e\monish-k\tit-5kch IInd 4-01-1 A TEXTBOOK OF ENGINEERING MATHEMATICS
A TEXTBOOK OF ENGINEERING MATHEMATICS For BTECH (5 th Semester) Computer Science and Information Technology FOR MAHATMA GANDHI UNIVERSITY, KOTTAYAM, KERALA (Strictly According to the Latest Syllabus) NP BALI Former Principal SB College, Gurgaon Haryana By JAYASREE TG Asst Professor in Mathematics Adishankara Institute of Engineering and Technology Kalady, Kerala UNIVERSITY SCIENCE PRESS (An Imprint of Lami Publications Pvt Ltd) BANGALORE l CHENNAI l COCHIN l GUWAHATI l HYDERABAD JALANDHAR l KOLKATA l LUCKNOW l MUMBAI l RANCHI NEW DELHI l BOSTON, USA e\monish-k\tit-5kch IInd 4-01-1
Copyright 013 by Lami Publications Pvt Ltd All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Published by : UNIVERSITY SCIENCE PRESS (An Imprint of Lami Publications Pvt Ltd) 113, Golden House, Daryaganj, New Delhi-11000 Phone : 011-43 53 5 00 Fa : 011-43 53 5 8 wwwlamipublicationscom info@lamipublicationscom Price : ` 1500 Only First Edition : 013 OFFICES India & Bangalore 080-6 61 15 61 & Jalandhar 0181-1 7 & Chennai 044-4 34 47 6 & Kolkata 033-7 43 84 & Cochin 0484-37 70 04, 405 13 03 & Lucknow 05-0 95 78 & Guwahati 0361-54 36 69, 51 38 81 & Mumbai 0-4 91 54 15, 4 9 78 69 & Hyderabad 040-4 65 3 33 & Ranchi 0651-1 47 64 UEM-9681-15-ENGG MATH V MGU (KE)-JAY Typeset at : Goswami Associates, Delhi C 5443/01/07 Printed at : Ajit Printer, Delhi e\monish-k\tit-5kch IInd 4-01-1
CONTENTS Preface Syllabus (vii) (viii) Pages Module 1: Finite Differences 1 35 11 The Forward Difference Operator D 1 1 Differences of Factorial Polynomial 5 13 Backward Difference Operator Ñ 6 14 The Displacement (or Shift) Operator E 6 15 (a) Relations Between D, Ñ and E 7 15 (b) Other Operators 9 15 (c) Relations Between the Operators 9 16 Interpolation and Etrapolation 13 17 Newton-Gregory Formulae for Equal Intervals 13 18 Lagrange s Formula for Unequal Intervals 17 19 (a) Divided Differences 1 19 (b) Newton s Divided Difference Formula is 110 Numerical Differentiation 3 111 Numerical Integration 9 11 (a) Newton-cote s Quadrature Formula 30 11 (b) Trapezoidal Rule (n = 1) 30 11 (c) Simpson s One-Third Rule (n = ) 31 11 (d) Simpson s Three-Eighth Rule (n = 3) 31 Module : Z-Transforms 36 55 1 Definition 36 Some Standard Z-Transforms 36 3 Properties 37 4 Convolutions Theorem 46 5 Evaluation of Inverse Z-Transforms 47 6 Applications to Difference Equations 51 Module 3: Discrete Numeric Functions and Generating Functions 56 7 31 Introduction 56 3 Manipulation of Numeric Functions 56 33 Generating Functions 60 34 Properties of Generating Functions 60 ( v ) e\monish-k\tit-5kch IInd 4-01-1
( vi ) 35 Recurrence Relations 64 36 Linear Recurrence Relation with Constant Coefficients 65 37 Homogeneous Solution 65 38 Particular Solution 66 39 Solution by the Method of Generating Functions 69 310 Simultaneous Difference Equations 71 Module 4: Comple Integration 73 1 41 Introduction 73 4 Function of a Comple Variable 73 43 Limit of f (z) 74 44 Continuity of f(z) 74 45 Derivative of f(z) 74 46 Analytic Function 74 47 Necessary and Sufficient Conditions for f(z) to be Analytic 74 48 Cauchy-Riemann Equations in Polar Coordinates 76 49 Harmonic Functions 77 410 Orthogonal System 78 411 Application of Analytic Functions to Flow Problems 78 41 Comple Integration 89 413 Simply and Multiply