19485352 Advanced Mathematics

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No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to [email protected]

PUBLISHING FOR ONE WORLD

NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002

Visit us at www.newagepublishers.com ISBN (13) : 978-81-224-2719-6

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Preface

We feel happy and honoured while presenting this book “Advanced Mathematics” for engi-neering students studying in B. Tech. IV Semester (EE and EC Branch) of Rajasthan Technical University and all Indian Universities. In this book we have presented the subject matter in very sim-ple and precise manner. The treatment of the subject is systematic and the exposition easily under-standable. All standard examples have been included and their model solutions have also been given.

This book falls into five units:

In first and second unit we have discussed the Numerical Analysis. The unit I deals with Finite Difference—Forward, Backward and Central difference, Newton’s formula for Forward and Backward differences, Interpolation, Stirling’s formula and Lagrange’s interpolation formula. Solution of non-linear equations in one variable by Newton-Raphson method, Simultaneous algebraic equation by Gauss and Regula-Falsi method, Solution of simultaneous equations by Gauss elimination and Gauss Seidel methods, Fitting of curves (straight line and parabola of second degree) by method of least squares are also discussed.

In unit II, we have discussed Numerical differentiation, Numerical Integration, Trapezoidal rule, Simpson’s one-third and three-eighth rules. Numerical solution of ordinary differential equations of first order, Picard’s method, Euler’s and modified Euler’s methods. Miline’s method and Runga-Kutta fourth order method, Simple linear difference equations with constant coefficients are also discussed in the unit.

Unit III deals with the special functions, Bessel’s functions of first and second kind, Simple recurrence relations, Orthogonal property of Bessel’s transformation and generating functions. Legendre’s function of first kind, Simple recurrence relations, Orthogonal property and generating function are also discussed.

In unit IV, the basic principles of probability theory is given in order to prepare the background for its application to various fields. Baye’s theorem with simple applications, Expected value, Theo-retical probability distributions—Binomial, Poisson and Normal distributions are discussed.

Unit V deals with Lines of regression, concept of simple Co-relation and Rank correlation. Z-transforms, its inverse, simple properties and application to difference equations are also discussed.

We are grateful to New Age International (P) Limited, Publishers and the editorial department for their commitment and encouragement in bringing out this book within a short span of period.

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Acknowledgement

We are thankful to Prof. L. K. Maheshwari, Vice-Chancellor, Prof. R. K. Mittal, Deputy Director (Administration), Prof. G. Raghurama, Deputy Director (Academic) of Birla Institute of Technology & Science (BITS), Pilani for their encouragement and all over support in completing the book. The authors are also highly thankful to Dr. P. S. Bhatnagar, Director, BK Birla Institute of Engineering & Technology (BKBIET), Pilani for his motivation, time to time support and keen interest in the project. Dr. S. R. Singh Pundir, Reader, D. N. College Meerut, deserves for special thanks.Thanks are also due to Mr. Anil and Mr. Rahul of BKBIET for providing necessary help during the project.

We also place our thanks on record to all those who have directly or indirectly helped us in completion of the project.

At the last but not in the least we are very much indebted to our family members without whom it was not possible for us to complete this project in time. Thanks are also due to M/s NEW AGE INTERNATIONAL (P) LTD. PUBLISHERS and their editorial department.

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Preface v

Acknowledgement vii

UNIT I : NUMERICAL ANALYSIS-I

CHAPTER 1 CALCULUS OF FINITE DIFFERENCES . . . 3–20

1.1. Finite Differences ... 3

1.2. Forward Differences ... 3

1.3. Backward Differences ... 4

1.4. Central Differences ... 5

1.5. Shift Operator E ... 6

1.6. Relations Between the Operators ... 6

1.7. Fundamental Theorem of the Difference Calculus ... 7

1.8. Factorial Function ... 8

1.9. To Show that Δn_x(n)_{= n ! h}n _and Δn+1_x(n)_{= 0 ... 8}

1.10. To Show that f(a + nh) = f(a) + n_C 1 Δf(a) + nC2 Δ2 f(a) + ... + nCn Δn f(a) ... 9

Solved Examples ... 10

Exercise 1.1 ... 18

Answers ... 20

CHAPTER 2 INTERPOLATION . . . 21–41 2.1. To Find One Missing Term ... 21

2.2. Newton-Gregory’s Formula for Forward Interpolation with Equal Intervals ... 22

2.3. Newton-Gregory’s Formula for Backward Interpolation with Equal Intervals ... 23

2.4. Lagrange’s Interpolation Formula for Unequal Intervals ... 24

2.5. Stirling’s Difference Formula ... 25

Exercise 2.1 ... 39

Answers ... 41

*CHAPTER 3 SOLUTION OF LINEAR SIMULTANEOUS EQUATIONS . . . 42–51 3.1. Linear Equations ... 42

3.2. Gauss Elimination Method ... 43

3.3. Gauss-Seidel Method ... 44

UNIT II : NUMERICAL ANALYSIS-II

CHAPTER 1 NUMERICAL DIFFERENTIATION . . . 75–89 1.1. Derivatives Using Forward Difference Formula ... 75

1.2. Derivatives Using Backward Difference Formula ... 76

1.3. Derivatives Using Stirling Difference Formula ... 77

1.4. Derivatives Using Newton’s Divided Difference Formula ... 78

Exercise 1.1 ... 88

Answers ... 89

CHAPTER 2 NUMERICAL INTEGRATION . . . 90–101 2.1. A General Quadrature Formula for Equally Spaced Arguments ... 90

2.2. The Trapezoidal Rule ... 91

2.3. Simpson’s One-third Rule ... 92

2.4. Simpson’s Three-Eighth Rule ... 92

Exercise 2.1 ... 100

Answers ... 101 *Not for EC branch students.

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CHAPTER 3 ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER . . . 102–118

3.1. Euler’s Method ... 102

3.2. Euler’s Modified Method ... 103

3.3. Picard’s Method of Successive Approximation ... 103

3.4. Runge-Kutta Method ... 104

3.5. Milne’s Series Method ... 105

Exercise 3.1 ... 117

Answers ... 118

*CHAPTER 4 DIFFERENCE EQUATIONS . . . 119–135 4.1. Difference Equations ... 119

4.2. Order of Difference Equation ... 119

4.3. Degree of Difference Equation ... 120

4.4. Solution of Difference Equation ... 120

4.5. Formation of Difference Equation ... 120

4.6. Linear Difference Equation ... 121

4.7. Homogeneous Linear Difference Equation with Constant Coefficient ... 122

4.8. Non-Homogeneous Linear Difference Equation with Constant Coefficient ... 123

Exercise 4.1 ... 134

Answers ... 134

UNIT III : SPECIAL FUNCTION

CHAPTER 1 BESSEL’S FUNCTIONS . . . 139–155 1.1. Bessel’s Equation ... 139

1.2. Solution of the Bessel’s Functions ... 139

1.3. General Solution of Bessel’s Equation ... 141

1.4. Integration of Bessel’s Equations in Series for N = 0 ... 142

1.5. Generating Function for J_N(X) ... 143

1.6. Recurrence Relations for J_N(X) ... 144

1.7. Orthogonal Property of Bessel’s Functions ... 146

Exercise 1.1 ... 154

CHAPTER 2 LEGENDRE’S FUNCTIONS . . . 156–176 2.1. Introduction ... 156

2.2. Solution of Legendre’s Equation ... 156

2.3. General Solution of Legendre’s Equation ... 159

2.4. Generating Function of Legendre’s Polynomial P_N(X) ... 159 *Not for EC branch students.

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2.5. Orthogonal Properties of Legendre’s Polynomials ... 160

