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Department of Engineering Mathematics and Physics, Faculty of Engineering Zagazig University 2009-2010 Numerical solution of non linear equations. Numerical solution of linear equations. Numerical integrations. Curve fitting Numerical solution of ordinary differential equations. Numerical solution of partial differential equations. Next The Main Menu اPrevious Linear equation A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and the first power of a single variable. Linear equations can have one or more variables. The standard form Ax + By = C. a- Cannot have exponents (or powers). b - Cannot multiply or divide each other. c - Cannot be found under a root sign or square root sign (sqrt) System of linear equations In mathematics, a system of linear equations (or linear system) is a collection of linear equations involving the same set of variables. i.e. A system of linear equations is simply a two or more linear equations using the same variables. a11x1 a12 x 2 ... a1n x n b1 The equation may be written in a 21x1 a 22 x 2 ... a 2 n x n b 2 compact form [A]. ........................................... Next [x] = [b] a m1x1 a m 2 x 2 ... a m n x n b n . The Main Menu اPrevious The augment matrix The system of linear equations is [A]nxn. [x]nx1 = [b]nx1 a11 a12 a 21 a 22 ... ... a n1 a n 2 ... a1n x1 b1 ... a 2 n x 2 b 2 ... ... ... ... ... a nn x n b n The augment matrix is a11 a12 a 21 a 22 ... ... a n1 a n 2 Next ... a1n b1 ... a 2 n b 2 ... ... ... ... a nn b n The Main Menu اPrevious It is iterative method for solving system of linear equations when the systems of linear equations are very large (large number of unknowns). Jacobi method converge for diagonal dominant matrices A (diagonal entries of A larger than the others). In this method we covert the system of linear equations [A] .[ x] = [b] to [x] = [C] . [x] + d Generate a sequence of approximation x1, x2,….. where xn = Cxn-1 + d a13 a12 b x 1n x 2(n 1) x 3(n 1) 1 , a11 a11 a11 a11x1 a12 x 2 a13 x 3 b1 a a b a 21x1 a 22 x 2 a 23 x 3 b 2 x 2 n 21 x1(n 1) 23 x 3(n 1) 2 , a 22 a 22 a 22 a 31x1 a 32 x 2 a 33x 3 b 3 . a a b x 3 n 31 x1(n 1) 32 x 2 (n 1) 3 a 33 a 33 a 33 Next The Main Menu اPrevious It is iterative method for solving system of linear equations [A] .[ x] = [b]. It is a modification to Jacobi’s method. a11x1 a12 x 2 a13 x 3 b1 a 21x1 a 22 x 2 a 23 x 3 b 2 a 31x1 a 32 x 2 a 33x 3 b 3 . Next x 1n a a12 b x 2(n 1) 13 x 3(n 1) 1 , a11 a11 a11 x 2n a a 21 b x1(n 1) 23 x 3(n 1) 2 , a 22 a 22 a 22 x 3n a 31 a b x1(n 1) 32 x 2 (n 1) 3 a 33 a 33 a 33 The Main Menu اPrevious Solve the linear system by Jacobi’s method 4x -y + z = 7 4x -8y+ z = -21 -2x + y + 5z = 15. From the system of linear equation we get: 7 y z x , 4 21 4 x z y , 8 15 2 x y z . 5 xn yn zn 7 y (n 1) z (n 1) , 4 21 4 x (n 1) z (n 1) 8 15 2 x n 1 y n 1 . 5 If we start with (x0, y0, z0) = (0, 0, 0), Next , The Main Menu اPrevious xn yn zn n 7 y (n 1) z (n 1) , 4 21 4 x (n 1) z (n 1) 8 , 15 2 x n 1 y n 1 . 