Engineering Mathematics

Transcription

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2 Elementary Engineering Mathematics 15th Edition

3 Also available by the same author : HIGHER ENGINEERING MATHEMATICS for B.E., B. Tech., M.E., M. Tech & equivalent Professional Exams. (43st edition) Contents : Algebra & Geometry : Solution of Equations ; Linear Algebra ; Solid Geometry. Calculus : Differentiation and Integration of functions with Applications ; Vector Calculus. Series : Infinite series & ourier series. Differential Equations : Solution of Ordinary and Partial differential equations with Applications ; Series solution and Special functions. Complex Analysis : Complex numbers and functions ; Analytic functions, Conformal mapping ; Complex integration and Calculus of Residues. Transforms : Laplace transforms ; ourier transforms & Z-transforms. Numerical Techniques : Empirical Laws & Curve-fitting ; Statistical methods ; Probability & Distributions ; Sampling & Inference, Numerical Solution of Equations ; inite differences & Interpolation ; Numerical Differentiation & Integration ; Difference equations ; Numerical Solution of Ordinary & Partial differential equations ; Linear Programming. Special Topics : Calculus of Variations ; Integral equations ; Discrete Mathematics ; Tensor Analysis. Objective Type of Questions. Index NUMERICAL METHODS IN ENGINEERING AND SCIENCE (with Programs in C & C ++ & MATLAB) for B.E., B. Tech., M.E., M.C.A., B. Sc. (Computer Science), M.Sc. (Physics/Maths) (10th Edition) Contents : Approximations and Errors in Computation Numerical Solution of Algebraic, Transcendental and Simultaneous Equations Matrix Inversion and Eigen-value Problems Empirical Laws and Curve itting inite Differences and Interpolation Numerical Differentiation and Integration Difference equations Numerical Solution of Ordinary and Partial differential equations Linear Programming Use of Computers in Numerical Methods Numerical techniques using C, C ++ & MATLAB

4 Elementary Engineering Mathematics for I & II Semesters of B. Tech. & Diploma Courses B.S. GREWAL, Ph.D. Professor of Applied Mathematics Principal Scientific Officer (Ex.) Defence Research & Development Organisation, New Delhi ormerly of : College of Military Engineering, Poona Delhi College of Engineering, Delhi J.S. GREWAL, M.I.E., I. Engr. (U.K.), M.I. Mar. E. (London) ifteenth Edition 015 KP KHANNA PUBLISHERS 4575/15, Onkar House, Ground loor Opp. Happy School, Darya Ganj, New Delhi Phones : ; Mobile : ax :

5 Published by : Romesh Chander Khanna & Vineet Khanna for KHANNA PUBLISHERS -B, Nath Market, Nai Sarak, Delhi (India) All Rights Reserved [This book or part thereof cannot be translated or reproduced in any form (except for review or criticism) without the written permission of the Authors and the Publishers.] ISBN No. : Price : ` Typesetting at : Goswami Printers, Delhi Printed at : Saras Graphics Pvt. Ltd., Rai.

6 Preface to the ifteenth Edition The book has been thoroughly revised and a number of sections have been rewritten. Some elementary topics have been included to meet the requirement of beginers in engineering studies deleting some sections of lesser utility. A variety of new illustrative examples and problems selected from various examination papers which sharpen ones skill to use various methods, have been added. The list of Objective Type of Questions appended to each chapter, has also been updated. It is hoped that the book in its revised form will serve a move useful purpose. The authors take this opportunity to thank fellow professors for their suggestions and patronage of the book. In particular, they are grateful to Prof. B.K. Yadav, Chouksey Engg. College, Bilaspur (C.G.) ; Dr. Jeevragi Phakirappa, RBYM Engg. College, Bellary (Kar.) ; Prof. Pankaj Kumar, Lovely Professional Univ., Phagwara (Pb.) ; Dr. Hemant Kumar Nashine, Disha Inst. of Management & Technology, Raipur (C.G.) ; Prof. Amarapu Ramesh Babu, A.N. Inst. of Science & Technology, Visakhapatnam ; Prof. Pankaj S. Gholap, Dr. D.Y. Patel School of Engg., Lohegaon, Pune; Prof. P.C. Pillai, NSS College of Engg.,(Ker.); Dr. A.P. Burnwal, R.I.T., Koderma (Jh. K.) and Dr. Saroj Panigrahi, J.P. Inst. of Engg. & Tech., Guna (MP). Suggestions for improvement of the text and intimation of misprints will be thankfully acknowledged. New Delhi B.S. GREWAL J.S. GREWAL