Connected Regions 94 414 Cauchy's Integral Theorem 94 415 Cauchy s Integral Formula 96 416 Series of Comple Terms 101 417 Taylor s Series 10 418 Laurent s Series 103 419 Singular Points, Residues 108 40 Residue Theorem 109 41 Calculation of Residues 109 4 Application of Residues to Evaluate Real Integrals 114 Module 5: Queuing Theory 13 14 51 Introduction 13 5 Characteristic of Queuing Model 13 53 Transient and Steady State of the System 15 54 Waiting Time and Idle Time 15 55 Kendall s Notation for Queuing Models 15 56 Model 1 {(M/M/1): ( /FCFS)} 16 57 Model {(M/M/1): (N/FCFS)} 17 58 Queuing Formula 17 59 Little s Formula 19 510 Model 3 {(M/M/1): (N/FCFS)}: Single Server, Finite (or Limited) Queue Model 137 e\monish-k\tit-5kch IInd 4-01-1
PREFACE The objective of this book is to provide the readers with the thorough understanding of the topics included in the courses of Computer Science in an easy and simple way The topics which are very relevant with respect to university syllabus are fully covered by this book and will support in self study Each module of this book covers the latest syllabus prescribed by Mahatma Gandhi University (MGU) for the fifth semester of BTech Courses in Computer Science Almost all problems are worked out and additional problems are incorporated from latest Mahatma Gandhi University Question papers to increase the flavor of the book All efforts have been made to keep the book free from errors Although suggestions and recommendations for the improvement of the book, are invited We wish our readers good luck for brilliant success in life Authors ( vii ) e\monish-k\tit-5kch IInd 4-01-1
SYLLABUS EN010501 B ENGINEERING MATHEMATICS IV (CS, IT) Teaching sheme Credits 4 hours lecture and hour tutorial per week Objective: To use basic numerical techniques for solving problems and to know the importance of learning theories in mathematics and in queueing system Module 1 : Finite differences (1 hours) Finite difference operators D, Ñ, E, m, d-interpolation using Newtons forward and backward formula Newton s divided difference formula Numerical differentiation using Newtons forward and backward formula Numerical integration Trapezoidal rule Simpsons 1/3 rd and 3/8 th rule Module : Z transforms (1 hours) Definition of Z transforms transform of polynomial function and trignometric functions shifting property, convolution property-inverse transformation solution of 1 st and nd order difference equations with constant coefficients using Z transforms Module 3 : Discrete numeric functions (1 hours) Discrete numeric functions Manipulations of numeric functions-generating functions Recurrence relations Linear recurrence relations with constant coefficients Homogeneous solutions Particular solutions Total solution solution by the method of generating functions Module 4 : Comple integration (1 hours) Functions of comple variable analytic function Line integral Cauchy s integral theorem Cauchy s integral formula Taylor s series, Laurent s series Zeros and singularities types of singularities Residues Residue theorem evaluation of real integrals in unit circle Contour integral in semi circle when poles lie on imaginary ais Module 5 : Queueing Theory (1 hours) General concepts Arrival pattern service pattern Queue disciplines The Markovian model M/M/1/, M/M/1/N steady state solutions Little s formula ( viii ) e\monish-k\tit-5kch IInd 4-01-1
