Engineering Mathematics Volume-IIIA by B.K.Pal and K.Das [Volume-IIIA]

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Revised 8th Edition According to the Latest Syllabus .J

ENGINEERING MATHEMATICS Volum.e-IIIA for 3rd Semester (For CSE & IT) B.K.PAL, K.DAS,

M.Sc.,Ph.D.

M.Sc.,B.Ed.,Ph.D

•.

COMPUMENTAR'fC Y ~01 FOR SALE

https://www.instagram.com/ranjan_mahato_rkm/ Syllabus

Preface

MAULANA ABUL KALAM AZAD UNIVERSITY OF TECHNOLOGY, W. B.

Engineering Mathematics being one of the most essential subjects of the B.Tech., course offered by the Engineering or Technological Institutes. This text book is written according to the syllabus of WBU Tech covering the entire course and also as reference book to the all India Engineering Courses in four separate volumes. The present volume (Vol IlIA) has been designed for B. Tech 3rd semester, i.e. Second year 1st semester for the sudents of Computer Science & Information Technology under WBU Technology syllabus . Sincere attempts have been made to present the topics in a simple and lucid manner to create interest into the subject along with various types of worked out problems for securing high marks in the examination . Valumes I, II and IV, published previously, have achived acceceptation to the Professors and have become popular to the students considering the utility of the same. We take this opportunity to express our gratitude to all the Teachers & Professors who recomended our books to the Students and sent their valuable comments. We are thankful to Dr. Purnendu Dhar, M. Sc.(Chem)., Ph.D.,Director, Mis. U. N. Dhur & Sons Pvt. Ltd. for publishing the book in time. Any positive suggestion shall be acknowledged with thanks. This work would not also be possible without the constant encouragement and enthusiasm of our family members. Kalyani Govt. Engg. College Nadia, West Bengal. Dated: July, 2005.

Bidyut Kumar Pal Kalidas Das

Syllabus for B. Tech in CSE & Information Technology (Applicable from the academic session 2018-2019) Subject Code: BSC 301, Category: Basic Science Course Subject Name: Mathematics- III Semester : Third [L-T-P : 2-0-0 ] Credit:2: Pre- Requisites: No-prerequisiteCourse Content: MODULE1: Sequences and series [8L] Convergence of sequence and series, tests for convergence, power series, Taylor's series. Series for exponential, trigonometric and Logarithmic functions. MODULE 2: Multivariable Calculus (Differentiation) [7L] Limit, continuity and partial derivatives, Chain rule, Implicit function, Jacobian, Directional derivatives, Totalderivative; Maxima, minima and saddle points; Gradient, curl and divergence and related problems. MODULE3: Multivariable Calculus (Integration) [8L] Double and triple integrals (Cartesian and polar), change of order of integration in double integrals, Change of variables (Cartesian to polar). Theorems of Green, Gauss and Stokes (Statement only) and related problems. MODULE4: Ordinary Differential Equation [9L] First Order Differential Equation, Exact, Linear and Bemoulli 's equations, Equations of first order but not of first degree: equations solvable for p, equations solvable for y, equations solvable for x and Clairaut's form, general & singular solution. [5L] Second order linear differential equations with constant coefficients, D-operator method, method of variation of parameters, Cauchy Euler equation. [4L] MODULE5: Graph Theory [8L] Basic Concept of graph, Walk, Path Circuit, Euler and Hamiltonian graph, diagraph. Matrix Representation: Incidence & Adjacency matrix. Tree: Basic Concept of tree, Binary tree, Spanning Tree, Krus Kal and Prim's algorithm for finding the minimal spanning tree.

Conten.ts

https://www.instagram.com/ranjan_mahato_rkm/ CONTE TS [II]

(MODULE-II

1-73 1-75 1-76 1-82

QUESTIONS

ANSWERS

I[J]

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

Ib;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;S;.E;;;;Q;.v;;;;E;;;;N

C;;;;E;;;;;;;;;;;;;;;&;;;;S;;;;E;;;;R;;;;IE;;;;;;;S

1.1.1. Introduction. 1.1.2. Sequence. 1.1.3. Monotone Sequence. 1.1.4. Bounded Sequence. 1.1.5. Limit of a Sequence. 1.1.6. Convergent, Divergent and Oscillating sequences. 1.1.7. A few Basic Theorems. 1.1.8. A useful sequence. 1.1.9. Theorems on Limits of Sequences. 1.1.10. Illustrative Examples. I.I~ II. Infinite Series. 1.1.12. Two Important Infinite Series. 1.1.13. Some Properties of an Infinite Series. 1.1. 14. Power series. 1.1.15. Interval of Convergence. 1.1.16. Properties of power series. 1.1.17. Test of Convergence of Series of non-negative terms. 1.1.18. Comparison Test. 1.1.19. Illustrative Examples. 1.1.20. D' Alembert's Ratio test. 1.1.21. Illustrative Examples. 1.1.22. Cauchy's Root test. 1.1.23. Illustrative Examples. '1.1.24. Raabe's test. 1.1.25. Illustrative Examples. 1.1.26. Alternating Series. 1.1.27. Illustrative Examples. 1.1.28. Absolute and Conditional Convergence. 1.1.29. Illustrative Examples. EXERCISE

[I]

LONG ANSWER

IX

SHORT ANSWER ANSWERS

QUESTIONS

1-1 1-\ 1-2 1-2 1-3 1-3 1-4 1-6 1-9 1-10 1-19 1-21 1-22 \-23 1-23 1-24 1-24 1-26 1-27 1-40 1-41 1-48 1-48 1-5\ 1-52 1-62 1-63 1-65 1-66 \-69 1-69 1-72

1.2.1. 1.2.2. 1.2.3. 1.2.4. 1.2.5.

Introduction. Taylor's Infinite Series. Different Forms of Taylor's Infinite Series. Maclaurin's Infinite Series. Illustrative Examples.

1-83 \-83 1-84 1-84 1-85 1-89 1-92 1-93 1-93

EXERCISE ANSWERS

[ I]

MULTIPLE

CHOICE

QUESTION

ANSWERS

IMODULE-2I

I~.ll ~_

SEVERAL

VARIABLE

--;;;;;;;;;;;;;;;;;D;;;;;;;;I F;;;;F;;;;E;;;;R;;;;E;;;;N;;;;T;;;;I;;;;A;;;;T;;;;IO;;;;N;;;;;;;;O;;;;F;;;;;;;;F;;;;U;;;;N;;;;;C;;;;;;T;;;;;I O;;;;;N;;;;;;;;O;;;;;;;;;F

2.1.1. 2.1.2. 2.1.3. 2.1.4. 2.1.5. 2.1.6. 2.1.7. 2.1.8. 2.1.9. 2.1.10. 2.1.l1. 2.1.12. 2.1.13. 2.1.14.

Introduction. Function of Two or More Variables. Geometrical Interpretation of a Function of Two Variables. Limit and continuity of a Function of Two Variables. Illustrative Examples. Repeated Limit. Partial Derivative. Illustrative Examples. Higher Order Partial Derivatives. Illustrative Examples. Function of functions: Chain Rule Illustrative Examples. Chain Rule for Functions of three or more Variables. Illustrative Examples.

2-1 2-1 2-1 2-2 2-3 2-5 2-7 2-8 2-14 2-15 2-23 2-23 2-25 2-25

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x

ENGINEERING

2.1.15. 2.1.16. 2.1.17. 2.1.18.

Differentiation of Implicit Function. Total Differentiation. Second Order Differential. lllustrative Examples.

[I]

SHORT ANSWER

MATIIEl\IATICS-III.\

EXERCISE QUESTIONS

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

2-31 2-35 2-36 2-37 2-46 2-46 2-49 2-49 2-55 2-56 2-59

I~I~~J_A_CO_B_IAN~_&~H_O_M_O_G_E_NE~OU_S~F;;;U~N;;;;;CT;;;I~O~N

2.2.1. 2.2.2. 2.2.3. 2.2.4. 2.2.5.

Jacobians. lllustrative Examples. Homogeneous Functions. Euler's Theorem. lllustrative Examples. EXERCISE

[I]

SHORT ANSWER

QUESTIONS

ANSWERS

[II]

Lone

ANSWER

QUESTIONS

ANSWERS

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

IMI

MAxiMA

&

MINIMA LAGRANGE MULTIPUER

2.3.1. 2.3.2. 2.3.3. 2.3.4. 2.3.5.

Introduction. Maximum and Minimum of a function of two variables: Illustrative Examples. Lagrange's Method of Undetermined Multipliers. Illustrative Examples.

[I]

SHORT ANSWER

EXERCISE

ANSWERS

2-60 2-61 2-65 2-65 2-66 2-78 2-78 2-79 2-79 2-80 2-81 2-83

QUESTIONS

2-84 2-84 2-88 2-97 2-98 2-116 2-116 2-117

CONTENTS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

XI

2-118 2-120 2-121 2-122

IMI;;;;;;;;;;;;;;;;;;;;;G;;;;;R;;;;;AD;;;;;;;;;;I;;;;;E;;;;;N;;;;;T;;;,D;;;;I;;;;VE;;;;;;;;;;R;;;;G;;;;;E;;;;N;;;;;C;;;;E;;;;A;;;;;N;;;;;D;;;;;;;C;;;;;U;;;;R;;;;;;;;;;;L

2.4.1. 2.4.2. 2.4.3. 4.4.4. 2.4.5. 2.4.6. 4.4.7. 2.4.8. 2.4.9. 2.4.10. 2.4.11.

Introduction. Scalar and Vector Fields. Gradient. Directional Derivative of a Scalar Point Function. Tangent Plane and Normal to a Level Surface. Illustrative Examples. Divergence of a Vector Point Function. Curl of a Vector Point Function. The Laplacian Operator V2• Important Identities. Illustrative Example.

[I]

SHORT ANSWER

EXERCISE QUESTIONS

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[Ill]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

2-123 2-123 2-123 2-127 2-130 2-131 2-137 2-138 2-140 2-140 2-143 2-151 2-151 2-153 2-153 2-155 2-156 2-158

IMODULE-3I I~I~;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;D;;;;;O;;;;;;;U;;;;;;;B;;;;;;;L;;;;;;;E;;;;;;;I;;;;;;;N;;;;;;;T;;;;;;;EG;;;;;;;R;;;;;;;A;;;;;;;L;;;;;;S

3.1.1. 3.1.2. 3.1.3. 3.1.4. 3.1.5.

Introduction. Definition of Double Integral. Evaluation of Double Integral by Repeated Integral. Transportation of Double Integral. Application of Double Integrals.

[I]

SHORT ANSWER

EXERCISE QUESTIONS

3-1 3-1 3-2 3-14 3-19 3-29 3-29

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XII

ENGINEERING

ANSWERS LONG ANSWER QUESTIONS ANSWERS MULTIPLE CHOICE QUESTIONS ANSWERS

[II] [III]

I~I 3.2.1. 3.2.2. 3.2.3. 3.2.4. [I] [II] [III]

IMI 3.3.1. 3.3.2. 3.3.3. 3.3.4. 3.3.5. 3.3.6. 3.3.7. [I] [II] [III]

MATHEMATICS-iliA

3-32 3-33 3-38 3-39 3-41

TRIPLE

Introduction. Definition of Triple Integral. Evaluation of Triple Integral by Repeated Applications of triple integrals. EXERCISE SHORT ANSWER QUESTIONS ANSWERS LONG ANSWER QUESTIONS ANSWERS MULTIPLE CHOICE QUESTIONS ANSWERS

THEOREM

OF GREEN,

Introduction. Line Integral in the Plane. Line Integral in space. Surface Integrals. Green's Theorem in the Plane. The Divergence Theorem of Gauss. Stoke's Theorem. EXERCISE SHORT ANSWER QUESTIONS ANSWERS LONG ANSWER QUESTIONS ANSWERS MULTIPLE CHOICE QUESTIONS ANSWERS

INTEGRALS

Integral.

GAUSS

&

3-42 3-42 3-42 3-56 3-63 3-63 3-64 3-64 3-66 3-66 3-67

STOKES 3-68 3-68 3-73 3-77 3-82 3-89 3-100 3-108 3-108 3-110 3-111 3-114 3-114 3-117

XIII

CONTENTS

IMODULE-4I

I~I 4.1.1. 4.1.2. 4.1.3. [I] [ II] [III]

~ 4.2.1. 4.2.2. [I] [II] [Ill]

19] 4.3.1. 4.3.2.

4.3.3.

EXACT EQUATIONS

Introduction. Exact Equations. Integrating factor (I. F). EXERCISE SIIORT ANSWER QUESTIONS ANSWERS LONG ANSWER QUESTIONS ANSWERS MULTIPLE CI IOICE QUESTIONS ANSWERS

LINEAR AND BERNOULLI'S Linear equations. Bernoulli's equation. EXERCISE SHORT ANSWER QUESTIONS ANSWERS LONG ANSWER QUESTIONS ANSWERS MULTIPLE CHOICE QUESTIONS ANSWERS

4-1 4-1 4-5 4-14 4-14 4-15 4-15 4-17 4-18 4-20

EQUATIONS 4-21 4-24

4-32 4-32 4-33 4-33 4-35 4-36 4-38

DIFFERENTIAL EQUATION OF FIRST ORDER AND HIGHER DEGREE 1ntroduction. Equations solvable EXERCISE-A ANSWERS Equations solvable EXERC/SE-B

for p.

for y.

4-39 4-39 4-42 . 4-43 4-43 4-47

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ENGINEERI

G MATHEMATICS-iliA

4-47 4-48 4-53 4-61 4-61 4-61 4-62 4-62 4-63 4-66

ANSWERS

4.3.4. 4.3.5.

Equations solvable for x. Clairaut's Equation. EXERCISE D

[I]

SHORT ANSWER

QUESTIONS

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

xv

CONTE

TS

4.6.1.

Cauchy-Euler Homogeneous Linear Equations. EXERCISE

[I]

SHORT ANSWER

QUESTIONS

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[III]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

IMODULE-5I 4.4.1. 4.4.2. 4.4.3.

Introduction. Theorems. Rules for finding CF,

4-67 4-67 4-68 4-71 4-71

EXERCISE-A ANSWERS

4.4.4.

Rules for finding the particular integral (P.I.) by operator method. 4.4.4.1. The inverse operator. 4.4.4.2. General method for finding P. I. 4.4.4.3. Short methods for finding P. I. in some special cases. EXERCISE B [I] SHORT ANSWER QUESTIONS ANSWERS

[II]

LONG ANSWER

[III]

MULTIPLE

QUESTIONS

ANSWERS

•...~ 4.5.1.

..

CHOICE

QUESTIONS

ANSWERS

METHOD

OF

VARIATION

OF

EXERCISE SHORT ANSWER

[II]

LoNG

QUESTIONS

ANSWERS ANSWER

ANSWERS

QUESTIONS

I~

I~;;;;G;;;;RAP=H;;;;' ;;;;D;;;;I-G=RAP=H=&;;;;I;;;;T;;;;S ;;;;F;;;;U;;;;ND;;;;;;;;AM;;;;;;;;;;;;;;;E;;;;:;NT;

5.1.1. 5.1.2. 5.1.3. 5.1.4. 5.1.5. 5.] .6.

Introduction to Graph Graph and Related Terms Isomorphic Graphs Di-Graph (Directed Graph) Theorems of Graph. Illustrative Examples EXERCISE

[I]

SHORT ANSWER

QUESTIONS

ANSWERS

[I I]

LONG ANSWER

QUESTIONS

ANSWERS

[Ill]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

5-] 5-] 5-]6 5-23 5-26 5-34 5-42 5-42 5-44 5-45 5-55 5-57 5-65

I~I~==;;;;;M=A;;;;;TRIX=;;;;;R=EP;;;;;R;;;;;E;;;;;S;;;;;E;;;;;N;;;;;TA;;;;;T;;;;;IO;;;;;N=O;;;;;F;;;;;G;;;;;RA=P;;;;;;;;;H

PARAMETERS

Method of variation of parameters to find P. I.

[I]

4-72 4-72 4-73 4-74 4-89 4-89 4-90 4-90 4-92 4-96 4-100

4-115 4-122 4-122 4-123 4-123 4-124 4-126 4-127

4-100 4-]1] 4-111 4-112 4-112 4-113

5.2.1. 5.2.2. 5.2.3 5.2.4 5.2.5. 5.2.6.

Introduction Incidence Matrix of a Connected Graph. Incidence Matrix of a connected Di-Graph. Incidence Matrix of Disconnected Graph / Di-graph Permutations of Rows/Columns of an Incidence Matrix. Adjacency Matrix ofa Graph (or, Connection Matrix)

5-66 5-66 5-68 5-70 5-72 5-75

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G MATII£l\1ATICS-IIIA

XVI

5.2.7. 5.2.8.

Adjacency Matrix of a Di-Graph Adjacency Matrix for Disconnected Graph / Di-Graph.

5-77 5-79 5-82 5-82 5-86 5-87 5-92 5-93 5-99

EXERCISE

[I]

SIIORT ANSWER

QUESTIONS

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[II I]

MULTIPLE

CHOICE

QUESTIONS

ANSWERS

1MI 5.3.l. 5.3.2. 5.3.3. 5.3.4. 5.3.5. 5.3.6. 5.3.7. 5.3.8. 5.3.9. 5.3.10. 5.3.11. 5.3.12. 5.3.13. 5.3.14. 5.3.15.