7 Contents Chap. Pages 1. Preliminaries Progressions Permutations and Combinations Logarithms Binomial Theorem Partial ractions Angle Trigonometric Ratios Sum and Difference ormulae Double Angle ormulae Products to Sums or Difference of t-ratios Sum or Difference to Product of t-ratios Sine ormula Cosine ormula Projection ormula Area of a Triangle Heights and Distances.... Solution of Equations & Curve itting Introduction Transformations of Equations Solution of Quadratic & Cubic Equations Curve itting, Method of Least Squares Objective Type of Questions Determinants Introduction Definition ; Expansion of a Determinant Properties of Determinants Rule for Multiplication of Determinants Solution of Linear Equations Cramer s Rule Matrices Definition ; Special Matrices Matrix Operations Related Matrices Rank of a Matrix Elementary Transformations of a Matrix Normal orm of a Matrix Solution of Linear System of Equations ( vii )

8 ( viii ) Chap. Pages 4.8. Consistency of Linear System of Equations Vectors ; Orthogonal Transformations Eigen Values Reduction to Diagonal orm Objective Type of Questions Complex Numbers Complex Numbers Geometric Representation of Complex Numbers De Moivre s Theorem Roots of a Complex Number To Expand sin n θ, cos nθ and tan nθ To Expand sin mθ, cos nθ or sin mθ cos nθ Exponential unctions of a Complex Variable Hyperbolic unctions Real and Imaginary Parts of Hyperbolic unctions Objective Type of Questions Analytical Plans Geometry Coordinates of a Point Straight of a Line ; Equations of a Line Circle Conics Parabola Ellipse Hyperbola Objective Type of Questions Analytical Solid Geometry Space Co-ordinates Direction Cosines ; Angle between Two Lines Projection of the Join of Two Points on a Line Equation of a Plane Perpendicular Distance of a Point from a Plane Equations of a Straight Line Conditions for a Line to lie in a Plane Conditions for Two Lines to Coplanar Shortest Distance Between Two Planes Sphere Equation of the Tangent Plane Cone Cylinder Ellipsoid, Hyperboliod, Paraboliod Cylindrical Coordinates Spherical Coordinates Objective Type of Questions

9 Chap. ( ix ) 8. Differentiations and its Applications unctions; Limit of a unction Differential Coefficient ; Standard Result Successive Differentiation ; Leibritz Theorem undamental Theorems and Expansions Indeterminate orms Tangents and Normals Curvature ; Evolute ; Envelope Increasing and Decreasing unctions ; Maxima & Minima Asymptotes Objective Types of Questions Integration and its Applications Integration ; Standard Results Integration by Substitution Integration by Parts Integration of Rational Algebraic ractions by Partial ractions z n n 9.5. Reduction ormulae for sin xdx cos xdx... 3 andz m n Pages 9.6. Reduction ormulae for zsin x cos x dx Definite Integrals Properties Curve Tracing Areas of Cartesian Curves Lengths of Curves Volumes of Revolution Surface Areas of Revolution Objective Type of Questions Partial Differention and its Applications unctions of Several Variables ; Partial Derivatives Homogeneous unctions ; Eulers Theorem Total Derivatives ; Change Variables Jacobians Taylor s Theorem for unctions of Two Variables Errors and Approximations Maxima and Minima of unctions of Two Variables Lagrange s Method of Undetermined Multipliers Objective Type of Questions Multiple Integrals and Beta, Gamma unctions Double Integrals Change of Order of Integration Double Integrals in Polar Co-ordinates Area Enclosed by Plane Curves

