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Mathematical Methods for
Numerical Analysis and Optimization
(BCA Part-II)
Varsha Gupta
M.Sc. (Maths)
Lecturer
Deptt. of Information Technology
Biyani Girls College, Jaipur
Hastlipi, Jaipur
Preface am glad to present this book, especially designed to serve the needs of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concept of the topic. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach. I have made a meaningful effort to summarize the complete syllabus. This includes various methods which are explained in simplest way. Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the reader for which the author shall be obliged. I acknowledge special thanks to Mr. Rajeev Biyani, Chiarman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who is the backbone and main concept provider and also have been constant source of motivation throughout this endeavour. We also extend our thanks to M/s. Hastlipi, Omprakash Agarwal/Sunil Kumar Jain, Jaipur, who played an active role in co- ordinating the various stages of this endeavour and spearheaded the publishing work. I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and the students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address. Author
I
Syllabus
B.C.A. Part-II
Mathematical Methods for Numerical Analysis and Optimization Computer arithmetics and errors. Algorithms and programming for numerical solutions. The impact of parallel computer : introduction to parallel architectures. Basic algorithms Iterative solutions of nonlinear equations : bisection method, Newton- Raphson method, the Secant method, the method of successive approximation. Solutions of simultaneous algebraic equations, the Gauss elimination method. Gauss- Seidel Method, Polynomial interpolation and other interpolation functions, spline interpolation system of linear equations, partial pivoting, matrix factorization methods. Numerical calculus : numerical differentiating, interpolatory quadrature. Gaussian integration. Numerical solutions of differential equations. Euler's method. Runge-Kutta method. Multistep method. Boundary value problems : shooting method.
Content S.No. Name of Topic Page No.
- Computer Arithmetic and Errors 9- 11
- Bisection Method 12- 17
- Regula Falsi Method 18- 24
- Secant Method 25- 28
- Newton – Raphson Method 29- 33
- Iterative Method 34- 39
- Gauss Elimination Method 40- 43
- Gauss – Jordan Elimination Method 44- 45
- Matrix Inversion Method 46- 50
- Matrix Factorization Method 51- 59
- Jacobi Method 60- 63
- Gauss – Seidel Method 64- 68
- Forward Difference 69
- Backward Difference 70- 71
- Newton – Gregory Formula for Forward Interpolation 72- 75 S.No. Name of Topic Page No.