;;;T;;;R;;;EE;;;S __ A;;;N;;;D;;;S;;;P;;;A;;;N;;;N;;;IN;;;G__

Introduction Trees and Related terms: Theorems on Trees. Theorems on Binary Trees. Spanning Tree and Co-Tree. Breadth First Search (BFS) Algorithm for construction of a spanning tree. Depth First Search Algorithm for construction of a spanning Tree. Theorems on Spanning Tree. Illustrative Examples. Finding all Spanning Trees in a Graph. Weight of an edge and Weighted Graph. Minimal Spanning Tree (or, Shortest Spanning Tree). Kruskal's Algorithm of finding Minimal (or, shortest) Spanning Tree. Prim's Algorithm of finding Minimal (or, shortest) Spanning Tree. Miscelleneous Examples. EXERCISE

[I]

SHORT ANSWER

QUESTIO

S

ANSWERS

[II]

LONG ANSWER

QUESTIONS

ANSWERS

[III]

T;;;R;;;E;;;;;;E

MULTIPLE ANSWERS

CIIOICE

QUESTIONS

5-100 5-100 5-103 5-108 5-109 5-113 5-114 5-116 5-117 5-119 5-12 I 5-123 5-125 5-127 5-131 5-136 5-136 5-140 5-141 5-150 5-152 5-159

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1.1.1. Introduction. The present chapter introduces briefly the theory of sequence and infinite series. Unlike finite series the sum of an arbitrary infinite series may not. be obtained. When it is obtained the infinite series is named convergent series. We begin with- definitions of convergence and divergence and oscillatory series. We introduce a number of useful tests for the convergence of series. In first few sections we deal only with series of positive terms. In the later sections we consider series whose terms are not restricted to be positive, introducing the notion of absolute convergence. We then demonstrate a theorem, called Leibnitz's theorem which is useful for testing convergency of alternating series. By its use we .exhibit such an infinite series which converges but which fails to converge absolutely. But first of all, we discuss some basic concepts on 'Sequence' which is very useful for studying Infinite Series. 1.1.2 Sequence. A collection S of real numbers (not necessarily distinct) is said to be a sequence of numbers if corresporiding to every positive integer n, there exist a unique element of S. Note. (1) The element of S corresponding to the positive integer n is usually denoted by an' (2) A sequence of numbers can be written in the form -a1 , a2, a3 . Notation. -A sequence consisting of ai' a2' {ai' a2,

••••••

}

is usually denoted by

or briefly by { an}.

Illustration. (i)

{5, 7, 9, ll} is a finite sequence.

--Ill}. _ I (ii)" { 1,-,-,-, ... 1.e.,{an}wherean=-, 2 3 4 n known as harmonic sequence. (Hi) The sequence {an} where an It is {5, 5, 5, }. E.M:3A-l

=5

IS

a sequence.

It is

for all n is a constant sequence.

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E GlNEERlNG MATHEMATICS-lllA

1-2

(iv) . The sequence

{an}

where an = ~ {I + (_1)"-I}, where n is

natural number, i.e., the sequence is {I, 0, 1,0,

}.

3.1.3. Monotone Sequence. Monotonic Increasing Sequence A sequence {an} of real numbers is said to be monotonic increasing ifal Sa2 Sa3 Sa4 S ,i.e.,aj Saj+1 forl=I,2,3, Monotonic Decreasing .Sequence.

.

A sequence {an} of real numbers is said to be monotonic decreasing if al

a2 ~ a3 ~

~

Monotonic

0,

.

Sequence.

~ sequence sequence

, i.e., aj ~ aj+1 for i = 1,2,3,

{an}

{an}

of real numbers is said to be monotonic

if the

is either monotonic increasing or monotonic decreasing.

ote. A sequence can neither be increasing nor decreasing e.g. {I, 0, 1, } is neither monotonic increasing nor monotonic decreasing.

1.1.4. Bounded Sequence. A sequence {an} of real numbers is said to be bounded above if there exist a real number M such that all S M, for n = 1,2, . A sequence {an} of real numbers ~s said to be bounded below if there exist a real number m such that an ~ m, for n = 1,2, 3, . Definition. A sequence {all} of real numbers is said to be bounded if the sequence is bounded above as well as bounded below. ote. Obviously a sequence of real numbers is said to be bounded if and only if there exist a positive number k such that I an I S k, for n = 1, 2, . i.e., ..-k San S k, for n = 1,2, Illustration. (i) . The' sequence

{I, 2\ ' ·3\ ,.'.} (ii) 2

{I, 3, 3 , below).

. where

is bounded because

The sequence ••• }

{an}

{a n}

is not bounded,

.

°

an

< an

where an although

=

1 n2

i.e.,

the

sequence

S 1 for all +ve integer n. = 3n-l,

that is the sequence

all ~ 1 'if n EN (i.e.,

bounded

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SEQUENCE & SERIES

1.1.5.

1-3

Limit of a Sequence.

{a n}

A sequence

of real numbers is said to tend to a limit I, if for

every e ~ 0, there exist a positive integer N depending on e, such that for

Ian :- II < e.

every n > N,

,,~oo - = I or simply

Then we write lim a" that is -a" ~ I as 11 ~ 1.1.6.

Convergent,

lim a"

=I

00,

Divergent

Convergent Sequence. Jim an is finite,

and Oscillating

{a,,}

A sequence

sequences.

is said to be Convergent if

n......• 00

Illustration.

-{-II-I}

The sequence _ _ hI, were

1,-, -2'''' -, 3 3 3" an

=n 3

IS

i.e. the sequence

..,

'1' convergent, smce

{an}

Divergent Sequence. A sequence

1111 ,,--+00

{a,,}

1 = 0, a" = I'1111 n ,,--+00

3

is said to be Divergent if lim

a"

11-+00

is not finite i.e. if lim a" .

= + 00 or -

Illustration. (i) The sequence

{an}

where

lim an n--+ 00

_(ii)

00,

n-+ct'J

The sequence {- 4 lim

n-.oo

an =

n}

=

a"

3

=n

lim n3

n --+ 00

is divergent, .as

=+

00,

is divergent as

lim

(-4") = -

00,

11-+00

Oscillating- Sequence. A. sequence which is neither convergent divergent is called an Oscillating sequence, Illustration. The sequence {-I, 1,-1, 1,

} that is, {(-I)"} is oscillatory,

Here «, =(_I)n, Jim

Ozn

n~oo

n400

-I'1m a

+1

211

11--+00

= -lim (_1)2/1 = 1 = I'1m (1)2n+1 = - 1, 11-+00

nor

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1-4

ENGINEERING MATHEMATICS -ilIA

lim

does not exist and hence the sequence is not convergent

all

n --. 00

and not divergent also. 1.1.7. A few Basic Theorems. In this article six important theorems on sequence will be discussed; these theorems are useful in solving problems of convergence of sequences. Theorem I. A convergent sequence determines its limit uniquely. Proof: If possible, II and 12 be two distinct limits of a convergent sequence {xn}' Since II and 12 are distinct, II

* 12, we can take

II - 12 = 8 , where 0

is a non-zero positive number. Now, let us choose a positive number suchthat

E
xn for all n

EN. and as

say

xIIi,

{XII:

is not bounded above, there exists at least one number,

of the sequence such that

XIII>

M,

M being a large positive

number, and since {xn} 'is monotonic increasing

Xm+2,

•••••

are all

Theorem VI. A monotonic decreasing sequence diverges to bounded below:

-00,

if not

XI/HI'

greater than M .Therefore, xn > M, for all n ~ m i.e. lim xn

= 00

n-+a:>

Hence, the sequence

{xn}

diverges to

+00

The proof is similar to that of Theorem V, taken above.

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1-6

ENGINEERING MATHEMATICS -ilIA

r

1.1.8. A useful sequence. Toprove that the sequence

where

{xn},

xn =

(1 +.;

is convergent.

Proof: It will be enough to show that the given sequence is monotonic increasing and bounded above. By the Binomial theorem, we have x =( n

1+-n

l)n

n

...+

=

n(n-l)

1 =l+n.-+

1

n

2!

2

+n(n-l)(n-2)

1 '-3

n

3!

+

n(n -l)(n - 2)...3.2.1 1 .nn

n!

1+1+~(I-.!.)+~(I-.!.) (1-~)+... + ~!(I-';)(I-;)...(Ii) (1- n-I) + I( 1- ;;1)(1-;;... "2!

n

3!

n

n

n:l)

n!

Replacing

n by (n + I),

xn+1 =

-n-

(I)

we have,

1+1+~(1__1 )+~(I __l )(1__2 )+... 2!

+

n+1

3!

n+1

n+l

1 (1__"n+l 1 )(1 __n+I" 2 ) (I __n+1n )

.(2)

(n+l)!

From (I)" and (2), we observe that: (i) the first two terms of 1

xn

1

and 1"

xn+1

are equal, each being 1;

1

(ii) also .: -< -, 1- -> 1- - ,and so on, thus except the first n+1 n n+1 n two terms, every term of xn+1 is greater than the corresponding term of xn

(iii) xn involves (n + I) terms, while all the terms are positive. Hence," xn+l

~ xn

for "all n

i.e. the sequence

{xn}

xn+l

involves

EN"

is monotonic increasing:

(n + 2)

terms and

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SEQUENCE

& SERIES

1-7

Next, we note that xn ~ 2 for all n EN .

{xn}

i.e. the sequence Also,

Xn

=

is bounded below, and 2 is a lower bound.

I+I+~(l-~)+~(l-~)(l-~)+ 2!

n

3!

n

n

1(

...

1)(02)( n-l) 1- -; ... 1- -n-

+ n! 1- -; 1 1 1 0, there exists a positive integer N1, such that 1 IXn - AI < - E for all n > Nl .2

Also, since lim y" 11-+00

. . N2, such that IYn

If N

±

E

E,

for all n > N2

then for n> N ,

IYn - BI
0, there exists a positive integer

1

- BI 1,

XII

> 1, and then we can write

I

So,

x; = 1+ h .

II

:. x = (1 + hn

. :. hn

x-I

< --

n

r

. ( h; > 0 )

> 1+ nhll

[Bernoulli's inequality]

and hn' ~ 0 , as n ~

00

I

conesequently, x; ~ 1 ' as n~

00

case (3) .. When O c x c l , ..!.

'.

-

x""< 1 and we write .

. Then

. 1 X=

(l+hll)n

I

I

xn =-. -

1 +hn

(hn

> 0)

1

0

Il~C:O

3n Example 5.If un = --1 ' show that {un} . n+

is monotonic

increasing and

bounded above. Find its limit. . Solution: .

. Here un

3n = --

n+l

=

3(n + 1) - 3 n+l

.

= 3---

3 n+l

< 3 forall n EN.

. 3 A gam un+1 = 3--n+2

S

UI

n+

0,

-u 11

=(3 __n +3 2 ) _(3 __n 3+ 1)

"=

Hence

{Un}

3(_1 n+l

1_) n+2

=

3 (n+l)(n+2).

> 0, for all n EN

is monotonic increasing and bounded above.

=

Now, "lim u; n ...• eo

lim(3 - _3_) = 3 n +1

n...• co

So, the limit of un is 3. Example 6. Using definition of convergent sequence of real numbers show that ~he sequence {~} ~onverges to O. Solution:

Let

E

I.!.-

We have , n

be any pre-assigned very small positive number.

01 = I.!.I = .!. < E n

n

whenever n > '

.!. E

I If G be the greatest integer in -, not exceeding it, E

then

I~01

G

Hence {~} converges to O.

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E GI EERING MATHEMATICS -iliA

1-14

Example

7. Consider

the sequence

{;:::}

Examirie (i) whether it is monotonic

increasing

or decreasing;

(ii) whether it is bounded above or below, and (iii) State whether it is convergent. Solution: .

.

4

and

U

So,

1

-37[I3n+4 -

n

n+1 -

7

'3 - 3' 3( n + 1) + 4

Un+ I =

• U

.•

4n + 3 1 ( I 2n + 16 - 7 ) 4 7 == -3 - -3 . 311+ 4 3n+ 4 3 3n +4

u; = --

Here, ..

=

(3n+4)

7

(3n+7)

>

0

for n

E

N

> Un for all n EN. [N = set of natural numbers]

Un+1

Hence, the sequence Un =

is monotonic

{Un}

4

(ii) Again.

:. {un}

1]

- 3n+7

7

increasing.

4

3- 3(3n + 4) < 3 for all

n EN

is bounded above .

. (iii) Since the sequence

{un}

is monotonic

increasing

and bounded

above, it is convergent. Example

8. If

monotonic

decreasing

.

Solution

Xn =

X

I 2n + 1 + 1 2n + I

1 1=

= ..!..[1+ 2 2(n+l)+1

. I( 1

n=2'

{xn}

is strictly

and hence prove that it is convergent.

n+1

x,,+1

Xn+1-

show that the sequence .

xn = -= -. 2n + 1 2

: Here, .

Similarly, .

n+l -2 I' n+

2n+3-2n+l

I)'

1(

1)

= - 1+-2 2n + I

_1_]

..!..[1+ 2 2n+3

-I

--'-":'" log2

(1)

log Thus, by, choosing an integer' N >

E,

log2 We have IXn - 21 XI

X2

or,

'

Here, obviously,

x2

< XI

X2

> xI

Therefore, the sequence is monotonic increasing. From (1) we have,

X;+I

xn - 2 = 0

-

or, x~ -xn -211

Vn be .two series of positive numbers and

v~ is divergent. Then the series (i)

lIn ~ Vn

L:

u; is divergent if

there exists a positive integer N such that

Un ~

kv; for all values

of n ~ N where k is constant" or if lim!!J:L = / where / is a non-zero nwnber (l may be oc) .

(ii)

n~

00 VII

Proof Beyond the scope. ote : (i) of the above theorem gives that if

L vn

is divergent if Theorem . Then (i) if ~

3. Let

>~

Un+1

series

L:

(ii) if ~ U,,+l

L:

Un ~ V n

for all n then

L un

is so.

L:

U/l

and

L:

v" be two series of positive numbers .

and

L:

v" is convergent, then the

for all values of nand

L:

v" is divergent, the series

for all values of

11

V/l+1

un

is also convergent 1). Therefore the series L IIn i.e. the given series is convergent. L

VII

i.e.,

L

-2

log 11

00

Example. 4. Test the series ,

L n=2

rrrt:

"\In

+I

Solution. Let u; (n = 2, 3, ... ) denote the n-th term of the given series. Then for n

2, 3, ... , logn ull = .In+l . 7'

Consider a series

L Vn

' 'un }nn -

N ow,

n--'fOCJ

and 'the series'

LVII

Vn

=

where }' 1m

VII

= ~.

..rniog n

n--'foo..r;+T'

=

n1m

log n

n=v «: ~1

+~

= 00

,

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& SERIES

SEQUENCE

i.e.,

L~

1-29

=~
O.

LU"

i.e.the given series is divergent.

J2 Y2

+

.J3

Y2

+

+ b a.2 2 + b

2

1).

aJ

2

+b

+ ." [W.B. UT 2004]

Un

(n= 1,2, ... ) denote the n-th term of the given series.

Then for n

=

J, 2, ... ,

Un

=

Solution. Let

..In 3/" 0.n/2

+ b

We see in the fraction U," (the degree of the expression in denominator) - (the degree of the expression in numerator) So we consider a series

= ~ - ~ = l.

L v n where v n = ~. II

,.- . B ut liun ,un -. ,,-too vn.

Now

LVn

i.e.

liun

=

n-too

n3 e n

anYz + b

Yz =

liim n-too

n3

U'I

. 3

6. Test the convergence 1/ 3' -

={n + IV

a fini illite num b er.

L ~ is a divergent series (being ap-series

Therefore by comparison test the series divergent. Example.

-,1

anYz + b a

LU

withp

= 1).

i.e., the given series is

n

of the series

n.

LlIn

where

[W.B. UT.2003,2007J

Solution. Here I

={n (l+ nl3)V-n 3

=17

=-,

1·1 -~(~-I).(1) 1+-.-.: [ 3

+-.JJ

1/"

1 (1 : 1

1/-

2!

---,-:-",oo,uptooo.

3

9

n"

-

1/3

)

2

]

+oo'Uptooo-11

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1-30

ENGIN~ERING MATHEMATICS-UIA

Consider a series :. ,

L vn where

~._I- + ... upto 9 n3

lim un = -lim (~Vn n-+oo 3

n-+oo

LUll

Hence by comparison test, both diverge. ,But ~ vn i.e., withp

=

vn =~. n

'and

LVII

(0)

=~,

3

a finite number.

will either both converge or

L nl2 is a convergent

series (being a p-seties

2>" I).

LUll

Therefore the series

i.e., the given series is convergent.

. . 1 22 33 44 Example. 7. Test the convergence 01 the senes 1 + -2 + -3 + 4 + -5 + .... 234 5 [WB.UT. 2005,2016] Solution. Omitting the frrst term, let un (n = 1, 2, ... ) be the n-th term of the given series. Then for n = I, 2; .:.. nil

U

----7"

II -

(n+1)n+1'

Since the difference of degree of the denominator and numerator is 1,

LVn

so.we considera series . , lim un

vn

nn+1

=

=~. n

11m ---_

n-+covn

n-+co(n+1)11+1

., 1 1un -------:-

= ,

n-+co

I' II

~moo

1

1

=-

e

(

1 )11+1

1+n 1

r.

(1 + ~ (1 + ;) 1

=: -;. (l

LV

where

+ 0) [

':

lim (I +

II-' 1, is convergent.

1·rm n~oo

L

+ ~ + ~ + ~+ ....

is so. 1 logn

[c.p 1998]

>! , for all >2 11

11

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SEQUE CE & SERIES

1-35

Let us compare the given series with the divergent series

L:>" =!+!+!+

...-:-!+...

2 3 4 n Each term of the given series exceeds the corresponding term ofthe

LVII .

diverge~t series

Hence, the given series is also divergent. 1 J L00(1 -srn + vn-l ,------;

Example. 19. Test the convergence of the series. Solution:

Denoting the given series by 1

1

u; = r + ,J

We have

vn .

I

"n(n-l)

1'1-1

LVII , where

vn

=

l

(~.+ Fn)Fn

U

lim _"

=

VI1

11••••• 00

L t!" ,

~+Fn

=

Let us introduce a comparison series .

,,=2

lim -'--~;=====-00 ~n(n -1)

11•••••

I

1

1

1

Example 20. Examine whether the series - - - +- - -+ 11 2.3 3.4 4.5 convergent. Solution : Den~ting the given series by . 1

1

1

Iunl=n(n+l) =-;;- n+l s, =·ilunl= (1_!)+(!_!)+

21~,we have, un=(_I)n-l.~),\n+l

So

.

1

2

.

So, lim

00

SI1

tt •...•

2

....:+(! __ 1 ) =1__ 1

3

n.

n+l

=1

Thus the series )

"'-J

u; is absolutely convergent, so it is convergent.

Example 21. Obtain the sum of the series .

Solution:

n+l

Denoting the given series by ..

