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1 Virtual University of Pakistan File Version v.0.0 Prepared For: Final Term Note: Use Table Of Content to view the Topics, In PDF(Portable Document Format) format, you can check Bookmarks menu Disclaimer: There might be some human errors, if you find please let me know at duplication of data may be possible but at least possible level Your Feed Back is Highly Appreciated. Compiled and Prepared by: File Version updated on 7/4/0

2 Page No. ----:Table of Content:---- Table of Content TABLE OF CONTENT... INTRODUCTION TO NUMERICAL ANALYSIS... 6 COURSE CONTENT:... 6 EACH TOPIC ONE SOLVED QUESTION WITH STEPS:COMPILED FROM PAST ASSIGNMENT... 6 PRELIMINARY MATERIAL:... 6 Binary Number System (Base )=0,... 6 Octal Number System (Base 8)=0,,,3,4,5,6, ERROR... 6 Decimal Number System (Base 0)=0,,,3,4,5,6,7,8, Hexa Number System (Base 6)=0,,,3,4,5,6,78,9,A,B,C,D,E,F... 6 The initial approximation... 6 NON-LINEAR EQUATIONS... 6 Bisection Method... 6 Regula Falsi Method... 0 Method of Iterations... 3 Newton-Raphson Method... 5 Secant Method... 6 Muller s Method... 9 Graeffe s Root Square Method... 0 LINEAR EQUATIONS... 0 Gaussian Elimination Method... 0 Guass-Jordan Elimination Method... 3 Jacobi s Iterative Method... 3 Gauss-Seidel Iteration Method... 5 Relaxation Method... 8 Matrix Inversion... 8 EIGEN VALUE PROBLEMS... 8 Power Method... 8 Jacobi s Method... 3 INTERPOLATION Forward Differences Backward Differences Divided Differences By Newton s Forward Difference Formula b) By Lagrange s Formula Langrange s Interpolation Differentiation Using Difference Operators NUMERICAL INTEGRATION... 5 Trapezoidal Rule... 5 Simpson s /3 and 3/8 rules DIFFERENTIAL EQUATIONS Taylor Series Method Euler Method Euler Modified Method: Runge-Kutta Method... 6

3 Page No. Milne s Predictor Corrector Method Adam Moultan s Predictor Corrector Method FAQ UPDATED VERSION QUESTION: WHAT IS BRACKETING METHOD? QUESTION: WHAT IS AN OPEN METHOD? QUESTION: EXPLAIN MULLER S METHOD BRIEFLY QUESTION: EXPLAIN THE DIFFERENCE BETWEEN THE LINEAR AND NON-LINEAR EQUATIONS QUESTION: EXPLAIN WHICH VALUE IS TO BE CHOOSED AS X0 IN N-R METHOD QUESTION: DEFINE ITERATIVE METHOD OF SOLVING LINEAR EQUATIONS WITH TWO EXAMPLES QUESTION: DEFINE PIVOTING QUESTION: WRITE THE TWO STEPS OF SOLVING THE LINEAR EQUATIONS USING GAUSSIAN ELIMINATION METHOD. 67 QUESTION: DESCRIBE GAUSS-JORDAN ELIMINATION METHOD BRIEFLY QUESTION: DESCRIBE BRIEFLY CROUT S REDUCTION METHOD QUESTION: DESCRIBE BRIEFLY THE JACOBI S METHOD OF SOLVING LINEAR EQUATIONS QUESTION: WHAT IS THE DIFFERENCE BETWEEN JACOBI S METHOD AND GAUSS SEIDAL METHOD? QUESTION: WHAT IS THE BASIC IDEA OF RELAXATION METHOD? QUESTION: HOW THE FAST CONVERGENCE IN THE RELAXATION METHOD IS ACHIEVED? QUESTION: WHICH MATRIX WILL HAVE AN INVERSE? QUESTION: WHAT ARE THE POPULAR METHODS AVAILABLE FOR FINDING THE INVERSE OF A MATRIX? QUESTION: EXPLAIN GAUSSIAN ELIMINATION METHOD FOR FINDING THE INVERSE OF A MATRIX QUESTION: WHAT ARE THE STEPS FOR FINDING THE LARGEST EIGEN VALUE BY POWER METHOD QUESTION: WHAT IS THE METHOD FOR FINDING THE EIGEN VALUE OF THE LEAST MAGNITUDE OF THE MATRIX [A]? 69 QUESTION: WHAT IS INTERPOLATION? QUESTION: WHAT IS EXTRAPOLATION? QUESTION: WHAT HAPPENS WHEN SHIFT OPERATOR E OPERATES ON THE FUNCTION QUESTION: WHAT IS THE BASIC CONDITION FOR THE DATA TO APPLY NEWTON S INTERPOLATION METHODS?.. 69 QUESTION: WHEN IS THE NEWTON S FORWARD DIFFERENCE INTERPOLATION FORMULA USED? QUESTION: WHEN IS THE NEWTON S FORWARD DIFFERENCE INTERPOLATION FORMULA USED? QUESTION: WHEN THE NEWTON S BACKWARD DIFFERENCE INTERPOLATION FORMULA IS USED? QUESTION: IF THE VALUES OF THE INDEPENDENT VARIABLE ARE NOT EQUALLY SPACED THEN WHICH FORMULA SHOULD BE USED FOR INTERPOLATION? QUESTION: TO USE NEWTON S DIVIDED DIFFERENCE INTERPOLATION FORMULA, WHAT SHOULD THE VALUES OF INDEPENDENT VARIABLES BE? QUESTION: WHICH DIFFERENCE FORMULA IS SYMMETRIC FUNCTION OF ITS ARGUMENTS? QUESTION: IS THE INTERPOLATING POLYNOMIAL FOUND BY LAGRANGE S AND NEWTON S DIVIDED DIFFERENCE FORMULAE IS SAME? QUESTION: WHICH FORMULA INVOLVES LESS NUMBER OF ARITHMETIC OPERATIONS? NEWTON OR LAGRANGE S? 70 QUESTION: WHEN DO WE NEED NUMERICAL METHODS FOR DIFFERENTIATION AND INTEGRATION? QUESTION: IF THE VALUE OF THE INDEPENDENT VARIABLE AT WHICH THE DERIVATIVE IS TO BE FOUND APPEARS AT THE BEGINNING OF THE TABLE OF VALUES, THEN WHICH FORMULA SHOULD BE USED? QUESTION: WHY WE NEED TO USE RICHARDSON S EXTRAPOLATION METHOD? QUESTION: TO APPLY SIMPSON S /3 RULE, WHAT SHOULD THE NUMBER OF INTERVALS BE? QUESTION: TO APPLY SIMPSON S 3/8 RULE, WHAT SHOULD THE NUMBER OF INTERVALS BE? QUESTION: WHAT IS THE ORDER OF GLOBAL ERROR IN SIMPSON S /3 RULE? QUESTION: WHAT IS THE ORDER OF GLOBAL ERROR IN TRAPEZOIDAL RULE? QUESTION: WHAT IS THE FORMULA FOR FINDING THE WIDTH OF THE INTERVAL?... 70