5 xn yn zn 0 0 0 0 1 1.75 2.625 3 2 1.656 3.875 3.175 3 1.925 3.85 2.887 4 1.99 3.948 3 5 1.99 3.995 2.997 3.995 2.997 6 1.9995 The iteration appears to converge to the solution (2, 4, 3) Next The Main Menu اPrevious Solve the linear system by Gauss seidle method 4x -y + z = 7 4x -8y+ z = -21 -2x + y + 5z = 15. From the system of linear equation we get: 7 y z x , 4 21 4 x z y , 8 15 2 x y z . 5 xn yn zn 7 y (n 1) z (n 1) , 4 21 4 x (n 1) z (n 1) 8 15 2 x n 1 y n 1 . 5 If we start with (x0, y0, z0) = (0, 0, 0), Next , The Main Menu اPrevious xn yn zn 7 y (n 1) z (n 1) , 4 21 4 x (n 1) z (n 1) , n xn 15 2 x n 1 y n 1 . 5 0 0 1.75 8 1 yn zn 0 3.5 2 1.875 3.9375 2.9625 3 1.994 3.9922 2.887 4 1.998 2.999 5 3.999 1.9999 3.9998 2.9999 The iteration appears to converge to the solution (2, 4, 3) Next 0 3 The Main Menu اPrevious Next The Main Menu اPrevious Non Linear equation An equation in which one or more terms have a variable of degree 2 or higher is called a nonlinear equation. The non linear equation f (x) = 0 and we find the roots of this equation. System of non linear equations A nonlinear system of equations contains at least one nonlinear equation. A system has no solution. A system has many solutions. Next A system has one solution. The Main Menu اPrevious A system has two solutions It is one of the first numerical methods to find the root of a nonlinear equation f(x) = 0 also called binary-search method. The method is based on finding the root between two points. At least one root exists between the two points if the function is real, continuous, and changes sign. If the function does not change sign between the two points then there may not be any root for the equation between the two points. The steps to apply the bisection method to find the root of the equation are: 1- Choose a and b for the root such that f(a) f(b) < 0, or in other words, f (x) changes sign between a and b. 2- Estimate the root xn such that xn = (an + bn) / 2. If f(xn) = 0, then xn is the root of the equation. If f(xn ) 0, f(xn) > 0 ,find the new estimate such that the mid point between a , xn. If f(xn) 0, f(xn) < 0, find the new estimate such that the mid point between xn , b. Next The Main Menu اPrevious It is the numerical method to find the root of a nonlinear equation f(x) = 0 also called method. The steps to apply the simple iteration method to find the root of the equation are: 1- Choose x0. 2- The function f(x) = 0 is transformed into equation x = g(x). 3- Test the convergence g(x 0 ) 1 4- Compute xn+1 = g(xn) 5- If x n 1 x n ε ; end Next The Main Menu اPrevious It is the numerical method to find the root of a nonlinear equation f(x) = 0. The secant method is defined by the recurrence relation: x n 1 x n x n x n 1 f(x n ) f(x n ) f(x n 1 ) It requires two initial values, x0 and x1, which should ideally be chosen to lie close to the root. The steps to apply the Secant method to find the root of the equation are: 1. Select x0 and x1 2. Evaluate f(x0) and f(x1) 3. Evaluate xn+1 4. If x n 1 x n ε ; end Next The Main Menu اPrevious It is the best numerical method and fast way to find the root of a nonlinear equation f(x) = 0. This method is defined by the recurrence relation: f(x n ) f (x n ) The steps to apply the Newton-Raphson method to find the root of the equation are: x n 1 x n 1- From f (x) get f\ (x) f(x 0 ) f (x 0 ) 1. 