Numerical Analysis for Engineers: Methods and Applications, Second Edition (Solutions, Instructor Solution Manual) [2 ed.] 1482250357, 9781482250350

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Instructors’ and Solution Manual to Accompany

NUMERICAL ANALYSIS FOR ENGINEERS: Methods and Applications Second Edition Bilal M. Ayyub Richard H. McCuen CRC PRESS A CHAPMAN & HALL BOOK K23768 ISBN 978-1-4822-5035-0

COPYRIGHT NOTICE Ayyub and McCuen 2015 This Instructors’ and solution manual is the sole property of Professors Bilal M. Ayyub and Richard H. McCuen (the authors), and no portion or portions of it, or the entire document shall be used in any other documents without a written permission of the authors.

Department of Civil and Environmental Engineering University of Maryland College Park, MD 20742, USA

AUGUST 17, 2015

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Preface This is the instructors’ and solution manual for the book “Numerical Analysis: Methods and Engineering Applications, Second Edition,” by Ayyub and McCuen. The book consists of the following chapters: (1) introduction; (2) matrix analysis; (3) introduction to numerical methods; (4) the identification of the roots of equations; (5) the solution of linear simultaneous equations; (6) the interpolation of values of a dependent variable for a given set of discrete measurements; (7) approximation of the differential or integral of an unknown function for given set of discrete measurements from the function; (8) solutions to differential equations; (9) describing data graphically and computing important characteristics of sample measurements, and making basic statistical computations, (10) fitting a curve or a model to data, and (11) numerical optimization. The book was designed for an introductory course in numerical analysis with emphasis on applications. The book can be covered in one semester or two semesters in conjunction with a programming language. The chapter sequence can be followed as a recommended sequence. However if needed, instructors can choose a subset of the chapters of the book for courses that do not permit a complete coverage of the materials, or a coverage that cannot follow the presented order. After completing Chapters 1, 2, and 3, the readers will have sufficient background to follow and understand the materials in Chapters 4, 5, 6, 7, and 8 in any sequence. However, before reading Chapter 10, Chapter 9 should be studied. Chapter 11 can be covered after completing at least up to Chapter 7. The following figure illustrates the logical sequence of these chapters in terms of their interdependencies:

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Acknowledgments We would like to acknowledge all of the students who participated in the testing of the book for their comments, input and suggestions. The students who took courses on computational methods in civil engineering in the semesters of 1992 to 1994 contributed to this endeavor. Their feedback was very helpful and greatly contributed to the final product. Also, we would like to acknowledge the instructors who used the book during this testing period of the first edition. They include Drs. I. Assakkaf, P. Johnson, and R. Muhanna. The assistance of Mr. R.-J. Chao, Dr. M. Hassan, and Dr. I. Assakkaf in critically reviewing the problems at the end of the chapters, developing some of the example problems and problem solutions is gratefully acknowledged. We would like to invite users of the book or the manual that include readers and instructors to send us any comments on the book or the manual by E-mail at [email protected]. These comments will be used in developing future editions of the book and this manual.

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Table of Contents Acknowledgments ............................................................................................................................ iii  Table of Contents .............................................................................................................................. iv  CHAPTER 1. INTRODUCTION .......................................................................................................1  1.2 Analytical Versus Numerical Analysis .....................................................................................1  1.4 Applications...............................................................................................................................1  Computer Programs .........................................................................................................................6  CHAPTER 2. MATRICES .................................................................................................................9  2.1 Introduction ...............................................................................................................................9  2.2 Matrix Operations....................................................................................................................11  2.3 Vectors.....................................................................................................................................14  2.4 Determinants............................................................................................................................17  2.5 Rank of a Matrix......................................................................................................................18  2.6 Applications.............................................................................................................................19  CHAPTER 3. INTRODUCTION TO NUMERICAL METHODS..................................................20  3.1 Introduction .............................................................................................................................20  3.2 Accuracy, Precision, and Bias .................................................................................................20  3.3 Significant Figures ..................................................................................................................22  3.4 Analysis of Numerical Errors ..................................................................................................23  CHAPTER 4. ROOTS OF EQUATIONS ........................................................................................27  4.1 Introduction .............................................................................................................................27  4.2 Eigenvalue Analysis ................................................................................................................30  4.3 Direct-Search Method .............................................................................................................30  4.4 Bisection Method.....................................................................................................................32  4.5 Newton-Raphson Iteration.......................................................................................................35  4.6 Secant Method .........................................................................................................................50  4.8 Synthetic Division ...................................................................................................................55  4.9 Multiple Roots .........................................................................................................................70  4.10 Systems of Nonlinear Equations ...........................................................................................70  CHAPTER 5. SIMULTANEOUS LINEAR EQUATIONS.............................................................72  5.2 Gaussian Elimination...............................................................................................................72  5.3 Gauss-Jordan Elimination .......................................................................................................74  5.5 LU Decomposition ..................................................................................................................76  5.6 Iterative Equation-Solving Methods........................................................................................81  5.6.1 Jacobi Iteration ............................................................................................................................................... 81  5.6.2 Gaussian-Seidel Iteration ............................................................................................................................... 85  5.6.3 Convergence Consideration of the Iterative Methods .................................................................................... 90 

5.7 Use of Determinants ................................................................................................................94  5.8 Matrix Inversion ......................................................................................................................99  5.9 Applications...........................................................................................................................101  Computer Programs .....................................................................................................................103  CHAPTER 6. NUMERICAL INTERPOLATION.........................................................................105  6.2 Method of Undetermined Coefficients ..................................................................................105  6.3 Gregory-Newton Interpolation Method .................................................................................109  6.4 Finite Difference Interpolation ..............................................................................................112  6.5 Newton’s Method ..................................................................................................................114  6.6 Lagrange Polynomials ...........................................................................................................119  6.7 Interpolation Using Splines ...................................................................................................124  6.9 Multi-Dimensional Interpolation ...........................................................................................133  CHAPTER 7. DIFFERENTIATION AND INTEGRATION ........................................................135  iv

7.1 Numerical Differentiation .....................................................................................................135  7.2. Numerical Integration...........................................................................................................142  CHAPTER 8. Differential Equations..............................................................................................150  8.1 Introduction ...........................................................................................................................150  8.2 Taylor Series Expansion ........................................................................................................150  8.3 Euler’s Method ......................................................................................................................154  8.4 Modified Euler’s Method ......................................................................................................157  8.5 Runge-Kuta Methods ............................................................................................................159  8.6 Predictor-Corrector Methods .................................................................................................164  8.7 Least-Squares Method ...........................................................................................................167  8.8 Garlekin Method....................................................................................................................170  8.9 Higher-Order Differential Equations .....................................................................................172  8.10 Boundary Value Problems ...................................................................................................172  8.11 Integral Equations................................................................................................................176  CHAPTER 9. Data Description and Treatment ..............................................................................177  9.2 Classification of Data ............................................................................................................177  9.3 Graphical Description of Data ...............................................................................................177  9.5 Histograms and Frequency Diagrams ...................................................................................185  9.6 Descriptive Measures ............................................................................................................187  CHAPTER 10. Curve Fitting and Regression Analysis .................................................................190  10.1 Introduction .........................................................................................................................190  10.2 Correlation Analysis ............................................................................................................190  10.3 Introduction to Regression ..................................................................................................200  10.4 Principle of Least Squares ...................................................................................................201  10.5 Reliability of the Regression Equation ................................................................................204  10.8 Correlation Versus Regression ............................................................................................207  10.9 Application of Bivariate Regression Analysis ....................................................................209  10.8 Multiple Regression Analysis .............................................................................................213  10.9 Regression Analysis of Nonlinear Models ..........................................................................220  CHAPTER 11. Numerical Optimization ........................................................................................238  11.1 Introduction .........................................................................................................................238  11.2 The Response Surface Analysis ..........................................................................................238  11.3 Numerical Least Squares .....................................................................................................239  11.4 Steepest Descent Method ....................................................................................................247 

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CHAPTER 1. INTRODUCTION 1.2 Analytical Versus Numerical Analysis Problem 1-1. Solution not provided. Problem 1-2. The two methods differ on the basis of their respective algorithms. The analytical method is based on analytical calculus while the numerical method is based on finite differences arithmetic. Analytical approaches provide direct solutions and will result in exact solutions if they exist. Analytical methods usually require less time to find a solution. Analytical solution procedure becomes considerably more complex when constraints are involved. Numerical analysis, on the other hand, can be used to find solutions of moderately complex problems, and it is quite easy to include constraints on the unknowns in the solutions. However, numerical methods most often require a considerable number of iterations in order to find a solution with a reasonable accuracy. The solution provided by the numerical methods is usually not exact. Therefore, error analysis and error estimations are required.

1.4 Applications Problem 1-3. cos( x )  1 

x2 x4  ....... 2! 4!

For h = 0.1 x = x0 + h = 0 + 0.1 = 0.1 cos(0.1)  1.00000000 (one term) 2 (01 .)  0.99500000 (two terms) cos(0.1)  1  2 (01 . ) 2 (01 . )4 cos(0.1) 1    0.99500417 (three terms) 2 24 True value = 0.99500417 The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1:

1

f (x 0+h ) f (x 0+h ) f (x 0+h ) True value one term two terms three terms 0.1 0.1 1.00000000 0.99500000 0.99500417 0.99500417 0.2 0.2 1.00000000 0.98000000 0.98006667 0.98006658 0.3 0.3 1.00000000 0.95500000 0.95533750 0.95533649 0.4 0.4 1.00000000 0.92000000 0.92106667 0.92106099 0.5 0.5 1.00000000 0.87500000 0.87760417 0.87758256 0.6 0.6 1.00000000 0.82000000 0.82540000 0.82533561 0.7 0.7 1.00000000 0.75500000 0.76500417 0.76484219 0.8 0.8 1.00000000 0.68000000 0.69706667 0.69670671 0.9 0.9 1.00000000 0.59500000 0.62233750 0.62160997 1.0 1.0 1.00000000 0.50000000 0.54166667 0.54030231 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is specially true as the separation distance increases. x

h

Problem 1-4. sin( x )  x 

x3 x5  ....... 3! 5!

For h = 0.1 x = x0 + h = 0 + 0.1 = 0.1 sin(0.1)  0.100000000 (one term) 3 (01 .) sin(0.1)  01 .   0.09983333 (two terms) 6 (01 . ) 3 (01 . )5 sin(0.1)  01 .    0.09983342 (three terms) 6 120 True value = 0.09983342 The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1: f (x 0+h ) f (x 0+h ) f (x 0+h ) x h True value one term two terms three terms 0.1 0.1 0.10000000 0.09983333 0.09983342 0.09983342 0.2 0.2 0.20000000 0.19866667 0.19866933 0.19866933 0.3 0.3 0.30000000 0.29550000 0.29552025 0.29552021 0.4 0.4 0.40000000 0.38933333 0.38941867 0.38941834 0.5 0.5 0.50000000 0.47916667 0.47942708 0.47942554 0.6 0.6 0.60000000 0.56400000 0.56464800 0.56464247 0.7 0.7 0.70000000 0.64283333 0.64423392 0.64421769 0.8 0.8 0.80000000 0.71466667 0.71739733 0.71735609 0.9 0.9 0.90000000 0.77850000 0.78342075 0.78332691 1.0 1.0 1.00000000 0.83333333 0.84166667 0.84147098 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is specially true as the separation distance increases.

2

Problem 1-5. ex  1 x 

x2 ....... 2!

For h = 0.1 x = x0 + h = 0 + 0.1 = 0.1 e(0.1)  1.00000000 (one term) (0.1) e  10 .  01 .  110000000 . (two terms) 2 (01 .) .  01 .   110500000 . (three terms) e(0.1) 10 2 True value = 1.10517092 The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1: f (x 0+h ) f (x 0+h ) f (x 0+h ) x h True value one term two terms three terms 0.1 0.1 1.00000000 1.10000000 1.10500000 1.10517092 0.2 0.2 1.00000000 1.20000000 1.22000000 1.22140276 0.3 0.3 1.00000000 1.30000000 1.34500000 1.34985881 0.4 0.4 1.00000000 1.40000000 1.48000000 1.49182470 0.5 0.5 1.00000000 1.50000000 1.62500000 1.64872127 0.6 0.6 1.00000000 1.60000000 1.78000000 1.82211880 0.7 0.7 1.00000000 1.70000000 1.94500000 2.01375271 0.8 0.8 1.00000000 1.80000000 2.12000000 2.22554093 0.9 0.9 1.00000000 1.90000000 2.30500000 2.45960311 1.0 1.0 1.00000000 2.00000000 2.50000000 2.71828183 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is specially true as the separation distance increases. Problem 1-6.

f ( x ) = x 3  3x 2  5x  10

f (2) = 16

f ' ( x ) = 3x 2  6 x  5

f ' (2) = 5

f '' ( x ) = 6 x  6

f '' (2) = 6

f ''' ( x ) = 6 f ''' (2) = 6 f '''' ( x ) = 0 f '''' (2) = 0 Referring to Eq. 1-1 in the textbook, the value of the function, f(x), can be approximated by: f(x0 + h)  16 + 5h + 3h2 + h3 where x0 = 2 For h = 0.1 f(2.1)  16 (one term) f(2.1)  16 + 5(0.1) = 16.5 (two terms) 2 f(2.1)  16 + 5(0.1) + 3(0.1) = 16.530 (three terms) 2 3 f(2.1)  16 + 5(0.1) + 3(0.1) + (0.1) = 16.531 (four terms) The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1:

3

f (x 0+h ) f (x 0+h ) f (x 0+h ) f (x 0+h ) True value one term two terms three terms four terms 2.1 0.1 16.0000 16.5000 16.5300 16.5310 16.5310 2.2 0.2 16.0000 17.0000 17.1200 17.1280 17.1280 2.3 0.3 16.0000 17.5000 17.7700 17.7970 17.7970 2.4 0.4 16.0000 18.0000 18.4800 18.5440 18.5440 2.5 0.5 16.0000 18.5000 19.2500 19.3750 19.3750 2.6 0.6 16.0000 19.0000 20.0800 20.2960 20.2960 2.7 0.7 16.0000 19.5000 20.9700 21.3130 21.3130 2.8 0.8 16.0000 20.0000 21.9200 22.4320 22.4320 2.9 0.9 16.0000 20.5000 22.9300 23.6590 23.6590 3.0 1.0 16.0000 21.0000 24.0000 25.0000 25.0000 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is specially true as the separation distance increases. x

h

Problem 1-7.

f ( x ) = x 5  5x 4  x 2  6

f (2) = -38

f ' ( x ) = 5x 4  20 x 3  2 x

f ' (2) = -76

f '' ( x ) = 20 x 3  60 x 2 + 2

f '' (2) = -78

f ' ' ' ( x ) = 60 x 2 - 120 x

f ' ' ' (2) = 0

f ' ' ' ' ( x ) = 120 x  120 f ' ' ' ' (2) = 120 f ' ' ' ' ' ( x ) = 120 f ' ' ' ' ' (2) = 120 Referring to Eq. 1-1 in the textbook, the value of the function, f(x), can be approximated by: f(x0 + h)  -38 - 76h - 39h2 + 5h4 + h5 where x0 = 2 For h = 0.1 f(2.1)  -38 (one term) f(2.1)  -38 - 76(0.1) = -45.6 (two terms) 2 f(2.1)  -38 - 76(0.1) - 39(0.1) = -45.990 (three terms) f(2.1)  -38 -76(0.1) - 39(0.1)2 + 5(0.1)4 = -45.98950 (four terms) 2 4 5 f(2.1)  -38 -76(0.1) - 39(0.1) + 5(0.1) + (0.1) = -45.98949 (five terms) The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1:

4

f (x 0+h ) f (x 0+h ) f (x 0+h ) f (x 0+h ) f (x 0+h ) True value one term two terms three terms four terms five terms 2.1 0.1 -38.00000 -45.60000 -45.99000 -45.98950 -45.98949 -45.98949 2.2 0.2 -38.00000 -53.20000 -54.76000 -54.75200 -54.75168 -54.75168 2.3 0.3 -38.00000 -60.80000 -64.31000 -64.26950 -64.26707 -64.26707 2.4 0.4 -38.00000 -68.40000 -74.64000 -74.51200 -74.50176 -74.50176 2.5 0.5 -38.00000 -76.00000 -85.75000 -85.43750 -85.40625 -85.40625 2.6 0.6 -38.00000 -83.60000 -97.64000 -96.99200 -96.91424 -96.91424 2.7 0.7 -38.00000 -91.20000 -110.31000 -109.10950 -108.94143 -108.94143 2.8 0.8 -38.00000 -98.80000 -123.76000 -121.71200 -121.38432 -121.38432 2.9 0.9 -38.00000 -106.40000 -137.99000 -134.70950 -134.11901 -134.11901 3.0 1.0 -38.00000 -114.00000 -153.00000 -148.00000 -147.00000 -147.00000 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is especially true as the separation distance increases. x

h

Problem 1-8. f ( x ) = x 2  5x 0.5  6

f (2) = 2.92893

f ' ( x ) = 2 x  2.5x 0.5

f ' (2) = 2.23223

f ' ' ( x ) = 2  125 . x 1.5

f ' ' (2) = 2.44194

f ' ' ' ( x ) = 1875 . x 2.5

f ' ' ' (2) = -0.33146

f ' ' ' ' ( x ) = 4.68750 x 3.5 f ' ' ' ' ' ( x ) = -16.40625x

f ' ' ' ' (2) = 0.41432 4 .5

f ' ' ' ' ' (2) = -0.72506

.

.

. .

.

f n ( x ) .................................................................... f (n) (2)  ....... Referring to Eq. 1-1 in the textbook, the value of the function, f(x), can be approximated by: f(x0 + h)  2.92893 + 2.23223h + 1.27097h2 - 0.05524h3 + 0.01726h4 where x0 = 2 For h = 0.1 f(2.1)  2.92893 (one term) f(2.1)  2.92893 + 2.23223(0.1) = 3.15215 (two terms) 2 f(2.1)  2.92893 + 2.23223(0.1) + 1.27097(0.1) = 3.16486 (three terms) 2 3 f(2.1)  2.92893 + 2.23223(0.1) + 1.27097(0.1) - 0.05524(0.1) = 3.16481 (four terms) f(2.1)  2.92893 + 2.23223(0.1) + 1.27097(0.1)2 - 0.05524(0.1)3 + 0.01726(0.1)4 = 3.16481 (five terms) The following table summarizes the results for h = 0.1 to 1.0 in an increment of 0.1:

5

f (x 0+h ) f (x 0+h ) f (x 0+h ) f (x 0+h ) f (x 0+h ) True value one term two terms three terms four terms five terms 2.1 0.1 2.928930 3.152153 3.164863 3.164807 3.164809 3.164312 2.2 0.2 2.928930 3.375376 3.426215 3.425773 3.425800 3.423802 2.3 0.3 2.928930 3.598599 3.712986 3.711495 3.711635 3.707125 2.4 0.4 2.928930 3.821822 4.025177 4.021642 4.022084 4.014033 2.5 0.5 2.928930 4.045045 4.362788 4.355883 4.356961 4.344306 2.6 0.6 2.928930 4.268268 4.725817 4.713885 4.716122 4.697742 2.7 0.7 2.928930 4.491491 5.114266 5.095319 5.099463 5.074162 2.8 0.8 2.928930 4.714714 5.528135 5.499852 5.506922 5.473400 2.9 0.9 2.928930 4.937937 5.967423 5.927153 5.938477 5.895307 3.0 1.0 2.928930 5.161160 6.432130 6.376890 6.394150 6.339746 From the table above, it is clear that as the terms of the Taylor series are added incrementally, the accuracy improves as compared with the true values. This is specially true as the separation distance increases. x

h

Problem 1-9. Solution not provided. Problem 1-10. Solution not provided.

Computer Programs Problem 1-11. The following is a computer program in FORTRAN to evaluate the Taylor series expansion for cos(x) using one term, two terms, and three terms. x2 x4  f(x0+h) = 1  2 24 where x0 = 0.0

C****************************************************** PROGRAM TAYLOR8 H=0.1 WRITE(*,1) 1 FORMAT(1X,' X 1 TERM 2 TERMS 3 TERMS ') PRINT*,'_______________________________________________________' DO 10 I=1,10 FX1=1.0 FX2=FX1-(H**2)/2 FX3=FX2+(H**4)/24 X=H H=H+0.1 6

WRITE(*,2)X,FX1,FX2,FX3 2 FORMAT(1X,4F12.8) 10 CONTINUE END Problem 1-12. The following is a computer program in FORTRAN to evaluate the Taylor series expansion for ex using one term, two terms, and three terms. x2 f(x0+h) = 1  x  2 where x0 = 0.0

C****************************************************** PROGRAM TAYLOR9 H=0.1 WRITE(*,1) 1 FORMAT(1X,' X 1 TERM 2 TERMS 3 TERMS ') PRINT*,'_______________________________________________________' DO 10 I=1,10 FX1=1.0 FX2=FX1+H FX3=FX2+(H**2)/2 X=H H=H+0.1 WRITE(*,2)X,FX1,FX2,FX3 2 FORMAT(1X,4F12.8) 10 CONTINUE END Problem 1-13. The following is a computer program in FORTRAN to evaluate the Taylor series expansion: f(x0+h) = -38 - 76h - 39h2 +5h4 + h5 where x0 = 2

C****************************************************** PROGRAM TAYLOR10 H=0.1 WRITE(*,1) 1 FORMAT(1X,' X 1 TERM 2 TERMS 3 TERMS 4 T +ERMS 5 TERMS') PRINT*,'__________________________________________________________ +_____________________' DO 10 I=1,10 FX1=-38 FX2=FX1-76*H FX3=FX2-39*H**2 FX4=FX3+5*H**4 FX5=FX4+H**5 7

X=2+H H=H+0.1 WRITE(*,2)X,FX1,FX2,FX3,FX4,FX5 2 FORMAT(1X,6F12.5) 10 CONTINUE END Problem 1-14. The following is a computer program in FORTRAN to evaluate the Taylor series expansion f(x0+h) = 2.92893 + 2.23223h + 1.27097h2 - 0.05524h3 + 0.01726h4 where x0 = 2

C****************************************************** PROGRAM TAYLOR11 H=0.1 WRITE(*,1) 1 FORMAT(1X,' X 1 TERM 2 TERMS 3 TERMS 4 T +ERMS 5 TERMS') PRINT*,'__________________________________________________________ +_____________________' DO 10 I=1,10 FX1=2.92893 FX2=FX1+2.23223*H FX3=FX2+1.27097*H**2 FX4=FX3-0.05524*H**3 FX5=FX4+0.01726*H**4 X=2+H H=H+0.1 WRITE(*,2)X,FX1,FX2,FX3,FX4,FX5 2 FORMAT(1X,6F12.5) 10 CONTINUE END

8

CHAPTER 2. MATRICES 2.1 Introduction Problem 2-1. Daily maximum and minimum temperatures in Fahrenheit: Maximum Minimum   Monday   Tuesday  Day Wednesday   Thursday  Friday   Saturday  Sunday

45

30

55 61 57

32 29 36

51 39

29 21

35

19

          

37.5 34.5   45.0   Daily Means = 46.5  40     30   27   

Means for daily maximum and daily minimum = 49 28 Sample Calculations: 51 + 29 Mean temperature for Thursday =  40 o F 2 45 + 55 + 61 + 57 + 51 + 39 + 35 Mean for daily maximum = = 49 o F 7 Problem 2-2. Solution not provided. Problem 2-3.  Student A Score  Student B Score  My Score

Test 1

Test 2

80

68

100 40

85 55

Test 3 91  70  68 

 0.3 Weight =  0.3   0.4  80 68 91  0.3  80x0.3 + 68x0.3 + 91x0.4  80.8 100 85 70 .  0.3 = 100x0.3 + 85x0.3 + 70x0.4 = 83.5          40 55 68 0.4   40x0.3 + 55x0.3 + 68x0.4  55.7 

9

The resulting vector provides the course final scores for each student.

Problem 2-4. Solution not provided. Problem 2-5. Solution not provided. Problem 2-6.

Mashining Time Short Moderate Long Cutting Time

Short  Long 

010 . 015 .

0.35 0.25

0.55 0.60

  

Problem 2-7. Matrix 1.

Future Rainfall Low Average High  0.3 Past Rainfall Average 0.6  High 0.7

0.3

Low

0.2 0.2

0.4  0.2   0.1

Note: row summation = 1.0 Matrix 2.

Future Rainfall Low Average High 0.7 Past Rainfall Average 0.2   0.3 High

0.2

Low

0.6 0.3

0.1  0.2   0.4 

Note: row summation = 1.0

Problem 2-8.

2 -3 a) AT =   6 4 

,

 -1 3  b) BT = -9 0  ,    6 -7 

Problem 2-9. Solution not provided.

10

c)

 0.2 -0.8 -0.3 CT =   -0.5 0.1 0.7 

2.2 Matrix Operations Problem 2-10. Addition:

 3  2 2  3 1 1  2  3 3  2  1 1 A+B=  =  , B+A =     0  1 4  2  1 6  1  0 2  4  1 6 Hence, A + B = B + A ====> and matrix addition is not directional Subtraction:  3  (2) 2  3  5 5  2  3 3  (2)   5 5   A-B= , B-A= =      4  2   1 2  24   01  1 0  1 2  ====> and matrix subtraction is directional Hence, A - B  B - A

Problem 2-11. Solution not provided. Problem 2-12.

 0.1 0.8 0.9  0.4 0.3 0.5   0.3 .5 0.4  D = E - C =    0.6 0.3 0.4 0.0 0.7 0.2 0.6 0.4 0.2

Problem 2-13. Solution not provided. Problem 2-14. 3 2  2 3 6  2 9  4  A.B =  .    0 4   1 2   0  4 0  8    2 3  3  2  B.A =  .    1 2  0 4 

8 5  4 8  

6  0 4  12   3  0  2  8   

6 16  3 6  

A.B  B.A  Matrix multiplication is not commutative. Problem 2-13. Solution not provided. Problem 2-16. . 0.7  0.2 0.8   01 F.G =  .   0.3 0.4   0.5 0.6

0.33 0.34  0.26 0.48  

11

. 0.7 0.26 018 .  0.2 0.8   01 G.F =   .    .   0.5 0.6 0.3 0.4  0.23 011 F.G  G.F  Matrix multiplication is not commutative.

Problem 2-17. Solution not provided. Problem 2-18. Solution not provided. Problem 2-19.

a.

b.

c.

d.

1 . 0.3 0.2   17 .   01 C = A.B =  .  4      . 0.4 18 . 0.6 01  2    .   0.3   0.06 0.21 015    C = A.B = 12 0.6   .  .  0.2 0.7 0.5  0.24 0.84 . 0.35 0.25 0.5  010  0.3   C = A.B =  0.2 0.7 0.5 .  12  .   0.53 0.5 0.6   0.4   . 2 3 1 4   C = A.B = 13 .  0 . 2    6 4 0 5    0.8 15 . 

. 0.4 4.6 4.4 12 14 4.7 13 . 4.2   . 7.4 8.4 0.8 4.3

Problem 2-20. Solution not provided. Problem 2-21. c11 c12  2 3 1 0 c21 c22  .  1 2   0 1      

-2c11 +c12

=1

3c11 +2c12

=0

-2c21 +c22

 0.286 0.429   B-1 =  0.286 .  0143

=0

3c21 +2c22 = 1

12

Problem 2-22. Solution not provided. Problem 2-23. Solution not provided. Problem 2-24. c11 c12 c13   3 1 2  1 0 0 c 21 c 22 c23 . 1 4 -3  0 1 0       c 31 c 32 c33 2 5 3  0 0 1

3c11 + c12 + 2c13 -c11 +4c12 +5c13 2c11 -3c12 +3c13

=1 =0 =0 =0 3c21 + c22 + 2c23 =1 -c21 +4c22 +5c23 =0 2c21 -3c22 +3c23 =0 3c31 +c32 +2c33 =0 -c31 +4c32 +5c33 =1 2c31 -3c32 +3c33 0155 . 0.060  0.321   Solving the above equations yields the inverse = 0107 . 0.060 0131  .  0.036 0.202 0155  . Problem 2-25. Let C be the inverse of X-1 ,then C.X-1 = I Post-multiply both sides of above equation with X, provides C . X-1. X = I . X Therefore, C = X.I = X Proof Problem 2-26. In both (a) and (b) , the matrices are singular. The inverse in both cases does not exist. Also , the determinant is zero for (a) and (b). The determinant can be computed as a. |A| = (0.1).(1.2) - (0.3).(0.4) = 0 b. |B| = -2[ (-3 x -2) - (2 x -6) ] + 6[ (1 x -2) - (2 x 2) ] + 2[ (1x-6) - (-3 x 2) ] = 0 Problem 2-27. Solution not provided. Problem 2-28. Solution not provided.

13

Problem 2-29. Solution not provided. Problem 2-30. Solution not provided. Problem 2-31. The correlation matrix is a square matrix with diagonal elements that are equal to one. Therefore, the trace of a correlation matrix of size n is n. Problem 2-32. tr(B) = -2 + (-3) + (-2) = -7 Problem 2-33. Solution not provided. Problem 2-34. Solution not provided.

2.3 Vectors Problem 2-35. (a) V1 + V2 = 2 6 -3 + -3 1 2 = -1 7 -1

(b)

-0.4   0.2  0.6  0.1   0.3  0.2    =  V1 + V2 =   +   -0.5 -0.5 0.0        0.2   0.5   0.3

Problem 2-36. Solution not provided. Problem 2-37. (a) V2 - V1 = -3 1 2 - 2 6 -3 = -5 -5 5

(b)

.  10 -0.4  0.6  01  0.1  0.2  .   -   =  V2 - V1 =  0.5  -0.5 0.0       .  01  0.2   0.3

14

Problem 2-38. Solution not provided. Problem 2-39. Solution not provided. Problem 2-40.

(a)

(b)

 6 2 4  18 6 12     9 3 6 0.4   01 .  T     016 V1 V2 = 0.6 0.2 0.0 0.3 . .   0.5    0.2  2 V1 V2 =  6  .  3 1 2    3 T

Problem 2-41.

(a)

(b)

3 V1 V2 = 2 6 3 .  1    6    2  0.6 0.2  V1 V2T =   .  0.4 01 . 0.5 0.2  0.0    0.3 T

Problem 2-42. Solution not provided. Problem 2-43. Solution not provided. Problem 2-44. Solution not provided. Problem 2-45. Solution not provided.