2.6. Laplace’s First Integral for P_N(X) ... 162

2.7. Laplace’s Second Integral for P_N(X) ... 163

2.8. Rodrigue’s Formula ... 164

2.9. Recurrence Relations ... 166

Exercise 2.1 ... 175

Answer ... 176

UNIT IV : STATISTICS AND PROBABILITY

CHAPTER 1 THEORY OF PROBABILITY . . . 179–209 1.1. Terminology and Notations ... 179

1.2. Definitions ... 181

1.3. Elementary Theorems on Probability ... 182

1.4. Addition Theorem of Probability ... 184

Exercise 1.1 ... 187

Answers ... 188

1.5. Independent Events ... 188

1.6. Conditional Probability ... 189

1.7. Multiplicative Theory of Probability or Theorem of Compound Probability ... 189

Exercise 1.2 ... 194

Answers ... 194

1.8. Theorem of Total Probability ... 195

1.9. Baye’s Theorem ... 195 Solved Examples ... 196 Exercise 1.3 ... 201 Answers ... 201 1.10. Binomial Theorem ... 202 1.11. Multinomial Theorem ... 202 1.12. Random Variable ... 202 1.13. Expected Value ... 203 Solved Examples ... 203 Exercise 1.4 ... 208 Answers ... 209

CHAPTER 2 THEORETICAL DISTRIBUTIONS . . . 210–247 2.1. Terminology and Notations ... 210

2.2. Binomial Distribution ... 211

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Exercise 2.1 ... 220

Answers ... 222

2.4. Poisson’s Distribution ... 222

2.5. Constants of Poisson distribution ... 224

2.6. Recurrence Formula for Poisson Distribution ... 226

2.7. Mode of the Poisson Distribution ... 226

Exercise 2.2 ... 232

Answers ... 234

2.8. Normal Distribution ... 234

2.9. Constants of Normal Distribution ... 234

2.10. Moment About the Mean M ... 236

2.11. Area Under the Normal Curve ... 237

2.12. Properties of the Normal Distribution and Normal Curve ... 238

Exercise 2.3 ... 247

Answers ... 247

CHAPTER 3 CORRELATION AND REGRESSION . . . 248–280 3.1. Frequency Distribution ... 248

3.2. Bivariate Frequency Distribution ... 248

3.3. Correlation ... 249

3.4. Positive Correlation ... 249

3.5. Negative Correlation ... 249

3.6. Linear or Non-Linear Correlation ... 249

3.7. Coefficient of Correlation ... 250

3.8. Measurement of Correlation ... 250

3.9. Karl Pearson’s Coefficients of Correlation ... 250

3.10. Probable Error ... 254

3.11. Correlation Coefficient for a Bivariate Frequency Distribution ... 255

3.12. Rank Correlation or Spearman’s Coefficient of Rank Correlation ... 260

3.13. Rank Correlation Coefficient for Repeated Ranks ... 261

3.14. Scatter Diagram or Dot Diagram ... 261

Exercise 3.1 ... 267

Answers ... 269

3.15. Regression ... 269

3.16. Line of Regression ... 270

3.17. Standard Error of Estimate ... 271 *Not for EC branch students.

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Exercise 3.2 ... 278

Answers ... 280

UNIT V : CALCULUS OF VARIATIONS AND TRANSFORMS

*CHAPTER 1 CALCULUS OF VARIATIONS . . . 283–302 1.1. Functional ... 283

1.2. Euler’s Equation ... 283

1.3. Equivalent Forms of Euler’s Equation ... 285

1.4. Solution of Euler’s Equations ... 286

1.5. Strong and Weak Variations ... 287

1.6. Isoperimetric Problems ... 288

1.7. Variational Problems Involving Several Dependent Variables ... 288

1.8. Functionals involving Second Order Derivatives ... 289

Exercise 1.1 ... 290

Answers ... 302

*CHAPTER 2 Z-TRANSFORM . . . 303–323 2.1. Z-Transform ... 303

2.2. Linearity Properties ... 303

2.3. Change of Scale Property or Damping Rule ... 304

2.4. Some Standard Z-Transforms ... 304

2.5. Shifting U_N to the Right ... 306

2.6. Shifting U_N to the Left ... 307

2.7. Multiplication by n ... 307

2.8. Division by n ... 308

2.9. Initial Value Theorem ... 308

2.10. Final Value Theorem ... 309

Solved Examples ... 309 Exercise 2.1 ... 315 Answers ... 316 2.11. Inverse Z-Transform ... 317 2.12. Convolution Theorem ... 317 Solved Examples ... 317 Exercise 2.2 ... 320 Answers ... 320

2.13. Solution of Difference Equation by Z-Transform ... 321

Exercise 2.3 ... 323

Answers ... 323 *Not for EC branch students.

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In this unit, we shall discuss finite differences, forward, backward and central differences. Newton’s forward and backward differences interpolation formula, Stirling’s formula, Lagrange’s interpolation formula.

The unit is divided into five chapters:

The chapter first deals with the forward, backward, central differences and relation between them, fundamental theorem of the difference calculus, factorial notation and examples.

Chapter second deals with interpolation formula of Newton’s forward, Newton’s backward, Stirling’s for equally width of arguments, and Lagrange’s formula for unequally width of arguments.

Chapter third deals with solution of linear simultaneous equation by Gauss elimination and Gauss-Seidel method.

Chapter fourth deals with solution of algebraic and transcendental equation by Regula-Falsi and Newton-Raphson method.

Chapter fifth deals with fitting of curves for straight line and parabola of second degree by method of least squares.

U

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NUMERICAL ANAL

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3 INTRODUCTION

Numerical analysis has great importance in the field of Engineering, Science and Technology etc. In numerical analysis, we get the result in numerical form by computing methods of given data. The base of numerical analysis is calculus of finite difference which deals with the changes in the dependent variable due to changes in the independent variable.

1.1. FINITE DIFFERENCES

Suppose the function y = f(x) has the values y₀, y₁, y₂, ... y_n for the equally spaced values x = x₀, x₀+ h, x₀ + 2h, ... x₀ + nh. If y = f (x) be any function then the value of the independent variable ‘x’ is called argument and corresponding value of dependent variable y is called entry. To determine the value of y and dy

dx for some intermediate values of x, is based on the principle of finite difference. Which requires

three types of differences.

1.2. FORWARD DIFFERENCES

The differences y₁– y₀, y₂– y₁, ... y_n – y_{n – 1} are called the first forward differences of the function y = f (x) and we denote these difference by ∆ y₀, ∆ y₁ ... ∆ y_n, respectively, where ∆ is called the descending or forward difference operator.

In general, the first forward differences is defined by

∆y_x = y_{x + 1} – y_x

The differences of the first forward differences are called the second forward differences and denoted by ∆2_y

0, ∆2y, etc.

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Therefore, we have ∆2_y 0 = ∆ [y1 – y0] = ∆y₁ – ∆y₀ = (y₂ – y₁) – (y₁ – y₀) = y₂ – 2y₁ + y₀ Similarly, ∆2_y 1 = ∆[y2 – y1] = (y3 – y2) – (y2 – y1) = y3 – 2y2 + y1 In general, we have ∆2_y x = ∆yx + 1 – ∆yx

Again, the differences of second forward differences are called third forward differences and denoted by ∆3_y 0, ∆ 3_y 1 etc. Therefore, we have ∆3_y 0 = ∆2y1 – ∆2y0 = (y₃ – 2y₂ + y₁) – (y₂ – 2y₁ + y₀) = y₃ – 3y₂ + 3y₁ – y₀ and so on In general, the nth forward difference is given by

∆n _y

x = ∆n –1 yx + 1 – ∆n –1 yx

Forward Difference Table

Argument Entry First Differences Second Differences Third Differences Fourth Differences

x y = f(x) ∆y ∆2_y _∆3_y _∆4_y x₀ y₀ y₁– y₀ = ∆y₀ x₀ + h y₁ ∆y₁ – ∆y₀ = ∆2_y 0 y₂– y₁ = ∆y₁ ∆2_y 1 – ∆2 y0 = ∆3 y0 x₀ + 2h y₂ ∆y₂ – ∆y₁ = ∆2_y 1 ∆ 3_y 1 – ∆ 3_y 0 = ∆ 4 _y 0 y₃– y₂ = ∆y₂ ∆2_y 2 – ∆ 2 _y 1 = ∆ 3 _y 1 x₀ + 3h y₃ ∆y₃ – ∆y₂ = ∆2_y 2 y₄– y₃ = ∆y₃ x₀ + 4h y₄ 1.3. BACKWARD DIFFERENCES

The differences y₁ – y₀, y₂ – y₁, ..., y_n – y_{n – 1} are called the first backward differences of the function y = f (x) and we denote these differences by ∇y₁, ∇y₂, ..., ∇y_n , respectively, where ∇ is called the ascending or backward differences operator.