10 Chap. ( x ) Pages Triple Integrals Volumes of Solids Change of Variables Beta function ; Gamma unction Objective Type of Questions Vector Analysis Vectors Algebra: Resolution of Vectors Products of Two Vectors Scalar or Dot Product Right handed and Left Handed Systems Vector or Gross Product Physical Applications Scalar Product of Three Vectors Vector Product of Three Vectors Differentiation of Vectors Scalar and Vector Point unctions Del Applied to Scalar Point unctions : Gradient Del Applied to Vector Point unctions : Divengence, Curl Del Applied Twice to Point unctions Del Applied to Products of Point unctions Integration of Vectors Line Integral ; Conservative Vector ields Surface Integral Green s Theorem in the Plane Stoke s Theorem Volume Integral Gauss Divergence Theorem Objective Type of Questions Differential Equations of irst Order with Applications Definitions ; ormation of a Differential Equation Equations of the irst Order and irst Degree Variables Separable Homogeneous Equations Equations Reducible to Homogeneous orm Leibnitz s Linear Equations Bernoulli s Equation Exact Differential Equations Equations Reducible to Exact Equations Equations of the irst Order and Higher Degree Orthogonal Trajectories Physical Applications Simple Electric Circuits Newton s Law of Cooling Rate of Decay of Radio-active Materials

11 Chap. ( xi ) Pages Chemical Reactions and Solutions Objective Type of Questions Linear Differential Equations with Applications Definitions ; Complementary unction and Particular, Integral Procedure to Solve the equation Methods of Variation of Parameters Cauchy s Homogeneous Linear Equations ; Legendre s Linear Equations Simultaneous Linear Equations with Constant Coefficients Simple Harmonic Motion Oscillations of a Spring Oscillatory Electrical Circuits Objective Type of Questions Partial Differential Equations with Applications Introduction ; ormation of Partial Differential Equations Equations Solvable by Direct Integration Lagrange s Linear Equations Non-Linear Equations of irst Order Homogeneous Linear Equations with Constant Coefficients Method of Separation of Variables Partial Differential Equations of Engineering Vibrations of a Stretched String Wave Equation One-Dimensional Heat-low Laplace s Equation in Two Dimensions Objective Type of Questions Infinite Series Convergence, Divergence, Oscillation of a Series Comparison Test ; Integral Test D Alembert s Ratio Test Raabe s Tests and Lagarithmic Test Cauchy s Root Test Alternating Series Series of Positive and Negative Terms Power Series Procedure for Testing a Series for Convergence Objective Type of Questions ourier Series and Harmonic Analysis Euler s Coefficients unctions having Points of Discontinuity Change of Interval Even and Odd unctions Half-Range Series Typical Waveforms

12 ( xii ) Chap. Pages Practical Harmonic Analysis Objective Type of Questions Laplace Transforms with Applications Introduction ; Definition Laplace Transforms of Standard unctions Properties of Laplace Transforms Laplace Transforms of Periodic unctions Laplace Transforms of Derivatives Multiplication by t n ; Division by t Inverse Laplace Transforms Convolution Theorem Application to Differential Equations Unit Step unction Unit Impulse unction Objective Type of Questions Simple Numerical Methods Numerical Solution of Equations Bisection Method Method of alse Position Newton-Raphson Method inite Differences Differences of a Polynomial ; actorial Notation Shift Operator, Relation between Oprators To ind Missing Terms Newton s orward Interpolation ormula Lagrange s Interpolation ormula Trapezoidal Rule ; Simpson s Rule Elements of Statistics and Probability Classification of Data Measures of Central Tendency Measures of Dispersion Probability ; Difinition Addition Law of Probability Mulitiplication Law of Probability Introduction to Linear Programming Introduction ormulation of the Problem Graphical Method Some Exceptional Cases Appendix Useful Results Index 61 63