15

0.24  0.08   0.00  .  012

0.06 0.30 012 .  . 0.02 010 0.04  0.00 0.00 0.00  0.03 015 . 0.06

Problem 2-46. .  0.5883   0.5921 01830       V1 = 0.7845 , V2 = 01502 , V3 = 0.3649     .    01961   0.7917   0.9129  .

Length of V1 = 1.00, length of V2 = 1.00, and length of V3 = 1.00 V1, V3, and V3 are normalized. V1 and V3 are orthonormal since the product V1.V3 = 0. V1 and V2, and V2 and V3 are not orthonormal. Problem 2-47. Solution not provided. Problem 2-48. Solution not provided. Problem 2-49. Solution not provided. Problem 2-50. Solution not provided. Problem 2-51. Length of line connecting the two points = [(-1-3)2+(3-4)2+(3+2)2]0.5 = 6.481  1  3  o 1 = cos-1    128.1 ( 2.236 rad)  6.481   3 4 o 2 = cos-1  . rad)   98.9 (1726  6.481  3 2 o 3 = cos-1    39.5  6.481

( 0.690 rad)

Problem 2-52. For A to be normal, its length must equal 1; [(0.7)2+((0.4)2+(-0.2)2+(-0.3)2+x2]0.5 = 1.0  0.78 + x2 = 1.0  x = 0.469 Problem 2-53. b22 = [ 1- (0.66)2 - (0.26)2 ]0.5 = 0.705 The matrix is not orthonormal because

16

 0.66  0.23 -0.58 0.78 . 0.705 = - 0.0543  0  0.26  and  0.66  0.23 -0.58 0.78 . -0.705 = 0.7635  0  0.26 

Problem 2-54. The normal vector condition results in a122 + a322 =1-(0.6)2 = 0.64 (1) The vector orthogonality condition results in 0.5a12 - 0.843a32 = 0.12 (2) From (1) and (2), a12 = 0.742428, -0.617512 and a32 = 0.298000, -0.508607 Hence, there are two pairs of values for a12 and a32 that are necessary for matrix A to be orthonormal in the columns. These pairs are: a12  0.742428 a  = 0.298000  32   

and a12  -0.617512 a  = -0.508607  32   

2.4 Determinants Problem 2-55. A  12  0  12

B  4  3  7

Problem 2-56. 3 1 2 1 4 3  3 [ (4 x 3) - (-3 x 5) ] + 1 [ (1 x 3) - (-3 x 2) ] + 2 [ (1 x 5) - (4 x 2) ] = 84 2 5 3

17

Problem 2-57. 2 6 2 1 3 2  - 2 [ (-3 x - 2) - (2 x - 6) ] - 6 [ (1 x - 2) - (2 x 2) ] + 2 [ (1 x - 6) - (-3 x 2) ] = 0.0 2 6 2

Problem 2-58. Solution not provided. Problem 2-59. Solution not provided. Problem 2-60. Solution not provided.

2.5 Rank of a Matrix Problem 2-61. 0.1 -0.3 = 0.17  0 ,  r(A) = 2 0.6 -0.1

a.

| -4 | = -4  0 ,



r(B) = 1

b.

| -0.5 | = -0.5  0,  | 0.7 | = 0.7  0 , 

r(A) = 1 r(B) = 1

c.

| -0.2| = -0.2  0,  |1.2 | = 1.2  0 , 

r(A) = 1 r(B) = 1

0.4 0.6 = - 0.86  0 ,  r(A) = 2 1.3 -0.2

d.

3 -1 = - 4  0 ,  r(B) = 2 -4 0

Problem 2-62.

a.

b.

1 r(  . 15 3 r( 1  2

2 4 2 )  2, because  4 4 6 2 1 3  3 2 )  3, because 1 1 3 2

4 = -4  0 6 2 1 3 2 = 18  0 1 3

18

c.

2.0 8.0 8.0   15 . 4.5 6.0  )  2 because of the following r(   35 . 24.5 14.0   . 4.5 6.0   15

All combination of 3 by 3 matrices have a determinant equal to zero. The second combination of matrices that need to be tested is the 2 by 2 . At least one of the 2 by 2 matrices produces a determinant of -3, namely 2.0 8.0 1.5 4.5   Hence, the rank = 2.

2.6 Applications Problem 2-63. Xm = [4.3333 0.076667 S = [2.0817 0.032146

12.3333] 1.52753]

. 0.7258 0.8728  11209  Z =  0.3203 0.4148 10911 .    0.8006 11406 0.2182  . . 0.80 0.58  100 1  T R= Z Z = 0.80 100 . 0.03  nr  1  0.58 0.03 100 .  where nr = 2.

19

CHAPTER 3. INTRODUCTION TO NUMERICAL METHODS 3.1 Introduction Problem 3-1. x0 + x = get:

3

x , or (x0 + x)3 = x ; neglecting the term that involve (x)2 or higher, we

x03 + 3x2 x = x, x 

or x =

x  x 03 , In general : 3x 02

x  x i3 3x i2

31  33  0148148 . 3(3) 2 31  (3148148 . )3  0.006753 with x = 3 + 0.148148 = 3.148148, x = 3(3148148 . )2 A third iteration will give x = -0.000015 and x3 = 3.141381 , which is within the desired accuracy of 0.00005.

If x = 31, and x0 = 3, then x =

Problem 3-2. Solution not provided. Problem 3-3. Solution not provided.

3.2 Accuracy, Precision, and Bias Problem 3-4. Data Guess True value Error

127 134 7

183 196 13

98 101 3

154 147 -7

113 121 8

222 219 -3

67 75 8

142 154 12

122 127 5

165 176 11

Sum 1393 1450 57

The following figure shows that the employee provides biased estimates by under-estimating the weight, with therefore low accuracy:

20

True value vs Guess 250

True value

200 150

True value 45 Degree line

100 50 0 0

50

100

150

200

250

Guess

Problem 3-5. Data

Actual Predicted Error

47.2 45.8 1.4

53.3 51.2 2.1

64.4 61.3 3.1

58.7 58.3 0.4

43.8 42.8 1

Sum 267.4 259.4 8

Graph of predicted values vs actual values 65

Predicted

60 55 50

Predicted

45

45 degree line

40 35 30 40

45

50

55 Actual

Conclusion: low bias and high accuracy.

21

60

65

Problem 3-6. 60 50

Results

40

Lab 1 Lab 2

30

Lab 3 Exact

20 10 0 0

1

2

3

Lab

Conclusions: Bias Precision Accuracy

Lab 1 very small very small small

Lab 2 small high high

Lab 3 high small small

Problem 3-7. Solution not provided. Problem 3-8. Solution not provided.

3.3 Significant Figures Problem 3-9. V  2 gh = 2(32. 2)(9. 7)  624. 68  24. 993599  25.0

Problem 3-10. V = Lwh = (3.217)(0.7924)(1.302) = 3.319 (4 significant figure) Problem 3-11. (A) = (2r2 + 2rh) = 2(r2 + rh) = 2[ (3.172)2 + (3.172)(7.41)] = 2 (10.061584 + 23.504520) = 2 (33.566104) = 2(33.6) = 67.1  Problem 3-12. Solution not provided.

22

Problem 3-13. Solution not provided.

3.4 Analysis of Numerical Errors Problem 3-14. Results Field X Y Approx. Model Y' Error = |Y' - Y|

1 0 6.1 6 -0.1

2 0.5 7.9 7 -0.9

3 1.4 12.3 8.8 -3.5

4 2.9 12.5 11.8 -0.7

5 4.1 16.2 14.2 -2

6 6.2 17.5 18.4 0.9

7 7.8 16.8 21.6 4.8

8 9.3 17.7 24.6 6.9

7 7.8 16.8 17.748 0.948

8 9.3 17.7 19.488 1.788

Approximate Model 25 20

Data

15 Y 10

Approximate Model

5 0 0

2

4

6

8

10

X

Results Field X Y Statistical Model Y'

Error = |Y' - Y|

1 0 6.1 8.7 2.6

2 0.5 7.9 9.28 1.38

3 1.4 12.3 10.324 -1.976

4 2.9 12.5 12.064 -0.436

23

5 4.1 16.2 13.456 -2.744

6 6.2 17.5 15.892 -1.608

Statistical Model 20 18 16 14 12 Y 10 8 6 4 2 0

Data Statistical Model

0

2

4

6

8

10

X

The following summary provides a comparison between the two models in terms of bias, precision, and accuracy: Assessment Approximate Model Statistical Model Bias High Low Precision Low High Accuracy Medium Medium

Problem 3-15. Solution not provided. Problem 3-16. Solution not provided. Problem 3-17. The errors for the two models were computed using e = |Y’ - Y| in Problem 3-9. The relative errors can be computed as Y  Y' . 100 e = Y As an example, consider the statistical model with the data for field 3: 12.3  10.324 e . 100 = 16.1% 12.3 For the case of statistical model , sample calculations for the average error and the standard deviation for the errors are as follows: 2.6 + 1.38 + 1.976 + 0.436 + 2.744 + 1.608 + 0.948 + 1.788 Average error = = 1.685 8

Standard deviation =

(2.6  1685 . ) 2  (138 .  1685 . ) 2  (1978 .  1685 . ) 2  (....) 2 ........ = 0.78 81

24

SUMMARY: Approximate Model Error = |Y' - Y| 0.1 0.9 3.5 Relative Error (%) 1.63934 11.3924 28.4553 Average error = 2.475 Standard deviation = 2.39091

Statistical Model Error = |Y' - Y| 2.6 1.38 Relative error (%) 42.623 17.4684 Average error = 1.685 Standard deviation = 0.77975

1.976 16.065

0.7 2 0.9 4.8 6.9 5.6 12.3457 5.14286 28.5714 38.9831

0.436 2.744 1.608 0.948 1.788 3.488 16.9383 9.18857 5.64286 10.1017

Problem 3-18.

8 8  3( 4 )  6   4  0.00  i 2.00 x 4 8 x  3( 0.00  i 2.00 )  6   6.00  i10.00  2.97  i1.68 0.00  i 2.00   xi 1  xi  (2.97  i1.68)  (0.00  i 2.00)  2.97  i 0.32 x  3x  6 

8  12.88  i6.20  3.69  i 0.84 2.97  i1.68   (3.69  i 0.84 )  (2.97  i1.68)  0.71  i 0.84 x  3(2.97  i1.68)  6 

8  14.99  i 2.99  3.89  i 0.38 3.69  i 0.84

x  3(3.69  i 0.84)  6    0.21  i 0.46 x  3(3.89  i 0.38)  6 

8  15.64  i1.36  3.96  i 0.17 3.89  i 0.38

  0.07  i 0.21 x  3(3.96)  i 0.17)  6 

8  15.86  i 0.60  3.98  i 0.08 3.96  i 0.17

  0.02  i 0.10 One more trial will yield, x = 3.99 +i0.03 and  = 0.01-i0.04 e = xc - xt = 3.99 + i0.03-(4.00+i0.00) = -0.01+i0.03

25

Summary i xi 0 -4 1 0.00+i2.00 2 2.97+i1.68 3 3.69+i0.84 4 3.89+i0.38 5 3.96+i0.17 6 3.98+i0.08 7 3.99+i0.03

26

xi 2.97-i0.32 0.71-i0.84 0.21-i0.46 0.07-i0.21 0.02-i0.10 0.01-i0.04

CHAPTER 4. ROOTS OF EQUATIONS 4.1 Introduction Problem 4-1. The first plot of f(x) suggests that the roots lie in the interval x = -2 and x= 2. However the second plot gives a more precise locations for the roots. In particular it suggests that there is only one real root for f(x).   Plot of f (x ) with a large interval (-10,10) 800 600 400 200 f (x )

0 -200 -400 -600 -800 -1000 -10

-8

-6

-4

-2

0

2

4

6

8

x

  Plot of f (x ) with a smaller subinterval 2 1.5 1 f (x ) 0.5 0 -0.5 -1 -0.60

-0.10

0.40

0.90

1.40

x

From the second plot, an initial estimate of x can be taken as 0.9. The polynomial f ( x )  x 3  0.2 x 2  0.67 x  0.364  0 can be rearrange and solved for x as follows: 0.364 x  0.2 x  0.67  x With the initial estimate of xi = 0.9, 0.364 x i 1  0.2(0.9)  0.67   0.946 0.9 Substituting the value of xi+1 = 0.946 in the equation x3 + 0.2x2 - 0.67x - 0.364 yields 27

f(x) = -0.088 With the revised estimate of x = 0.946 , a second trial gives xi+1 = 0.929 and f(x) = -0.012. A fourth trial gives x = 0.934 f(x)= 0.001 Problem 4-2. The following two plots of f(x) show the locations of the roots of the equation f ( x )  x 3  17.2 x 2  8197 . x  83.79  0 The first plot suggests that there is definitely one real root in the interval (20,25). The second plot shows that there are two real roots lie in the interval (-2.5,-1.5).   Plot of f (x ) for the interval (-5,23) 1500 1000 500 0 f (x ) -500 -1000 -1500 -2000 -5

0

5

10

15

20

x

  Plot of f (x ) for the interval (-3,-1) 5 0 -5 f (x ) -10 -15 -20 -25 -3.00

-2.50

-2.00

-1.50

-1.00

x

From both plots, initial estimates for the three distinct real roots can be taken as 20, -1.5, and -2.5. The polynomial f ( x )  x 3  17.2 x 2  81.97 x  83.79  0 can be rearranged and solved for x as follows: 83.79 x  17.2 x  81.97  x With an initial estimate xi =20 for the first root 83.79 x i 1  17.2(20)  81.97   20.740 20 and f ( x )  (20.740) 3  17.2(20.740) 2  81.97(20.740)  83.79  261.125 With the revised value of xi = 20.740 28

x i 1  17.2(20.740)  81.97 

83.79  21.041 20.740

and f ( x )  (21.041) 3  17.2(21.041) 2  81.97(21.041)  83.79  108.019 Iterating in the same manner until the expression for xi+1 equals the right side, yields x1 = 21.244 For initial estimates of -1.50 and -2.50 for the second and third root respectively, and by adopting the same approach, the answers are x2 = -1.642 x3 = -2.402 Problem 4-3. The plot the figure below suggests that there exist three real roots for the polynomial f(x) = x3 + 7.2x2 +13.55x + 3.444 From the plot, the initial estimates for the roots can be taken as -0.5, -3.0, and -4.0. Rearranging the polynomial x3 + 7.2x2 + 13.55x + 3.444 = 0 as 3.444 x  7.2 x  13.55  ; x and following the same approach that was used in Problems 4-1 and 4-2, the results are x1 = -0.30 x2 = -2.80 x3 = -4.10 Plot of f (x ) for the interval (0,-5) 4 2 0 -2 -4 -6 -8 -5.00

-4.00

-3.00

-2.00

-1.00

x

Problem 4-4. X = { -3.714, -1.3077, 2.5} Problem 4-5. X = (-A * X ** 2 - B * X - C - D / X) ** (1 / 3)

29

-10 0.00

f (x )

Problem 4-6. X = (-6. -0.5, 0.3333, 1.5)

4.2 Eigenvalue Analysis Problem 4-7. Off-diagonal corr = .0 E.value = 1.0000 Off-diagonal corr = .7 E.value = .7141 Off-diagonal corr = .2 E.value = .9798 Off-diagonal corr = 1.0 E.value = .0000 As the off-diagonal correlation increases, the e.value decreases. Therefore, large e.values indicate low intercorrelation. Problem 4-8. L^3 - 3 L^2 + 1.75 L - 0.23 Problem 4-9. Solution not provided.

4.3 Direct-Search Method Problem 4-10. (a) x

f(x) 0 -0.924 0.25 -0.30713 0.5 -0.034 0.75 -0.01088 1 -0.144 1.25 -0.33963 1.5 -0.504 1.75 -0.54338 2 -0.364 2.25 0.12788 The above table gives the value of x and f(x) using the direct-search method with an increment of 0.25. The table shows that there exists one root in the interval (2, 2.25). The function changes sign in this interval. In fact, the function has three real roots. These roots are 2.2, 0.6, and 0.7. The direct-search method only identified one root and already missed two, and probably it will not find them unless the increment is reduced.

30

(b) Graph of f (x ) 0.2 0 -0.2 0

0.5

1

1.5

2

f (x ) -0.4

-0.6 -0.8 -1 x

The most feasible intervals to search for roots are (0.50, 0.75) and (2.0, 2.25). Problem 4-11.

f(x) = x4 - 1.74 x3 + 0.9419 x2 - 0.4252 x + 0.0027 = 0

(a) f(x) 0 0.0027 0.1 -0.03204 0.2 -0.05698 0.3 -0.07897 0.4 -0.10244 0.5 -0.12943 0.6 -0.15958 0.7 -0.19013 0.8 -0.21592 0.9 -0.2294 1 -0.2206 The table above shows that one real root exists in the interval (0, 0.1). The function changes sign in this interval. There could be other roots in the interval (0, 1) that the direct-search method could not identify because either the increment is high or the rest of the roots are not within the interval (0, 1). x

(b)

31

Graph of f (x ) 0.005 0 -0.0050.00

0.02

0.03

0.05

0.06

0.08

0.09

-0.01 f (x ) -0.015

-0.02 -0.025 -0.03 -0.035 x

From the above graph of f(x), the most feasible interval to search for the roots are (0,0.02) and probably some other intervals beyond x > 0.1 and x < 0. Problem 4-12. Lower Upper ROOT x x Fl(x) Fu(x) ------------ ------- ------- ------- -------6.0002790 -6.0100 -6.0000 2.5484 -.0762 -.4999938 -.5099 -.4999 -.0926 .0004 .3333238 .3301 .3401 .0202 -.0415 1.4999380 1.4901 1.5001 -.1705 .0020

4.4 Bisection Method Problem 4-13. (a). f ( x )  x 3  35 . x 2  3.28 x  0.924  0 From Problem 4-5, we know that there is one root in the interval (2, 2.5), hence for i = 0, xs = 2, and xe = 2.5. Therefore, x  x e 2  2.5 xm  s   2.25 2 2 f ( x s ) = -0.364

f ( x e ) = 1026 . f ( x m ) = 01279 . f ( x s ) . f ( x m ) = negative f ( x m ) . f ( x e ) = positive The root lies between 2 and 2.25 Estimate of the root = x m = 2.25 For i = 1, xs = 2, and xe = 2.25,

32

x s  x e 2  2.25   2.125 2 2 f ( x s ) = -0.364 . f ( x e ) = 01279 . f ( x m ) = 016298 f ( x s ) . f ( x m ) = positive f ( x m ) . f ( x e ) = negative The root lies between 2.125 and 2.25 Estimate of the root = 2.125 xm 

.  d  2.125  2.25  0125 > 0.02 Since the tolerance criterion of 0.02 is not met, we continue the iterative procedure until the required tolerance is satisfied. The following table summarizes the results. The first root was found to be 2.203 (to 3 significant figures) as shown in the following table. i 0 1 2 3 4

xs 2.000 2.000 2.125 2.188 2.188

xm 2.250 2.125 2.188 2.219 2.203

xe 2.500 2.250 2.250 2.250 2.219

f (x s ) f (x m ) f (x e ) f (x s ).f (x m ) f (x m ).f (x e ) d -0.364 0.128 1.026 + ----0.364 -0.163 0.128 + 0.13 -0.163 -0.030 0.128 + 0.06 -0.030 0.046 0.128 + 0.03 -0.030 0.008 0.046 + 0.02

(b) Same as part (a) except that the error was computed as relative percent error as follows: 2.125  2.25 r  x 100 = 6 % > 1 % 2.125 Since the tolerance criterion of 1 % is not met, we continue the iterative procedure until the required tolerance is satisfied. The following table summarizes the results. The first root was found to be 2.219 (to 3 significant figures).

i 0 1 2 3

xs 2.000 2.000 2.125 2.188

xm 2.250 2.125 2.188 2.219

xe 2.500 2.250 2.250 2.250

f (x s ) -0.364 -0.364 -0.163 -0.030

f (x m ) 0.128 -0.163 -0.030 0.046

f (x e ) f (x s ).f (x m ) f (x m ).f (x e ) 1.026 + 0.128 + 0.128 + 0.128 +

r ---6 3 1

Problem 4-14. (a) f ( x ) = x 4  174 . x 3  x 2  0.4252 x  0.0027 = 0 From the figure of Problem 4-6, we know that there is a root within the interval (0, 0.02), hence for i = 0, xs = 0, and xe = 0.02. Therefore,

33

x s  x e 0  0.02   0.01 2 2 f ( x s ) = 0.0027

xm 

f ( x e ) = 0.005441 f ( x m ) = 0.0014595 f ( x s ) . f ( x m ) = negative f ( x m ) . f ( x e ) = positive The root lies between 0 and 0.01 Estimate of the root = x m = 0.01 For i = 1, xs = 0, and xe = 0.01,

x s  x e 0  0.01   0.005 2 2 f ( x s ) = 0.0027 f ( x e ) = 0.0014595 f ( x m ) = 0.00059733 f ( x s ) . f ( x m ) = positive f ( x m ) . f ( x e ) = negative The root lies between 0.005 and 0.01 Estimate of the root = 0.005 xm 

 d  0.005  0.01  0.005 > 0.002 Since the criterion for the tolerance of 0.002 is not met, we continue the iterative procedure until the required tolerance is satisfied. The following table summarizes the results. The first root was found to be 0.0063 (to 4 significant figures). i 0 1 2 3

xs 0.0000 0.0000 0.0050 0.0050

xm 0.0100 0.0050 0.0075 0.0063

xe 0.0200 0.0100 0.0100 0.0075

f (x s ) 0.0027 0.003 0.001 0.001

f (x m ) -0.001 0.001 -4E-04 8E-05

f (x e ) f (x s ).f (x m ) f (x m ).f (x e ) -0.005 + -0.001 + -0.001 + -4E-04 + -

d ---0.005 0.003 0.001

(b) Same as part (a) except that the error was computed as relative percent error as follows: 0.005  0.01 r  x 100 = 100 % > 0.1 % 0.005 Since the criterion for the tolerance of 0.1% is not met, we continue the iterative procedure until the required tolerance is satisfied. The following table summarizes the results. The first root was found to be 0.0064 (to 4 significant figures).

34

i 0 1 2 3 4 5 6 7 8 9 10 11

xs 0.0000 0.0000 0.0050 0.0050 0.0063 0.0063 0.0063 0.0064 0.0064 0.0064 0.0064 0.0064

xm 0.0100 0.0050 0.0075 0.0063 0.0069 0.0066 0.0064 0.0065 0.0064 0.0064 0.0064 0.0064

xe 0.0200 0.0100 0.0100 0.0075 0.0075 0.0069 0.0066 0.0066 0.0065 0.0064 0.0064 0.0064

f (x s ) 0.0027 0.0027 0.0006 0.0006 0.0001 0.0001 8E-05 1E-05 1E-05 1E-05 6E-06 2E-06

f (x m ) -0.0015 0.0006 -0.0004 0.0001 -0.0002 -0.0001 1E-05 -2E-05 -2E-06 6E-06 2E-06 1E-07

Problem 4-15. BISECTION METHOD Lower Upper ROOT x x Fl(x) Fu(x) ------------ ------- ------- ------- -------3.7142720 -3.7200 -3.7100 .1243 -.0936 -1.3076950 -1.3100 -1.3000 -.0305 .1030 2.5000000 2.4900 2.5000 .3425 -.0006 Problem 4-16. BISECTION METHOD Lower Upper ROOT x x Fl(x) Fu(x) ------------ ------- ------- ------- -------.4999997 -.5100 -.5000 -2.2291 .0042 .3333223 .3300 .3400 .4890 -.9913 1.5000000 1.4900 1.5000 -4.1303 .0076

4.5 Newton-Raphson Iteration Problem 4-17.

f(x) = x3 - 0.39x + 0.07 f ' ( x ) = 3x 2  0.39 f ( xi ) x i+1 = x i  f ' ( xi )

35

f (x e ) f (x s ).f (x m ) f (x m ).f (x e ) -0.0054 + -0.0015 + -0.0015 + -0.0004 + -0.0004 + -0.0002 + -5E-05 + -5E-05 + -2E-05 + -2E-06 + -2E-06 + -2E-06 + -

r(%) ---100.0 33.3 20.0 9.1 4.8 2.4 1.2 0.6 0.3 0.2 0.1

r (%) i xi f (x i) f '(x i) x i+1 0 1 0.68 2.61 0.73946 ----1 0.7395 0.1860 1.2504 0.5908 35.2 2 0.5908 0.0458 0.6570 0.5211 25.2 3 0.5211 0.0083 0.4246 0.5016 13.4 4 0.5016 0.0006 0.3648 0.5000 3.9 5 0.5000 0.0000 0.3600 0.5000 0.3 From the above table, the first root is 0.5. The polynomial can be reduced as _ x2 + 0.50x - 0.14 3 x - 0.5 | x - 0.39x + 0.07 x3 - 0.50x2 _ 2 0.50x - 0.39x + 0.07 0.50x2 - 0.25x _ - 0.14x + 0.07 - 0.14x + 0.07 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 + 0.50x - 0.14 The following table summarizes the results. i xi f (x i) f '(x i) 0 1 1.36 2.5 1 0.4560 0.29594 0.91587 2 0.1329 -0.0559 0.53531 3 0.2373 0.03497 0.61263 4 0.1802 -0.0174 0.56496 5 0.2110 0.01005 0.58907 6 0.1940 -0.00539 0.57525 7 0.2033 0.00302 0.5827 8 0.1982 -0.00165 0.57854 9 0.2010 0.00092 0.58081 From the above table, a second root of 0.2010  0.2 is found. technique once again, the third root can be found as follows: x + 0.70 x - 0.2 | x2 + 0.50x - 0.14 x2 - 0.20x 0.70x - 0.14 0.70x - 0.14 0 = error Summary of roots: x1 = 0.5 x2 = 0.2 x3 = -0.7

36

r (%) x i+1 0.456 ----0.1329 119.3 0.2373 243.2 0.1802 44.0 0.2110 31.7 0.1940 14.6 0.2033 8.8 0.1982 4.6 0.2010 2.6 0.1994 1.4 Using the polynomial reduction

_ _ _

Problem 4-18.

f(x) = x3 - 1.7x2 + 0.84x - 0.108 f ' ( x ) = 3x 2  3.4 x  0.84 f ( xi ) x i+1 = x i  f ' ( xi ) r (%) i xi f (x i) f '(x i) x i+1 0 1 0.032 0.44 0.92727 ----1 0.9273 0.00649 0.26678 0.9029 7.8 2 0.9029 0.00063 0.21591 0.9000 2.7 3 0.9000 8.5E-06 0.21008 0.9000 0.3 From the above table, the first root is 0.9. Reducing the polynomial results in x2 - 0.8x + 0.12 _ x - 0.9 | x3 - 1.7x2 + 0.84x - 0.108 x3 - 0.9x2 _ - 0.8x2 + 0.84x - 0.108 - 0.8x2 + 0.72x _ 0.12x - 0.108 0.12x - 0.108 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 -0.8x + 0.12 The following table summarizes the results.

r (%) i xi f (x i) f '(x i) x i+1 0 1 0.32 1.2 0.73333 ----1 0.7333 0.07111 0.66667 0.6267 36.4 2 0.6267 0.01138 0.45333 0.6016 17.0 3 0.6016 0.00063 0.40314 0.6000 4.2 4 0.6000 2.4E-06 0.40001 0.6000 0.3 From the above table, a second root of 0.6 is found. Using the polynomial reduction technique once again, the third root can be found as follows: x - 0.2 _ x - 0.6 | x2 - 0.8x + 0.12 x2 - 0.6x _ - 0.2x + 0.12 - 0.2x + 0.12 _ 0 = error Summary of roots: x1 = 0.9 x2 = 0.6 x3 = 0.2

Problem 4-19.

f(x) = x3 - 2x2 - 5x + 6 f '( x ) = 3x 2  4 x  5 37

x i+1 = x i 

f ( xi ) f ' ( xi )

r (%) i xi f (x i) f '(x i) x i+1 0 4 18 27 3.33333 ----1 3.3333 4.14815 15 3.0568 20.0 2 3.0568 0.59066 10.8047 3.0021 9.0 3 3.0021 0.02126 10.0297 3.0000 1.8 From the above table, the first root is 3.0021  3. Reducing the polynomial results in x2 + x - 2 _ 3 2 x - 3 | x - 2x - 5x + 6 x3 - 3x2 _ 2 x - 5x + 6 x2 - 3x _ - 2x + 6 - 2x + 6 _ 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 + x - 2 The following table summarizes the results. xi f (x i) i 0 3 10 1 1.5714 -2.91545 2 0.8195 1.10939 3 0.9967 0.01995 4 1.0000 1.1E-05 From the above table, a second root of 1 is found. again, the third root can be found as follows: x +2 _ 2 x-1|x + x-2 x2 - x _ 2x - 2 2x - 2 _ 0 = error Summary of roots: x1 = 3 x2 = 1 x3 = -2

r (%) x i+1 7 1.57143 -----3.87755 0.8195 90.9 -6.26321 0.9967 91.7 -6.00661 1.0000 17.8 -6 1.0000 0.3 Using the polynomial reduction technique once f '(x i)

Problem 4-20.

f(x) = x3 + x2 - 17x +15 f '( x ) = 3x 2  2 x  17 f ( xi ) x i+1 = x i  f ' ( xi ) 38

r (%) i xi f (x i) f '(x i) x i+1 0 2 -7 -1 -5 ----1 -5.0000 0 48 -5.0000 140.0 2 -5.0000 0 48 -5.0000 0.0 From the above table, the first root is -5. Therefore, _ x2 - 4x + 3 x + 5 | x3 + x2 - 17x + 15 x3 + 5x2 _ - 4x2 - 17x + 15 - 4x2 - 20x _ 3x + 15 3x + 15 _ 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 - 4x + 3 The following table summarizes the results: i xi f (x i) 0 5 8 1 3.6667 1.77778 2 3.1333 0.28444 3 3.0078 0.01575 4 3.0000 6.1E-05 From the above table, a second root of 3 is found. more, the third root can be found as follows:

r (%) x i+1 6 3.66667 ----3.33333 3.1333 36.4 2.26667 3.0078 17.0 2.01569 3.0000 4.2 2.00006 3.0000 0.3 Using the polynomial reduction technique once f '(x i)

x-1 _ x - 3 | x2 - 4x + 3 x2 - 3x _ -x + 3 -x + 3 _ 0 = error Summary of roots: x1 = -5 x2 = 3 x3 = 1 Problem 4-21.