In general, the first backward difference is defined by

∇y_x = y_x – y_{x – 1}

The differences of the first backward differences are called second backward differences and denoted by ∇2_y

2, ∇2y3, etc.

Therefore ∇2_y

2 = ∇(y2 – y1) = ∇y2 – ∇y1

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In general, we have ∇2_y

x = ∇yx – ∇yx – 1

Again the differences of second backward differences are called third backward differences and denoted by ∇3_y

3, ∇3y4 etc.

Therefore, we have ∇3_y

x = ∇2yx – ∇2yx – 1

In general, the nth backward differences is given by

∇n_y

x = ∇n–1yx – ∇n–1yx –1

Backward Differences Table

Argument Entry First Diff. Second Diff. Third Diff. Fourth Diff.

x y = f(x) ∇y ∇2_y _∇3_y _∇4_y x₀ y₀ y₁– y₀ = ∇y₁ x₀ + h y₁ ∇y₂ – ∇y₁ = ∇2_y 2 y₂– y₁ = ∇y₂ ∇2_y 3 – ∇ 2_y 2 = ∇ 3 _y 3 x₀ + 2h y₂ ∇y₃ – ∇y₂ = ∇2 _y 3 ∇ 3_y 4 – ∇ 3_y 3 = ∇ 4_y 4 y₃– y₂ = ∇y₃ ∇2_y 4 – ∇2 y3 = ∇3 y4 ∇y4 – ∇y3 = ∇ 2_y 4 x₀ + 3h y₃ y₄– y₃ = ∆y₄ x₀ + 4h y₄ 1.4. CENTRAL DIFFERENCES

The differences y₁ – y₀ = δ y_1/2, y₂ – y₁ = δ y_3/2,..., y_n – y_{n –1} = δy_n–1/2 are called central differences and

δ is called central difference operator.

Similarly δy_3/2 – δy_1/2 =δ2_y 1 δy5/2 – δy3/2 =δ2y2 and δ2_y 2 – δ 2_y 1 =δ 3_y 3/2 and so on.

The Central Difference Table

Argument Entry First Diff. Second Diff. Third Diff. Fourth Diff.

x y = f(x) δy δ2_y _δ3_y _δ4_y x₀ y₀ δy1/2 x₁ y₁ δ2_y 1 δy3/2 δ 3_y 3/2 x₂ y₂ δ2_y 2 δ4y2 δy5/2 δ 2_y 5/2 x₃ y₃ δ2_y 3 δy7/2 x₄ y₄

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1.5. SHIFT OPERATOR E

The shift (increment) operator E is defined as Ey_x = y_{x + h} E2_y x = yx + 2h M M En_y x = yx + nh

The inverse operator E–1_{is defined as} E–1 _y

x = yx+ (– h) = yx – h.

1.6. RELATIONS BETWEEN THE OPERATORS

(i) We know that ∆y_x = y_x+h– y_x = Ey_x– y_x

∆y_x = (E – 1) y_x

∴ ∆ = E – 1

or E = ∆ + 1

(ii) We have ∇y_x= y_x – y_{x – h} = y_x – E–1_y x ∇y_x = (1 – E–1_{) y}

x

∴ ∇ = 1 – E–1

or E–1_{= 1 –}_∇

(iii) We have E∇y_x = E ( y_x– y_{x –h}) = Ey_x – Ey_x–h = y_x+h– y_x

E∇y_x = ∆y_x

E ∇ = ∆ ...(i)

Again ∇Ey_x = ∇y_x+h = y_{x +h} – y_x

∇Eyx = ∆yx

or ∇E = ∆ ...(ii)

By (i) and (ii) E∇ = ∇E = ∆

(iv) Since δ y_x = y_{x + h/2} – y_{x – h/2} = E1/2y_x – E–1/2 y_x δ y_x= (E1/2 – E–1/2) y_x Thus δ = E1/2_{– E}–1/2 (v) We have ∆y_x = y_{x + h} – y_x = Ey_x – y_x = (E – 1) y_x = (E1/2_{– E}–1/2_{) E}1/2 _y x

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∆yx = δ E1/2 yx

⇒ ∆ = δ E1/2

(vi) We have E f(x) = f(x + h)

= f (x) + hf′(x) + h

2

2 ! f″ (x) + ... [By Taylor’s Theorem] = f (x) + hD f(x) + h 2 2 !D 2_{f(x) + ....} = 1 2 2 2 + +

F

HG

hD h D

I

KJ

! ... f (x) E f (x) = ehD_f(x) E = ehD (vii) We have (1 + ∆) (1 – ∇) f(x) = (1 + ∆) (f(x) – ∇ f(x)) = (1 + ∆) [f(x) – (f(x) – f(x – h))] = (1 + ∆) f(x – h) = E f(x – h) (1 + ∆) (1 – ∇) f(x) = f(x) Thus (1 + ∆) (1 – ∇) = 1

1.7. FUNDAMENTAL THEOREM OF THE DIFFERENCE CALCULUS

If f(x) be a polynomial of nth degree in x, then the nth difference of f(x) is constant and ∆n+1_{f(x) = 0.}

Proof. Consider the nth degree polynomial

f (x) = A₀ + A₁x + A₂x2_{+ ... + A} nx

n

where A₀, A₁, A₂, ... A_n are constants and n is a positive integer. By the definition, we have

∆f(x) = f(x + h) – f(x) = [A₀ + A₁ (x + h) + A₂(x + h)2_{+ ... A} n (x + h) n_{] – [A} 0 + A1x + A2x 2_{+ ... + A} n x n_] = A₁ h + A₂ [(x + h)2_{– x}2_{] + A} 3 [(x + h) 3_{– x}3_{] + ... + A} n [(x + h) n_{– x}n_] = A₁ h + A₂ [x2₊2_C 1x h + h 2_{– x}2_{] + A} 3[x 3₊3_C 1 x 2_{h +}3_C 2 x h 2_{+ h}3_{– x}3_{] + ...} + A_n [xn₊n_C 1 x n–1_{h +}n_C 2 x n – 2_h2_{+ ... +}n_C n h n_{– x}n₎ ∆f(x) = B₁ + B₂x + B₃x2_{+ ... + B} n–1 x n–2_{+ n A} nh x n– 1 _...(1.1)

where B₁, B₂ ... B_{n – 1} are constants

By (1.1), we see that the first difference of a polynomial of degree n is again a polynomial of degree (n – 1).