13 ( xiii ) Note : The references given alongside the problems pertain to the various Engineering Examinations of the various universities and professional bodies. The abbreviations used for some of these are given below : Andhra stands for Andhra University, Waltair Anna Anna University, Chennai Bhopal Rajiv Gandhi Technical University, Bhopal B.P.T.U. Biju Patnaik Technical University, Rourkela C.S.V.T.U. Swami Vivekanand Technical University, Chhatisgarh Coimbatore Bharathiyar University, Coimbatore Delhi Guru Gobind Singh Indraprastha University, Delhi D.T.U. Delhi Technical University J.N.T.U. Jawahar Lal Nehru Technological University, Hyderabad Kottayam Mahatama Gandhi Memorial University, Kottayam Kurukshetra National Institute of Technology, Kurukshetra Madurai Madurai Kamaraj University, Madurai Marathwada B.A.M. University, Aurangabad P.T.U. Punjab Technical University, Jalandhar Rohtak Maharishi Dayanand University, Rohtak Tirupati Sri Venkateswara University, Tirupati U.P.T.U. UP Technical University, Lucknow U.T.U. Uttarakhand Technical University, Dehradun V.T.U. Visveswaraiah Technological University, Belgaum W.B.T.U. West Bengal University of Technology, Kolkata

14 Preliminaries 1 Algebra 1.1. Progressions (1) Arithmetic Progression. Numbers a, a + d, a + d,... are said to be in arithmetic progression (A.P.), where a is its first term and d is the common difference. (i) Its nth term T n = a + n 1 d (ii) Sum of n terms S n = n (a + n 1d) Proof. We have S n = a + (a + d) + ( a + n d) + ( a + n 1 d) Reversing the order of terms S n = ( a + n 1 d) + ( a + n d) (a + d) + a Adding S n = ( a + n 1d) + ( a + n 1d) + + ( a + n 1d) + ( a + n 1d) i.e., S n = n( a + n 1d) or S n = n a n d (iii) Arithmetic mean between two numbers a and b (A.M.) = a+ b () Harmonic Progression. Numbers 1/a, 1/(a + d), 1/(a + d),... are said to be in Harmonic progression (H.P.) i.e. a sequence is said to be in H.P. if its reciprocals are in A.P. Its nth 1 term T n =. a+ ( n 1) d Example 1.1. ind the first term and the number of terms of an A.P. whose second term is 7.75, 31st term is 0.5 and last term is 6.5. Sol. If a is the first term and d is the common difference, then T = a + d = (i) and T 31 = a + 30d = (ii) Subtracting (i) from (ii), we get 9d = = 7.5 or d = 0.5 Then (i) gives a = 7.75 d = 7.75 ( 0.5) = 8 Also T n = a + n 1 d = 6.5 (given) or 8 + (n 1) ( 0.5) = 6.5 or n 1 = 14.5/( 0.5) = 58. Hence, n = 59. Example 1.. ind the sum of 35 terms of an A.P. whose third term is 1 and 6th term is 11. Sol. If a is the first term and d the common difference, then T 3 = a + d = 1 and T 6 = a + 5d = 11 1

15 ELEMENTARY ENGINEERING MATHEMATICS Solving these equations, we get d = 4 and a = 9 Hence, sum of 35 terms = 35 [ 9 + (35 1) ( 4)] = 065. (3) Geometric Progression. Numbers a, ar, ar,... are said to be in Geometric Progression (G.P.) where a is its first term and r is the common ratio. (i) Its nth term T n = ar n 1 (ii) Sum of n terms S n = a ( 1 r ) 1 r Proof. We have, S n = a + ar + ar ar n 1 Multiplying by r, we get rs n = ar + ar + ar ar n Subtracting, S n rs n = a ar n or S n = a ( 1 r ) 1 r a (iii) Sum to infinity S = 1 r Since r n 0 for r < 1 when n for r < 1, Lt n S n (i.e. S ) = n a. 1 r where r < 1 (iv) Geometric mean between two numbers a and b i.e. G.M. = ( ab) Since G.M. (= G) between a and b is order that a, G, b are in G.P. i.e. G a = b a or G = ( ab) Example 1.3. If the common ratio of a G.P. is 3, its last term is 486 and sum of these terms is 78, find its first term. Sol. Let a be the first term and n the number of terms in the given G.P. Then T n = ar n 1 = 486. or a(3) n 1 = (i) [ r = 3] Also S n = a r n n ( 1 ) a( 1 3 ) = 78 i.e., 1 r 1 3 = 78 or using (i) a 3 (a3 n 1 ) = 78 = 1456 or a 3 (486) = 1456 or a =. Example 1.4. ind two numbers whose sum is 10 and the ratio between arithmetic mean and geometric mean is 1.5. Sol. Let the numbers be a and b so that a + b = 10...(i) n Also A.M. = 1 (a + b) = 60 ; G.M. = ab. A.M. G.M. = 60 = 1.5 (given) i.e. ab = 60/1.5 = 48. ab or ab = 304, from (i), a + 304/a = 10 or a 10a = 0 a = [( 10) 4 304] = 1 (10 + 7) = 96. (4) Natural numbers. 1,, 3, 4,... are called natural numbers and Σ, pronounced as sigma, denotes the summation sign. (i) n (= Σ n) = nn ( + 1)

