Mathematical Methods for Numerical Analysis and Optimization

Biyani's Think Tank Concept based notes Mathematical Methods for Numerical Analysis and Optimization (MCA) Varsha Gupta Poonam Fatehpuria M.Sc. (Maths) Lecturer Deptt. of Information Technology Biyani Girls College, Jaipur

Published by : Think Tanks Biyani Group of Colleges Concept & Copyright : Biyani Shikshan Samiti Sector-, Vidhyadhar Nagar, Jaipur-0 0 (Rajasthan) Ph : 04-87, 859-95 Fa : 04-8007 E-mail : acad@biyanicolleges.org Website :www.gurukpo.com; www.biyanicolleges.org ISBN : 978-9-854-4-4 First Edition : 009 While every effort is taken to avoid errors or omissions in this Publication, any mistake or omission that may have crept in is not intentional. It may be taken note of that neither the publisher nor the author will be responsible for any damage or loss of any kind arising to anyone in any manner on account of such errors and omissions. Leaser Type Setted by : Biyani College Printing Department

Mathematical Methods for Numerical Analysis and Optimization Preface I 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 concepts of the topics. The book is selfeplanatory and adopts the Teach Yourself style. It is based on questionanswer pattern. The language of book is quite easy and understandable based on scientific approach. This book covers basic concepts related to the microbial understandings about diversity, structure, economic aspects, bacterial and viral reproduction etc. 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 readers for which the author shall be obliged. I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this Endeavour. They played an active role in coordinating 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 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

4 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 approimation. 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, matri 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.

Mathematical Methods for Numerical Analysis and Optimization 5 Content S.No. Name of Topic. Computer Arithmetic and Errors. Bisection Method. Regula Falsi Method 4. Secant Method 5. Newton Raphson Method 6. Iterative Method 7. Gauss Elimination Method 8. Gauss Jordan Elimination Method 9. Matri Inversion Method 0. Matri Factorization Method. Jacobi Method. Gauss Seidel Method. Forward Difference 4. Backward Difference 5. Newton Gregory Formula for Forward Interpolation

6 S.No. Name of Topic 6. Newton s Formula for Backward Interpolation 7. Divided Difference Interpolation 8. Lagrange s Interpolation 9. Spline Interpolation 0. Quadratic Splines. Cubic Splines. Numerical Differentiation. Numerical Integration 4. Euler s Method 5. Euler s Modified Method 6. Rungs Kutta Method 7. Shooting Method

Mathematical Methods for Numerical Analysis and Optimization 7 Chapter- Computer Arithmetic and Errors Q.. An approimate value of is given by = /7 =.4857 and its true value is =.4596. Find the absolute and relative errors. Ans.: True value of () =.4596 Approimate value of () =.4857 Absolute error is given by Ea = =.4596.4857 = 0.00645 Relative error is given by Er = =.4596.4857.4596 = 0.00645.4596 = 0.000405 Q.. Let = 0.0045859 find the absolute error if is truncated to three decimal digits.

8 Ans.: = 0.0045859 = 0.45859 0 - [in normalized floating point form] = 0.458 0 - [after truncating to three decimal places] Absolute error = = 0.45859 0-0.458 0 - = 0.00059 0 - = 0.00059 E = 0.59 E 5 Q.. Let the solution of a problem be a = 5.5 with relative error in the solution atmost % find the range of values upto 4 decimal digits, within which the eact value of the solution must lie. Ans.: We are given that the approimate solution of the problem is (a) = 5.5 and it has relative error upto % so 5.5 < 0.0 = -0.0 < 5.5 < 0.0 Case-I : if -0.0 < 5.5-0.0 < 5.5 5.5 < + 0.0 5.5 < ( + 0.0) 5.5 < (.0) 5.5 <.0 5.5.0 < > 4.5588 _ ()

Mathematical Methods for Numerical Analysis and Optimization 9 Case-II: if 5.5 < 0.0 5.5 < 0.0 0.0 < 5.5 0.98 < 5.5 < 5.5 0.98 < 5.969 _ () From equation () and () we have 4.5588 < < 5.969 The required range is (4.5588, 5.969)

0 Chapter- Bisection Method Q.. Find real root of the equation - 5 + upto three decimal digits. Ans.: Here ƒ() = 5 + ƒ(0) = 0 0 + = ƒ(0) (say) ƒ() = 5 + = = ƒ() (say) Since ƒ(0), ƒ() < 0 so the root of the given equation lies between 0 and 0 + 0 + So, = = = 0.5 Now, ƒ() = ƒ(0.5) = (0.5) 5 (0.5) + = 0.5.5 + = 0.65 (which is positive) ƒ().ƒ() < 0 + + 0.5 So, = = = 0.75 Now, ƒ() = ƒ(0.75) = (0.75) 5 (0.75) + = 0.48.75 + = 0.8 (which is negative) ƒ().ƒ() < 0

Mathematical Methods for Numerical Analysis and Optimization + 0.5 0.75 So, 4 = = = 0.65 Now, ƒ(4) = ƒ(0.65) = (0.65) 5 (0.65) + = 0.44.5 + = 0.9 (which is positive) ƒ().ƒ(4) < 0 + 4 0.75 + 0.65 So, 5 = = = 0.687 Now, ƒ(5) = ƒ(0.687) = (0.687) 5 (0.687) + = 0.08 (which is negative) ƒ(4).ƒ(5) < 0 4 + 5 0.65 + 0.687 So, 6 = = = 0.656 Now, ƒ(6) = ƒ(0.656) = (0.656) 5 (0.656) + = 0.00 (which is positive) ƒ(5).ƒ(6) < 0 5 + 6 0.687 + 0.656 So, 7 = = = 0.67 Now, ƒ(7) = ƒ(0.67) = (0.67) 5 (0.67) + = 0.058 (which is negative) ƒ(6).ƒ(7) < 0 6 + 7 0.656 +0.67 So, 8 = = = 0.66 Now, ƒ(8) = ƒ(0.66)

ƒ(6).ƒ(8) < 0 = (0.66) 5 (0.66) + = 0.90.5 + = 0.0 (which is negative) 6 + 8 0.656 + 0.66 So, 9 = = = 0.659 Now, ƒ(9) = ƒ(0.659) ƒ(6).ƒ(9) < 0 = (0.659) 5 (0.659) + = 0.0089 (which is negative) 6 + 9 0.656 + 0.659 So, 0 = = = 0.657 Now, ƒ(0) = ƒ(0.657) ƒ(6).ƒ(0) < 0 6 + 0 So, = = (0.657) 5 (0.657) + = 0.0040 (which is negative) = Now, ƒ() = ƒ(0.656) ƒ().ƒ(0) < 0 0 + So, = 0.656 + 0.657 = (0.656) 5 (0.656) + = 0.8.8 + = 0.656 = 0.000 (which is positive) = 0.657 + 0.656 = 0.656 Since and both same value. Therefore if we continue this process we will get same value of so the value of is 0.565 which is required result.

Q.. Mathematical Methods for Numerical Analysis and Optimization Find real root of the equation cos - e = 0 correct upto four decimal places. Ans.: Since, ƒ() = cos - e So, And ƒ(0) = cos0 0e 0 = (which is positive) ƒ() = cos e = -.779 (which is negative) ƒ(0).ƒ() < 0 Hence the root of are given equation lies between 0 and. let ƒ(0) = ƒ(0) and ƒ() = ƒ() 0 + 0 + So, = = = 0.5 Now, ƒ() = ƒ(0.5) ƒ(0.5) = cos(0.5) (0.5)e 0.5 ƒ().ƒ() < 0 = 0.05 (which is positive) + + 0.5 So, = = Now, ƒ() = ƒ(0.75) ƒ().ƒ() < 0 = cos(0.75) (0.75)e 0.75 =.5 = 0.75 = 0.856 (which is negative) + 0.5 + 0.75 So, 4 = = = 0.65 ƒ(4) = ƒ(0.65) ƒ().ƒ(4) < 0 = cos(0.65) (0.65)e (0.65) = 0.56 (which is negative) + 4 0.5 + 0.65 So, 5 = = = 0.565

