# UNIT - I LESSON - 1 The Solution of Numerical Algebraic and Transcendental Equations

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1 UNIT - I LESSON - 1 The Solution of Numerical Algebraic and Transcendental Equations Contents: 1.0 Aims and Objectives 1.1 Introduction 1.2 Bisection Method Definition Computation of real root Illustrations 1.3 Newton Raphson Method Definition Computation of Real root by Newton-Raphson Method Illustration 1.4 The Method of False Position Definition Computation of Real root by method of false position Illustrations Check your progress 1.5 Lesson End Activities 1.6 Let us Sum Up 1.7 References 1.0 Aims and Objectives In this Lesson, we have discussed about the solution of equations of the form f(x) =0, f(x) is polynomial of degree two or three or four or more. We get the solution of the equations by using Bisection method, Newton Raphson method and method of false position. After reading this lesson, you should be able to To compute solution of equations by using the Bisection Method. To compute real root by Newton-Raphson Method. To find out the real root of equation by false position method

2 1.1 Introduction The solution of the equation of the form f(x) = 0 occurs in the field of science, engineering and other applications. If f(x) is a polynomial of degree two or more, we have formulae to find solution. But, if f(x) is a transcendental function, we do not have formulae to obtain solutions. When such type of equations are there, we have some methods like Bisection method, Newton-Raphson Method and The method of false position. Those methods are solved by using a theorem in theory of equations, i.e., If f(x) is continuous in the interval (a,b) and if f(a) and f(b) are of opposite signs, then the equation f(x) = 0 will have atleast one real root between a and b. 1.2 Bisection Method Let us suppose we have an equation of the form f(x) = 0 in which solution lies between in the range (a,b). Also f(x) is continuous and it can be algebraic or transcendental. If f(a) and f(b) are opposite signs, then there exist atleast one real root between a and b. Let f(a) be positive and f(b) negative. Which implies atleast one root exits between a and b. We assume that root to be x o = (a+b)/2. Check the sign of f(x o ). If f(x 0 )is negative, the root lies between a and x o. If f(x 0 )is positive, the root lies between x o and b. Subsequently any one of this case occur. a + x 0 x 0 + b x1 = or x1 = Z 2 When f(x 1 ) is negative, the root lies between xo and x1 and let the root be x 2 = (x 0 +x 1 ) / 2. Again f(x 2 ) negative then the root lies between x o and x 2, let x 3 = (x 0 +x 2 )/2 and so on. Repeat the process x o, x 1,x 2,. Whose limit of convergence is the exact root. Steps: 1. Find a and b in which f(a) and f(b) are opposite signs for the given equation using trial and error method. 2. Assume initial root as x o = (a+b)/2. 3. If f(x 0 )is negative, the root lies between a and x o and take the root as x 1 = (x o +a)/2. 4. If f(x o ) is positive, then the root lies between x o and b and take the root as x 1 = ( x o +b)/ If f(x 1 ) is negative, the root lies between x o and x 1 and let the root be x 2 = (x 0 +x 1 ) / If f(x 2 ) is negative, the root lies between x o and x 1 and let the root be x 3 = (x 0 +x 2 ) / Repeat the process until any two consecutive values are equal and hence the root.

3 Illustrations: Find the positive root of x 3 x = 1 correct to four decimal places by bisection method. Solution: Let f(x) = x 3 x 1 f(0) = 0 3 f(1) = 1 3 f(2) = = -1 = -ve = -1 = -ve = 5 = +ve So root lies between 1 and 2, we can take (1+2) /2 as initial root and proceed. i.e., f(1.5) = = +ve and f(1) = -1 = -ve So root lies between 1 and 1.5, Let x o = (1+1.5) /2 as initial root and proceed. f(1.25) = So root lies between x 1 between 1.25 and 1.5 Now x 1 = ( ) /2 = f(1.375) = = +ve So root lies between x 2 between 1.25 and Now x 2 = ( )/2 = f(1.3125) = = -ve Therefore, root lies between 1.375and Now x 3 = ( ) / 2 = f(1.3438) = = +ve So root lies between and Now x 4 = ( ) / 2 = f(1.3282) = = +ve So root lies between and Now x 5 = ( ) / 2 = f(1.3204) = = -ve So root lies between and Now x 6 = ( ) / 2 = f(1.3243) = -ve So root lies between and Now x 7 = ( ) / 2 = f(1.3263) = +ve

4 So root lies between and Now x 8 = ( ) / 2 = f(1.3253) = +ve So root lies between and Now x 9 = ( ) / 2 = f(1.3248) = +ve So root lies between and Now x 10 = ( ) / 2 = f(1.3246) = -ve So root lies between and Now x 11 = ( ) / 2 = f(1.3247) = -ve So root lies between and Now x 12 = ( ) / 2 = Therefore, the approximate root is Illustration : Find the positive root of x cos x = 0 by bisection method. Solution : Let f(x) = x cos x f(0) = 0 - cos (0) = 0-1 = -1 = -ve f(0.5) = 0.5 cos (0.5) = = -ve f(1) = 1 cos (1) = = +ve So root lies between 0.5 and 1 Let x o = (0.5 +1) /2 as initial root and proceed. f(0.75) = 0.75 cos (0.75) = = +ve So root lies between 0.5 and 0.75 x 1 = ( ) /2 =0.625

5 f(0.625) = cos (0.625) = So root lies between and x 2 = ( ) /2 = f(0.6875) = So root lies between and x 3 = ( ) /2 = f( ) = cos( ) = So root lies between and x 4 = ( ) /2 = f( ) = = - ve So root lies between and x 5 = f( ) = = + ve x 6 = ( ) /2 = f( ) = So root lies between and x 7 = ( ) = f(0.7402) = cos(0.7402) = So root lies between and The root is x 8 = f( ) = x 9 = Check Your Progress : 1. Find a positive root of the following equation by bisection method : (i) x 3-4x -9 (Ans: ) 1.3 Newton-Raphson method (or Newton s method)

6 Let us suppose we have an equation of the form f(x) = 0 in which solution is lies between in the range (a,b). Also f(x) is continuous and it can be algebraic or transcendental. If f(a) and f(b) are opposite signs, then there exist atleast one real root between a and b. Let f(a) be positive and f(b) negative. Which implies atleast one root exits between a and b. We assume that root to be either a or b, in which the value of f(a) or f(b) is very close to zero. That number is assumed to be initial root.. Then we iterate the process by using the following formula until the value is converges. Steps: f(x n ) X n+1 = Xn f (Xn) Illustration 1: 1. Find a and b in which f(a) and f(b) are opposite signs for the given equation using trial and error method. 2. Assume initial root as X o = a i.e., if f(a)is very close to zero or Xo = b if f(a)is very close to zero 3. Find X 1 by using the formula f(x o ) X 1 = Xo f (Xo) 4. Find X 2 by using the following formula f(x 1 ) X 2 = X 1 f (X 1 ) 5. Find X 3,X 4, X n until any two successive values are equal Find the positive root of f(x) = 2x 3-3x-6 =0 by Newton Raphson method correct to five decimal places. Solution: Let f(x) = 2x 3-3x 6 ; f (x) = 6x 2 3 f(1) = = -7 = -ve f(2) = = 4 = +ve So, a root between 1 and 2. In which 4 is closer to 0 Hence we assume initial root as 2.