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Again ∆2_{f(x) =}∆f (x + h) – ∆f(x) = B₁ + B₂ (x + h) + B₃ (x + h)2 + ... + B_n–1 (x + h)n –2 + nA_nh (x + h)n–1 – [B₁ + B₂x + B₃ x2_{+ ... + B} n – 1 xn – 2 + n Anh xn–1] = B₂ h + B₃ [(x + h)2 – x2] + B₄ [(x + h)3 – x3] + ... + B_n–1[(x + h)n–2_{– x}n–2_{] + n A} nh [(x + h)n–1 – xn –1] = B₂ h+ B₃ [ x2 + 2C₁x h + h2 –x2] + B₄ [x3 + 3C₁ x2 h + 3_C 2 x h2 + h3 – x3] + ... + Bn – 1 [xn – 2 + n–2C1 xn–3 h + n–2_C 2 xn –3 h2 + ... + n–2Cn–2 hn – 2 – xn – 2 ] + nA_nh [xn–1 ₊ n–1_C 1 xn – 2 h + n–1C2 xn–3 h2 + ... + n–1Cn–1 hn – xn–1] = C₂ + C₃ x + C₄ x2_{+ ... + C} n – 1 xn–3 + n(n – 1) h2 An xn –2 ...(1.2)

where C₂, C₃, ..., C_{n – 1} are constants

By (1.2) we see that the second difference of a polynomial of degree n is again a polynomial of degree (n – 2)

Proceeding in the same way, we will get a zero degree polynomial for the nth difference i.e., ∆n_{f(x) = n(n – 1) (n – 2) .... 1 h}n_a

n xn – n = n ! hn an

Thus, nth difference is constant

Now ∆n+1 _{f (x) =}_∆[∆n_{f(x)] =}_{∆ [n ! h}n _a

n] = 0 [Q ∆C = 0]

1.8. FACTORIAL FUNCTION

A product of the form x (x – h) (x – 2h) ... (x – (n – 1) h) is called factorial function and denoted by x(n)_.

Thus x(n) _{= x(x – h) (x – 2h) ... (x – (n – 1) h).}

If interval of differencing is unity.

Then x(n)_{= x(x – 1) (x – 2) (x – 3) ... (x – (n – 1))}

1.9. TO SHOW THAT ∆n _x(n)_{= n ! h}n _{AND ∆}n+1_x(n)_{= 0}

Proof. By the definition of ∆ we have

∆x(n)_{= (x + h)}(n)_{– x}(n) = (x + h) (x + h – h) (x + h – 2h) ... (x + h – (n – 1)h) – x(x – h) (x – 2h) ... (x – (n – 1)h) = (x + h)x (x – h) (x – 2h) ... (x – (n – 2)h) – x(x – h) (x – 2h) ... (x – (n – 2)h) (x – (n – 1)h) = x (x – h) (x – 2h) ... (x – (n – 2)h) ((x + h) – (x – (n – 1)h) = x(n – 1)_{nh = nh x}(n –1)

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Again ∆2_x(n)₌∆∆x(n) ₌∆[ nhx(n – 1)_{] = nh}∆ x(n – 1)_{= nh [(x + h)}(n – 1)_{– x}(n – 1)_] = nh [(x + h) (x + h – h) (x + h – 2h) ... (x + h – (n – 2)h) – x (x – h) (x – 2h) ... (x – (n – 2) h] = nh [(x + h) x (x – h) (x – 2h) ... (x – (n – 3)h) – x (x – h) (x – 2h) ... (x – (n – 3) h) (x – (n – 2) h)] = nh x(x – h) (x – 2h) ... (x – (n – 3)h) [x + h – (x – (n – 2) h) = nh x(n – 2)_{(n – 1) h} = n (n – 1)h2_x(n – 2)

Proceeding in the same way, we get

∆n_x(n)_{= n (n – 1) (n – 2) ... 1 h}n_x(n–n)_{= n ! h}n

Again ∆n + 1_x(n)₌_∆(∆n_xn_{) =}_{∆(n ! h}n_{) = 0}

1.10. TO SHOW THAT f(a + nh) = f(a) + n_C

1 ∆f(a) + nC2 ∆2 f(a) + ... + nCn ∆n f(a)

We shall prove this by the method of mathematical induction

We have ∆f(a) = f(a + h) – f (a)

∴ f (a + h) = ∆f(a) + f (a) = f(a) + ∆f(a) it is true for n = 1 Again ∆f (a + h) = f(a + 2h) – f (a + h)

∴ f(a + 2h) = ∆f(a + h) + f(a + h)

= ∆ [∆ f(a) + f(a) ] + ∆f(a) + f(a) = f(a) + 2∆f(a) + ∆2_f(a)

f (a + 2h) = f(a) + 2C₁∆ f(a) + ∆2 f(a) It is true for n = 2

Similarly f (a + 3h) = ∆f (a + 2h) + f (a + 2h)

= ∆ [f(a) + 2∆f(a) + ∆2_{f(a)] + [f(a) + 2}∆f(a) + ∆2_{f (a)]}

= f(a) + 3∆ f(a) + 3∆2 f(a) + ∆3f(a) f(a + 3h) = f(a) + 3_C

1 ∆ f(a) + 3C2∆2 f(a) + ∆3 f(a)

It is true for n = 3

Now Assume that it is true for n = k then f(a + kh) = f(a) + k_C

1∆f(a) + kC2∆2 f(a) + ... + kCk∆k f(a)

Now we shall show that this result is true for n = k + 1 Now f(a + (k + 1) h) = f(a + kh) + ∆f (a + kh)

= [f(a) + k_C

1∆f (a) + kC2∆2 f(a) + ... + kCk∆k f(a)]

+ ∆ [ f(a) + k_C

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= f(a) + [k_C

1 + 1] ∆f(a) + [kC2+ kC1] ∆2 f(a)

+ [kC₃+ kC₂] ∆3 f(a) + ... + ∆k +1f(a). f (a + (k + 1) h) = f (a) + k +1_C

1∆f(a) + k +1C2∆2f (a) + k +1C3∆3f (a) + ... + ∆k + 1f (a)

Hence the result is true for n = k + 1 [Q kC_r + kC_r_{+ 1}= k +1C_{r + 1}] So by the principle of mathematical induction it is true for all n, we have

f (a + nh) = f (a) + nC₁∆f (a) + nC₂∆2 f(a) + ... + nC_n∆n f (a)

SOLVED EXAMPLES

Example 1. Prove that ∆3_{≡ E}3_{– 3E}2_{+ 3E – 1.}

Solution. By the definition we have

∆f(x) = f(x + h) – f(x), E f (x) = f(x + h) and En_{f(x) = f(x + nh)} _...(i) ∴ ∆2_{f(x) =}_{∆[f(x + h) – f(x)] = f(x + 2h) – 2f(x + h) + f(x)} and ∆3_{f(x) =}_{∆[f(x + 2h) – 2f (x + h) + f(x)]} = f(x + 3h) – 3f (x + 2h) + 3f(x + h) – f (x) = E3_{f(x) – 3E}2_{f (x) + 3E f (x) – f(x)} _{[Using (i)]} ∆3_{f(x) = (E}3_{– 3E}2_{+ 3E – 1) f(x)} or ∆3_{≡ E}3_{– 3E}2_{+ 3E – 1.} _{Hence proved.} Example 2. Evaluate

(i)∆ cosh (a + bx) (ii) ∆ tan–1 ax Solution. (i) By the definition of ∆, we have

∆ cosh (a + bx) = cosh (a + b(x + h)) – cosh (a + bx)

= 2 sinh a+b x( +h)+ +a bx 2 sinh a+b x( +h)− −a bx 2 = 2 sinh

FH

a+bx+bh

IK

2 sinh bh 2

FH IK

Ans. (ii) By the definition of ∆, we have

∆ tan–1_{ax = tan}–1_{a(x + h) – tan}–1_ax

= tan–1 a x h ax a x h ax ( ) ( ) + − + + 1 = tan–1 ah a x a xh 1+ 2 2 + 2

F

HG

I

KJ

. Ans. Example 3. Evaluate:

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Solution. (i) ∆2_[3ex_{] = 3}∆2_[ex_{] = 3}∆ . ∆ [ex_{] = 3}∆ [ex+ 1_{– e}x_]

= 3[ex + 2 – ex + 1 – ex + 1 + ex] = 3[e2_{– 2e + 1]e}x_{= 3(e – 1)}2_ex

(ii) By the definition of ∆, we have

∆[sin (ax + b)] = sin (a (x + h) + b) – sin (ax + b)

= 2 cos

FH

a x( +h)+ +b ax+b

IK

2 × sin a x( +h)+ −b ax+b

FH

2

IK

= 2 cos

FH

ax+ +b ah

IK

2 sin ah 2 . Example 4. Evaluate ∆ (3x + e2x_{+ sin x).}

Solution. By the definition of ∆, we have

∆(3x + e2x_{+ sin x) = [3(x + h) + e}2(x + h)_{+ sin (x + h)] – [3x + e}2x _{+ sin x]}

= 3h + e2x [(e2h – 1) + 2 cos

FH

x+ +h x

IK

x+ −h x

2 sin 2

= 3h + e2x_(e2h_{– 1) + 2 cos}

FH IK FH IK

_x+h h

2 sin 2 .

Example 5. Evaluate ∆ [e2x log3x].

Solution. We have ∆ [f(x) g(x)] = f(x + h) ∆g(x) + g(x) ∆f(x)

Take f(x) = e2x, g(x) = log 3x

then ∆ (e2x_{log 3x) = e}2(x + h) _{∆ log 3x + log 3x ∆ e}2x

= e2(x + h)_{[ log 3(x + h) – log 3x] + log 3x . (e}2(x + h) _{– e}2x₎

= e2x e h x e x h h 2 2 1 1 3 log

FH IK

+ +( − ) log

L

NM

O

QP

. Example 6. Evaluate ∆n_(eax + b_{), h = 1.} Solution. We have ∆ f(x) = f(x + 1) – f(x) ∴ ∆eax + b_{= e}a (x + 1) + b_{– e}ax + b_{= e}ax + b _(ea_{– 1)} Again ∆2 eax + b = ∆(∆eax + b) = ∆ [eax + b (ea – 1)] = (ea_{– 1)}∆eax + b_{= (e}a_{– 1) e}ax + b _(ea_{– 1)} = (ea – 1)2 eax + b Proceeding in the same way, we get

∆n_(eax+b_{) = (e}a_{– 1)}n_eax + b_.

Example 7. Show that ∆r_y

x = ∇ ryx+r .

Solution. ∇r_y

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= E E r −

FH IK

1 y_x+r = (E – 1)r_E–r_y x +r = (E – 1)r_y x [Q ∆ = E – 1] = ∆r_y x. Hence proved.

Example 8. Prove that ∆ log f(x) = log 1 f(x) f(x)

+

L

NM

∆

O

QP

. Solution. By the definition of ∆, we have

∆ log f(x) = log f(x + h) – log f(x)

= log f x h f x E f x f x ( ) ( ) log ( ) ( ) + ₌

L

NM

O

QP

= log ( ) ( ) ( ) 1+

L

NM

f x∆ f x

O

QP

[Q E = 1 + ∆] = log f x f x f x ( ) ( ) ( ) +

L

NM

∆

O

QP

= log

L

1+

NM

∆f xf x

O

QP

( ) ( ) Hence proved. Example 9. Evaluate ∆ 2 E

F

HG

I

KJ

x3 , h = 1. Solution. We have ∆ 2 E

F

HG

I

KJ

x3₌ (E I) E −

L

NM

O

QP

2 x3₌ E I E E 2 2 + −

L

NM

O

QP

x3 = [E + E–1_{– 2I] x}3_{= (x + 1)}3_{+ (x – 1)}3_{– 2x}3 = x3 + 3x2 + 3x + 1 + x3 – 3x2 + 3x – 1– 2x3 = 6x. Example 10. Evaluate ∆3_{(1 – x) (1 – 2x) (1 – 3x), h = 1.} Solution. Here f(x) = (1 – x) (1 – 2x) (1 – 3x) = 1 – 6x + 11x2_{– 6x}3

which is a polynomial of degree 3 in x

∴ ∆3_{f(x) =}_∆3_{(1 – 6x + 11x}2_{– 6x}3₎

= 0 – 6.0 + 11.0 – 6 . 3! = – 36. Example 11. If

x 1 2 3 4 5

y 2 5 10 20 30

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Solution. First, we form forward difference table x y = f(x) ∆y ∆2_y _∆3_y _∆4_y 1 2 3 2 5 2 5 3 3 10 5 – 8 10 – 5 4 20 0 10 5 30

By above we observe that ∆4_{y(1) = – 8.}

Example 12. Represent the function f(x) = x4_{– 12x}3_{+ 42x}2 _{– 30x + 9 and its successive}

differ-ences into factorial notation.

Solution. Let x4_{– 12x}3_{+ 42x}2_{– 30x + 9 = Ax}(4)_{+ Bx}(3)_{+ Cx}(2) _{+ Dx}(1)_{+ E}

= Ax (x – 1) (x – 2) (x – 3) + Bx(x – 1) (x – 2) + Cx (x – 1) + Dx + E ...(i) where A, B, C, D and E are constants . Now, we will find the value of these constants

Putting x = 0 in (i) we get, E = 9

Again putting x = 1 in (i), we get 1 – 12 + 42 – 30 + 9 = D + E

⇒ D = 1

Putting x = 2 in (i), we get 16 – 12 × 8 + 42 × 4 – 30 × 2 + 9 = 2C + 2D + E

⇒ C = 13

Putting x = 3 in (i), we get 81 – 12 × 27 + 42 × 9 – 30 × 3 + 9 = 6B + 6C + 3D + E

⇒ B = – 6

Equating the coefficient of x4_{on both sides, we get A = 1. Putting the values of A, B, C, D, E in}

(i), we get f(x) = x4_{– 12x}3_{+ 42x}2_{– 30x + 9 = x}(4)_{– 6x}(3)_{+ 13x}(2)_{+ x}(1)_{+ 9} Now ∆f(x) = 4x(3)_{– 18x}(2)_{+ 26x}(1)_{+ 1} ∆2_{f(x) = 12x}(2)_{– 36x}(1)_{+ 26} ∆3_{f(x) = 24x}(1)_{– 36} ∆4_{f(x) = 24} ∆5_{f(x) = 0} Aliter: Let f(x) = x4_{– 12x}3_{+ 42x}2_{– 30x + 9} = Ax(4)_{+ Bx}(3)_{+ Cx}(2)_{+ Dx}(1)_{+ E}

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Now, we obtain the values of A, B, C, D, E by synthetic division. The procedure is as follows: 1 1 – 12 42 – 30 9 = E 0 1 – 11 31 2 1 – 11 31 1 = D 0 2 – 18 3 1 – 9 13 = C 0 3 4 1 – 6 = B 0 1 = A Hence f(x) = x(4)_{– 6x}(3)_{+ 13x}(2)_{+ x}(1)_{+ 9.}

Example 13. Find the function whose first difference is 9x2_{+ 11x + 5.}

Solution. Let f(x) be the required function then ∆f(x) = 9x2_{+ 11x + 5}

First, we change ∆ f(x) in factorial notation

Let f(x) = 9x2_{+ 11x + 5 = Ax}(2)_{+ Bx}(1)_{+ C}

= Ax(x – 1) + Bx + C ...(i)

Putting x = 0 we get C = 5

Putting x = 1 we get 9 + 11 + 5 = B + C ⇒ C = 20 On comparing like term in (i) we get A = 9

On putting in (i), we get

∆f (x) = 9x(2)_{+ 20x}(1)_{+ 5} Integrating, we get f(x) = 9 3 20 2 3 2 x( ) x( )

+ + 5x + C₁where C₁ is constant of Integration = 3x(x – 1) (x – 2) + 10x(x – 1) + 5x + C₁

f(x) = 3x3_{+ x}2_{+ x + C.}

Example 14. Find the function whose first difference is eax + b_.