1

un

=

(n+3) (n+4)

f I

Lilli

1 ) [c.p 2006] (1'1+3)(11+4 ,

we have

. IS

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1-36

ENGINEERING MATHEMATICS-ilIA

s;

If Sn

denote th~ nth partial sum of

I

UII

,

+ U2·+·····+ull I I I 1. =-+-+-+ .... +----4.5 5.6 ·6.7

::·UI

(n+3)(n+4) =(±-±)+(±-i)+(i-~}"""+(n~3 . So . lim -~

S;

.I

- n~4) =±- n~4

.. I Hence the sum~f series is _.

=-

··4

4

Example 22. Test for convergence of the series: . I· I. ~ 1 (i) 1+ - +-+... (ii) L..J J, . 2! . l! 11=1 -Jn). Solution:

(i) Ifwe denote the given series by

I

UII

,

.I .Un=·~,

.

n. we notice ·that Ii! ~ 2n-1 for all n ~ 2 . 1 I·· for all n~2 n! .2n.

:.-:5--1

~ Now, L..J

1

2/1-1 =1

I

1

1

+2" +22 +23. +

being an infinite geometric se-

11=1

ries with common ratio .

.2·

J.. < I,

Hence the given series:

is convergent.

I ~! is also convergent,

(ii) we know, .n! ~ 2n-1 for all n ~ 2 .

or,

II-I

../n! ~ 2 "2 , for all

< _1_. .i.e, ._:_1_ r., -.. vn! -

II-I

.

2

I·./11_

2

n~2

,for all n ~ 2 .

~ < 1 is con1 bei!}g a geometric series with common ratio . v2 2 2. . I . vergent, Hence, J, is also convergent. . vn!

I

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,-

SEQUENCE & SERIES

1-37

Example 23. Test each of the following series for convergence: 1 J 5 (i) 123 + 2.3.4 + 3.4.5 +..... . Solution:

1

1

.Jli + .J2j + .Ji4 +....

(i) Denoting the given series by

L u; ,

2ri -1 =------,-

U

n(n + 1)(n + 2)

n

Let us compare

LU

with

n

2

.

un.

Now,

;,;-=

So, Jim u" .vn·

n-+oo

L

The~ But

1

.(ii)

U".

and

L vn = ~,n

Hence

L

Vn,

where

VII

=~

n

(11 )(211-1) = n.+l ~

(2n-l)n n(n+l)(n+2)

= 2 =t 0,

L

=

(

1 J(2-lJ

1+~

1+~.

and finite

LV

.

n

converge or diverge together.

being a p-seriers with p

=2>1

is convergnet.

i~ convergent.

UII

(ii) If the given series is represented by

L

Un ,

1

'.In=~n(n+l) . Let, us introduce a comparison series ~ . ~ 'I'UD-= u; I'im n-."9

Vn " u-...,

.

L

Vn

R

where vll .

=~

n

n liun 1 == 1 =t 0 an d fimi't e. ~n(n + 1) IHOO. • 1 . L--

n

So, either both But .

Vn ,

L u, and LV

n

converge or both diverge.

= L~ ,being a p-series n

with p

= 1, diverges.

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1-38

ENGINEERING MATHEMATICS - IIIA

Hence' ".

L u; = L

1 is divergent. ~n(n +I) .

. . . 1 1 1 1 Example 24; Testfor convergence of the series.-+-+-+-+ '. . '. . 3.7 4.9. 5.4 . 6.13 Solution : If the given se~ies is represented by . I = ...,-----:-:,-----,-

U

n .

. Let, v n Now

1· = .

(n+2)

.

1')2

,

u

'. IHOOVil

(2n+5) .

LVII is convergent

thus. ".

lirn -II

. n2 .

= 11m

= 2" "* 0

L u;

(p = 2) I'

.

= lim -,----,--,.--.,-

1H0)(1+~)(2+~)

n~0)(n+2)(2n+5) I.

Hence

LUll ,

.

.

and finite.

is .convergent.

.

.

Example 25. Testfor convergece of theseries:

f(Vn4 + 1 ~FnCl) fI=l

Lu; , we have ~FnCl)( V+i +FnCl)

Solution : Denoting the given series by

,( ~ u -

'V+i+FnCl

n -

. .' '" we shall compare ~ Since

.'

ullTI-= Un

fI~OOV • 1/

'" u; with ~

. ,.

rrn

II~'"

"

. and finite and .

- V+i +FnCl V" ,

where

VII

-;:r

== 1

2

1

R="*

n4

n4

2n

I' Im Vrr=: n' + 1 + Vrr=:> n' - 1 tHOO

LV"

1 ="""T

n

2

-

2

R

1+-+

I

I 1--

is convergent, '" ~ u" is convergent.

0

.

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'SEQUENCE & SERIES

1-39

1·2 3·4 5·6 Example. 26. Show that the series 32.42 + 52.62 + 72 .82 + ..... is convergent. Solution, : If we denote the given series by

I u; ,

(2n-I)2n, Un

=

2

, I Let vn = 2' then n ull '2n

I v, 3()2n -

is convergent. 2 4 ' -- n

1

Now, v" = (2n+I)2(2n+2)2 so that Jim~=

, But

I

(2+~)'(2+;)"

=

J

Vn

I1-+CO

L u;

Therefore,

2

I) (2n+2)

' (2n+

I v"

and

v,,' 'is convergent,

either both converge or both diverge.

Lun

Hence,

is convergent

Example. 27. Test for convergence of the series: (1) Isin~;

··) ~ .1 C II ~S102

, . n

, Solution:

,

(i) Let

lI,i

. I = SIn- and n

, Now, Jim 11--+00

I But I

,

till

VII

, ."

VII

SIO-

Un

= lim -'

VfI

11--+00

I

_11

n So,

111 ~

.

= lim ~ 9--+0

-y11

1 =n

• J

,

I 1 r tan-

'C···)'~

n

8

e

, where

=1;tO

I both converge or diverge together. =I~ is divergent. and

VfI

11

,

'

I'

'

Hence "'-' u" =,~~ sin-n is divergent. ' ~,

17

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ENGINEERING MATHEMATICS -iliA

1-40

") H ,,1 C 11 ere, un = SIn2, ,

Iet vn I)

. thus lim·""-= lim . ( ") 1 VII

I1~OO

, ,'2

So,

I

1

n'

, (Sin n-4C1'J

=2

n

=.j" 0, and finite as before . .

n

'1 Here, I ,

I

and

UI1

Vn

both converge or both diverge,

Vn

'

=I

2, being a p-series with p = 2, is convergent.

'n

'

I>n =Isin

, Hence

"') , C 111 Let, ,

un

1 = r '",n

( 1 ) ,n2

, is convergent.'

tan- 1 n

We consider 'a comparison series

,

,'

r:

liri1:un = lim' n~

'Now,

11--+'"

v II

11--+00

'"

n

L

'(

VI1 ,

1)

tan-' tan(1.) = lim ( )

n

1.

11--+""

n So,

L un

But Iv n . Hence 1.1.20.

a~d

L vn

LV

fiX ' being a p-series "I

n

I

n n

= 1~ 0,

'

11

2

and finite

'

are both convergent or both divergent.

1

I

=

I

..In = X

where vn =

with

p

> I , is convergent.

..II) ,

= L.. ..In ta,\

-;; is convergent.

D' Alembert's Ratio test. 00

Let

L

u;

be a series of positive numbers such that lim

n=1

11-t 00

1In+1

exist

ul1

aJ

finitely and let Urn n-too

un+l Un

= I, The series L

divergent if l» 1. When 1 = 1, the test falls, Proof

Beyond scope,

Un

is convergent if I < 1 and

n=l

[WB. lJ.T,2005,2016]

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& SERIES

SEQUENCE

1.1.21. Illustrative

1-41

Examples.

Example. 1. Examine the series for convergence .

.

So

un+1 ,

(n+l)2

_n2

Solution : Here,

1In+1 =

un - -n'

2

. 217

+1.

= (n+ \)2 .~=.!.(1+.!.)2 2'!+1

Un··

n2.

2·

n

Therefore, .lim U,,+l =.!. < 1 u; 2 ,,~!X)

2

IU = I ~ is convergent. 2

Hence the given series

n

. 1 2 3 n Example. 2. Show that the series-+-2 +-3 +...+-n is convergent. . 2 2 2 2 Solution : Denoting the given series ,

. n + 1 2" n+1 lun--).= lim--= n ,,~OO 2 + n ,,-+"" 2n

U,,+)

:. hm--=

,,~oo

by I·u"

Un

.}lence the series

I u;

,

( 1 1) 1 lim -+_. =- I).

LU"

Hence by comparison test the series

is convergent when x

= I.

Thus the given series is convergent when x :::;1. Example. 8. Discuss the convergence of the series 2P 3P 4P 1+. -. + - + - + ... (p > 0). 2! 3! 4!. . Solution, Let

lI."

[W.B. U. T. 2008,2014]

(n = 1,2, ... ) denote the n-th term of the given series.

Then for n = 1,2, ... nP it =-

n!

n

U

-

. ,,+1 -

(n + l)P (n + 1)!

,.

· --= 1111+1 I1m n~a:>

lIn

I'1m n~ 1.

when

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1-46

ENGlNEERING MATHEMATICS -111A

Example. 11. Discuss the convergency of the power series 1

x2

X4

~6

--+--+--+--+

. 2../1 312 4./3 Solution. Let

UI1

...

5../4

.

(n = 1, 2, ... ) denote the n-th term of the given series.

Then for n = 1, 2, ... X2n-2

un

=

en + 1).Jn· X211

U

11+1

=---r===

(n

+ 2)..Jn+f·

.un+I·=·n+l 'n+2

Un

li

~_n_' n+l

=_1_+_;;_1 ~_l_

x2

.

1+ 0

u n+1

n~~

.

2

1+n ~

1+n

2

--;;:= 1+2.0VT+O·,X

x2

1.

2

=x

.

Hence by D' Alembert's ratio test, the given series is convergent when x2

1, when

1

= ----=

(n+l).Jn

n

x2

n

=

1 the test fails.

~2( 1)' _

1+-

/1

LV

Consider a series

. n

where

1

VII= -3

.

/12

I·1m -Un = I'1m -- 1 n~oovn'

= -- 1 = 1 a filn1it e num b er. 1+0 '

n~ool+~

n Now

LVn

i.e.,

L

1 -3

is a convergent series 3

n2

(being a p-series with p So by comparison test 2 X

=.1.

.

L un

=-

2

> 1).

i.e., the given series is convergent when

Hence the given series is convergent if

X2

s 1 and divergent if

X2

> l.

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1-47

CE & SERIES

SEQUE

Example. 12. Test the convergence of the positive term series u+1 1+--+ ~+ 1

(u+1)(2u+1)· .

+

(u+1)(2u+1)(3u+1)

(~+1)(2~+ 1) .

+ ...

(~+ 1)(2~+ 1) (3~+ 1)

Solution. Omitting the first term, let

Un

(n = 1,2, ... ) denote the n-th term

of the given series. Then for n = 1, 2, ... (u+1)(2u+1) . Un =

... (nu·+1)

(nu + 1) ((n + 1)u + 1) (n~+1)((n+1)~+1)·

(u + 1) (2u + 1)

un+1

.

(J3 + 1)(2~+ 1) ... (n~+ 1)

(~+1)(2~+1)

=

lim ulI+1 = (1+0)u+O = U. UII (1+O)~+0 ~

IHex>

Hence by D' Alembert's ratio test, the given series is convergent when u.< 1 i.e., when a < 13 and divergent when, u > 1

p

.

When u

=

13

i.e., when a>

13.

1 i.e., u =~, the test fails.

13 When u

= 13,

un = 1.

Then lim u; = 1 :t; O. n

--+ 00

So, by Note (2) of a previous Theorem, LUll does not converge. Therefore being a series of positive terms, it must diverge. Hence the given series is convergent if f3 > a> 0 and divergent if

a ~f3 > O. Example. 13. Test the convergence of the series

(~)2 + (.!2)2 + (~)2 3

3.5

+ ...

[WB. U. T.2002,2007,201

0,2012]

3.~.7

Solution. Let un(n = 1, 2, ... ) denote the n-th term of the given series. Then for

11 =

1, 2, ... .(

Mil

=

1.2.3...

3.5.7 ...

(211

11

+ 1)

l.2.3 ... n (n lIn+1

=(

)2 .

+ 1)

,2

. 3.5.7... (211 + 1)(2n + 3»)

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1-48

ENGI EERING MATHEMATICS-iliA 2 1 +-1 + 1 ), = 2n + 3 = ( 2 + ~ .

Un+1

-;:

n J2

(n

(1+0)2 --

U';+I I'1m --= ,,--> U"

2+

00

1 =-nwhere. v" .= .!.. n

= lim

I1~OO

< 1 and

l)n = (1 + ~)n ..~

+

= (n

r =

r

L.!.

+.!)n.= e, a fmite number. n

is a divergent series.

n

. So by comparison test the given series i.e., Lun·is divergent when r 1.1.24 .: Raabe's

= l.

test.

co

Let

LU

be a series of positive numbers and suppose that

n

11=1

.

lim n(~ - 1) = I .

n~oo

un+1

.

co

Then the series .

L »:

·n=1

is convergent if I> 1 and divergent if I < 1.:

when 1:= 1, the test fails

Proof Beyond scope. Note. Raabe's test is used when D , Alembert's ratio test fails.

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1-52

ENGINEERING MATHEMATICS -IlIA

. 1.1.25. Illustrative

Examples.

Example. 1. Discuss the convergent of the series Solution. Let for n'= 1,2...

UIl(~ =

I 1.3 -+-+--+

2

1.35

2.4 2.4.6

....

1,2, ...) denote the.';-th term of the given series. Then

.

- ·1.35 ... (2n - I) =

U n

:. UII~I

=

2.4.6 ... 2n

1.35 :...(2n - 1)(2n + I) 2.4.6 ... 2n(2n + 2)

1 . I'Im~= un+1 I· .2n+ 1 I'Im-- n Iffi--= =2+0=1 n-to> U n-too 2n + 2 tI-too 2 . n +_2 2+0 n Thus D' Alembert's ratio test fails 2+-

0) . '. . 1.2 . 1.2.3 Solution. Omitting the first term, let the given series ..

u,;(n =.1, 2, ...) be the n-th term of

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SEQUENCE & SERIES

1-53

Then for n = 1,2, ... u" .

.

:.

Un+1

=

=

•. U,,+I

. I' .. 1m ,,--+00

1.2.3

+ n) .

a+n n+ 1

I' nn

U,,+I = U

n

a( a + 1)(a + 2) (a . 1.2.3 ... n()n + 1

--=--

u;

a(a+l)(a+2) ...(a+n-l)

,,--+00

"

a -+1

a + n __. lirn --n n+1

n--+oo

1

1+n

0+ 1_ -_ -- 1 1+ 0

So D' Alembert's ratio test fails. Now n(~-I) ,.

I·

=n(~-I)= a +n

u,,+1

(Un

.", 1m n -n--+OO

-

1) = lirn ---'----'n(l-a)

U,,+I·

n--+oo

a +n

n--+oo

. =(l-a)lim

n(l-a) a +n

I

1 .. =(l-a).-=l-aO,

1+ 0

.

given]

00

:. By Raabe's test,

LU

n

is divergent.

n=1

Hence the given series is divergent. Ex~m·ple. 3. Test for convergence the power series .J 3.6 1 +-:-x+--x 7 7.10

(

0)

Solution. Omitting the first term, let u,,(n

= 1,2,

given series. Then for 3.6.9

.un =

2

11

3.6.9 7.10.13

+--x

= 1,2, ...

(3~1)

7.l0.l3 (3n+4)x

".

3

+ ... x>

...) be n-th term of the

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ENGINEERING MATHEMATICS - IlIA

I-54 .

. :. un+l

3.6.9 (3n)(3n+3) n+l 7.l0.l3 (3n + 4)(3n + 7) x

=

UII+l _ 3n+ 3 -----x 3n+7 II

.•

•• ·U

=x

. lilIn--un+l· -1'· 3n+3· Im.--x 11-+.00 un . n-+oo 3n + 7

..

3 3+lim--n- 7 =x--=x 3+0 3+ .3 + 0

n-+ 1 . When x = 1,the ratio test fails For x=1, . 3n+.7 ~.-=--

. un

3n+ 3·

,un+l

.. n(~-I) .•

=n(3n+7 _1)=~=_4 3n+3 3n+3

UII+1

. u

.,. linn(Un --11-+00 U 1 . . . n+. .

3+l n

1) =hm--=--=-> . 4 3 3 +4 0 43 11-+00

.

:. By Raabe's test

n

1

3 +-: n

is convergent.

Hence..the given serie~ is convergent if x ~ 1 and divergent if x> 1. Example: 4. Test the convergencet of the power series x 1 x3 1.3 x5 -+-.-+-.-+--.-+ ·1 2 3 2.4 5

1.35

7

x

2.4.6 7

... (x>O)

Solution. Omittirig the first term, let ulI(n = 1,2, ...) denote the n-th term of the given series. Then for n = 1,2, ... UII

=

1.35... (2n

x2n+l

2.4.6... 2n

2n + 1

-1)

(211 + 1)

1.35 ... (2n .~.ulI+l =

-1)

2.4.6... 2n

()

211

+2

211

+3

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& SERIES

SEQUENCE

. U':+I

. (2n

=

.. un

]-55

+ 1)2

(2+~J

x2

(2n+2)(2n+3)'

.

= (2+~)(2+~)'X

. (2+!)2

.. --!!±!. u. = lim :. lim IHoo ":

2

n

2

.x2 =

(2 + 0)

.

(2+0)(2+0).

Hoo(2+~)(2+~)

.x2 = X2

00

if 1

L

:. By Ratio test, the series X 2 > 1. .But test fails when For X2 = 1, UM1

«;

:. Jim n~oo

=

_ 4n2+4n+l

(2n+2)(2n+3)""

Un+1

.•

is convergent if X2 < 1 and divergent

1.

(2n+l/

_

-1) =

. n(~

X2 =

U"

,,=1

4n2 +10n+6

6_1]-

6n2 + 5n - 4n2 + 4n + 1

n(4n2 + IOn + 4n2 + 4n + 1·

n(-.5L - 1)

\.UTI n~CXl

ulI+1

5 6'+2 6n +5n \' i1. 6+0 3 = UTI =. =_ > 1 4n2 + 4n + 1 I/~CXl 4 + + _1_ 4 + 0 + 0 2 n n2

.± 00

:. By Raabe's test, the series

L

U"

is convergent.