7 Consider x 0 =2 So X 1 = X 0 f(x 0 )/f (X 0 ) = X 0 - ((2X 0 3-3X 0-6) / 6α 0-3) = (4X )/(6X 0 2-3) X i+1 = (4X i 3 + 6)/(6X i 2-3) X 1 = (4(2) 2 +6)/(6(2) 2-3) = 38/21 = X 2 = (4( ) 3 +6)/(6( ) 2-3) = / = X 3 = (4( ) 3 +6)/(6( ) 2-3) = / = X 4 = (4( ) 3 +6)/(6( ) 2-3) = / = Since X 3 and X 4 are equal, hence root is Illustrations 2: Using Newton s method, find the root between 0 and 1 of x 3 = 6x 4 correct to 5 decimal places. Solution : Let f(x) = x 3-6x+4; f(0) = 4 = +ve; f(1) = -1 = -ve So a root lies between 0 and 1 f(1) is nearer to 0. Therefore we take initial root as X 0 = 1 f (x) = 3x 2-6 = x f(x) f (x) = x - (3x 3-6x+4)/(3x 2-6) = (2x 3-4)/(3x 2-6) X 1 = (2X )/(3X 0 2-6) = (2-4)/(3-6) = 2/3 =

8 X 2 = (2(2/3) 3 4 )/(3(2/3) 2-6) = X 3 = (2( ) 3 4 )/(3( ) 2-6) = ( / ) = X 4 = (2( ) 3 4 )/(3( ) 2-6) = ( / ) = The root is correct to 5 decimal places. Check Your Progress : Solve the following by using Newton Raphson Method : x 3 -x -1 (Ans : ) 1.4 Method of False Position ( or Regula Falsi Method ) Consider the equation f(x) = 0 and f(a) and f(b) are of opposite signs. Also let a< b. The graph y = f(x) will Meet the x-axis at some point between A(a, f(a)) and B (b,f(b)). The equation of the chord joining the two points A(a, f(a)) and B (b,f(b)) is y f(a) f(a) - f(b) = x - a a - b The x- Coordinate of the point of intersection of this chord with the x-axis gives an approximate value for the of f(x) = 0. Taking y = 0 in the chord equation, we get f(a) f(a) - f(b) = x - a a - b x[f(a) - f(b)] - a f(a) + a f(b) = -a f(a) + b f(b) x[f(a) - f(b)] = b f(a) - a f(b) This x 1 gives an approximate value of the root f(x) = 0. (a < x 1 < b) Now f(x 1 ) and f(a)are of opposite signs or f(x 1 ) and f(b) are opposite signs. If f(x 1 ), f(a) <0. then x 2 lies between x 1 and a.

9 a f(x 1 ) x 1 f(b) Therefore x 2 = f(x 1 ) - f(a) This process of calculation of ( x 3, x 4, x 5,.) is continued till any two successive values are equal and subsequently we get the solution of the given equation. Steps: 1. Find a and b in which f(a) and f(b) are opposite signs for the given equation using trial and error method. 2. Therefore root lies between a and b if f(a)is very close to zero select and compute x 1 by using the following formula: a f(b) - b f(a) x 1 = f(b) - f(a) 3. If f(x 1 ), f(a) <0. then root lies between x 1 and a. Compute x 2 by using the following formula: a f(x 1 ) x 1 f(b) x 2 = f(x 1 ) - f(a) 4. Calculate the values of ( x 3, x 4, x 5,.) by using the above formula until any two successive values are equal and subsequently we get the solution of the given equation. Illustrations 1: Solve for a positive root of x 3-4x+1=0 by and Regula Falsi method Solution : Let f(x) = x 3-4x+1 = 0 f(0) = (0)+1 = 1 = +ve f(1) = 1 3-4(1)+1 = -2 = -ve So a root lies between 0 and 1 We shall find the root that lies between 0 and 1. Here a=0, b=1 a f(b) - b f(a)

10 x 1 = f(b) - f(a) (0 x f(1) 1 x f(0) ) = ( f(1) f(0)) - 1 = (-2-1) = f(x 1 ) = f(1/3) = (1/27)-(4/3) +1 = Now f(0) and f(1/3) are opposite in sign. Hence the root lies between 0 and 1/3. (0 x f(1/3) 1/3 x f(0) ) x 2 = ( f(1/3) f(0)) x 2 = (-1/3)/ ( ) = Now f(x 2 ) = f( ) = = -ve So the root lies between 0 and x 3 = (0 x f( ) x f(0)) / (f( ) f(0)) = / = f(x 3 ) = f( ) = So the root lies between 0 and x 4 = (0 x f( ) x f(0)) / (f( ) f(0)) = / =

11 f(x 4 ) = f( ) = The root lies between 0 and x 5 = (0 x f( ) x f(0)) / (f( ) f(0)) = / = Hence the root is Illustration 2. Find an approximate root of x log 10 x 1.2 = 0 by False position method. Solution : Let f(x) = x log 10 x 1.2 f(1) = -1.2 = -ve; f(2) = 2 x = f(3) = 3x = = +ve So, the root lies between 2 and 3. 2f(3) 3f(2) 2 x x ( ) x 1 = = = f(3) f(2) f(x 1 ) = f(2.7210) = The root lies between x 1 and 3. x 1 x f(3) 3 x f(x 1 ) x x ( ) x 2 = = = f(3) f(x 1 ) = =

12 f(x 2 ) = f(2.7402) = x log(2.7402) 1.2 = So the root lies between and x f(3) 3 x f(2.7402) x x ( ) x 3 = = f(3) f(2.7402) = = f(2.7406) = So the root lies between and x f(2.7406) x f(2.7402) x 4 = f(2.7406) f(2.7402) x x = = = Hence the root is Check Your Progress 1.Solve the following by method of false position (Regula Falsi Method) : (i) x 3 +2x 2 +10x 20 (Ans :1.3688)