4 Page No.3 QUESTION: WHAT TYPE OF REGION DOES THE DOUBLE INTEGRATION GIVE? QUESTION: COMPARE THE ACCURACY OF ROMBERG S INTEGRATION METHOD TO TRAPEZOIDAL AND SIMPSON S RULE. 7 QUESTION: WHAT IS THE ORDER OF GLOBAL ERROR IN SIMPSON S 3/8 RULE?... 7 QUESTION: WHICH EQUATION MODELS THE RATE OF CHANGE OF ANY QUANTITY WITH RESPECT TO ANOTHER? 7 QUESTION: BY EMPLOYING WHICH FORMULA, ADAM-MOULTON P-C METHOD IS DERIVED?... 7 QUESTION: WHAT ARE THE COMMONLY USED NUMBER SYSTEMS IN COMPUTERS?... 7 QUESTION: IF A SYSTEM HAS THE BASE M, THEN HOW MANY DIFFERENT SYMBOLS ARE NEEDED TO REPRESENT AN ARBITRARY NUMBER? ALSO NAME THOSE SYMBOLS QUESTION: WHAT IS INHERENT ERROR AND GIVE ITS EXAMPLE QUESTION: WHAT IS LOCAL ROUND-OFF ERROR?... 7 QUESTION: WHAT IS MEANT BY LOCAL TRUNCATION ERROR?... 7 QUESTION: WHAT IS TRANSCENDENTAL EQUATION AND GIVE TWO EXAMPLES QUESTION: WHAT IS MEANT BY INTERMEDIATE VALUE PROPERTY?... 7 QUESTION: WHAT IS DIRECT METHODS OF SOLVING EQUATIONS?... 7 QUESTION: WHAT IS ITERATIVE METHOD OF SOLVING EQUATIONS?... 7 QUESTION: IF AN EQUATION IS A TRANSCENDENTAL, THEN IN WHICH MODE THE CALCULATIONS SHOULD BE DONE? 7 QUESTION: WHAT IS THE CONVERGENCE CRITERION IN METHOD OF ITERATION?... 7 QUESTION: WHEN WE STOP DOING ITERATIONS WHEN TOLL IS GIVEN?... 7 QUESTION: HOW THE VALUE OF H IS CALCULATED IN INTERPOLATION?... 7 QUESTION: WHAT IS AN ALGEBRAIC EQUATION?... 7 QUESTION: WHAT IS DESCARTES RULE OF SIGNS?... 7 QUESTION: WHAT ARE DIRECT METHODS?... 7 QUESTION: WHAT IS MEANT BY ITERATIVE METHODS?... 7 QUESTION: WHAT IS GRAPHICALLY MEANT BY THE ROOT OF THE EQUATION?... 7 QUESTION: Q. WHAT IS THE DIFFERENCE BETWEEN OPEN AND BRACKETING METHOD?... 7 QUESTION: CONDITION FOR THE EXISTENCE OF SOLUTION OF THE SYSTEM OF EQUATIONS QUESTION: SHOULD THE SYSTEM BE DIAGONALLY DOMINANT FOR GAUSS ELIMINATION METHOD? QUESTION: WHAT IS MEANT BY DIAGONALLY DOMINANT SYSTEM? QUESTION: STATE THE SUFFICIENT CONDITION FOR THE CONVERGENCE OF THE SYSTEM OF EQUATION BY ITERATIVE METHODS QUESTION: THE CALCULATION FOR NUMERICAL ANALYSIS SHOULD BE DONE IN DEGREE OR RADIANS QUESTION: HOW WE CAN IDENTIFY THAT NEWTON FORWARD OR BACKWARDS INTERPOLATION FORMULA IS TO BE USED. 73 QUESTION: WHAT IS MEANT PRECISION AND ACCURACY? QUESTION: WHAT IS THE CONDITION THAT A ROOT WILL LIE IN AN INTERVAL QUESTION: HOW THE DIVIDED DIFFERENCE TABLE IS CONSTRUCTED? QUESTION: WHAT IS GAUSS-SEIDEL METHOD QUESTION: WHAT IS PARTIAL AND FULL PIVOTING? QUESTION: HOW THE INITIAL VECTOR IS CHOOSE IN POWER METHOD? QUESTION: WHAT IS THE RELATION SHIP BETWEEN P=0 AND NON ZERO P IN INTERPOLATION QUESTION: WHAT IS CHOPPING AND ROUNDING OFF? QUESTION: WHEN THE FORWARD AND BACKWARD INTERPOLATION FORMULAE ARE USED? QUESTION: WHAT IS FORWARD AND BACKWARD DIFFERENCE OPERATOR AND THE CONSTRUCTION OF THEIR TABLE. 75 QUESTION: WHAT IS JACOBI S METHOD? QUESTION: WHAT IS SIMPSON'S 3/8TH RULE QUESTION: WHAT IS CLASSIC RUNGE-KUTTA METHOD QUESTION: WHAT IS MEANT BY TOL?... 76

5 Page No.4 QUESTION: WHAT IS MEANT BY UNIQUENESS OF LU METHOD QUESTION: HOW THE VALU OF H IS CALCULATED FROM EQUALLY SPACED DATA GLOSSARY (UPDATED VERSION) Absolute Error : Accuracy : Algebraic equation : Bisection method : Bracketing Method : Crout s Method : Direct methods : Gauss Elimination Method : Gauss Seidel iterative method : Graeffee s root squaring method : Intermediate value property : Inverse of a matrix : Iterative methods : Jacobie s iterative method: : Muller s Method : Newton Raphson Method. : Non singular matrix : Open methods : Pivoting : Precision : Regula Falsi method : Relaxation method : Secant Method : Significant digits : Singular matrix : Transcendental equation : Truncation Error : IMPORTANT FORMULA FOR MTH Bisection Method Muller Method Regula Falsi Method (Method of False position) Newton Rophson method Secant Method Newton s Formula Graffee root squaring method the truncation error (TE) is given by SHORT QUESTIONS PAPER SET SET SET Question: MULTIPLE CHOICE QUESTION SET