2. Choose x0and test the convergence 2 [f (x n )] 3. Evaluate x n 1 x n f(x n ) f (x n ) 4. If x n 1 x n ε ; end Next The Main Menu اPrevious Solve, using the Bisection method to find the positive root of the equation: f(x) = x3 + 4 x2 – 10 = 0. Let x =a = 1 x = b =2 n f(1) = –, f(2) = +, then the root of this equation lies between [1, 2]. an a b xn n n 2 bn f(xn) 0 1 2 1.5 2.375 1 1 1.5 1.25 -1.79687 2 1.25 1.5 1.375 0.16211 3 1.25 1.375 1.3125 -0.84839 12 1.36476094 1.365234375 Next The Main Menu 1.364990235 اPrevious -0.003396 Solve, using the simple iteration method to find the positive root : f(x) = x – sin x = 0.25 correct to three decimal places. f(x) = x –sin x - 0.25 = 0. x 0 1 2 1.5 f(x) - 0.25 - 0.091 0.841 0.253 Let x0 =1.2, x = sin x + 0.25. Then x = g(x) = sin x + 0.25. g(x) cosx Test the convergence g ( x0 ) cos1.2 0.362 1 Then xn+1= g(xn) =sin xn +0.25. n 0 1 2 3 4 xn+1 1.182 1.175 1.173 1.172 1.172 Next The Main Menu اPrevious Solve, using the Secant method to find the positive root : f(x) = 0.18x3 – 0.5 x2 -2x +1 correct to four decimal places, where x0 = 4, x1 = 6. x n x n 1 f(x n ) f(x n ) f(x n 1 ) x n x n 1 x n 1 x n (0.18 x 3 n 0.5 xn2 2 xn 1) 3 2 3 2 (0.18 x n 0.5 xn 2 xn 1) (0.18 x n 1 0.5 xn 1 2 xn 1 1) x n 1 x n n xn-1 f(xn-1) xn f(xn) xn+1 1 4 -3.4800 6 9.8800 4.5210 2 6 9.8800 4.5210 -1.6287 4.7303 3 4.5210 -1.6287 4.7303 -0.5967 4.8513 4 4.7303 -0.5967 4.8513 -0.0815 4.8368 5 4.8513 -0.0815 4.8368 -0.0032 4.8373 Next The Main Menu اPrevious Solve, using the Newton - Raphson method to find the positive root : f(x) = 0.18x3 – 0.5 x2 -2x +1 correct to four decimal places, x 1 2 3 4 5 f(x) -1.32 -3.56 -4.64 -3.48 1 Let x0 = 4.5, f (x) 0.54 x 2 x 2, f (x) 1.08 x 1. Test the convergence f(x 0 ) f (x 0 ) (1.7225)(3.86) 0.33 1 [f (x n )]2 (4.435) 2 f(x n ) 0.18 x 3n 0.5 x 2n 2 x n 1 . x n 1 x n x n 1 x n 2 0.54 x n x n 2 f (x n ) n 0 1 2 xn+1 4.8884 4.8382 4.8373 Next The Main Menu اPrevious Let there be given a system of two equations in two unknowns, F (x, y) = 0, G (x, y) = 0. Let x = x0, y = y0 be the approximate values of the root of this system. By using Newton-Raphson’s method The general formula to get the required root is: F x n 1 x n Fy G Gy Fx Fy Gx Gy Next (x n , y n ) , y n 1 y n (x n , y n ) The Main Menu Fx F Gx G Fx Fy Gx Gy (x n , y n ) اPrevious (x n , y n ) . Find the root of the system, F (x, y) = 4 – x2 – y2 = 0, G (x, y) = 1 – ex – y = 0. Given that (x0, y0 ) = (1, -1.7). Two steps are required. Correct to 4D. We get: F Fx 2 x, x Gx Fy F 2 y, y G G 1. ex , G y y x For (x0, y0) = (1, -1.7), we get: Fx 2, Fy 3.4, G x 2.7183, G y 1, F 0.11, G 0.0813. Next The Main Menu اPrevious Then, x1 x 0 F Fy 0.11 G Gy (x 0 , y 0 ) Fx Fy Gx Gy y1 y 0 1 F Gx G Fx Fy Gx Gy 0.0183 1 1.0043 2 3.4 2.7183 1 (x n , y n ) Fx 3.4 2 (x 0 , y 0 ) (x 0 , y 0 ) 0.11 2.7183 0.0183 1.7299. 2 3.4 2.7183 1 For (x1, y1) = (1.0043, -1.7299), we get: Fx 2.0086, G x 2.7300, Fy 3.4598, G y 1,F 0.0012, G 0.0001. 0.0012 3.4598 2.0086 0.0012 0.0001 1 x 2 1.0043 1.0042, 2.7300 0.0001 2.0086 3.4598 y 2 1.7299 1.7296 2.0086 3.4598 2.7300 1 2.7300 1 Next The Main Menu اPrevious