11=1

. Hence the given series is convergent when X2 > 1. .

X2 ~

1 and divergent when

Example, 5~Apply Raabe's test to examine the convergence of : 1+ !.!+ Q .!+ 1.3.5 ..!_ +.... 2 3 2.4 5 2.4.6 7 Solution: Denoting the given series, by UII

=

l.3.5

(2n

-1)

2.4.6 .... 211

2n .,·1

L u; ,

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E

1-56

(2n + 1)

l.35 and

un+1

so,

Un+1 =

=.

Therefore

1

).-_.

2.4.6..2n 2n + 2 2n + 3

(2n + 2)(2n + 3)

un

.

(

(2n + 1)2 .

{Un

}

lirn n --

- 1

Un+1

'/1~ao

Hence the series

=

. n(6n + 5) lim ( ) 2 2n + 1

3 >I 2

=-

/1~ao

L u; .is convergent. L

Example.6. 'Examine the convergence of the' series h were

.

Un =

11/1 ,

. 3.6.9.... .3n

7.1 0.13. .. (3n + 4)

· H 3.6.9.... .3n S0I ution : ere, un = () . 7.10.13 ... 3n+4 and

and

3.6.9

un+1

U

so' that

_n_

7.1 0.13 ... (3n + 4 )(3n + 7)

Iim n{~-l}= n~ao

3n(3n + 3)

= ----'-----,-----'-:-.,.-~-

un+1

Iim n{3n+7 -I} 3n + 3

Un+1

=

n~ao

.Hence by Raabe's test, the series

L u"

3n+7

= __

3n+ 3

Iim ~=i>l 3n + 3

3

n~oo

is convergent.

Example. 7. Examine the convergence of the series:

±+(fJ +(%r ~(i)\······ SOlutiO~ : Denoting the given seriesby

So .

'

· ( Un ).!.n = I"lm--= n 1im /1~ao 2n + 1

,,~oo

.

LU" , here u;

I'un--=-< I 2 1 +n

,,~ao

1 2

=

(2n:

I

Hence, by Cauchy's Root test, the given series is convergent. Example. 8. Examine the convergence of the power series: . 3x2 2x+-+-+

8

4x3

27

(x >'0)

1)"

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SEQUENCE.&

SERIES

I-57

L un is positive

Solution: Since x> 0, each term of the given series

. Here,

(n + 1)xn ==

Un

~

n

3

n+2

==

SO'U +1 n

n3

n+1

(n+l)3x

3

1

n (n+2)

·(n+l)'-;n-;::: (n+lt

.x

2

1+n

- (1+~J.x-->x,as _

~Y D'Alembert's

n-->OO

L

ratio test,

is convergent,

Un

x < 1, and'

if

divergent when x > 1 For,. x == i,the test fails. When

x

n+ 1 n3

.

= 1,

U

n

==--

LV

Let us consider another series

L v" Now,.

1·1m -.un'

Since

L

VII

+I ,,~oo n

n == 1· 1m --

Vn

is convergent,

== 1· lffi n-+oo

(1 + -1) n

LUll

Thus, the given series is convergent Example.

9. Examine 2

2.. 2 x +-.x 3.4

4

2

2

if

+

fini irute number.

x ~ 1 , and divergent if x > 1 . of

2

6

== . I , a

is also convergent.

the convergence

2.4 +--.x. 3.4.5.6

n

> 1, is convergent.

being a p-series with p== 2

n-t'OO

1 v n == 2

where

n ,

262

2.4. 8 x + 3.45.6.7.8

Solution: Denoting the first term by uo' 2 2 .42 .62 ..... ( 2n )2 Un

. and Un+l ==

==

(

) (

3.4.5... 2n + 1 2n + 2

2 2 .42 .62 .... ( 2n )2 ( 2n + 2 )2 ()

3.4.5..... 2n+3

(

2n+4 ).x

211+2 ) .x 2n+4

(

)

x>O

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1-58

ENGINEERI1'lG

SO,

Un+l _ . - ( Un

:. lim UII+l 11-+00U II ,

(2n + 2)2 )( 2n+32n+4

= 11-+00 lim (

-iliA

2

).x

( 1+-1)2

3 1+2n

MATHEMATICS

)n(. 2 )

2

= x2

.X

1+n

Therefore, the series is convergent, if x2 < 1, i.e. if -I < x < 1 The ratio test fails when x2 test. When x =.± 1 we have

= 1, i.e. when

x

= ±l,

and we apply Raabe's

!~Hu::,-ll!~H(2n(:~~~;2+4) -I}] 8

6+-

2

= lim

6n + 8n = lim n-+00(2n+2)2 1H00(2+~r

n=

2> 1 2

,Hence the given series is convergent, when x = ± 1 Thus the given series is convergent, if -1 ~ x ~ 1, and divergent, if x> 1 or x < -1· Example. 10. Examine the convergence of the following series: x2 x3 x4 x+,-+~+-+ ... , 2 3 4 Solution: De~oting the series by Un+l h ere, I'un -n-+oo UII

=

1·

n.x

lffi --

11-+00n + 1

=

LUll ,

1·

1

im --I

n-+oo 1

'

.x

=X

+-

n So the series is convergent if x < i and divergent, if x > 1 . The ratio test fails, if, x When x

= 1, the

=

1 1

1

1

series becomes 1+ - + - + -+ ..... 234 .which is divergent by p-test (here p = 1) Hence the series is convergent for 0 < x < 1, and divergent for x ~ 1

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1-59

SEQUENCE & SERlES

Example. 11. Test for convergence of the series: . 2 6 2 14 3 30 4 1+-x+-x +-x +-x +... 5 9 17 .. 33 Solution : Ignoring the first term, U

n

=

2n+1 211 -t-

_

1

2

+ l'

n

.x

Un+1

2n+2

=

2

-

2 2n+

n+1

+1

.x

So, ,

By Ratio test the series converges, if x < 1, and diverges, if x> 1. Ratio test fails, when x

For x

=

. 1, lim u"

= 1.

..

= lim

n-+oo

2n+1

n-+oo

n 1

2

2

_

+

+

So, the series diverges when x

1

=

. lim 11-+00

(l)n

1-"2

(1 )"+1

. = 1-:t; 0

1+ 2

= 1.

Hence, the series converges if x < 1 and diverges if x ~ 1 . Example 12. Examine the convergence of the series: a

a(a+l)

. b + b( b + 1) +

a(a+l)(a+2) b( b + 1)(b + 2) +

where a and b are non-negative integers and b -:t; 0 . Solution : If we denote the given series by

(b) n 1,. ..as n a n (1 +-

L u; ,

n 1+-

. here,

un

b+n

un+1

a + 11

--.= --

=

) --t

.

n

--t

00

, [C.P. 1997]

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ENGINEERING MATHEMATICS - IJIA

)-60

So, D'Alernbert's Ratio test fails, and we proceed to Raabe's test. . I·Imn{Un ---1 11-+00

}

UII+I

=

I} = limn(b-a) a +n

I·1mn {b+n --a +n

=b-a

n-+oo

n-+oo

Hence, the series is convergent, if b - a > 1 , and divergent, if b - a .< 1 , when b == a, the series becomes 1+ 1+ 1+.... which is obviously divergent. Example 13. Examine the convergence of the series: I+a (l+u)(2+u) 1+ 1+~ + (1+~)(2+~)

(l+u)(2+u)(3+u) + (1+~)(2+~)(3+u)

[C.P. 1965]

+.....

where u· i= ~ Solution : Ignoring the first term, here (l+u) Un =

(2+u)

(n+u)

(n +~)

(1 + ~)(2 + ~)

1

-+1+-

u

I·Hfl 1 + n + u = I·Hfl n n =1 1 + n + ~. n-+oo lIP . -+ +n n So Ratiotest fails, and we proceed to Raabe's test.

Thus

I·1m

11->00

Un+1 --

u;

=

n-+OO

n{~ -

. limo n-+OO

I}

=

.UII+I

lim n-+oo

n{ 1+~ u- u+ n } = lim 1+P -1+u u

=P-

U

n->oo

,

.

n

Hence the given series is convergent if

P-

u > 1 and divergent.

When ~ - u < 1 and u > 0, ~ > 0 Raabe's test is inconclusive, when

u

1

P-

=

when ~ - u = 1 ,i.e. ~ = n + 1 , we have.

=

u n

u(u + l)(u+ (00 VII

I·im . 11->00 U

n

+ 1/ + l

·n= I·un n->OO U

u +1

--+ II

n

wh'ich is divergent.

= 1] . 1

=u> 0

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1-61

SEQUENCE & SERIES

se.

L

UtI

is' divergent.

p-

Hence the given series is convergent for

a> 1, and divergent for

p-a$1. Example. 14. Determine the region of convergence ofthe following series: 2

3

2!

3!

..) 1 x x X ( II +-+-+-+

1!

"

2

.

3

X' x . x ... ~ ... (1.1.1.)'1 +-+-+-+ 2! 3! 4!

. x-3 -1 (x-3f (v) --+2 .3 2.3 Solution:

1

(x-3)3

3

3

+-.

+

3

.

(i) The series: 1 +x +x2 +x3 +.... , being a geometric series, is

Ix! < 1, and divergent if Ix! > 1 .

absolutely convergent for When x

=

Again, if ;

1 , the series is obviously divergent.

= -1 , the

series becomes 1- 1+ 1-1 +.....and it oscillates

finitely between 0 and 1, and the series is not convergent. From above discussions it is clear that the; interval of convergence is

!x!< 1, i.e. -1 < x < + 1. x

x3

x2

(ii) The series is: 1 +-+-+-+ ..... 1! 2! 3! .

Ignoring the first term, .'

n

Un

n+l

= ~,

Un+1

n!

= _x_

n+ 1

By D'Alembert's ratio test,

. I~n+ll= .

lim :"-+00

lun (

UtI

n-+oo

x,,+1

I I -70 0 for all n.

un > un+1 for all n.

;.

So the sequence {un} is a monotonic decreasing. · . 1un un -

Now

1· 1 n 1 nn -. 2· = _ lim n~oo 2 (n+ 1) 2 n~oo

1

n

= 0\ 1)2 1+n Hence by Leibnitz's theorem, the given series is convergent. n~oo

.

(

.

L: CDS2 n7r • 00

Example. 4. Discuss the convergence of the series .

11=1

n +1

. . can bee wri (-1)11 [ .; CDS ntt S o I·ution, Th·e given series written as ~ L. -2 11=1 n + 1 . ~ nIl .i.e., L... (-1) ull WlereU =-2-. I1 n=1 n .+ 1 So the given series is an alternating series. 1

Now,

U~·~ Un+l·

.

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

= -2 -1

=

Il~oo

(n + l)(n2 + 2n + 2)

{un}

So the sequence Also· lim

2n+ 1

.

un

2

= lim Il~oo

> 0 for all n.

is a monotonic decreasing. I

-2 -

n +I

.

I 2

= lim _n_ = O. n~oo

·1. 1

+-;:;-. n~

= ( - 1)11]

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1-65

CE & SERIES

SEQUE

Hence by Leibnitz's theorem the given series is convergent.

ttre convergence of the series

Example. 5. Examine 1

1

'1

1

----+-----+···(a>O, a a + b a + 2b a + 3b

b>O). '

Solution. The given series is an alternating series which can be written , ~ II_I 1 as L.. (-1) un where Un = ' . 11=1 a+(n-l)b Now,

=

Un - un+f

,

=

1

.

+ (n - l)b

a

----

1

a + nb

b

(a+nb){a+(n-l)b}

>

°

for all n.

Un > U,,+i for all n.

So the sequence .

Now lim 'n-too,

Un

{tI,,}

is a monotonic decreasing. 1

= lim

n =0. II-too!!..+b_!!. , n 11

1

= lim

Il-tooa+(n-l)b

Hence by Leibnitz's theorem, the given series is convergent. l.l.28.

Absolute and Conditional

Convergence.

00

L u,

An infinite series

of real numbers is said to be absolutely

n=1 00

L Iunl

convergent if the series

is convergent.

n=1 00

An infinite series

L

un of real numbers is said to be conditionally

11=1 00

convergent, if the series

L ull is convergent

but not absolutely convergent

n=1

[which is known as non-absolute convergence]. 00

Theorem.

If the series

L ull

of real numbers is absolutely convergent,

/1=1

then it is c.onvergent. Proof. Beyond the scope of the book. E.M.3A-5

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1-67

SEQUENCE· & SERIES '.

.

2

.

X

3

n

+-X - ...+ ()n+l -1

X

+... is 2 3 n absolutely convergent when [x] < 1 and conditionally convergent when [WB. U.Tech.2001,2007,2016] x=l.

Example. 3. Prove that the senes

X - -

.

-

00

Solution.

The

given

series

can

be written

as

n

Un =(~I)n+l .

..

L u"

where

,,~

~ . .

n

lu~1= Ixln . n

So,

Ixln+1 lu +11=--1 n+ n

lim

U,,+1 U"

. 11-+00

= .

.

lim ,,-+00

_n_lxl = n+l·

lim _1-1 Ixl = Ix], 1+n

n-+oo

00

By D' Alembert's ratio test the series

L Iunl is

convergent if [x] < 1

/1=1

and divergent if [x] > 1. But the test fails when [x]

=

1.

Hence the given series is absolutely convergent when Ixl 1

, L..

is convergent, by.Leibnitz's test.

[see Ex.24.]

!x!~ 1 and divergent

Hence the given series is convergent when

.

X2

Example. 5. Test the convergence of the senes x - -

12

The given series can be written as'

I

x3

+-

when

x4

--

J3 .,f4 + ...

(_1),,-1 x" Un

where

r:

1I" =

vl1

Ixln+1

:. !un+d= -cr=rt : n +1 lim ",••, ~ lim ~ n U

n~oo

n +1

II~OO II

Ixl ~

lim ~ II~OO

.

1

11

Ixl = Ix].

+n

2: Iulll

Therefore by D' Alembert's r~tio test, the series if

But

Ixl < 1i.e. if -I < x < 1 and are divergent Ixl > 1. i. e. if x > 1 or x < -l. the test fails if [x] = 1 i.e. when x = ±l.

When x

.'

= 1 the

111 series becomes 1- - + -

Evidently the sequence { );;} Also .},; ~ 0 as n ~

L

. (-1)"

00.

1

J;i

if

12J3.,f4 -

alternating series and can be written

is convergent.

as ;2: (~1)"

is convergent

-

+ ... which is an

~.

is a monotonic decreasing.

Hence by Leibnitz's

theorem

the series

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SEQUENCE

1-69

& SERIES

Again when x = -1, the series becomes 1·

.- ( 1 +

1

.fi + .J3 + ...

)

.

=-

1

L ..r;;

which is divergent being a p-series, with p = ~ < 1. Hence the given series is convergent for -1 < x ~ 1. EXERCISE [I]

SHORT ANSWER QUESTIONS

1. Show that the sequence {xn} defined by 1 I I xn =--+-- .. +n+ I n +2 2n is monotonic increasing and also show that it is bounded Is it convergent? 2. If

Xn

.

=

n+l ( ) 2n + I' n EN,

show that the sequence

{xn}

is

strictly

monotonic decreasing and hence prove that it is convergent. Find limx.,

.

.

.

3. Show that the sequence {xn} defined by xn

( =

)n+1 1+ ~

is monotonic

decreasing. 4. Show that the sequence

3/1

{x

l1}

defined by

XI1

=

+I

n +2

' is strictly

monotonic incerasing sequence. Is the sequence convergent? Justify your answer. Also find its limit. .. .. . 3n + 1 5. (i) Show that the sequence {xn}, where xn = ~ is bounded Evaluate limx; (ii) Show t~at the sequence

{:n:;}

is bounded and monotone

increasing. Is it convergent. Find the limit (iii) Show that the sequence {x n} ,where

1 x" = 5 + 6n is decreasing

and bounded. Is the sequence convergent? Find limx,.

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1-70

ENGINEERING MATHEMATICS- I1JA

3n° (iv) Ifun = --1' show that n+ above. Find its limit. °

o

{un}

is monotonic increasing and bounded

°

(v) Show that the sequence {~;:

~} is increasing. Find its limit

6.(i) Discuss the bOehaviourof the sequence number.

{xn} , where x is any real

°

(ii) Prove that the sequence °

{rn}

lim r" {I} (iii) Show that the sequence -;;

is convergent, if °

Irl < 1.

°

Test, the convergence ,

1

7. 1+ u

1

+;

4/3

0

of the following series :-

1

u + u + '"

9/3

16/3

1

°

3

5

8. -+--+--+ ... 1.2.3 2.3.4 3.4.5

00. °

f fn +,In+l' 1

9

converges to O.

2 3 4 5 10. 15+25+)5+7+'"

°

• n=1

00 .

1 1 I 11.1+-+-2 +-3 + ... 1.2 2.2 3.2 °

1 I ,-sin-.

°

13. ~

"n

14.

I

1

O). 14 24 3

2·4 2·4·6 {xv)-+--+ 3·5 3·5·7'

'

2·4·6·8 3·5·7·9

2J:s + 3A

+ ...

[WB.UT. 2002]

1 1·3 ... (xvi) 1+-+-+--+ 2 2·4

Ln

3

+ 4$

1·3·5

...

2·4·6

00

7. Discuss the convergencey

of

4e-1/2

1/=1

[WB.UT. 2005]

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SEQUENCE & SERIES

1-75

8. Examine the convergence of the following series :(ii) "

L...

(iv)

(l+nx)" n"

1 L ;"=-x n +1 2

(x> 0) .

n,

[WB. U. T 2006] 9. Examine the' convergency and conditional convergency of the following senes :1 1 1 1 -----+-----+ ... (ii) ~-~+~-~+ ../2+1 ../3+1 ./4+1./5+1 2 5 10

(i)

. 1

1

1

1

log3

log4

-log5

(iii) -----+-----+ log2

.....

•

00

(IV)~

(1~+?

(-1)")

10. For what values of x are the following series convergent .

Xl

XS

x7

3 5-7+

. (1) x-

+

••

X2

Xl

X4

11) x-2T+Y-42+

...

...

11. Prove that the following series converges absolutely

ANSWERS 5. (ii) 2.. 6. (i) Convergent. (:.) 11 C onvergent Iif

p - -1 > 1

(iii) Convergent if

p >~

.22

.2.