Solution. Let f(x) be the required function

Then ∆f(x) = eax +b ...(i)

Let us consider f(x) = Aeax + b

so that ∆f(x) = ∆[Aeax + b] = A∆ eax + b

= A [ea(x + 1) + b _{– e}ax + b_]

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On comparing (i) and (ii) we get A = 1 1 ea − ∴ f(x) = e e ax b a + − 1.

Example 15. What is the lowest degree polynomial which takes the following values

x 0 1 2 3 4 5

f(x) 0 3 8 15 24 35

Solution. First we prepare the forward difference table

x f(x) ∆f(x) ∆2_f(x) _∆3_f(x) 0 0 3 1 3 2 5 0 2 8 2 7 0 3 15 2 9 0 4 24 2 11 5 35

We know that f (a + nh) = f(a) + n_C

1∆f (a) + n_C 2∆ 2_{f (a) +}n_C 3∆ 3_{f (a) + ...} Putting a = 0, h = 1, n = x, we get f(x) = f(0) + x_C 1∆f (0) + x_C 2∆ 2_{f(0) +}x_C 3∆ 3_{f(0) + ...} _...(1)

Putting the value of f(0), ∆f(0), ∆2_{f(0) and}_∆3_{f(0) in (1) from Difference table, we get} f(x) = 0 + x . 3 + x x( ) x x x ! ( ) ( ) ! −1 ₊ − − 2 2 1 2 3 . 0 + 0 = 3x + x (x– 1) = x2_{+ 2x.}

Example 16. A second degree polynomial passes through the points (0, 1) (1, 3), (2, 7), and (3, 13). Find the polynomial.

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Solution. First we prepare the forward difference table x f(x) ∆f(x) ∆2_f(x) _∆3_f(x) 0 1 2 1 3 2 4 0 2 7 2 6 3 13 We know that f(a + nh) = f(a) + n_C

1∆f(a) + nC2 ∆2 f(a) + nC3∆3 f(a) ... nCn∆n f(a)

Putting a = 0, h = 1, n = x, we get f(x) = f(0) + x_C

1 ∆ f(0) + xC2∆2 f(0) + xC3∆3 f(0) + ... ...(1)

Putting the values of f(0), ∆f(0), ∆2_{f(0) and}_∆3_{f(0) from difference table in (1) we get} f(x) = 1 + x . 2 + x x( ) x x x ! ( ) ( ) ! −1 ₊ − − 2 2 1 2 3 . 0 = 1 + 2x + x (x – 1) = x2_{+ x + 1.}

Example 17. Given u₀ = 1, u₁ = 11, u₂ = 21, u₃ = 28 and u₄ = 29 find ∆4_u

0 without forming difference table. Solution. We have ∆4_u 0 = (E – I)4 u0 = (E4 – 4C1 E3 + 4C2 E2 – 4C3 E + I) u0 = E4_u 0 – 4E3u0 + 6E2u0 – 4E u0 + u0 = u₄ – 4u₃ + 6u₂ – 4u₁ + u₀ = 29 – 4 × 28 + 6 × 21 – 4 × 11 + 1 = 29 – 112 + 126 – 44 + 1 = 0. Example 18. Prove that

(i) f(4) = f(3) + ∆f(2) + ∆2_{f(1) +}_∆3_f(1)

(ii) f(4) = f(0) + 4∆f(0) + 6∆2_{f(–1) + 10}_∆3_{f(– 1)}

as for as third difference. Solution. (i) We have

∆f (3) = f(4) – f(3) or f(4) = f(3) + ∆f(3) = f(3) + ∆[f(2) + ∆f(2)] [Q ∆f(2) = f(3) – f(2)] = f(3) + ∆f(2) + ∆2_f(2) = f(3) + ∆f(2) + ∆2[f(1) + ∆f(1)] [Q ∆f(1) = f(2) – f(1)] f(4) = f(3) + ∆f(2) + ∆2 f(1) + ∆3 f(1) Hence proved.

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(ii) We have

f(4) = f(– 1 + 5) = E5 f(– 1) = (1 + ∆)5 f(– 1) = (1 + 5_C

1∆ + 5C2∆2 + 5C3∆3 + 5C4∆4 + 5C5∆5) f(–1)

= f (– 1) + 5∆f(– 1) + 10 ∆2 f(– 1) + 10 ∆3 f(– 1) taking upto 3rd difference = [f(– 1) + ∆f(– 1)] + 4 [∆f(– 1) + ∆2_{f(– 1)] + 6}∆2_{f (– 1) + 10}∆3_{f(– 1)}

= [f (– 1) + ∆f(– 1)] + 4∆[f (– 1) + ∆f(– 1)] + 6∆2 f(– 1) + 10∆3f(– 1)

= f(0) + 4∆f(0) + 6∆2_{f(– 1) + 10}_∆3_{f(– 1)} _[_{Q ∆f(–1) = f(0) – f(– 1)]} ⇒ f(4) = f(0) + 4∆f(0) + 6∆2_{f(– 1) + 10}_∆3_{f(– 1)} _{Hence proved.}

Example 19. Prove that u_x = u_{x –1} + ∆u_x–2 + ∆2_u

x–3 + ... + ∆n–1 ux–n + ∆n ux–n. Solution. Consider u_x – ∆n_u x–n = ux – ∆ n_E–n_u x =

F

HG

1−

I

KJ

∆n n E ux = E E n n n −

F

HG

∆

I

KJ

u_x = 1 1 2 3 2 1 E E E E E u E n n n n n x ( ) [ ... ] ( ) − + + + + − − − − − ∆ ∆ ∆ ∆ ∆ (Q E = 1 + ∆) = (E–1₊∆E–2₊∆2_E–3_{+ ...}+ ∆n – 1 _E–n_)u x u_x – ∆n u_{x – n} = u_{x – 1} + ∆u_x–2 + ∆2 u_x–3 + ... + ∆n – 1 u_{x – n} ∴ u_x = u_{x– 1} + ∆u_{x– 2}+ ∆2_u x– 3 + ... + ∆n–1 ux – n + ∆n ux – n. Hence proved.

Example 20. Prove that u₁x + u₂ x2_{+ u}

3x3 + ... = x 1−x u1 + x 1 x 2 2 ( − ) ∆u1 + x 1 x 3 3 ( − ) ∆ 2 _u 1 + ... and 0 < x < 1. Solution. R.H.S. = x x 1− u1 + x x 2 2 1 ( − ) ∆u1 + x x 3 3 1 ( − ) ∆ 2 _u 1 + ... = x xu x x E u x x E u 1 1 1 1 1 1 2 2 1 3 3 2 1 − +( − ) ( − ) +( − ) ( − ) +... = x x x x x x 1 1 1 2 2 3 3 − − − + −

F

HG

( ) ( ) ...

I

KJ

u1 + x x x x 2 2 3 3 1 2 1 ( − ) −( − ) +...

F

HG

I

KJ

Eu₁ + x x 3 3 1 ( − ) +...

F

HG

I

KJ

E2u₁ + ... = x x x x u x x x x u 1 1 1 1 1 1 1 1 2 2 2 2 −

F

HG

+ −

I

KJ

+ −

FH IK

+ − − − ( ) ( ) + x x x x u 3 3 3 3 1 1 1 ( − )

FH IK

+ − +... − = u₁x + u₂ x2_{+ u} 3 x3 + ... .