13 1.5 Lesson End Activities Find a positive root of the following equation by bisection method : (i) 3x = cos x + 1 (ii) x 3 +3x -1 (iii) e x -3x (iv) cos x -2x Solve the following by using Newton Raphson Method : (i) x 4 -x -9 (ii) x 3 +2x 2 +50x +7 (iii) cos x -x e x (iv) x 2 sin x Solve the following by method of false position (Regula Falsi Method): (i) 2x 3 sin x =5 (ii) e x -3x (iii) e -x = Sin x (iv) cos x -2x Let us Sum Up In this lesson we have dealt with the following: 1. We have discussed the Bisection method to find solution of numerical algebraic and transcendental equations. 2. We have discussed iterative formulae by name Newton-Raphson method to find solution of algebraic and transcendental equations. 3. We have also discussed the method of false position, which gives solution of numerical algebraic and transcendental equations. Model Answers Answer (Bisection Method) (i) (ii) (iii) (iv) Answer (Newton Raphson Method) (i) (ii) (iii) (iv) Answer (Regula Falsi Method) (i) (ii)

14 (iii) (iv) Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi., S.Chand &Company Ltd., Revised Edition Numerical Methods A. Singaravelu, Meenakshi Publication, Arupakkam Po, Numerical Methods E. Balagurusamy, Tamil Nadu, Tata McGraw Hill.

15 Contents: LESSON - 2 The Solution of Simultaneous Linear Algebraic Equation - Direct Methods 2.0 Aims and Objectives 2.1 Introduction 2.2 Gauss Elimination Method Introduction Gauss Elimination Method for System of equation Illustrations 2.3 Gauss-Jordon Method Introduction Gauss-Jordon Method for System of equation Illustrations 2.4 Lesson End Activities 2.5 Let us Sum Up 2.6 References 2.0 Aims and Objectives In this Lesson, we have discussed about the solving of simultaneous linear algebraic equations, which occurs in the field of science and engineering. Early studies of solving the equations are tedious. With help computer we solve by using numerical methods. These numerical methods are of two types namely Gauss Elimination and Gauss-Jordon method After reading this lesson, you should be able To Solve the system of equations by using the Gauss Elimination method To Solve the system of equations by using the Gauss-Jordon method 2.1 Introduction Simultaneous linear algebraic equations occur in various fields of science and engineering. We solve such type of equation by Cramer s rule. These methods are time consuming and tedious. To solve such equations, we go to numerical methods. The numerical methods are of two types name (i) direct (ii) iterative. Now, we will study these methods and detailed below : 2.2 Gauss Elimination Method This is direct method based on number of unknowns, by eliminating the same by combining the equations to a triangular form. To illustrate the method consider the following system equations Consider the n linear equations in n unknowns,

16 a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 a 21 x 1 + a 22 x 2 + +a 2n x n = b 2. a n1 x 1 + a n2 x 2 + +a nn x n = b n Where a ij and b i are unknown constants and x i s are unknowns. a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 The above system of equation is written in the matrix form as AX = B Now our aim is to reduce the given matrix (A,B) to upper triangular matrix. The system of equation can be solved simply thus : a 11 a 12 a 1n b a 21 a 22 a 2nn b a n1 a n2 a nn b n - Now, multiply the first row of above matrix by - a i1 / a 11 and add to the i th row (A,B), where i = 2, 3,., n. By doing this to all elements in the first column of (A, B) except first row. Now the above matrix is reduced to a 11 a 12 a 1n b b 22 b 2nn c b n2 b nn c n - Now, multiply the Second row of the above matrix by - b i1 / b 22 and add to the i th row (A,B), where i = 3,4., n. By doing this to all elements in the second column of (A, B) except first and second row. Now the above matrix is reduced to a 11 a 12 a 1n b b 22 b 2n c c 33 c 3n d c n3 c nn d n - Now, multiply the third row of the above matrix by - c i1 / c 33 and add to the i th row (A,B), where i = 4., n. By doing this to all elements in the third column of (A, B) except first, second and third row. Continuing the process, all elements below the leading diagonal elements of A, and the above matrix is reduced to a 11 a 12 a 1n b 1 -

17 -0 b 22 b 2n c c 33 c 3n d n nn n n - The above system of linear equations is equivalent to a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 b 22 x 2 + +a 2n x n = c 2. n nn x n = m n Going from the bottom of these equation, we get x n = m n / n nn and subsequently we solve the equation x n-1, x n-2,. x 2, x 1. Steps to solve the system of three equation with three unknowns : Let us consider the system of equation a 11 x 1 + a 12 x 2 +a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 +a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 1. To eliminate x 1 from the second equation, multiply the first row of the equation matrix by - a 21 / a 11 and add it to second equation. Similarly eliminate x 1 from the third equation and subsequently all other equations. We get new equation of the form a 11 x 1 + a 12 x 2 +a 13 x 3 = b 1 +b 22 x 2 + b 23 x 3 = c 2 + b 32 x 2 + b 33 x 3 = c 3 Where b 22 = a 22 (a 21 / a 11 ) X a 12 b 23 = a 23 (a 21 / a 11 ) X a 13 ) c 2 = b 2 (a 21 / a 11 ) x b 1 b 32 = a 32 (a 31 / a 31 ) X a 12 b 33 = a 33 (a 31 / a 11 ) X a 13 c3 = b 3 (a 31 / a 11 ) x b 1 2. To eliminate x 2 from the third equation, multiply the second row of the equation matrix by - b 32 / b 22 and add it to third equation. Similarly eliminate x 2 from the third equation and subsequently all other equations. a 11 x 1 + a 12 x 2 +a 13 x 3 = b 1 +b 22 x 2 + b 23 x 3 = c 2 b 33 x 3 = d 3 where c 33 = b 33 (b 32 / b 22 ) X b 23

18 d3 =c 3 (b 32 / b 22 ) x c 2 3. From the above reduced system of equation substitute the values x 3, x 2 and x 1 by backward substitution we get the solution of the given equations. Illustration 1 : Solve the system of equation by Gauss elimination method x+ 2y + z = 3 2x+ 3y + 3z = 10 3x - y + 2z = 13 Solution : The given system of equation is equivalent to Now, we have to make the above matrix as upper triangular By using the following modifications R 2 = R 2 + (-2) R 1 ; R 3 = R 3 + (-3) R Now we have to take b 22 = -1 as the key element and reduce b 32 as 0 By using the following modifications R 3 = R 3 + (+7) R From the above matrix x + 2y + z = 3 - y + z = 4-8z = -24 Therefore z = 3, y = -1, x =2 by back substitution. Illustration 2 : Solve the system of equation by Gauss elimination method 2x + y + 4z = 12 8x - 3y + 2z = 20 4x + 11y -z = 33

