 17Cauchy’s test, 24Centre of gravity, 100Change of variable, 81Comparison test, 23convergence, 2converges

pointwise, 108, 115uniformly, 109, 115

critical point, 65

de’Alembert’s test, 24decreasing, 63

strictly, 63Definite Integral, 68derivative, 52differentiable at x0, 52Dirichlet’s Theorem, 127divergence, 6domain of convergence, 121dominated series, 115

eventually constant, 4

exponential function, 44

Fourier coefficients, 126Fourier series, 126Fourier Series of

2π-periodic functions, 125Fourier series of

2`-periodic functions, 130even and odd expansions, 128even and odd function, 127functions on arbitrary intervals,130

Fundamental theorem, 78, 79

gamma function, 90geometric series, 22greatest lower bound, 8

property, 9

improper integral, 85increasing, 63

strictly, 63indefinite integral, 78infimum, 8Integrability, 71integrable, 71integral, 71integral of

continuous function, 83Integral test, 26

least upper bound, 8property, 9

limit, 2local extremum, 64local maximum, 64local minimum, 64logarithm, 44, 47logarithm function, 47

145

146 Index

lower bound, 8lower integral, 70lower sum, 70

Mean-value theorem, 76generalized, 77

Moment of inertia, 103monotonically decreasing, 63monotonically increasing, 63

natural logarithm, 44, 47natural logarithm function, 47

parametric form, 97parametrization, 94partial sum, 115Partition, 69periodic, 125pointwise

limit, 108Polar coordinates, 94–96power series, 120primitive, 79Product formula, 80

radius of convergence, 121ratio test, 24real sequence, 1Riemann integrable, 71Riemann integral, 71Riemann sum, 73root test, 24

Sandwitch theorem, 5sequence, 1

alternating, 6bounded, 6monotonically decreasing, 7monotonically increasing, 7strictly decreasing, 7strictly increasing, 7

series, 21of functions, 115convergent, 21divergent, 21

Series of functions, 115Solid of revolution, 99

subsequence, 9supremum, 8Surface of revolution, 99

tag, 73trigonometric series, 125

unbounded function, 49unbounded set, 48uniform convergence, 109upper bound, 8upper integral, 70upper sum, 70

Volume, 98

calculus - i1 sequence and series of real numbers 1.1 sequence of real numbers suppose for each...

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