f(x) = x3 - 6.1 x2 + 11.26x - 6.336 f ' ( x ) = 3x 2  12.2 x  1126 . f ( xi ) x i+1 = x i  f ' ( xi )

39

r (%) i xi f (x i) f '(x i) x i+1 0 1 -0.176 2.06 1.08544 ----1 1.0854 -0.022 1.55219 1.0996 7.9 2 1.0996 -0.00057 1.47217 1.1000 1.3 From the above table, the first root is 1.1. Therefore, _ x2 - 5x+5.76 x - 1.1 | x3 - 6.1x2 + 11.26x - 6.336 x3 - 1.1x2 _ - 5.0x2 + 11.26x - 6.336 - 5.0x2 + 5.50x _ 5.76x - 6.336 _ 5.76 - 6.336 _ 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 - 5x + 5.76 The following table summarizes the results: r (%) i xi f (x i) f '(x i) x i+1 0 1 1.76 -3 1.58667 ----1 1.5867 0.34418 -1.82667 1.7751 37.0 2 1.7751 0.0355 -1.44983 1.7996 10.6 3 1.7996 0.0006 -1.40086 1.8000 1.4 From the above table, a second root of 1.8 is found. Using the polynomial reduction technique once more, the third root can be found as follows: x - 3.2 _ x - 1.8 | x2 - 5 x + 5.76 x2 - 1.8x _ - 3.2x + 5.76 - 3.2x + 5.76 _ 0 = error Summary of roots: x1 = 1.1 x2 = 1.8 x3 = 3.2

Problem 4-22.

f(x) = x3 + 3x2 - 4 f '( x ) = 3x 2  6 x f ( xi ) x i+1 = x i  f ' ( xi )

40

 

r (%) i xi f (x i) f '(x i) x i+1 0 -3 -4 9 -2.55556 ----1 -2.5556 -1.09739 4.25926 -2.2979 17.4 2 -2.2979 -0.29268 2.05368 -2.1554 11.2 3 -2.1554 -0.07619 1.00478 -2.0796 6.6 4 -2.0796 -0.01949 0.49636 -2.0403 3.6 5 -2.0403 -0.00493 0.2466 -2.0203 1.9 From the above table, the first root is -2.0203  -2. Therefore, _ x2 + x - 2 x + 2 | x3 + 3x2 -4 x3 + 2x2 _ 2 x -4 x2 + 2x _ - 2x - 4 - 2x - 4 _ 0 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 + x - 2 The following table summarizes the results:  

r (%) i xi f (x i) f '(x i) x i+1 0 2 4 5 1.2 ----1 1.2000 0.64 3.4 1.0118 66.7 2 1.0118 0.03543 3.02353 1.0000 18.6 3 1.0000 0.00014 3.00009 1.0000 1.2 From the above table, a second root of 1 is found. Using the polynomial reduction technique once more, the third root can be found as follows: x+2 _ x - 1 | x2 + x - 2 x2 - x _ 2x - 2 2x - 2 _ 0 = error Summary of roots: x1 = -2 x2 = 1 (Note: the double root) x3 = -2

Problem 4-23.

f(x) = x4 - 15x2 - 6x + 24 f '( x ) = 4 x 3  30 x  6 f ( xi ) x i+1 = x i  f ' ( xi )

41

 

r (%) i xi f (x i) f '(x i) x i+1 0 1 4 -32 1.125 ----1 1.1250 -0.13257 -34.0547 1.1211 11.1 2 1.1211 -0.00011 -33.9968 1.1211 0.3 From the above table, the first root is 1.1211. Therefore, _ x3 + 1.1211x2 - 13.7431x - 21.4074 4 2 x - 1.1211 | x - 15x - 6x + 24 x4 - 1.1211x3 _ 1.1211x3 - 15.0000x2 - 6x + 24 1.1211x3 - 1.2569x2 _ - 13.7431x2 - 6.0000x + 24 - 13.7431x2 + 15.4074x _ - 21.4074x + 24 - 21.4074x + 23.9998 0.0002 = error The Newton-Raphson method can be used now to find a root for the reduced polynomial x3 + 1.1211x2 - 13.7431x - 21.4074 The following table summarizes the results:  

r (%) i xi f (x i) f '(x i) x i+1 0 1 -33.0294 -8.5009 -2.8854 ----1 -2.8854 3.5582 4.76386 -3.6323 134.7 2 -3.6323 -4.62039 17.6937 -3.3712 20.6 3 -3.3712 -0.64881 12.7927 -3.3205 7.7 4 -3.3205 -0.023 11.8882 -3.3185 1.5 From the above table, a second root of -3.3185 is found. Using the polynomial reduction technique again, the third root can be found as follows: x 3  11211 . x 2  13.7431x  214074 .  x 2  2.1974 x  214074 . x  3.3185 The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 - 2.197x - 21.4074 The following table summarizes the results:  

r (%) i xi f (x i) f '(x i) x i+1 0 1 -7.6484 -0.1974 -37.7457 ----1 -37.7457 1501.23 -77.6888 -18.4221 102.6 2 -18.4221 373.402 -39.0415 -8.8578 104.9 3 -8.8578 91.4745 -19.9131 -4.2641 108.0 4 -4.2641 21.102 -10.7257 -2.2967 107.7 5 -2.2967 3.87076 -6.79085 -1.7267 85.7 6 -1.7267 0.3249 -5.65085 -1.6692 33.0 7 -1.6692 0.00331 -5.53586 -1.6686 3.4 8 -1.6686 3.6E-07 -5.53467 -1.6686 0.0 From the above table, a third root of -1.6686 is found. Using the polynomial reduction technique once more, the fourth root can be found as follows:

42

x 2  2.1974 x  6.4510  x  38660 . x  16686 . Hence, the fourth root is 3.8660 Summary of roots: x1 = x2 = x3 = x4 =

1.1211 -3.3185 -1.6686 3.8660

Problem 4-24.

f(x) = x4 - 10x2 + 9 f '( x ) = 4 x 3  20 x f ( xi ) x i+1 = x i  f ' ( xi ) r (%) i xi f (x i) f '(x i) x i+1 0 2 -15 -8 0.125 ----1 0.1250 8.84399 -2.49219 3.6737 1500.0 2 3.6737 56.1816 124.846 3.2237 96.6 3 3.2237 13.0749 69.5299 3.0356 14.0 4 3.0356 1.7668 51.1816 3.0011 6.2 5 3.0011 0.05347 48.098 3.0000 1.2 From the above table, the first root is 3. Therefore, x 4  10 x 2  9  x 3  3x 2  x  3 x3 The Newton-Raphson method can be used now to find a root for the reduced polynomial x3 + 3x2 - x - 3 The following table summarizes the results: r (%) i xi f (x i) f '(x i) x i+1 0 2 15 23 1.34783 ----1 1.3478 3.55059 12.5369 1.0646 48.4 2 1.0646 0.54223 8.7879 1.0029 26.6 3 1.0029 0.02335 8.03497 1.0000 6.2 4 1.0000 5.1E-05 8.00008 1.0000 0.3 From the table, a second root of 1 is found. Using the polynomial reduction technique again, the third root can be found as follows: x 3  3x 2  x  3  x2  4x  3 x 1 The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 + 4x + 3 The following table summarizes the results:

43

r (%) i xi f (x i) f '(x i) x i+1 0 1 8 6 -0.33333 ----1 -0.3333 1.77778 3.33333 -0.8667 400.0 2 -0.8667 0.28444 2.26667 -0.9922 61.5 3 -0.9922 0.01575 2.01569 -1.0000 12.6 4 -1.0000 6.1E-05 2.00006 -1.0000 0.8 From the above table, a third root of -1 is found. Using the polynomial reduction technique once more, the fourth root can be found as follows: x 2  4x  3  x3 x 1 Hence, the fourth root is -3.

Summary of roots: x1 = x2 = x3 = x4 =

3 1 -1 -3

Problem 4-25.

f(x) = x4 - 2x3 -15x2 - 4x + 20 f '( x ) = 4 x 3  6 x 2  30 x  4 f ( xi ) x i+1 = x i  f ' ( xi ) r (%) i xi f (x i) f '(x i) x i+1 0 2 -48 -56 1.14286 ----1 1.1429 -5.44273 -40.1516 1.0073 75.0 2 1.0073 -0.26369 -36.2188 1.0000 13.5 3 1.0000 -0.00079 -36.0007 1.0000 0.7 From the above table, the first root is 1. Therefore, x 4  2 x 3  15x 2  4 x  20  x 3  x 2  16 x  20 x 1 The Newton-Raphson method can be used now to find a root for the reduced polynomial x3 - x2 - 16x - 20 The following table summarizes the results: i 0 1 2 3 4 5

xi 1 -1.5714 -1.7434 -1.8298 -1.8777 -1.9067

f (x i) -36 -1.207 -0.44416 -0.19787 -0.10279 -0.06006

f '(x i) x i+1 -14 -1.57143 -7.02041 -1.7434 -5.13878 -1.8298 -4.12583 -1.8777 -3.54444 -1.9067 -3.18618 -1.9256

44

r (%) ----163.6 9.9 4.7 2.6 1.5

From the above table, a second root of -1.9256  -2 is found. Using the polynomial reduction technique again, the third root can be found as follows: x 3  x 2  16 x  20  x 2  3x  10 x2 The Newton-Raphson method can be used now to find a root for the reduced polynomial x2 - 3x - 10 The following table summarizes the results: r (%) i xi f (x i) f '(x i) x i+1 0 1 -12 -1 -11 ----1 -11.0000 144 -25 -5.2400 109.1 2 -5.2400 33.1776 -13.48 -2.7788 109.9 3 -2.7788 6.05773 -8.55751 -2.0709 88.6 4 -2.0709 0.5011 -7.14174 -2.0007 34.2 5 -2.0007 0.00492 -7.00141 -2.0000 3.5 6 -2.0000 4.9E-07 -7 -2.0000 0.0 From the above table, a third root of -2 is found. Using the polynomial reduction technique once more, the fourth root can be found as follows: x 2  3x  10  x 5 x2 Hence, the fourth root is 5. Summary of roots: x1 = x2 = x3 = x4 =

1 -2 -2 5

( Note: double root)

Problem 4-26.

f(x) = x4 - 5.6x3 -39.97x2 - 12.33x + 24.288 f ' ( x ) = 4 x 3  16.8 x 2  79.94 x  12.33 f ( xi ) x i+1 = x i  f ' ( xi ) r (%) i xi f (x i) f '(x i) x i+1 0 1 -32.612 -105.07 0.68962 ----1 0.6896 -4.83395 -74.1357 0.6244 45.0 2 0.6244 -0.20626 -67.8219 0.6214 10.4 3 0.6214 -0.00044 -67.5293 0.6214 0.5 From the above table, the first root is 0.6214. Therefore, x 4 - 5.6 x 3 - 39.97 x 2 - 12.33x + 24.288  x 3  4.9786 x 2  43.0637 x  39.0898 x  0.6214 The Newton-Raphson method can be used now to find a root for the reduced polynomial x 3  4.9786 x 2  43.0637 x  39.0898 The following table summarizes the results:

45

r (%) i xi f (x i) f '(x i) x i+1 0 1 -86.1321 -50.0209 -0.72192 ----1 -0.7219 -10.9721 -34.3119 -1.0417 238.5 2 -1.0417 -0.76326 -29.4359 -1.0676 30.7 3 -1.0676 -0.00547 -29.0136 -1.0678 2.4 4 -1.0678 -2.9E-07 -29.0105 -1.0678 0.0 From the above table, a second root of -1.0678 is found. Using the polynomial reduction technique again, the third root can be found as follows: x 3  4.9786 x 2  43.0637 x  39.0898  x 2  6.0464 x  36.6074 x  10678 . The Newton-Raphson method can be used now to find a root for the reduced polynomial x 2  6.0464 x  36.6074 The following table summarizes the results: r (%) i xi f (x i) f '(x i) x i+1 0 1 -41.6538 -4.0464 -9.29404 ----1 -9.2940 105.967 -24.6345 -4.9925 110.8 2 -4.9925 18.5036 -16.0313 -3.8382 86.2 3 -3.8382 1.33222 -13.7229 -3.7412 30.1 4 -3.7412 0.00942 -13.5287 -3.7405 2.6 5 -3.7405 4.9E-07 -13.5273 -3.7405 0.0 From the above table, a third root of -3.7405 is found. Using the polynomial reduction technique once more, the fourth root can be found as follows: x 2  6.0464 x  36.6074  x  9.7869 x  3.7405 Hence, the fourth root is 9.7869.

Summary of roots: x1 = x2 = x3 = x4 =

0.6214 -1.0678 -3.7405 9.7869

Problem 4-27. ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 3.0000 1 2.5800 -21.0560 -50.13600 .162781 2 2.5026 -2.8511 -36.81534 .030946 3 2.5000 -.0889 -34.52728 .001030 ROOT = 2.50000

ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -1.0000 46

1 -1.3001 4.2560 14.18400 .230803 2 -1.3077 .1020 13.37305 .005832 3 -1.3077 .0001 13.34219 .000007 ROOT = -1.30769 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -4.5000 1 -3.9216 25.5640 -44.19600 .147498 2 -3.7351 5.0661 -27.16638 .049927 3 -3.7145 .4586 -22.29912 .005536 4 -3.7143 .0053 -21.78100 .000066 ROOT = -3.71429 Problem 4-28. ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 1.0000 1 .6988 -.0212 -.07028 .431015 2 .7200 .0017 -.07914 .029408 3 .7200 .0000 -.07896 .000031 ROOT = .72000

ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -2.0000 1 -1.4399 -.2513 .44872 .388985 2 -1.2071 -.0580 .24933 .192828 3 -1.1617 -.0082 .18031 .039104 4 -1.1600 -.0003 .16778 .001467 ROOT = -1.16000 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 3.0000 1 2.5737 .1423 .33372 .165647 2 2.4214 .0301 .19729 .062908 3 2.4004 .0033 .15515 .008736 4 2.4000 .0001 .14962 .000162 ROOT = 2.40000 Problem 4-29. ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 1.0000 1 .8241 .2056 1.16920 .213407 2 .7231 .0564 .55850 .139714 3 .6751 .0140 .29162 .071095

47

4 .6612 .0026 .18629 .020970 5 .6600 .0002 .15843 .001859 ROOT = .66001 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 .0000 1 -.2153 .0364 .16920 1.000000 2 -.1391 -.0563 .73893 .548318 3 -.1210 -.0091 .50535 .149255 4 -.1200 -.0005 .45514 .008356 5 -.1200 .0000 .45241 .000025 ROOT = -.12000 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 .5000 1 .4509 -.0040 -.08080 .108915 2 .4598 .0011 -.12267 .019280 3 .4600 .0000 -.11619 .000531 ROOT = .46000 Problem 4-30.

f(x) = x3 - 2x2 - 5x + 6 From the figure below, the intervals are: Root Interval 1 (-3,-1) 2 (0,1.5) 3 (2.5,4) Graph of f (x ) 20 15 10 5 0 f (x ) -5 -10 -15 -20 -25 -3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

x

Problem 4-31.

f(x) = x3 - 3.5x2 + 3.28x - 0.924 From the two figures below, the intervals are:

48

4.0

Root 1 2 3

Interval (2,2.5) (0.5,0.65) (0.65,0.8)

Graph of f (x ) with large interval 1 0.5 f (x ) 0 -0.5 -1 0.0

1.0

2.0 x

Graph of f (x ) with a smaller interval 0.02 0 f (x ) -0.02 -0.04 -0.06 0.48

0.56

0.64

0.72

0.80

x

Problem 4-32.

f(x) = x4 - 10x2 + 9 From the figure below, the intervals are:

Root 1 2 3 4

Interval (-3.5,-2.5) (-1.5,-0.5) (0.5,1.5) (2.5,3.5)

Graph of f (x ) 40 20 f (x )

0 -20 -3.5

-2.5

-1.5

-0.5

0.5

1.5

x

Problem 4-33.

f(x) = x4 - 15x2 -6x + 24 49

2.5

3.5

From the figure below, the intervals are: Interval (-4,-3) (-2,-1) (1,2) (3,4)

Root 1 2 3 4

Graph of f (x ) 100 50 f (x )

0 -50 -4.0 -3.0 -2.0 -1.0 0.0

1.0

2.0

3.0

4.0

x

4.6 Secant Method Problem 4-34.

f(x) = x3 - 1.7x2 + 0.84x - 0.108 f ( x i )[ x i-1  x i ] x i+1 = x i  f ( x i-1 ) - f ( x i ) Using initial estimates of x0 = 0, and x1 = 1, the following table summarizes the results: r (%) i xi f (x i-1) Comments initial condition 0 0 -0.108 100 initial condition 1 1 0.032 23 2 0.77143 -0.01259 6 3 0.83598 -0.00961 21 4 1.04362 0.05375 18 5 0.86747 -0.00581 2 6 0.88465 -0.00299 2 7 0.9029 0.00062 0 ---8 0.89978 -4.7E-05 First root = 0.89978  0.9. Therefore, the reduced polynomial (see Problem 4-10) is x2 - 0.8x + 0.12 Using the secant method with the same initial conditions the following table summarizes the results:

50

i xi f (x i-1) Comments r (%) 0 0 0.12 100 initial condition 1 1 0.32 160 initial condition 2 -0.6 0.96 240 3 1.8 1.92 480 4 -3 11.52 576 5 2.76 5.5296 532 6 8.07692 58.8951 587 7 2.20907 3.23275 34 8 1.86828 2.11585 65 9 1.22268 0.63681 28 10 0.94472 0.25672 19 11 0.75698 0.08743 10 12 0.66001 0.02761 4 13 0.61527 0.00634 1 14 0.60193 0.00077 --With a second root of 0.60193  0.6, and by the polynomial reduction, the third root was found to be 0.2.

Summary of roots: x1 = 0.9 x2 = 0.6 x3 = 0.2 Problem 4-35.

f(x) = x3 + x2 - 17x + 15 f ( x i )[ x i-1  x i ] x i+1 = x i  f ( x i-1 ) - f ( x i ) Using initial estimates of x0 = 0, and x1 = 1, the following table summarizes the results: r (%) i xi f (x i-1) Comments 0 0 15 100 initial condition 1 1 0 0 initial condition 2 1 0 ---The first root = 1. Therefore, the reduced polynomial (see Problem 4-12) is x2 + 2x - 15 Using the secant method with the same initial conditions, the following table summarizes the results:

51

i xi f (x i-1) Comments r (%) 0 0 -15 100 initial 1 1 -12 400 initial 2 5 20 250 3 2.5 -3.75 39 4 2.89474 -0.83102 11 5 3.00712 0.05699 1 6 2.99991 -0.00076 --With a second root of 2.99991  3, and by polynomial reduction, the third root was found to be -5 (see Problem 4-12).

Summary of roots: x1 = 1 x2 = 3 x3 = -5 Problem 4-36.

f(x) = x3 - 6.1x2 + 11.26x - 6.336 f ( x i )[ x i-1  x i ] x i+1 = x i  f ( x i-1 ) - f ( x i ) Using initial estimates of x0 = 0, and x1 = 1, the following table summarizes the results: r (%) i xi f (x i-1) Comments 0 0 -6.336 100 initial condition initial condition 1 1 -0.176 3 2 1.02857 -0.11965 6 3 1.08924 -0.01615 1 4 1.0987 -0.00191 ---The first root = 1.0987  1.1. Therefore, the reduced polynomial (see Problem 4-13) is x2 - 5x + 5.76 Using the secant method with the same initial conditions, the following table summarizes the results: i xi f (x i-1) Comments r (%) 0 0 5.76 100 initial condition initial condition 1 1 1.76 44 2 1.44 0.6336 25 3 1.6875 0.17016 9 4 1.77837 0.03075 2 5 1.79841 0.00222 0 6 1.79998 3.4E-05 --With a second root of 1.79998  1.8, and by polynomial reduction, the third root was found to be 3.2 (see Problem 4-13).

52

Summary of roots: x1 = 1.1 x2 = 1.8 x3 = 3.2 Problem 4-37.

f(x) = x4 - 5.6x3 - 39.97x2 - 12.33x + 24.288 f ( x i )[ x i-1  x i ] x i+1 = x i  f ( x i-1 ) - f ( x i ) Using initial estimates of x0 = 0, and x1 = 1, the following table summarizes the results: r (%) i xi f (x i-1) Comments 0 0 24.288 100 initial condition initial condition 1 1 -32.612 57 2 0.42685 11.3398 15 3 0.57473 3.04495 5 4 0.62901 -0.51926 1 5 0.6211 0.01759 ---The first root = 0.6211. Therefore, the reduced polynomial (see Problem 4-18) is x3 - 4.9786x2 - 43.0637x -39.0898 Using the secant method with the same initial conditions, the following table summarizes the results: i xi f (x i-1) Comments r (%) initial condition 0 0 -39.0898 100 1 1 -86.1321 183 initial condition 2 -0.83095 -7.31738 17 3 -1.00094 -1.97639 6 4 -1.06384 -0.11536 0 5 -1.06774 -0.00211 ---The second root = -1.06774. Therefore, the reduced polynomial (see Problem 4-18) is x2 - 6.0464x -36.6074 Using the secant method with the same initial conditions, the following table results:

i xi f (x i-1) Comments r (%) initial condition 0 0 -36.6074 100 1 1 -41.6538 825 initial condition 2 -7.25416 59.877 487 3 -2.38633 -16.4841 105 4 -3.43715 -4.01099 34 5 -3.77507 0.46927 4 6 -3.73967 -0.01071 0 7 -3.74046 -2.7E-05 ---With a third root of -3.74046, and by polynomial reduction, the fourth root was found to be 9.7869 (see Problem 4-18).

53

Summary of roots: x1 = x2 = x3 = x4 =

0.6211 -1.06774 -3.74046 9.78690

Problem 4-38. Root 4 = .0064407 1.00000 -1.73365 .93089 -.41929 Root 3 = 1.2587390 1.00000 -.47490 .33311 Root 2 = .2683893 1.00000 .32045 Root 1 = -.3204473 Problem 4-39. ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 1.0000 1 1.0000 .0000 -16.00000 .000000 ROOT = 1.00000

ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -1.0000 1 -1.0000 .0000 16.00000 .000000 ROOT = -1.00000 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 2.0000 1 .1250 -15.0000 -8.00000 15.000000 2 3.6737 8.8440 -2.49219 .965974 3 3.2237 56.1816 124.84630 .139594 4 3.0356 13.0749 69.52986 .061947 5 3.0011 1.7668 51.18160 .011502 6 3.0000 .0535 48.09798 .000371 ROOT = 3.00000 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -2.0000 1 -.1250 -15.0000 8.00000 15.000000 2 -3.6737 8.8440 2.49219 .965974 3 -3.2237 56.1816-124.84630 .139594 4 -3.0356 13.0749 -69.52986 .061947 5 -3.0011 1.7668 -51.18160 .011502 54

6 -3.0000 .0535 -48.09798 .000371 ROOT = -3.00000 Problem 4-40. ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 5.0000 1 4.2907 244.0000 344.00000 .165312 2 3.9539 61.0358 181.24750 .085169 3 3.8709 10.1829 122.64030 .021450 4 3.8661 .5343 109.87910 .001258 ROOT = 3.86605

ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 1.0000 1 1.1250 4.0000 -32.00000 .111111 2 1.1211 -.1326 -34.05469 .003472 ROOT = 1.12111 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -1.0000 1 -1.8000 16.0000 20.00000 .444444 2 -1.6661 -3.3024 24.67200 .080336 3 -1.6686 .0626 25.48321 .001472 ROOT = -1.66860 ROUND X f(X) f`(X) ERROR ----- ------- ------- ------- ------0 -7.0000 1 -5.5171 1732.0000********** .268777 2 -4.4882 527.0344-512.22140 .229250 3 -3.8249 154.5493-232.99500 .173421 4 -3.4640 41.5329-115.08220 .104185 5 -3.3356 8.7770 -68.34113 .038503 6 -3.3188 .9110 -54.37854 .005048 7 -3.3185 .0145 -52.65564 .000083 ROOT = -3.31853

4.8 Synthetic Division Problem 4-41.

f(x) = x3 - 0.6x2 + 0.11x + 0.006 R (x ) x i+1 = x i  0 i R1 ( x i )

55

i xi r(%) R0 R1 b3 b2 b1 b0

0

1 0.736 35.8 0.149 0.852 1 -0.6 0.11 -0.01

1 --0.504 1.91 1 -0.6 0.11 -0.01

2 0.562 31.1 0.044 0.382 1 -0.6 0.11 -0.01

3 0.447 25.5 0.013 0.174 1 -0.6 0.11 -0.01

4 0.374 19.5 0.004 0.081 1 -0.6 0.11 -0.01

5 0.331 13.3 9E-04 0.041 1 -0.6 0.11 -0.01

6 0.308 7.3 2E-04 0.025 1 -0.6 0.11 -0.01

7 8 0.301 0.3 2.4 0.3 2E-05 2E-07 0.021 0.02 1 1 -0.6 -0.6 0.11 0.11 -0.01 -0.01

1 1 1 1 1 1 1 0.4 0.136 -0.04 -0.15 -0.23 -0.27 -0.29 0.51 0.21 0.088 0.042 0.026 0.021 0.02

c3 c2 c1

1 -0.3 0.02

9 0.3

1 -0.3 0.02

1 1 1 1 1 1 1 1 1 d3 d2 1.4 0.872 0.523 0.295 0.149 0.061 0.016 0.002 2E-05 d1 1.91 0.852 0.382 0.174 0.081 0.041 0.025 0.021 0.02 From the above table, the first root is 0.3. Using synthetic division for the second order polynomial x2 -0.3x + 0.2, the following table summarizes the results: i xi r(%) R0 R1 b2 b1 b0 c2 c1

0 1 --0.72 1.7 1 -0.3 0.02

1 2 3 4 5 6 0.5765 0.3662 0.2639 0.2179 0.2023 0.2 73.5 57.4 38.8 21.1 7.7 1.2 0.1794 0.0442 0.0105 0.0021 0.0002 5E-06 0.8529 0.4323 0.2277 0.1358 0.1047 0.1001 1 1 1 1 1 1 -0.3 -0.3 -0.3 -0.3 -0.3 -0.3 0.02 0.02 0.02 0.02 0.02 0.02

1 1 1 1 1 1 0.7 0.2765 0.0662 -0.036 -0.082 -0.098

7 0.2

1 -0.1

1 1 1 1 1 1 1 d2 d1 1.7 0.8529 0.4323 0.2277 0.1358 0.1047 0.1001 From the above table, a second root of 0.2 is found. The values c2 = 1, and c1 = -0.1 give the polynomial x - 0.1 = 0 Thus, the third root is 0.1. Summary of roots: x1 = 0.3 x2 = 0.2 x3 = 0.1

56

Problem 4-42.

f(x) = x3 + 0.9x2 - 10.56x - 16.94 R (x ) x i+1 = x i  0 i R1 ( x i ) i xi r(%) R0 R1 b3 b2 b1 b0 c3 c2 c1

0 1.00 ---25.60 -5.76 1.00 0.90 -10.56 -16.94

1 2 3 4 5 6 7 -3.444 -2.873 -2.554 -2.382 -2.292 -2.247 -2.223 129.0 19.9 12.5 7.2 3.9 2.0 -10.75 -2.89 -0.757 -0.194 -0.049 -0.012 18.833 9.037 4.406 2.1718 1.0776 0.5367 1 1 1 1 1 1 0.9 0.9 0.9 0.9 0.9 0.9 -10.56 -10.56 -10.56 -10.56 -10.56 -10.56 -16.94 -16.94 -16.94 -16.94 -16.94 -16.94

1.00 1 1 1 1 1 1 1.90 -2.544 -1.973 -1.654 -1.482 -1.392 -1.347 -8.66 -1.796 -4.89 -6.337 -7.031 -7.368 -7.535

1.00 1 1 1 1 1 1 d3 d2 2.90 -5.989 -4.847 -4.207 -3.864 -3.685 -3.593 d1 -5.76 18.833 9.037 4.406 2.1718 1.0776 0.5367 From the above table, the first root is -2.223. Using synthetic division for the second order polynomial x2 -1.347x -7.535, the following table summarizes the results: i 0 1 2 xi 1 13.061 7.1894 r(%) --92.3 81.7 R0 -7.882 145.46 34.472 R 1 0.6535 24.775 13.032 b2 1 1 1 b1 -1.347 -1.347 -1.347 b0 -7.535 -7.535 -7.535 c2 c1

3 4.5443 58.2 6.9967 7.7421 1 -1.347 -7.535

4 3.6406 24.8 0.8167 5.9347 1 -1.347 -7.535

5 3.503 3.9 0.0189 5.6594 1 -1.347 -7.535

6 7 3.4996 3.4996 0.1 1E-05 5.6527 1 -1.347 -7.535

1 1 1 1 1 1 1 -0.347 11.714 5.8429 3.1978 2.2941 2.1565 2.1531

1 1 1 1 1 1 1 d2 d1 0.6535 24.775 13.032 7.7421 5.9347 5.6594 5.6527 From the above table, a second root of 3.4996 = 3.5 is found. The values c2 = 1, and c1 = 2.1531  2.2 give the polynomial x + 2.2 = 0 Thus, the third root is -2.2.