16 PRELIMINARIES 3 (ii) n (= Σ n ) = nn ( + 1)( n+ 1) 6 L NM O QP (iii) n 3 (= Σ n 3 nn ( + 1) ) =. Example 1.5. ind the sum of the series (i) to n terms (ii) to n terms Sol. (i) nth term of the given series = (nth terms of 3, 5, 7...) (nth term of 5, 7, 9,...) i.e. T n = (3 + n )(5 + n ) = (n + 1) (n + 3) = 4n + 8n + 3 S n = Σ (4n + 8n + 3) = 4Σn + 8Σn + 3n = 4. nn + 8. nn 6 + 3n = n 3 [(n + 3n + 1) + 1(n + 1) + 9] = n 3 (4n + 18n + 3). (ii) Here T n = (nth term of 1,, 3...) (nth term of 3, 5, 7...) (nth term of 6, 9, 1,...) = n(n + 1)(3n + 3) = 6n 3 + 9n + 3n S n = 6 Σn 3 + 9Σn + 3Σn = 6 = 3 (n3 + 4n + 5n + 3) L NM O L NM nn ( + 1) nn ( + 1)( n+ 1) nn ( ) + QP QP Example 1.6. ind the sum of the series (i) (ii) to n terms. Sol. (i) We have S = = ( ) ( ) = (Σn ) n = 5 (Σn ) n = 4 ( + )( + ) ( + )( + ) = = = (ii) Let S = T n 1 + T n Also S = T n 1 + T n Subtracting, we get 0 = (T n T n 1 ) T n or T n = (T n T n 1 ) = 3 + [ (n 1) terms] = 3 + = 1 (3n n + 4) L NM O L NM n 1 { 4 ( ) + ( n 1 13 ) } S n = 1 [3Σn Σn + 4Σ 1] = 1 3 nn ( + 1)( n+ 1) nn ( + 1) + 4n 6 = n 4 (n + 3n + 1 n 1 + 8) = n (n + n + 4). L NM O QP O QP O QP