(iv) Convergent.

an d di vergent

and diver~ent if

rr

1 p - -1 ~ 1.

p

s ~. 2

(v) Divergent.

(vi) Convergent for a < 1,a > 1 but divergent for a = 1 .(vii) Divergent (viii)Convergent if x < 1 and divergent if (ix) Convergent jf x < 1 and divergent if (x)Convergent.if

x ~ I. X ~

... 17

I.

x s 1 and divergent if x> 1.

".

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1-76

ENGINEERING

(xi) Convergent if

s

x

and divergent if x > 1.

1

(xii) Convergent if x < 1 and divergent if (xiii) Convergent if

x

(xiv) Divergent

MATHEMATlCS-

x ~ I.

s 1 and divergent if x > 1. (xv) Divergent

(xvi) Divergent

7. Convergent. 8. (i) Convergent if x < 1 and divergent-if

x ~ I.

(ii) Convergent if x < 1 and divergent if

x ~ I.

(iii) Convergent if x < 1 and divergent if

x ~ I.

(iv) Conv for 0 < x < 1 and div for x> 1. 9. ·(i) Convergent.

(ii) Convergent.

(iii) Convergent; conditional convergent (iv) Convergent 10. (i)-l 0

16. Art infinite geometric (a)

-l 1

(b) p> 1

(c)

p-l.

4. Expand .log, x in power series in (x - I). 5. Expand cos x in power of x in infinite series. 6. Expand (1 + xf, x in infinite series.

n being a negative integer or a fraction, in power of

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EXPANSION OF FUNCTION

7. Assuming

the possibility

1-91

of expansion,

expand tanx, as far as the term

in x5.

8. Assuming the possibility of expansion, expand e' log(l + x) in ascending ofx

powers

9. Assuming

validity of expansion, ~in(:+8)=

10. Assuming

show that

~{1+8-~~-~:+

the validity of expansion,

....

smx

.. -} show that

(x-%r (x-~r

+~---'--

=j-

2!

4!

[Hint: sinx = sin(2: + x - 2:) = sin(a + h),a + 2:"h = x - ~ ]

.

2

2

2

2

11. Expand in power series in h:

(i)

ex+h,

(ii) eh sin(x + 11)

12. Show that 1ogx=(x-l)--(x-l) ·23 13. Expand

1

2

1 +-(x-l)

3

. -····IS true

for Ob if exists and suppose lim (y) = 1 y->b

=1

Thus lim lim {(X,y) .

y->bx->a.

This limit is called repeated limit of the function {(x,y)

as x ~ a and

then as y ~ b . Similarly the other repeated limit

lim lim {(x,y)

x-oa y-sb

can be constructed.

This is not ensured that the two repeated limits liin lim {(x,y) lim lim {(x,y).

x->ay->b

and

y->bx->a

will be always equal.

Moreover the simultaneous limit lim {(x,y) x->a y->b

may not be equal to the

repeated limits. Even it may happen that the two repeated limits exist and become equal but that equal value is not the limit lim {(x, y) . x->a y->b

However the following result is very interesting. Theorem.

For a function {(x,y)

.

.

if

lim

{(x,y)

(x,y)->(a,b)

andif both the two repeated limits Jim lim {(x,y) .

•

x->a y-l,b

exist then lim lim {(x,y) .

Proof.

x->a y->b

.

exists and equals I .

and lim lim {(x,y) y-sb x-w

= lim lim {(x,y) = 1. y-vb x->a

.

Kept beyond the scope of this book.

ote. From this theorem it is obvious that lim lim {(x,y);t:

x->a y->b

lim lim ((x,y)

y->b x->a

Illustration (1) Consider the function ':\

{(

2 3

x,y, =x y

+

logxsiny y

if

then (x.yl)i~(a.b){(x,y) does not exist.

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2-6

ENGINEERING

MATHE:VIATICS-IIIA

Let x ~ 2 and y is kept constant. Then ""

""

11ml(x,y)

?

= lun(x-y

x~2

3

+

logxsiny

= 22

=4y

i + "log2 sinyy

3

+log2--

)

Y

x~2

.: y is kept constant

siny y

"Next let y~O" Then

Jim lim f(x,y) Y--70

=

2

X~

Jim(4y3

+ log2 sin y)

Y

y~O

= 4x 0 + log 2 x 1= log 2 and

(2 3

"" f( x,y ) = I"un I"rrn x y +---==----'10gXSiny)" li m Iim x~2y~0

Y

"x~2y~0

2

=lim(x xO+logxxl)

.: xisconstant

x~2

= 0+ log 2 = 10g2

Here we see the two repeated (2) Consider

the function

"Here we see for y

limits are same"

f(x,y)=

*0 = 3y

Iimf(x,y)=lim2x+3Y x~o

Then

x~o

lim ~im f(x,y)

X"#

y~O

:. lim Urn f(x.,y) x:-+o y~O

4

-4y

=_~

4

=-~

4

0",

lim f(x,y) y~O

x -4y

= iim (~~)

y~O x-..o

Again) for

2x+3y x-4y

"

. 2x+3y 2x = lim =-=2 y~O x ~ 4 Y x

= Urn 2 = 2 x~o

Here we see the two repeated limits are not same.

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-7

Moreover we see if (x,y) ~ (0,0) along the line y =mx then lim

f(x,y)=

limf(x,mx)

x~o

(x,y)~(O,O)

= lim x~o

2x+3mx X - 4mx

2+3m

2+3m

= lim--=-x~o 1- 4m

1- 4m

which depends on m i.e.,

lim'

f(x, y) has different value as (x, y) ~ (0,0) along different

(x,y)~(O,O)

"

direction. So

'lim

f(x, y) does not exist

(x,y)~(O,O)

2.1.7. Partial Derivative. Let z = f(x, y) be a function. The derivative of f(x, y) with respect to x, keeping y constant is called the Partial Derivative of f(x, y) w.r.t x. It is denoted by

:

or

fx(x,y)

etc.

f(x + h,y) - f(x,y) . ..' . IS the partial derivative at (x, y). h , Similarly, the derivative of f(x,)I) with respect to y, keeping x constant, I'

8f So, -

ax = h~Oun

,

of

is c~lIed the Partial Derivative of f(x, y) w.r.t. y. It is denoted by' or ,t;,(x,y)." S0,

tly

,

aj 'I'

= im

f(x,y+k)-

f(x,y)

.

'" the partialderivative

at (x, y). 8y k~O k This definition of partial derivative can be extended for function of more variables. -' -

IS

= I(x}, x2, 'oz of Then --=--=/x e.g., let z

oX2 =

•••

iJX2

: f(x) I1m

2

'~2

xn) be a function of n number of variables. (x},x2"""Xn)

+ h, X3

.•. Xn) - f(X»

h~O

is the partial derivative of f(x}, x2,

X2, X3,

: •• Xn)

h •••

xn) w.r.t. x2 at (x,; x2,

..•

XII)'

Note. Unlike function of single variable existence of partial derivatives does not ensure continuity of a function f(x,y,···). See an Example in "Illustrative Examples".

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2-8

ENGINEERING

2.1.8. Illustrative

MATHEMATICS

-lIIA

Examples.

"

1"

Example. 1. Let f(x,y)=x+ysin-, "

X

x #0

=0 Prove that

lim

f(x,y)

= O.

(X,y)-+(O,O)

Then find, if possible,

the repeated" limits

lim lim f(x,y)

and

y-+Ox-+O

lim iim f(x,y) %-+Oy-+O

Solution. We see !t(x,y)-Ol =

Ix + YSin~-ol

=

~I" C

Ix + YSi~l~1

~ Ixl+ Iyljsin

s Ixj"+lyl

ISin~1~

1)

e'e " e E < 2 +2 whenever Ixl < 2 and Iyl < 2 .. If{x, y) -

Thus "

01 -:

e whenever Ix"-

01 < .:: and 2

Iy -

01 < .:: 2

lim" f(x,y) exists and equals 0". (%,y)-+(O,O)

"

Now keepingy constant lim f(x,y)= %-+0

lim(x+ ysin~)which %-+0

does not

X

exist as we know lim sin ~ does not ex..ist.So we conclude the repeated " %-+0 x limit lim Jim f(x,y) does not exist." y-+Ox-+O

Again fo~'costant x lim f(x,y) "

:. lim lim f(x,y) %~O y~O.

y-+O

= lim(x + ysin-l-) = x + 0 = x X

y-+O

= lim x = 0 %-.0

Example. 2. Find, from definition, the partial derivative of the function f(x,y)=

x12e>' w.r.t. x at the point (1,2)

" Solutlon, Here we have to find

"

Ix (1,2) = (If\ -)

""

ox

. (1,2)

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-9

According to definition fx(1,2)=(aj) ax

="Iim f(1+h,2)-f(1,2) h~O h

(1,2)

"3/

=

3/

. (1+h)12e2_e2 11m , u-:» h

"" 2.z2-12 1i

=e

IIm

"

1i

3

3

3 2

--I

= e2• -12

=-e

2

Example. 3. Find"

[Putting z

z-1

"z~l

81

=

(I+h)12_1 e lim ~---'-_ h~O h 2.

and

oy"

= 1+ h]

2

(81) oy

where f(x,y)

= X3y2 + exy2.

(1,.0)

According to definition of partial derivative we would keep x constant at th~ time of finding iJ f " dy Solution. So, -8f oy

=-

.

a { x 3y 2 + e~y2}

oy

= ; (x3y2)+ = x3

;

a "

(exi) ?

a

Oy (y2) + eXY- Oy (xy2)

= x3 .2y + e~2 .x~(y2) " oy

" (8f) "

oy

=2x

[.: x constant]

[.: x is constant]

13 x 0 + 2 xl;' 0 x e1x02

= 0 + 0 = o.

(1,0)

auoz and (au) Oz

p

x +y 2 +z 2 )--

2

Example. 4. Find

1

where u(x,y,z)=

2

(

and P=(1,O,-I).

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2-10

ENGINEERING .

- IlIA

I

0 (2 2 Solution. -. =- x + Y +z OZ OZ 1 ?? =--(x2+y-+z-) 2. .

MATHEMATICS

OU

1.)--

2

_3.. 0 ?? 2_._(x2+y+z_)

oz

[Here x and yare constant] 3

_~(x2

=

+ y2 + z2)""2 (0 + 0 + 2z) [.: x,y are constant] 3

.= -z (x2 +/

=(ou) . (ou) OZ OZ p

.

+z2) -"2. =1(12+02+12)-%=2-%.

(1,0,-\)

.

Example. 5. If

2

ft x, y) = x - xy , (x, y)

#-

(0, 0)

x+y

=0

= (0, 0)

, (x, y)

Find fx(O, 0) and 1;,(0,0). . . By de fiiruuon .. So Iution, .

f x (0, 0) = I'Im-~/.:.-(h_+_O~,O...:..)_~..::..../...:..(O~, 0...:..) h

h~O

=

lim I(h, h-+O

0) - 1(0,0) h

h2 -h x h+O = lim ._!.!..2.~ h-+O h h-O = lim --=1. h-+O h Similarly !y(0, 0) can be evaluated. Example. 6. Given {(x,y)

.

0

__

(2,3)

= 2xy,

(x,y)

:;5,

(x,y) = (2,3)

#-

°_

Prove that this function is not continuous at (2,3) and

exist. Solution. :. f(x~y)

.

1im

(x,y)->(2,3)

f(x,y)=

lim

(x,y)->(2,3)

2xy=2x2x3=12~f(2,3)

is not continuous at (2, 3).

of, of do not

ax ay

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DIFFERENTIATION

(

of) ax

OF FUNCTION

OF SEVERAL

t(2 + h, 3) - f(2, 3) h--+O h

(2,3)

2·(2+h)·3-5 h

h--+o

.

6h+7

= lim h ...• O

does not exist finitely,

h--+O

Of) (

Similarly

.

av

,

does not exist also.

(2.3)

J

.... Example.

6(h+2)-5 h ..

= lim -h-

x3+/

.' f(x,y)

7. For the function

=

Prove that f(x,y)

x

·' Iurn

f(

(X,y)--+(o.o)

)

3

I'

= im

x--+o

3 3

2

lirn

f(x,y)

3

{I +(l-mx

x--+o

=~

m

ax

at (0,0)

= limo f(O+h,O)-f(O,O) 00

,

h

h--+u

h3 +03 . . ---0 = lim f(h,O)-f(O,O) = lim h-O h--+o h . . h--+O h 2

= lim h h--+O

of) 8y

=

(

(0,0)'

u

lm

k--+O

h

= lim h = h--+O

°

.f_(.:..-O-=-,O_+_k~) --.!f...:.,(O--:.., O~) k

r:

then 2)3}

mx3

which depends on m.

does not exist

is.not continuous

Now, (Of)

.

x

(for different value of m)

(X,y)--+(O,O)

:. f(x,y)

an d

.

exist at (0, 0).

=lim-~-.....,--~

tends to different value as (x,y) ~.(O,o)

- mx3

.

)3

m

x--+o

..

ax oy

x +(x-mx ) 3 "X - (x - mx)

= lim1+(1-mx So f(x,y)

y

(0,0) along the curve y = x - mx3

4

.

X,y

=

x", Y

at (0,0) but of, of

'is not continuous

Solution. If (x, y)

,

x-y

=0,

=' x

2-11

= lim

= lim

y

VARIABLES

.

along different

curve

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2-12

u =

IIn

k-)O

of

f(O,k)-f(O,O) k

.af

03+e ---0

,.

---,O,,---.:..k,,--'_ = 1111= filii k-)O k k-)O

k )=0 .

( -

.

:. -

ax and -.oy

exist at (0 0)

,

,.2

Example.

8. If

e = 'I

n

e

find what value of n will make

41,

(TV.B.U. T 2012] r2

Solution. Here 8 = t 11 e

(I)

.41

or 08) or or

~(r2

.{ 1.

2

3

3r 8 + r

=

2t .

.

. = -21.

8)

3 r

= ~(_

(8)} -21 r

+ r3

-~(3r20 2t

=-

08) or

(

1 2 r8 4 ) 2t 3r 8 - 2:t

.

r12:r(r2~~)=-~(%-~:J .0s . Again -

= ntn-1e

--r2

+tne

41

--r r2 ( 41

2)

S

=.-

4t2

ot 1 a ( 2 as) as ? Or r ar =a;

-'

gives

( n +_r2 )

t

4t

- ~(~-C) = ~ (n C) t

2

41

+

t

4t

3

n=-"2'

i.-e. Example.9.Ifz(x+y)=x · . G iven z (x + y)

=

2

2

+y 2

2

x +Y .

(OZ

Oz)2

ox oy

,.showthat ---

(

OZ OZ)

ox

=4 1----.

0)'

( 1)

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-13

Differentiating partially w.r.t. x we get oz -(x+y)+z(l+O)=2x "ox " d z 2y-z Similarly we get - = --. " dy x+ y

_ 2y_z)i x+y

So, (OZ _ Oz)2 =(ix-z ox dy x+y "Also

4(1-

oz 2x-z or, -=--. ox x+y

=4(X_y)2 x+y

"oz _ oz),;,_" _4_{x+ y-2(x+ ox oy x+y

=_4_{x x+y

+ y-2(x

+ y)+2. X2 + x+y

l}

Hence proved.

"

Example. 10. If x

="rcose

ax a

,y

/F

ax

1

ar

8r

x

ax

Again ae ~ ae

(rcose)

From the given relation,

ax

!(X2

=rcosS=cosS r

-:1=-.

" ax" a"

1 .', ae

"I

=-

r

sinS .

"" = -rsine

.

tanS = E :. e = tan-IE x x

Hence

ax

1" ae

-:1=-.

ae

ax

1 :1= ae

ax

l

:.: =~(XZ + /f"i ~x2 + /

and as

ax

I

=~(xZ+/r2.2x= 2 " Hence

"

ax

1

prove -;::--:1=-a & r

.

I

"

= 4 (x _ y)2 x+y

ax

= rsine

Solution. Here -=-(rcose)=coss "" 8r 8r From the given ~elation we have rZ = XZ+ or, "r =(x2+

y)+2z}

+/)"

.

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E CI EERINC MATHE:vtATICS-llIA

2-14

2.1.9.

Ilighcr Order Partial Derivatives.

For simplicity we consider a function lex, y) of two variables. 0J"U\\ ix(x, y) and !y(x,y) are also function of x and y (Because the value of the partial derivatives vary from point to point). So they may also possess partial derivatives. The partial derivatives of and /.v(x, y) are called second order partial derivative of lex, y). Thus there are four kinds of second order partial derivatives of lex, y) which are:

f.(x, y)

The definition of the above second order partial derivatives would be r ( Jx x

a,

b)=l'

.

1;,),(a, b) = .

im

fx(a+h,b)-fx(a,b)

,,~O

h

. !yea, b + k) - fv(a, b) lim .

k

k~O

'. '. !y(a + h, b) - !yea, b) fx y (a, b) = lun --'------"--,,~O h r (

Jyx a,

.

'b)= I'

. 1m

k~O

fx(a,b+k)-

fx(a, b) k

.

ote. (1) If any of the above limits does not exist, we say the corresponding partial derivative does not exist. . (2) fxy(a •.b) may not be equal to t;'x(a,b) for every function f(x,y) .

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2.~.1 O. Illustrative

Examples.

,

;iu

if1u

??

Example, 1. If u = log (x" + Y-) find --, ,--, ,-d x: d y' d Solution. Here u is a function of x and y.

2 + y 2})

OU {} { =log (x

(}x

ox.

1 =?

x- + Y

-du = - 0 oy

·oy

=

2

(2x)

1 . (2y) x2 +.y2

02U =~(ou)=~( ,ox oy OX d y

=

=

1 2

x +y

2

0 (}x

-(x-

,

(iu. x

,

+ y-)

2x 2 x~ + Y

=,

{ 2 2} log (x + y

=

2

1

.

2

x +yoy

0

-(x-

,

+Y2)

2y .xl + y2

ox

2-15

2y )=2Y~(X2+y2)-1 x2 + y2 ox

·';"·2y(":"1)(x2+y2)-2.2x=

4xy (x2 + y2)2

(}Y

(}lU

and -,.,-,.,-. cya

x

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ENGINEERING MATHEMATICS -iliA

2-16

Example. 2. Given the function I(x,y)=

.