4 Now, ƒ() = ƒ(0.565) = cos(0.565) 0.565e 0.565 = 0.49 (which is negative) ƒ().ƒ(5) < 0 + 5 0.5 + 0.565 So, 6 = = = 0.5 Now, ƒ(6) = ƒ(0.5) = cos(0.5) (0.5)e 0.5 = 0.045 (which is negative) ƒ().ƒ(6) < 0 + 6 0.5 + 0.5 So, 7 = = = 0.556 Now, ƒ(7) = ƒ(0.556) = cos(0.556) (0.556)e 0.556 = 0.00655 (which is positive) ƒ(6).ƒ(7) < 0 6 + 7 0.5 + 0.55 So, 8 = = = 0.5 Now, ƒ(8) = ƒ(0.5) = cos(0.5) (0.5)e 0.5 = 0.074 (which is negative) ƒ(7).ƒ(8) < 0 7 + 8 0.55 + 0.5 So, (9) = = = 0.59 Now, ƒ(9) = ƒ(0.59) = cos(0.59) (0.59)e 0.59 = 0.005 (which is negative)

Mathematical Methods for Numerical Analysis and Optimization 5 ƒ(7).ƒ(9) < 0 7 + 9 0.55 0.59 So, (0) = = = 0.575 Now, ƒ(0) = ƒ(0.575) ƒ(9).ƒ(0) < 0 9 + 0 So, = = cos(0.575) (0.575)e 0.575 = 0.000607 (which is positive) = Now, ƒ() = ƒ(0.585) ƒ(0).ƒ() < 0 0 + So, = 0.595 + 0.575 = cos(0.585) (0.585)e 0.585 = 0.585 = 0.0060 (which is negative) = 0.575 + 0.585 = 0.580 Hence the root of the given equation upto decimal places is = 0.58 Thus the root of the given equation is = 0.58

6 Chapter- Regula Falsi Method Q.. Find the real root of the equation log0. = 0 correct upto four decimal places. Ans.: Given ƒ() = log0. _ () In this method following formula is used - ( n- n ) f ( n) n+ = n ( f( ) f( )) Taking = in eq.() ƒ() =. log0. Taking = in eq.() n n = (which is negative) ƒ() =. log0. Taking = in eq.() = 0.5979 (which is negative) ƒ() =. log0. ƒ().ƒ() < 0 = 0. (which is positive) _ () So the root of the given equation lies between and. let = and = ƒ() = ƒ() = 0.5979

Mathematical Methods for Numerical Analysis and Optimization 7 And ƒ() = ƒ() = 0. Now we want to find so using eq.() ( - ) f ( ) = f( ) f( ) = = 0. 0.89 (- ) (0.) 0. ( 0.5979) = 0.789 =.7 ƒ() = ƒ(.7) ƒ().ƒ() < 0 =.7 log0.7. = 0.070 (which is negative) Now to find 4 using equation () Now ( - ) f ( ) 4 = f( ) f( ) =.7 =.7 0.00474 0.48 (.7 - ) (-0.070) (-0.070-0.) =.7 + 0.090 =.740 ƒ(4) = ƒ(.740) ƒ().ƒ(4) < 0 =.740 log0.740. = 0.000890 (which is negative) Now to find 5 using equation ()

8 ( 4- ) f ( 4) 5 = 4 [ f( ) f( )] 4 (.740 ) =.740 ( 0.000476) ( 0.000476 0.) =.740 + =.740 + ( 0.598)( 0.000476) 0.7 (0.0007) 0.7 =.740 + 0.00054 =.7406 ƒ(5) = ƒ(.7406) ƒ().ƒ(5) < 0 =.7406 log0.7406. = 0.000040 (which is negative) To find 6 using equation () ( 5- ) f ( 5) 6 = 5 f( ) f( ) =.7406 + 5 (.7406 - ) ( 0.000040) ( 0.00004) (0.) =.7406 + 0.00000 =.7406 The approimate root of the given equation is.7406 which is correct upto four decimals. Q.. Find the real root of the equation 5 = 0 correct upto four decimal places. Ans.: Given equation is ƒ() = 5 _ ()

Mathematical Methods for Numerical Analysis and Optimization 9 In this method following formula is used :- ( n- n ) f ( n) n+ = n [ f( ) f( )] n n _ () Taking = in equation () ƒ() = 5 = 6 (which is negative) Taking = in equation () ƒ() = 8 4 5 = (which is negative) Taking = ƒ() = 7 6 5 = 6 (which is positive) Since ƒ().ƒ() < 0 So the root of the given equation lies between and. Let = and = ƒ() = ƒ() = and ƒ() = ƒ() = 6 Now to find using equation () = ( - ) f( ) - f( ) = ( - ) 6 6 + = 6 7 =.0588 f ( ) ƒ() = (.0558) (.0588) 5 = 8.765 4.76 5 = 0.9 (which is negative) ƒ().ƒ() < 0

0 Now to find 4 using equation () ( - ) 4 = [ f( ) f( )] f ( ) =.0588 (.0588 - ) 6-0.9-6 (-0.9) =.0588 + ( 0.94) ( 0.9) 6.9 ƒ(4) = 9.044 4.64 5 So ƒ(). ƒ(4) < 0 = 0.48 (which is negative) Now using equation () to find 5 ( 4- ) 5 = 4 [ f( ) f( )] 4 f ( ) =.08 (.08 - ) (-0.48) (-0.48-6) =.08 + =.08 + 8.40 =.0896 ( 0.988) ( 0.48) 6.48 ƒ(5) = 9.40 4.79 5 (5 - ) f ( 5) f( ) - f( ) 4 5 = 0.055 (which is negative) ƒ().ƒ(5) < 0 Now using equation () to find 6 ( 5- ) f ( 5) 6 = 5 f( ) f( ) =.0896 5 (.0896 ) ( 0.055 6) (-0.055) =.08 0 -

Mathematical Methods for Numerical Analysis and Optimization =.0896 + (0.0505) 6.055 =.097 ƒ(6) = 9.647 4.854 5 So ƒ().ƒ(6) < 0 = 0.007 (which is negative) Now using equation () to find 7 ( 6- ) 7 = 6 f( ) f( ) =.097 =.097 + 6 f ( ) (.097 - ) (-0.007) (-0.007-6) (-0.907) (-0.007) 6.007 =.097 +.7 0 - =.098 Now ƒ(7) = 9.79 4.876 5 So ƒ().ƒ(7) < 0 = 0.0084 (which is negative) Now using equation () to find 8 ( 7- ) 8 = 7 f( ) f( ) =.098 =.098 + 7 6 f ( ) (.098 - ) (-0.0084) (-0.0084-6) (-0.906) (-0.0084) 6.0084 =.098 + 4.755 0-4 =.0947 7

ƒ(8) = 9.85 4.8854 5 So ƒ().ƒ(8) < 0 = 0.004 (which is negative) Now using equation () to find 9 ( 8- ) 9 = 8 f( ) f( ) =.0947 =.0947 =.0944 8 f ( ) 8 (.0947 - ) (-0.004) (-0.004-6) (-0.9057) (-0.004) 6.004 The real root of the given equation is.094 which is correct upto three decimals.

Mathematical Methods for Numerical Analysis and Optimization Chapter-4 Secant Method Note : In this method following formula is used to find root Q.. n+ = n ( n n ) f ( n ) f ( ) f ( ) n n _ () Find the root of the equation 5 7 + 0 [use Secant Method] correct upto four decimals. Ans.: Given ƒ() = 5 7 + 0 _ () Taking = 0 in equation () ƒ(0) = 0 Now taking = ƒ() = 5 7 + 0 = Since ƒ(0) = 0 (positive) and ƒ() = (which is negative) so the root of the given equation lies between 0 and. Let = 0 and = ƒ() = 0 and ƒ() = using equation () to find ( - ) = f( ) f( ) f ( )

4 = = + = 0.95 ( - 0) (-) (-) - 0 () = =- (- ) - ƒ() = ƒ(0.95) = (0.95) 5 (0.95) 7 (0.95) + 0 = 0.866 4.54 6.89 + 0 = 0.40 (which is positive) Using equation () to find 4 ( - ) 4 = f( ) f( ) = 0.95 = 0.95 f ( ) (0.95 - ) [0.40 - (-)] (-0.0477) (0.40) (.40) 0.40 = 0.95 + 0.005865 = 0.958 ƒ(4) = (0.958) 5 (0.958) 7 (0.958) + 0 = 0.8794 4.5897 6.877 + 0 = 0.000 (which is positive) ( 4- ) 5 = 4 f( ) f( ) = 0.958 = 0.958 4 f ( ) (0.958-0.95) (0.000) - (0.40) 4 0.000 Hence the root of the given equation is 0.958 which is correct upto four decimal.