6 Page No.5 SET SET SET SET SET SET SET SET SET SET SET SET =========================================================>

7 Page No.6 Introduction To Numerical Analysis Course Content: Number systems, Errors in computation, Methods of solving non-linear equations, Solution of linear system of equations and matrix inversion, Eigen value problems, power method, Jacobi s method, Different techniques of interpolation, Numerical differentiation and integration, Numerical integration formulas, different methods of solving ordinary differential equations. Each Topic One Solved Question with Steps:Compiled from Past assignment Preliminary material: Representation of numbers Binary Number System (Base )=0, Used in Computer Internal Operations and Calculation Octal Number System (Base 8)=0,,,3,4,5,6,7 ERROR Types of error are, Inherent errors, Local round-off errors Local truncation errors Decimal Number System (Base 0)=0,,,3,4,5,6,7,8,9 World Wide used and the most common system Hexa Number System (Base 6)=0,,,3,4,5,6,78,9,A,B,C,D,E,F Used in Processor Register Addressing Errors in computations The initial approximation It may be found by two methods either by graphical method or Analytical method Graphical method Non-Linear Equations Bisection Method Procedure in Detail: Step,Take two Initial Approximation such that f(x).f(x)<0.means both must have opposite signs.take their mean by x3=(x+x)/ Next Take two element from which will be x3 and another from x or x such that both x3 and the other element should have opposite sign. Repeat the above process to the required numbers of iterations. Question Find the root of the equation given below by bisection method. 3 x x x 7 0

8 Page No.7 (Note: accuracy up to three decimal places is required.) Marks: 0 SOLUTION Let 3 f ( x) x x x 7 3 Now f () () () and f 3 () () () 7 0 f 3 (3) (3) (3) As f () f (3) 4 0 Therefore a real root lies between Iteration Let x and x 3 0 then x0 x x ( ) 3 ( ).5 Now f and 3 3 (.5) (.5) (.5) As f () f (.5) (4.875) Therefore a real root lies between and.5 Iteration x and x.5 0 then x0 x x3 ( ).5 ( ).5 Now f 3 (.5) (.5) (.5) As f () f (.5) (.5785) Therefore a real root lies between and.5 Iteration 3

9 Page No.8 x and x then x0 x3 x4 ( ).5 ( ).5 Now f 3 (.5) (.5) (.5) As f () f (.5) ( ) Therefore a real root lies between and.5 Iteration 4 x and x then x0 x4 x5 ( ).5 ( ).065 Now f 3 (.065) (.065) (.065) As f (.065) f (.5) ( ) Therefore a real root lies between.065 and.5 Iteration 5 x.065and x then x5 x4 x6 ( ) ( ) Now f 3 (.09375) (.09375) (.09375) As f (.09375) f (.5) 0.48( ) Therefore a real root lies between Iteration 6 x.09375and x then x6 x4 x7 ( ) ( ) Now f and.5 3 (.09375) (.09375) (.09375) As f (.09375) f (.09375) 0.48(0.0455) Therefore a real root lies between and Iteration 7

10 Page No.9 x.09375and x then x6 x7 x8 ( ) ( ).0565 Now f 3 (.0565) (.0565) (.0565) As f (.0565) f (.09375) (0.0455) Therefore a real root lies between Iteration 8 x.0565and x then x8 x7 x9 ( ) ( ) Now f.0565 and ( ) ( ) ( As f (.0565) f ( ) (0.006) Therefore a real root lies between.0565 and Iteration 9 x.0565and x then x8 x9 x0 ( ) ( ) Now f 3 ( ) ( ) ( ) As f ( ) f ( ) (0.006) Therefore a real root lies between and Iteration 0 x and x then x0 x9 x ( ) ( ) Now f 3 ( ) ( ) ( ) As f ( ) f ( ) (0.006) Therefore a real root lies between and Iteration

11 Page No.0 x and x then x x9 x ( ) ( ) Now f 3 ( ) ( ) ( ) As f ( ) f ( ) (0.00) Therefore a real root lies between and As the next root lies between and and these roots are equal up to three decimal places. So, the required root up to three decimal places is.04 Example From Handout at page # 4 Solve x 3 9x+=0 for the root between x= and x=4 by bisection method Solution: Here we are given the interval (,4) so we need not to carry out intermediate value property to locate initial approximation. 3 Here f(x) =x 9x 0 Now f()=-9 and f(4)=9. Here f().f(4)<0 So the root lies between and 4. So, x, x 4 o xo x and x =3 (This formula predicts next iteration) Now f(3)=,here f().f(3)<0 So the root lies between and 3. Repeat above process to required number of iteration. Step,Take two Initial Approximation such that f(x).f(x)<0.means both must have opposite signs.take their mean by x3=(x+x)/ Next Take two element from which will be x3 and another from x or x such that both x3 and the other element should have opposite sign. Repeat the above process to the required numbers of iterations. Regula Falsi Method Formula for the Regula Falsi Method is x x x3 x f ( x) f ( x) f ( x) x x x x f x n n ( n) n n f xn f xn Steps: If the interval is given, check whether the root lies in between the interval or not by the principle that states if both number have opposite sign, then the root lies in between interval. Find the next approximation with the help of the formula. or

12 Page No. Question# Marks 0 x Use Regula-Fasli method to compute the root of the equation f ( x) cos x xe In the interval [0, ] after third iteration. Solution: x As f ( x) cosx xe f f 0 (0) cos 0 0e 0 () cose So root of the eq. lie between 0 and. Let x 0 and x Thus f ( x ) and f ( x ) The formula for finding the root of the function f(x) by Regula-Fasli is given by. xn xn x( n) xn f xn f x f x n n For n= we have, x xo x x f x f x f x o f x cos(0.3467) e = Since f (x) and f (x) are of opposite signs, the root lies between x and x. Therefore, for n= we have, x x x3 x f x f x f x f x cos( ) e = Since f (x) and f (x3) are of opposite signs, the root lies between x and x3. Therefore, for n=3 we have, x3x x4 x3 f x f x f x The required root after 3 rd iteration using Regula-Falsi method is Question 3 Use the Regula Falsi (method of false position) to solve the equation x 4x9 0 (Note: accuracy up to four decimal places is required) Marks: 0 SOLUTION

13 Page No. Let f x x x 3 ( ) 4 9 Now f f f f 3 (0) (0) 4(0) () () 4() () () 4() (3) (3) 4(3) Now, Since f () and f (3) are of opposite signs, therefore, the real root lies between and 3 Now, let x and x 3 First iteration x x x3 x f ( x) f ( x ) f ( x ) 3 3 (6) 6 ( 9) 3 (6) f x 3 ( 3) f (.6) (.6) 4(.6) 9.84 Second iteration x3 x x4 x3 f ( x3) f ( x ) f ( x ) (.84) f x 3 ( 4) f (.6935) (.6935) 4(.6935) Third iteration

14 Page No.3 x x x x f ( x ) f ( x4) f ( x3) (.84) f x 3 ( 5) f (.70536) (.70536) 4(.70536) Fourth iteration x5 x4 x6 x5 f ( x5 ) f ( x ) f ( x ) ( ) ( 0.375) f x 3 ( 6) f (.70653) (.70653) 4(.70653) Fifth iteration x x x x f ( x ) f ( x6) f ( x5) ( ) ( ) Hence, The required root of the given equation correct to 4 decimal places is Method of Iterations Question# Marks 0 x Use the method of iterations to determine the real root of the equation e 0x in the interval [0, ], correct to four decimal places after four Iterations. Solution: x x Since e 0 x. Let f ( x) e 0 x. we see that f (0) and f () 9.63 x As e 0 x, we can easily find the value of x, thus x x e e x, Let x 0 0 By taking its derivative we get x e x, we see that x for all values in [0,]. 0 Therefore we can apply method of iterations to the given function.