XY(X2 _ y2) ~ 2,(X,y):;t(0,0) X- + Y = 0, (x, y) = (0,

Ix y (0, 0) and /y x (0, 0).

find, from definition,

Ix/o,

0)

0) =/yx(O,

Verify whether

0).

[WB.

o.t:

2003,2015,2016]

Solution. Now; according to definition, . ly(O+h,O)-I),(O,O) 11m

Ixy(O,O)=

=

h

11--.0

. ly(h,O}-ly(O,O) 11m . h

h--.O

. ..

Now,

1 (h, 0) ~ lim I(h, 0+ k) - I(h, 0). y k--.O k hk (h

=

lim

2

k

-

2

lim

f ih, k)

k--.O.

°

(h

2 -

02

- 1(17, 0) k

)

h2 + k2 h2 + 02 --:-=--.:.....:..:..---""""'-'-------=---"----

k

k40

h(h2_k2)

.

=

h x

)

=

(1)

11m ---'--=-----=--'k--.O h2+k2

h xh2 =--=h h2

/y

(0,0)

= 0.

h-O So, from (1) f~yCO, 0) = lim -. . h--.O h Again Jy1r .

J.

v

.,

(0 0) _ 1·1m .;c..:clx,-CO_, _0+_k....;..)_:-..:...:lx:....;.(O_, -,-0) --k--.O k ,

-

=

lim

Ix (0, k) - Ix (0, 0) . k

k--.O

ow,

= 1.

Ix (O~Ii)

== lim

1 (0 + h, k) - 1 (0, k) h

h-+O

=

lim ICh,k)11--.0

ICO,k) .

h

(2)

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-17

hk(h2.-k2)

0.k(02_k2) 2 = lim _h:..:..2---'-+....:ck::.... __ ---'0::....2_+.:......:..::k_2_ h~O h

=.

linTI -~--:-'hk(h 2 - k2 ) h~O h (h2 + k2)

.

:.

I, (0,0)

2

=

Jim k(h -k h-.+O h2+k2

=

0. .

fyx (0, 0) =

So, from (2),

.

hm

2

) =-k. '. -k-(-O)

k~O

.Therefore,

1:0,(0,0)"#

= -1.

k

fyx(O, 0).

Note. From the above illustration

we see that f'xy

(x,y) may not be

t;x

equal to. (x,y) for every function f( x,y) . There are some theorem's which state the conditions under which the above equality hold. Among those theorems, Schwarz's theorem is well known. Example. 3. Show that z=log{(x-a)2 02 Z 02 Z --2 + --2 = except at (a, b). ox oy

+(y_b)2}

°

•.

Solution. z oz

~

ox

=

log (x2 + y2 - 2ax - 2by + a2 + b2). . 2x - 2a

= -".--;(2V oV _ 2X) + ot/>(2V OV _ oa . ox ob ox

be thr ee functi unctions ot .

side)

0) + ot/>(2V OV _ 0) = 0 OC ox

v°t/>.ov _x0t/> +v°t/>.ov +v°t/>.ov =0 oa ox oa ob ox OC ox ov(ot/> ot/> ot/»· ot/> v oxoa + ob + oC =X oa ~ ov (at/> + at/>+ at/» = o.if> x ox oa db OC d a at/> v dv da = -::-,----~--:-:X ox ot/>+ ot/>+ ot/>. d a db OC

Similarly

we can show

04>

&

vov and ~

oz

=

04>

04>

-+-+--

04>

oa ob oc

Adding these we get

~ OV +~ ov +~ OV =1 d x yay Z oz

.x or,

ov yoy

1 dv : 1

OV ZOZ 1

~-+--+--=-. x dx

1 v

of

2005.2014]

t/>(a, b, c) = o.

at/>= 0 (Differentiating w.r.t. x both ox .. ot/>.oa +ot/>.ob +ot/>.oc = 0 oa ox ob ox oc ox

2-27

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2-28

ENGINEERING

MATHEYlATICS

Example. 5. If x=u+v+w,y=vw+wu+uv,z=lIVW and function of x, y, z show that

of of of of of of u-+v-+w-=x-+2y-+3z-~-. ou OV ow ox oy az Solution. After transformation F becomes a function of . "oF oFox oFoy By chain rule -" -=--+--+-du ox du d y ou or,

of

-

ou or,

of"

of

= -.1 +-.

ox

of

(w

4-

v) + -(

oy

.".

of ov

of ox

if F is a

v and w also.

oFoz oz ou vw)

oz

of of" of" "-=-+(w+v)-+vwou ox dy

Similarly,

1I,

- IlIA

(

of oz

)oF oy

of oz

-"-=-+ w+u -+ wu-

of of of of -=-+(u+v)-+uvow ox oy oz of of of L.H.S. =u-+v-+w, du dv ow " of of of =u-+u(w+v)-+uvw-+v-+ " ox oy oz

and

of ox

of

v(w+u)cy

of of of of +vwu- + w- + w(u+ v)- + wuvoz ox .d y oz of =(u + v+w)-+2 "ox " of of =x-+2y-+3z-. ox " dy Example.

of of (uv+wu+vw)-+3uvw-" dy oz of OZ

6. Prove that; by the transformation

. I . partia. Id'ffi" 1 erentia equation

2

0 Z --2

or,

2

ff Z re duces to --0 = C2 --2

01 OZ 8z ou oz ov -=-.-+-.-=-c-+cot OU 01 ov 01 2

ct, v = x + ct

u =x -

ox oz "ou

Z

du dv

oz ov

z)

0 Z 0 (0 Z 0 Z ) { 0 (0 0 (0 Z )} 012= c 01 OV- 01.1= C 01 ov - 01 ou

= 0.

the

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. DIFFERENTIATION

OF FUNCTION

If

02

OU { tlUOV·-;3I+

=C

Z

VARIABLES

2-29

OV 02 Z OU C2 Z OV} .-;31- C1I2 ·Of - OUOV·Ot

Z

oi

. {iiz

OF SEVERAL

Ifz

ciz}

Ifz

--(-C)+-) (C)--, (-C)---C OU OV OVOUOU OV

=c

.

{

2

=C

,

-2~+~+ OU OV

2

0 S·lilli-Iar Iy .--,

2}

OV-

0 ~ . OU-

2

Z

ox-

_.

..,2

2

2

0 Z + 2 -_.0 Z- + --. 0 Z 2 2 Ou OU OV ,? 0- Z

== --

ov

0- Z

Putting these in the given relation __ , - = ("2 ;::)2

. -or

-

v z

.

.

--,

OX-

we get --=0.

OU OV

.

Example.

7.

If

Solution.

In fact

X)/2

z=e

,x=fcost~y=tsint,

dz . az dx

= eX/

:

dz dt at

1=2

7r

z becomes function of t only. az dy

:. By chain rule dt = & . dt + . or,

obtain

0' . dt

./(COsf-tsint)+exi n

rt

.x2y.(sinc+lcost)

7r

Now when t =-,X

=-

:.(\...dZ) . = /Hr dt I=l:. _

.(~)2(cos~_~sin~)+eo{~r

2

2

0 and y=-

cos-=

2

2

2

2

2

rt 2 .0.2.~

2

2

. 7r 7r or,

..,

(~~

}=l:. = (~

.

2

Example. 8. .)

Ox

222

=-

u(x,Y)=f(x2

r -zx -au + (?X- -

? (

If

J(-~)

rr; .

+2yz,y2 +2zx) prove that

yz ) -au + (2z - xy ) -all

0'

7r

(SIll-+-COS-

oz

=0

)

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ENGINEERING

2-30

Solution.

Let

v=x2+2yz

.

.

..

and

w=/+2zx

au auOv au ow -=-:-+-·-=2x-+2z'Ox Ov Ox Ow ox

by chain rule

.

MATHEYIATICS

:.u=f(v,w) au av

-iliA

. au Ow

au = au . ov + ~. Ow = 2z au + 2 y ~ oy .irv oy away ov Ow

= 2 x 0 x au + 2·x 0 x au = 0 .

ov

Ow

Example. 9. It u =

show that

.:

J( ax2 +2hxy+by2),V = ~(ax2+ 2hxy+ b/)

(u :)

=

!(u ~)

[C.H.1934]

:.~(uOV)=u02V +Ou.ov . Oy ax . OyOx ay ax and

~(uav)=u ax Oy.

02V + au. ov oxoy Ox Oy

(2)

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DIFFERENTIATlO

OF FUNCTIO

OF SEVERAL

Solution. . Let G = ax2 + 2hxy + by' and H

:.u=/(G)

VARIABLES

2-31

= ax2 + 2hxy +by2

and v=(H)

.:.Ov = d.'oH = '(H){2ax+2hy) = 2'(H)(ax+ hy)

ax dH

and

&

8v ~2'(H)(by+lq)

By.

au ~ 1'( G) 8G = (2ax + 2hy )I'( G) = 2(ax + hy )I'{ G) Ox &

au =1'(G)8G =2(by+hx)f'(G) By By au·Ov

:.By .ax = 2 (by +hx)J'(G)·2{ax

+hy){H)

,,;4(ax + hy){by + hx)/'( G)'(H) and

au oV

&.8; = 2(ax+hy)f'(G).2(by+hx)'(H) . = 4(ax

+ hy)(by +hx)l'( G) ' (H)

Putting these in (1) and (2) "" get;

(u :)

=

!(u ~)

2.1.15 .. Differ.entiation of Implicit Function. Implicit Functions. An equation such as F (x, y, z) = 0 defines one variable, say z, in terms of a function of other two variables x and y. In that case z is called implicit function of x and y. This can be displayed as z F (x, y, I(x,

= ft x, y).

This will give

y») = o.

Illustration. (i)

X2

+ / - t2 = O·is an equation. 1

t

= (x 2 + /)2".

From this we can express .

Here t becomes an implicit function of .r, y.

(ii) x + y2 = 5 is an equation .. From this we get x = 5 - y2. Here x bec?mes implicit funtion of y.

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2-32'

ENGI EERING i\IATHE:\lATlCS-ll:A

Theorem

1.. (Two variables, One equation)

If i(x,

y) =.0 gives y as an implicit function of x in a region on x-y

oP . d

plane then

-

"F

.z = _ 0 x dx

provided ~

.

.

dF so -=.0. dx

Proof;'J'. Since F(x,y)=O,

ButdF dx

=oF+oF dy .0 x 0 y . dx ' of

dy -..

dx

= -

~ O.

oy .

O~ dy

i.e. of +oF.dy=o. 0x

0 Y dx

ox er oy

Theorem 2. (Three var iables, One equation) If F (x, y, z) gives z as ari implicit function ofx and y in a region on X-)' plane then Fy (a) -=--, (b) -=--, where Fz~O.

oz

r,

ax

oz

.Fz

oy

F;

Proof Beyond the scope of the book. Illustrative Examples . . Example. 1. Find dy if(cosxV

= (sin j-)".

d"

Solution.

Let F(x,y)

F;

.

= (cos x)"

= y (cOSX)y-1(-

. ~ = (cosx)"

-(siny)x.

sinx) - (sinv)" log siny

logcosx-x(siny)""-I

.By Theorem (1) above,

cosy.

dy °f/ax dx = - of / . /oy

=

.=

-y (cosxy-I

sinx - (siny)"'" log siny

. (cosx)" log cosx - ~ (sin v):"" cosy Y· (cos x}V-1 sin x

.

+ (cos x)Y log sin y

. (cosx)" (cosx)" log cosx - x . .cosy .

Y tan x + log siny

= ~----="--~ log cosx - x coty

SillY

, from

.

.

the given relation

~

0

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES 2

E~ample.

2. Find dy and d

.;

dx

dxSolution. Here F (x, y) = x3 +

2-33

= 0.

if x3 + yJ - 3xy

i -3.xy.

From the given relation y can be expressed

as function of x.

of . B)I

or,

.

Theorem

dy ox = -, dx of

1 -

oy

o _(x3

dy 'dx

+ y3 -3.xy)

ox

3x2

-

3/

-3x

{i (y + x2 - 2xi)(i

y-x 2 y2 -x'

3y

_x)2

x2)(/_

-x) -(y-

2

2yx

+ x)

(/_ x)3 -2xy

(Y 2 -x )3 . .

.

'.

Example .. 3. From the equation

...

From the equation

ox oy

2x2 - yz + xz2 - 4

Solution. Here F(x,y,z)=2x2 022 -(2x FjI

Now-.=--·

dy

2

= 0,

F,

=

oy

022 -(2x

ox

-yz+xz2

-4.

-yz+xz

-4)

-yz+xz

-4)

.

ox

allti-.

oz

x can be evaluated

z.

function of y and

. oX"'

2

2x - yz + xz = 4 find -

-z 4X+Z2

as

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MATHE~ATICS -iliA

E GINEERING

2-34

and

13x F.. -=--'"' 13z F;;

13 -(2x ...!O...!=-13z 13 -(2x 13x

-=

2

2

- yz+xz

2

- yz+xz

2

..

-4)

=

-4)

y-2xz . 4X+Z2

-y+2xz 4X+z2

.

Example. 4. If a, 13, yare the angles of a triangle such that .

2

. 2

Sill a + Sill

. 2

13 + Sill Y

=

da constant, prove that df3

=

tan 13 - tan y tany-tana

By problem ·a + 13 + y = 1'- Given sin 2 a + sin 2 ,8 + sin 2 y

=k

.

(constant)

9f, sin 2 a + sin 2 ,8 + sin 2 { n - (a + ,8)} = k or, sin2· a + sin2 ,8 + sin2 (a

+ sin

or, 'sin2 a

2

,8 + sin2 (a +,8) - k

= sin2

Let F(a,,8)

+ ,8) = k = O.

a + sin2 ,8 + sin 2 (a +,8) - k.

By Theorem 1, 13F -!p{Sin2 a + sin2 ,8 +sin2(a +,8) ·da 13,8 d,8=-OF = {SID2 a + sin 2 + sin 2 (a + ,B) .13a 13a

p

..£..

=

2 sin 13 cos,8 + 2 sin (a + 13) cas (a + 13) 2 sina cas a + 2 sin (a + ,8) cos (a + ,8)

=

sin 213 + sin 2 (a +13) sin 2a + sin 2 (a + 13) 2 cas

=

'2

= = =

k} k}

sin 2~ - sin 2y = ----=---~ sin 2a - sin 2y

2~+2y. 2~-2y . sm' .

~2

~2~_

2a+2y. 2a-2y cas sin 2 ~ cas (13 + y) sin (,8 - y) cos (a + y) sin (a - r) cos (Jr - a)( sin 13 cas y - cas 13 sin r ) (Jr - ,8) (sin a cosy - cosa sin r)

COS

-cosa (sinfJ cosy - cosfJ sinr) - cos ,8 (sin a cos y - cos a sin y)

tanfJ- tany tan r - tan a .

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DIFFERENTIATION

OF FUNCTION

Example.s.·IfF(x,y,z)=O,

OF SEVERAL

prove that (ox) 8y

.

OX

Applying Theorem 2 we get oy

2.1.16.

F = _--..L

F:

z

VARIABLES

2-35

X(?7) x(~=)

=-1.

r [WB. U.T.200],200Jj 0_

x

ox

when z constant.

Total Differentiation.

Increment.

Let z

and y. Let ~

= f(x,

= dx

x and y. Then

y) be a function of two independent variables x

6y = dy

and Llz

=

increment of the function z

be the increments given to the variables

+ Llx, y + Lly) - f(x,

f(x

= f(x,

y)

= 4f

is called

y)

Differentiability and Total Differential. Let z

=

f(x,y)

be a function of two independent variables x and y.

The function f(x, (x,y) if fx and expressed as

y)

is said. to be' differentiable or total differentiable at

fy exists at

(x.y) and if its increment Llz (or

4/)

.

Llz= of Llx+

aj 6Y+E1Llx+E26y

ox oy oz oz =-. dx+-dY+E1dx+E2dy ox 8y

where E[,E2 are variable depending on Sx, 6y and

"[ ~ 0, "2 ~

°

as L1xand Lly~ 0.

oz

oz

The .expression dz = - dx + - dy ox oy

.

of

or df=-dx+-dy ox

8f dy

is called the total differential orsimply differential ofz or f(x,y). E.M ..3A-9·

c~n be

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E GINEERING

2-36

MATHEYIATICS.-IIIA

If 4z cannot be expressed in above way we say z = f(x, y) is not differentiable .: Obviously f(x,y)

is total differentiable at

E, AX+ E2 AY ~(AX)2

..

-; 0 as AX and Ay -; 0 .

.t(Ay)2

IAZ -dzi

I.e. If

-; 0 as

~(d;)2 + (dy)2

oz oX,

·oz oX2

dz = --dx,

dx, dy -; 0

.

.

1. If z = I(x\> x2,

Theorem

(x,y) if and only if

.••

then

xn)

oz . ox"

+ --dx2

+ ... + --dx

, where the variables x" x2,

.••

XI/

n

may not be independent and dz is total differential of z. Proof. Beyond the scope of the book. Theorem 2. If Jt»,

X2' •••

xn)

= c,

a constant then df

= O.

Proof Beyond the scope of the book.

fx(x,y),fy(x,y) do not ensures at (x,y). See Examples regarding this

[ote. Existence of partial derivatives the total differentiability of

f(x,y)

in the next Article 'Illustrative Examples'. 4.1.17. Second Order Differential. Let z = f(x, y) be a function of two independent variables x and y. The second ·orderdifferential

of z = f(x, y)

is defined by d2 z where

d2z = d(dz): Theorem 3. If. z = ftx, y) is assumed to have continuous second order 2 . 2 ;:)J . 2 0 Z 2 0 Z a: Z J derivative then (i) d z = --2 (dx) + 2--dx dy + --) (cryt

ox

(ii)d2z

ox oy

= (~dx

ox

oy-

+ ~dy)2 z. dy

Proof. Beyond the scope of this book. Note. (1) The above theorem can be generalised

z

= ft x,

x2,

Here

? d-z=

.•• XII)

•

(0oX

--dx1

I

0ox .