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Example 21. Prove that u₀ + u₁ + u₂ + ...+ u_n = n +1_C 1 u0 + n + 1_C 2 ∆ u0 + n + 1_C 3 ∆ 2_u 0+ ... + ∆ n_u 0. Solution. We have u₀ + u₁ + u₂ + ... + u_n= u₀ + Eu₀ + E2_u 0 + ... + En u0 = (1 + E + E2_{+ ... + E}n_{) u} 0 = E E n₊ − − 1 1 1 u0 [Sum of n term in G.P.] = (1 ) 1 1 +_∆ + − ∆ n u₀ = 1 ∆[1 + n + 1C1∆ + n + 1_C 2 ∆ 2₊n + 1_C 3 ∆ 3_{+ ... +} n + 1_C n + 1 ∆ n + 1_{– 1] u} 0 = 1 ∆ [n+1 C1 ∆u0 + n+1_C 2∆ 2_u 0 + n+1_C 3 ∆ 3_u 0 + ... + n+1_C n+1 ∆ n+1_u 0] = n+1_C 1u0 + n+1C2 ∆ u0 + n+1C3 ∆2 u0 + ... + ∆n u0 = R.H.S. Example 22. Prove that u₀ + n_C

1 u1x + nC2 u2x2 + ... + unxn = (1 + x)n _u 0 + nC1 (1 + x)n–1 x ∆u0 + nC2 (1 + x)n–2x2∆2 u0 + ... + xn∆n u0 Solution. R.H.S. (1 + x)n _u 0 + nC1 (1 + x)n–1 x∆u0 + nC2 (1 + x)n–2 x2∆2 u0 + ... + xn∆n u0 = ((1 + x) + x∆)n_u 0 = (1 + x ( 1 + ∆)) n_u 0 = (1 + xE) n _u 0 = (1 + n_C 1 xE + n_C 2 x 2_E2₊n_C 3 x 3_E3_{+ ... + x}n _En_{) u} 0 = u₀ + n_C 1 u1 x + n_C 2 u2 x 2₊n_C 3 u3 x 3_{+ ... + x}n _u n = L.H.S. Example 23. ∆n_u x = ux+n – n_C 1ux+n–1 + n_C 2 ux + n–2 – ...+ (– 1) n_u x. Solution. R.H.S. u_{x +n} – n_C 1 ux + n – 1 + n_C 2 ux + n–2 – ... + (– 1) n_u x = (En_–n_C 1 E n –1₊n_C 2 E n–2_{– ... + (– 1)}n_{) u} x = (E – 1)nu_x = ∆ny_x = L.H.S.

EXERCISE 1.1

1. Prove that ∆2_{≡ E}2_{– 2E + 1.}

2. Prove that if f(x) and g(x) are the function of x then

(i)∆[f(x) + g(x)] = ∆f(x) + ∆g(x) (ii)∆[af(x)] = a ∆f(x) (iii)∆[f(x) g(x)] = f(x) ∆ g(x) + g(x + 1) f(x) = f(x + 1) ∆ g(x) + g(x) ∆f(x) (iv)∆ f x g x g x f x f x g x g x g x h ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

L

NM

O

QP

= − + ∆ ∆ _. 3. Evaluate

(i)∆[sinh (a + bx)] (ii)∆[tan ax] (iii)∆ [cot 2x_] _(iv)_{∆ (x + cos x)}

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(v)∆ (x2_{+ e}x_{+ 2)} _(vi)_{∆ [log x]}

(vii)∆ [eax_{log bx]} _(viii)_∆ x

x 2 2 cos

L

NM

O

QP

4. Evaluate

(i)∆2_{cos 2x} _(ii)_∆2_(abcx₎

(iii) ∆ 2 E

F

HG

I

KJ

x3 _(iv)_∆2 5 12 5 6 2 x x x + + +

L

NM

O

QP

. 5. Evaluate (i)∆n 1 x

FH IK

(ii)∆n_{[sin (ax + b)]}

(iii)∆6 _{(ax – 1) (bx}2_{– 1) (cx}3_{– 1)} _(iv)_∆n_[axn_{+ bx}n – 1_]

6. Prove that ex₌ ∆ 2 E

F

HG

I

KJ

ex_.Ee e x x

∆2 ; the interval of differencing being h. 7. Prove that ∇n_y

x = ∆nyx–n.

8. Evaluate

(i) (2∆2₊_{∆ – 1) (x}2_{+ 2x + 1)} _{(ii) (}_{∆ + 1) (2∆ – 1) (x}2_{+ 2x +1)}

(iii) (E + 2) (E + 1) (2x + h_{+ x)} _{(iv) (E}2_{+ 3E + 2) 2}x+h_{+ x}

9. Write down the polynomial of lowest degree which satisfies the following set of number 0, 7, 26, 63, 124, 215, 342, 511.

10. A third degree polynomial passes through the points (0, – 1) (1, 1) (2, 1) and (3, – 2). Find the polynomial.

11. Construct a forward difference table for

x 0 5 10 15 20 25

f(x) 7 11 14 18 24 32

12. If f(0) = – 3, f(1) = 6, f(2) = 8, f(3) = 12 prepare forward difference table.

13. Given f(0) = 3, f(1) = 12, f(2) = 81, f(3) = 200, f (4) = 100 and f(5) = 8. Form a difference table and find

∆5_f(0).

14. Given u₀ = 3, u₁ = 12, u₂ = 81, u₃ = 200, u₄ = 100, u₅ = 8 find ∆5 _u

0 without forming difference table.

15. If f(0) = – 3, f(1) = 6, f(2) = 8, f (3) = 12 and the third difference being constant, find f(6).

16. Represent the function f(x) = 2x3_{– 3x}2_{+ 3x – 10 and its successive differences into factorial notation.}

17. Find the function whose first difference is x3_{+ 3x}2_{+ 5x + 12.}

18. Obtain the function whose first difference is:

(i) ex _{(ii) x(x – 1)} _{(iii) a}

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19. Prove that u₀ + x_C 1∆u1 + xC2 ∆2 u2 + .... = ux + xC1∆2 ux–1 + xC2∆4 u x–2+ ... 20. Prove that u_x+n = u_n+ x_C 1 ∆ux–1 + x+1_C 2∆ 2 _u x–2 + x+2_C 3∆ 3 _u x–3 + ... 21. Prove that ∆n_u x–n = ux – n_C 1 ux–1 + n_C 2 ux–2 – n_C 3 ux–3 + ... 22. Prove that u₀ + u x1 u x2 u x 2 3 3 1! + 2! + 3! +... = e x _u _{x u} x _u x _u 0 0 2 2 0 3 3 0 2 3 + + + +

L

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∆ ∆ ∆

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! ! ...

ANSWERS

3. (i) 2 sinh b 2

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cosh

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2 (ii) sin cos cos ( ) a ax a x+ 1

(iii) – cosec 2x+1 _{(iv) h – 2 sin}

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_x₊h

2 sin h 2

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(v) 2hx + h2_{+ e}x_(eh_{– 1)} _{(vi) log} 1+

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(viii) h x h x x h x h x h x

( ) cos sin sin ( )

cos ( ) cos 2 2 2 2 2 2 2 2 + + + +

4. (i) – 4 sin2_{h cos (2x + 2h)} _{(ii) (b}c_{– 1)}2_abcx

(iii) 6x (iv) 4 2 3 4 6 3 4 5 (x+ ) (x+ ) (x+ )+(x+ ) (x+ ) (x+ ) 5. (i) ( ) ! ( ) ( ) ... ( ) − + + + 1 1 2 n n x x x x n (ii) 2 2 2 sin a sin ax b n a n

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(iii) 720 abc (iv) a (n !).