MODULE 1 Finite Differences 11 THE FORWARD DIFFERENCE OPERATOR D Let y = f() The values, which the independent variable takes, are called arguments and the corresponding values of f() are called entries The difference between consecutive values of is called the interval of differencing If the interval of differencing be h and the first argument be a, then Arguments : a, a + h, a + h, a + 3h, Entries f() : f(a), f(a + h), f(a + h), f(a + 3h), For brevity, these entries are denoted by y 0, y 1, y, y 3, y 1 y 0 = f(a + h) f(a) is called the first forward difference of y 0 and is denoted by D y 0 or D f(a) Thus ÿÿd y 0 = y 1 y 0 or ÿdÿf(a) = f(a + h) f(a) Similarly, D y 1 = y y 1, D y = y 3 y In general, D y n = y n + 1 y n or D f() = f( + h) f() The differences of the first forward differences are called second forward differences Thus D (D y 0 ) = D (y 1 y 0 ) or D y 0 = D y 1 D y 0 is called the second forward difference of y 0 Similarly, D y 1 = Dy Dy 1, D y = Dy 3 Dy In general, ÿd y n = Dy n+1 D y n Similarly, we can define differences of higher order The table showing the various forward differences is called forward differences table and is given below Argument Entry First Diff Second Diff Third Diff y D y D y D 3 y a y 0 y 1 y 0 = Dy 0 a + h y 1 D y 1 D y 0 = D y 0 y y 1 = D y 1 D y 1 D y 0 = D 3 y 0 a + h y D y D y 1 = D y 1 y 3 y = D y a + 3h y 3 y 0 is called the leading term and Dy 0, D y 0, D 3 y 0, are called the leading differences 1
A TEXTBOOK OF ENGINEERING MATHEMATICS The operator D has the following properties : (i) Dc = 0, c being a constant (ii) Dcf() = c Df() (iii) D [a f() + bg()] = a Df() + b Dg() (iv) The n th difference of an n th degree polynomial is a constant = (co-eff of n ) n h n and hence higher order differences are zero Eample 1 Prove that : ILLUSTRATIVE EXAMPLES f( ) g( ) f( ) f( ) g( ) (i) D [f() g()] = f( + h) Dg() + g() Df() (ii) D g ( ) % % g ( h) g ( ) Sol (i) D [f() g()] = f( + h) g( + h) f() g() = f( + h) g( + h) f( + h) g() + f( + h) g() f() g() = f( + h) [g( + h) g()] + g() [f( + h) f()] = f( + h) Dg() + g() Df() (ii) D f( ) g ( ) f( h) g ( h) f( ) g ( ) f( h) g( ) f( ) g( h) g ( h) g ( ) = f ( h ) g ( ) f ( ) g ( ) f ( ) g ( ) f ( ) g ( h ) g ( h) g ( ) = g ( ) [ f ( h ) f ( )] f ( )[ g ( h ) g ( )] = g ( ) % f ( ) f ( ) % g ( ) g ( h) g ( ) g ( h) g ( ) Eample Evaluate the following, interval of differencing being unity (i) D tan 1 e a (ii) D (iii) D ( 1) e e Sol (i) D tan 1 a = tan 1 a( + 1) tan 1 a a( a = tan 1 1) a = tan 1 1 a ( 1) a 1 a a (ii) ÿd (iii) D e e e 1 = ( ) ( 1) ( ) ( 1) ( ) e 1 e e 1 ( 1) e e e 1 1 1 ( ) (iv) D (e log 3) 1 1 1 e e e e e e = 1 ( 1) 1 1 [ e e ]( e e ) ( e e )( e e ) (iv) D (e log 3) = e (+1) log 3( + 1) e log 3 = e (+1) log 3( + 1) e (+1) log 3 + e (+1) log 3 e log 3 = e (+1) [log 3( + 1) log 3] + [e (+1) e ] log 3 1 ( ) e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1