19 Solution : The given system of equation is equivalent to Now, we have to eliminate x from the second and third equation By using the following modifications R 2 = R 2 + (-4 ) R 1 ; R 3 = R 3 + (-2) R Second step we eliminate y from the third equation. Taking (b 23 = 9 /7) as the key element multiply the second equation by key element and add it to the third equation By using the following modifications R 3 = R 3 + (9/7) R From the above matrix 2x + y + 4z = 12-7y -14 z = z = -27 By back substitution, we get the solution of the equation z = 1, y = 2. x = 3 Therefore z = 3, y = -1, x =2 by back substitution. Check Your Progress Solve the system of equation by Gauss elimination method 20x + y + 4z = 25 8x +13y + 2z = 23 4x - 11y + 21z = 14 (Ans: x= y= z = 1) 2.3 Gauss Jordon Method This method is a slightly modification of the above Gauss Elimination method. Here elimination is performed not only in the lower triangular but also upper triangular. This leads to unit matrix and hence solution is obtained. This is

20 Jordon s modification of the Gauss elimination and hence the name is Gauss- Jordon Method. Consider the n linear equations in n unknowns, a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 a 21 x 1 + a 22 x 2 + +a 2n x n = b 2. a n1 x 1 + a n2 x 2 + +a nn x n = b n Where a ij and b i are unknown constants and x i s are unknowns. a 11 x 1 + a 12 x 2 + +a 1n x n = b 1 The above system of equation is written in the matrix form as AX = B Now our aim is to reduce the given matrix (A,) to unit matrix. The system of equation can be solved simply thus : a b b c c 33 0 d n nn m n - The above system of linear equations is equivalent to a 11 x = b 1 b 22 x = c 2. n nn x n = m n From the above the equation we get solution directly.. Steps to solve the system of three equation with three unknowns : Let us consider the system of equation a 11 x 1 + a 12 x 2 +a 13 x 3 = b 1 a 21 x 1 + a 22 x 2 +a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 2. To eliminate x 1 from the second equation, multiply the first row of the equation matrix by - a 21 / a 11 and add it to second equation. Similarly eliminate x 1 from the third equation and subsequently all other equations. We get new equation of the form a 11 x 1 + a 12 x 2 +a 13 x 3 = b 1

21 +b 22 x 2 + b 23 x 3 = c 2 + b 32 x 2 + b 33 x 3 = c 3 Where b 22 = a 22 (a 21 / a 11 ) X a 12 b 23 = a 23 (a 21 / a 11 ) X a 13 ) c 2 = b 2 (a 21 / a 11 ) x b 1 b 32 = a 32 (a 31 / a 31 ) X a 12 b 33 = a 33 (a 31 / a 11 ) X a 13 c3 = b 3 (a 31 / a 11 ) x b 1 2. In this method, we eliminate x 2 from the first and third equation, Multiply the second row of the equation matrix by - b 32 / b 22 and add it to third equation. Similarly eliminate x 2 from the first equation. a 11 x a 13 x 3 = b 1 +b 22 x 2 + b 23 x 3 = c 2 b 33 x 3 = d 3 where c 33 = b 33 (b 32 / b 22 ) X b 23 d 3 =c 3 (b 32 / b 22 ) x c 2 b 13 = a 13 a 12 / b 22 x b 23 \ b 1 = b 1 a 12 / b 22 x c 2 Similarly eliminate x 3 from first and second equation obtained. 3. From the above reduced system of equation the values x 1, x 2 and x 3 are Illustration 1 : Solve the system of equation by Gauss-Jordon method 10x+ y + z = 12 2x + 10y + z = 10 x + y + 5z = 13 Solution : The given system of equation is rearranged for computation convenience, Interchange the first and last equation, since coefficient of the x in the last equation is unity (1) : Now, we have to make the above matrix as upper triangular By using the following modifications

22 R 2 = R 2 + (-2) R 1 ; R 3 = R 3 + (-10) R Now we have to take b 22 = 1/8 as the key element and reduce b 32 as 0 By using the following modifications R 2 = R 2 /8 ; R 3 = R 3 + (+9/8) R /8-1/ /8-473/8- Now we have to make b 33 = 1 as the key element and reduce b 32 as 0 By using the following modifications R 3 = R 3 x - 8 /473 ; R 1 = R 1 + (-1) R /8 57/ /8-1/ Now we have to make b 23 = 0 and b 13 = 0. By using the following modifications R 2 = R 2 + ( - 9/8 )R 3 ; R 1 = R 1 + (-49/8) R Therefore x = 1, y = 1, z= 1. Illustration 2 : Solve the system of equation by Gauss-Jordon method 2x + y + 4 z = 12 8x- 3y + 2z = 20 4x + 11y - z = 33 Solution : The given system of equation is equivalent to

23 Now, we have to eliminate x from the second and third equation By using the following modifications R 2 = R 2 + (-4 ) R 1 ; R 3 = R 3 + (-2) R Second step we eliminate y from the third equation. Taking (b 23 = 9 /7) as the key element multiply the second equation by key element and add it to the third equation By using the following modifications R 3 = R 3 + (9/7) R At this stage, we eliminate y from the first equation. Z from the first and second equation. By using following modifications ; R 1 = R 1 /2 ; R 2 = R 2 /-7 ; R 3 = R 3 / (-27) R 1-1 1/ R 2 = R 2 + (-2) R 3 ; R 1 = R 1 + ( -1/2 ) R 2 ; R 1 = R 1 + ( -1 ) R 1 ; Therefore x = 3, y = 2, z= 1. Check Your Progress Solve the system of equation by Gauss-Jordon method 10x+ y + z = 13 2x + 10y + z = 14 x + y + 15z = 32 (Ans: x=y=1, z =2) 2.4 Lesson End Activities 1. Solve the system of equation by Gauss Elimination method 3.15x 1.96 y z = 12.95

24 2.13x y z = x y z = Solve the system of equation by Gauss elimination method 3x + 4 y + 6 z = 18 2x- y + 8z = 13 5x -2y +7 z = Solve the system of equation by Gauss-Jordon method 2x + y + 4 z = 9 8x- 3y + z = 12 4x + 11y - z = Solve the system of equation by Gauss-Jordon method 2x - y + 4 z = 5 8x- 3y + z =6 x + 11y - z = Let us Sum Up In this lesson we have dealt with the following: We have discussed Gauss Elimination method to solve the system of linear equations, which occurs in the field of science and engineering We have discussed the Gauss-Jordon method to solve the system of equations. Model Answer For Lesson End Activities 1. (Ans : x =1.7089, y = , z = ) 2. (Ans: x = 3, y = 1, z= 1) 3. (Ans: x = 2, y = 1, z= 1) 4. (Ans: 1 = 1, y = 1, z= 1) 2.6 Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi, S.Chand &Company Ltd., Revised Edition 2005.