8. (i) 5h2_{+ 2hx + 2h – x}2_{– 2x – 1} _{(ii) 5h}2_{+ 2hx + 2h – x}2_{– 2x – 1} (iii) h (iv) h 9. x3_{+ 3x}2_{+ 3x} _{10. –}1 6 (x 3_{+ 3x}2_{– 16x + 6)} 13. 755 14. 755 15. 126 16. 2x(3)_{+ 3x}(2)_{+ 2x}(1)_{– 10, 6x}(2)_{+ 6x}(1)_{+ 2, 12x + 6, 12} 17. 1 4x (4)_{+ 2x}(3)₊9 2 x (2)_{+ 12x}(1) _{+ C} 18. (i) e e x h ( − 1) + C (ii) x( )3 3 + C (iii) ax + C (iv) x x ( )3 ( )2 3 5 2 + + C (v) – 1 2 sin x (vi) 1 4 5 x_.

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21 INTRODUCTION

Suppose y = f (x) be a function of x and y₀, y₁, y₂, ..., y_n are the values of the function f (x) at x₀,x₁, x₂, ...., x_n respectively, then the method to obtaining the value of f (x) at point x = x_i which lie between x₀ and x_n is called interpolation.

Thus, interpolation is the technique of computing the value of the function outside the given interval.

If x = x_i does not lie between x₀ and x_n then computing the value of f (x) at this point is called the extra polation.

The study of interpolation depends on the calculus of finite difference.

In this chapter, we shall discuss Newton-Gregory forward and backward interpolation, Lagrange’s, Stirling’s interpolation formula and method of finding the missing one and more term.

2.1. TO FIND ONE MISSING TERM

Method 1. Suppose one value of f (x) be missing from the set of (n + 1) values (i.e., n values are given)

of x, the values of x being equidistant. Let the unknown value be X. Now construct the difference table. We can assume y = f (x) to be a polynomial of degree (n – 1) in x, since n values of y are given. Now equating to zero the nth difference, we get the value of X.

Method 2. Suppose one value of f (x) be missing from the set of (n + 1) values (i.e., n values are

given) of x, the values of x being equidistant. Then we can assume y = f (x) to be a polynomial of degree (n – 1) in x ∴ ∆n_{f (x) = 0} or (E – I)n _{f (x) = 0} or (En_–n_C 1 E n–1₊n_C 2 E n–2_{– ... + (–1)}n_{I) f (x) = 0} or En _{f (x) –}n_C 1 E n–1 f (x) + nC2 E n–2 f (x) – ... + (– 1)n f (x) = 0

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2

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or f (x + nh) – n_C

1 f (x + (n – 1)h) +

n_C

2 f (x + (n – 2)h) – ... + (– 1)

n_{f (x) = 0} _...(2.1) If x = x₀ is the first value of x then putting x = x₀ in (2.1) and solving we get the missing term.

To find two missing term. Suppose two value X₁ and X₂ of f (x) be missing from the set of

(n + 2) values (i.e., n values are given) of x, the values of x being equidistant. Then, we can assume y =

f (x) be a polynomial of degree (n – 1) in x

∴ ∆n _{f (x) = 0}

or f (x + nh) – n_C

1 f (x + (n – 1) h) + nC2 f (x + (n – 2)h) – ... + (– 1)n f (x) = 0 ...(2.2)

If x = x₀ is the first value of x then putting x = x₀, and x = x₁ successively in (2.2), we get two equation in missing X₁ and X₂. On solving we get X₁ and X₂.

2.2. NEWTON-GREGORYS FORMULA FOR FORWARD INTERPOLATION WITH EQUAL INTERVALS

Let y = f (x) be a function which assumes the values f (a), f (a + h), f (a + 2h), ..., f (a + nh) for x = a,

a + h, a + 2h, ..., a + nh respectively where h is the difference of the arguments. Let f (x) be a

polyno-mial in x of degree n. So f (x) can be written as

f (x) = a₀ + a₁(x – a) + a₂ (x – a) (x – a – h) + a₃ (x – a) (x – a – h) (x – a – 2h) + ... + a_n(x – a) (x – a – h) ... (x – a – (n – 1) h) ...(2.3) where a₀, a₁, a₂, ..., a_n are constants

Putting successively the values x = a, a + h, a + 2h, ..., a + nh in (2.3), we get

f (a) = a₀ or a₀ = f (a) f (a + h) = a₀ + a₁(a + h – a) or f (a + h) = a₀ + a₁ h ⇒ a₁ = f a h a h f a h f a h f a h ( + )− ( ) ( ) ( ) = + − = 0 ∆ i.e., a₁ = ∆f a h ( ) f (a + 2h) = a₀ + a₁ (a + 2h – a) + a₂ (a + 2h – a) (a + 2h – a – h) = a₀ + a₁2h + a₂ 2h.h 2h2_a 2 = f (a + 2h) – 2ha1 – a0 a₂ = f a h f a h f a f a h ( ) ( ( ) ( )) ( ) ! +2 −2 + − − 2 2 = f a h f a h f a h f a h ( ) ( ) ( ) ! ( ) ! +2 −2 + + ₌ 2 2 2 2 2 ∆

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Proceeding in the same way, we get a₃ = ∆ 3 3 3 f a h ( ) ! ... a f a n h n n n = ∆ ( ) !

Putting the values a₀, a₁, a₂, ...., a_n into (1), we get

f (x) = f (a) + (x – a) ∆f a h ( ) + (x – a) (x – a – h) ∆ 2 2 2 f a h ( ) ! + (x – a) (x – a – h) (x – a – 2h) ∆ 3 3 f a hn ( ) ! + ... + (x – a) (x – a – h) (x – a – 2h) + ... + (x – a – (n – 1)h) ∆n n f a n h ( ) ! ! ...(2.4)

This is Newton-Gregory formula for forward interpolation putting x = a + hu in (2.4), we get

f (a + hu) = f (a) + u∆ f (a) + u u( ) ! − 1 2 ∆ 2_{f (a) + ... +}u u u u n n ( ) ( ) ... ( ( )) ! −1 −2 − −1 ∆n_{f (a).}

2.3. NEWTON-GREGORYS FORMULA FOR BACKWARD INTERPOLATION WITH EQUAL INTERVALS

Let y = f (x) be a function which assumes the values f (a), f (a + h), f (a +2h), ..., f (a + nh) for x = a, a + h, a + 2h, ...., a + nh, respectively where h is the difference of arguments. Let f (x) be a polynomials in x of degree n. So f (x) can be written as

f (x) = a₀ + a₁(x – a – nh) + a₂(x – a – nh) (x – a – (n – 1)h)

+ a₃ (x – a – nh) (x – a – (n – 1)h) (x – a – (n – 2)h) + ...

+ a_n (x– a – nh) (x – a – (n – 1)h) ... (x – a – h) ...(2.5) where a₀, a₁, a₂, ..., a_n are constants

Putting successively the values x = a + nh, a + (n – 1)h, a + (n – 2) h, ..., a + h in (2.5), we get

f (a + nh) = a₀ ⇒ a₀ = f (a + nh) f (a + (n – 1)h) = a₀ + a₁ (a+ (n – 1)h – a – nh) a₁ = f a nh f a n h h f a nh h ( + )− ( +( − ) ) ( ) =∇ + 1 f (a + (n – 2)h) = a₀ + a₁(a + (n – 2) h – a – nh) + a₂ (a + (n – 2)h – a – nh) . (a + (n – 2) h – a – (n – 1)h) a₂ = ∇ + 2 2 2 f a nh h ( ) !

19485352 Advanced Mathematics

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Preface

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Acknowledgement

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UNIT I : NUMERICAL ANALYSIS-I

Contents

UNIT II : NUMERICAL ANALYSIS-II

UNIT III : SPECIAL FUNCTION

UNIT IV : STATISTICS AND PROBABILITY

UNIT V : CALCULUS OF VARIATIONS AND TRANSFORMS

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SOLVED EXAMPLES

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