+--dx2

2

0ox"

+·..+--dx"

J2 z.

for the function

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DIFFERENTIATION

then d2 x = 0 because

(2) Ifx is independent

= Ii ~ (d~)2 = ~(ox)(dx)2

d2x .

ox-

ox

2.1.18 .. Illustrative

Liz = f(x

=

x2y - 3y, find Liz. Hence find whether

= 2.xy.1x =

as

.1x = -0.01,

dz

tu~O

t.y~O.

and

= 2xy.1x ~ (x2 = 2xy dx + (x2

.1y

-' 3) L1y - 3) dy.

2

= Txydx +(x

3. Using

approximate

dz when x = 4, y = 3,

= 0.02.

From the previous

Example.

L1"

txx ~ 0, t.y ~ 0

2. If z = f(x, y) = x2 Y - 3y. determine

Example.

Lly

- 3) L1y + L).1x + c2L1y.

Sx.y + 2xt.y ~ 0 as

The total differential,

Solution.

2x.1x.1y + (.1x)2

the function f(x, y) is differentiable.

Therefore

d::

y +

+ (x2 - 3) ~y + {L1x.y + 2x. L1y}L1.~+ (L1.\:) 2 L1y

c2=(~x)2~0

Solution.

3 (y + .1y)} - (x2)' - 3y)

- 3).1y + (L1x/

= 2..xy.1x + (x2

and

z is

+ L1x, y + .1y) - f(x, y)

= 2xyL1x + (x2

C)

= O.

ox

If so find dz.

= {(X + L1x)2 (y + .1y) -

We see

= ~(1)(dX)2

ox

Examples.

1. If z = f(x,y)

Example.

differentiable. Solution.

2-37

OF FU

-3) dy

problem,

= 2 (4)

the concept

we have (42

(3) (-0.01)+ of total

-

3) (0.02)=

differential

0.02.

compute

the

value of (5.12)2 x 6.85 - 3 x 6.85. Con~ider

the function

jt», y) ~

x2

Y - 3y. We choose

x =

5,

= 0.12; y = 7 and L1y= - 0.15. Since .1x, .1y are small we think dz == Liz. Now ow;

f(x dz

L1,~,y + .1y) - f(x, y) = Liz == dz. 2x; dx + (x2 - 3) dy [see Ex. 1 above]

+

=

= 2 (5)(7)(0.12)

+ (52 - 3)(-0.15)

= 5.1.

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2-38

ENGINEERING MATHD-IATlCS-IIIA f(5 + 0.12,7 - 0.15) - f(5, 7);:;:5.1

So, from above

or, . f(5.12) 6.85);:;: f(5, 7) + 5.1 (5.12)2. x (6.85) - 3 x 6.85;:;: (52 x 7 - 3 x 7) + 5.1

or,

Example, 4. Prove that the

.

(Of)· ox

=

/im

f(O+h,O)-f(o,o).=

(Of)· oy

at, af

ax oy df

=

h--+O

h

=0

(Of) ~ x + (aayf) ~y = ° . ox (0,0)

(mind that for independent variable x, y

,dx

= ~ x,

dy = ~ y )

N=f(o+~~,o+~Y)-f(o,o) = f(~,~Y)-

f(o,o)

=)I~~yl-JQ;O

IN- dfl Now,

0

exist at (0,0) with equal value.

(0.0)

. Now,

.

tun -

=0

.

.

h

(0.0)

~hese show

:. a~(0,0)

=

h

h-+O

f(h,O)-f(O,O)

h--+O

. #-01-..)(0.0)

.

/im

h.

h--+O

(0.0)

= IIm

Similarly

is not total differentiable at

of and of exist at (0,0) and they are equal. ox Oy

(0,0) though Solution.

f(x,y)=~lxYI

= 159.1.

~I~II~YI

+~y2

~~2

=f~I~~YI

+~y2

~~2

~rcose·rsine vr.2 cos2 + r2 sin2 I

. =

.• because as

~sinecose

e

e '.

putting

Sx

= r cos e, ~y = r sill e

which does not tend to

(~,~y)~(O,O),

r~O,emay

°

as

(~,~y) ~ (0,0)

be kept constant.

:. f(x·,y) is not total differentiable at (0,0).

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DIFFERENTIATION 'OF FUNCTION OF SEVERAL VARIABLES

. Example.S,

For the function f(x.y)

2-31)

= 2xy prove that f is differentiable

(i.e .• total differentiable) at a point (a•.b). . . af af . Solution. Obviously ax. and By exist at (a, b). af -=2y. ax f ·:.at(a,b)dz=(a )

af -=2x· ay

f dX+(a ) ax (a.b) iJy ;= Zbdx « 2ady = 2(bdx+ady)

dy (a.b)

= t:. x,

(mind that for single independent variable x, y, dx Now, at. (a,

dy

=

t:. Y )

b),. t:.z = f( a + Sx.b « t:.y) - f( a,b) = 2(a + t:.x )(

b + t:.1) - 2ab

= 2bt:.x + 2at:.y + 2t:.xt:.y = 2bdx + 2ady + 2dx dy :·.at (a, b), t:.z-dz=2dxdy.

. It:.z - dzl Now, . . . ~(dx)2 +(dy)2 :. f(x,y)

2dxdy

=

~O

.

as dx dy=s

~(dX)2 + (dy)2

is differentiable at (a, b) and df = 2(b dx + a dy) .

the total differential

. (xy Example. 6. Show that the function f x,y) .

=~

•

X2 + y2 .

= 0, is not differentlabl~ at-E6, 0) though fx Sol

io~. fx(~;O)=

(0. 0)

(x,y) ~ (0,0)

(x.y)

=

.

lim f(h.O)- f(O,Q) -Jim 0-0 1i40

h

11-+0

=

0

h

Similarly, fy(O.O)=O

At (0,0)

(0. 0)

and fy

(0. 0)

exist.

the total differential df

= Ix

=0

= (0.0)

and fy(O,O) exist.

lim f(O+h,~-f(O.O) h-+O

Thus fx

t)

(0, 0) t:.x+ fy (O.O)"t:.y

.

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2-40

= f(

and increment '6f _ ,

Llx . !!.y

_

0_

- ~Llx2 +Lll

Idf - LVI So, ~ !!.x2 + 6i

0.0) = f(Llx.Lly) - f( 0.0)

0+ Llx.O + Lly) - f(

Llx . Lly

- ~6x2 +Lll

ILlx 6yl (Llx)2

+

(6y)2

.jrcos9rsiliSI. 2

2

r cos S + r = IcosSsin91

.

2,2 SI/1

puttmg 6x=rcos9,6y=rsm9

S

which does not tend to 0 as (Llx.Lly)~(O.O)

when (Llx.6y)~(0.0).

r~Oand9

beC311S\!

'may be kept constant.

.', f(x. y) is not total differentiable (i.e., differentiable) at (0, 0) , Example.7. Show that the function f(x.y)=

is not differentiable at (0,0) though fx(O.O)

-i 2'

x3

(x,y);e(O,O)

2

x +y =0

. (x,y)=(O.O)

and fy(O.O) exist.

h) olution. fx

,,

f(O+h litn "

,

h-tO

(0,0) =

0)-

f(O

h

0)

'=

12-0 lilll-'/,-It

h-+u

= liml = 1 1I-t0

_k3 and

f (00) = /im f(O,O+k),Y'

:.fx(O.O)

k-+o

f(O,O) k

= lim k-tO

=-1

and fy(O.O) exist.

At (0,0), df;::;fx(0.0)6X+fy(0.0)6y At (0, 0),

ek

=6X-6y

N = f(0+6X.0+Lly)-~L1X.LlY)-0 _(Lly)3 (Llx)2 + (!!.y)2

, '(6X)3 :.at(0,0),6f-df=

_(6y)3 ,

f

.~ ,-~6X-6Y)

(~xt +(c.y)"

(6X)3 _(6y)3

-(6X-LlY)(Llx)2 (6X)2 +(6y)2

+(Lly)2)

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DIFFERENTIATION

OF FUNCTION

OF SEVERAL

-t..X(t..y)2 +t..y (t..X)2

(t..x/ +(t..y)2

It..x t..yll( t..x - t..y)1

IN-dfl

3f?

.. ~(t..X)2 + (tW)2

(t..X)2 + (t..y)2 ) -

(

Ir cos 9 r sin911r cas 9 .

(r2 cos2 9 + r2 sin2)

r sill 91 3/2

putting t..x = r cos 9 .. r3lsin9cos91lcos9-sin91 =

r3

Isin9cos911cos9 -sin91

which does not tend to

0 as (t..x,t..y) ~(O,O) because as

(t..x.,t..y) ~ (0,0), r ~ 0 but 9 may be kept constant :. f(x,y)

is not total differentiable at (0,0).

Example. 8. If z

- 3y, determine d2z

= x2 Y

,d-z= , -.a dx+-dya . .

( ox

)2 z

ay

.

az 2 aZz =-, (dx) .+2--dxdY+-2 2

. a2z

(dy)-.

ox

2

.az

a·

ai

=

;i z axay So,

ay·

= -(2xy) = 2y

Here -, ax-

~

a2z,

ax Oy e :

ax-

and

2-41

t..X·t..y(t..X:-t..y)

(t..X)2 + (t..y)2 .

VARIABLES

By (x =

2

-3)= 0

{} ( a.z) axlay

d2z=2y(dx)2 = 2y (dx)2

=

a

2

.

ax (x -3)=2x.

+4xdxdy+O.(dy)2

+ 4x dx dy.

,::.,.y

=

r s i1/ 0

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. ENGI EERING l\1ATHE:\1ATICS- IlIA

2-42, ,

2

Example.

9. If the Kinetic Energy K

= WV , find approximately

the 2g change in the K, E. as W changes from 49 to 49.5 and V changes from 1600 to 1590. Solution" Here K is function of the independent variables Wand V. 2

= oK

dV + oK dW= 2WV dV + V dW oV oW 2g 2g

Now

dK

or, .

. WV V dK=-dV+-dW=. g . 2g

V( WdV+-dW. V)

2

g

Now at W = 49, V = 1600, dW dV

2

= 49.5 - 49 = 0.5 and

= 1590 -1600 = -10. So, from above we get 1600 (49 x -10 + 1600 x 0.5) = -4500. 32 2 Therefore approximate decrease of K is 4500. dK

=

Example. 10 .. The height h and semi-vertical angle a of a cone are measured and from them A, the total area of the surface of the cone including the base is calculated. If h and a are in error by small quantities db and da respectively, find the corresponding error in the area. Show further th~t if a

.

.

=

;r, an error of 1% in h will be approximately

6

...

compensated by an error of - 0.33 degrees in a. Solution. If I be the lateral height of the cone then A =;rrl +;rr2. (1) . . hr. Now, cosa = -. :. 1= hseca and tana = -I.e., r I h •. From (1) A =;r htana hseca +;r h2 tan2 a or,·

..

A=fth

2

= ;rh2

secatana+;rh sina

.

. 1- sina

Here A depends on h and a.

2

2

tan a

2

e

= h tana.

nh tana(seca+tana)

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DIFFERENTIA TION OF FlJNCTION OF SEVERAL VARIABLES

dA

=

aA dci+ aA dh= 1th2 cosa (1- sina)-(-cosa) aa' all (1- sina)2

da + 2Jih sin,a ,.I-sma,

dh

= Jih2 COS(1- sin a~ + si~ a

2-43

sina x

cosa da

(I-slOat'

+27rh sina I-sina

dh=mi

cO,sa ?da+2Jih (l-sma),

sin,a I-sma

dh

trh (h' cosa d a + 2',sm a dh) w h:ICh iISt'th e correspon dimg , I- SIn a I- sm a error in area, Now if, due to an error} % in hand - 0.33 degree in a error in area, dA = ,0 then we understand the error of h is compensated by error of a, or, dA =,

.

,

Puttin g

a

h· ,dh=-we

get

=

,

1(.

. 6'.

da

=

-0.33

degree

= ~0.33 x ~radian . 180"

= -,0027r,

'100

Jd ' [hCOS 7r(-,0027r)

tiA.»

,6

1

I-sin

7r

I-sin

'6

7r 6

' ,

,

=,2,7rh(-..fjhx,0027r+ =

h ] '+2sin7r_ 6100

1~0)=2Jih2(-,Ol+,OI)

(apnroximately)

0, Hence proved,

, Example .. 11. If ~ be the area of a triangle ABC, prove that d~ = R(cosAda+cosBdb+cosCdc) where R is the circum radius of the triangle. Solution. We know, ~

'

= ~s(s-a)(s-'b)(s~c)

or; t'i = s(s~a)(s-b)(s-c)

(%+%-~)

=~(a+b+c>(,-~+%+~)(~-%+~)

. 1 =

"

,

T6(a+b+c)(-a+b+c)(a-b+.c)(a+b-c)

,}

??

= -(2b-:c-

16

.

??

'

??

+2c-a-' +2a-b- -a '. "

444

-b

-c)

f(a

="

b c)

.

(say)

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E Gl EERING MATHE:\1ATICS -iliA

2-44

Taking differential we get 2~ d6.

or

=

8f da+ 8f db+ 8f de 8a 8b 8e

2~ d ~ =J..(4c2a+4ab2 '.

)d

16

+J..(4b2e+4ea2 16

or, 2~·d~=

I~{4a(b2 +e2-J)ch+4l{e2 == -

I

b.

(4be2 +4a2b-4b3

-4a3)da+J..

16

-4c3)dc

.

+J -b2)th+4c{i

)tt}

2

+b2 -e

{4a .2becos A da + 4blea cos B db + 4e.2ab cos C de}

16

.

I = -abe(cosAda . 2.

+ cosBdb + cosCdc) .

I

== -.4~R(cosAda

+cosBdb+cosCde)

.2 .. d6.=R(cosAda+cosBdb+cosCde). sm a

2

Example. 12. If A ~ tth . .' I-sIna variables.

find dA where h and a are independent

Solution. 8~d 8A dA =- a+- dh fu 8h = tih

.

=.

2

_1_2cosa(l-sina)+cosasinad (1-sina)2·

a+

=nn

cosu 2

(I-sina)

do. + Znh

xh (h coso l= sinu i-sina

sinu l=-sincc

2-' IlJ1

sino; l=-sinu

''/ ((,/

dh

d a+ 2·Slfl O,d'/1)

.

Example. 13. If P dx + Qdy + Rdz can be made a perfect differential of some function x, y, z on multiplication by a factor, prove that [C.fl.195-1]

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DIFFERENTIATION

OF FUNCTION

OF SEVERAL

VARIABLES

2-~5

Let j (x,y,z) be the factor such that

Solution.

f Pdx + f Qdy+ I Rdz = a perfect differential (1)

=dU·

o

.

0 So, oy (f R) = oz (f Q) oR Of or, I oy + R oy

=

oQ Of I oz T Q oz

~(f P) =~(f ozox

R)

or I oP + P Of , oz oz

=I

~(fQ) ox

P)

=~(f oy

or I oQ + Q Of , ox· ax

=I

(2)

oR + R 01 ox oX

(3)

oP + P Of oy oy

(4)

(5)

From (3),

oR

OP j-= OXOZ

f--

or IQ(OR _ oP)= , ax oz .~P _

From (4), j 0 ' y

fa

oj P--R-, OZ

ox

PQol _RQol oz oX

(6)

oQ _ of _pol -Qa 0 x x y

.. R( 8P _ aQ} = Q~ of ._P R of ay' ox ax ay

or, j

01

(7)

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2-46

ENGINEERING

MATHEMATICS

-IlIA

Adding (5). (6), (7) we get

I{P(OQ _ Og)+)OR _ OP)+R(OP _ OQ1}=0 , oz oy ~lox oz ,oy, ox) o~,

JOQ_OR)+JOR_OP)+R(OP J

loz

oy

_OQ)=o

~ ox oz

oy ox

EXERCISE

rI]

SHORT ANSWER

QUESTIONS

I. Evaluate the following limits

(a) "lim(2y

- 3x)

(b)

x-+4 'y-+-I

(')c

I'. un (x,y)-)

(1'.2)

. Iirn

(e)

(x.y)-)

lim

(.\y-3x+4)

tx,y) -) (2.1)

3-x+y + x - 2y

(d)

4

lim (x,y)-)

(0.0)

,

~x - )' x " +)'2

xsin(x2+y2) x2 + y2,

(0, 0)

2. Investigate the continuity of the following functions at the given points (a)

x

.

at (0, 0) 3x + 5y , (c) l(x,y)=(x2

=

!b) xy

+ 6x at (1, 2)

2 1 2,(x,y):;t:(0,0) x +y 0, (x, y) ::!(O,0) at (0, 0) +y2)sin

2xy 2 ;(x,y):;t: (0,0) x +y = 0, (x,y) = (0,0) is not continuous at (0,0)

3.(a) Show that I(x,y) ,

. [Hint: TIle limit (b) Verify that (c) Show that .

=

2

lim I(x,y)

{x,y) ..• (o.o)

li1~

(x,)') ..• (O,o)

J(x:y)

=x -

does not exist at all] ,

x + Y does not exist. x - Y

+ )'3 x-y

= Xl

.= 0, [Hint: Consider y

[W.B. U. T. 2006]

,x:;t: y x =Y

mix) and y

=x -

is not continuous at (0,0) . m2x3

]

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-.J7

4. Find ~he following partial derivatives from first principle (a) af,.a.r at(o,~)whenf(x'Y)=YSinX ax ay. 2

(b) af, af whenf(x,y)=x3y2

(c) .

ax

ay.

.

f a, ax

af when f(x, y) ay

= eXY·

. ar aj. 5. Evaluate _. and when . ax ay 2

(i) ft x, y) = loge (y2 + X 2)

(ii)

ft»; y)

= tan -I

2

+Y

X

x+y (iii) x + y + I (X, y) 6. Let ft x, ;). =

2XY

2'

= loge

I (x, y).

(x, y) # (0, 0)

+y

.x

=.0, f?therwise.

Ix (0,0)

Evaluate •.

= x - y, find af and af at (2, - 1) from the definition.

7. If f(x, y) .

8. If x

x +.y

= rcos8,y

.

X4

9. If z

=

rCi\t~.(O, 0).

log

ax

sin8 find or anci°8 in terms of rand ox oy

=r

+ y4

. 8z find x-+ Ox

x+y

= Iog ( tanx + tany

1.O If u

.ay

8.

OZ

y-. 8y

ou = 2 .

. 2 y-;::) , prove t hat at si sm2x- 8u + sm ox oy

[WB. 11.'IfV=~2tan-l~,fllld(02V)

.