25 Contents: LESSON - 3 Gauss-Jacobi Method 3.0 Aims and Objectives 3.1 Introduction 3.2 Gauss Jacobi Method for System of equation 3.3 Illustrations 3.4 Lesson End Activities 3.5 Let us Sum Up 3.6 References 3.0 Aims and Objectives In this Lesson, we have discussed about the solving simultaneous linear algebraic equations, which occurs in the field science and engineering. Early study of solving the equations is not applicable for all the problem, even if we apply its required tedious calculations. With help of Gauss-Jacobi iteration process, we solve by linear equations with minimal steps. After reading this lesson, you should be able to To know about Gauss-Jacobi method iteration procedure. To Solve the system of equations by using the Gauss-Jacob method 3.1 Introduction Early methods of study in solving algebraic linear equations are direct methods. Now we will discuss some indirect methods or iterative methods. This method is not always successful to all systems of equations. If this method is to succeed, each equation must satisfy a condition. i.e., when diagonal elements are exceeding all other elements in the respective equations. We will discuss two methods of this category namely Gauss-Jacobi and Gauss-Seidel method. 3.2 Gauss-Jacobi Method Let us consider this method in the case of three equations in three unknowns. Consider the 3 linear equations in 3 unknowns, a 1 x + b 1 y +c 1 z = d 1 a 2 x + b 2 y +c 2 z = d 2 a 3 x + b 3 y +c 3 z = d 3

26 This method is applied only when diagonal elements are exceeding all other elements in the respective equations i.e., a 1 > b 1 + c 1 = d 1 a 2 > b 2 + c 2 = d 2 a 3 > b 3 + c 3 = d 3 Let the above condition is true we apply this method or we have to rearrange the equations in the above form to fulfill the above condition. We start with initial values of x,y and z as zero. Solve x, y,z in terms of other variables. That is, 1 x = (d 1 b 1 y c 1 z) a 1 1 y = (d 2 a 2 x c 2 z) b 2 1 z = (d 3 a 3 x b 3 y) c 3 The above values are initial values x (0), y (0), z (0) of x, y z respectively, then 1 x (1) = (d 1 b 1 y (0) c 1 z (0) ) a 1 1 y (1) = (d 2 a 2 x (0) c 2 z (0) ) b 2 1 z (1) = (d 3 a 3 x (0) b 3 y (0) ) c 3 Again using new values of x (1), y (1), z (1) of x, y z respectively, then 1 x (2) = (d 1 b 1 y (1) c 1 z (1) ) a 1 1 y (2) = (d 2 a 2 x (1) c 2 z (1) ) b 2 1 z (2) = (d 3 a 3 x (1) b 3 y (1) ) c 3 Repeating the process in the same way, and the r th iterates are x (r), y (r), z (r) and given below 1 x (r+1) = (d 1 b 1 y (r) c 1 z (r) ) a 1

27 1 y (r+1) = (d 2 a 2 x (r) c 2 z (r) ) b 2 1 z (r+1) = (d 3 a 3 x (r) b 3 y (r) ) c 3 The above iteration is continued until any two successive values are equal. 3.3 Illustration 1: 1. Solve the system of equation by Gauss-Jacobi method 27x + 6 y - z = 85 6x +15y + 2z = 72 x +6y + 54 z = 110 Solution: To apply this method, first we have to check the diagonal elements are dominant. i.e., 27 > 6+ 1 ; 15 > 6+ 2 ; 54 > 1+1. So iteration method can be applied x = 1/27 (85-6 y + z ) y =1 /15 (72-6x - 2z ) z = 1/54 ( 110 x - y ) First iteration : From the above equations, we start with x = y = z = 0 x (1) = 85/27 = (1) y (1) = 72/15 = 4.8..(2) z (1) = 110/54 = (3) Second iteration :Consider the new values of y (1) = 4.8 and z (1) = in the first equation x (2) = 1/27( 85 6 x ) = y (2) = 1/15 ( 72-6 x x ) = z (2) = 1/54 ( ) ) = Fourth iteration : Consider the new values of x (2) = , y (2) = and z (2) = in the first equation x (3) = 1/27( 85 6 x ) = y (3) = 1/15 ( 72-6 x x ) = z (3) = 1/54 ( ) ) =

28 Thus, we continue the iteration and result is noted below Iteration No. x y z From the above table 9 th and 10 th iterations are equal by considering the four decimal places. Hence the solution of the equation is x = y = z = Illustration 2. Solve the system of equation by Gauss-Jacobi method 10x - 5 y - 2 z = 3 4x - 10y + 3z = -3 x +6y + 10 z = 3 Solution: To apply this method, first we have to check the diagonal elements are dominant. i.e., 10 > 5+ 2 ; 10 > 4+ 3 ; 10 > 1+6. So iteration method can be applied x = 1/10 (3 + 5 y + 2z ) y =1 /10 (3 + 4x + 3z ) z = 1/10 ( -3 x - 6y ) First iteration : From the above equations, we start with x = y = z = 0 x (1) = 3/10 = 0.3.(1) y (1) = 3/10 = 0.3..(2) z (1) = -3/10 = -0.3 (3) Second iteration :Consider the new values of y (1) =0.3 and z (1) = -0.3 in the first equation x (2) = 1/10( x.3 +( -0.3)) = 0.39 y (2) = 1/10 ( x x( -0.3) ) = 0.33 z (2) = 1/10 [ -3 (0.3) 6(0.3) ] = Third iteration : Consider the new values of x (2) = 0.39, y (2) = 0.33 and z (2) =-0.51 in the first equation