57

Summary of roots: x1 = -2.2 x2 = 3.5 x3 = -2.2 Problem 4-43.

f(x) = x3 + 0.6x2 - 4.77x + 2.43, and x i+1 = x i 

i xi r(%) R0 R1 b3 b2 b1 b0

0 1 2 3 4 1.00 -0.298 0.4998 0.5935 0.6 --- 435.3 159.7 15.8 1.1 -0.74 3.8795 0.3206 0.0193 1E-04 -0.57 -4.861 -3.421 -3.001 -2.97 1.00 1.00 1.00 1.00 1.00 0.60 0.60 0.60 0.60 0.60 -4.77 -4.77 -4.77 -4.77 -4.77 2.43 2.43 2.43 2.43 2.43 1.00 1 1 1 1.60 0.3018 1.0998 1.1935 -3.17 -4.86 -4.22 -4.062

c3 c2 c1

R0 ( x i ) R1 ( x i )

5 0.6

1 1.2 -4.05

1.00 1 1 1 1 d3 d2 2.60 0.0035 1.5997 1.7871 1.7999 d1 -0.57 -4.861 -3.421 -3.001 -2.97 From the above table, the first root is 0.6. Using synthetic division for the second order polynomial x2 + 1.2x - 4.05, the following table summarizes the results: i xi r(%) R0 R1 b2 b1 b0 c2 c1

0

1 2 3 1 1.5782 1.5014 1.5 --36.6 5.1 0.1 -1.85 0.3343 0.0059 2E-06 3.2 4.3563 4.2028 4.2 1 1 1 1 1.2 1.2 1.2 1.2 -4.05 -4.05 -4.05 -4.05 1 1 1 2.2 2.7781 2.7014

4 1.5

1 2.7

1 1 1 1 d2 d1 3.2 4.3563 4.2028 4.2 From this table, a second root of 1.5 is found. The values c2 = 1, and c1 = 2.7 give the polynomial x + 2.7 = 0 58

Thus, the third root is -2.7. Summary of roots: x1 = 0.6 x2 = 1.5 x3 = -2.7 Problem 4-44.

f(x) = x3 - 13.1x2 + 48.48x - 46.62 R (x ) x i+1 = x i  0 i R1 ( x i ) i xi r(%) R0 R1 b3 b2 b1 b0

0 1.00 ---10.24 25.28 1.00 -13.10 48.48 -46.62

1 1.4051 28.8 -1.591 17.59 1.00 -13.10 48.48 -46.62

2 3 1.4955 1.5 6.0 0.3 -0.072 -2E-04 16.008 15.93 1.00 1.00 -13.10 -13.10 48.48 48.48 -46.62 -46.62

1.00 1 1 -12.10 -11.69 -11.6 36.38 32.048 31.126

c3 c2 c1

4 1.5

1 -11.6 31.08

1.00 1 1 1 d3 d2 -11.10 -10.29 -10.11 -10.1 d1 25.28 17.59 16.008 15.93 From the above table, the first root is 1.5. Using synthetic division for the second order polynomial x2 - 11.6x + 31.08, the following table summarizes the results: i xi r(%) R0 R1 b2 b1 b0

0 1 --20.48 -9.6 1 -11.6 31.08

1 2 3 4 3.1333 3.9867 4.1875 4.200 68.1 21.4 4.8 0.3 4.5511 0.7282 0.0403 0.0002 -5.333 -3.627 -3.225 -3.2 1 1 1 1 -11.6 -11.6 -11.6 -11.6 31.08 31.08 31.08 31.08

c2 c1

1 1 1 1 -10.6 -8.467 -7.613 -7.413

1 -7.4

d2 d1

1 1 1 1 -9.6 -5.333 -3.627 -3.225

1 -3.2

59

5 4.200

From the above table, a second root of 4.2 is found. The values c2 = 1, and c1 = -7.4 give the polynomial x - 7.4 = 0 Thus, the third root is 7.4. Summary of roots: x1 = 1.5 x2 = 4.2 x3 = 7.4 Problem 4-45.

f(x) = x4 - 8.37x2 - 10.336x - 1.734 R (x ) x i+1 = x i  0 i R1 ( x i ) i xi r(%) R0 R1 b4 b3 b2 b1 b0 c4 c3 c2 c1

0 1 2 3 4 5 1.00 0.1576 -0.118 -0.193 -0.2 -0.20 --- 534.7 233.6 39.0 3.3 0.03 -19.44 -3.57 -0.631 -0.047 -4E-04 0.00 -23.08 -12.96 -8.369 -7.128 -7.021 -7.02 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 -8.37 -8.37 -8.37 -8.37 -8.37 -8.37 -10.34 -10.34 -10.34 -10.34 -10.34 -10.34 -1.73 -1.73 -1.73 -1.73 -1.73 -1.73 1.00 1.00 1.00 0.16 -7.37 -8.35 -17.71 -11.65

1.00 -0.12 -8.36 -9.35

1.00 -0.19 -8.33 -8.72

1.00 -0.20 -8.33 -8.67

6 -0.2

1.00 -0.20 -8.33 -8.67

1.00 1.00 1.00 1.00 1.00 1.00 d4 2.00 0.32 -0.24 -0.39 -0.40 -0.40 d3 -5.37 -8.30 -8.33 -8.26 -8.25 -8.25 d2 -23.08 -12.96 -8.37 -7.13 -7.02 -7.02 d1 From the above table 4-48, the first root is -0.2. Using synthetic division for the third order polynomial x3 - 0.20x2 - 8.33x - 8.67, the following table summarizes the results:

60

i xi r(%) R0 R1 b3 b2 b1 b0 c3 c2 c1

0 1 2 3 4 1.00 -1.82723 -1.73719 -1.70524 -1.70014 --154.7 5.2 1.9 -16.20 -0.21762 -0.04532 -0.00549 -5.73 2.41715 1.41842 1.07565 1.00 1.00 1.00 1.00 -0.20 -0.20 -0.20 -0.20 -8.33 -8.33 -8.33 -8.33 -8.67 -8.67 -8.67 -8.67 1.00 1 1 1 0.80 -2.02723 -1.93719 -1.90524 -7.53 -4.6258 -4.96471 -5.0811

1.00 1 1 1 d3 d2 1.80 -3.85445 -3.67439 -3.61048 d1 -5.73 2.41715 1.41842 1.07565 From the above table, a second root is -1.7. Using synthetic division for the second order polynomial x2 - 1.90524x - 5.0811, the following table summarizes the results: i xi r(%) R0 R1 b2 b1 b0

0 ---5.986 0.0948 1 -1.905 -5.081

c2 c1

1 -0.905

1

1 64.175 98.4 3991.1 126.44 1 -1.905 -5.081

2 32.611 96.8 996.28 63.317 1 -1.905 -5.081

3 4 5 6 7 16.876 9.103 5.395 3.848 3.434 93.2 85.4 68.7 40.2 12.0 247.58 60.433 13.746 2.3937 0.1709 31.848 16.3 8.8848 5.7904 4.9637 1 1 1 1 1 -1.905 -1.905 -1.905 -1.905 -1.905 -5.081 -5.081 -5.081 -5.081 -5.081

8 3.400 1.0 0.0012 4.8948 1 -1.905 -5.081

9 3.400

1 1 1 1 1 1 1 1 62.27 30.706 14.971 7.1973 3.4898 1.9426 1.5292 1.4948

1 1 1 1 1 1 1 1 1 d2 d1 0.0948 126.44 63.317 31.848 16.3 8.8848 5.7904 4.9637 4.8948 From the above table, a third root of 3.4 is found. The values c2 = 1, and c1 = 1.4948 =1.5 give the polynomial x + 1.5 = 0 Thus, the third root is -1.5. Summary of roots: x1 = -0.2 x2 = -1.7 x3 = 3.4 x4 = -1.5 61

Problem 4-46.

f(x) = x4 - 7.8x3 + 22.08x2 - 26.738x + 11.6127 R (x ) x i+1 = x i  0 i R1 ( x i ) i xi r(%) R0 R1 b4 b3 b2 b1 b0

0 1.0000 --0.1547 -1.9780 1.0000 -7.8000 22.0800 -26.7380 11.6127

1 1.0782 7.2537 0.0268 -1.3138 1.0000 -7.8000 22.0800 -26.7380 11.6127

2 1.0986 1.8600 0.0016 -1.1618 1.0000 -7.8000 22.0800 -26.7380 11.6127

c4 c3 c2 c1

1.0000 -6.8000 15.2800 -11.4580

1.0000 -6.7218 14.8325 -10.7454

1.0000 -6.7014 14.7176 -10.5686

3 1.1000

1.0000 1.0000 1.0000 d4 -5.8000 -5.6436 -5.6027 d3 9.4800 8.7475 8.5622 d2 -1.9780 -1.3138 -1.1618 d1 From the above table, the first root is 1.1. Using synthetic division for the third order polynomial x3 - 6.7014x2 + 14.7176x - 10.5686, the following table summarizes the results:

i xi r(%) R0 R1 b3 b2 b1 b0

0 1 2 3 4 5 1.00 1.35977 1.57179 1.67372 1.70213 1.70436 --26.5 13.5 6.1 1.7 -1.5524 -0.43251 -0.10834 -0.01957 -0.00133 4.3149 2.03992 1.0629 0.68921 0.59619 1.0000 1.00 1.00 1.00 1.00 -6.7014 -6.70 -6.70 -6.70 -6.70 14.7176 14.72 14.72 14.72 14.72 -10.5686 -10.57 -10.57 -10.57 -10.57

c3 c2 c1

1.0000 1 1 1 1 -5.7014 -5.34159 -5.12956 -5.02763 -4.99923 9.0162 7.45427 6.65498 6.30272 6.20827

d3 d2 d1

1.0000 1 1 1 1 -4.7014 -3.98182 -3.55777 -3.35391 -3.2971 4.3149 2.03992 1.0629 0.68921 0.59619 62

From the above table, a second root is 1.704. Using synthetic division for the second order polynomial x2 - 5x + 6.20827, the following table summarizes the results: i xi r(%) R0 R1 b2 b1 b0

0 1 --2.209 -2.999 1 -4.999 6.2083

1 1.7365 42.4 0.5425 -1.526 1 -4.999 6.2083

2 3 4 5 2.092 2.247 2.295 2.300 17.0 6.9 2.1 0.2 0.1264 0.024 0.0023 3E-05 -0.815 -0.505 -0.41 -0.399 1 1 1 1 -4.999 -4.999 -4.999 -4.999 6.2083 6.2083 6.2083 6.2083

6 2.300

1 1 1 1 1 1 -3.999 -3.263 -2.907 -2.752 -2.705 -2.699

c2 c1

1 1 1 1 1 1 d2 d1 -2.999 -1.526 -0.815 -0.505 -0.41 -0.399 From the above table, a third root of 2.3 is found. The values c2 = 1, and c1 = -2.699 = -2.7 give the polynomial x - 2.7 = 0 Thus, the third root is 2.7. Summary of roots: x1 = x2 = x3 = x4 =

1.1 1.7 2.3 2.7

Problem 4-47.

f(x) = x4 + 4.5x3 + 2.64x2 - 4.528x + 1.0752 R (x ) x i+1 = x i  0 i R1 ( x i )

63

i xi r(%) R0 R1 b4 b3 b2 b1 b0

0 1.0000 --4.6872 18.2520 1.0000 4.5000 2.6400 -4.5280 1.0752

1 0.7432 34.5541 1.3205 8.4946 1.0000 4.5000 2.6400 -4.5280 1.0752

2 0.5877 26.4487 0.3588 4.0509 1.0000 4.5000 2.6400 -4.5280 1.0752

3 0.4992 17.7464 0.0945 1.9688 1.0000 4.5000 2.6400 -4.5280 1.0752

4 0.4511 10.6444 0.0244 0.9689 1.0000 4.5000 2.6400 -4.5280 1.0752

5 0.4260 5.9019 0.0062 0.4804 1.0000 4.5000 2.6400 -4.5280 1.0752

6 7 0.4131 0.4066 3.1194 1.6053 0.0016 0.0004 0.2392 0.1193 1.0000 1.0000 4.5000 4.5000 2.6400 2.6400 -4.5280 -4.5280 1.0752 1.0752

c4 c3 c2 c1

1.0000 5.5000 8.1400 3.6120

1.0000 5.2432 6.5367 0.3301

1.0000 5.0877 5.6303 -1.2188

1.0000 4.9992 5.1354 -1.9646

1.0000 4.9511 4.8737 -2.3293

1.0000 4.9260 4.7385 -2.5094

1.0000 1.0000 4.9131 4.9066 4.6697 4.6349 -2.5989 -2.6435

8 0.4033

1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 d4 6.5000 5.9864 5.6755 5.4983 5.4023 5.3520 5.3262 5.3132 d3 d2 14.6400 10.9858 8.9660 7.8799 7.3108 7.0184 6.8700 6.7952 18.2520 8.4946 4.0509 1.9688 0.9689 0.4804 0.2392 0.1193 d1 From the above table, the first root is 0.4033  0.4. Using synthetic division for the third order polynomial x3 + 4.9066x2 + 4.6349x - 2.6435, the following table summarizes the results:

i xi r(%) R0 R1 b3 b2 b1 b0 c3 c2 c1

0 1 2 3 4 5 1.00 0.54734 0.40729 0.39352 0.39339 0.39339 --82.7 34.4 3.5 0.0 7.8980 1.5273 0.12571 0.00116 1E-07 17.4481 10.9049 9.12934 8.96114 8.95956 1.0000 1.0000 1.0000 1.0000 1.0000 4.9066 4.9066 4.9066 4.9066 4.9066 4.6349 4.6349 4.6349 4.6349 4.6349 -2.6435 -2.6435 -2.6435 -2.6435 -2.6435 1.0000 1 1 1 1 5.9066 5.45393 5.31387 5.3001 5.29997 10.5415 7.62011 6.7992 6.72061 6.71988

1.0000 1 1 1 1 d3 d2 6.9066 6.00127 5.72115 5.69362 5.69336 d1 17.4481 10.9049 9.12934 8.96114 8.95956 From the above table, a second root is 0.39339  0.4. Using synthetic division for the second order polynomial x2 +5.29997x + 6.71988, the following table summarizes the results:

64

i xi r(%) R0 R1 b2 b1 b0 c2 c1

0 1 --13.02 7.3 1 5.3 6.7199

1 -0.784 227.6 3.181 3.7329 1 5.3 6.7199

2 -1.636 52.1 0.7262 2.0285 1 5.3 6.7199

3 -1.994 18.0 0.1282 1.3126 1 5.3 6.7199

4 -2.091 4.7 0.0095 1.1173 1 5.3 6.7199

5 6 -2.100 -2.100 0.4 7E-05 1.1002 1 5.3 6.7199

1 1 1 1 1 1 6.3 4.5164 3.6643 3.3063 3.2086 3.2001

d2 1 1 1 1 1 1 d1 7.3 3.7329 2.0285 1.3126 1.1173 1.1002 From the above table, a third root of -2.1 is found. The values c2 = 1, and c1 = 3.2001  3.2 give the polynomial x + 3.2 = 0 Thus, the third root is -3.2. Summary of roots: x1 = x2 = x3 = x4 =

0.4 0.4 -2.1 -3.2

Problem 4-48.

f(x) = x4 + 7.4x3 + 17.99x2 + 15.31x + 2.4 R (x ) x i+1 = x i  0 i R1 ( x i )

65

i xi r(%) R0 R1 b4 b3 b2 b1 b0

0 1.0000 --4.6872 18.2520 1.0000 4.5000 2.6400 -4.5280 1.0752

c4 c3 c2 c1

1.0000 5.5000 8.1400 3.6120

1 2 3 4 5 6 0.7432 0.2596 -0.0349 -0.1689 -0.1986 -0.2000 34.5541 186.2556 844.133 79.3460 14.9450 0.6950 27.0576 7.7216 1.8874 0.2923 0.0125 0.0000 55.9541 26.2178 14.0815 9.8463 9.0085 8.9701 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 7.4000 7.4000 7.4000 7.4000 7.4000 7.4000 17.9900 17.9900 17.9900 17.9900 17.9900 17.9900 15.3100 15.3100 15.3100 15.3100 15.3100 15.3100 2.4000 2.4000 2.4000 2.4000 2.4000 2.4000 1.0000 8.1432 24.0420 33.1779

7 -0.2000

1.0000 1.0000 1.0000 1.0000 1.0000 7.6596 7.3651 7.2311 7.2014 7.2000 19.9786 17.7330 16.7685 16.5598 16.5500 20.4970 14.6913 12.4774 12.0211 12.0000

1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 d4 6.5000 8.8864 7.9193 7.3302 7.0621 7.0028 7.0000 d3 14.6400 30.6463 22.0347 17.4773 15.5755 15.1689 15.1500 d2 18.2520 55.9541 26.2178 14.0815 9.8463 9.0085 8.9701 d1 From the above table, the first root is -0.2. Using synthetic division for the third order polynomial x3 + 7.2x2 + 16.55x + 12, the following table summarizes the results:

i xi r(%) R0 R1 b3 b2 b1 b0 c3 c2 c1

0 1.00 --36.7501 33.9500 1.0000 7.2000 16.5500 12.0000

1 -0.08248 1312.5 10.6835 15.3828 1.0000 7.2000 16.5500 12.0000

2 -0.77699 89.4 3.01854 7.17255 1.0000 7.2000 16.5500 12.0000

3 -1.19783 35.1 0.78783 3.60564 1.0000 7.2000 16.5500 12.0000

4 -1.41633 15.4 0.16175 2.17283 1.0000 7.2000 16.5500 12.0000

5 -1.49077 5.0 0.01594 1.7501 1.0000 7.2000 16.5500 12.0000

6 7 -1.49988 -1.50001 0.6 0.00023 1.70066 1.0000 7.2000 16.5500 12.0000

1.0000 1 1 1 1 1 1 8.2000 7.11753 6.42302 6.00217 5.78367 5.70923 5.70012 24.7500 15.963 11.5594 9.36043 8.35843 8.03886 8.00052

1.0000 1 1 1 1 1 1 d3 d2 9.2000 7.03505 5.64603 4.80434 4.36734 4.21846 4.20024 d1 33.9500 15.3828 7.17255 3.60564 2.17283 1.7501 1.70066 From the above table, a second root is -1.5. Using synthetic division for the second order polynomial x2 +5.70012x + 8.00052, the following table summarizes the results:

66

i xi r(%) R0 R1 b2 b1 b0

0 --14.701 7.7001 1 5.7001 8.0005

1 -0.909 210.0 3.6448 3.8818 1 5.7001 8.0005

c2 c1

1 6.7001

1 4.791

1

2 -1.848 50.8 0.8816 2.0039 1 5.7001 8.0005

3 -2.288 19.2 0.1935 1.1241 1 5.7001 8.0005

4 -2.460 7.0 0.0296 0.7797 1 5.7001 8.0005

5 6 -2.498 -2.500 1.5 0.0014 0.7036 1 5.7001 8.0005

1 1 1 1 3.852 3.4121 3.2399 3.2019

d2 1 1 1 1 1 1 d1 7.7001 3.8818 2.0039 1.1241 0.7797 0.7036 From the above table, a third root of -2.5 is found. The values c2 = 1, and c1 = 3.2019  3.2 give the polynomial x + 3.2 = 0 Thus, the third root is -3.2. Summary of roots: x1 = x2 = x3 = x4 =

-0.2 -1.5 -2.5 -3.2

Problem 4-49. Root 4 = .3333334 24.00000 120.007 -161.9625 -108.03500 Root 3 = -.5001968 24.00000 108.00220 -215.98520 Root 2 = 1.4998990 24.00000 143.99980 Root 1 = -5.9999930 Problem 4-50. Root 3 = -1.3076920 -1.45600 -1.76806 13.52001 Root 2 = -3.7143130 -1.45600 3.63998 Root 1 = 2.4999850 Problem 4-51. Root 3 = .7199747 .02500 -.03100 -.06960

67

Root 2 = 2.4000430 .02500 .02900 Root 1 = -1.1599630 Problem 4-52. Root 3 = -.1200000 1.00000 -1.12000 .30360 Root 2 = .4600120 1.00000 -.66001 Root 1 = .6600063 Problem 4-53. 1   0.73 0.24 0.73 1   0.74  0  3  32  18619 .   0.3974  0   1  1926 . ,  2  0166 . ,  3  1240 . 0.24 0.74 1  

Problem 4-54. 1   0.28 0.84 0.28 1   0.62  0  0.84 0.62 1  

3  32  18317   01232 . .  0   1  2.189,  2  0.0767,  3  0.734

Problem 4-55. 1   0.54 0.67 0.38 1   0.55  0  3  32  35432 .   18775 .  0   1  1449 . 0.92 0.24 1    2  0.7755  i 0.8333  3  0.7755  i 0.8333

Problem 4-56. 2.0   12 . 0.8 0.6 3.0   15 .  0  3  62  7.3  1092 .  0   1  4.396,  2  0174 . ,  3  1430 . 11 . 14 . 1 

Problem 4-57. Solution not provided.

68

Problem 4-58. Solution not provided. Problem 4-59.

f(x) = x3 - c (1) 2 (2) f '( x ) = 3x f ( xi ) (3) x i+1 = x i  f ' ( xi ) The algorithm is as follows: 1. Input an initial estimate of the root, x0 2. Evaluate f(x0) and f ' ( x 0 ) using equation 4-1, and 4-2 3. Compute a revised estimat of the ,x1, using equation 4-3. 4. Check for convergence according to a specified criterion. 5. If the specified criterion is not met, repeat step 1 through 4. The following table demonstrates the use of the algorithm with c = 15, x0 = 2, and a tolerance of 1%. The cube root to 4 significant figures is found to be 2.4662. i 0 1 2 3 4 5 6

xi 1 5.6667 3.9335 2.9455 2.5400 2.4683 2.4662

f (x i) -14 166.963 45.8602 10.5546 1.38641 0.03873 3.3E-05

f '(x i) 3 96.3333 46.417 26.0276 19.3543 18.278 18.2466

x i+1 5.66667 3.9335 2.9455 2.5400 2.4683 2.4662 2.4662

r (%) ----82.4 44.1 33.5 16.0 2.9 0.1

Problem 4-60. Solution not provided. Problem 4-61. The following is a computer program written in FORTRAN for the synthetic division method: PROGRAM SYNTHETIC INTEGER SIZE PARAMETER (SIZE=20) REAL B(SIZE), C(SIZE), D(SIZE) PRINT*, ' Enter N, x0, and Tolerance ?' READ*, N, X0, TOL N=N+1 PRINT*,' Input bj ? (b0, bn, ...... , bn)' READ*, (B(J), J=1,N) PRINT*, ' N=',N,' ','X0=',X0 I=0 100 I=I+1 J=N PRINT*, 'ITERATION NO. ',I

69

C(N)=B(N) D(N)=C(N) PRINT*, ' j b c d' WRITE(*,1)J,B(N),C(N),D(N) 1 FORMAT(1X,' ',I2,' ',F12.8,' ',F12.8,' ',F12.8) DO 10 J=N-1,2,-1 C(J)=B(J)+X0*C(J+1) D(J)=C(J)+X0*D(J+1) WRITE(*,2)J,B(J),C(J),D(J) 2 FORMAT(1X,' ',I2,' ',F12.8,' ',F12.8,' ',F12.8) 10 CONTINUE J=1 WRITE(*,3)J,B(J) 3 FORMAT(1X,' ',I2,' ',F12.8) R0=B(1)+X0*C(2) R1=C(2)+X0*D(3) WRITE(*,4)R0,R1 4 FORMAT(1X,' R0=',F12.8,' R1=',F12.8) XX0=X0 X0=X0-R0/R1 WRITE(*,5) X0 5 FORMAT(1X,' ESTIMATE OF ROOT = ',F12.8) IF(ABS(X0-XX0) .GT. TOL) GOTO 100 END

4.9 Multiple Roots Problem 4-62.

f(x) = x4 - 8x3 + 18x2 - 16x + 15 f ' ( x ) = 3x 3  24 x 2  36 x  16 f ( xi ) x i+1 = x i  f ' ( xi ) x = 1,1,1, 5

4.10 Systems of Nonlinear Equations Problem 4-63. The system of equations can be solved for x and y respectively as follows:



x  8 x 3 y  18 x 2 y  16 x

y

x 3  18 xy 2 16 70



1 4

With initial estimates that x = y = 3, the iterations are shown in the following table. The system of equations does not have a solution, therefore, the solution procedure does not converge. i 1 2 3 4 5 6 7 8

x

3 3.80675 9.34897 98.019 23017.8 7.8E+09 8.3E+22 5.1E+53

Problem 4-64. i x y 1 .988525 .754982 2 1.015361 .591495 3 1.042945 .495639 4 1.060007 .447429 5 1.068993 .426610 6 1.074273 .421688 7 1.079449 .430509 8 1.087623 .458852 9 1.100156 .514620 10 1.114001 .593616 11 1.123715 .674748 12 1.126612 .737398 13 1.123704 .774642 14 1.117600 .790129 15 1.110767 .790940 16 1.104850 .783668 17 1.100613 .773262 18 1.098150 .762991 19 1.097156 .754691 20 1.097173 .749077 21 1.097746 .746075 22 1.098517 .745149 23 1.099247 .745586 24 1.099809 .746710 25 1.100165 .747998

71

y

3 41.9912 18596.3 3.8E+10 3.8E+25 1.2E+61 1E+145 #NUM!

CHAPTER 5. SIMULTANEOUS LINEAR EQUATIONS 5.2 Gaussian Elimination Problem 5-1. Forward elimination: Step 1 2 -1 3.1  .  R1'  R1  1 2 1 31 1 3 1 1 14    ' ' R2  R2 / ( 3)  R1 .   0 -2.3333 0.6667 -3.5667  1 1 4 7.3 0 1 3 10.4  R3'  R3  R1' Forward elimination: Step 2 -1 3.1  2 1 31 .  R1'  R1 1 1 2 0 2.3333 0.6667 35667    ' R2  R2 / ( 2.3333) .   0 1 -0.285714 1.528571 0 0 0 3.285714 8.87144  1 3 10.4  R3'  R3  R2' Forward elimination: Step 3 -1 3.1  1 31 . R1'  R1 1 2  1 2 0 1 0.285714 1528571    ' R2  R2 .   0 1 -0.285714 1.528571 0 0 3.285714 8.87144  0 0 1 2.700000 R3'  R3 / ( 3.285714) The last matrix represents the following set of equations: X1 + 2X2 X3 = 3.100000 X2 - 0.285714X3 = 1.528571 X3 = 2.700000 The backward substitution can be used to obtain values for the unknowns: Backward Substitution: Step 1 1 31 . R1'  R1  R3' 1 2  1 2 0 5.8  0 1 0.285714 1528571  0 1 0 2.3 ' ' R2  R2  0.285714 R3 .     0 0 0 0 1 2.7  1 2.700000 R3'  R3 Backward Substitution: Step 2 .  R1'  R1  2 R2' 1 2 0 58 1 0 0 1.2  0 1 0 2.3 0 1 0 2.3 ' R2  R2     0 0 1 2.7  0 0 1 2.7  R3'  R3 The last matrix contains the solution: .   X 1  12  X    2.3  2    X 3  2.7  Problem 5-2. Following the procedure of Problem 5-1, the forward pass and backward substitution result into

72

. 0.5 0.3  1 15 0 1 0.6364 0.7636   0 0  1 15667 .

Forward Pass:

Backward Substitution:

. 1 0 0 01333  0 1 0 0.2333    0 0 1 15667  .

The solution is  X1  X    2  X 3 

. 01333   0.2333     15667  .

Problem 5-3. Following the procedure of Problem 5-1, the forward pass and backward substitution result into . 0.5 18500 . 1 15   Forward Pass: . 4.3667  0 1 16667 0 0 1 3.4000 

Backward Substitution:

1 0 0 2.1000 0 1 0 13000  .   0 0 1 3.4000 

The solution is  X1  X    2  X 3 

2.1 13   .   3.4 

Problem 5-4. Following the procedure of Problem 5-1, the forward pass and backward substitution result into . 1966000 . 1 0.666667 1333333   Forward Pass: 1 1423071 . 0.930766 0 0 0 1 0.400000

Backward Substitution:

.  1 0 0 15 0 1 0 15 .   0 0 1 0.4

The solution is  X1  X    2  X 3 

.   15  15   .  0.4

73

Problem 5-5. Following the procedure of Problem 5-1, the forward pass and backward substitution result into . 1 2 1150000 . 1 15  0 1 1142857 . 0 0.528571    Forward Pass: 0 0 1400000 . 2.300000 1   0 1 0 0 0  1 0 0 0 0.3  0 1 0 0 2.1   Backward Substitution: 0 0 1 0 2.3   0  0 0 0 1 The solution is  X 1   0.3   X   2.1  2     X 3  2.3     X4   0  Problem 5-6. There is an infinite number of solutions for this system of equations. The first and the fourth equation are not unique, since one equation can be produced by the other by multiplying it by a constant. Problem 5-7. Solution not provided. Problem 5-8. Solution not provided.

5.3 Gauss-Jordan Elimination Problem 5-9. Step 1 R1'  R1 / (3)  3 3 2 7.6  1 1 -0.6667 2.5333   2 4 1 14  0 -3 1.1667 -1.8222 ' ' R2  R2 / (2)  R1 .     1 2 5 6.3 0 1 -4.3333 3.7667  R3'  R3  R1' Step 2 R1'   ( R1  R2' ) 1 1 0.6667 2.5333  1 0 -0.2778 1.9222  0 3 11667  0 1 -0.3889 0.6111 ' R2  R2 / ( 3) . 18333 .     0 1 4.3333 3.7667  0 0 -3.9444 3.1556 R3'  R3  R2' Step 3

74

. R1'  R1  0.2778 R3' 1 0 0.2778 19222  0 1 0.3889 0.6111 R2'  R2  0.3889 R3'   0 0 3.9444 31556  . R3'  R3 / ( 3.9444) The last matrix contains the following solution: .   X 1   17  X    0.3    2   X 3  0.8

1 0 0 1.7  0 1 0 0.3    0 0 1 -0.8

Problem 5-10. There is an infinite number of solutions for this system of equations. The first and the third equation are not unique, since one equation can be produced by the other by multiplying it by a constant. Problem 5-11. Following the procedure of Problem 5-7, the Gauss-Jordan Elimination technique results into 1 0 0 2.2  0 1 0 3.3   0 0 1 0.6  From which  X 1   2.2   X   3.3  2    X 3   0.6  Problem 5-12. Following the procedure of Problem 5-7, the Gauss-Jordan Elimination technique results into .  1 0 0 13 0 1 0 2.4   0 0 1 0.6 From which .   X 1   13  X   2.4   2    X 3   0.6  Problem 5-13. Following the procedure of Problem 5-7, the Gauss-Jordan Elimination technique results into 1 0 0 0 2.1  0 1 0 0 16 .    0 0 1 0 15 .    0 0 0 1 0.6 

75

From which  X1  X   2  X3   X4 

 2.1  16 .     15 .     0.6 

Problem 5-14. Following the procedure of Problem 5-7, the Gauss-Jordan Elimination technique results into .  1 0 0 0 13 0 1 0 0 2.9    0 0 1 0 0.4   .  0 0 0 1 11 From which .   X 1   13  X   2.9   2     X 3  0.4     .   X 4   11 Problem 5-15. Solution not provided. Problem 5-16. Solution not provided.