Q.. Mathematical Methods for Numerical Analysis and Optimization 5 Given that one of the root of the non-linear equation cos e = 0 lies between 0.5 and.0 find the root correct upto three decimal places, by Secant Method. Ans.: Given equation is ƒ() = cos e And = 0.5 and =.0 ƒ() = cos (0.5) (0.5) e 0.5 = 0.87758 0.846 = 0.05 Now ƒ() = cos () () e = 0.5400.788 =.780 Now to calculate using equation () ( - ) = f( ) f( ) = = f ( ) ( 0.5) (.780) (.780 0.05) = 0.48807 = +0.59 (0.5) (.780). ƒ() = ƒ(0.59) = cos (0.59) (0.59)e 0.59 = 0.8750 0.854 = 0.0767 Now for calculating 4 using equation () ( - ) 4 = f( ) f( ) f ( )

6 = 0.59 = 0.59 (0.59- ) (0.767) - (-.780) = 0.59 + 0.00864.9567 = 0.59 + 0.0097 = 0.5584 (-0.48808) (0.0767).9567 ƒ(4) = cos (0.5584) (0.5584)e 0.5584 = 0.86987 0.86405 = 0.00584 (which is positive) Now for calculating 5 using equation () ( 4- ) 5 = 4 f( ) f( ) = 0.5584 = 0.5584 4 f ( ) (0.5584-0.59) (0.00584-0.0767) 0.009 (-0.085) = 0.5584 +0.009 = 0.5776 = 0.578 4 (0.00584) Now ƒ(5) = cos (0.578) (0.578)e 0.578 = 0.8689 0.8690 = 0.0000 0.0767 0.00584 = 0.0000 (upto four decimals) Hence the root of the given equation is = 0.578 (which is correct upto four decimal places) This process cannot be proceed further because ƒ(5) vanishes.

Mathematical Methods for Numerical Analysis and Optimization 7 Chapter-5 Newton Raphson Method Hint : Formula uses in this method is Q.. n+ = n f ( n ) f '( ) n Find the root of the equation 5 + = 0 correct upto 5 decimal places. (use Newton Raphson Method.) Ans.: : Given ƒ() = 5 + = 0 Taking = 0 Taking = ƒ(0) = (which is positive) ƒ() = 5 + = (which is negative) ƒ(0). ƒ() < 0 The root of the given equation lies between 0 and Taking initial approimation as = Since = 0.5 0 = 0.5 ƒ() = 5 + f' () = 5 ƒ() = (0.5) 5(0.5) +

8 = 0.5.5 + = 0.5 f' () = (0.5) 5 = 5 = 4 Now finding = 0.5 ( 0.5) 4 = 0.5 0.5 4 = 0.475 ƒ() = (0.475) 5(0.475) + = 0.940.875 + = 0.00906 f' () = (0.475) 5 Now finding = 4.5 f ( ) = f ( ) = 0.475 0.00906 ( 4.5) = 0.475 + 0.0009469 = 0.4844 ƒ() = (0.4844) 5(0.4844) + = 0.9.9 + = 0.0000 f'() = (0.4844) 5 = 4.

Mathematical Methods for Numerical Analysis and Optimization 9 f ( ) 4 = f ( ) = 0.4844 0.0000 ( 4.) = 0.4844 + 0.00000485 = 0.4844 Hence the root of the given equation is 0.4844 which is correct upto five decimal places. Q.. Apply Newton Raphson Method to find the root of the equation cos = 0 correct the result upto five decimal places. Ans.: Given equation is Taking = 0 ƒ() = cos ƒ(0) = (0) cos 0 = Now taking = ƒ() = () cos () = 0.540 =.4597 Taking initial approimation as = At = 0.5 0 = 0.5 ƒ() = cos f'() = + sin ƒ() = (0.5) cos (0.5) =.5 0.8775

0 = 0.7758 f'() = sin (0.5) =.4794 Now to find using following formula f ( ) = f '( ) = 0.5 ( 0.7758) (.4794) = 0.5 + 0.085 = 0.6085 ƒ() = (0.6085) cos (0.6085) =.8556 0.80494 = 0.005066 f'() = + sin (0.6085) Now finding =.5765 = 0.6085 (0.005066) (.5765) = 0.6085 0.0048 = 0.6070 ƒ() = (0.6070) cos (0.6070) =.8 0.805884 = 0.00000588 f' () = + sin (0.6070) = + 0.57048 =.5704 Now to find 4 using following formula

Mathematical Methods for Numerical Analysis and Optimization f ( ) 4 = f '( ) = 0.6070 ( 0.00000588).5704 = 0.6070 + 0.0000064 = 0.6070 Which is same as Hence the root of the given equation is = 0.6070 which is correct upto five decimal places.

Chapter-6 Iterative Method Q.. Find a root of the equation + = 0 in the interval (0,) with an accuracy of 0-4. Ans.: Given equation is ƒ() = + = 0 Rewriting above equation in the form = () The given equation can be epressed in either of the form : (i) + = 0 + = ( + ) = = = + ( ) _ () (ii) + = 0 = = ( + ) (/) _ () (iii) + = 0 = = ( ) / _ ()

Mathematical Methods for Numerical Analysis and Optimization Comparing equation () with g () = 0 we find that g() = ( ) g() = ( + ) -/ g'() = ½ ( + ) -/ g'() = ½ ( + ) / = / ( + ) < Now comparing equation () with g() = 0 We find that g() = ( ) / g'() = ½ ( + ) -/ (- ) = - g'() = 0 (- ) Which is not less than one. / Now comparing equation () with g() = 0 g() = ( ) / g'() = ( ) -/ ( ) = ( - ) / g'() = / ( ) Which is not less than one. Hence this method is applicable only to equation () because it is convergent for all (0, )

4 Now taking initial approimation = 0 = 0.5 So = (+ ) [using iteration scheme n+ = ( n ) ] = Similarly = 4 = 5 = 6 = 7 = 8 = 0.5 =.5 = ( ) = ( ) = ( ) 4 = ( ) 5 = ( ) 6 = ( ) 7 = 0.8649 0.8649 = 0.749 0.749 = 0.7576 0.7576 = 0.754 0.754 = 0.7550 0.7550 = 0.7548 0.7548 = 0.7548 Hence the approimate root of the given equation is = 0.7548 Q.. Find the root of the equation = cos + correct upto decimal places. Ans.: Given equation is ƒ() = cos = 0 Rewriting above equation in the form = g() = cos + = cos + _ ()

Mathematical Methods for Numerical Analysis and Optimization 5 Comparing above equation with the following equation = g() we find the g() = cos + g'() = For (, ) sin g'() = sin sin < = cos Hence the iterative scheme n+ = cos ( ) + n Now taking initial approimation =.5 = = 4 = 5 = 6 = 7 = 8 = cos + = cos ( ) + = cos ( ) + = cos ( 4) + = cos ( 5) + = cos ( 6) + = cos ( 7) + = Which is same as 7 cos (.5) + =.55 cos (.55) + =.577 cos (.577) + =.565 cos (.565) + =.5 cos (.5) + =.54 cos (.54) + =.50 cos (.50) + =.5 is convergent. Hence the root of the given equation is =.5 (which is correct upto decimals)

6 Q.. Find the root of the equation e = in the internal (0, ) (use iterative Method) Ans.: Given equation is e = 0 Rewriting above equation in the form of = g () e = 0 e = = e - Comparing it with the equation = g () we find that g() = e - g' () = -e - g' () = e - < Hence the iterative scheme is n+ = e n Now taking initial approimation = 0.5 = e = = e = 4 = e = 4 5 = e = 5 6 = e = 6 7 = e = 7 8 = e = 8 9 = e = 9 0 = e = (0.5) e = 0.6065 (0.6065) e = 0.545 (0.545) e = 0.5797 0.5797 e = 0.5600 0.5600 e = 0.57 (0.57) e = 0.5648 (0.5648) e = 0.5684 (0.5684) e = 0.5664 (0.5664) e = 0.5675

Now Mathematical Methods for Numerical Analysis and Optimization 7 0 = e = = e = = e = 4 = e = 0.5675 e = 0.5669 0.5669 e = 0.567 (0.567) e = 0.567 (0.567) e = 0.567 Hence the approimate root the given equation is = 0.567

8 Chapter-7 Gauss Elimination Method Q.. Use gauss elimination method to solve : + y + z = 7 + y + 4z = 4 + y + z = 6 Ans.: Since in the first column the largest element is in the second equation, so interchanging the first equation with second equation and making as first pivot. + y + 4z = 4 _ () + y + z = 7 _ () + y + z = 6 _ () Now eliminating form equation () and equation () using equation () - equation () + equation (), equation () equation () we get - - y - z = - + y + 4z = 4 and z + y + 4z = 4 _ (4) z = _ (5) y z = 0 _ (6) 6 y 9z 48 6 6y 8z 48 y z 0 = y - z = 0

Mathematical Methods for Numerical Analysis and Optimization 9 Now since the second row cannot be used as the pivot row since a = 0 so interchanging the equation (5) and (6) we get + y + 4z = 4 _ (7) y z = 0 _ (8) z = _ (9) Now it is upper triangular matri system. So by back substitution we obtain. z = From equation (8) y = 0 y = y = From equation (7) + () + 4 () = 4 + + = 4 + 5 = 4 = 9 = Hence the solution fo given system of linear equation is =, y =, z = Q.. Solve the following system of linear equation by Gauss Elimination Method : + 4 + = + = + = 6 Ans.: Since in the first column the largest element is in the second row, so interchanging first equation with second equation and making as first pivot.