15 Page No.4 Take any value within [0,]. Let x e x , f ( x) e x , f ( x) e x , f ( x3) x e 5 x , f ( x4) x0 0 x Example From the Handout at page # 4

16 Page No.5 Newton-Raphson Method Procedural Detail: Find the limit if not provided by starting from x=0 to, Find two consecutive numbers for f(x) should have opposite sign. Use the Newton Raphson Formula to find next approximation. Question#3 Marks 0 Find the real root of the equation interval [,3] after third iteration. Solution: x 3 3x5 0 using Newton-Raphson method in the As f x x x f f 3 ( ) () (3) So root of the eq. will lie in,3 / // now f ( x) 3x 3 and f ( x) 6x f f () and f (3) / / () 6. and f (3) // // // Since f (3) and f (3) are of same sign. So we c so by Newton ' s method we have x f( x ) x0 / f ( x0 ) 4 f x 3 ( ) (.4583) 3(.4583) / / x.4583 / f x 0 3 f ( x ) f (.4583) 3(.4583) x f x f( x ) ( ) 5.97 hoose x 3 ( ) (.943) 3(.943) f ( x ) f (.943) 3(.943) x / / f( x ) x / f ( x).794 Example From the Handout at page # Value of e=.78

17 Page No.6 Secant Method Question Do Four iterations of Secant method, with an accuracy of 3 decimal places to find the. root of 3 f ( x) x 3x 0, x0, x 0.5 Marks: 0 Solution: FORMULA OF SECANT METHOD x ( ) ( ) n xn fxn xn f xn ITERATION n 0 f ( x ) f o ( fx ) f ( x ) n f ( x ) f x x ( fx ) x f ( x ) x x x x x ( fx ) f ( x ) ()( 0.375) (0.5)( ) ( 0.375) ( ) ()( 0.375) (0.5)() ( 0.375) () ( 0.375) (0.5) o o n ITERATION

18 Page No.7 n f( x ) f x x ( fx ) x f ( x ) x 3 3 x x 3 3 x 3 ( fx ) f ( x ) (0.5)(0.408) (0.)( 0.375) (0.408) ( 0.375) (0.04) (0.075) (0.408) (0.375) ITERATION 3 n 3 3 f( x ) f x x f ( x ) x f ( x ) f ( x ) f ( x ) 3 x (0.) f (0.3563) (0.3563) f (0.408) x x x 4 x f(0.3563) f(0.408) (0.)( ) (0.3563)(0.408) ( ) (0.408) ITERATION 4

19 Page No.8 n 4 3 f ( x ) f x f ( x ) x x f ( x ) x f ( x ) x x x x x f ( x ) f ( x ) 4 3 (0.3563)( ) (0.3563)( ) ( ) ( ) f ( x ) Hence, the root after four iterations is ================================================================ Question Use the secant method to solve the equation Solution x e 3x for0x. (Perform only 3 iterations.) f x e x x x x ( ) 3, 0 0, both aretheinitialapproximations f f 0 (0) e 3(0) () e 3() 0.8 nowwe calculatethesec ondapproximation x x f ( x ) x f ( x ) (0)( 0.8) () f ( x0) f ( x) 0.8 f (0.7806) e nowx and x f ( x) f ( x) (0.7806) f() 0.8 f(0.786) x now x x x x f ( x ) x f ( x ) (0.3546) (0.786)( 0.8) 0.905f(0. 905) f(0.786) x f ( x ) x f ( x ) (0.0074) (0.905)(0.3546) f ( x3) f ( x) sorootis

20 Page No.9 Muller s Method Question Solve the equation 3 x x x by using Muller s method only perform three Iterations.( x0 0.5, x, x 0 ) Solution st iteration x0 0.5, x and x 0 f x f f 3 ( 0) 0 (0.5) (0.5) 7(0.5) 4(0.5) f x f f 3 ( ) () 7() 4() 6 f x f f 3 ( ) (0) (0) 7(0) 4(0) 6 6 c f h x x h x0 x h f ( h h ) f h f a h h ( h h ) 0 (0.5)() ( )( 0.65) (0.5)( 6) ()( 0.65) (0.5)( 6) a (0.5)(0.5)( ) (0.5)(0.5)() = 5.5 b f f ah h 0 ( 0.65) ( 5.5)(0.5) 0.65) (5.5)(0.5) b c x x0 b b 4ac 0.5 ( 0.65) x 8 (8) (4)( 5.5)( 0.65) x.87 nd iteration x0 0.5, x.87 and x f x f f 3 ( 0) 0 (0.5) (0.5) 7(0.5) 4(0.5) f x f f 3 ( ) ().87 7(.87) 4(.87) 6.8

21 Page No.0 f x f f 3 ( ) (0) (0) 7(0) 4(0) 6 6 c f h x x h x0 x h f ( h h ) f h f a h h ( h h ) 0 ( 0.5)(.8) ( )( 0.65) (.37)( 6) a 4.70 (.37)( 0.5)( ) b f f ah h 0.8 ( 0.65) (4.70)(.37) b c x x0 b b 4ac x x.87 For 3rd iteration x0 0.9, x.87, x.6 (7.7) (4)(4.7)( 0.65) For third iteration we will proceed in the same manner. Graeffe s Root Square Method Linear Equations Gaussian Elimination Method Question Using Gaussian Elimination Method, solve the following system of equations x x x 9 Solution: 3 x 3x x 3 3 3x x 3x 4 3 The Augmented Matrix of the given system of equations is Marks: 0

22 Page No. A b R by R R and R3 3R R 0 0 by R R 0 0 by R 3 ( ) R R 0 0 by R Which shows that from the third and second rows Z=3, y = And from the first row X+y+z=9 Using the values of y and z, we get x = Hence the solution of the given system is X =, y =, z = 3 Question Using Gaussian Elimination Method, solve the following system of equations x y z Marks: 0 x 0y 3z 5 x y z 3 Solution: The Augmented Matrix of the given system of equations is