5.5 LU Decomposition Problem 5-17. 1 1   3 A  C    4 2  2  Using Eq. 5-52a of the textbook, the following results l11  a11  1

l21  a 21  4 Using Eq. 5-52b of the textbook, the following results a 1 u12  12   1 l11 1 Using Eq. 5-52e of the textbook, the following results 2 1

l22  a 22   l2 k u k 2  a 22  l21u12  2  (4)(1)  6 k 1

Therefore, the L and U matrices are 76

1 0  L   4  6 1 1 U  0 1 Equations 5-53 and 5-54 of the textbook result in C 3 e1  1   3 l11 1 2 1

e2 

C2   l 2 j e j j 1

l22 5 X 2  e2  3 X 1  e1 

2

u

1j

j 11



C2  l21e1 2  (4)(3) 5   6 3 l22

 5 4 X j  e1  u12 X 2  3  (1)    3 3

Summary of the solution: 4   X1        3      X 2   5   3  Problem 5-18. 2 5  3 A C   2  3 1 Using Eq. 5-52a of the textbook, the following results l11  a11  2

l21  a 21  3 Using Eq. 5-52b of the textbook, the following results a 5 u12  12   2.5 2 l11 Using Eq. 5-52e of the textbook, the following results 2 1

l22  a 22   l2 k u k 2  a 22  l21u12  1  ( 3)( 2.5)  6.5 k 1

Therefore, the L and U matrices are 2 0 L  3 6.5 1 2.5 U  0 1  Equations 5-53 and 5-54 of the textbook result in Summary of the solution: 77

 X1       X 2 

1      1 

Problem 5-19. .   1 2 1  31    A   3 1 1  C  14 .   1 1 4  7.3 Using Eqs. 5-52a of the textbook, the following results l11  a11  1 l21  a 21  3 l31  a 31  1 Using Eq. 5-52b of the textbook, the following results a 2 u12  12   2 l11 1 a 1 u13  13   1 l11 1 Using Eq. 5-52c of the textbook, the following results 2 1

l22  a 22   l2 k u k 2  a 22  l21u12  1  ( 3)(2)  7 k 1 2 1

l32  a 32   l3k u k 2  a 32  l31u12  1  ( 1)(2)  1 k 1

Using Eq. 5-52d of the textbook, the following results 2 1

a 23   l2 k u k 3

a 23  l21u13 1  ( 3)( 1)   0.285714 7 l22 l22 Using Eq. 5-52e of the textbook, the following results u23 

k 1



3 1

l33  a 33   l3k u k 3   a 33  l31u13  l32 u23  4  ( 1)( 1)  (1)( 0.285714)  3.285714 k 1

Therefore, the L and U matrices are 0  1 0   0 L   3 7   1 1 3.285714 1  1 2  U  0 1 0.285714  0 0 1

Equations 5-53 and 5-54 of the textbook result in

78

e1 

C1 31 .   31 . 1 l11 2 1

e2 

C2   l 2 j e j j 1

l22



C2  l21e1 14 .  ( 3)(31 .)   1528571 . 7 l22



C3  l31e1  l32 e2 7.3  ( 1)(31 . )  (1)(1528571 . )   2.7 3.285714 l33

31

e3 

C3   l 3 j e j j 1

l33 X 3  e3  2.7 X 2  e2 

3

u

2j

.  ( 0.285714)(2.7)  2.3 X j  e2  u23 X 3  1528571

j  2 1

X 1  e1 

3

u

1j

X j  e1  u12 X 2  u13 X 3  31 .  (2)(2.3)  ( 1)(2.7)  12 .

j 11

Summary of the solution: .   X 1  12  X   2.3  2    X 3  2.7 Problem 5-20. 0.6 2 3 1   A   3 1 2  C   3.3  19  1 2 1  .  Using Eqs. 5-52a of the textbook, the following results l11  a11  2 l21  a 21  3 l31  a 31  1 Using Eq. 5-52b of the textbook, the following results a 3 u12  12   15 . l11 2 a 1 u13  13   0.5 l11 2 Using Eq. 5-52c of the textbook, the following results 2 1

l22  a 22   l2 k u k 2  a 22  l21u12  1  (3)( 15 . )  55 . k 1

2 1

l32  a 32   l3k u k 2  a 32  l31u12  2  (1)( 15 . )  35 . k 1

Using Eq. 5-52d of the textbook, the following results

79

2 1

a 23   l2 k u k 3

a 23  l21u13 2  (3)(0.5)   0.636364 l22 l22 55 . Using Eq. 5-52e of the textbook, the following results u23 

k 1



3 1

l33  a 33   l3k u k 3   a 33  l31u13  l32 u23  1  (1)(0.5)  (35 . )( 0.636364)  2.727274 k 1

Therefore, the L and U matrices are 0  2 0    L   3 55 0 .   1 35 . 2.727274  0.5 . 1 15   U  0 1 0.636364  0 0  1 Equations 5-53 and 5-54 of the textbook result in C 0.6  0.3 e1  1  2 l11 2 1

e2 

C2   l 2 j e j j 1

l22



C2  l21e1 3.3  (3)(0.3)   0.763636 55 . l22



C3  l31e1  l32 e2 19 .  (1)(0.3)  (35 . )( 0.763636)   1566665 . 2.727274 l33

31

e3 

C3   l3 j e j j 1

l33 . X 3  e3  1566665 X 2  e2 

3

u

2j

. )  0.233333 X j  e2  u23 X 3  0.763636  ( 0.636364)(1566665

j  2 1

X 1  e1 

3

u

1j

. )(0.233333)  (0.5)(1566665 . )  0133333 . X j  e1  u12 X 2  u13 X 3  0.3  ( 15

j 11

Summary of the solution: .  X 1  0133333   X    0.233333   2    X 3   1566665  . Problem 5-21. Solution not provided. Problem 5-22. Solution not provided.

80

5.6 Iterative Equation-Solving Methods 5.6.1 Jacobi Iteration Problem 5-23. The system of equations can be rearranged as follows: ( 3.7  X 2  2 X 3 ) X1  3 (17.9  2 X 1  2 X 3 ) X2  3 ( 6.2  X 1  2 X 2 ) X3  4 With initial estimates of X1 = X2 = X3 = 1 substituted into above equations, we obtain a new set of values for the unknowns. These values are X 1  156667 . X 2  5.96667 X 3  180000 . These new values for X1, X2, and X3 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 |  X 1| X2 |  X 2| X3 |  X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 -1.567 2.567 5.967 4.967 -1.800 2.800 2 1.956 3.522 5.811 0.156 -4.925 3.125 3 3.987 2.031 1.380 4.431 -3.967 0.958 4 1.871 2.116 0.664 0.715 -1.243 2.724 5 -0.183 2.054 3.891 3.226 -1.414 0.171 6 1.006 1.190 5.146 1.255 -3.541 2.127 7 2.843 1.836 2.935 2.211 -3.871 0.330 . . . . . . . . . . . . . . 68 1.600 0.001 3.100 0.001 -2.700 0.000 69 1.600 0.000 3.099 0.000 -2.700 0.001 70 1.600 0.000 3.100 0.001 -2.700 0.000 71 1.600 0.000 3.100 0.000 -2.700 0.000 Summary of the solution: .  X 1   1600   X    3100   2  .   X 3  2.700 Problem 5-24. In order to avoid divergence problems, the system of equations must be rearranged as follows:

81

4 X 1  3 X 2  X 3  2.1 2 X 1  2 X 2  3 X 3  12.9 3 X 1  2 X 2  5 X 3  12.3

From which

( 2.1  3 X 2  X 3 ) 4 (12.9  2 X 1  3 X 3 ) X2  2 ( 12.3  3 X 1  2 X 2 ) X3  5 With initial estimates of X1 = X2 = X3 = 1 substituted into above equations, we obtain a new set of values for the unknowns. These values are X 1  0.475 X 2  5.950 X 3  2.660 These new values for X1, X2, and X3 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 0.475 0.5250 5.950 4.9500 -2.660 3.6600 2 3.273 2.7975 10.915 4.9650 -0.365 2.2950 3 7.570 4.2975 10.270 0.6450 -0.058 0.3075 4 7.163 0.4069 14.106 3.8363 -2.894 2.8365 5 9.331 2.1681 17.954 3.8479 -1.115 1.7786 . . . . . . . . . . . . . . 135 30.200 0.0001 42.200 0.0000 -3.700 0.0000 136 30.200 0.0000 42.200 0.0001 -3.700 0.0000 137 30.200 0.0000 42.200 0.0001 -3.700 0.0000 138 30.200 0.0000 42.200 0.0000 -3.700 0.0000 Summary of the solution:  X 1  30.200  X   42.200  2    X 3  3.700 X1 

Problem 5-25. In order to avoid divergence problems, the system of equations must be rearranged as follows: 5 X 1  3 X 2  2 X 3  2 X 4  6.4

X 1  5 X 2  3 X 3  4 X 4  4.1 2 X 1  X 2  3 X 3  2 X 4  3.7 3 X 1  X 2  4 X 3  5 X 4  3.3 From which 82

(6.4  3 X 2  2 X 3  2 X 4 ) 5 (4.1  X 1  3 X 3  4 X 4 ) X2  5 (3.7  2 X 1  X 2  2 X 4 ) X3  3 (3.3  3 X 1  X 2  4 X 3 ) X4  5 With initial estimates of X1 = X2 = X3 = X4 = 1 substituted into above equations, we obtain a new set of values for the unknowns. These values are X 1  1880 . X1 

X 2  0.420 X 3  0.433 X 4  2.260 These new values for X1, X2, X 3, and X4 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 X4 | X 1| | X 2| | X 3| | X 4| Iteration 0 1.000 ----1.000 ----1.000 ----1.000 ----1 1.880 0.880 0.420 0.580 0.433 0.567 2.260 1.260 2 2.263 0.383 -1.104 1.524 1.667 1.233 2.219 0.041 3 0.838 1.424 -0.407 0.697 1.386 0.280 3.130 0.911 4 1.733 0.895 -1.020 0.613 1.277 0.110 2.191 0.940 5 1.034 0.700 -0.513 0.507 1.042 0.234 2.517 0.327 . . . . . . . . . . . . . . . . . . 22 1.401 0.002 -0.600 0.001 0.900 0.000 2.100 0.001 23 1.400 0.001 -0.600 0.001 0.900 0.000 2.100 0.001 24 1.400 0.001 -0.600 0.000 0.900 0.000 2.100 0.001 25 1.400 0.001 -0.600 0.000 0.900 0.000 2.100 0.000 26 1.400 0.000 -0.600 0.000 0.900 0.000 2.100 0.000 Summary of the solution: .  X 1   1400   X  0.600  2     X 3   0.900       X 4   2.100  Problem 5-26. In order to avoid divergence problems, the system of equations must be rearranged as follows:

83

5 X 1  2 X 2  3 X 3  2 X 4  18.0 2 X 1  4 X 2  X 3  2 X 4  81 . 2 X 1  X 2  3 X 3  X 4  12.5 3 X 1  3 X 2  2 X 3  4 X 4  3.3 From which

(18  2 X 2  3 X 3  2 X 4 ) 5 ( 81 .  2 X1  X 3  2 X 4 ) X2  4 ( 12.5  2 X 1  X 2  X 4 ) X3  3 (7.9  3 X 1  3 X 2  2 X 3 ) X4  4 With initial estimates of X1 = X2 = X3 = X4 = 1 substituted into above equations, we obtain a new set of values for the unknowns. These values are X 1  4.200 X1 

X 2  2.275 X 3  2.833 X 4  1475 . These new values for X1, X2, X 3, and X4 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 X4 | X 1| | X 2| | X 3| | X 4| Iteration 0 1.000 ----1.000 ----1.000 ----1.000 ----1 4.200 3.200 -2.275 3.275 -2.833 3.833 -1.475 2.475 2 1.580 2.620 1.521 3.796 -2.617 0.217 1.465 2.940 3 2.053 0.473 -1.313 2.834 -2.118 0.498 -3.239 4.704 4 3.099 1.047 1.150 2.463 -4.316 2.198 -0.510 2.729 5 1.675 1.425 0.859 0.292 -1.887 2.429 -2.671 2.161 . . . . . . . . . . . . . . . . . . 57 3.100 0.000 0.801 0.001 -2.300 0.000 -1.400 0.001 58 3.100 0.000 0.800 0.001 -2.300 0.001 -1.400 0.001 59 3.100 0.001 0.800 0.000 -2.300 0.001 -1.400 0.001 60 3.100 0.001 0.800 0.000 -2.300 0.001 -1.400 0.000 61 3.100 0.000 0.800 0.000 -2.300 0.000 -1.400 0.000 Summary of the solution: .  X 1   3100   X   0.800   2     X 3  2.300     .   X 4   1400

84

Problem 5-27. Solution not provided. Problem 5-28. Solution not provided. 5.6.2 Gaussian-Seidel Iteration Problem 5-29. The system of equations can be rearranged as follows: (13 .  2 X 2  3X 3 ) X1  5 ( 14 .  2 X1  X 3 ) X2  4 (13 .  2 X1  X 2 ) X3  3 With initial estimates of X1 = X2 = X3 = 1 substituted into above equations, we obtain a new set of values for the unknowns. Note that the updated value for X1 was used in the second equation. Also the updated values for both X1 and X2 were used in the third equation. The new set of values for the unknowns are X 1  0.060 X 2  0.070 X 3  0.370 These new values for X1, X2, and X3 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 0.060 0.940 -0.070 1.070 0.370 0.630 2 0.010 0.050 -0.253 0.183 0.343 0.028 3 -0.047 0.057 -0.288 0.035 0.368 0.026 4 -0.076 0.030 -0.296 0.008 0.385 0.017 5 -0.090 0.014 -0.298 0.003 0.394 0.008 6 -0.096 0.006 -0.299 0.001 0.397 0.004 7 -0.098 0.003 -0.300 0.000 0.399 0.002 8 -0.099 0.001 -0.300 0.000 0.399 0.001 9 -0.100 0.000 -0.300 0.000 0.400 0.000 Summary of the solution: .   X 1   0100  X   0.300  2    X 3   0.400 

85

Problem 5-30. In order to avoid divergence problems, the system of equations must be rearranged as follows: 2 X 1  3 X 2  X 3  5.7 X 1  4 X 2  2 X 3  17.1 3 X 1  2 X 2  5 X 3  2.1 From which ( 5.7  3 X 2  X 3 ) X1  2 (17.1  X 1  2 X 3 ) X2  4 (2.1  3 X 1  2 X 2 ) X3  5 With initial estimates of X1 = X2 = X3 = 1 substituted into above equations, we obtain a new set of values for the unknowns. Note that the updated value for X1 was used in the second equation. Also the updated values for both X1 and X2 were used in the third equation. The new set of values for the unknowns are X 1  3850 . X 2  2.813 X 3  0.765 These new values for X1, X2, and X3 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 3.850 2.850 2.813 1.813 0.765 0.235 2 6.686 2.836 2.221 0.592 2.703 1.938 3 4.830 1.857 1.716 0.505 1.791 0.912 4 4.528 0.302 2.247 0.531 1.398 0.394 5 5.522 0.994 2.196 0.052 2.015 0.617 . . . . . . . . . . . . . . 22 5.100 0.001 2.100 0.000 1.800 0.000 23 5.100 0.000 2.100 0.000 1.800 0.000 24 5.100 0.001 2.100 0.000 1.800 0.000 25 5.100 0.000 2.100 0.000 1.800 0.000 Summary of the solution: .   X 1  5100  X   2.100   2   X 3  1800 .  Problem 5-31. There is an infinite number of solutions for this system of equations. The first and the third equation are not unique, since one equation can be produced by the other by multiplying it by a constant.

86

Problem 5-32. The system of equations can be rearranged as follows: ( 25.2  2 X 2  3 X 3 ) X1  5 (19.2  3 X 1  4 X 3 ) X2  4 (2.8  2 X 1  2 X 2 ) X3  3 With initial estimates of X1 = X2 = X3 = 1 substituted into above equations, we obtain a new set of values for the unknowns. Note that the updated value for X1 was used in the second equation. Also the updated values for both X1 and X2 were used in the third equation. The new set of values for the unknowns are: X 1  4.840 X 2  0170 . X 3  4.273 These new values for X1, X2, and X3 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 -4.840 5.840 0.170 0.830 4.273 3.273 2 -2.544 2.296 -1.381 1.551 1.708 2.565 3 -3.462 0.918 0.495 1.876 3.571 1.863 4 -3.095 0.367 -1.093 1.587 2.268 1.303 5 -3.242 0.147 0.100 1.193 3.162 0.893 . . . . . . . . . . . . . . 22 -3.200 0.000 -0.401 0.001 2.800 0.001 23 -3.200 0.000 -0.400 0.001 2.800 0.001 24 -3.200 0.000 -0.400 0.001 2.800 0.000 25 -3.200 0.000 -0.400 0.000 2.800 0.000 Summary of the solution:  X 1   3.200  X   0.400  2    X 3   2.800  Problem 5-33. In order to avoid divergence problems, the system of equations must be rearranged as follows: . 5 X 1  X 2  X 3  2 X 4  81

2 X 1  3 X 2  3 X 3  X 4  7.9 2 X 1  3 X 2  4 X 3  3 X 4  9.2 .  X 1  2 X 2  2 X 3  3 X 4  118 From which 87

( 81 .  X2  X3  2X4 ) 5 ( 7.9  2 X 1  3 X 3  X 4 ) X2  3 ( 9.2  2 X 1  3 X 2  3 X 4 ) X3  4 ( 118 .  X1  2 X 2  2 X 3 ) X4  3 With initial estimates of X1 = X2 = X3 = X4 = 1 substituted into above equations, we obtain a new set of values for the unknowns. Note that the updated value for X1 was used in the second equation. Also the updated values for both X1 and X2 were used in the third equation, and the updated values for X1, X2, and X3 were used in the fourth equation. The new set of values for the unknowns are . X 1  1620 X1 

X 2  4.380 X 3  5.645 X 4  3.290 These new values for X1, X2, X 3, and X4 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 X4 | X 1| | X 2| | X 3| | X 4| Iteration 0 1.000 ----1.000 ----1.000 ----1.000 ----1 1.620 0.620 4.380 3.380 -5.645 6.645 -3.290 4.290 2 2.683 1.063 -0.126 4.506 -6.014 0.369 -0.886 2.404 3 0.746 1.937 -2.588 2.462 -1.397 4.617 4.478 5.365 4 -0.968 1.715 -0.902 1.686 2.220 3.617 6.337 1.859 5 -0.651 0.317 2.306 3.208 1.049 1.171 3.312 3.025 . . . . . . . . . . . . . . . . . . 69 0.585 0.001 0.772 0.003 -1.509 0.003 2.217 0.001 70 0.586 0.001 0.776 0.003 -1.511 0.002 2.213 0.004 . . . . . . . . . 99 0.587 0.000 0.773 0.000 -1.514 0.000 2.213 0.000 100 0.587 0.000 0.773 0.000 -1.513 0.000 2.213 0.000 Summary of the solution:  X 1   0.587   X   0.773   2     X 3  1513 .       X 4   2.213  Problem 5-34. In order to avoid divergence problems, the system of equations must be rearranged as follows:

88

4 X 1  2 X 2  2 X 3  3 X 4  2.4 3 X 1  4 X 2  2 X 3  X 4  29.9 2 X 1  3 X 2  4 X 3  X 4  29.6 2 X1

 2 X 3  3 X 4  2.6

From which

(2.4  2 X 2  2 X 3  3 X 4 ) 4 (29.9  3 X 1  2 X 3  X 4 ) X2  4 ( 29.6  2 X 1  3 X 2  X 4 ) X3  4 (2.6  2 X 1  2 X 3 ) X4  3 With initial estimates of X1 = X2 = X3 = X4 = 1 substituted into above equations, we obtain a new set of values for the unknowns. Note that the updated value for X1 was used in the second equation. Also the updated values for both X1 and X2 were used in the third equation, and the updated values for X1, X2, and X3 were used in the fourth equation. The new set of values for the unknowns are . X 1  0150 X1 

X 2  8.338 X 3  0.972 . X 4  1615 These new values for X1, X2, X 3, and X4 are then used as the new solution estimate. This process is repeated until the difference between the previous values and the new values are small. The following table summarizes the results: X1 X2 X3 X4 | X 1| | X 2| | X 3| | X 4| Iteration 0 1.000 ----1.000 ----1.000 ----1.000 ----1 -0.150 1.150 8.338 7.338 -0.972 1.972 -1.615 2.615 2 6.466 6.616 1.736 6.601 -3.269 2.297 1.265 2.879 3 2.154 4.312 4.541 2.805 -2.601 0.668 -1.165 2.429 4 5.045 2.891 2.100 2.441 -3.594 0.993 0.100 1.265 5 3.372 1.673 3.174 1.074 -3.308 0.286 -0.824 0.925 . . . . . . . . . . . . . . . . . . 18 4.101 0.003 2.399 0.002 -3.700 0.001 -0.599 0.001 19 4.099 0.002 2.401 0.001 -3.700 0.000 -0.600 0.001 20 4.100 0.001 2.400 0.001 -3.700 0.000 -0.600 0.001 21 4.100 0.001 2.400 0.000 -3.700 0.000 -0.600 0.000 22 4.100 0.000 2.400 0.000 -3.700 0.000 -0.600 0.000

89

Summary of the solution:  X 1   4.100   X   2.400   2     X 3   3.700      X 4  0.600 Problem 5-35. Solution not provided. Problem 5-36. Solution not provided. 5.6.3 Convergence Consideration of the Iterative Methods Problem 5-37.

5 X 1  2 X 2  9.6 2 X 1  3 X 2  6.5

X1  

9.6  2 X 2 5

6.5  2 X 1 3 The following tables shows the results for both the Jacobi and Gauss-Seidel iterations: JACOBI ITERATION | X 1| | X 2| X2 X1 Iteration 0 1.000 ----1.000 ----1 1.520 0.520 -1.500 2.500 2 2.520 1.000 -1.153 0.347 3 2.381 0.139 -0.487 0.667 4 2.115 0.267 -0.579 0.092 5 2.152 0.037 -0.757 0.178 6 2.223 0.071 -0.732 0.025 7 2.213 0.010 -0.685 0.047 8 2.194 0.019 -0.691 0.007 9 2.197 0.003 -0.704 0.013 10 2.202 0.005 -0.702 0.002 11 2.201 0.001 -0.699 0.003 12 2.200 0.001 -0.699 0.000 13 2.200 0.000 -0.700 0.001 14 2.200 0.000 -0.700 0.000 X2 

90

GAUSS-SEIDEL ITERATION X2 X1 | X 1| | X 2| Iteration 0 1.000 ----1.000 ----1 1.520 0.520 -1.153 2.153 2 2.381 0.861 -0.579 0.574 3 2.152 0.230 -0.732 0.153 4 2.213 0.061 -0.691 0.041 5 2.197 0.016 -0.702 0.011 6 2.201 0.004 -0.699 0.003 7 2.200 0.001 -0.700 0.001 8 2.200 0.000 -0.700 0.000 Comparing the results of the two tables, the Gauss-Seidel method has a faster rate of convergence than that of the Jacobi method. The solution for the system of equations is  X 1   2.200   X   0.700   2  Problem 5-38.

6 X 1  5 X 2  36.3 3 X 1  4 X 2  4.5

X1  

36.3  5 X 2 6

4.5  3 X 1 4 The following tables shows the results for both the Jacobi and Gauss-Seidel iterations: JACOBI ITERATION X2 X1 | X 1| | X 2| Iteration 0 1.000 ----1.000 ----1 -5.217 6.217 -1.875 2.875 2 -7.613 2.396 2.788 4.663 3 -3.727 3.885 4.584 1.797 4 -2.230 1.497 1.670 2.914 . . . . . . . . . . 39 -4.300 0.001 2.101 0.000 40 -4.300 0.000 2.100 0.001 41 -4.300 0.001 2.100 0.000 42 -4.300 0.000 2.100 0.000 X2 

91

GAUSS-SEIDEL ITERATION X1 X2 | X 1| | X 2| Iteration 0 1.000 ----1.000 ----1 -5.217 6.217 2.788 1.788 2 -3.727 1.490 1.670 1.117 3 -4.658 0.931 2.369 0.698 4 -4.076 0.582 1.932 0.436 . . . . . . . . . . 17 -4.300 0.001 2.100 0.001 18 -4.300 0.001 2.100 0.001 19 -4.300 0.001 2.100 0.000 20 -4.300 0.000 2.100 0.000 Comparing the results of the two tables, the Gauss-Seidel method has a faster rate of convergence than that of the Jacobi method. The solution for the system of equations is  X 1  4.300  X    2.100    2  Problem 5-39.

X1 

15.7  2 X 2  3 X 3 4

X2 

14.1  2 X 1  X 3 4

4 X 1  2 X 2  3 X 3  15.7 2 X 1  4 X 2  X 3  14.1



3 X 1  X 2  3 X 3  4.2

4.2  3 X 1  X 2 3 The following tables shows the results for both the Jacobi and Gauss-Seidel iterations: JACOBI ITERATION X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 3.675 2.675 -2.775 3.775 2.733 1.733 2 0.488 3.188 -1.004 1.771 4.150 1.417 3 0.310 0.177 -2.244 1.240 1.553 2.597 4 1.639 1.328 -2.982 0.738 0.963 0.590 . . . . . . . . . . . . . . 20 1.300 0.001 -2.401 0.001 1.899 0.001 21 1.300 0.000 -2.400 0.000 1.900 0.001 22 1.300 0.001 -2.400 0.000 1.900 0.000 23 1.300 0.000 -2.400 0.000 1.900 0.000 X3 

92

GAUSS-SEIDEL ITERATION X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 3.675 2.675 -1.438 2.438 4.596 3.596 2 -0.241 3.916 -2.496 1.059 0.327 4.269 3 2.431 2.672 -2.227 0.269 3.089 2.762 4 0.495 1.937 -2.505 0.278 1.059 2.029 . . . . . . . . . . . . . . 26 1.299 0.001 -2.400 0.000 1.899 0.001 27 1.300 0.001 -2.400 0.000 1.900 0.001 28 1.300 0.001 -2.400 0.000 1.900 0.001 29 1.300 0.000 -2.400 0.000 1.900 0.000 Comparing the results of the two tables, the Gauss-Seidel method has a slower rate of convergence than that of the Jacobi method. The solution for the system of equations is .  X 1   1300   X   2.400  2   .  X 3   1900  Problem 5-40.

X1  4 X 1  2 X 2  3 X 3  10.8 3 X 1  5 X 2  2 X 3  119 .



10.8  2 X 2  3 X 3 4

X2 

2 X 1  2 X 2  5 X 3  0.2

119 .  3X1  2 X 3 5

0.2  2 X 1  2 X 2 5 The following tables shows the results for both the Jacobi and Gauss-Seidel iterations: JACOBI ITERATION X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 -2.950 3.950 2.180 1.180 0.760 0.240 2 -2.180 0.770 4.454 2.274 -0.348 1.108 3 -0.212 1.968 3.549 0.905 0.870 1.218 4 -1.578 1.366 2.855 0.694 1.295 0.425 . . . . . . . . . . . . . . 35 -1.500 0.001 3.600 0.000 0.800 0.000 36 -1.500 0.000 3.600 0.001 0.800 0.000 37 -1.500 0.001 3.600 0.000 0.800 0.000 38 -1.500 0.000 3.600 0.000 0.800 0.000 X3 

93

GAUSS-SEIDEL ITERATION X1 X2 X3 | X 1| | X 2| | X 3| Iteration 0 1.000 ----1.000 ----1.000 ----1 -2.950 3.950 4.550 3.550 0.600 0.400 2 -0.875 2.075 3.145 1.405 0.868 0.268 3 -1.779 0.904 3.794 0.649 0.766 0.102 4 -1.378 0.401 3.513 0.281 0.814 0.048 . . . . . . . . . . . . . . 10 -1.499 0.003 3.599 0.002 0.800 0.000 11 -1.500 0.001 3.600 0.001 0.800 0.000 12 -1.500 0.001 3.600 0.000 0.800 0.000 13 -1.500 0.000 3.600 0.000 0.800 0.000 Comparing the results of the two tables, the Gauss-Seidel method has a faster rate of convergence than that of the Jacobi method. The solution for the system of equations is .   X 1  1500  X    3.600   2    X 3   0.800  Problem 5-41. Solution not provided. Problem 5-42. Solution not provided.

5.7 Use of Determinants Problem 5-43.

A A1  A2 

2 3  (2)(4)  (3)( 3)  17 3 4 0.3

3

12.3 4

 35.7

2 0.3  255 . 3 12.3

Thus, the solution is

94

X1 

A1

X2 

A2

A



35.7  2.1 17



. 255 .  15 17

A

Problem 5-44.

A

6 5  (6)(4)  ( 5)(3)  39 3 4 19.6 5  23.4 110 4 .

A1  A2 

6 19.6 3

110 .

X1 

A1

X2 

A2

 124.8

Thus, the solution is A

A



23.4  0.6 39



124.8  3.2 39

Problem 5-45.

3

2

4

A  2 2

5 3

0  4[( 2)(3)  (5)(2)]  4[(3)(5)  ( 2)( 2)]  108 4

12 2 4 . 0  1512 A1  10.7 5 . 8.5 3 4 3 12 . A2  2 10.7

4 0  2916 . 4

2

8.5

3

2

12 .

A3  2 2 Thus, the solution is

5 3

10.7  64.8 8.5

95

X1 

A1

X2 

A2

X3 

A3

A

A

A



. 1512 .  14 108



2916 .  2.7 108



64.8  0.6 108

Problem 5-46.

6 5

2

A  3 2 7  6[( 2)( 3)  (7)(4)]  3[( 5)( 3)  ( 2)( 4)]  153 0 4 3 20.3 5 2 A1  2.1 2 7  397.8 11 4 3 . 6 20.3 A2  3 2.1 0

11 .

2 7  2601 . 3

6 5 20.3 A3  3 2 0 4 Thus, the solution is

2.1  290.7 11 .

X1 

A1

X2 

A2

X3 

A3

A

A

A



397.8  2.6 153



. 2601 .  17 153



290.7  19 . 153

96

Problem 5-47.

2 3 1 2 2 3 2 2  3 2 3 4 1 3 4 5 1 1  (1) 2 2 4  (5) 2 2 4  (3) 3 4 A 2 2 0 4 1 1 1 1 1 1 2 2 4 1 1 3 1  (1)(6)  (5)( 30)  (3)(74)  78 10.3 3 1 2 4.3 4 5 1  46.8 A1  5.4 2 0 4 1 3 1 18 . 2 10.3 1 2 3 4.3 5 1  273 A2  2 5.4 0 4 1 18 3 1 . 2 3 10.3 2 3 4 4.3 1 A3   124.8 2 2 5.4 4 1 1 1 . 18 2 3 A4  2 1 Thus, the solution is

3 1 10.3 4 5 4.3  7.8 2 0 5.4 1 3 18 .