29 x (3) = 1/10 [ 3 5 x0.33 +(-0.51)] = y (3) = 1/10 ( x x (-0.51) ) = z (3) = 1/10[ x (0.33) ] = Thus, we continue the iteration and result is noted below Iteration No. X y z From the above table 8 th and 9 th iterations are equal by considering the 3 decimal places. Hence the solution of the equation is x =0.342, y = 0.285, z = Check Your Progress 1. Solve the system of equation by Gauss-Jacobi method 3.15x 1.96 y z = x y z = x y z = 6.88 (Ans : x =1.7089, y = , z = ) 3.4 Lesson End Activities Solve the following system of equations by using Gauss-Jacobi Method 1. 8x -3y + 2z=20 ; 4x +11y z = 33; 6x +3y +12 z = x+4y -z= 32 ; x +3y +10 z = 24; 2x +3y +10 z = x -2y +z = -4 ; x + 6y -2z = -1; 3x+y+5z = x +y+z = 8 ; 2x+4y + z = 4 ; x +3y + 3z = Let us Sum Up In this lesson we have dealt with the following: We have discussed Gauss Jacobi method to solve the system of linear equations, which occurs in the field of science and engineering. This method is an iterative method and it is widely applied. Model Answer For Lesson End Activities 1. (Ans: 3.017, 1.986, 0.912) 2. (Ans: 0.994, 1.507, 1.849) 3. (Ans:-1.0,.999, 3 )

30 4. (Ans: 0.83,.32, 1.07) 3.6 Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi, S.Chand &Company Ltd., Revised Edition 2005.

31 Contents: LESSON - 4 Gauss-Seidel Method 4.0 Aims and Objectives 4.1 Introduction 4.2 Gauss Seidel Method 4.3 Illustrations 4.4 Lesson End Activities 4.5 Let us Sum Up 4.6 References 4.0 Aims and Objectives In this Lesson, we have discussed about the solving simultaneous linear algebraic equations, which occurs in the field of science and engineering. Early study of solving the equations are time consuming when compared to this method. With help of Gauss-Seidel iteration process, we solve by linear equations with minimal iterations After reading this lesson, you should be able To know about Gauss-Seidel method iteration procedure. To Solve the system of equations by using the Gauss-Seidel method 4.1 Introduction Early methods of study in solving algebraic linear equations are direct methods. Now we will discuss some indirect methods or iterative methods. This method is not always successful to all systems of equations. If this method is to succeed, each equation must satisfy a condition. i.e., when diagonal elements are exceeding all other elements in the respective equations. We will discuss an iterative and self correcting method, namely Gauss-Seidel method. 4.2 Gauss-Seidel Method This method is only an enhancement of Gauss-Jaobi Method. In the previous method Let us consider this method in the case of three equations in three unknowns. Consider the 3 linear equations in 3 unknowns, a 1 x + b 1 y +c 1 z = d 1 a 2 x + b 2 y +c 2 z = d 2 a 3 x + b 3 y +c 3 z = d 3

32 This method is applied only when diagonal elements are exceeding all other elements in the respective equations i.e., a 1 > b 1 + c 1 = d 1 a 2 > b 2 + c 2 = d 2 a 3 > b 3 + c 3 = d 3 Let the above condition is true we apply this method or we have to rearrange the equations in the above form to fulfill the above condition. We start with initial values of x,y and z as zero. Solve x, y,z in terms of other variables. That is, 1 x = (d 1 b 1 y c 1 z) a 1 1 y = (d 2 a 2 x c 2 z) b 2 1 z = (d 3 a 3 x b 3 y) c 3 We proceed with the initial values y (0), z (0) for y, z and get x (1) from the first equation, i.e., 1 x (1) = (d 1 b 1 y (0) c 1 z (0) ) a 1 When we calculate y (1), we use new values of x i.e., x (1) and z (0) 1 y (1) = (d 2 a 2 x (1) c 2 z (0) ) b 2 Similarly, while we calculate z (1), we use new values of x,y i.e., x (1) and y (1) 1 z (1) = (d 3 a 3 x (1) b 3 y (1) ) c 3 Again using new values of x (1), y (1), z (1) of x, y z respectively, then 1 x (2) = (d 1 b 1 y (1) c 1 z (1) ) a 1 1 y (2) = (d 2 a 2 x (2) c 2 z (1) ) b 2 1 z (2) = (d 3 a 3 x (2) b 3 y (2) ) c 3 Repeating the process in the same way, and the r th iterates are x (r), y (r), z (r) and given below 1

33 x (r+1) = (d 1 b 1 y (r) c 1 z (r) ) a 1 1 y (r+1) = (d 2 a 2 x (r+1) c 2 z (r) ) b 2 1 z (r+1) = (d 3 a 3 x (r+1) b 3 y (r+1) ) c 3 The above iteration is continued until any two successive values are equal. Note : 1. For all systems of equation, this method will not work 2.Iteration method is self correcting method. Any error made in computation is corrected automatically in subsequent iterations 3. Iteration is stopped when any two successive iteration values are equal 4.3 Illustration : 1. Solve the system of equation by Gauss-Seidel method 10x - 5 y - 2 z = 3 4x - 10y + 3z = -3 x +6y + 10 z = 3 Solution: To apply this method, first we have to check the diagonal elements are dominant. ie., 10 > 5+ 2 ; 10 > 4+ 3 ; 10 > 1+6. So iteration method can be applied x = 1/10 (3 + 5 y + 2z ) y =1 /10 (3 + 4x + 3z ) z = 1/10 ( -3 x - 6y ) First iteration : From the above equations, we start with x = y = z = 0 x (1) = 3/10 = 0.3.(1) New value of x is used for further calculation ie., x = 0.3 y (1) = 1/10 (3 + 4x (0)] = (2) New values of x and y is used for further calculation ie., x = 0.3 and y = 0.42 z (1) = 1/10 ( (0.42) = (3) Second iteration : Consider the new values of y (1) =0.42 and z (1) = in the first equation x (2) = 1/10( x ( )) = y (2) = 1/10 ( x x( ) ) = z (2) = 1/10 [ -3 (0.3936) 6( ) ] =

