VED e\monish-k\tit-5kch IInd Kerala (Semester V)

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1 e\monish-k\tit-5kch IInd A TEXTBOOK OF ENGINEERING MATHEMATICS

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3 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 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 Phone : Fa : wwwlamipublicationscom info@lamipublicationscom Price : ` 1500 Only First Edition : 013 OFFICES India & Bangalore & Jalandhar & Chennai & Kolkata & Cochin , & Lucknow & Guwahati , & Mumbai , & Hyderabad & Ranchi UEM 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

5 CONTENTS Preface Syllabus (vii) (viii) Pages Module 1: Finite Differences 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 Newton-Gregory Formulae for Equal Intervals Lagrange s Formula for Unequal Intervals (a) Divided Differences 1 19 (b) Newton s Divided Difference Formula is 110 Numerical Differentiation Numerical Integration 9 11 (a) Newton-cote s Quadrature Formula (b) Trapezoidal Rule (n = 1) (c) Simpson s One-Third Rule (n = ) (d) Simpson s Three-Eighth Rule (n = 3) 31 Module : Z-Transforms 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 Introduction 56 3 Manipulation of Numeric Functions Generating Functions Properties of Generating Functions 60 ( v ) e\monish-k\tit-5kch IInd

6 ( vi ) 35 Recurrence Relations Linear Recurrence Relation with Constant Coefficients Homogeneous Solution Particular Solution Solution by the Method of Generating Functions Simultaneous Difference Equations 71 Module 4: Comple Integration Introduction 73 4 Function of a Comple Variable Limit of f (z) Continuity of f(z) Derivative of f(z) Analytic Function Necessary and Sufficient Conditions for f(z) to be Analytic Cauchy-Riemann Equations in Polar Coordinates Harmonic Functions Orthogonal System Application of Analytic Functions to Flow Problems Comple Integration Simply and Multiply Connected Regions Cauchy's Integral Theorem Cauchy s Integral Formula Series of Comple Terms Taylor s Series Laurent s Series Singular Points, Residues Residue Theorem Calculation of Residues Application of Residues to Evaluate Real Integrals 114 Module 5: Queuing Theory Introduction 13 5 Characteristic of Queuing Model Transient and Steady State of the System Waiting Time and Idle Time Kendall s Notation for Queuing Models Model 1 {(M/M/1): ( /FCFS)} Model {(M/M/1): (N/FCFS)} Queuing Formula Little s Formula Model 3 {(M/M/1): (N/FCFS)}: Single Server, Finite (or Limited) Queue Model 137 e\monish-k\tit-5kch IInd

7 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

8 SYLLABUS EN 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

9 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

10 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 ( ) (iv) D (e log 3) 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 IIIrd IVth 7-1-1

11 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 IIIrd IVth 7-1-1

12 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 IIIrd IVth 7-1-1

13 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() = 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 = A = 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 Separate the remainder from the quotient by drawing a vertical line Divide 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 IIIrd IVth 7-1-1

14 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) (3) (1) + c = 1 ( 1) ( ) ( 3) + 3( 1) ( ) c = 1 [( ) + 6( 3 + ) + 8] + c = 1 ( ) + 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 IIIrd IVth 7-1-1

15 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 ( + ) = [( + ) + ( + )] ( + ) + [( ) + ( )] = ( ) ( + ) + ( 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 IIIrd IVth 7-1-1

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