23 Page No. Ab R by R R by R R, R3 R R by R 0 R by R R 3 Using Gaussian Elimination method, by backward substitution, we get as follows From the third row, we get z z From the second row, we get 8 8 y z 8 8 y z Putting thevalueof z, we get 8 8 y ( ) And finally from the first row, we get x 0y 3z 5 Putting the values of y and z, we get x 0(0) 3( ) 5 x 3 5 x

24 Page No.3 So, the solution is x, y 0, z Guass-Jordan Elimination Method Question# Marks 0 By using Gauss Jordan elimination method, solve the following system of equations, x y z 7 3x 3y 4z 4 x y 3z 6 Solution: The given system in matrix form is x y 4 3 z 6 A X B : 7 A B : 4 3 : 6 : 7 3 : 6 R : 4 : 7 0 : R R R, R3 R3 3R 0 0 : 3 0 : 9 0 : R R R 0 0 : : : R R R3, R R R3 0 0 : : : R 0 0 : 3 So, x 3, y, z 3 Jacobi s Iterative Method

25 Page No.4 Solve the following system of equations 0x y z 7 3x 0y z 8 x 3y 0z 5 By Jacobi s iterative method taking the initial starting of solution vector as (0,0,0) T and perform the first three iterations. Solution: startingwith (0,0,0) Iteration#0 Iteration#0 x 0.85, y 0.9, z.5 Iteration#03 x.0, y 0.965, z.03 0x y z 7 3x 0y z 8 x 3y 0z 5 7 yz x 0 8 3x z y 0 5 x3y z x y z x y z

26 Page No.5 Gauss-Seidel Iteration Method x y z Question#3 Marks 0 Solve Question No. # by Gauss-Seidel iterative method and perform first three iterations. What you see the difference after solving the same question by two different iterative methods? Give your comments. Solution: The above system of linear equations is diagonally dominant; therefore, Gauss-Seidel iterative method could be applied to find out real roots 0x y z 7 3x 0y z 8 x 3y 0z 5 The above system of equations could be written in the form startingwith (0,0,0) Iteration#0 7 yz x 0 8 3x z y 0 5 x3y z x y z Iteration#0 x 0.85, y.075, z.00875

27 Page No.6 Iteration# x y z x.00465, y , z x y z In Gauss Siedal Method, the newly computed values in each iteration are directly involved to find the other value of the system of equation and save memory for computation and hence results are more accurate. ============================================================== Question Do five iterations to solve the following system of equations by Gauss-Seidal iterative method 0x y 3z 305 x 0y z 54 x y 0z 0 Marks: 0 Solution: Since, the given system is diagonally dominant; hence we can apply here the Gauss- Seidal method. From the given system of equations r 305 r 3 r x 0 y z r 54 r r y x z 0 r 0 r r z x y 0 ITERATION For r =0 Taking y=z=0 on right hand side of first equation. In second equation we take z=0 and current value of x. In third equation we take current value of both x and y.

28 Page No x 305 (0) 3(0) y 54 (30.5) (0) z 0 (30.5) (.5) ITERATION Similar procedure as used in Iteration will be used for Iterations, 3, 4 and x 305 (.5) 3(0.5) y 54 (40.875) (0.5) z 0 (40.875) (7.65) ITERATION x 305 (7.65) 3(.938) y 54 (4.906) (.938) z 0 (4.906) (8.569) ITERATION x 305 (8.569) 3(3.438) y 54 (43.45) (3.438) z 0 (43.45) (8.737) ITERATION x 305 (8.737) 3(3.53) y 54 (43.304) (3.53) z 0 (43.304) (8.765) Above Results are summarized in tabular form as Variables Iterations x y z

29 Page No Hence the solution of the given system of equations, after five iterations, is x y z Relaxation Method Matrix Inversion Eigen Value Problems Power Method Question Find the largest eigen value of the matrix MARKS 0 And the corresponding eigenvector, by Power Method after fourth iteration starting with the initial vector v (0) =(0,0,) T SOLUTION 4 0 Let A = Choosing an initial vector as ITERATION v (0) =(0,0,) T then () (0) u [ A] v Now we normalize the resultant vector to get 0 () () u 4 qv 4

30 Page No.9 Continuing this procedure for subsequent Iterations, we have ITERATION u u [ A] v q v 3 () () () () ITERATION (3) () u [ A] v (3) (3) u q3v 0.9 ITERATION (4) (3) u [ A] v (4) (4) u 0.85 q4v 0.06 Therefore, the largest eigen value and the corresponding eigen vector accurate to three decimals places are ( X ) 0.06 ======================================================================= Question marks 0

31 Page No.30 Find the largest eigen value of the matrix And the corresponding eigen vector, by Power Method after fourth iteration starting with the initial vector v (0) =(0,,0) T SOLUTION Let A = Choosing an initial vector v (0) =(0,,0) T then ITERATION () (0) u [ A] v Now we normalize the resultant vector to get 0.5 () () u 0 0. qv Continuing this procedure for subsequent Iterations, we have ITERATION u u [ A] v q v () () () () ITERATION 3

32 Page No (3) () u [ A] v (3) (3) u q3v ITERATION (4) (3) u [ A] v (4) (4) u q4v Therefore, the largest eigen value and the corresponding eigen vector accurate to four decimals places are 0.73 and ( X ) Jacobi s Method Question Using Jacobi s method, find all the eigenvalues and the corresponding eigenvectors of the following matrix, Note: Give results at the end of third rotation. Solution. Let A t Then, A Marks: 0

33 Page No.3 Hence, Matrix A is real and Symmetric and Jacobi s method can be applied. Rotation In Matrix A, all the off-diagonal elements are found be. So, the largest off-diagonal element is found to be a a a a a a Therefore, we can choose any one of them as the largest element. Suppose, we choose a as the largest element Then, we compute the rotation angle as, tan a a a Therefore tan π θ= 4 Therefore, we construct an orthogonal matrix S such that, cos 4 sin 4 0 S sin 4 cos Now the first rotation gives, D S AS

34 Page No.33 Here, So, S = D S AS i.e. 0 0 D To check that we are right in our calculations, we can see that the sum of the diagonal elements is, 3 3, which is same as the sum of the diagonal elements of the original matrix A. Rotation For the second rotation we choose the largest off-diagonal element d3 d3.88, then d d d 3 3 tan.88 So, 33 tan Therefore, we construct an orthogonal matrix S such that,

35 Page No.34 cos sin S 0 0 sin cos S = Now the rotation gives, D S D S D Again to check that we are right in our calculations, we can see that the sum of the diagonal elements is 5 3 which is same as the sum of the diagonal elements of the original matrix A. Rotation 3 We can see that in above iteration that D is a diagonal matrix, so we stop here and take the Eigen Values as λ = 5, λ = -, λ = - 3 Now the Eigenvectors are the columns vectors of the matrix S SS, which are,