- Newton’s Formula for Backward Interpolation 76- 77
- Divided Difference Interpolation 78- 81
- Lagrange’s Interpolation 82- 83
- Spline Interpolation 84
- Quadratic Splines 85- 86
- Cubic Splines 87
- Numerical Differentiation 88- 90
- Numerical Integration 91- 93
- Euler’s Method 94- 95
- Euler’s Modified Method 96- 99
Case-I : if - 0.02x < ⇒ - 0.02x < x – 35. ⇒ 35.25 < x + 0.02x ⇒ 35.25 < x (1 + 0.02) ⇒ 35.25 < x (1.02) ⇒ 35.25 < 1.02x ⇒ < x ⇒ x > 34.5588 _ _ _ (1) Case-II : if < 0. ⇒ x – 35.25 < 0.02x ⇒ x – 0.02x < 35. ⇒ 0.98x < 35. ⇒ x <
⇒ x < 35.9693 _ _ _ (2) From equation (1) and (2) we have 34.5588 < x < 35. The required range is (34.5588, 35.9693)
x – 35. x x – 35. x x – 35. x
x – 35. x Chapter- 2 Bisection Method Q.1. Find real root of the equation x 3 - 5x + 3 upto three decimal digits. Ans.: Here ƒ(x) = x 3 – 5x + 3 ƒ(0) = 0 – 0 + 3 = 3 ƒ(x 0 ) (say) ƒ(1) = 1 – 5 + 3 = – 1 = ƒ(x 1 ) (say) Since ƒ(x 0 ), ƒ(x 1 ) < 0 so the root of the given equation lies between 0 and 1 So, x 2 = = 0+ 2 = 0. Now, ƒ(x 2 ) = ƒ(0.5)
= 0.625 (which is positive) ƒ(x 1 ).ƒ(x 2 ) < 0 So, x 3 = = = 0. Now, ƒ(x 3 ) = ƒ(0.75) = (0.75) 3 – 5 (0.75) + 3 = 0.4218 – 3.75 + 3 = – 0.328 (which is negative) ƒ(x 2 ).ƒ(x 3 ) < 0 So, x 4 = = 0.5 0. 2 = 0. Now, ƒ(x 4 ) = ƒ(0.625) = (0.625) 3 – 5 (0.625) + 3 = 0.244 – 3.125 + 3 0 1 + 2 x x ∵ 1 2 + 2 x x 1 + 0. 2 ∵ 2 3 + 2 x x = 0.119 (which is positive) ƒ(x 3 ).ƒ(x 4 ) < 0 So, x 5 = = = 0. Now, ƒ(x 5 ) = ƒ(0.687) = (0.687) 3 – 5 (0.687) + 3 = – 0.1108 (which is negative) ƒ(x 4 ).ƒ(x 5 ) < 0 So, x 6 = = = 0. Now, ƒ(x 6 ) = ƒ(0.656) = (0.656) 3 – 5 (0.656) + 3 = 0.0023 (which is positive) ƒ(x 5 ).ƒ(x 6 ) < 0 So, x 7 = = = 0. Now, ƒ(x 7 ) = ƒ(0.671) = (0.671) 3 – 5 (0.671) + 3
So, x 11 = 6 10 x +x 2 = = 0. Now, ƒ(x 11 ) = ƒ(0.656) = (0.656) 3 – 5 (0.656) + 3 = 0.2823 – 3.28 + 3 = 0.00230 (which is positive) ƒ(x 11 ).ƒ(x 10 ) < 0 So, x 12 = = = 0. Since x 11 and x 12 both same value. Therefore if we continue this process we will get same value of x so the value of x is 0.565 which is required result. Q.2. Find real root of the equation cos x - xex = 0 correct upto four decimal places. Ans.: Since, ƒ(x) = cosx - xex So, ƒ(0) = cos0 – 0e 0 = 1 (which is positive) And ƒ(1) = cos1 – 1e 1 = - 2.1779 (which is negative) ƒ(0).ƒ(1) < 0 Hence the root of are given equation lies between 0 and 1. let ƒ(0) = ƒ(x 0 ) and ƒ(1) = ƒ(x 1 ) So, x 2 = = = 0. Now, ƒ(x 2 ) = ƒ(0.5) ƒ(0.5) = cos(0.5) – (0.5)e 0. = 0.05322 (which is positive) ƒ(x 1 ).ƒ(x 2 ) < 0 ∵ 6 9 + 2 x x 0.656 + 0. 2 ∵ 0.656 + 0. 2 ∵ 10 11 + 2 x x 0.657 + 0. 2 ∵ 0 1 + 2 x x 0 + 1 2 ∵ So, x 3 = = = = 0. Now, ƒ(x 3 ) = ƒ(0.75) = cos(0.75) – (0.75)e 0.