17 4 ELEMENTARY ENGINEERING MATHEMATICS 1.. Permutations and Combinations The process of arranging certain objects in a particular order is called a permutation and the process of selecting the objects (without the older of occurrence) is called a combination. or instance the permutations of out of 3 things a, b, c are AB, BA, BC, CB, CA, CB (i.e. 6) while the combinations (groups) of out of 3 things A, B, C are AB, BC, CA (i.e. 3). Permutations of n things taken r at a time is n n! P r = ( n r)! Combinations of n things taken r at a time is n n! C r = r!( n r)! Also n C n r = n C r ; n C 0 = n C n = 1 Example 1.7. (a) If n C 6 = n C 8, find n C 3. (b) If 4 P = k 4 C, find k. Sol. (a) We have n C 6 = n C 8 = n C n 8 i.e. 6 = n 8 or n = 14. Now n C 3 = 14 14! C 3 = = = 364 3!(14 3)! 3..1 (b) 4 P = 4!! = 1, 4 4! C =!! = 6 4 P = k 4 C gives 1 = k. 6 i.e. k = Logarithms (1) Definition. If a and x are two positive numbers and p is any other number such that a p = x, then p is called the logarithm of x to the base a and is written as log a x. or practical purposes, base is taken 10 and logarithms to base 10 is called common logarithms and is written as log 10 x. Logarithm to the base e is called natural logarithm and is simply written as log x. () Basic Laws: (i) log (mn) = log m + log n, (ii) log (m/n) = log m log n; (iii) log m n = n log m. (3) Particular cases: (i) log a 1 = 0, (ii) log a 0 = (a > 1), (iii) log a a = 1. Problems ind the sum of all odd numbers between 100 and 00.. If the first term of an A.P. is and last term is 14. ind the common difference if the sum of the series is The second term of an A.P. is and seventh term is. ind the sum of its first 35 terms. 4. ind the sum of the G.P. series: (i), 1/, 1/8,... to 1 terms (ii) ind the first term of a G.P. whose common ratio is 4/5 and the sum to infinity is 80/9. 6. ind two positive numbers whose difference is 1 and whose A.M. exceeds the G.M. by. 7. ind the sum to n terms the series: (i) (ii) ind the sum of the following series: (i) (ii) ind the sum to n terms of the series: (i) (ii)

18 PRELIMINARIES If n C r = n C r 1 and n P r = n P r + 1 find the values of n and r. 11. If n C r / n C r + 1 = 1// and n C r + 1 / n C r + = /3 find n and r Binomial Theorem (1) When n is a positive integer (a + x) n = n C 0 a n x 0 + n C 1 a n 1 x 1 + n C a n x n C x a n r x r n C n a 0 x n When a = 1, (1 + x) n = n C 0 + n C 1 x + n C x n C r x r n C n x n Expansion of (1 + x n ) has n + 1 terms. () Binomial coefficients. Coefficients of terms equidistant from the beginning and the end are equal and are known as the Binomial coefficients. i.e., n C 0 = n C n, n C 1 = n C n 1, n C = n C n... [ n C r = n C n r ] Also sum of the binomial coefficients i.e., n C 0 + n C 1 + n C n C n = n (3) Middle term in Binomial expansion If n is even, then the middle term is If n is odd, then the middle terms are n + 1 n + I K J th term 1I K J th and Example 1.8. Using binomial theorem, expand (i) (x 1/x) 4 (ii) (1 + x 3x ) Sol. (i) x x 1I K J 4 = 4 C 0 (x) x I K J n + 3 th terms. I K J + 4 C 1 (x) 3 1 I 1 x K J + 4 C (x) 1 I x K J + 4 C 3 (x) 1 1 I 3 x K J + 4 C 4 (x) 0 1 I 4 x K J = 1(16x 4 ) + 4(8x 3 ) ( 1/x) (4x ) (1/x ) (x) ( 1/x3 ) + 1(1/x 4 ) = 16x 4 3x 4 1/x + 1/x 4 (ii) (1 + x 3x ) 5 = (1 + y) 5 where y = x 3x = C 1 y + 5 C y + 5 C 3 y C 4 y C 5 y 5 = (x 3x ) (x 3x ) (x 3x ) (x 3x ) 4 + (x 3x ) 5 = x + 5x 40x 3 190x 4 + 9x x 6 360x 7 675x x 9 43x 10 Example 1.9. Evaluate (3 + ) 5 (3 ) 5. Sol. We have (a + x) n (a x) n = [ n C 1 a n 1 x + n C 3 a n 3 x ] Putting a = 3 and x =, we get (3 + ) 5 (3 ) 5 = [ 5 C 1 (3) 4 ( ) C 3 (3) ( ) C 5 (3) 0 ( ) 5 ] = [5(81) + 10(9)( ) + 5(4 )] = 110 Example Write the middle terms in the following expansions (i) (x/ + 7y) 10 (ii) (3x x /6) 9 Sol. Here n = 10 is an even number, therefore term I K J th term i.e., 6th term is the middle