36 Page No S Therefore, the corresponding EigenVectors are X X, X, Interpolation For a given table of values (, ), 0,,,..., k k x y k= nwith equally spaced abscissas of a function y= f(x),we define the forward difference operator Δ as follows, These differences are called first differences of the function y and are denoted by the symbol Δyi Here, Δ is called the first difference operator. Similarly, rth Difference operator would be Leading term =yo Leading Diffenrence= y Backward Difference Operators: Central Difference is given by,

37 Page No.36 Shift operator, E Let y = f (x) be a function of x, and let x takes the consecutive values x, x + h, x + h, etc. We then define an operator having the property E f( x) = f(x+h) Thus, when E operates on f (x), the result is the next value of the function. Here, E is called the shift operator. If we apply the operator E twice on f (x), we get Thus, in general, if we apply the operator E n times on f (x), we get E n f(x)= f (x+nh)

38 Page No.37 Newton Forward Difference Interpolation. Forward Differences Question Construct a forward difference table from the following values of x and y. x y=f(x) Solution. Forward-difference table Marks: x y y y 3 y 4 y 5 y 6 y.0.5

39 Page No Backward Differences Question Construct a backward difference table from the following values of x and y. x y=f(x) Solution. Backward-difference table Marks: x y y y 3 y 4 y 5 y 6 y

40 Page No ===================================================================== Question 0 marks Form a table of forward and backward differences of the function 3 f ( x) x 3x 5x 7 For x,0,,,3,4,5 SOLUTION For the given function, the values of y for the given values of x are calculated as for x f 3 ( ) ( ) 3( ) 5( ) 7 6 for x 0 f 3 (0) (0) 3(0) 5(0) 7 7 for x f 3 () () 3() 5() 7 4 for x f 3 () () 3() 5() 7 for x 3 f 3 (3) (3) 3(3) 5(3) 7 for x 4 f 3 (4) (4) 3(4) 5(4) 7 for x 5 f 3 (5) (5) 3(5) 5(5) 7 8 So, the table of values of x and y is x y f ( x) Forward difference table Forward difference table for the table of values of x and y f ( x) is shown below x y y y 3 y 4 y 5 y 6 y

41 Page No Backward difference table Backward difference table for the table of values of x and y f ( x) is shown below x y y y y y y y Divided Differences Question marks 0 For the following table of values, estimate f (.5) using Newton s forward difference interpolation formula. x y f ( x)

42 Page No.4 SOLUTION Forward difference table Forward difference table for the given values of x and y is shown below x y y y 3 y 4 y 5 y 6 y Newton s forward difference interpolation formula is given by p( p ) p( p )( p ) 3 yx y0 py0 y0 y0! 3! p( p )( p )( p 3) p( p )( p )...( p n ) 4! n! 4 n y0 y0 Here x x0.5.5 p.5 h And y, y 7, y, y 6, y y y So, by putting the above values in Newton s forward difference interpolation formula, We have

43 Page No.4 p( p ) p( p )( p ) 3 yx y0 py0 y0 y0! 3!.5(.5 ).5(.5 )(.5 ).5(7) () (6)! 3! i.e. y =========================================================== Question marks 0 Compute f (.5) for the following data by using Newton s divided difference interpolation formula. x f( x ) SOLUTION The divided difference table for the given data is given by x y st D.D. nd D.D. 3 rd D.D. 4 th D.D / /3-9/6 7 -/ /3 47/3 8 9

44 Page No.43 Newton' s Divided Difference formula is f ( x) y ( x x ) y[ x, x ] ( x x )( x x ) y[ x, x, x ] ( x x )( x x )( x x ) y[ x, x, x, x ] ( x x )( x x )( x x )( x x ) y[ x, x, x, x, x ] f (.5) 5 + (.5 - ) 9 + (.5-)(.5-)(- ) + (.5-) (.5-)(.5-4) 3 9 +(.5-) (.5-)(.5-4)(.5-5)(- ) 6 = =0.795 Question 3 0 marks Find y / (6) and y // (6) from the following table of values. x y SOLUTION Backward difference table Backward difference table for the given values of x and y is shown below x y y 3 y y y y By backward difference formula for first derivative, we have

45 Page No.44 y y y y n yn h n n n y y y y y n yn h y (6) n n n n By backward difference formula for second derivative, we have y n yn yn yn yn h 6 Therefore y // 5 (6) [ ( ) ( ) ( 3)] Question# Marks 0 Find an equation of a cubic curve which passes through the points (4,-43), (7, 83), (9,37) and (, 053) using Divided Difference Formula. Solution: Newton s divided difference table is X Y st D.D nd D.D 3 rd D.D

46 Page No.45 Newton' s Divided Difference formula is y f ( x) y ( x x ) y[ x, x ] ( x x )( x x ) y[ x, x, x ] ( x x )( x x )...( x x ) y[ x, x,... x ] 0 n 0 now putting values in formula y f ( x) 43 ( x 4)(4) ( x 4)( x 7)(6) ( x 4)( x 7)( x 9)() 43 ( x 4){4 6x x 6x 63} 43 ( x 4)( x 7) 3 43 x 7x 4x 8 3 x x x Henceit is required polinomial n Question marks 0 Form a table of forward and backward differences of the function 3 f ( x) x 3x 5x 7 For x,0,,,3,4,5 SOLUTION For the given function, the values of y for the given values of x are calculated as for x f 3 ( ) ( ) 3( ) 5( ) 7 6 for x 0 f 3 (0) (0) 3(0) 5(0) 7 7 for x f 3 () () 3() 5() 7 4 for x f 3 () () 3() 5() 7 for x 3 f 3 (3) (3) 3(3) 5(3) 7 for x 4 f 3 (4) (4) 3(4) 5(4) 7 for x 5 f 3 (5) (5) 3(5) 5(5) 7 8 So, the table of values of x and y is x y f ( x) Forward difference table Forward difference table for the table of values of x and y f ( x) is shown below

47 Page No.46 x y y y 3 y 4 y 5 y 6 y Backward difference table Backward difference table for the table of values of x and y f ( x) is shown below x y y y 3 y 4 y y 6 y

48 Page No Question Marks 0 Find the interpolating polynomial for the following data by a) Newton s Forward Difference Formula b)lagrange s Formula Hence show that both the methods give raise to the same polynomial. Solution: x 0 3 y By Newton s Forward Difference Formula The Newton s Forward difference table is given as x y y y 3 y We have Newton s forward difference interpolation formula as p( p ) p( p )( p ) 3 y y0 py0 y0 y0... ()! 3! Since, the third and higher order differences are zero; Therefore, Newton s forward difference Interpolation formula reduces to pp ( ) y y0 py0 y0 ()! Here, x 0 =0, y 0 =, y0 =, And p is given by y 0 =, h =