X1 

A1

X2 

A2

X3 

A3

X4 

A4

A

A

A

A



46.8  0.6 78



273  35 . 78



124.8  16 . 78



7.8 .  01 78

97

Problem 5-48. 5 1 1 2 3 1 A 3 1 3 2 4 1

3 3 1 2 2 1 2 2 3 2  2 3 1 2  (5) 1 3 5  ( 1) 3 3 5  (1) 3 1 5  ( 3) 3 1 3 5 4 1 2 2 1 2 2 4 2 2 4 1 2  (5)( 33)  (1)(36)  (1)( 60)  ( 3)(9)  216

119 . 1 1 2.0 3 1 A1  14.1 1 3 4.4 4 1

3 2  496.8 5 2

5 119 1 . 2 2.0 1 A2  3 14.1 3 2 4.4 1

3 2 0 5 2

. 5 1 119 2 3 2.0 A3  3 1 14.1 2 4 4.4

3 2  302.4 5 2

. 5 1 1 119 2 3 1 2.0 A4   129.6 3 1 3 14.1 2 4 1 4.4 Thus, the solution is X1 

A1

X2 

A2

X3 

A3

X4 

A4

A

A

A

A



496.8  2.3 216



0 0 216



302.4  14 . 216



129.6  0.6 216

Problem 5-49. Solution not provided.

98

Problem 5-50. Solution not provided.

5.8 Matrix Inversion Problem 5-51.

P 1 P = I -0.45 1 0 q11 q12   1  q   1  0 1  21 q 22  -0.45 The above equation produces two sets of simultaneous equations. These two sets and their solutions are . q11  0.45q12  1   q11  1254     q12  0 0.45q11  q12  0.564 q 21  0.45q 22  0  q 21     q 22  1 0.45q 21  q 22 

0.564   1254  .

Thus, the inverse matrix is . 0.564  1254 P 1    . 0.564 1254 

Problem 5-52.

P 1 P = I q11 q12   2.1 -1.4 1 0 q     21 q 22  0.7 1.8  0 1 The above equation produces two sets of simultaneous equations. These two sets and their solutions are 2.1q11  0.7q12  1  q11  0.378     . q11  1.8q12  0 14 q12  0.294 2.1q 21  0.7q 22  0  q 21     . q 21  1.8q 22  1 14 q 22 

Thus, the inverse matrix is  0.378 0.294  P 1   . 0.441 0147

99

.  0147  0.441   

Problem 5-53.

P 1 P = I -0.62 0.37  1 0 0 q11 q12 q13   1   q 1 -0.51  0 1 0  21 q 22 q 23  -0.62 q 31 q 32 q 33   0.37 -0.51 1  0 0 1 The above equation produces three sets of simultaneous equations. These three sets and their solutions are q11  0.62q12  0.37q13  1  q11   1.635  q12  0.51q13  0  q12    0.953  0.62q11  q13  -0.119  q13  0 0.37q11  0.51q12  q 21  0.62q 22  0.37q 23  0  q 21  q 22  0.51q 23  1  q 22   0.62q 21  q 23  q 23  0 0.37q 21  0.51q 22 

0.953 1.907    0.620

q 31  0.62q 32  0.37q 33  0  q 31  q 32  0.51q 33  0  q 32   0.62q 31  q 33  q 33  1 0.37q 31  0.51q 32  Thus, the inverse matrix is . 0.953 0119 .   1635  1 P   0.953 1907 . 0.620  0119  . 0.620 1360 .

-0.119   0.620     1.360 

Problem 5-54.

P 1 P = I q11 q12 q13   7 -4 3  1 0 0     q  21 q 22 q 23  -2 2 -3  0 1 0 q 31 q 32 q 33   3 2 -4 0 0 1 The above equation produces three sets of simultaneous equations. These three sets and their solutions are 7q11  2q12  3q13  1  q11   -0.083  4q11  2q12  2q13  0  q12   -0.417  q13   0.250  3q11  3q12  4q13  0 7q 21  2q 22  3q 23  0  q 21   4q 21  2q 22  2q 23  1  q 22   q 23  3q 21  3q 22  4q 23  0

100

-0.708 -1.542    0.625 

7q 31  2q 32  3q 33  0  q 31   4q 31  2q 32  2q 33  0  q 32   q 33  3q 31  3q 32  4q 33  1

-0.417  -1.083    0.250 

Thus, the inverse matrix is P

1

 0.083 0.417 0.250   0.708 1542 . 0.625 0.417 1083 . 0.250

Problem 5-55. Solution not provided. Problem 5-56. Solution not provided.

5.9 Applications Problem 5-57. n5

X

i

Problem 5-58. n6

X

i

Problem 5-59. n6

X

i

 15  X i2  59  Yi  11  X i Yi  40 Hence, the set of the two simultaneous equations and its solution are given by 5a  15b  11 a  0.700     15a  59b  40 b  0.500

 20  X i2  106  Yi  18  X i Yi  43 Hence, the set of the two simultaneous equations and its solution are given by 6a  20b  18 a   4.4407      20a  106b  43 b  0.4322 

 99  X i2  2067  Yi  9  X i Yi  304 Hence, the set of the two simultaneous equations and its solution are given by 6a  99b  9 a  4.4187      99a  2067b  304 b   0.3587  Problem 5-60. Solution not provided.

101

Problem 5-61.

I1  I 2  I 3  0

I1  I 2  I 3  0

 6  I 1 (2) + I 3 (4) 6  I 1 (2)  I 2 (2  3)

2 I1  4I3  6  2 I 1  5I 2 6

 I1  I    2  I 3 

.  14211  0.6316    0.7895 

Problem 5-62.

I1  I 2  I 3  0

I1  I 2  I 3  0

3  I 1 (5) + I 3 ( 2)  3  I 1 (5)  I 2 (4)

5I 1  2I3  3  5I 1  4 I 2 3

 I1  I    2  I 3 

 0.4737  01579   .   0.3158 

Problem 5-63.

I1  I 2  I 3  0

I1  I 2  I 3  0

 3  6  I 1 (2) 3  I 1 (2)  I 2 (4  3)

2 I1 2 I1  7 I 2

 3  3

 I1  I    2  I 3 

102

.  15000   0.8571   2.3571

Problem 5-64.

9  I 1 (3)  ( I 1  I 3 )(4) 6  I 2 (5 + 6) 6   ( I 3  I 1 )(4)

 4I3  9

7 I1



11I 2 4 I1

6   4I3  6

 I1  I    2  I 3 

 1   0.5455    0.5000

Computer Programs Problem 5-65. The following is a computer program written in FORTRAN for the Gauss elimination method. It should be noted that this program is intended only for demonstration purposes for the Gauss elimination technique. It can solve a relatively small system of equations that has a solution. With the recommendations outlined in Sections 5.4.1 to 5.4.3 of the textbook, this program can be modified to handle general sets of simultaneous equations.

C******************************************************************** PROGRAM LINEQ INTEGER SIZE1, SIZE2 PARAMETER (SIZE1=15, SIZE2=SIZE1+1) REAL A(SIZE1,SIZE2),X(SIZE1) INTEGER N,J,I PRINT*, ' INPUT NUMBER OF EQUATIONS TO BE SOLVED?' READ*, N PRINT*, ' INPUT COEFFICIENT MATRIX, A, ROWWISE ?' READ*, ((A(I,J), J=1,N), I=1,N) PRINT*, ' INPUT CONSTANT VECTOR, C ?' READ*, (A(I,N+1),I=1,N) CALL GAUSS (A,SIZE1, SIZE2, N, X) PRINT*, ' THE SOLUTION VECTOR, X, IS:' DO 10 I=1,N WRITE(*, 1) I, X(I) 1 FORMAT(1X,'X(',I2,') =',F8.4) 10 CONTINUE END C ***************************************************** C SUBROUTINE GAUSS C ***************************************************** 103

SUBROUTINE GAUSS(A,SIZE1,SIZE2,N, X) INTEGER SIZE1, SIZE2, N,I REAL A(SIZE1,SIZE2), X(SIZE1),DIVIDE I=1 C ****************FORWARD PASS DO 100 J=I+1,N DIVIDE=-A(J,I)/A(I,I) DO 200 K=I,N+1 A(J,K)=A(J,K)+DIVIDE*A(I,K) 200 CONTINUE 100 CONTINUE C ***************** BACK SUBSTITUTION X(N)=A(N,N+1)/A(N,N) DO 300 J=N-1,1,-1 X(J)=A(J,N+1) DO 400 K=J+1,N X(J)=X(J)-A(J,K)*X(K) 400 CONTINUE X(J)=X(J)/A(J,J) 300 CONTINUE END Problem 5-66. Same as Problem 5-65 except that the subroutine GAUSS is to be modified to handle the elimination of all coefficients in A except for the pivot elements. See Section 5.3 of the text.

104

CHAPTER 6. NUMERICAL INTERPOLATION 6.2 Method of Undetermined Coefficients Problem 6-1. 1 0.4  a  0.064  1 0.5 b  =  0.125     

a  0.180 b    0.610      The interpolating polynomial is 

f ( x )  0.180  0.610 x and f (0.47)  0.180  0.61(0.47)  0.1067

Problem 6-2. 1 0.6 a  0.216 1 0.7 b  = 0.343     

a  0.546 b    1.270      The interpolating polynomial is 

f ( x )  0.546  1.270 x and f (0.62)  0.546  1.27(0.62)  0.241

Problem 6-3. .  a  0.064  .  a   0120 1 0.4 016      1 0.5 0.25 b    0125      .   b   0.740   c   1500 1 0.6 0.36  c  0.216 . Thus, the interpolating polynomial is f ( x )  0120 .  0.740 x  15 . x2 and f (0.47)  0120 . . (0.47) 2  0104 .  0.740(0.47)  15

Problem 6-4. .  a   0.210  1 0.5 0.25 a   0125  1 0.6 0.36 b   0.216  b   1070    .         c   1800 1 0.7 0.49  c  0.343 . Thus, the interpolating polynomial is f ( x )  0.210  1.070 x  1.800 x 2 and f (0.58)  0.210  1070 . (0.58)  1800 . (0.58) 2  0195 .

105

Problem 6-5. . 0.064  a  0.064  1 0.4 016 1 0.5 0.25 0125 .   b   0125 .       1 0.6 0.36 0.216  c  0.216      1 0.7 0.49 0.343 d   0.343 Thus, the interpolating polynomial is f ( x)  x 3 and f (0.47)  (0.58) 3  0195 .



 a   0  b   0     c   0      d   1

Problem 6-6. Solution not provided. Problem 6-7. Solution not provided. Problem 6-8. Solution not provided. Problem 6-9. Solution not provided. Problem 6-10. Solution not provided. Problem 6-11. Let F(x) be the function used to derive the data point F ( x )  a 0  a1 x  a 2 x 2 ......... a n x n Let f(x) be the polynomial with the same order as the function F(x) f ( x )  b0  b1 x  b2 x 2 ........ bn x n If the data set is given by the following table x c1 c2 ......... cn f 1(x ) f 2(x ) ......... f n(x ) F(x ) the method of undetermined coefficient will result in

or, using a matrix format

106

c n+1 f n+1(x )

1 c1 c12 ..... c1n  b0   f 1 ( x )    c22 ..... c2n  b1   f 2 ( x )  1 c2     . . . . .  .  .  (1) . . . . . .  .       cn2 ..... cnn  b2   f n ( x )  1 cn 2 n  1 c cn+1 ..... cn+1 n+1  bn   f n+1 ( x )   Since the data set is generated by the F(x), the following matrix equation can be obtained 1 c1 c12 ..... c1n  a 0   f 1 ( x )    c22 ..... c2n   a1   f 2 ( x )  1 c2     . . . . .  .  .  (2) . . . . . .  .       cn2 ..... cnn  a 2   f n ( x )  1 cn 2 n  1 c cn+1 ..... cn+1 n+1  a n   f n+1 ( x )   Comparing Eqs. (1) and (2) results in a 0  b0   a  b   1  1  .  =  .   F(x) = f(x)  and the polynomial provide error free interpolated values . . a  b   2  2 a n  bn  Problem 6-12. Second-Order polynomials: f ( x )  b0 0  b0  100 1 0 1 100 10000  b    91      1   1 200 40000 b2   44  Hence, the first polynomial is . f 1 ( x )  100  01000 x  0.0019 x 2 and

 b1 x  b2 x 2 b0   100  b    010000   1  .  b2  0.0019 

f 1 (50)  100  01000 . (50)  0.0019(50) 2  100.25 %  % error 

100  100.25 x100 = 0.25% 100

In a similar manner, the second polynomial is derived as follows: b0   89  1 300 90000  b0  20 1 400 160000 b    9   b   0.3200   1     1   b2   0.0003  1 500 250000 b2   4  Hence, the second polynomial is f 1 ( x )  89  0.32 x  0.0003x 2 and f 1 (450)  89  0.32(450)  0.0003(450) 2  5.75 %

107

 % error 

6  5.75 x100 = 4.2% 6

Problem 6-13. Solution not provided. Problem 6-14. Linear Interpolating Polynomial: f(x) = b0 + b1x b0  0.003850 1 12  b0   0.20791 1 14  b  = 0.24192   b   0.017005   1     1   f ( x )  0.003850  0.017005x f (12.5)  0.22261 f (13)  0.22492 The following table compares the accuracy achieved by the method: x f (x ) f True(x ) % error 12.5 0.22261 0.21644 2.851 13 0.22492 0.22495 0.013

Quadratic Interpolating Polynomial: f(x) = b0 + b1x + b2x2 . 1 10 100 b0   017365  b0  0.00140000 1 12 144  b    0.20791  b    0.01781750     1    1   1 14 196 b2  0.24192 b2   0.00003125 f ( x )  0.0014  0.0178175x  0.00003125x 2 f (12.5)  0.21644 f (13)  0.22495 The following table compares the accuracy achieved by the method: x f (x ) f True(x ) % error 12.5 0.21644 0.21644 0.000 13 0.22495 0.22495 0.000 Cubic Interpolating Polynomial: f(x) = b0 + b1x + b2x2 + b3x3 . 0 1 10 100 1000  b0   017365  b0    b   0.0174608333  1 12 144 1728  b   0.20791    1       1   b2  0.0000012500 1 14 196 2744 b2  0.24192           b4   0.0000008333 1 16 256 4096 b4  0.27564  f ( x )  0.0174608333x  0.00000125x 2  0.0000008333x 3 f (12.5)  0.21644 f (13)  0.22495 The following table compares the accuracy achieved by the method: x f (x ) f True(x ) % error 12.5 0.21644 0.21644 0.000 13 0.22495 0.22495 0.000

108

Problem 6-15. Solution not provided.

6.3 Gregory-Newton Interpolation Method Problem 6-16. The interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of x. The calculations are summarized to four significant figures in the following table: i x f (x ) ai Eq. 6-10 (Textbook) 1 0.50 0.4621 0.4621=a 1 0.4621 0.7680 2 0.55 0.5005 0.5005=a 1+a 2(0.55-0.50) -0.3800 3 0.60 0.5370 0.5370=a 1+a 2(0.60-0.50)+a 3(0.60-0.50)(0.60-0.55) 0.13333 4 0.65 0.5717 0.5717=a 1+a 2(0.65-0.50)+a 3(0.65-0.50)(0.65-0.55)+ a 4(0.65-0.50)(0.65-0.55)(0.65-0.60) . This yields the following interpolating polynomial: f ( x )  0.4621  0.7680( x  0.5)  0.3800( x  0.5)( x  0.55)  0.1333( x  0.5)( x  0.55)( x  0.60) From which f (0.475)  0.4422  To four significant figure, there is no error generated as compared f (0.525)  0.4815 with true values. Problem 6-17. Solution not provided. Problem 6-18. The interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of x. The calculations are summarized to five significant figures in the following table: i D A ai Eq. 6-10 (Textbook) 1 0.00 100 100=a 1 100 -0.0900000 2 100 91 91=a 1+a 2(100-0) 3 200 44 44=a 1+a 2(200-0)+a 3(200-0)(200-100) -0.001900 . 4 300 20 20=a 1+a 2(300-0)+a 3(300-0)(300-100)+ a 4(300-0)(300-100)(300-200) 0.0000102 . 5 400 9 9=a 1+a 2(400-0)+a 3(400-0)(400-100)+ a 4(400-0)(400-100)(400-200)+ . -2.958E-08 a 5(400-0)(400-100)(400-200)(400-300) . 6 500 4 4=a 1+a 2(500-0)+a 3(500-0)(500-100)+ a 4(500-0)(500-100)(500-200)+ . a 5(500-0)(500-100)(500-200)(500-300)+ . 6.167E-11 a 6(500-0)(500-100)(500-200)(500-300)(500-400)

109

This yields the following interpolating polynomial: f ( x )  100  0.09 x  0.0019 x ( x  100)  0.0000102 x ( x  100)( x  200)

. * 10 11 - 2.958 * 108 x ( x  100)( x  200)( x  300)  6167

Problem 6-19. Solution not provided. Problem 6-20. A second-order interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of t. The calculations are summarized to three significant figures in the following table: i t (0F) h (Btu/lb) ai Eq. 6-10 (Textbook) 1 800 1303.6 1303.6=a 1 1303.6 2 1000 1458.4 1458.4=a 1+a 2(1000-800) 0.774 3 1200 1585.3 1585.3=a 1+a 2(1200-800)+a 3(1200-800)(1200-1000) -0.000349 This yields the following second-order interpolating polynomial:

Problem 6-21. The fifth-order interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of D. The calculations are summarized to three significant figures in the following table:

This yields the following fifth-order interpolating polynomial: A( D)  100  0.09 D  0.0019 D( D  100)  0.0000102 D( D  100)( D  200) - 2.958 * 10 8 D( D  100)( D  200)( D  300) . * 10 11 D( D  100)( D  200)( D  300)( D  400)  6167 From which 100  108.87 A(50)  108.87 %  % error  x100  8.87 % 100

Problem 6-22. Solution not provided. Problem 6-23. Solution not provided. Problem 6-24. (a). The fourth-order interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of the angle . The calculations are summarized to three significant figures in the following table:

110

This yields the following fourth-order interpolating polynomial: . * 10 7 (  5)(  10)(  15) sin()  0.0174  0.00000267(  5)(  10)  1333 (b). To three significant figures, no errors are generated. sin( 7)  0.122  (c). A third-order interpolating polynomial can be formed by applying Eq. 6-10 of the textbook for each pair of values in increasing order of the angle  using an increment of 100. The calculations are summarized to three significant figures in the following table: i ai angle sine Eq. 6-10 (Textbook) 1 0 0 0=a 1 0 0.017400000 2 10 0.174 0.174=a 1+a 2(10-0) 3 20 0.342 0.342=a 1+a 2(20-0)+a 3(20-0)(20-10) -0.000030000 4 30 0.500 0.500=a 1+a 2(30-0)+a 3(30-0)(30-10)+ -0.000000667 a 4(30-0)(30-10)(30-20) . This yields the following third-order interpolating polynomial: sin()  0.0174  0.00003(  10)  0.000000667(  10)(  20) and sin(7)  0.0174(7) - 0.00003(7)(7 - 10) - 0.000000667(7)(7 - 10)(7 - 20) = 0.122 As in part (b), no error is generated to three significant figures as compared to the true value of 0.122. Problem 6-25. If there is an error of 5% in the sine of 100, this means that sin(10) = 0.1653 instead of 0.174, and the resulting fourth-order and third-order polynomials as dictated in part (a) and (b) of problem 613 will be, respectively, sin()  0.0174  0.000174(  5)  0.00003213(  5)(  10)

 0.33466 * 10 5 (  5)(  10)(  15) and

sin()  0.01653  0.000057(  10)  0.000005017(  10)(  20) The estimate for sine of 70 is given by For the Fourth-Order Polynomial: 0122 .  0117 . error = sin( 7)  0.117  x100  4.1% 0122 . For the Third-Order Polynomial 0122 .  0113 . error = sin( 7)  0.113  x100  7.4% 0122 . Obviously, the fourth-order polynomial gives a better estimate for sin(7) than that of the thirdorder.

Problem 6-26. Solution not provided. Problem 6-27. Solution not provided.

111

Problem 6-28. First-Order Polynomial (two data points): sin()  0.0174 sin(2.5)  0.043500 Second-Order Polynomial (three data points): sin()  0.0174 sin(2.5) = 0.043500

Third-Order Polynomial (four data points): sin()  0.0174  0.266667 * 10 5 (  5)(  10) sin(2.5)  0.043375 Fourth-Order Polynomial (five data points): sin()  0.0174  0.266667 * 10 5 (  5)(  10)  0133336 . * 10 6 (  5)(  10)(  15) sin(2.5) = 0.043297 Fifth-Order Polynomial (six data points): sin()  0.0174  0.266667 * 10 5 (  5)(  10)  0133336 . * 10 6 (  5)(  10)(  15)

 0.533366 * 10 8 (  5)(  10)(  15)(  20) sin(2.5)  0.043242 Sixth-Order Polynomial (seven data points): sin() = 0.0174 - 0.266667 *10 -5 (  5)(  10)  0133336 . * 10 6 (  5)(  10)(  15)

 0.533366 * 10 8 (  5)(  10)(  15)(  20)  0.255803 * 10 13 (  5)(  10)(  15)(  20)(  25) sin(2.5)  0.043242 The following table shows the change in error for an interpolated value of the sine of 2.50 as the order of the polynomial is increased. i No.of data points Polynomial Order sin(2.5) |sini+1(2.5)-sini(2.5)| 1 2 1 0.043500 0.000000 2 3 2 0.043500 0.000125 3 4 3 0.043375 0.000078 4 5 4 0.043297 0.000055 5 6 5 0.043242 0.000000 6 7 6 0.043242 ----------

6.4 Finite Difference Interpolation Problem 6-29. The following is the finite difference table:

112

D 0

A 100

A

2A

3A

4A

5A

-9 100

91

-38 -47

200

61

44

23 -24

300

-10

20

13 -11

400

-71 74 3 -7

9

6 -5

500 4 A second-order polynomial can be derived as follows: Using Eq. 6-37b of the textbook results in 2 A  2b2 ( D) 2

38  2b2 (100) 2 b2  0.00190 Using Eq. 6-36b of the textbook results in A  2b2 ( D) D  b2 ( D) 2  b1 ( D) 9  2( 0.0019)(100)(0)  0.0019(100) 2  b1 (100) b1  0100 . Using Eq. 6-33 of the textbook results in A  b2 D 2  b1 D  b0

100  0.0019(0) 2  01 . (0)  b0 b0  100 Thus, the interpolating polynomial is A  0.0019 D 2  01 . D  100 The estimated value of A(50) and the error are A(50)  0.0019(50) 2  01 . (50)  100  100.25 %

% error 

100  100.25 x100  0.25 % 100

Problem 6-30. Solution not provided. Problem 6-31. Solution not provided. Problem 6-32. The following is the finite difference table:

113

degrees) 0

sin 

sin () 0.000

2sin 

3sin 

4sin 

5sin 

6sin 

0.087 5

0.087

0.000 0.087

10

0.174

-0.002 -0.002

0.085 15

0.259

0.000 -0.002

0.083 20

0.342

-3E-17

-0.002

0.423

-0.002

0.000

0.081 25

0.002

-0.002 -0.002

-0.004 0.077

30 0.500 A second-order polynomial can be derived as follows: Using Eq. 6-37b of the textbook results in 2 sin  2b2 ( ) 2

0.002  2b2 (5) 2 b2  0.000040 Using Eq. 6-36b of the textbook results in  sin  2b2 ( )  b2 ( ) 2  b1 ( ) 0.087  2( 0.000040)(5)(5)  (0.000040)(5) 2  b1 (5) b1  0.0180 Using Eq. 6-33 of the textbook results in sin()  b2  2  b1  b0

0.087  0.00004(5) 2  0.018(5)  b0 b0  0.002 Thus, the interpolating polynomial is sin(  )  0.00004 2  0.018  0.002 The estimated value of sin(70) and the error are sin(7 0 )  0.00004(7) 2  0.018(7)  0.002  0122 .

% error 

0122  0122 . . x100  0 % 0122 .

6.5 Newton’s Method Problem 6-33. (a) The following is the finite difference table:

114

0.000 -0.002

Latitude(0N) 0

R 891

R

 2R

 3R

 4R

-35 20

856

-102 -137

40

719

14 -88

-225 60

494

24 38

-50 -275

80

(b)

219 n(n  1) n(n  1)(n  2) n(n  1)(n  2)(n  3) R ( L)  891  n( 35)  ( 102)  (14)  (24) 2! 3! 4! L  L0 where n  L 0 For L = 35 , L0 = 0.0, and L = 200 35  0.0 . n  175 20 175 . (175 .  1) . (175 .  1)(175 .  2) 175 R(35)  891  175 . ( 35)  ( 102)  (14) 2 6 . (175 .  1)(175 .  2)(175 .  3) 175  (24) 24 g.cal  762.5 / day cm 2

Problem 6-34. Solution not provided. Problem 6-35. Solution not provided. Problem 6-36. The following is the finite difference table: h (Btu/lb) t (0F) 800 1303.6

h

 2h

 3h

 4h

154.8 1000

1458.4

-27.9 126.9

1200

1585.3

21.8 -6.1

120.8 1400

1706.1

-0.7 120.1

1600

1826.2

115

-16.4 5.4

h(t )  1303.6  n(154.8) 

n(n  1) n(n  1)(n  2) n(n  1)(n  2)(n  3) ( 27.9)  (218 . ) ( 16.4) 2! 3! 4!

t  t0 t 0 For t = 1100 F, t0 = 800 0F, and t = 200 0F 1100  800 . n  15 200 15 . (1..5  1) . (15 .  1)(15 .  2) 15 h(1100)  1303.6  15 . (154.8)  ( 27.9)  (218 . ) 2 6 . (15 .  1)(15 .  2)(15 .  3) 15 ( 16.4)  24  1523.6 Btu / lb

where n 

Problem 6-37. The table for the times of sunrise and sunset is reconstructed here to show the times in minutes in an increment of 28 days: date day sunrise (minutes) sunset (minutes) May 1 1 299 1133 May 29 29 274 1159 June 26 57 272 1172 July 24 85 290 1161 (a). Sunrise Finite Difference Table day Rise Rise 2Rise 3Rise 1 299 -25 29 274 23 -2 -3 57 272 20 18 85 290 n(n  1) n(n  1)(n  2) Rise  299  n( 25)  (23)  ( 3) 3! 2! day  day 0 where n  day Sunset Finite Difference Table day Set Set 2Set 3Set 1 1133 26 29 1159 -13 13 -11 57 1172 -24 -11 85 1161

116

n(n  1) n(n  1)(n  2) ( 13)  ( 11) 2! 3! day  day 0 where n  day 12  1 May 12  day = 12  n  0.3929 28 0.3929(0.3929  1) Rise( May 12)  299  0.3929( 25)  (23) 2 0.3929(0.3929  1)(0.3929  2) ( 3)  6  286.24 minutes  4 hr 46 m 77  1 July 16  day = 77  n  2.7143 28 2.7143(2.7143  1) Set(July 16)  1133  2.7143(26)  ( 13) 2 2.7143(2.7143  1)(2.7143  2)  ( 11) 6  1167.23 minutes  19 hr 27 m Set the derivative of Rise with respect to n equal to zero. This will result in: dRise . n 2  26n  37.5  0 .  15  n  15877 dn day  1 . Date = June 14 15877   day  45  28 . )  270.23 min Time = 4 hr 30 m Rise(15877  Note: Alternatively, values for the function (Rise) can be generated for the whole range of day = 1 to 85, and by using a spread sheet or a program code the minimum value of the time and its corresponding value of the date can be determined easily. As in part (d), set the derivative of Set with respect to n equal to zero. This will result in dSet . n 2  2n  28.8333  0  55  n  2.1150 dn day  1 2.1150   day  60  Date = June 29 28 Rise(2.1150)  1172.16 min  Time = 19 hr 32 m Note: Alternatively, values for the function (Set) can be generated for the whole range day = 1 to 85, and by using a spread sheet or a program code the maximum value of the time and its corresponding value of the date can be determined easily. Set the derivative of (Set - Rise) with respect to n equal to zero. This will result in Set  1133  n(26) 

(b).

(c).

(d).

(e).

(f).

Note: Alternatively, values for the function (Set - Rise) can be generated for the whole range day = 1 to 85, and by using a spread sheet or a program code the maximum value of the time and its corresponding value of the date can be determined easily. Problem 6-38. Solution not provided.

117

Problem 6-39. The table for the times of sunrise and sunset is reconstructed here to show the times in minutes in an increment of 28 days: date day sunrise (minutes) sunset (minutes) May 1 1 330 1112 May 29 29 319 1114 June 26 57 322 1122 July 24 85 332 1120 (a). Sunrise Finite Difference Table day Rise Rise 2Rise 3Rise 1 330 -11 29 319 14 3 -7 57 322 7 10 85 332 n(n  1) n(n  1)(n  2) Rise  330  n( 11)  (14)  ( 7 ) 2! 3! day  day 0 where n  day Sunset Finite Difference Table day Set Set 2Set 3Set 1 1112 2 29 1114 6 8 -16 57 1122 -10 -2 85 1120 n(n  1) n(n  1)(n  2) Set  1112  n(2)  (6)  ( 16) 3! 2! day  day 0 where n  day 12  1 (b). May 12  day = 12  n  0.3929 28 0.3929(0.3929  1) Rise( May 12)  330  0.3929( 11)  (14) 2 0.3929(0.3929  1)(0.3929  2)  ( 7) 6  32356 . minutes  5 hr 24 m 77  1 (c). July 16  day = 77  n  2.7143 28

118

2.7143(2.7143  1) (6) 2 2.7143(2.7143  1)(2.7143  2)  ( 16) 6  1122.52 minutes  18 hr 43 m Set the derivative of Rise with respect to n equal to zero. This will result in dRise . n 2  21n  20.3333  0 .  35  n  12138 dn day  1 . 12138   day  35  Date = June 4 28 . )  318.7 min Rise(12138  Time = 5 hr 19 m Note: Alternatively, values for the function (Rise) can be generated for the whole range of day = 1 to 85, and by using a spread sheet or a program code the minimum value of the time and its corresponding value of the date can be determined easily. As in part (d), set the derivative of Set with respect to n equal to zero. This will result in dSet  8n 2  22n  6.3333  0  n  2.4233 dn day  1 2.4233   day  69  Date = July 8 28 Rise(2.4233)  1123.3 min  Time = 18 hr 43 m Note: Alternatively, values for the function (Set) can be generated for the whole range day = 1 to 85, and by using a spread sheet or a program code the maximum value of the time and its corresponding value of the date can be determined easily. Set the derivative of (Set - Rise) with respect to n equal to zero. This will result in d( Set  Rise)  4.5n 2  n  14  0  n  18784 . dn day  1   day  54  Date = June 23 18784 . 28 Rise(18784 . )  Set (18784 . )  112124 .  32112 .  80012 . min Longest perid of daylight = 800.12 min = 13 h 20 m Note: Alternatively, values for the function (Set - Rise) can be generated for the whole range day = 1 to 85, and by using a spread sheet or a program code the maximum value of t the time and its corresponding value of the date can be determined easily. Set(July 16)  1112  2.7143(2) 

(d).