34 Third iteration : Consider the new values of x (2) = , y (2) = and z (2) = in the first equation x (3) = 1/10 [ 3 5 x ( )] = y (3) = 1/10 ( x x ( ) = z (3) = 1/10[ x ( ) ] = Thus, we continue the iteration and result is noted below Iteration No. X Y Z The values correct to 3 decimal places are x = 0.342, y = 0.285, z = Note : Check the above equations by substituting values of x,y and z Illustration 2 : 1. Solve the system of equation by Gauss-Seidel method 28x + 4 y - z = 32 4x +3y + 10 z = 24 2x +17y + 4z = 35 Solution: To apply this method, first we have to rewrite the equation in such way that to fulfill diagonal elements are dominant. 28x + 4 y - z = 32 2x +17y + 4z = 35 4x +3y + 10 z = 24 ie., 28 > 4+ 1 ; 17 > 2+ 4 ; 10 > 4+3. So iteration method can be applied x = 1/28 (32-4 y + z ) y =1 /17 (35-2x -4z ) z = 1/10 ( 24 x - 3y ) First iteration : From the above equations, we start with y = z = 0, we get x (1) = 32/28 = New value of x is used for further calculation ie., x = y (1) = 1/17 ( (0)] = New values of x and y is used for further calculation ie., x =

35 and y = z (1) = 1/10 [ ( ) ] = ) Second iteration : Consider the new values of y (1) = and z (1) = x (2) = 1/28[ 32 4 (1.9244) +( )] = y (2) = 1/17 [ 35-2 (0.9325) - 4 ( ) ] = z (2) = 1/10 [ 24 (0.9325) 3(1.5236) ] = Third iteration : Consider the new values of x (2 ) = , y (2) = and z (2) = x (3) = 1/28[ 32 4 (1.5236) +( )] = y (3) = 1/17 [ 35-2 (0.9913) - 4 ( ) ) = z (3) = 1/10 [ 24 (0.9913) 3(1.5070) ] = Thus, we continue the iteration and result is noted below Iteration No. X y Z Therefore x = , y = , z = Check Your Progress 1. Solve the system of equation by Gauss Seidel method 3.15x 1.96 y z = x y z = x y z = 6.88 (Ans : x =1.7089, y = , z = ) 4.4 Lesson End Activities 1. 8x -6y +z =13.67; 3x +y -2z =17.59; 2x -6y +9z = x 2y +3z =75; 2x+ 2y +18z = 30 ; x + 17y -2z =48 3. y x + 10z=35.61; x + z + 10y =20.08; y- z +10x = x -2y +z = 12 ; x + 9y -z =10; 2x y + 11z = x y +z =18; 2x +5y -2z = 3; x+y 3z = x + y + z =4; x + 2y z = 4; x + y + 2z = 4

36 4.5 Let us Sum Up In this lesson we have dealt with the following: We have discussed Gauss-Seidel method to solve the system of linear equations, which occurs in the field of science and engineering. This method is an iterative method and it is widely applied. Model Answer For Lesson End Activities , 0.32, , , , 1.522, , , , , , 1, Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi, S.Chand &Company Ltd., Revised Edition 2005.

37 UNIT-II LESSON - 5 Contents : Numerical Differentiation 5.0 Aims and Objectives 5.1 Introduction 5.2 Newton s forward difference formula 5.3 Illustrations 5.4 Lesson end activities 5.5 Let us Sum Up 5.6 References 5.0 Aims and Objectives In this Lesson, we have discussed about Newton s forward difference formula for finding derivatives. If the derivative occurs closer to the beginning of the table, we use this method. After reading this lesson, you should be able 5.1 Introduction To know about construction of the difference table. To find derivatives using Newton s forward difference formula. Let y=f(x) be a function taking the values y 0, y 1, y 2, y n corresponding to the values x 0, x 1, x 2,.. x n of the independent variable x. Now we are trying to find the derivative value of y = y k for the given x = x k.. If the derivative is required at a point nearer to starting value in the table., i.e., If the value occur between x o to x 1 or beginning of the table, we use Newton s forward interpolation formula. 5.2 Newton s forward difference formula : Suppose the following table represents a set of values of x and y. x: x 0 x 1 x 2 x 3.. x n y: y 0 y 1 y 2 y 3.. y n

38 From the above values, we want to find the derivative of y = f(x) passing through (n+1) points, at a point closer to the starting value x = x 0 Newton s forward difference interpolation formula is given below y(x 0 + uh) = y u = y 0 + u y 0 + u (u-1) 2 y 0 + u (u-1)(u-2) 3 y ! 3! Where y(x) is a polynomial of degree n in x..(1) x- x 0 where u = and h y 0, 2 y 0, 3 y 0 are obtained from difference table Differentiating with respect to x, finally it reduced to the following way dy / dx = Dy 0 = (1/ h) { y 0 - (1/2) 2 y 0 + (1/3) 3 y } d 2 y / dx 2 = D 2 y 0 = (1/ h 2 ) { 2 y 0-3 y 0 +(11/12) 4 y 0.. } 5.3 Illustration 1. Find the first two derivatives of (x) (1/3) at x = 50 given table below : x : y : Solution : Step 1. Write down the formula : dy / dx = Dy 0 = (1/ h) { y 0 - (1/2) 2 y 0 + (1/3) 3 y } d 2 y / dx 2 = D 2 y 0 = (1/ h 2 ) { 2 y 0-3 y 0 +(11/12) 4 y 0.. } Step 2. Construct the difference table to find various s Difference table x y y 0 2 y 0 3 y ( )

39 By applying Newton s forward formula : [ dy / dx] x=xo =[ dy / dx] u=o = (1/ h) { y 0 - (1/2) 2 y 0 + (1/3) 3 y } = (1/ 1) [ (1/2) ( ) + (1/3) 0 ] = [d 2 y / dx 2 ] x=50 = D 2 y 0 = (1/ h 2 ) [ 2 y 0-3 y 0 + ] = 1 [ ] = Illustration 2. The population of a certain town is given below. Find the rate of growth of the population in 1931 and 1941 Year x : Population y : Solution. Construct the difference table x y y 0 2 y 0 3 y 0 4 y By applying Newton s forward formula : To find (i) f (1931) x- x where u = = = 0

40 h 10 [ dy / dx] x=1931 =[ dy / dx] u=o To find (i) f (1941) = (1/ h) [ y 0 - (1/2) 2 y 0 + (1/3) 3 y 0 - (1/4) 4 y 0 ] = (1/ 10) [ (1/2) (-1.03) + (1/3) (5.49) (-4.47) ] = (1/ 10) [ ] = x- x where u = = = 1 h 10 [ dy / dx] x=1941 =[ dy / dx] u=1 = (1/ h) { y 0 + [(2u-1)/2] 2 y 0 + [(3u 2-6u+2)/6] 3 y 0 +[(4u 3-18u 2 +22u- 6)/24 ] 4 y 0 } = (1/ 10) [ (1/2) (-1.03) (1/6) (5.49) +(1/12)(-4.47) ] = (1/ 10) [ ] = Check Your Progress 4.Find the first two derivatives of t5he function x = 1.5 from the table below x y (Ans : 4.75, 9.0) 5.4 Lesson End Activities 1. Find the first and second derivative of the function tabulated below at x = 3 x : f(x) : Find second derivative of y at x = 0.96 from the data x : y : Find the value of cos(1.74) from the following table.