49 Page No.48 x x0 x 0 p x h Substituting these values in Eq.(), we have xx ( ) y x() () x x( x ) x x x x x Hence, the required polynomial is y x x. b) By Lagrange s Formula Lagrange s Interpolation formula is given by ( x x )( x x )( x x ) ( x x )( x x )( x x ) y y y ( x x )( x x )( x x ) ( x x )( x x )( x x ) ( x x )( x x )( x x ) ( x x )( x x )( x x ) y y ( x x )( x x )( x x ) ( x x )( x x )( x x ) By putting the values in the formula, we get ( x )( x )( x 3) ( x 0)( x )( x 3) y () (3) (0 )(0 )(0 3) ( 0)( )( 3) ( x 0)( x )( x 3) ( x 0)( x )( x ) (7) (3) ( 0)( )( 3) (3 0)(3 )(3 ) y 3 3 x 6x x 6 3x 5x 8x ( )( )( 3) ()( )( ) ( x 4x 3 x) 3( x 3x x) ()()( ) (3)()() y 3 3 x 6x x 6 3x 5x 8x + 6 x x x x x x x 6x x 6 3x 5x 8x 7x 8x x 3x 39x 6x y x x x x x x x x x x x x

50 Page No.49 y x x x x x x x x x x x x 6x 6x6 6 6( x x) 6 x x Thus we have the equation y x x This is the same polynomial as obtained in Newton s Forward difference interpolating formula. Hence it is proved that both the methods give rise to the same polynomial Langrange s Interpolation Find interpolation polynomial by Lagrange s Formula, with the help of following table, x 3 5 f(x) And hence find value of f (6). Solution:

51 Page No.50 Lagrange' s int erpolation formula is ( x x )( x x )( x x ) ( x x )( x x )( x x ) y f ( x) f ( x ) f ( x ) ( x x )( x x )( x x ) ( x x )( x x )( x x ) ( x x )( x x )( x x ) ( x x )( x x )( x x ) f( x ) ( x x )( x x )( x x ) ( x x )( )( ) x3 x x3 x x x 3 x 5 x x 3x 5 x x x x x x x x x x x x x x x f( x ) y f ( x) x 9x 3x 5 3 x 8x 7x 0 3 x 6x x x x x x x x x x x x x x x x x x 34 x x 6 3 ( ) 3 (6) x 3 (6) 3 6 x y f x x nowwe haveto find f (6) so f 6 5 Differentiation Using Difference Operators Question#3 Marks 0 Find / (.5) / / f and f (.5) from the following table, Solution: x f(x) Since x=.5 occur at the end of the table, it is appropriate to use the backward difference formula for the derivation.

52 Page No.5 Backward difference table is as x f( x) y y 3 y 4 y u sin g thebackward difference formula for y( x) and y( x), we have 3 4 yn yn y n y n yn h 3 4 now putting value frombackward diff. table we have 0. ( ) (.4) y y y y y n yn yn yn h y.5 0. (.4) (0.3) y (0.09) y.5 3 (0.09) y Numerical Integration Trapezoidal Rule Question Marks 0

53 Page No.5 x e dx Evaluate the integral 0 Using (I) Trapezoidal rule () Simpson s /3 rule By dividing the interval of integration into eight equal parts. Also calculate the percentage error from its true values in both cases. Solution: a) By Trapezoidal Rule Since the no. of intervals are eight, so, n =8 And a0, b So, the width of the interval is given by b a h n Since the integrand is given by x y f ( x) e, Therefore, the table of values will be x y f ( x) Now applying the Trapezoidal rule, we have 0 x h e dx y y y y y y y y y Therefore x e dx 3. 0

54 Page No.53 b) By Simpson s Rule Applying Simpson s rule, we have 0 x h e dx y 4 y y y y y y y y Therefore x e dx True value Bu integration, we have x x e e dx 0 0 x e 0 0 e e Therefore the true value is x e dx 0 Percentage error in Trapezoidal rule Exact value Computed value Error = So, Error

55 Page No.54 And % error is given by error % error 00% exact value % % Percentage error in Simpson s /3 rule Exact value Computed value Error = Error And % error is given by error % error 00% exact value % % Question # Marks 0 Evaluate the following double integral, dxdy 3 3 x y By using trapezoidal rule, with h = k = 0.5. Solution: Taking X =,.5,.50,.75,.00 and y =,.5,.50,.75,.00, the following table is obtained using the integrand f ( x, y) 3 3 x y X, Y Now using trapezoidal, keeping variable x fixed and changing the variable y. 0.5 f (, y) dy [0.5 ( ) 0.]

56 Page No.55 = f (.5, y) dy [ ( ) 0.005] = f (.50, y) dy [0.86 ( ) ] = f (.75, y) dy [0.57 ( ) ] = f (.00, y) dy [0. ( ) 0.065] = h dxdy [ f (, y) { f (.5, y) f (.50, y) f (.75, y)} f (.00, y)] 3 3 x y 0.5 [575 ( ) ] = Simpson s /3 and 3/8 rules A Simpson s /3 Rule: Question: Evaluate the following integral, x x.8 e e dx.0 Using Simpson s /3 rule by taking h=0. and e=.783 Solution: First of all table of the function is created, x x x e e y f ( x) Simpson ' s rdrule for the given function is h f f f f f 0. I I I Question marks 0

57 Page No.56 A river is 80 feet wide. Depth d in feet at a distance of x feet from one bank is given by the following table. x Find d approximately the area of the cross-section by Simpson s /3 rule. SOLUTION Let y d, depth in feet, then the table is given by x Here y d h 0 Let A be the area of the cross-section, then the area of the cross-section is given by 80 A ydx 0 And the area of cross-section by Simpson s /3 rule is given by 80 h A ydx [ y0 4( y y3 y5) ( y y4 y6) y8] [0 4( ) (7 4) 3] 3 0 [0 4(36) (33) 3] 3 0 [ ] 3 0 (3) 3 70 sq. ft A Simpson s 3/8 Rule: Use Simpson s 3/8 rule to estimate the number of square feet of hand in given lots when x and y are measured in feet (Use accuracy up to three places of decimal) x y Solution Simpson s 3/8 th rule may be written as

58 Page No.57 3 h I [ f0 3( f f ) f3 3( f4 f5 ) f6 3( f7 f8 ) f9 f0 ] 8 I I I I 3(00) [5 3(5 0) () 3(90 90) (95) 3(88 75) (35) 0] 8 3(00) [5 3(45) () 3(80) (95) 3(63) (35)] 8 (37.5)[ ] (37.5)[375] Differential Equations Taylor Series Method Question marks 0 Apply Taylor s Series algorithm of order on the initial value problem in the interval [0,0.]. Sol. / As y x y Since x 0and y so y x y 0 / / / / Now y x y y (0) ()() As Taylor's Series method algorithm of order is / y x y ; y(0) ; h 0.05 ( t t0) y( t) y( t0) ( t t0) y( t0) y( t0)! h y( t) y( t0) hy( t0) y( t0)! (.05) y(. 05) y.05() ().055! / Now we find y, so y x y (.05) (.055) // Now y x y y (.05) (.055)(.05) h Now y( t) y y( t) hy( t) y( t)! y(0.) y (.05).055 (.05)(.05) (.437)! Question marks 0 / Given y 3x y and y(0), find by Taylor s series y(0.) and y(0.) SOLUTION First few derivatives from the given differential equation are as follows taking h 0..