40 + 4 = _ () + 4 + = _ () + = 6 _ () Eliminating form equation () and equation () using equation () - equation () + equation () and + equation () equation () 6 = 9 6 + 4 4 = 4 and 8 7 = + = 8 + = 5 + 5 = 0 8 + 7 = = -4 So the system now becomes : + = _ (4) 8 + 7 = _ (5) = 4 _ (6) Now eliminating from equation (6) using equation (5) {8 equation (6) equation (5)} 8-8 = - 8 +7 = - -5 = -45 = So the system of linear equation is + = _ (7) 8 + 7 = _ (8) = _ (6) Now it is upper triangular system so by back substitution we obtain =

Mathematical Methods for Numerical Analysis and Optimization 4 From equation (8) 8 + 7() = 8 = 8 = 8 = From equation (9) +( ) () = = + + 6 = 6 = Hence the solution of the given system of linear equation is : =, =, =

4 Chapter-8 Gauss-Jordan Elimination Method Q.. Solve the following system of equations : 0 + + = 9 _ () + 0 = 44 _ () - + + 0 = _ () Use Gauss Jordan Method. Ans.: Since in the given system pivoting is not necessary. Eliminating from equation () and equation () using equation () 5 equation () equation (), 5 equation () + equation () 0 00 0 = 0 0 + + = 9 98 = 9 and Now the system of equation becomes 0 + + = 9 _ (4) 98 = 9 _ (5) + = 7 _ (6) 0 + 5 + 50 = 0 0 + + = 9 7 + 5 = 9 = + = 7 Now eliminating from equation (4) and (6) using equation (5)

Q.. Mathematical Methods for Numerical Analysis and Optimization 4 98 equation (6) equation (5), 49 equation (4) - equation (5) 98 +94 = 686 98 = 9 05 = 95 490 + 98 + 49 = 44 98 = 9 490 + 60 = 670 = = 49 + 6 = 67 Now the system of equation becomes : 49 +0 + 6 = 67 _ (7) 98 = 9 _ (8) = _ (9) Hence it reduces to upper triangular system now by back substitution. = From equation (8) 98 = 9 98 = 9 + 98 = 96 = From equation (7) 49 + 6() = 67 49 = 67 8 49 = 49 = Thus the solution of the given system of linear equation is =, =, = Solve the following system of equation using Gauss-Jordan Elimination Method. + 5 = _ () + + 4 = 0 _ () + = 0 _ () Ans.: Solve this question like question no. 7.

44 Chapter-9 Matri Inversion Method Q.. Solve the given system of equation using Matri inversion Method. 6 + + 7 = 7 + 5 + = -7 7 + + 0 = Ans.: The given system of equations cab be written in the form of AX = B A = 6 7 5 7 0, X =, B = 7 7 The solution can be given by X = A - B so to find the solution first we have to find A - using Gauss-Jordan Method. The inverse of matri A that is A - is obtained by reducing the argumented matri [A/I] into the matri [I/A - ] The argumented matri is given by 6 7 0 0 5 0 0 7 0 0 0 R R

Mathematical Methods for Numerical Analysis and Optimization 45 7 0 0 0 5 0 0 6 7 0 0 R 7R - R, R 7 6 R - R 7 0 0 0 0 4 0 7 7 0 0 6 6 R R + R, R R + R 5 66 0 7 0 4 0 7 77 0 0 0 R R 6 + R, R 4 R + R 04 77 66 0 0 69 8 46 77 847 85 0 0 4 77 0 0 7 R 04 R, R 4 46 R, R R

46 0 0 0 0 0 0 46 6 9 57 57 57 4 5 57 57 57 9 9 9 9 R 4 04 R, R 46 R, R R Hence A - = 57 46 6 9 4 5 9 7 Thus the matri A is reduced to identity matri Hence the solution of the given system of equations is X =A - B 57 46 6 9 7 4 5 7 9 7 + - 77 8-77 - 65 57-6 + 5 57 57 4 57 Q.. Solve the following system of linear equations using matri inversion method. + + 4 = 7 + + = 7 + + 4 =

Mathematical Methods for Numerical Analysis and Optimization 47 Ans.: The given system of linear equations can be written in the form of AX = B 4 4 7 7 The solution can be given by X = A - B. for this we have to first find the value of A - using Gauss Jordan Method. The inverse of the matri A using Gauss Jordan method is obtained by reducing the argumented matri [A/I] in the form of[i/a - ]. The argumented matri is given as follows : 4 0 0 0 0 4 0 0 Here pivoting is not necessary. R = R R, R = R R 4 0 0 5 0 0 0 7 8 0 R R = 4 R + R, R = R + R 4 0 0 4 4 5 0 0 7 5 0 0 4 4 4 9 7 R + R, R 7 5 R + R

48 = Now R = Hence 7 0 0 8 8 7 4 7 0 0 70 0 5 4 7 5 0 0 4 4 4 0 0 0 0 0 0 A - = 9 8 70 4 7 R, R 7 R, R 7 R 4 9 9 9 7 8 5 9 9 9 5 7 9 9 9 4 7 8 5 5 7 Thus the solution of given matri is given by i.e. X = A - B = 9 4 7 8 5 5 7 7 7 = 9 7 8 4 49 56 0 5 49 = 9 7 6 Hence = 9, = 7 9, = 6 9

Mathematical Methods for Numerical Analysis and Optimization 49 Chapter-0 Matri Factorization Method Q.. Solve the following system of linear equation using Matri Factorization Method. + 5 + = 8 8 + = -7 6 + + 8 = 6 Ans.: Above system of equation can be written in the form AX = B where A = 5 0 8 6 8 ; B = Let us assume that A = LU 8 7 6 and X = 0 0 Where L = l 0 l l and U = U U U 0 U U 0 0 U 0 0 LU = l 0 l l U U U 0 U U 0 0 U U U U l U l U +U l U +U l U l U + l U l U + l U +U

50 Since A = LU so comparing both matrices. U = _ () U = 5 _ () U = _ () l U = 0 _ (4) l U + U = 8 _ (5) l U + U = _ (6) l U = 6 _ (7) l U + l U = _ (8) l U+ l U + U = 8 _ (9) l U = 0 l = 0 l = 0 _ (0) l U = 6 l = 6 l = 6/ l = _ () Now from equation (5) l U + U = 8 0 U + U = 8 U = 8 _ () From equation (6) l U + U = 0 U + U = U = _ () From equation (8) l U + U =

Mathematical Methods for Numerical Analysis and Optimization 5 5 + L 8 = l 8 = 0 = 8 l = From equation (9) l U+ l U + U = 8 + ( ) + U = U = 8 4 + U = 6 L = 0 0 0 0 and U = 5 0 8 0 0 6 Since the given system of equation can be written as AX = B [Here A = LU] LUX = B _ (4) Now let UX = Y _ (5) LY = B _ (6) 0 0 0 0 y y y 8 = 7 6 y y y y y = 8 7 6 On comparing both matrices, we get y = 8, y = 7 and y y + y = 6 8 + 7 + y = 6 y = 6 6 7