FINITE DIFFERENCES 3 = e + log 3 ( 1 ) + e (e 1) log 3 3 = e 1 e log 1 ( e 1) log 3 Eample 3 Evaluate the following, interval of differencing being h: (i) D ( + sin ) (ii) D (sin cos 4) (iii) D cot a (iv) D Sol (i) D ( + sin ) = D + D sin = [( + h) ] + [sin ( + h) sin ] = h + h + cos h h sin (ii)ÿd(sin cos 4) = D ( 1 = h(h + ) + sin h cos h cos 4 sin ) sin = 1 D (sin 6 sin ) = 1 (D sin 6 D sin ) = 1 [{sin 6( + h) sin 6} {sin ( + h) sin }] = 1 [ cos (6 + 3h) sin 3h cos ( + h) sin h] = sin 3h cos 3( + h) sin h cos ( + h) (iii) D cot a = cot a +h cot a = cos sin a a h h cos a sin a (iv) ÿd sin h h sin a cos a cos a sin a = h sin a sin a = ( h) = sin ( h) sin h = sin ( a a ) sin a ( 1 a ) h h sin a sin a sin a sin a ( h) sin sin ( h) sin ( h) sin ( h) sin sin sin sin ( h) sin ( h) sin h = [( h) ] sin [sin ( h) sin ] sin ( h) sin = hh ( )sin sin h cos( h ) sin ( h) sin Eample 4 Evaluate the following, the interval of differencing being h: (i) D (cos ) (ii) D (ab ) (iii) D n a c+d (iv) D n cos (c + d) Sol (i) D cos = cos ( + h) cos = sin ( + h) sin h = sin h sin ( + h) e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1
4 A TEXTBOOK OF ENGINEERING MATHEMATICS ÿ D cos = D (D cos ) = D ( sin h sin ( + h)) = sin h D sin ( + h) = sin h [sin {( + h) + h} sin ( + h)] = sin h cos ( + h) sin h = h sin h cos ( + h) (ii) ÿd (a b ) = a D (b ) = a (b +h b ) = a b (b h 1) = a (b h 1) b ÿÿ ÿ ÿd (ab ) = D (D ab ) = D[a(b h 1) b ] = a (b h 1) Db = a (b h 1) b (b h 1) = a (b h 1) b (iii) ÿd a c+d = a c(+h)+d a c+d = a c+d (a ch 1) = (a ch 1) a c+d ÿd a c+d = D (D a c+d ) = D (a ch 1) a c+d = (a ch 1) D a c+d = (a ch 1) (a ch 1) a c+d = (a ch 1) a c+d D 3 a c+d = (a ch 1) 3 a c+d and so on Similarly, Generalising, D n a c+d = (a ch 1) n a c+d (iv) ÿd cos (c + d) = cos [c( + h) + d] cos (c + d) = sin ch ch c d sin = sin ch c d ch Q cos Q ' sin R cos R = sin ch c d ch Q cos Thus the first difference of cos (c + d) is obtained by multiplying by the constant factor sin ch and increasing the angle by ch Q ÿÿd cos (c + d) = D [D cos (c + d)] ch = D sin cos c d ch Q Similarly, D 3 cos (c + d) = 3 ch sin cos c d ch = sin cos c d ch % Q = sin ch sin ch cos c d ch Q ch = sin cos c d 3 ch Q ch Q n Generalising, D n ch cos (c + d) = sin cos c d n ch Q e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1
FINITE DIFFERENCES 5 1 DIFFERENCES OF FACTORIAL POLYNOMIAL If n is a positive integer, then the epression ( h) ( h) ( n 1h) involving n factors, beginning with and decreasing by h every time, is called a factorial polynomial of degree n and is denoted by (n) For eample, (1) =, () = ( h), (3) = ( h) ( h) D (n) = ( + h) (n) (n) = [( + h) () ( h) ( n h)] = ( h) ( n h) [( + h) ( n 1h)] = nh ( h) ( n h) = nh (n 1) Similarly, D (n) = D ( (n) ) = nh D (n-1) [( h) ( n h) ( n 1 h)] = nh (n 1) h (n ) = n(n 1) h (n ) ÿd 3 (n) = n(n 1) (n ) h 3 (n 3) D n (n) = n(n 1)(n ) 1 h n = n h n = constant D n+1 (n) = 0 Note If h = 1, D (n) = n (n 1) Þ Differencing is analogous to differentiation Þ The process of getting the function whose first differences are given is analogous to integration Eample 5 Epress the function f() = 3 + 3 5 + 4 and its successive differences in factorial notation Also obtain a function whose first difference is f() Sol We first epress f() in factorial notation Let f() = A 0 (3) + A 1 () + A (1) + A 3 1 3 5 4 = A 3 5 5 0 = A 4 9 = A 1 = A 0 [Eplanation Write the co-efficients of f() in the first row Divide f() by The remainder is 4 = A 3 and the quotient is + 3 5 Separate the remainder from the quotient by drawing a vertical line Divide + 3 5 by 1 For this, draw a vertical line to the left of the leading co-efficient and write 1 to its left In the quotient, the leading co-efficients is Multiply by 1 and add the product to 3, thus getting 5 Multiply 5 by 1 and add the product to 5, thus getting 0 The remainder is 0 = A and the quotient is + 5 Divide ( + 5) by For this, write to the left of the vertical line In the quotient, the leading co-efficient is Multiply by and add the product 4 to 5, thus getting 9 The remainder is 9 = A 1 and the quotient is This last quotient, which is a constant, is A 0 e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1