59 Page No.58 / y 3x y y 6x y y // / 6y /// // ( iv) /// y y The initial condition is given as x 0, y 0 0 So, using the given initial condition, we have y 3x y 3(0) () 4 / y 6x y 6(0) (4) 8 y y // / y 6 (8) /// // 0 0 y () 44 ( iv) /// 0 0 ( v) ( iv) y0 y0 (44) 88 Now, using Taylor s series method up to the fifth term, we have 3 ( x x0) ( x x0) y( x) y0 ( x x0) y 0 y 0 y ( x x0) IV ( x x0) V y0 y0 4 0 Substituting the above values of the derivatives, and the initial condition, we obtain 3 (0.0) (0.0) y(0.) (0. 00)(4) (8) () (0.0) (0.0) (44) (88) 4 0 y(0.) y(0.) y(0.).4439 So, Now, we have x 0., y.4439 Using the above condition derivatives are calculated as follows y 3x y 3(0.) (.4439) / y 6x y 6(0.) (4.978) y y y // / 6 y 6 (0.4356) 6.87 /// // y (6.87) ( iv) /// y (53.744) ( v) ( iv)

60 Page No.59 Substituting the value of y and its derivatives into Taylor s series expansion we get, after retaining terms up to fifth derivative only, we have 3 ( x x) ( x x) y(0.) y ( x x ) y y y ( x x) IV ( x x) V y y (0. 0.) (0. 0.) y(0.).4439 (0. 0.)(4.978) (0.4356) (6.87) (0. 0.) (0. 0.) (53.744) ( ) 4 0 y(0.) So, y(0.).999 Euler Method Question marks 0 Use Euler s method to approximate y when x=, given that dy y x, y (0), taking h 0. dx y x Sol. dy y x, y (0), taking h 0. dx y x herey y hf ( x, y ) m m m m y x f ( x, y) y x x 0, y 0 0 y y hf ( x, y ) (0.)

61 Page No.60 x 0., y. y y hf ( x, y ). 0.. (0.). 0. x 0.4, y.3484 y y hf ( x, y ) (0.) x 0.6, y y y hf ( x, y ) (0.) x 0.8, y y y hf ( x, y ) (0.) x, y y y hf ( x, y ) (0.) Euler Modified Method: Question marks 0 / Given that y log( x y) with y(0), Use Modified Euler s method to find y (0.), taking h 0. SOLUTION The given differential equation is dy / y log( x y), y 0 dx, x 0, y So, here 0 0, h 0. At first, we use Euler s method to get

62 Page No.6 y y hf x, y y hlog( x y ) [log(0 )] 0. log() 0.(0) 0 x So 0., y Then, we use modified Euler s method to find f x0, y0 f x, y y 0. y y0 h log(0 ) log log() log [log() log. ] 0. ( ) 0.(0.079) Runge-Kutta Method Question marks 0 Given y / x y, y(0), Find y() and y() using the Runge-Kutta method of fourth order taking h SOLUTION Given differential equation is / y x y, y(0) Here h x 0, y and 0 0 The Fourth-Order Runge-Kutta Method is described as :-

63 Page No.6 y y k k 6 k k where, n n 3 4 k hf x, y h k k hf xn, yn h k k3 hf xn, yn k hf x h, y k n 4 n n 3 n First Iteration:- x 0, y 0 0 k hf ( x, y ) 0 0 f (0, ) 0 () (5) 5 h k k hf x0, y0 5 f (0, ) f (0.5, 4.5) 0.5 (4.5) 0.5 h k k3 hf x0, y0 0.5 f (0, ) f (0.5, 7.5) 0.5 (7.5) 6 k hf x h, y k f ( 0, 6) f (,8) (8) 38 Now

64 Page No.63 y n y n k k k k y y0 k k k3 k4 6 5 (0.5) (6) 38 6 (5 3 38) 6 (96) 6 8 Second Iteration:- x, y 8, h k hf ( x, y ) f (,8) (8) (38) 38 h k k hf x, y 38 f (,8 ) f (.5,37).5 (37) 76.5 h k k3 hf x, y 76.5 f (,8 ) f (.5,56.5).5 (56.5) 5 k hf x h, y k 4 3 f (,8 5) f (,33) (33) 69

65 Page No.64 So, y n y n k k k k y y k k k3 k (76.5) (5) ( ) 6 8 (690) So, y() 8 y() 33 Milne s Predictor Corrector Method Question Marks 0 Find the solution of / y y x y y ( ), (0) Using Milne s P-C method at x 0.4 given that y (0.).689, y(0.3).504 Solution: y(0.).7739 and

66 Page No.65 Here, x 0, x 0., x 0., x 0.3, x and y, y.689, y.7739, y / y ( x y) y y ' ( x y ) y (0..689)(.689) y ' ( x y ) y ( )(.7739) y ' ( x y ) y (0..504)(.504) Now, using Predictor Formula 4h y4 y0 y ' y ' y3 ' 3 4*0. y4 (.35933) (.7363) 3 y Adam Moultan s Predictor Corrector Method Question marks 0 Using Adam-Moulton Predictor-Corrector Formula find f(0.4) from Ordinary Differential Equation with the help of following table. Solution: Here, / y xy y h ; (0) 0 ; 0. X Y

67 Page No.66 h 0. f ( x, y) xy y ' x y 0 0 y ' x y y ' x y y ' x y Now, Using Adam's P-C Pair Formula:- h yn yn 55 y ' n 59 y ' 4 Putting the values; 4 37 y' 9 y' n n n3 h y4 y3 55 y ' 3 59 y ' 37 y ' 9 y ' y y Computing y ' for the Corrector Formula; y ' x y y ' Now Applying the Corrector Formula; h yn yn 9 y ' n9 y ' n 5 y ' n y ' n 4 h y4 y3 9 y ' 4 9 y ' 3 5 y ' y ' 4 0. y y FAQ updated version. Question: What is Bracketing method? Answer: Methods such as bisection method and the false position method of finding roots of a nonlinear equation f(x) = 0 require bracketing of the root by two guesses. Such

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