5 y = Y = 8 7 From equation (5) UX =Y 5 0 8 0 0 6 = 8 7 5 0 8 0 0 6 = 8 7 Comparing both matrices + 5 + = 8 _ (7) 8 + = -7 _ (8) 6 = _ (9) From equation (9) 6 = = From equation (8) 8 + = 7 8 + 8 = 7 8 = 8 = = 7

Mathematical Methods for Numerical Analysis and Optimization 5 From equation (7) + 5 + = 8 = 8 + 5 = = 4 Thus the solution of given system of equation is = 4, = and = Q.. Solve the following system of linear equation using Matri Factorization Method. + y + z = 4 + y + z = + y + z = Ans.: Above system of equation can be written in the form of AX = B where A = ; B = Let us assume that A = LU where 4 and X = y z 0 0 L = l 0 l l and U = U U U 0 U U 0 0 U LU = U U U l U l U +U l U +U l U l U + l U l U + l U +U Since A = LU so comparing both sides we get U = _ () U = _ ()

54 U = _ () l U = _ (4) l U + U = _ (5) l U + U = _ (6) l U = _ (7) l U + l U = _ (8) l U+ l U + U = _ (9) From equation () l U = l = l = _ (0) From equation (5) l U + U = + U = U = 6 U = 5 _ () From equation (6) l U + U = + U = U = 9 U = 7 _ () From equation (7) l U = l = l = _ () From equation (8) l U + l U =

Mathematical Methods for Numerical Analysis and Optimization 55 + l (- 5) = l ( 5) = 4 l = 5 _ (4) From equation (9) l U + l U + U = + 5 (- 7) +U = 6-7 5 + U = U = 6 + 7 5 = 5 + 7 5 = 5 7 5 U = L = 8 5 0 0 0 5 We know that AX = B and U = 0 5 7 0 0 8 5 _ (5) LUX = B where [A = LU] _ (6) Now let UX = Y _ (7) So LY = B _ (8) 0 0 0 5 y y + y y y y = y + y y 5 4 = 4

56 Comparing both sides we get y = 4 _ (9) y + y = _ (0) y + y + y = _ () 5 From equation (0) y + y = 4 + y = y = 4 y = From equation () (4) + (-) + y = 5 8 5 + y = y = 8 + 5 y = 7 + 5 y = y = 85 5 54 5 Y = 4 54 5

Mathematical Methods for Numerical Analysis and Optimization 57 Since UX = Y 4 0 5 7 y = 0 0 8 5 z 54 5 y z 0 5y 7z 8 0 0 z 5 = 4 54 5 + y + z = 4 _ () 5y 7z = _ () 8 5 z = 54 5 z = 54 5 5 8 z = From equation () 5y + 7z = 5y + 7 = 5y = = 0 y = From equation () + y + z = 4 + 4 + 9 = 4 = 4 = Thus the solution of the given system of equation is = ; y = and z = _ (4)

58 Chapter- Jacobi Method Q.. Solve the following system of equation by Jacobi Method. 8 + 4 = 95 7 + 5 + =04 + 8 + 9 = 7 Ans.: Since the given system of equation is 8 + 4 = 95 _ () 7 + 5 + =04 _ () + 8 + 9 = 7 _ () The diagonal elements in the given system of linear equations is not zero so the equation (), () and () can be written as : (n+) = 8 (n+) = 5 (n+) = 9 (n) [95 (n) [04 7 (n) [7 + 4 (n) ] (n) 8 (n) ] ] and Now taking initial approimation as : (0) = 0 ; (0) = 0 and (0) = 0

Mathematical Methods for Numerical Analysis and Optimization 59 Now for first approimation : () = 8 () = 5 () = 9 [95 (0) [04 7 (0) [7 (0) + 4 (0) ] =.446 (0) ] = 8 (0) ] =.448 Similarly second approimation : () = 8 [95 () + 4 () ] = [95 () + 4(.448)] = 0.9975 8 () = 5 [04 7 () () ] = [04 7(.446) (.448)] =.8 5 () = 9 [7 () 8 () ] = [7 (.446) 8 ] =.778 9 Now the third iteration : () = 8 [95 () + 4 () ] = [95 (.8) + 4(.778)] =.0668 8 () = 5 [04 7 () () ] = [04 7 (0.9975) (.778)] = 5 5 [7.90] =.4

60 () = 9 [7 () 8 () ] = [7 (0.9975) 8 (.8)] =.0047 9 Similarly other iterations are : (4) =.058 (4) =.55 (4) =.9459 (5) =.0588 (5) =.78 (5) =.9655 (6) =.0575 (6) =.66 (6) =.960 (7) =.0580 (7) =.676 (7) =.960 (8) =.0579 (8) =.67 (8) =.966 (9) =.0579 (9) =.67 (9) =.966

Mathematical Methods for Numerical Analysis and Optimization 6 Thus the values obtained by successive iteration is given by following table : (n) (n) (n) (n+) (n+) (n+) 0 0 0 0.446.448.446.448 0.9975.8.778 0.9975.8.778.0667.4.0047.0667.4.0047.058.55.9459 4.058.55.9459.0587.78.9655 5.0587.78.9655.0575.66.960 6.0575.66.960.0580.676.960 7.0580.676.960.0579.67.966 8.0579.67.966.0579.67.966 Thus the solution is =.0579 ; =.67 and =.966

6 Chapter- Gauss Seidel Method [This method is also called the method of successive displacement] Q.. Solve the following linear equation : + = 5 + + =0 + = 7 (Use Gauss Seidel Method) Ans.: Above system of equations can be written as : + = 5 _ () + = 7 _ () + + =0 _ () Iterative equations are : (n+) = (n) [5 + (n) ] _ (4) (n+) = (n+) [7 + (n) ] _ (5) (n+) = (n+) [0 (n+) ] _ (6) Taking initial approimations as : (0) = 0 ; (0) = 0 and (0) = 0

Mathematical Methods for Numerical Analysis and Optimization 6 First approimation is : () = [5 + (0) (0) ] = [5 + 0 0] = 5 =.5 () = [7 () + (0) ] = [7.5 + 0] = (4.5) =.5 () = () [0 () ] = [0.5.5] =.5 Now second approimation : () = [5 + () () ] = [5 + (.5).5] =.5 () = [7 () + () ] = [7.5 + (.5)] =.5 () = () [0 () ] = [0.5.5] = 0.8 () = [5 + () () ] = [5 +.5 0.8] =.

64 () = [7 () + () ] = [7. + 0.8] =.7777 () = () [0 () ] = [0..7777] =.07 () =., (4) = [5 + () () ] () =.7777, () =.07 = [5 +.7777.07] =.870 (4) =.0679 (4) = 0.9980 (4) =.870, (4) =.0679, (4) = 0.9980 Now (5) =.05 (5) =.9870 (5) = 0.9970 (6) =.9950 (6) =.9997 (6) =.009

Mathematical Methods for Numerical Analysis and Optimization 65 (7) =.9989 (7) =.006 (7) = 0.999 (8) =.00 (8) =.999 (8) =.000 (9) =.9994 (9) =.000 (9) = (0) =.000 (0) =.9999 (0) = () =.9999 () = () = () = () = () =

66 () = () = () = Hence the solution of the given system of linear equation is : =, =, =

Mathematical Methods for Numerical Analysis and Optimization 67 Chapter- Forward Difference Q.. Ans.: Construct a forward difference table for the following given data..60.65.70.75 y 6.598 8.475 40.447 4.5.60.65.70.75 y 6.598 8.475 40.447 4.5 y.877.97.074 y y 0.095 0.007 0.0

68 Chapter-4 Backward Difference Q.. Construct a backward difference table form the following data : sin 0 = 0.5000, sin 5 = 0.576, sin 40 = 0.648, sin 45 = 0.707 Assuming third difference to be constant find the value of sin 5. Ans.: y y 5? y y 0 5 40 45 0.5000 0.576 0.648 0.707 y0 =? 0.076 0.069 0.064 y5 =? -0.0044-0.0049 y40 =? -0.0005 Since we know that y should be constant so y40 = 0.0005 y40 y5 = 0.0005-0.0044 y5 = 0.0005

Mathematical Methods for Numerical Analysis and Optimization 69 y5 = +0.0005 0.0044 = -0.009 Again y5 = 0.009 y5 y0 = -0.009 0.076 y0 = 0.009 y0 = 0.009 + 0.076 = 0.0775 Again y0 = 0.0775 y0 y5 = 0.0775 0.5000 y5 = 0.0775 y5 = 0.5000 0.0775 = 0.45 Hence sin 5 = 0.45