(e).

(f).

Problem 6-40. Solution not provided.

6.6 Lagrange Polynomials Problem 6-41. The table for the times of sunrise and sunset is reconstructed here to show the times in minutes and the date in days:

119

date Nov. 1 Dec. 1 Jan. 1 Feb. 1 Mar. 1

day 1 31 62 93 121

sunrise (minutes) sunset (minutes) 441 964 514 902 542 904 493 973 418 1046

(a).

Sunrise: Rise( d 0 )  441w1 ( d 0 )  514 w2 ( d 0 )  542 w3 ( d 0 )  493w4 ( d 0 )  418w5 ( d 0 ) Sunset: Set ( d 0 )  964 w1 ( d 0 )  902 w2 ( d 0 )  904 w3 ( d 0 )  973w4 ( d 0 )  1046w5 ( d 0 ) where (d  31)(d 0  62)(d 0  93)(d 0  121) w1 (d 0 )  0 (1  31)(1  62)(1  93)(1  121) (d  1)(d 0  62)(d 0  93)(d 0  121) w2 ( d 0 )  0 (31  1)(31  62)(31  93)(31  121) (d  1)(d 0  31)(d 0  93)(d 0  121) w3 ( d 0 )  0 (62  1)(62  31)(62  93)(62  121) (d  1)(d 0  31)(d 0  62)(d 0  121) w4 ( d 0 )  0 (93  1)(93  31)(93  62)(93  121) (d 0  1)(d 0  31)(d 0  62)(d 0  93) w5 (d 0 )  (121  1)(121  31)(121  62)(121  93) (b). Feb. 20 is equivalent to d0 = 112 (112  31)(112  62)(112  93)(112  121) 692550 w1 (112)   0.034279  (1  31)(1  62)(1  93)(1  121) 20203200 (112  1)(112  62)(112  93)(112  121) 949050 w2 (112)  .   0182882 (31  1)(31  62)(31  93)(31  121) 5189400 (112  1)(112  31)(112  93)(112  121) 1537461 w3 (112)   0.444528  (62  1)(62  31)(62  93)(62  121) 3458639 (112  1)(112  31)(112  62)(112  121) 4045950 w4 (112)    0.817187 (93  1)(93  31)(93  62)(93  121) 4951072 (112  1)(112  31)(112  62)(112  93) 8541450 w5 (112)    0.478738 (121  1)(121  31)(121  62)(121  93) 17841600 Rise(112)  441( 0.034279)  514(0182882 . )  542( 0.444528)  493(0.817187)  418(0.478738)  440.79 min = 7 hr 21m (c). Nov. 11 is equivalent to d0 = 11 (11  31)(11  62)(11  93)(11  121) 9200400 w1 (11)    0.455393 (1  31)(1  62)(1  93)(1  121) 20203200 (11  1)(11  62)(11  93)(11  121) 4600200 w2 (11)    0.886461 (31  1)(31  62)(31  93)(31  121) 5189400 (11  1)(11  31)(11  93)(11  121) 1804000 w3 (11)   0.521592  (62  1)(62  31)(62  93)(62  121) 3458639 (11  1)(11  31)(11  62)(11  121) 1122000 w4 (11)    0.226618 (93  1)(93  31)(93  62)(93  121) 4951072 120

(11  1)(11  31)(11  62)(11  93) 836400  0.046879  (121  1)(121  31)(121  62)(121  93) 17841600 Set (11)  964(0.455393)  902(0.886461)  904( 0.521592)  973( 0.226618)  1046( 0.046879)  938.53 min  15 hr 39 m (d). Latest sunrise Using a spread sheet, values for the function Rise were generated for the whole range of d0 = 1 to 121. The maximum value was found to be 542.819 min which corresponds to d0 = 58, or Dec. 28, hence: Date = Dec. 28 Time = 542.819 min = 9 hr 3 m Note: Alternatively, set the derivative of Rise with respect to d0 equal to zero and to find the value of d0 that maximizes the function Rise. However, this approach is can be tedious. dRise  441w1'  514 w2'  542 w3'  493w4'  418w5'  0 d0 (e). Earliest sunset Using a spread sheet, the values for the function Set were generated for the whole range of d0 = 1 to 121. The minimum value was found to be 893.261 min which corresponds to d0 = 46, or Dec. 16, hence: Date = Dec. 16 Time = 893.261 min = 14 hr 53 m Note: Alternatively, set the derivative of Set with respect to d0 equal to zero and to find the value of d0 that minimizes the function Rise. However, this can be tedious. dRise  964 w1'  902 w2'  904 w3'  973w4'  1046w5'  0 d0 (f). Shortest period of daylight Using a spread sheet, the values for the function (Set-Rise) were generated for the whole range of d0 = 1 to 121. The minimum value was found to be 353.429 min, which corresponds to d0 = 52, or Date = Dec. 22 Time = 353.429 min = 5 hr 53 m Note: Alternatively, set the derivative of (Set-Rise) with respect to d0 equal to zero and to find the value of d0 that minimizes the function (Set-Rise). However, this approach can be tedious. w5 (11) 

Problem 6-42. D( A0 )  17.4 w1  15.0w2  13.3w3  10.8w4  9.0w5  7.7 w6  6.8w7  5.7 w8 where ( A0  10 . )( A0  15 . )( A0  2.5)( A0  4.0)( A0  6.0)( A0  10.0)( A0  16.0) w1 ( A0 )  (0.75  10 . )(0.75  15 . )(0.75  2.5)(0.75  4.0)(0.75  6.0)( 0.75  10.0)(0.75  16.0) ( A  0.75)( A0  15 . )( A0  2.5)( A0  4.0)( A0  6.0)( A0  10.0)( A0  16.0) w2 ( A0 )  0 (1  0.75)(1  15 . )(1  2.5)(1  4.0)(1  6.0)(1  10.0)(1  16.0) ( A  10 . )( A0  0.75)( A0  2.5)( A0  4.0)( A0  6.0)( A0  10.0)( A0  16.0) w3 ( A0 )  0 (15 .  10 . )(15 .  0.75)(15 .  2.5)(15 .  4.0)(15 .  6.0)(15 .  10.0)(15 .  16.0) ( A  10 . )( A0  0.75)( A0  15 . )( A0  4.0)( A0  6.0)( A0  10.0)( A0  16.0) w4 ( A0 )  0 (2.5  10 . )(2.5  0.75)(2.5  15 . )(2.5  4.0)(2.5  6.0)(2.5  10.0)(2.5  16.0)

121

( A0  10 . )( A0  0.75)( A0  15 . )( A0  2.5)( A0  6.0)( A0  10.0)( A0  16.0) (4.0  10 . )(4.0  0.75)(4.0  15 . )(4.0  2.5)(4.0  6.0)(4.0  10.0)(4.0  16.0) ( A  10 . )( A0  0.75)( A0  15 . )( A0  2.5)( A0  4.0)( A0  10.0)( A0  16.0) w6 ( A0 )  0 (6.0  10 . )(6.0  0.75)(6.0  15 . )(6.0  2.5)(6.0  4.0)(6.0  10.0)(6.0  16.0) ( A0  10 . )( A0  0.75)( A0  15 . )( A0  2.5)( A0  4.0)( A0  6.0)( A0  16.0) w7 ( A0 )  (10.0  10 . )(10.0  0.75)(10.0  15 . )(10.0  2.5)(10.0  4.0)(10.0  6.0)(10.0  16.0) ( A0  10 . )( A0  0.75)( A0  15 . )( A0  2.5)( A0  4.0)( A0  6.0)( A0  10.0) w8 ( A0 )  (16.0  10 . )(16.0  0.75)(16.0  15 . )(16.0  2.5)(16.0  4.0)(16.0  6.0)(16.0  10.0) 2 For A = 1.2 mm D(12 . )  17.4( 01729 . )  15.0(0.8091)  13.3(0.3939)  10.8( 0.03387)   9.0(0.00417)  7.7( 0.00039)  6.8(0.00000914)  5.7( 0.00000013) amps  14.036 mm 2 For A = 2.0 mm2 D(2.0)  17.4(0.283631)  15.0( 0.737449)  13.3(1077004 . )  10.8(0.401333)   9.0( 0.026591)  7.7(0.002116)  6.8( 0.000046)  5.7(0.00000062) amps  12.31 mm 2 w5 ( A0 ) 

Problem 6-43. Solution not provided. Problem 6-44. Solution not provided. Problem 6-45.

P( C0 )  1.35w1  1.91w2  2.99 w3  4.22 w4  5.40w5  12.38w6

where

(C0  0.05)(C0  0.02)(C0  0.0098)(C0  0.0059)(C0  0.00103) (01 .  0.05)(01 .  0.02)(01 .  0.0098)(01 .  0.0059)(01 .  0.00103) (C0  01 . )(C0  0.02)(C0  0.0098)(C0  0.0059)(C0  0.00103) w2 ( C0 )  (0.05  01 . )(0.05  0.02)(0.05  0.0098)(0.05  0.0059)( 0.05  0.00103) (C0  01 . )(C0  0.05)(C0  0.0098)(C0  0.0059)(C0  0.00103) w3 ( C0 )  (0.02  01 . )(0.02  0.05)(0.02  0.0098)(0.02  0.0059)(0.02  0.00103) (C0  01 . )(C0  0.05)(C0  0.02)(C0  0.0059)(C0  0.00103) w4 ( C0 )  (0.0098  01 . )(0.0098  0.05)(0.0098  0.02)(0.0098  0.0059)(0.0098  0.00103) (C0  01 . )(C0  0.05)(C0  0.02)(C0  0.0098)(C0  0.00103) w5 (C0 )  (0.0059  01 . )(0.0059  0.05)(0.0059  0.02)( 0.0059  0.0098)(0.0059  0.00103) (C0  01 . )(C0  0.05)(C0  0.02)(C0  0.0098)(C0  0.0059) w6 ( C0 )  (0.00103  01 . )(0.00103.05)(0.00103.02)(0.00103.0098)(0.00103  0.0059) For C0 = 0.015 moles/liter w1 (C0 ) 

122

P(0.015)  135 . (3.4429 * 10 5 )  191 . ( 0.00216)  2.99(0.30035)  4.22(14949 . )  5.40( 0.9723)  12.38(01793 . )  4.172 % For C0 = 0.003 moles/liter P(0.003)  135 . (9.2376 * 10 6 )  191 . ( 4.9194 * 10 4 )  2.99(0.0270)  4.22( 0.3500)   5.40(0.9342)  12.38(0.3892)  8.468 % Problem 6-46.

L( D0 )  100w1  45w2  39 w3  22 w4  5w5  0.5w6

where

( D0  1)( D0  2)( D0  10)( D0  50)( D0  100) (0  1)(0  2)(0  10)(0  50)(0  100) ( D  0)( D0  2)( D0  10)( D0  50)( D0  100) w2 ( D0 )  0 (1  0)(1  2)(1  10)(1  50)(1  100) ( D  0)( D0  1)( D0  10)( D0  50)( D0  100) w3 ( D0 )  0 (2  0)(2  1)(2  10)(2  50)(2  100) ( D  0)( D0  1)( D0  2)( D0  50)( D0  100) w4 ( D0 )  0 (10  0)(10  1)(10  2)(10  50)(10  100) ( D  0)( D0  1)( D0  2)( D0  10)( D0  100) w5 ( D0 )  0 (50  0)(50  1)(50  2)(50  10)(50  100) ( D0  0)( D0  1)( D0  2)( D0  10)( D0  50) w6 ( D0 )  (100  0)(100  1)(100  2)(100  10)(100  50) For D0 = 1.6 m L(16 . )  100( 0.0960)  45(0.5864)  39(0.5103)  22( 0.00071)  5(13495 . * 10 6 ) w1 ( D0 ) 

 0.5( 35759 . * 10 8 )  36.674 % Problem 6-47. Solution not provided. Problem 6-48.

where

S ( T0 )  9 w1  31w2  57 w3  64 w4  68w5  71w6

(T0  11)(T0  23)(T0  33)(T0  49)(T0  60) (5  11)(5  23)(5  33)(5  49)(5  60) (T0  5)(T0  23)(T0  33)(T0  49)(T0  60) w2 (T0 )  (11  5)(11  23)(11  33)(11  49)(11  60) (T  5)(T0  11)(T0  33)(T0  49)(T0  60) w3 (T0 )  0 (23  5)(23  11)(23  33)(23  49)(23  60) w1 (T0 ) 

123

(T0  5)(T0  11)(T0  23)(T0  49)(T0  60) (33  5)(33  11)(33  23)(33  49)(33  60) (T  5)(T0  11)(T0  23)(T0  33)(T0  60) w5 (T0 )  0 (49  5)(49  11)(49  23)(49  33)(49  60) (T  5)(T0  11)(T0  23)(T0  33)(T0  49) w6 (T0 )  0 (60  5)(60  11)(60  23)(60  33)(60  49) For T0 = 15 min S (15)  9( 0.01204)  31( 0.7470)  57( 0.5301)  64( 01840 . )  68( 0.0339)  71( 0.006613)  43.32 % For T0 = 30 min S (78)  9(0.03108)  31( 01015 . )  57(0.3909)  64(0.7122)  68( 0.03911)  71(0.0064)  62.79 % w4 (T0 ) 

Problem 6-49.

f ( 225)  14.6w1  10.7 w2  4.8w3  1.7 w4  0.3w5

where

(225  42)(225  128)(225  317)(225  433)  0.8820 (15  42)(15  128)(15  317)(15  433) w2(225) = -1.5612 w3(225) = 1.3128 w4(225) = 0.4258 w5(225)= -0.0593 f(225) = 14.6(0.8820)+10.7(-1.5612)+4.8(1.3128)+1.7(0.4258)+0.3(-0.0593) = 3.18 feet w1 (225) 

6.7 Interpolation Using Splines Problem 6-50. C P

0.1 1.35

0.05 1.91

0.02 2.99

124

0.0098 4.22

0.0059 0.00103 5.40 12.38

Linear Spline: i Ci (moles/liters) 1 0.1

2

3

4

5

6

P (C i) (%) 1.35

0.05

Linear Spline P i(C ) 1 .3 5 

1 .9 1  1 .3 5 ( C  0 .1 )  1 . 3 5  1 1 . 2 ( C  0 .1 ) 0 . 0 5  0 .1

1.9 1 

2 .9 9  1 .9 1 ( C  0 .0 5 )  1 .9 1  3 6 ( C  0 .0 5 ) 0 .0 2  0 .0 5

1.91

0.02

2.99

0.0098

2 .9 9 

4 .2 2  2 .9 9 ( C  0 .0 2 )  2 .9 9  1 2 0 .6 ( C  0 .0 2 ) 0 .0 0 9 8  0 .0 2

4 .22 

5.40  4 .22 ( C  0 .0098 )  4 .22  302 .6 ( C  0 .0098 ) 0 .0059  0 .0098

5 .4 0 

1 2 .3 8  5.4 0 ( C  0 .0 0 5 9 )  5.4 0  1 4 3 3.3 ( C  0 .0 0 5 9 ) 0 .0 0 1 0 3  0 .0 0 5 9

4.22

0.0059

5.4

0.00103

12.38 Linear Spline 14 12 10

P (% )

8 6 4 2 0.1

0.05

0.02

0.0098

0.0059

0.00103

0

C ( moles /liters )

Quadratic Spline: Using Eq. 6-71a of the textbook, the following equations can be developed: 0.01 . . a1 01 b1  c1  135

0.0025

a2

0.05

b2

 c2

 191 .

0.0004

a3

0.02

b3

 c3

 2.99

0.0098 b4

 c4

 4.22

0.00009604 a 4

0.00003481 a5 0.0059 b5  c5  5.40 Equation 6-71b of the textbook results in the following conditions: . 0.0025 a1 0.05 b1  c1  191 0.0004

a2

0.02

b2

 c2

 2.99

0.00009604

a3

0.0098

b3

 c3

 4.22

0.00003481

a4

0.0059

b4

 c4

 5.40

0.000001061 a5 0.00103 b5  c5  12.38 Equation 6-72 of the textbook results in the following conditions: 125

2a1 (0.05)

b1

 2a 2 (0.05)

b2

2a 2 (0.02)

b2

 2a 3 (0.02)

b3

2a 3 (0.0098)

b3

 2a 4 (0.0098) b4

2a 4 (0.0059) b4

 2a5 (0.0059) b5

or . 01

a1

b1

 01 .

a2

b2

0.04

a2

b2

 0.04

a3

b3

0.0196 a3

b3

 0.0196 a4

b4

0.0118 a4 b4  0.0118 a5 b5 Since a1 = 0, the resulting system of 14 equations can be summarized in a matrix format as .  01  0   0   0  0  . 005  0   0  0   0  1   0   0  0

0 b1   1.34  0 0.0025 0.05 1 0 0 0 0 0 0 0 0 0 c1   1.91      0 0 0 0 00004 002 1 0 0 0 0 0 0 a2   2.99  . .     0 0 0 0 0 0 0 000009604 00098 1 0 0 0 b2   4.22  . . 0 0 0 0 0 0 0 0 0 0 000003481 00059 1 c2   5.40  . .     1 0 0 0 0 0 0 0 0 0 0 0 0 a3   1.91  0 00004 002 1 0 0 0 0 0 0 0 0 0 b3   2.99  . .     0 0 0 0 000009604 00098 1 0 0 0 0 0 0 c3   4.22  . . 0 0 0 0 0 0 0 000003481 00059 1 0 0 0 a4   5.40  . .     0 0 0 0 0 0 0 0 0 0 0000001061 000103 1 b4  12.38 . . . 1 0 0 01 0 0 0 0 0 0 0 0 0 c4   0  . 004 . 1 0 0 004 1 0 0 0 0 0 0 0 a5   0      . . 1 0 0 0 0 0 00196 1 0 00196 0 0 0 b5   0  . . 1 0 c5   0  0 0 0 0 0 0 0 00118 1 0 00118 1

0

0

0

0

0

0

The solution of this system of equations is  b1  c   1 a 2    b2   c2    a 3  b   3   c3  a   4 b4  c   4  a5     b5   c5 

0

0

1140 .    2.48    820.00   93.40   4.53   .   588120  29585 .    6.55   31278.88    793.64   8.99   207127.08    2868.65  1511 . 

126

0

0

0

The following table summarizes the quadratic Spline: Ci P (C i) Quadratic Spline P i(C ) (%) (moles/liters) 0.1 1.35 -11.40C +2.48 0.05

1.91 820C 2 - 93.40C + 4.53

0.02

2.99 5881.20C 2 - 295.85C + 6.55

0.0098

4.22 31279C 2 - 793.64C + 8.99

0.0059

5.40 207127C 2 - 2868.65C + 15.11

0.00103

12.38 Quadratic Spline 14 12 10 P(%)

8 6 4 2 0.04

0.04

0.03

0.03

0.03

0.02

0.02

0.01

0.01

0.01

0.00

0.00

0

C (moles /liter)

Problem 6-51. Solution not provided. Problem 6-52. D (m) L (%) Linear Spline:

0 100

1 45

2 39

127

10 22

50 5.00

100 0.5

Di (meters) 0

L (D i) (%) 100

Linear Spline L i(D ) 45  100 ( D  0)  1 0 0  5 5( D  0) 1  0

100 

1

45 45 

2

39

10

39 

22  39 ( D  2 )  3 9  2 .1 2 5 ( D  2 ) 10  2

22 

5  22 ( D  1 0 )  2 2  0 .4 2 5 ( D  1 0 ) 50  10

22

50

5 5 

100

39  45 ( D  1)  4 5  6 ( D  1) 2  1

0 .5  5 ( D  5 0 )  5  0 .0 9 ( D  5 0 ) 100  50

0.5 Linear Spline 100 80 L (%)

60 40 20 0 0

1

2 10 D (meters)

50

100

Quadratic Spline: Using Eq. 6-71a of the textbook, the following equations can be developed: 0 a1 0 b1  c1  100 1

a2

1

b2

 c2

 45

4

a3

2

b3

 c3

 39

100

a4

10 b4

 c4

 22

2500 a5 50 b5  c5  5 Equation 6-71b of the textbook results in the following conditions: a1 1 b1  c1  45 1 4

a2

2

b2

 c2

 39

100

a3

10

b3

 c3

 22

2500

a4

50

b4

 c4

5

10000 a5 100 b5  c5  0.5 Equation 6-72 of the textbook results in the following conditions:

128

2a1 (1)

b1

 2a2 (1)

b2

2a2 ( 2)

b2

 2 a3 ( 2 )

b3

2a3 (10)

b3

 2a4 (10) b4

2a4 (50) b4

 2a5 (50) b5

or 2

a1

b1

2

a2

b2

4

a2

b2

4

a3

b3

20

a3

b3

 20

a4

b4

100 a4 b4  100 a5 b5 Since a1 = 0, the resulting system of 14 equations can be summarized in a matrix format as 0 0  0  0 0  1 0  0 0  0 1  0  0 0

1

0

0 0

0

0 0

0 0 0

1 0 0

1 1 0 0 0 0

0 4 0

0 0 0 2 1 0 0 0 100

0 1 0

0 0 4

0 0 0 0 2 1

0 0 0

0 0 0 0 0 0

0 0 0

0 0 0

0 0 100 10 1 0 0 0 0 0 0 2500 0 0 0 0 0 0

0 2 1 0 4 1 0 0 0 0 0 0

0 0 0 0 4 1 0 20 1 0 0 0

0

0 0 0

0 0 0 0 0 20 0 100

0 0 0 0 0 0  0 0 0 0 0  10 1 0 0 0 0 0 2500 50 1  0 0 0 0 0 0 0 0 0 0  0 0 0 0 0 50 1 0 0 0  0 0 10000 100 1 0 0 0 0 0 0 0 0 0 0  1 0 0 0 0 1 0 100 1 0 0 0

The solution of this system of equations is 55  b1   c   100  1   a 2   49     b2   153  c2   149     a3   5.64  b   6556 .   3     c3   69.56 a   117 .   4   b4   70.66  c   61156 .   4   a5   0.93      b5   139.38  c5  4639.5 The following table summarizes the Spline: 129

0

0

b1  100  c   45   1   a2   39      b2   22  c2   5      a3   45  b   39   3     c3   22  a   5   4   b4   0.5     c4   0  a5   0      b5   0   c5   0 

Di (meters) 0

L (D i) (%) 100

1

45

Quadratic Spline L i(D )

-55D +100

49D 2 - 153D + 149 2

39 -5.64D 2 + 65.56D - 69.56

10

22 1.17D 2 - 70.66D + 611.56

50

5.00 -0.93D 2 + 139.38D - 4639.50

100

0.5

Problem 6-53. T (min) S (%) Linear Spline: T i (min)

5

5 9

11 31

S (T i) (%)

33

49

60

49 68.00

9 31  9 ( T  5 )  9  3 .6 7 ( T  5 ) 11  5

31 31 

23

33 64

Linear Spline S i(T ) 9 

11

23 57

57  31 ( T  1 1 )  3 1  2 .1 7 ( T  1 1 ) 23  11

57 57 

64  57 ( T  2 3 )  5 7  0 .7 0 ( T  2 3 ) 33  23

64 

68  64 ( T  3 3 )  6 4  0 .2 5 ( T  3 3 ) 49  33

68 

71  68 ( T  4 9 )  6 8  0 .2 7 3 ( T  4 9 ) 60  49

64

68

71

130

60 71

Linear Spline 80 60 S (%) 40 20 0 5

11

23 33 T (min)

49

60

Quadratic Spline: Using Eq. 6-71a of the textbook, the following equations can be developed: 25 a1 5 b1  c1  9 121

a2

11 b2

 c2

 31

529

a3

23 b3

 c3

 57

1089 a 4

33 b4

 c4

 64

2401 a5 49 b5  c5  68 Equation 6-71b of the textbook results in the following conditions: 121 a1 11 b1  c1  31 a2

23 b2

 c2

 57

1089 a 3

33 b3

 c3

 64

2401 a 4

49 b4

 c4

 68

529

3600 a5 60 b5  c5  71 Equation 6-72 of the textbook results in the following conditions: 2a1 (11) b1  2a 2 (11) b2 2a 2 (23)

b2

 2a 3 (23)

2a 3 (33)

b3

 2a 4 (33) b4

2a 4 (49) b4

 2a5 (49) b5

or

b3

22 a1

b1

 22 a 2

b2

46 a 2

b2

 46 a 3

b3

66 a 3

b3

 66 a 4

b4

98 a 4 b4  98 a5 b5 Since a1 = 0, the resulting system of 14 equations can be summarized in a matrix format as

131

5 0  0  0 0  11 0  0 0  0 1  0  0  0

0 0 121 11 1 0 0 0 0 0 0 0 0 0  0 0 0 0 529 23 1 0 0 0 0 0 0  0 0 0 0 0 0 0 1089 33 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2401 49 1  1 0 0 0 0 0 0 0 0 0 0 0 0 0 529 23 1 0 0 0 0 0 0 0 0 0  0 0 0 0 1089 33 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2401 49 1 0 0 0  0 0 0 0 0 0 0 0 0 0 3600 60 1 0 22 1 0 0 0 0 0 0 0 0 0 0 0 46 1 0 46 1 0 0 0 0 0 0 0  0 0 0 0 66 1 0 66 1 0 0 0 0 0 0 0 0 0 0 0 98 1 0 98 1 0 1

0

0 0

0

0 0

The solution of this system of equations is  b1  c   1 a 2    b2   c2    a 3  b   3   c3  a   4 b4  c   4 a5     b5   c5  The following table summarizes the Spline:

0

0

0

 3.667  9.333    0125 .     6.417 24.458    0.0033  0.5133    43.430 0.0302     2.727  6.903    0.0460    4.742   189.893

132

0

0

b1   9   c  31  1   a2  57     b2  64 c2  68     a3  31 b  57  3  c3  64 a  68  4   b4  71     c4   0  a5   0      b5   0  c5   0 

S (T i) (%)

T i (min) 5

Quadratic Spline S i(T ) 9 3.667T - 9.333

11

31 -0.125T 2 + 6.417T- 24.458

23

57 0.00333T 2 + 0.5133T + 43.43

33

64 -0.0302T 2 + 2.727T + 6.903

49

68 -0.046T 2 - 4.742T + 189.893

60

71 Quadratic Spline

80.00 70.00 60.00 50.00 S (%) 40.00 30.00 20.00 10.00 0.00 5

9 13 17 21 25 29 33 37 41 45 49 53 57 T (min)

6.9 Multi-Dimensional Interpolation Problem 6-54.

(a).

0.9610  0.8123 (22.3  20)  0.846501 30  20 2.8901  2.1670 f (3,22.3)  2.1670  (22.3  20)  2.333313 30  20 2.333313  0.846501 f (2.3,22.3)  0.846501  (2.3  2)  12925446 . 3 2 f (2,22.3)  0.8123 

133

(b).

2.1670  0.8190 (10.6  10)  0.89988 20  10 31621 . .  18231 f (4,10.6)  18231 .  (10.6  10)  190344 . 20  10  0.89988 190344 . f (3.4,10.6)  0.89988  (3.4  3)  1301304 . 43 f (3,10.6)  0.8190 

134

CHAPTER 7. DIFFERENTIATION AND INTEGRATION 7.1 Numerical Differentiation Problem 7-1. Forward:

Backward: Two-Step:

dVc ( x ) V (58) - Vc (54 ) grain 5.420 - 4.725  0.174 3 o = Vc' (54 )  c = 58 - 54 4 dT ft . F dVc ( x ) V (54) - Vc (50) 4.725 - 4.108 grain = Vc' (54)  c =  0154 . dT 54 - 50 4 ft 3 . F o dVc ( x ) V (58) - Vc (50) 5.420 - 4.108 grain  0164 . = Vc' (54)  c = dT 58 - 50 8 ft 3 . F o

Problem 7-2. Solution not provided. Problem 7-3. Forward: Backward: Two-Step:

df ( x ) 1676 - 455 f (4) - f (3) = f ' (3)  =  1221 dx 4-3 1 df ( x ) f (3) - f (2) 455 - 152 = f ' (3)  =  303 dx 3- 2 1 df ( x ) 1676 - 152 f (4) - f (2) = f ' (54)  =  762 dx 4-2 2 f ' ( x ) = f ' ( 3) = 533

True Value: The accuracy of the solution is shown in the following table:

f '(x ) % Error

Forward Backward Two-Step 1221 303 762 129.1 43.2 43.0

TRUE 533 ---

Problem 7-4. Solution not provided. Problem 7-5. Forward:

Backward:

Two-Step:

d sin() sin(8 0 ) - sin(7 0 ) 0.13917 - 0.12187 = cos(7 0 )  =  0.99123 0 0 d 1(0.017453) 8 -7 d sin() sin(7 0 ) - sin(6 0 ) 0.12187 - 0.10453 = cos(7 0 )  =  0.99353 0 0 d 1(0.017453) 7 -6 d sin() sin(8 0 ) - sin(6 0 ) 0.13917 - 0.10453 = cos(7 0 )  =  0.99238 0 0 d 2(0.017453) 8 6

135

d sin() = cos(7 0 )  0.99255 d

True Value:

The accuracy of the solution is shown in the following table:

f' ( ) % Error

Forward Backward Two-Step TRUE 0.99123 0.99353 0.99238 0.99255 0.1330 0.0987 0.0171 ---

Problem 7-6. Solution not provided. Problem 7-7.

x 2

f

f (x ) 0.69315

2f

3f

4f

0.40546 3

1.09861

-0.11778 0.28768

4

1.38629

0.05325 -0.06453

0.22315 5

1.60944

-0.02955 0.0237

-0.04083 0.18232

6

1.79176

From the above finite difference table, the first derivative is calculated as follows: Forward:

Backward:

Two-Step:

df dx df dx df dx



f 0.22315   0.22315 5 4 x



f 0.28768   0.28768 5 4 x



f 160944 . .  109861   0.25542 5 3 x

x4

x4

x 4

The second derivative is computed as follows:

d2 f dx 2

 x 4

2 f 0.06453   0.06453 2 1 x

The following table compares estimates with the true values:

Forward Backward Two-Step TRUE f' (x ) 0.22315 0.28768 0.25542 0.25 f'' (x ) -------0.06453 -0.0625 % Error for f '(x ) 10.7 15.1 2.2 % Error for f ''(x ) ------3.248

Problem 7-8. Solution not provided.