..

x

.

ox 8y

(1.1)

.02 I 02 f 12. Show that -= -hold for the function . ox oj; 8yox

f(x,y)

..

= x y+e·l)'2 3

.

.

o.t: 2005]

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2-48

ENGINEERING

.

.

x

13. If lex, y) =

2

-1.Y 2 - - y tan

tan

-1 X

-,

x

prove that

Y

Y2

14~Ifz=tan(ax+y)-(y-ax)

'IATHKVIATIC

2, provethat~=a

.:... X

!,y -

o2z

x=rcos8,y=rsin8prove

x- + y

2

2 .

-2-'

oy

~ .., ov ov x-+y-=r-

that

Y

-

1

o2z

2

ox

15.. If

2

-iliA

ax

~

cv

oy

where

cr

v=v(x,y). 16. If z'=/(x,y),x=eu Oz

Oz

c::

ou

ov

c'.\:

---=x-. 17. Let z = /(x,y) where andy

18. If u =

8y

x = u2v,y = uvl , After putting these value or x

(OZx-+ ax

.u-+v-=3 of of ou . ov

1

y-

--,

x

xy

z - x) --

/I

Y

V=V{x.y}

-

. terms ,y=u- ? fi111d -dv 111 du

and

x=ucosa-vsina

2l. If z = / (u, v) and u = x2 - / ex

2 Oll

ox

.

02

oy

then prove that x

where a is c~nstant then prove that

+ y)- + (x

'oz )

y-

xz

. X I9 . If V=SIn-,x=e

(x

OZ -y-

let /(x,y) be changed to Ftu,v), Show that

20.If.

_ev, show that

+e-V;y=e-ll

oz..= 0

- y)-

. oy

IS

+y

0

2

OU2+;: -:;au = o.

-

oz

0)'

f 1I.

and y=lIi>inC1.+vco:-u

(OV)2 + (oV)2 = (~V)~ ax

ay

+(

cJV1~

\ov)'

\.ou

- 2xy, v = y prove that the equation

equivalent

oz

to -;- = O. uV

verify the fact -.du -dx = 1 for the implicit .... 22. Ven relation dx du

(i) ·1 XU-1 - lie -r. - II 5 = 0 23. If ftx.y, z, w) =0, prove that

.

(ii)

~. ay .~. 0la-awax

e:w -

S111

(x

+ II)

-

1 = O.

~v = 1 [WB.UTech.2005]

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DIFFERENTIA

nON

OF FUNCTION

OF SEVERAL

VARIABLES

2--tlJ

24. Find dy in the following: cases: . dx ~

=0

(I) x sin (x - y) - (x + y)

•

.(ii) e" + eY

(iii) xY + yX = ab (iv)xy sinx + cosy

= 2xy

= 0 at (0,

;).

ANSWERS 1. (a)

(b) 0

-14

(c) 4

2. (a) discontinuous

(b) continuous

(b) 3x2y2,2x3y

4.(a)~,o 2

2x

5.(i)

(d) does not exist

2'

2

(x. + y ). x

ye,T)',xe'"

(c)

+y

2

x2 + 2xy _ y2

(ll) {(x2 + y2)2

+

y2 + 2xy _ x2

ex + y)2}

, {(x2 + y2)2

...) both f(x, y) (III , 1- f(x,y)

.

6.0,0.,

cosO 8. cos8,--

7. -2, -4

+ (x + y)2}

1I- 2 -3-ell

'11. 1 19.

9; 3

r

y + x2 cos (x - y) 24.(i) "----::----"x + x2 cos (x - y)

(II]

(c) continuous

2y

2

..

(e) 0

ell

cos:,

II

eX -2y

...

(ii) ---=-

(iii) :

2x _ex

LONG ANSWER

y-I

yx' xy

x-I

l/

x

+ Y logy +x'

v

log ,v

(iv) 0

QUESTIONS

1. From definition show that (al lim (x-2y)=-1 , x-+I'

.

(b)

lim

(x,y)-+(2.1)

(xy-3x+4)=0.

y~1

2. Find the following partial derivatives from first principle (a)

f~(1, -1)

and fy(O, 2) when f(x, y)

= £y6

+ sin x)' 2

(b) !x(xo,yo)andfy(xo,Yo)whenf(x,y)=2x 3. If f(x,y)=x

3

,

,

.

y+eX)'

2

2

2

'a f

Verify whether -ax ay

= --.a f

ayax

2 a f

a2 f

af

af

ax

ay' ax ,ay

find-'-'--2

-.\y+

a2 f

a2 f

ax ay

ay ax

'-2-'--'--'

y

2

.

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2-50

ENGINEERING

4. (i) If V

=..r;y, find the value =

(ii) If Vex, y, z)

,

,

I

''\jx

02V ax:

(t)

02V

=

and

then prove that

2

(~~t

whereP

x2 + y2 x + Y , prove that

8. Calculate,

[WB.U.T200J]

8z

8y

6. For the functions (a) I(x, y) , Ix)' = !'vx· U

oy

02V

5. If f='(x - y) sin (3x + 2y) find

7. If

.o«:

+-2 +-2 =0.

-2

Find

02V

of -2- + -2-'

1 ., +:y +z

2

02V

MATHE:\1ATICS-IIIA

=

L, !y,

= (-1,

2,0).

Ixx,fyy,lxy'

2x - Y (b) ft x, y) x+y

Iyx at

=x

(0, ~).

tan xy; verify that

(auox - OU)2 ( Oil all) 8y = 4 1- ax: - 8y

r: !.V' t;(0, 0), I (0, 0) for the function y

XY

f(~, y) == ~

2 x

2

+ y2 ~ 0

ifx2

+y

= 0 if ,x = y =

o.

021' 021 9. Show that -=~ hold for the function [i», y) = x-".

,

oxoy,

vYox

,

,

10. Show that for the function I(x, y)

,

=

22, ~ y 2'

x +y

= 0, Ixy(O, 0) = lyxC0, 0). ,

11. If I(x,y)=xy

= 0,

+ 5y2 2'x x +y

x2

x =O,y=

= (0, 0)' [WB. U. Tech.2008]

2

2

+y

2

(x,y)

(x, y);c (0, 0)

°

prove that I).),(0,0) i:- Iyx (0, 0).

;cO

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OF SEVERAL

VARIABLES

2-51

12. Show that for the function

y) = (x2

f(x,

+ y2) tan -I X,

.

1t

='2y

x:;t:

0

X

2

x=O

,

.Jxy(O, 0) = 1and fyx(O, 0) =-:-1. 13. Prove that u;x + uyy

°

=

if

(i) u = tan -I a 'log (x2 + y2) +

P tan -I ~ X

(ii) 11

U

=

tan

_I

2xy 2 2 X -y

.

1

14.If z

=

(1 - 2xy + y2) -'2, show that

15. If V = ke-gx sin (mt - gx), where

k.

g, m are positive constants, and

2

if the equation .

av -aavt = )..l-2ax

_(x2 +i·+z2)

16. Ifu=3(ax+by+cz)2

. o2u o2u. the value of--+--+-and . ox2 oy2 where P

=

hold, prove that g =

o2u

2)..l

and a2 +b2 +c2 =1 find

[o2u o2u --+--+ox2 oy2

oz2

HEn -.

02u) oz2

p

(1, 1,4).

17. If z is a function of x and y and x

=

r cos 8, y

=r

sin 8

then prove that 02z

02z

02z

a,r2 + ay2 = or2

18. If u = (x2 +

au

i + z2) f(xy

(y-z)-+(z-x)-+(x. ox. E.M.3A-IO

1 Oz

+-;.or +;

au oy

1 fiz 002'

[W.B.U.Tech.2002,2008]

+ yi + zx), prove that ou y)-=O. . oz

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2-52

G l\lATHE:\1ATICS-IIIA

19. Ifv=f(x2 .2

(y

+2yz,y2 ~

2

-zx)-+(x Ox'

20. Let .r2

=' x2

+2zx), prove that ~

2

-yz)-+(z By.

+ y2 + z2 and U

222

= r cos e, y. = r sin e,

. 2

2

oro r (i) ox2 + oy2 .

2

.

ox

.

2

(ox)

)2}

. or +(By

(x;to 0, y

;to

0)

oy2

r r + ( :~

(iii) (.:: 22. If

1/1-1

2

e + 0 e =0 2

(ii) 0

2001,2012J

prove that

1 { or

=;

[WB.U.T

rill, prove that

=

o U o·U 0 U '-')-+-2-+-2-=I11(I11+1)r oxoy oz 21. If x

~

-xy)-=O. oz

Jf = ~.(2,x--

=

1.

3y, 3y - 4z, 4z - 2x), prove that

! oW ;! oW +!oW = O. 2 ox

.

23. If

1I

3 oy

4 oz

x

y

z

y ou ou x-+y-+z-=O. ox oy

z

x

= fer, s, I) and r = -, s = -,I = -

prove that

ou·· oz

24. If u = logr and r2 ~ x2 + y2 + z2, prove that r

2(o2u o2u -2-+-2 ox 8y

o2u)_

+-2

OZ .

-1.

25. 'If z = fex, y) and x = eU cos v, y = e" sin v then show that . oz Y ou

8z'

+x ~

=e

2u oz By

[W.B. U. Tech.2006

]

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DIFFERENTIA TION OF FUNCTION OF SEVERAL VARIABLES

26. (i)

2-53

. If the cartesian system of co-ordinate (x, y) is transformed

. polar co-ordinate system 2

(r,8)

to

then prove that

.

-_ - cos 29 . . (1·).~ ax ay r2 .. ax (n)

[H'mt. h ere x

=

r cos9,y

=r

sin e ]

1

a9"* ae . ax

.

.

u

27. If V= V(x,y)andx=e

u

cost,y=e

sint prove that

i;2v + o2V =e2U(02V + o2V). . .. ou2 at2 . ox2 ay2 28. (i) Find dF;dG if F=x3y_4xy2

+8i

and G=8xy2z3

(ii). Express the area of a triangle ABC, da C. Hence show that .

. t1

. (iii) If

x

z

= 0.9. Y = -1.2:

da

L1

-3x2yz.

as a function of a, band

db

= - a + - b . + cot C de.

= tan -I xy,

find

approximate

value

of z

when

. (iv) ind the percentage error in the area of an ellipse if 1% error is made in measuring the major and minor axis. (v) Find the upper bound for the magnitude of the error of the function f(x,y)

= x2 - 3xy+ 5 at (2, 1) when [x - 21:$ 0.1, Iy - 11:$ 0.1.

, (vi) Determine d2z where z=xy2 -x2y (vii) Determi·ne d2z where z (viii) Determine d2z where z 29. (i) If.x

= rcos9;y = rsin9

= xy(x =

xlx2x3

prove that xdy - ydx =·r2d9.

(ii) If U = x2 + 3y2 - xy, compute .1x = -·2. t1y =.1. . (iii) Evaluate ~(3.8)2

+ y)

t1U, dU

where

x

= 5, y = 4,

+ 2(2.1)3 approximately, using differential.

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2-54 .

ENGINEERING MATHEMATICS - IlIA

h change from initial value of

(iv) Suppose that the variables rand (ro, ho) = (1, 5) by the amounts

dr = 0·03 and dh = -0·1. Estimate

the

resulting absolute, relative and percentage changes in the values of the function

v

=

nr2 h.zZZZ

1b

.

(v) The area of a triangle is calculated by using the fOffi1ula"2 c sui

..!..%

If the lengths· band c are liable ·to an error .

.

A.

in measurement and the

4

angle A measured as 60° is liable to an error of 1 minute of angle, find the percentage' error in the value of the area. (vi) Determine d2z where z=sin(x,

+X2 +X3)

where

X"X2,X3

are

independent 30. Prove that

f(x,y) = Ixl + Iyl is not .differentiable at (0,0).

..

31. Prove that the function

f(x,y)

xy

=~

.

.

(x,y) ;t(O,O)

,

(x.y)

x2+i =0

is not differentiable at (0,0) though fx(O.O)

= (0,0)

and fy(O,O)

exist.

2

32. Prove th~t the function f(x.y) .

=

~Y

x +y

x;t 0

4'

=0

,

x=Q

is not differentiable at (0, 0) though it possess first order partial derivatives at (0, 0).

.

.

33. Prove that the function .

f(x,y)

. x3-i =

2

2'

x +y

=0

.

(x,y) ;t(o.o)

(x,y)=(o,o).

is continuous at (0, 0), has partial derivatives at (0, 0) but is not differentiable at (0,0). 34; Find dy I dx from the following: (i) (sin y)" - (cos x)"

=0

(ii) x

=y

log (xy)

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DIFFERENTIATION

OF FUNCTION OF SEVERAL VARIABLES

2-55

2

d --f dx

35. Find

in the following cases:

(i) ·.4 1 x·+y.=

4 axy 2

4

. (iii) x~ 36. If v

=. x

+ y~

=2

.

(log x +Iogy) where x

.

3'

(iv) y3 -3ax2

+y

3

au

2

=0.

= 1, find-.dv dx

+ 3xy

. 37. Evaluate ay from the relations x-y+u

+x3

2

3

+v =l,x+y+u

I'

e =2.

[Hint. apply Th. 3. of ] .;

ax

ax.ay

.

au av au

38.Fmd-,-,-

ay

and ~. whereu

i

uv

2

+v

2

=x

+yand

.

2

u+v=x

ANSWERS 2. (a) ~ - cosl, 0 2

(b) 4xo - Yo, - Xo + 2yo.

2,2

3. 3x y + Y eX) , x ,

3

2

4

2

2

2

3x2 + 2xy3 e~. + 2Y!eX)' 3x2 + 2 yt)3eX)'

'''/

4. (i)-±(x . 5.

-%~Y2+ y -%xY2)

~("+.j3)'~(2"-3.j3),%(,,.j3

+ 2yexl

(ii) -:' 52

. 2 2 3/~ (x: + y )

2

2

y eX)' + 2xe

aJ .-_ 2

'm8y

a2J

= __

8ym

x .2 .1

hold.

~ 52

-2),%(".j3 +3),~(2".j3 +1),~(2"J3

,

(x

2

2 3/2

+y )

28. (i) (3x2y-4y2)dX+(x3

(iii) -.835

,0,0.

+

16. 0, O.

-8xy+24y2)dy,(8y2z3

+ (l6xyz3 (iv) 2%

-3x2z)dY+(24xy2z2

(v) 7

(vi~ -2y (dx)2 + 4 (y - x) dx dy + 2x (dy) (vii) 2 {y(dx)2

2

x3'

. y3

8.

r 2

+ 2xyeX)' ,6.xj + Y e'Y ,4x

+ 2 (x + y)dx dy + x (dy)

(viii) '2xJdxJ'dx2

2

2}

+ 2X2 dx, dXJ + 2xJ dX2 dXJ.

-6xyz)dx -3x2y)dz

I).

_yo

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ENGI

2-56

34. (i)

ytanx

-+- logsiny

logcosx

(ii)

"

- xcoty

EERING

;VIATHE:\IATICS-

y(x - y) x (x + y)

2 (iii) ----::--:(l-x2)~

36. 1+ logxy - x (x2 + y) / Y (y2 + x). u3 + 2ve-v 37:

2u

4

2"

- 6u v

nrn

2u + 1 2v + 1 2u - 1 2v-I 38 • -- x ' -- 4x ' ----2 ' 2

.

MULTIPLE

.

CHOICE QUESTIO

S

. x 1. IIm-= x->0y y->O

(a) 0 2. f,:y

(b) 1

= f •.for x

all function

(c) f(x,y)

(b) no

" (a) yes

= u2 + 1 and

3. Let feu)

(d) does not exist

00

u = 3x2 + y then

Ix

=

(b) df dx 4. If 4>(x,y)

= 0 then

4>x (a)

=

4>y (b) ~

4>,,"

5. If 4>(x,y,z)

dy dx

=

(c) 4>y

(d) none of these

o then

(ay) (ax)" ay z=consl?< az (a) 1

(d) none

(b) -1

x=constau"1 x

(az) ax

(c) 0

= y=constatlI (d) none

IlIA

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DIFFERE

OF FUNCTlO

.

OF SEVERAL VARIABLES

2-57

ou = ax

6. If u -: log (x2 + y2) then -

.

1 (a) X2+y2 7. If f(x,

y)

(a)

x2 _ y2 (b)X2+y2

= Ixl + Iyl then

(d) none

1«0,0)

=

°

(b) 1

(c) does not exist 8. Which of the following f(x,y)

statements

is not true for the function

= X2 +.xy + y2 af

(a) f(x,y)

is continuous at (0,0) (b) -8 exists at (0.0)

x

of (c) 9. ~(xY)

ax

.

ay exists at (0,0)

,

.

10. If x

=

(d) j~y (x,y)

(c) xy logx

(b) yxY

.

,

ax =-or ar ax

rcosO,y =rsmO then -

(a) True

11. If x

,

,

:#=

fyx (x, y)

=

(a) 1

,

(d) none of these

=rcos,S,y =r

(b) False Ox sinS then -

=-

1

or or ax

(a) True

(b) False

(a) xr logx

(b) 0

(c) does not exist

(d) yxy-I

(d) yxy-I

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2-58 .

ENGINEERING

13. If I(x,y)

= X2 + y2 then I

(a) I

xy

[WE,

(x,y) =

(b) 0

,.

MATHEMATICS-iliA

(c) 2

(d) x + y

(c) 1

Cd) 2u

u. T.2007]

X2

14. If u = log-

then xu, + YUv

Y

,

(b) 0 Y

,x'

au

01,1

15. If u= sirr" - + cos-I - then x- + y- = y X ox Oy (a) sin-I'::' y

+ cos " y

(b) 2u

X

(d) 0

(c) 3u .

3322

16.Ifz=log(x

+y

-x

azaz

) then -+-=-OX oy

y-xy

(a) True

17. lim x~o

,

x2

IUTI . y~O x2

1. x+y

(b) False

-l + y2

=

' lirn ,

InTI y~O

x~o

x2 _

y

2

x2 + y2

(b) False

(a) True. , 18 . li1In . 2x Y 2 d'oes not exist x~ox + Y y~O

(a) True

(b) False

19. In the function V = ftu, v) if (a) 0

u

(b) constant

is independent variable then d2u = (c) I

(d) none

20. If the increment of the function z =