70 Chapter-5 Newton Gregory Formula for Forward Interpolation Q.. Ans.: Use Newton formula for interpolation to find the net premium at the age 5 from the table given below : Age 0 4 8 Annual net premium 0.047 0.058 0.077 0.0996 Age () ƒ() ƒ() ƒ() ƒ() 0 0.047 0.0054 4 0.058 0.0007 0.009-0.00004 8 0.077 0.000 0.0996 0.004 Here a = 0, h = 4 and = a + hu = a + hu 5 = 0 + 4 u 5 = 4u u =.5

Mathematical Methods for Numerical Analysis and Optimization 7 Using following Newton s Gregory forward interpolation formula : ƒ(a + hu) = ƒ(a) + u () ƒ(a) + () u! ƒ(a) + () u! ƒ(a) + _.5 (0.5) ( 0.75) ƒ(5) = 0.047 +.5 (0.0054) + (-0.00004).5 (0.5) (0.0007) + ƒ(5) = 0.047 + 0.0095 + 0.0000578 + 0.000005 = 0.0654 Q.. From the following table find the number of students who obtained less than 45 marks : Marks No. of Students 0 40 40 50 4 50 60 5 60 70 5 70 80 Ans.: Marks () No. of Students ƒ() Less than 40 Less than 50 7 9 Here a = 40, h = 0 and a + hu = 45 40 + 0 u = 45 ƒ() ƒ() ƒ() 4 ƒ() 4 5-5 Less than 60 4-6 7 5 Less than 70 59-4 Less than 80 90

7 0u = 5 u = using following forward interpolation formula : ƒ() = ƒ(a) + u () ƒ(a) + () u! ƒ(a) + () u! ƒ(a) + _ ƒ(45) = ƒ(40) + ƒ(40) +! ƒ(40) +! ƒ(40) + 4! 4 ƒ(40) Q.. = + (4) + (9) + = +.5.565.445 = 47.867 = 48 (approimately) (-5) + 5 - - - 4 (7) Hence the no. of students who obtained less than 45 marks are 48. Find the cubic polynomial which takes the following values 0 ƒ() 0 0 Find ƒ(4) Ans.: Here we know that a = 0, h = then form Newton s Gregory forward interpolation formula. Pn () = ƒ(0) + c ƒ(0) + c ƒ(0) + _ cn n ƒ(0) ()

Mathematical Methods for Numerical Analysis and Optimization 7 () ƒ() ƒ() ƒ() ƒ() 0-0 6 8 9 ƒ(4) 7 = 6 0 ƒ(4) 9 (it should be constant) ƒ(4) 0 4 ƒ(4) Substituting the values in equation () from above table : P () = + (-) + ( ) () + ( )( ) (6) P () = + + + = + Hence the required polynomial of degree three is + Again ƒ(4) 7 = 6 ƒ(4) =

74 Chapter-6 Newton s Formula for Backward Interpolation Q.. The population of a town in decennial census was as given below : Ans.: Year () Year 89 90 9 9 9 Population (in thousands) 46 66 8 9 0 Estimate the population for the year 95. Population (in thousand) ƒ() 89 46 ƒ() ƒ() ƒ() 4 ƒ() 0 90 66-5 5 9 8 - - - 9 9-4 9 0 8

Mathematical Methods for Numerical Analysis and Optimization 75 Here = 95, h = 0, a = 89 and a + nh = 9 (a + hn) + uh = 95 9 + uh = 95 uh = 95 9 = -0.6 0 Now using Newton s Backward interpolation formula : ƒ(a + nh + uh) = ƒ(a + nh) + u! 4 ƒ(a + nh) ƒ(a + nh) + + u(u +) (u +) (u +) 4! ƒ(95) = 0 + (-0.6) 8 + ( 0.6)(0.4)! + (-0.6) (0.4) (.4) (.4) (-) 4! = 0 4.8 + 0.48 +0.056 0.008 (- 4) + = 96.65 thousand (approimately) (-0.6) (0.4) (.4)! (- )

76 Chapter-7 Divided Difference Interpolation Q. Construct a divided difference table from the following data : 4 7 ƒ() 0 8 06 6 Ans.: ƒ() ƒ() ƒ() ƒ() 4 ƒ() 0 = 8 0 6 8 4 = 6 8 0 4 = 6 (.6 6) 7 = -.6 4 8 8 6 7 = -.6 0.55 (.6) = 0.94 06 8 7 4 = 8.75 (.6) 0.55 =

Mathematical Methods for Numerical Analysis and Optimization 77 ƒ() ƒ() ƒ() ƒ() 4 ƒ() 7 06 8 4 =.75 6 6 06 7 = Q.. By means of Newton s divided difference formula find the value of ƒ(), ƒ(8) and ƒ(5) from the following table : 4 5 7 0 ƒ() 48 00 94 900 0 08 Ans.: Newton s divided difference formula for 4, 5, 7, 0,, is : ƒ() = ƒ(4) + ( 4) 5 ƒ() + ( 4) ( 5) 5,7 ƒ(4) + ( 4) ( 5) ( 7) ƒ(4) 5,7,0 + ( 4) ( 5) ( 7)( 0) 4 5,7,0, ƒ(4) + () So constructing the following divided difference table : ƒ() ƒ() ƒ() ƒ() 4 ƒ() 4 48 00 48 5 4 = 5 5 00 97 5 7 4 = 5 94 00 7 4 = 97 5 0 4 =

78 ƒ() ƒ() ƒ() ƒ() 4 ƒ() 7 94 0 97 0 5 = 0 900 94 0 7 = 0 7 5 = 0 900 0 0 7 = 7 0 0 900 = 0 0 7 7 = 0 409 0 0 = 08 08 0 = 409 Substituting the values from above table in equation () ƒ() = 48 + 5 ( 4) + 5 ( 4) ( 5) + ( 4) ( 5) ( 7) = ( ) _() Now substituting =, 8 and 5 in equation () ƒ() = 4 ( -) = 4 ƒ(8) = 64 (8 ) = 448 ƒ(5) = 5 (5 ) = 50 Q.. Find the polynomial of the lowest possible degree which assumes the values,, 5, - when has values,,, - respectively.

Mathematical Methods for Numerical Analysis and Optimization 79 Ans.: Constructing table according to given data ƒ() ƒ() ƒ() ƒ() - - 8 5-7 - - -9 Substituting the values in Newton s divided difference formula : ƒ() = ƒ(0) + ( 0) ƒ(0, ) + ( 0) ( ) _ ( n-) + ƒ(0,,..n) = - + { (-)} 8 + { (-)} ( ) (-7) + { (-)} ( ) ( ) () = 9 + 7 + 6

80 Chapter-8 Lagrange s Interpolation Q.. Given that ƒ() =, ƒ() = 4, ƒ() = 8, ƒ(4) = 6, ƒ(7) = 8 Find the value of ƒ(5) with the help of Lagrange s interpolation formula. Ans.: According to question 0 =, =, =, = 4, 4 = 7, and ƒ(0) =, ƒ() = 4, ƒ() = 8, ƒ() = 6, and ƒ(4) = 8, Using Lagrange s formula for = 5 ƒ(5) = = (5 )(5 )(5 4)(5 7) (5 )(5 )(5 4)(5 7) + 4 ( )( )( 4)( 7) ( )( )( 4)( 7) + (5 )(5 )(5 4)(5 7) (5 )(5 )(5 )(5 7) 8 + 6 ( )( )( 4)( 7) (4 )(4 )(4 )(4 7) + (5 )(5 )(5 )(5 4) 8 (7 )(7 )(7 )(7 4) 8 8 494 4 5 5 5 =.9 Hence ƒ(5) =.9 Q.. Find the form of function given by the following table : - ƒ() 5 -

Mathematical Methods for Numerical Analysis and Optimization 8 Ans.: According to question 0 =, =, = and = - ƒ(0) =, ƒ() =, ƒ() = 5 and ƒ() = - Now substituting above values in Lagrange s formula : ƒ() = ( )( )( + ) ( )( )( + ) + ( )( )( ) ( )( )( ) + ( )( )( + ) ( )( )( ) 5 + ( )( )( ) ( )( )( ) = 8 ( + ) -4 ( + ) + 5 4 ( 4 + + 6) + 7 8 ( 6 + 6) ƒ() = 9 + 7 + 6 Q.. By means of Lagrange s formula prove that : y0 = (y + y-) - 8 [ (y y) - (y- y-)] Ans.: Here we are given y-, y- y and y and we have to evaluate y0. Using Lagrange s formula (0 )(0 )(0 ) (0 )(0 )(0 ) y0 = y + y ( )( )( ) ( )( )( ) Hence proved. (0 )(0 )(0 ) (0 )(0 )(0 ) y + y ( )( )( ) ( )( )( ) 9 9 = y y y y 6 6 6 6 = y y y- y y - y 6 = y y y- y y - y 8

8 Chapter-9 Spline Interpolation Q.. Given the set of data points (, 8), (, ) and (, 8) satisfying the function y = ƒ(). find the linear splines satisfying the given data. Determine the approimate values of y(.5) and y (.0). Ans.: Let the given points be A (, 8), B (, ) and C (, 8) equation of AB is s () = -8 + ( ) 7 = -8 + 7 7 = 7-5 And equation of BC is s () = - + ( ) (9) = - + 9-8 [si() = yi- + mi ( i-)] = 9 9 _ () Since =.5 belongs to the interval [, ] we have y (.5) = s (.5) = -9(.5) -9 = 8.5 And y () = +9 [from equation ()] Here we note that the splines si () are continuous in the interval [, ] but their slopes are discontinuous.