136

Problem 7-9.

x 1

f

f (x ) 1.00000

2f

3f

4f

5f

-0.08183 1.2

0.91817

0.05082 -0.03101

1.4

-0.01345

0.88716

0.03737 0.00636

1.6

-0.00587

0.89352

0.03150 0.03786

1.8

0.00758 -0.00245 0.00513 -0.00074

0.93138

0.03076 0.06862

2.0

1.00000

From the above finite difference table, the first derivative is calculated as follows:

df dx

Forward:

df dx

Backward:

df dx

Two-Step:



f 0.00636   0.03180 .  14 . x 16



f 0.03101   015505 . 14 .  12 . x



f 0.03101  0.00636   0.06163 16 .  12 . x

x 1.4

x 1.4

x 1.4

The second derivative is computed as follows:

d2 f dx 2

 x 1.4

2 f 0.03737   018685 . 2 0.2 x

Problem 7-10.

x 0.50

f (x ) 0.4621

f

 2f

 3f

0.0384 0.55

0.5005

0.60

0.5370

-0.0019 0.0365

0.0001 -0.0018

0.0347 0.65

0.5717

From the finite difference table, the first derivative at x = 0.55 is calculated (using two-step finite difference) as follows:

df dx

 x  0.55

f 0.0384  0.0365   0.7490 0.6  0.5 x

Problem 7-11. Solution not provided.

137

Problem 7-12.

t 0

D

D

2D

3D

4D

5D

0 10

2

10

30 40

4

30

50

60 100

6

20

150

80 180

8

100 280

(a)

(b)

V (6) = V (8) =

dD dt dD dt

V (10) =

610

 t 6

 t 8

dD dt

D t D t 

t 10



ft 180  100  70 sec 84



ft 280  180  115 sec 10  6

t 6

t 8

D t

 t 10

ft 280  140 sec 10  8

By linear interpolation between t = 8 sec and t = 10 sec, V(8.7) = 123.75

(c)

d2 D a (2) = 2 dt

2 D  t 2 t 2

 t 2

ft 30  15 2 sec 2

By linear interpolation between t = 0 sec and t = 2 sec, a(1) = 7.5

(d)

a (6) =

d2 D dt 2

 t 6

2 D t 2

 t 8

 D t 2

2

a (8) =

d D dt 2



ft 80  40 2 sec 2



ft 100  50 2 sec 2

t 6

2

t 8

ft sec 2

By linear interpolation between t = 6 sec and t = 8 sec, a(6.3) = 41.5

ft sec 2

138

ft sec

10 0

20

330

10

-10

Problem 7-13.

t 0

D

D

2D

3D

4D

5D

0 90

1

90

-20 70

2

-10

160

-30 40

3

10

200

-20 20

4

20 -25 -5 5

220

-15 5

5 (a)

(b)

dD dt

V (0) =

dD dt

V (3) =

dD dt

V (4) =

225

 t 0

 t 3

 t 4

D t D t D t



ft 90  90 sec 1 0



ft 40  20  30 sec 42



ft 20  5  12.5 sec 5 3

t 0

t 3

t 4

By linear interpolation between t = 3 sec and t = 4 sec, V(3.5) = 21.25

(c)

d2D a (3) = 2 dt

2 D  t 2 t 3

2

a (4) =

d D dt 2

 t 4

 D t 2



ft 20  20 1 sec 2



ft 15  15 1 sec 2

t 3

2

t 4

ft sec

By linear interpolation between t = 3 sec and t = 4 sec, a(3.5) = - 17.5

ft sec 2

Problem 7-14. By the method of undetermined coefficients, the interpolating polynomial and its first derivative are: Vc = -4.262 + 0.355T - 0.008T2 + 9.896x10-5T3 -3.255x10-7T4

Vc' = 0.355 - 0.016T + 2.969x10-4T2 - 1.302x10-6T3 Estimations of the first derivative at T = 540 and T = 640 : (a) (b)

grain ft 3 . F o grain Vc' (64) = 0.206 3 o ft . F

Vc' (54) = 0.152

Problem 7-15. Solution not provided.

139

Problem 7-16. Solution not provided. Problem 7-17. By the method of undetermined coefficients, the interpolating polynomial and its first derivative are: f(x) = 1100 - 2540x + 2011x2 - 643x3 + 77x4 f ' (x) = -2540 + 4022x - 1929x2 + 308x3 Estimations of the first derivative at x = 3 and x = 5.5: (a) f ' (3) = 481 (b) f ' (5.5) = 12472.25 The following table compares the estimates with the true values:

x

f '(x ) f '(x )True Error(%) 3 481 533 9.8 5.5 12472.3 14138 11.8

Problem 7-18. By the method of undetermined coefficients, the interpolating polynomial and its first derivative are: sin() = -0.00049 + 0.01780 -0.000092 + 0.000013 - 4.16667x10-74 ( in degrees)

dsin() 1  cos() = x10 7  3 0.01780  0.00018  0.00003 2  166667 . d 0.017453





( in degrees) Estimations of the first derivative at  = 70 and  = 5.50: (a)

(b)

dsin() d

 7

dsin() d

 5.5

 0.99916  0.99927

The following table compares the estimates with the true values:

deg)

f '() f '()True Error(%) 7 0.99916 0.99255 0.7 5.5 0.99927 0.99540 0.4

Problem 7-19. Solution not provided. Problem 7-20. By the method of undetermined coefficients, the interpolating polynomial and its first and second derivatives are: f(x) = -0.83186 + 1.12027x - 0.22618x2 + 0.026113x3 - 0.0012313x4 f ' ( x ) = 1.12027 - 0.45236x + 0.078339x2 - 0.0049252x3

f ' ' ( x ) = -0.45236 + 0.15668x - 0.014776x2 Estimations of the first and second derivatives at x = 4 and x = 1.7: (a)

f ' (4) = 0.24904 f ' ' (4) = -0.06206

140

(b)

f ' (1.7) = 0.55346 f ' ' (1.7) = -0.22871

The following table compares the estimates for part (a) with the true values:

TRUE %Error

f '(x ) f ''(x) 0.24904 -0.06206 0.25000 -0.06250 0.384 0.704

Problem 7-21. By the method of undetermined coefficients, the interpolating polynomial and its first and second derivatives are:

f ( x ) = 3.48150 - 5.84027 x  5.32299 x 2  2.54443x 3  0.64401x 4  0.063802 x 5  10.64598 x  7.63329 x 2  2.57604 x 3  0.31901x 4 f ' ( x ) = 584027 . f ' ' ( x ) = 10.64598  15.26658 x  7.72812 x 2  127604 . x3 Estimations of the first and second derivatives at x = 1.4:

f ' (1.4) = -0.05400 f ' ' (1.4) = 0.91843

Problem 7-22. Solution not provided. Problem 7-23. Fitting an interpolating polynomial to the data using the method of undetermined coefficients results in:

V  3.497 x10 -7  0.2863 y  36.631y 2  12715.6 y 3  1192074.5 y 4 dV  0.2863  73.26 y  38146.8 y 2  4768298 y 3 dy For y = 0.25Y = 0.25(0.016) = 0.004 ft,

dV dy

 0.274162 y 0.004



1 sec

dV lb  10 3 (0.274162)  0.002742 2 dy ft

The following table summarizes the results for the distribution of the shear stress across the space between the plates:

y (ft) 0 0.0040 0.0080 0.0120 0.0160

(lb/ft2)  0.000274162 0.000872353 0.003911900 0.011223828

Problem 7-24. Fitting an interpolating polynomial to the data using the method of undetermined coefficients results in: The angular velocity at the outer surface is given by:

141

dV dr

 765  18250(0.2)  225000(0.2) 2  625000(0.2) 3  1115 r  0. 2

rad sec

For r = 0.08 inches,

dV dr

 425 r  0.08



1 sec

dV lb  10 4 (425)  0.0425 2 dr ft

The following table summarizes the results:

r (in) 0 0.0400 0.0800 0.1200 0.1600 0.2000

(lb/ft2)  0.03550 0.04250 0.07350 0.10450 0.11150

Problem 7-25. Solution not provided.

7.2. Numerical Integration Problem 7-26.

The data table is re-produced here to show  in radians.

(rad) sin

0 0

True value = 1.0,

0.2618 0.5236 0.7854 1.0472 0.2588 0.50000 0.70701 0.8660 1  0.99424 % error = x100 = 0.576% 1

1.3090 0.9659

1.5708 1.0000

Problem 7-27.

F (t ) 

(1) 0.33  0.24  0.285 , and R(t) = 1-F(t) = 1 - 0.285 = 0.715. 2

(a)

For t = 1,

t f(t) F(t) R(t) (b)

Other values can be computed similarly. The following table can be constructed: 0 1 2 3 4 5 6 0.33 0.24 0.17 0.12 0.09 0.06 0.04 0 0.285 0.49 0.635 0.74 0.815 0.865 1 0.715 0.51 0.365 0.26 0.185 0.135 For t = 5 days, R(5) = 1-F(5) = 0.185.

Problem 7-28. 1.00 = radius i Xi Yi 1 .000000 1.000000

142

7 0.03 0.9 0.1

8 0.02 0.925 0.075

9 0.01 0.94 0.06

2 3 4 5 6 7

.166667 .986013 .333333 .942809 .500000 .866025 .666667 .745356 .833333 .552771 1.000000 .000000 .7654958 = numerical estimate .7853982 = true value

Problem 7-29. 3.00 = A 2.00 = B i Xi Yi 1 .000000 2.000000 2 .300000 1.989975 3 .600000 1.959592 4 .900000 1.907878 5 1.200000 1.833030 6 1.500000 1.732051 7 1.800000 1.600000 8 2.100000 1.428286 9 2.400000 1.200000 10 2.700000 .871780 11 3.000000 .000000 4.6567780 = numerical estimate 4.7123890 = true value Problem 7-30.

F (t ) 

2 0.10  0.23  0.0.33 , and R(t) = 1-F(t) = 1 - 0.33 = 0.67. 2

(a)

For t = 1,

t f(t) F(t) R(t) (b)

Other values can be computed similarly. The following table can be constructed: 0 2 4 6 8 10 12 0.02 0.05 0.06 0.07 0.08 0.07 0.06 0 0.07 0.18 0.31 0.46 0.61 0.74 1 0.93 0.82 0.69 0.54 0.39 0.26 The probability that the system will survive at least 2 hours but no more than 6 hours = 1 - (F(6) - F(2) = 1 - (0.31 - 0.07) = 1 - 0.24 = 0.76

Problem 7-31. Interval width = 1

x f (x ) 3

 f ( x)dx  0

0 0

1 1

2 1.2599

(1) . )  14422 . 0  2(1)  2(12599   2.9810 2

Interval width = 0.5

143

3 1.4422

14 0.05 0.85 0.15

16 0.02 0.92 0.08

x f (x )

3

 0

0 0.5 1.0 1.5 2.0 2.5 3.0 0 0.7937 1 1.1447 1.2599 1.3572 1.4422 . )  2(12599 . )  2(13572 . ) (0.5) 0  2(0.7937)  2(1)  2(11447 f ( x )dx  .   31383 . 2 14422 

Interval width = 0.25

x

3

 0

f (x ) 0 0.0000 0.25 0.6300 0.5 0.7937 0.75 0.9086 1 1.0000 1.25 1.0772 1.5 1.1447 1.75 1.2051 2 1.2599 2.25 1.3104 2.5 1.3572 2.75 1.4010 3 1.4422 . ) 0  2(0.63)  2(0.7937)  2(0.9086)  2(1)  2(10772 (0.25)    3.2022 f ( x )dx  2(11447 . )  2(12051 . )  2(12599 . )  2(13104 . )   2 2(13572  . )  2(14010 . )  14422 .

The following tables compares the estimates with the true value:

Interval Width 1 0.5 0.25

Integral %error True Value 2.981 8.1 3.2451 3.1383 3.3 3.2022 1.3

Problem 2-32. 200

Area under the curve, A =

 f ( x)dx 0

Area of channel,

A' = (22)(200) - A

(25) 22  2(15)  2(11)  2(0)  2(3)  2(8)  2(11)  2(16)  22  2150 ft 2 2 A'= (22)(200) - 2150 = 2250 ft 2 A

Q =  VdA = V  dA = V . A' = (6).(2250)  13500

Problem 7-33. Solution not provided.

144

ft 3 sec

Problem 7-34. Solution not provided. Problem 7-35. xi +1

Area under the curve at a specific location (xi,xi+1), A =

 f ( x)dx xi

Area of channel at a specific location (xi,xi+1), A' = (22)(xi+1 - xi) - A 8 Q= Vj A'j j=1 For a specific location of (25,50) and V = 4 ft/sec,



A

(25) 15 + 11  325 ft 2 2

A' = (22 )(50  25)  325 = 225 ft 2 Q j = ( 4 )(225) = 900

ft 3 sec

The following table summarizes the results:

j 1 2 3 4 5 6 7 8  8

Q=

 Vj A'j  14737.5 j=1

A j(ft2) 462.5 325 137.5 37.5 137.5 237.5 337.5 475 2150

A 'j(ft2) Q j(ft3/sec) 87.5 175.0 225 900.0 412.5 2887.5 512.5 4100.0 412.5 3300.0 312.5 1875.0 212.5 1275.0 75 225.0 2250 14737.5

ft 3 sec

Problem 7-36. Solution not provided. Problem 7-37.

The data table is re-produced here to show  in radians.

(rad) sin

0 0

0.2618 0.5236 0.7854 1.0472 1.3090 1.5708 0.2588 0.50000 0.70701 0.8660 0.9659 1.0000 1.5708 0.2618 (0  4(0.2588)  0.5)  (0.5  4(0.7071)  0.866)  .   100000 0 sind  3 (0.866  4(0.9659)  1  Comparison with true value: %error =

1  100000 . x100 = 0% 1

145

Comparison with trapezoidal rule: %error =

1  0.99424 x100 = 0.576% 1

Problem 7-38. For t = 6 days, R(5) = 1-F(5) = 0.135.

Problem 7-39. Solution not provided. Problem 7-40. 200

Area under the curve, A =

 f ( x)dx 0

Area of channel,

A' = (22)(200) - A

A' = (22)(200) - 2083.33 = 2316.67 ft 2 Q =  VdA = V  dA = V . A' = (6).(2316.67)  13900

ft 3 sec

Problem 7-41. Solution not provided. Problem 7-42. Interval width = 1.5

x f (x )

0 0

1.5 1.1447

3 1.4422

Interval width = 0.5

x f (x )

3

 0

0 0.5 1.0 1.5 2.0 2.5 0 0.7937 1 1.1447 1.2599 1.3572 . )  12599 . ) (0.5) (0  4(0.7937)  1)  (1  4(11447 f ( x )dx  .    31907  4(13572 . . )  14422 . ) 3 (12599 

146

3.0 1.4422

Interval width = 0.25

x

3

 0

f (x ) 0 0.0000 0.25 0.6300 0.5 0.7937 0.75 0.9086 1 1.0000 1.25 1.0772 1.5 1.1447 1.75 1.2051 2 1.2599 2.25 1.3104 2.5 1.3572 2.75 1.4010 3 1.4422 (0  4(0.63)  0.7937)  (0.7937  4(0.9086)  1)  (0.25)    3.2235 f ( x )dx   (1  4(10772  4(12051  12599 . )  11447 . )  (11447 . . . )  3   (12599  4(13104  4(14010 . . )  13572 . )  (13572 . . )  14422 . ) 

The following tables compares the estimates with the true value:

Interval Width 1.5 0.5 0.25

Integral %error True Value 3.0105 7.2 3.2451 3.1907 1.7 3.2235 0.7

Problem 7-43. Number of intervals = 2

x f (x ) 2

 0

dx

6  x 

3 2 2



0 1 2 0.06804 0.08944 0.35355

(1) (0.06804  4(0.08944)  0.35355)  0.25970 3

Number of intervals = 4

x f (x )

2

 0

dx

6  x 

3 2 2

0 0.5 1.0 1.5 2.0 0.06804 0.07253 0.08944 0.13771 0.35355 (0.5) (0.06804  4(0.07253)  0.08944)    0.24024 . )  0.35355)  3  (0.08944  4(013771

Number of intervals = 6

x f (x )

0 0.33333 0.66667 1.00000 1.33333 1.66667 2.00000 0.06804 0.06998 0.07637 0.08944 0.11526 0.17289 0.35355

147

2

 0

dx

6  x 

3 2 2

 1    3  3

(0.06804  4(0.06998)  0.07637)   (0.07637  4(0.08944)  011526 . )   0.23712   (011526 . . )  0.35355)   4(017289

Number of intervals = 8

x

2

 0

dx



6  x2



3 2

f (x ) 0 0.06804 0.25 0.06912 0.5 0.07253 0.75 0.07887 1 0.08944 1.25 0.10698 1.5 0.13771 1.75 0.19862 2 0.35355 (0.06804  4(0.06912)  0.07253)    (0.25)  (0.07253  4(0.07887)  0.08944)    0.23628 . )  013771 . ) 3  (0.08944  4( 010698   . . )  0.35355)   4(019862  (013771

Number of intervals = 10

x

2

 0

dx

6  x  2

3 2

f (x ) 0 0.06804 0.2 0.06873 0.4 0.07086 0.6 0.07466 0.8 0.08058 1 0.08944 1.2 0.1027 1.4 0.12315 1.6 0.15673 1.8 0.21809 2 0.35355 (0.06804  4( 0.06873)  0.07086)   (0.07086  4(0.07466)  0.08058)   (0.2)   (0.08058  4(0.08944)  01027  . )   0.23597 3    4(012315 . . )  015673 . )   (01027  (015673  4(0.21809)  0.35355  .

Problem 7-44. Number of intervals = 4

x f (x )

1 1.75 2.5 3.25 4 0.11853 0.06892 0.03654 0.01832 0.00891

148

Number of intervals = 8

x

4

dx

  3  2e  x

1

f (x ) 1 0.11853 1.375 0.09166 1.75 0.06892 2.125 0.05064 2.5 0.03654 2.875 0.02601 3.25 0.01832 3.625 0.01281 4 0.00891  4(0.09166)  0.06892)  . (011853  (0.06892  4(0.05064)  0.03654)  (0.375)    013744  . 3  (0.03654  4(0.02601)  0.01832)     (0.01832  4(0.01281)  0.00891) 

Number of intervals = 10

x 1 1.3 1.6 1.9 2.2 2.5 2.8 3.1 3.4 3.7 4

4

dx

  3  2e  x

1

f (x ) 0.11853 0.09672 0.07748 0.06108 0.04751 0.03654 0.02786 0.0211 0.01589 0.01192 0.00891

 4(0.09672)  0.07748)  . (011853  (0.07748  4(0.06108)  0.04751)   (0.3)   (0.04751  4( 0.03654)  0.02786)   013744  . 3    (0.02786  4(0.0211)  0.01589)   (0.01589  4(0.01192)  0.00891) 

Problem 7-45. Solution not provided. Problem 7-46. Solution not provided.

149

CHAPTER 8. Differential Equations 8.1 Introduction Problem 8-1. dy  2.5 x  y (a) dx dy (b)  2 x 2  3x  7 dx d2y  2 x 2  3y3 (c) 2 dx

y=2 at x=1 y=3 at x=0

y=1 and

dy  0 at x=1 dx

Problem 8-2. Solution not provided.

8.2 Taylor Series Expansion Problem 8-3. dy  5x 4  12 x 2  6 dx

such that y = 3 at x = 1

x0 = 1

g ( x 0 )  y (1)  g (1)  3 dy g' ( x)   5x 4  12 x 2  6 dx d2 y g ''( x )  2  20 x 3  24 x dx d3y g '''( x )  3  60 x 2  24 dx d4 y g ' ' ' ' ( x )  4  120 x dx d5 y g '''''( x )  5  120 dx d6 y g' ' ' ' ' ' ( x )  6  0 dx



g'(1)  1



g' ' (1)  4



g' ' ' (1)  36



g''''(1)  120



g' ' ' ' ' (1)  120



g' ' ' ' ' ' (1)  0

Using Eq. 8-9 of the textbook, the following six-terms Tayler series expansion can be obtained:

g ( x )  3  ( x  1)( 1) 

( x  1) 2 ( x  1) 3 ( x  1) 4 ( x  1) 5 ( 4)  (36)  (120)  (120) 2! 3! 4! 5!

For x = 1.05

150

(105 .  1) 2 (105 .  1) 3 (105 .  1) 4 (105 .  1) 5 g(105 . )  3  (105 .  1)  ( 4)  (36)  (120)  (120) 2 6 24 120  2.9458 The following table summarizes the estimated values of the function y(x) = g(x) for the range of x = 1.0 to 1.3 in an increment of 0.05. The table also compares the computed values with the true values. Practically, there is no error generated in the solution because all the derivatives were used in the solution and the derivatives beyond the fifth one equal zero. x Estimated y(x) True y(x) error(%) 1

3

3

0

1.05

2.945781563

2.945781563

0

1.1

2.88651

2.88651

-3.077E-14

1.15

2.827857188

2.827857188

-1.57E-14

1.2

2.77632

2.77632

1.5996E-14

1.25

2.739257813

2.739257813

-1.621E-14

1.3

2.72493

2.72493

-1.63E-14

Problem 8-4. dy  1 y dx

such that y = 2 at x = 0

x0 = 0, y0 = 2

g ( x 0 , y 0 )  y ( 0,2)  g (0,2)  2 dy  g' ( x, y)   1 y dx d2 y dy g ''( x , y )  2    dx dx d3y d2 y g' ' ' ( x, y)  3   2  dx dx

g'(0,2)  1 g''(0,2)  ( 1)  1 g' ' ' (0,2)  1

Using Eq. 8-12 of the textbook, the following four-terms Tayler Series expansion can be obtained:

g ( x )  2  ( x  0)( 1) 

( x  0) 2 ( x  0) 3 (1)  ( 1) 2! 3!

For x = 0.1

g(01 . )  2  (01 . )( 1) 

(01 . )2 (01 . )3 (1)  ( 1)  1904833 . 2 6

The following table summarizes the estimated values of the function y(x) = g(x) for the range of x = 0.0 to 0.5 in an increment of 0.1. It also compares the computed values with the true values. x Estimated y(x) True y(x) error(%) 0

2

2

0.00000

0.1

1.904833333

1.904837418

0.00021

0.2

1.818666667

1.818730753

0.00352

0.3

1.7405

1.740818221

0.01828

0.4

1.669333333

1.670320046

0.05907

0.5

1.604166667

1.60653066

0.14715

151

Problem 8-5. dy  3x 2 y dx

such that y = 1 at x = 0

x0 = 0, y0 = 1

g ( x 0 , y 0 )  y ( 0,1)  g ( 0,1)  1 dy g' ( x, y)   3x 2 y dx d2 y dy g ' ' ( x , y )  2  6 xy  3x 2 dx dx 3 dy d2 y d y g '''( x , y )  3  6 y  12 x  3x 2 2 dx dx dx



g'(0,1)  3(0) 2 (1)  0



g' ' (0,1)  6(0)(1)  3(0) 2 (0)  0



g'''(0,1)  6(1)  12(0)(0)  3(0) 2 (0)  6

Using Eq. 8-12 of the textbook, the following four-terms Tayler series expansion can be obtained:

g ( x )  1  ( x  0)(0) 

( x  0) 2 ( x  0) 3 (0)  (6)  1  x 3 2! 3!

For x = 0.1

g(01 . )  1  (01 . ) 3  1001 . The following table summarizes the estimated values of the function y(x) = g(x) for the range of x = 0.0 to 0.6 in an increment of 0.1. The table also compares the computed values with the true values. x Estimated y(x) True y(x) error(%) 0

1

1

0.00000

0.1

1.001000

1.001001

0.00005

0.2

1.008000

1.008032

0.00318

0.3

1.027000

1.027368

0.03580

0.4

1.064000

1.066092

0.19627

0.5

1.125000

1.133148

0.71910

0.6

1.216000

1.241102

2.02259

The six order Taylor series can be shown to be

g( x )  1  x 3  0.5 x 6

Problem 8-6.

d2 y  2x dx 2

such that y = 0 and

dy  0 at x = 0 dx

x0 = 0

g ( x 0 )  y ( 0)  g ( 0)  0 dy g' ( x)  0 dx d2 y g' ' ( x)  2  2 x dx d3y g '''( x )  3  2 dx



g'( 0)  0



g''(0)  2(0)  0



g' ' ' (0)  2

Using Eq. 8-9 of the textbook, the following four-terms Tayler series expansion can be obtained:

g ( x )  0  ( x  0)(0) 

( x  0) 2 ( x  0) 3 1 (0)  ( 2)  x 3 3! 3 2!

For x = 0.1

152

g(01 .)

1 (01 . ) 3  0.000333 3

The following table summarizes the estimated values of the function y(x) = g(x) for the range of x = 0.0 to 0.5 in an increment of 0.1. The table also compares the computed values with the true values. There is no error generated in the solution because all the derivatives were used in the solution and the derivatives beyond the third one equal zero. Also note that g(x) = x3/3 is identical to the true solution. x Estimated y(x) True y(x) error(%) 0

0

0.000000

0.00000

0.1

0.000333

0.000333

0.00000

0.2

0.002667

0.002667

0.00000

0.3

0.009000

0.009000

0.00000

0.4

0.021333

0.021333

0.00000

0.5

0.041667

0.041667

0.00000

Problem 8-7. dy  e x  xe x  1 dx

such that y = 0 at x = 0

x0 = 0

g ( x 0 )  y ( 0)  g ( 0)  0 dy g' ( x)   e x  xe x  1 dx d2 y g ' ' ( x )  2  2e x  xe x dx d3y g '''( x )  3  3e x  xe x dx d4 y g ' ' ' ' ( x )  4  4e x  xe x dx d5 y g '''''( x )  5  5e x  xe x dx



g'(0)  2



g''(0)  2



g' ' ' (0)  3



g''''(0)  4



g'''''(0)  5

Using Eq. 8-9 of the textbook, the following six-terms Tayler series expansion can be obtained:

g ( x )  0  ( x  0)(2) 

( x  0) 2 ( x  0) 3 ( x  0) 4 ( x  0) 5 ( 2)  (3)  (4)  (5) 2! 3! 4! 5!

For x = 0.2

g(0.2)  0  (0.2)(2) 

(0.2) 2 (0.2) 3 (0.2) 4 (0.2) 5 ( 2)  (3)  ( 4)  (5)  0.444280 2 6 24 120

The following table summarizes the estimated values of the function y(x) = g(x) for the range of x = 0 to 1 in an increment of 0.2. The table also compares the computed values with the true values. x Estimated y(x) True y(x) error(%) 0

0

0

0

0.2

0.444280

0.444281

0.00012

0.4

0.996693

0.996730

0.00367

0.6

1.692840

1.693271

0.02547

0.8

2.577920

2.580433

0.09738

1

3.708333

3.718282

0.26756

153

Problem 8-8. Solution not provided. Problem 8-9.

x=

0.1 = h

x

y-Taylor

y-exact

error(%)

0

3

3

0

0.1

2.996739

2.996665

0.002473

0.2

2.98723

2.986637

0.019872

0.3

2.971856

2.969848

0.067597

0.4

2.950959

2.946184

0.162062

0.5

2.924846

2.915476

0.321379

0.6

2.893792

2.877499

0.566224

8.3 Euler’s Method Problem 8-10. x= h

0.2

154

x

y'

y-Euler

y-exact

error(%)

y(0.0)=2

0

-1

2

2

0.00

y(0.2)=y(0.0)+h*y'(0.0)

0.2

-0.8

1.8

1.818731

-1.03

y(0.4)=y(0.2)+h*y'(0.2)

0.4

-0.64

1.64

1.67032

-1.82

y(0.6)=y(0.4)+h*y'(0.4)

0.6

-0.512

1.512

1.548812

-2.38

y(0.8)=y(0.6)+h*y'(0.6)

0.8

-0.4096

1.4096

1.449329

-2.74

y(1.0)=y(0.8)+h*y'(0.8)

1

-0.32768

1.32768

1.367879

-2.94

Problem 8-11. y'+y-1=0,

y'=1-y,

x= h

0 < x < 1.0 0.1

x

y'

y-Euler

y-exact

error(%)

0

-1

2

2

0.00 y(

0 )=

2

0.1

-0.9

1.9

1.904837

-0.25 y(

0.1 )=y(

0 )+h*y'(

0)

0.2

-0.8

1.81

1.818731

-0.48 y(

0.2 )=y(

0.1 )+h*y'(

0.1 )

0.3

-0.7

1.729

1.740818

-0.68 y(

0.3 )=y(

0.2 )+h*y'(

0.2 )

0.4

-0.7

1.6561

1.67032

-0.85 y(

0.4 )=y(

0.3 )+h*y'(

0.3 )

0.5

-0.6

1.59049

1.606531

-1.00 y(

0.5 )=y(

0.4 )+h*y'(

0.4 )

0.6

-0.5

1.531441

1.548812

-1.12 y(

0.6 )=y(

0.5 )+h*y'(

0.5 )

0.7

-0.5

1.478297

1.496585

-1.22 y(

0.7 )=y(

0.6 )+h*y'(

0.6 )

0.8

-0.4

1.430467

1.449329

-1.30 y(

0.8 )=y(

0.7 )+h*y'(

0.7 )

0.9

-0.4

1.38742

1.40657

-1.36 y(

0.9 )=y(

0.8 )+h*y'(

0.8 )

1

-0.3

1.348678

1.367879

-1.40 y(

1 )=y(

0.9 )+h*y'(

0.9 )

Problem 8-12. y'-5x+2=0, x=

x

y'=5x-2, 0.1 = h

y'

y=2.5x^2-2x+0.5 1