41 x : sin x : Let us Sum UPs In this lesson we have dealt with the following: We have discussed Newton s forward difference formulae to get the derivative by using the difference table. When the value of the derivative required at a point,which nearer to the starting value in the table. Model Answer For Lesson End Activities 1. f(3) =f (3) =18 2. f(0.96) = Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi, S.Chand &Company Ltd., Revised Edition LESSON - 6

42 Contents Newton s Backward Difference formula 6.0 Aims and Objectives 6.1 Introduction 6.2 Newton s backward difference formula 6.3 Illustrations 6.4 Lesson end activities 6.5 Let us Sum Up 6.6 References 6.0 Aims and Objectives In this Lesson, we have discussed about Newton s backward difference formula for finding derivatives. If the derivative occur closer to the end of the table, we use this method. After reading this lesson, you should be able to 6.1 Introduction To find derivatives using Newton s backward difference formula. Computation of derivatives. Let y=f(x) be a function taking the values y 0, y 1, y 2,.. y n corresponding to the values x 0, x 1, x 2,.. x n of the independent variable x. Now we are trying to find the derivative value of y = y k for the given x = x k.. If the derivative is required at a point nearer to end of the table., ie, If the value occur between x n-1 to x n or end of the table, we use Newton s Backward difference formula. 6.2 Newton s backward difference formula to compute derivative: Suppose the following table represents a set of values of x and y. x: x 0 x 1 x 2 x 3.. x n y: y 0 y 1 y 2 y 3.. y n From the above values, we want to find the derivative of y = f(x) passing through (n+1) points, at a point closer to the starting value x = x 0

43 Let us consider Newton s backward interpolation formula and is given below y(x n + vh) = y u = y n + v y n + v (v+1) 2y n + v (v+1)(v+2) 3y n 2! 3! Where y(x) is a polynomial of degree n in x..(1) where x n -x v = and h y n, 2y n, 3y n are obtained from difference table Differentiating with respect to x, finally it reduced to the following way dy / dx = Dy n = (1/ h) { y n + (1/2) 2y n + (1/3) 3 y n +.. } d 2 y / dx 2 = D 2 y n = (1/ h 2 ) { 2y n + 3 y n +(11/12) 4 y n.. } 6.3 Illustration 1. Find the first two derivatives of (x) (1/3) at x = 56 given table below : x : y : Solution : Step 1. Write down the formula : dy / dx = Dy n = (1/ h) { y n + (1/2) 2y n + (1/3) 3 y n +.. } d 2 y / dx 2 = D 2 y n = (1/ h 2 ) { 2y n + 3 y n +(11/12) 4 y n.. } Step 2. Construct the difference table to find various s Difference table x y y 0 2 y 0 3 y ( )

44 By applying Newton s backward difference formula : [ dy / dx] x=xn =[ dy / dx] v=o = (1/ h) { y n + (1/2) 2y n + (1/3) 3 y n +.. } = (1/ 1) [ (1/2) ( ) + (1/3) 0 ] = [d 2 y / dx 2 ] x=56 = D 2 y 0 = (1/ h 2 ) [ 2y n + 3y n + ] = 1 [ ] = Illustration 2. The population of a certain town is given below. Find the rate of growth of the population in 1961 and 1971 Year x : Population y : Solution. Construct the difference table x y y 0 2 y 0 3 y 0 4 y By applying Newton s forward formula : To find (i) f (1961)

45 x n - x where v = = = -1 h 10 [ dy / dx] x=1961 =[ dy / dx] v=-1 = (1/ h) { y n + ((2v+1) / 2) 2y n + ((3v 2 +6v+2)/6) 3 y n +.. } = To find (i) f (1971) = (1/ 10) [29.09 (1/2) (5.48) (1/6) (1.02) (1/12) (-4.47) ] = (1/ 10) [ ] x n - x where v = = = 0 h 10 [ dy / dx] x=1971 =[ dy / dx] v=0 = (1/ h) { y n + (1 / 2) 2y n + (1/3) 3 y n +(1/4) 4 y n.. } = (1/ 10) [ (1/2) (5.48) + (1/3) (1.02) + (1/4) (-4.47) ] = (1/ 10) [ ] = Lesson End Activities 1. find the first and second derivative of the function tabulated below at x = 4.0 x : f(x) : find second derivative of y at x = 1.04 from the data x : y : From the table below find y and y at x = 1.25 x : y :

46 4. Find first and second derivative of x at x =23 x_ : x : Model Answer for selected Lesson End Activities , Let us Sum Up In this lesson we have dealt with the following: We have discussed Newton s backward difference formulae to get the derivative by using the difference table. When the value of the derivative required at the end of the table. 6.6 Reference: Numerical Methods P.Kandasamy, K.Thilagavathi, K.Gunavathi, S.Chand &Company Ltd., Revised Edition 2005.

47 LESSON 7 Contents Numerical Integration 7.0 Aims and Objectives 7.1 Introduction 7.2 Trapezoidal rule 7.3 Illustrations 7.4 Let us Sum Up 7.5 Lesson end Activities 7.6 References 7.0 Aims and Objectives In this Lesson, we have discussed about numerical integration. Given set of paired values ( x i, y i ) I = 0, 1,2 n, even f(x) is unknown, it is possible to evaluate integration value. We have discussed about the Trapezoidal rule to evaluate integration. After reading this lesson, you should be able to To know about numerical integration. To evaluate integration using Trapezoidal Rule 7.1 Introduction Let b a f(x) dx represents the area between y = f(x), with the rand x =a and x =b. This integration is possible only when f(x) is explicitly given or otherwise it sis not possible to evaluate. In numerical integration can be detailed as follows ; Given set of (n+1) paired values of the function taking the values y 0, y 1, y 2,.. y n corresponding to the values x 0, x 1, x 2,.. x n.. where f(x) is not known explicitly, it is possible to compute a b f(x) dx by numerical integration by using various method. One of the simplest method is trapezoidal Rule that explained below 7.2 Trapezoidal Rule Suppose the following table represents a set of values of x and y. x: x 0 x 1 x 2 x 3.. x n y: y 0 y 1 y 2 y 3.. y n

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