Mathematical Methods for Numerical Analysis and Optimization 8 Chapter-0 Quadratic Splines Q.. Given the set of data points (, 8), (, ) and (, 8) satisfying the function y = ƒ(). find the quadratic splines satisfying the given data. Find also the approimate values of y(.5) and y' (.0). Ans.: Since we know that mi- + mi = we have h = taking i = taking i = h m0 + m = 4 m + m = 8 i (yi - yi-) [i =,. n] Since m0 = m we obtain m0 = m = 7 and m = using following equation si () = s () = = ( i ) ( ) h h i i- i m i- mi yi- mi- ( ) ( ) m m y m ( ) ( ) 7 (7) ()

84 = ( ) 5 (7) ( ) = 4 + Since.5 lies in the interval [, ] Hence y (.5) = s (.5) = (.5) 4 (.5) + = 6.5 4.5 + = 5.5 y' () = 4 4 y' () = 4 4 = 48 4 = 7.0

Mathematical Methods for Numerical Analysis and Optimization 85 Chapter- Cubic Splines Q.. Given the set of data points (, 8), (, ) and (, 8) satisfying the function y = ƒ(). find the cubic splines satisfying the given data. Determine the approimate values of y (.5) and y' (.0). Ans.: We have n = and p0 = p = 0 therefore from the following relation : 6 p i + 4p i + p i+= (y i y i + y i ) ( i,... n ) h gives p = 8 If s () and s () are respectively, the cubic splines in the intervals and, we obtain s () = ( - ) 8 ( ) 4 ( ) and s () = ( ) + 48 We therefore have y(.5) = s (.5) = 8 + 7 = 7.75 and y (.0) = s (.0) =.0

86 Chapter- Numerical Differentiation Q.. From the following table of values of and y obtain dy/d and d y/d at =...0..4.6.8.0 y 0.00 0.80 0.5440.960.40 4.00 Ans.: According to given question, h = 0., a = and =. Here. is close to the initial value so using Newton-Gregory forward difference formula. y = ƒ() y y y 4 y 0.00 0.80. 0.80 0.880 0.460 0.0480.4 0.5440 0.60 0 0.750 0.0480.6.960 0.840 0.60 0.0480.8.40 0.40.0 4.0000.5680

Mathematical Methods for Numerical Analysis and Optimization 87 Newton s Gregory forward formula is : ƒ(a + h) = ƒ(a) + c ƒ(a) + c ƒ(a) + c ƒ(a) +... or ƒ(a + h) = ƒ(a) + ƒ(a) + ƒ(a) + ƒ(a) +. _ () 6 Differentiating both sides of the equation () w. r. t. hf' (a + h) = ƒ(a) + ( ) ƒ(a) + Again differentiating equation () w. r.t. ƒ(a) + _ () 6 6 h f' (a + h) = ƒ(a) + ( ) ƒ(a) +. _ () Here we have to find f' (.) and f'' (.) Substituting a =, h = 0. and = in equation () and () 0.f' (.) = 0.80 + 0 + 6 6 4 (0.0480) + 0 Hence f' (.) = 0.60 _ (4) And (0.) f'' (.) = 0.880 + (0.0480) + 0 = 0.64 Hence f'' (.) = 6.60 _ (5) Q.. Using divided difference find the value of f' (8) given that : 6 7 9 ƒ().556.690.908.58 Ans.: y = ƒ() y y y 0 = 6.556 0.4

88 y = ƒ() y y y = 7.690-0.008 0.09 0.0005 = 9.908-0.005 0.08 =.58 Newton s divided difference formula is ƒ() = ƒ(0) + ( 0) ƒ(0) + ( 0) ( ) ƒ(0) + ( 0) ( ) ( ) ƒ(0) + ( 0) ( ) ( ) ( ) 4 ƒ(0) + _ () Differentiating both sides of equation () w. r.t. f'() = ƒ(0) + ( 0 ) ƒ(0) + [ (0 + + ) + 0 + + 0] ƒ(0) _ () Now substituting = 8, 0 = 6, = 7, = 9, = in equation () f (8) = 0.4 + [ 8 6 7 ] (-0.008) + [ 64 8 (6 + 7 +9) + 6 7 + 7 9 + 6 9](0.0005) = 0.4 0.049 + (9 5 + 59) (0.0005) = 0.0859

Mathematical Methods for Numerical Analysis and Optimization 89 Chapter- Numerical Integration Q.. Compute the value of following integral by Trapezoidal rule. 4. 0. (sin log +e ) d e Ans.: Dividing the range of integration in equal intervals in the interval [0.,.4].4 0.. 6 6 sin 0. h log e y = sin log +e e 0. 0.9867 -.6095.4 y0 =.096 0.4 0.894-0.96.498 y =.7975 0.6 0.5646-0.508.8 y =.8975 0.8 0.774-0..55 y =.66.0 0.845 0.0000.78 y4 =.5598. 0.90 0.8.0 y5 = 4.0698.4 0.9855 0.65 4.055 y6 = 4.704 Using following trapezoidal rule I = 4. 0. (sin log +e ) d e

90 = h [(y0 + y6) + (y + y + y +y4 +y5)] = 0. [7.78 + (6.4907)] = 4.075 Q.. Calculate the value of the integral Ans.: First of all dividing the interval [4 5.] in equal parts. 5. 4. 6 6 0. h i yi = 5. 4 log = e log d by Simpson s rule. e log.058 0 4.0 y0 =.86944 4. y =.450845 4.4 y =.486045 4.6 y =.56056 4.8 y4 =.568659 5.0 y5 =.604979 5. y6 =.6486586 Using following Simpson s rule : I = h [(y0 + y6) + 4 (y + y + y5) + (y +y4)] = 0. [.0495 + 8.5 + 6.004408] = 0. [7.47709] =.87847

Q.. Evaluate Mathematical Methods for Numerical Analysis and Optimization 9 0 d + using Simpson s 8 rule : Ans.: Dividing the interval [0, ] into si equal intervals. 0 h 6 6 y = h ( ) 0 = 0 y0 =.000 0 + h = /6 y = (6/7) = 0.9797 0 + h = /6 y = (6/40) = 0.90000 0 + h = /6 y = (6/45) = 0.80000 0 + 4h = 4/6 y4 = (6/5) = 0.69 0 + 5h = 5/6 y = (6/6) = 0.5906 0 + 6h = y6 = (/) = 0.50000 Using following Simpson s /8 rule. 0 +nh 0 0 h yd = (y 0+ y n )+(y + y + y 4+ y 5+...)+(y + y 6 +...) 8 yd = (+ 0.5) + (0.9797 + 0.9 + 0.69 + 0.5906) + (0.8) 6 = 6 [.5 + 9.466 +.6] = 0.78595

9 Chapter-4 Numerical Solution for Differential Equations [Euler s Method] Q.. Use Euler s Method to determine an approimate value of y at = 0. from initial value problem d = + 4y y(0) = taking the step size dy h = 0.. Ans.: Here h = 0., n =, 0 = 0, y0 = Given d = + 4y dy Hence y = y0 + hƒ(0, y0) = + 0. [ 0 + 4y0] = + 0. [ 0 + 4 ] = + 0. [ + 4] = + 0.5 5 =.5 Similarly y = y + hƒ(0 + h, y) =.5 + 0.[ 0. + 4.5] =.9