= – 0.856 (which is negative) ƒ(x 2 ).ƒ(x 3 ) < 0 So, x 4 = = = 0. ƒ(x 4 ) = ƒ(0.625) = cos(0.625) – (0.625)e (0.625) = – 0.356 (which is negative) ƒ(x 2 ).ƒ(x 4 ) < 0 So, x 5 = = = 0. Now, ƒ(x 3 ) = ƒ(0.5625) = cos(0.5625) – 0.5625e 0. = – 0.14129 (which is negative) ƒ(x 2 ).ƒ(x 5 ) < 0 So, x 6 = = = 0. Now, ƒ(x 6 ) = ƒ(0.5312) = cos(0.5312) – (0.5312)e 0. = – 0.0415 (which is negative) ƒ(x 2 ).ƒ(x 6 ) < 0 So, x 7 = = = 0. Now, ƒ(x 7 ) = ƒ(0.5156) = cos(0.5156) – (0.5156)e 0. = 0.006551 (which is positive) ƒ(x 6 ).ƒ(x 7 ) < 0 So, x 8 = = = 0. 1 2 + 2 x x 1 + 0. 2
2 ∵ 2 3 + 2 x x 0.5 + 0. 2 ∵ 2 4 + 2 x x 0.5 + 0. 2 ∵ 2 5 + 2 x x 0.5 + 0. 2 ∵ 2 6 +
x x 0.5195 + 0. 2 ∵ 10 11 + 2 x x 0.5175 + 0. 2 Chapter- 3 Regula Falsi Method Q.1. Find the real root of the equation x log 10 x – 1.2 = 0 correct upto four decimal places. Ans.: Given ƒ(x) = x log 10 x – 1.2 _ _ _ (1) In this method following formula is used - xn+1 = xn – _ _ _ (2) Taking x = 1 in eq.(1) ƒ(1) = 1. log 101 – 1. = – 2 (which is negative) Taking x = 2 in eq.(1) ƒ(2) = 2. log 10 2 – 1. = – 0.5979 (which is negative) Taking x = 3 in eq.(1) ƒ(3) = 3. log 10 3 – 1. = 0.2313 (which is positive) ƒ(2).ƒ(3) < 0 So the root of the given equation lies between 2 and 3. let x 1 = 2 and x 2 = 3 ƒ(x 1 ) = ƒ(2) = – 0. And ƒ(x 2 ) = ƒ(3) = 0. Now we want to find x 3 so using eq.(2) x 3 = x 2 – = 3 – 1 1 ( - ) ( ) ( ( ) ( )) n n n n n f f f x x x x x
2 1 2 2 1 ( - ) ( ) ( ) ( ) f f f x x x x x (3- 2) (0.2313) 0.2313 ( 0.5979) = 3 – = 3 – 0.2789 = 2. ƒ(x 3 ) = ƒ(2.7211) = 2.7211 log 10 2.7211 – 1. = – 0.01701 (which is negative) ƒ(x 2 ).ƒ(x 3 ) < 0 Now to find x 4 using equation (2) x 4 = x 3 – = 2.7211 – (2.7211 - 3)×(-0.0170) (-0.0170 - 0.2313) = 2.7211 – = 2.7211 + 0.01910 = 2. Now ƒ(x 4 ) = ƒ(2.7402) = 2.7402 log 10 2.7402 – 1. = – 0.0003890 (which is negative) ƒ(x 2 ).ƒ(x 4 ) < 0 Now to find x 5 using equation (2) x 5 = x 4 – = 2.7402 – (2.7402 3) ( 0.0004762) ( 0.0004762 0.2313) × = 2.7402 + ( 0.2598)( 0.0004762)
= 2.7402 +
Since ƒ(2).ƒ(3) < 0 So the root of the given equation lies between 2 and 3. Let x 1 = 2 and x 2 = 3 ƒ(x 1 ) = ƒ(2) = – 1 and ƒ(x 2 ) = ƒ(3) = 16 ∵ 5 2 5 5 2 ( - ) ( ) ( ) ( ) f f f x x x x x (2.7406 - 3) ( 0.000040) ( 0.00004) (0.2313) 1 1 ( - ) ( ) [ ( ) ( )] n n n n n f f f x x x x x Now to find x 3 using equation (2)
x 3 = x 2 – f
f f 2 1 2 2 1 (x - x ) (x ) (x ) - (x ) = 3 – = 3 – = 2. ƒ(x 3 ) = (2.0558) 3 – 2 (2.0588) – 5 = 8.7265 – 4.1176 – 5 = – 0.3911 (which is negative) ƒ(x 2 ).ƒ(x 3 ) < 0 Now to find x 4 using equation (2) x 4 = x 3 – = 2.0588 – × (-0.3911)