19 6 ELEMENTARY ENGINEERING MATHEMATICS Hence, middle term = T 6 = 10 C 5 (x/) 10 5 (7y) ! x = (7) 5 y 5 = x 5 y 5. 5!( 10 5)! 3 (ii) Here n = 9, therefore th and th i.e. 5th and 6th are two middle terms. T 5 = 9 C 4 (3x) 9 4 ( x /6) 4 = ( ) 4 x 1 = 3.65 x 1 ( 6) Example ind the term independent of x in the expansion of Sol. Let T r + 1 be the term independent of x. I K J H G I K J 6 r r 6 r Now T r + 1 = C r I x 3x K J H G I K J = C r 3 or T r + 1 to be independent of x, we must have 3r 6 = 0 i.e., r =. Hence, the term of independent of x = T 3 = 6 C Expand the following expressions: Problems 1. I KJ (x 3y) a b b a Evaluate the following: 4 1I 3 K J H G I K J = x x I K J x + x 5. ( 5 + ) 5 ( 5 ) 6 6. ( + 3 ) 6 ( 3 ) 6 ind the middle terms in the following expansions: 1I K J x I 9 x KJ x 7. a b ind the term independent of x in the expansion of (i) 3x 1 3x I KJ 9 (ii) 4 x I K J x 4 I K J 4 10 I K J 11 3 x + x 3 I KJ x 3 6 I x K J. r. x 3r Partial ractions (1) We are familiar with the process of combining two or more fractions into a single fraction. or instance, x + x 1 + x + 1 =...(i) ( x 1) ( x + )( x 1) Now we shall introduce the reader to the reverse process of breaking a given fraction into a sum of simple fractions. These simple fractions are called partial fractions of the given

20 PRELIMINARIES 7 fraction e.g., in (i), the fractions on the left hand side are the partial fractions of the fraction on the right. A fraction in which the degree of the numerator is less than that of the denominator is called a proper fraction. () Procedure to resolve a given fraction into partial fractions. Step I. Check whether the given fraction is a proper fraction. If the degree of the numerator is higher or equal to that of the denominator, divide the numerator by the denominator till the remainder is of lower degree than the denominator. Step II. actorize the denominator into real factors. These will be either linear or quadratic, and some factors repeated. Step III. Resolve the proper fraction into sum of partial fractions such that (i) to a non-repeated linear factor x a in the denominator corresponds a partial fraction of the form A/(x a); (ii) to a repeated linear factor (x a) r in the denominator corresponds the sum of r partial A1 A A3 Ar fractions of the form + + r x a ( x a) ( x a) ; ( x a) (iii) to a non-repeated quadratic factor (x + ax + b) in the denominator, corresponds a Ax + B partial fraction of the form ; x + ax + b (iv) to a repeated quadratic factor (x + ax + b) r in the denominator, corresponds the sum of Ax 1 + B1 Ax + B Ax r + Br r partial fractions of the form r x + ax + b ( x + ax + b) ( x + ax + b) Then we have to determine the unknown constants A, A 1, B 1, B 1 etc. as follows: Step IV. To obtain the partial fraction corresponding to the non-repeated linear factor x a in the denominator, put x = a everywhere in the given fraction except in the factor x a itself. In all other cases, equate the given fraction to a sum of suitable partial fractions in accordance with (i) and (iv) above, having found the partial fractions corresponding to the non-repeated linear factors by the above rule. Then multiply both sides by the denominator of the given fraction and equate the coefficients of like powers of x or substitute convenient numerical values of x on both sides. inally solve the simplest of the resulting equations to find the unknown constants. x + 3x 18 Example 1.1. Resolve into partial fractions ( x + x )( x 1). Sol. Since it is a proper fraction, we write it as x + 3x 18 ( x 1)( x + )( x 1) = A B C + + x 1 x + x 1 To find A, put x = 1 in the L.H.S. except in x 1 itself. Then A = ( 1+ )( 1) =. To find B, put x = in the L.H.S. except in factor x + itself. Then B = = = 4. ( 1)( 4 1) 15

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