Mathematical Methods for Numerical Analysis and Optimization 9 Q.. Using Euler s Method with step-size 0. find the value of y(0.5) from the following differential equation d dy = + y, y(0) = 0 Ans.: Here h = 0., n = 5, 0 = 0, y0 = 0 and ƒ(, y) = + y Hence y = y0 + hƒ(0, y0) = 0 + (0.) [0 + 0 ] = 0 Similarly y = y + hƒ(0 + h, y) = 0 + (0.) [(0.) +0 ] = (0.) = 0.00 y = y + hf[0 + h, y] = 0.00 + (0.) [(0.) +(0.00) ] = 0.00 + 0. [0.04 + 0.0000] = 0.00 + 0. [0.04000] = 0.005 y4 = y + hf[0 + h, y] = 0.005 + (0.) [(0.) +(0.005) ] = 0.005 + (0.) [0.09 + 0.00005] = 0.04 y5 = y4 + hf[0 + 4h, y4] = 0.04 + (0.) [(0.4) +(0.04) ] = 0.04 + (0.) [0.6 + 0.0096] = 0.0 Hence the required solution is 0.0

94 Chapter-5 Numerical Solution for Differential Equations [Euler s Modified Method] Q.. Using Euler s modified method, obtain a solution of the equation dy + y with initial conditions y = at = 0 for the range 0 d 0.6 in the step of 0.. Correct upto four place of decimals. Ans.: Here ƒ(, y) = + y 0 = 0, y0 =, h = 0. and n = 0 + nh (i) At = 0. First approimate value of y y () = y0 + hƒ(0, y0) = + (0.) [0 + ] =. Second approimate value of y y () = y0 + h [ƒ(0, y0) + ƒ(, y() )] = + 0. [(0 + ) + {0. +. }] =.95

Mathematical Methods for Numerical Analysis and Optimization 95 Third approimate value of y y () = y0 + h {ƒ(0, y0) + ƒ(, y() )} = + 0. [(0 + ) + {0. +.95 }] = + 0. [ +.0888] =.09 Fourth approimate value of y y (4) = y0 + h {ƒ(0, y0) + ƒ(, y() )} = + 0. [(0 + ) + (0. +.09 )] = + 0. [ +.0945] =.09 Since the value of y () and y (4) is same Hence at = 0., y =.09 (ii) At = 0.4 First approimate value of y y () = y + hƒ(, y) =.09 + (0.) {0. +.09 } =.497 Second approimate value of y y () = y + h [ƒ(, y) + ƒ(, y() )] =.09 + 0. [(0. +.09 ) + (0.4+.497 )] =.09 + 0. [.094598 + (.67604] =.540

96 Third approimate value of y y () = y + h [ƒ(, y) + ƒ(, y() )] =.09 + 0. [(.094598 + (0.4 +.540 )] =.09 + 0. [.094598 +.6450949] =.55 Fourth approimate value of y y (4) = y + h {ƒ(, y) + ƒ(, y() )} =.09 + 0. [(0. +.09 ) + (0.4 +.55 )] =.09 + 0. {.094598 +.650064] =.55 Hence at = 0.4, y =.55 (ii) At = 0.6 First approimate value of y y () = y + hƒ(, y) =.55 + 0. [0.4 +.55 ] =.85 Second approimate value of y y () = y + h {ƒ(, y) + ƒ(, y() )} =.55 + 0. [(0.4 +.55 ) + (0.6+.85 )] =.8849

Mathematical Methods for Numerical Analysis and Optimization 97 Third approimate value of y y () = y + h {ƒ(, y) + ƒ(, y() )} =.55 + 0. =.885 [(0.4 +.55 ) + (0.6 +.8849 )] Fourth approimate value of y y (4) = y + h {ƒ(, y) + ƒ(, y() )} =.55 + 0. =.885 Hence at = 0.6, y =.885 [(0.4 +.55 ) + (0.6 +.885 )]

98 Chapter-6 Numerical Solution for Differential Equations [Runge Kutta Method] Q.. Using Runge Kutta method find an approimate value of y for = 0. dy in step of 0. if + y given y = when = 0 d Ans.: Here ƒ(, y) = + y, 0 = 0, y0 = and h = 0. K = hƒ(0, y0) = 0.[0 + ] = 0.. _ () K = hƒ(0 + h, y0 + K) = 0. 0 (0.) 0.5 = 0.5 _ () K = hƒ(0 + h, y0 + K) = 0. 0 (0.) 0.5 = 0.68 _ ()

Mathematical Methods for Numerical Analysis and Optimization 99 K4 = hƒ(0 + h, y0 +K) = 0. 0 0. 0.68 and = 0.47 _ (4) K = (K +K +K +K4) 6 = 6 0. (0.5) (0.68) 0.47 {using equation (), (), () and (4)} = 0.65 Hence y = y0 + K = + 0.65 =.65 _ (5) Again = 0 + h = 0., y =.65, h = 0. Now K = hƒ(, y) = 0. 0. (.65) = 0.47 _ (6) K = hf h, y K = 0. 0. (0.).65 (0.47) = 0.55 _ (7) K = hf h, y K = 0. 0. (0.).65 (0.55) = 0.576 _ (8)

00 K4 = hf h, y K = (0.) 0. 0..65 0.576 and = 0.8 _ (9) K = (K +K +K +K4) 6 = 6 0.47 (0.55) (0.576) 0.8 {using equation (6), = 0.570 (7), (8) and (9)} Hence y(0.) = y = y + K =.65 + 0.570 =.75 which is required solution. Q.. Use Runge-Kutta method to solve y' = y for =.4. Initially =, y = (tale h = 0.). Ans.: (i) Here ƒ(, y) = y, 0 =, y0 =, h = 0. K = hƒ(0, y0) = 0.[ ] = 0.4 K = hƒ(0 + h, y0 + K ) = 0. 0. 0.4 [BCA Part II, 007]

Mathematical Methods for Numerical Analysis and Optimization 0 = 0. 0. 0. = 0... = 0.484 K = hƒ(0 + h, y0 + K ) = 0. = 0.494 0. 0.484 K4 = hƒ(0 +h, y0 +K) = 0. 0. 0.494 = 0.598776 K = (K +K +K +K4) 6 = 6 0.4 (0.484) (0.494) 0.598776 = 0.4949 y = y0 + K = + 0.4949 =.4949 (ii) = 0 + h = + 0. =., y =.4949 and h = 0. K = hƒ(, y) = 0.[(.) (.4949)] = 0.5984 K = hƒ( + h, y + K )

0 = 0. 0. 0.5984..49 y = y + K = 0.884 K = hƒ( + h, y + K ) = 0. = 0.7548 0. 0.884..49 K4 = hƒ(0 +h, y0 +K) = 0.. 0..49 0.754 = 0.90909 K = (K +K +K +K4) 6 = 0.775 =.49 + 0.775 =.675 y (.4) =.675

Mathematical Methods for Numerical Analysis and Optimization 0 Chapter-7 Boundary Valve Problem - Shooting Method Q. Solve the Boundary Value Problem y'' () = y () ; y(0) = 0 ; y() =.75 by the shooting method taking m0 = 0.7 and m = 0.8 Ans.: By Taylor s Series y() = y(0) + y'(0) + 4 y''(0) + y'''(0) + 6 4 y IV (0) + 6 + y VI (0) + 70 _ () Since y''() = y() we have y''() = y'() and y IV () = y''() = y() y V () = y'() y VI () = y''() = y(). Putting = 0 in above we get y''(0) = y(0) = 0, y'''(0) = y'(0) y IV (0) = 0, y''(0) = y'(0) Substituting these values in equation () 5 yv (0) 0 y() = y I (0) 5 7 9 6 0 5040 6800...

04 Since y(0) = 0 Hence y() = y'(0) 6 0 5040... _ () = y'(0)(.75) With y'(0) = m0 = 0.7 So equation () gives y() 0.86 Similarly y'(0) = m0 = 0.8 gives y() 0.940 Using linear interpolation, we obtain m = 0.7 + (0.) = 0.9998.75 0.86 0.940 0.86 Which is closer to eact value of y'(0) = with this value of m, we solve the initial value problem y''() = y(), y(0) = 0, y'(0) = m _ () and continue the process as above until the value of y() is obtained to the desired accuracy.