ƒ(x 4 ) = 9.0144 – 4.1624 – 5 = – 0.148 (which is negative) So ƒ(x 2 ). ƒ(x 4 ) < 0 Now using equation (2) to find x 5 x 5 = x 4 – = 2.0812 – = 2.0812 +
= 2.0812 + 8.4210 x
f f f 5 2 5 5 2 (x - x )× (x ) (x ) - (x ) 10 - 3 = 2. (3 - 2) × 16 16 + 1 16 17 3 2 3 3 2 ( - ) ( ) [ ( ) ( )] f f f x x x x x (2.0588 - 3)
- 0.3911 - 16 ( 0.9412) ( 0.3911)
4 2 4 4 2 ( - ) ( ) [ ( ) ( )] f
f f x x x x x (0.05025)
6 2 6 6 2 ( - ) ( ) ( ) ( ) f f f x x x x x (2.0927 - 3) × (-0.0207) (-0.0207 - 16) (-0.9073) (-0.0207)
7 2 7 7 2 ( - ) ( ) ( ) ( ) f f f x x x x x (2.0938 - 3) × (-0.0084) (-0.0084 - 16) = 2.0938 + = 2.0938 + 4.755 x 10 - 4 = 2. ƒ(x 8 ) = 9.1853 – 4.18854 – 5 = – 0.00324 (which is negative)
So ƒ(x 2 ).ƒ(x 8 ) < 0 Now using equation (2) to find x 9 x 9 = x 8 – = 2.09427 – = 2.09427 – = 2. The real root of the given equation is 2.094 which is correct upto three decimals.
8 2 8 8 2 ( - ) ( ) ( ) ( ) f f f x x x x x (2.09427 - 3) × (-0.00324) (-0.00324 - 16) (-0.90573) (-0.00324)
Chapter- 4 Secant Method Note : In this method following formula is used to find root –
xn+1 = xn –
1 1 ( ) ( ) ( ) ( ) n n n n n x x f x f x f x
Using equation (1) to find x 4 x 4 = x 3 – = 0.9523 – = 0.9523 – = 0.9523 + 0.005865 = 0. ƒ(x 4 ) = (0.9581) 3 – 5 (0.9581) 2 – 17 (0.9581) + 20 = 0.8794 – 4.5897 – 16.2877 + 20 = 0.0020 (which is positive) x 5 = x 4 – = 0.9581 – = 0. Hence the root of the given equation is 0.9581 which is correct upto four decimal. Q.2. Given that one of the root of the non-linear equation cos x – xex = 0 lies between 0.5 and 1.0 find the root correct upto three decimal places, by Secant Method. Ans.: Given equation is ƒ(x) = cos x – xex And x 1 = 0.5 and x 2 = 1. ƒ(x 1 ) = cos (0.5) – (0.5) e0. = 0.87758 – 0. = 0. Now ƒ(x 2 ) = cos (1) – (1) e 1 = 0.54030 – 2. = – 2. Now to calculate x 3 using equation (1) 3 2 3 3 2 ( - ) ( ) ( ) ( ) f f f x x x x x (0.9523 - 1) × 0. [0.1402 - (-1)] (-0.0477) (0.1402) (1.1402) 4 3 4 4 3 ( - ) ( ) ( ) ( ) f
f f x x x x x (0.9581 - 0.9523) × 0. (0.0020) - (0.1402) x 3 = x 2 – = 1 – (1 0.5) ( 2.1780) ( 2.1780 0.05321) × = 1 – = 1 – 0. = +0. ƒ(x 3 ) = ƒ(0.51192) = cos (0.51192) – (0.51192)e0. = 0.87150 – 0. = 0. Now for calculating x 4 using equation (1) x 4 = x 3 – = 0.51192 – = 0.51192 – = 0.51192 + = 0.51192 + 0. = 0. ƒ(x 4 ) = cos (0.51584) – (0.51584)e0. = 0.86987 – 0. = 0.005814 (which is positive) Now for calculating x 5 using equation (1) x 5 = x 4 – = 0.51584 – 2 1 2 2 1 ( - ) ( ) ( ) ( ) f f f