6 A TEXTBOOK OF ENGINEERING MATHEMATICS Thus A 3, A, A 1, A 0 are the successive remainders in the division of f() by, 1, ] \ f() = (3) + 9 () + 4 ; ÿÿdf() = 6 () + 9 (1) ÿÿd f() = 1 (1) + 9 ; D 3 f() = 1 Differences of higher order are zero Now, let F() be the function whose first difference is f() Then, ÿdf() = f() Þ F() = 1 % f() = 1 % [(3) + 9 () + 4] = ( 4) 4 + 9 (3) 3 + 4 (1) + c = 1 ( 1) ( ) ( 3) + 3( 1) ( ) + 4 + c = 1 [(3 6 + 11 6) + 6( 3 + ) + 8] + c = 1 (4 7 + 14) + c 13 BACKWARD DIFFERENCE OPERATOR Ñ The backward difference operator Ñ is defined by Ñy n = y n y n1 or Ñf(a) = f(a) f(a h) Thus Ñy 0 = y 0 y 1, Ñy 1 = y 1 y 0 Ñy = y y 1 etc ÿÿñ y 0 = Ñ(Ñy 0 ) = Ñ(y 0 y 1 ) = Ñy 0 Ñy 1 ÿñ y 1 = Ñy 1 Ñy 0, Ñ y = Ñy Ñy 1 etc The table showing the various backward differences is called backward difference table given below y Ñy Ñ y Ñ 3 y a 3h y 3 y y 3 = Ñy a h y Ñy 1 Ñy = Ñ y 1 y 1 y = Ñy 1 Ñ y 0 Ñ y 1 = Ñ 3 y 0 a h y 1 Ñy 0 Ñy 1 = Ñ y 0 y 0 y 1 = Ñy 0 a y 0 14 THE DISPLACEMENT (OR SHIFT) OPERATOR E The operator E increases the value of the argument by one interval If : a, a + h, a + h, and y 0 = f(a), y 1 = f(a + h), y = f(a + h), then Ef(a) = f(a + h) or Ey 0 = y 1 ; Ef(a + h) = f(a + h) or Ey 1 = y When the operator E is applied twice, the value of the argument increases by two intervals E y 0 = y, E y 1 = y 3, E y n = y n+ In general E r y n = y n+r, E r y n = y n r The operator E has the following properties : (i) Ecf() = cef() ; (ii) E[af() + bg()] = aef() + beg() (iii) E m [E n f()] = E m+n f() ; (iv) E and D are commutative, ie, EDf() = DEf() e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1
FINITE DIFFERENCES 7 15 (a) Relations Between D, Ñ and E (i) E º 1 + D and D º E 1 ÿÿdy n = y n+1 y n = Ey n y n = (E 1)y n Þ ÿ D º E 1 and E º 1 + D Note In general E n º (1 + D) n (ii) Ñ º 1 E 1 ÿñy n = y n y n 1 = y n E 1 y n = (1 E 1 )y n Þ Ñ º 1 E 1 (iii) Ñ º DE 1 ÿÿñy n = y n y = Dy n 1 n 1 = DE 1 y n ÿþ Ñ º DE 1 Eample 6 Prove that ÑE = EÑ = D = E 1 Sol ÑEy n = Ñy n+1 = y n+1 y n = Dy n EÑy n = E(y n y n 1 ) = y n+1 y n = Dy n (E 1)y n = y n+1 y n = Dy n Hence ÿÿ ÑE = EÑ = D = E 1 Eample 7 Evaluate (Ñ + D) ( + ), h = 1 Sol (Ñ + D) ( + ) = (1 E 1 + E 1) ( + ) = (E E 1 ) ( + ) = (E + E ) ( + ) = E ( + ) ( + ) + E ( + ) = [( + ) + ( + )] ( + ) + [( ) + ( )] = ( + 5 + 6) ( + ) + ( 3 + ) = 8 Eample 8 Eplain the difference between % E u() and % u() Eu() % (E 1) E E 1 Sol u ( ) E E $ # u ( ) u() E = (E + E 1 ) u() = u( + 1) u() + u( 1) % u ( ) = ( ) ( ) E 1 u ( E E 1) u ( ) u ( ) u ( 1) u ( ) Eu ( ) u ( 1) u ( 1) u ( 1) The difference is evident Eample 9 Prove that : % f() (i) D log f() = log 1 f() (ii) % E 3 = 6 interval of differencing being unity Sol (i) ÿd log f() = log f( + 1) log f() f( 1) = log = log f( ) = log E f( ) f( ) ( 1 %) f( ) ( ) ( ) ( ) log log 1 ( ) f % f ( ) % f f f f( ) e\l-kerala\5kch1-1 IInd 4-01-1 IIIrd 3-1-1 IVth 7-1-1