A textbook of Computer Based Numerical and Statiscal Techniques part 10. By joining statistical analysis with computer-based numerical methods, this book bridges the gap between theory and practice with software-based examples, flow charts, and applications. Designed for engineering students as well as practicing engineers and scientists, the book has numerous examples with in-text solutions. | 76 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES First approximation The first approximation is given by x1 x0 x0 sin x0 cos x0 x0 cos x0 x1 n - n sin n cos n n cos n Similarly successive iterations are x2 x3 x4 . Since x3 x4 hence the required root is correct to three places of decimal. Example 9. Find the real root of the equation x e-x using the Newton-Raphson s method. Sol. We have f x xex - 1 then f x 1 x ex Let x0 1 then First approximation x 1 - 11 1 11 1 2e j 2 e j Now f x1 and f x1 So that Second approximation A x2 - I 3 337012 Third approximation x3 Similarly x4 Hence the required root is correct to 4 decimal places. Example 10. Using the starting value 2 1 i solve x4 - 5x3 - 20x2 - 40x 60 0 by Newton-Raphson s method given that all the roots of the equation are complex. Sol. Let f x x4 - 5x3 20x2 - 40x 60 So that f x 4x3 - 15x2 40x - 40 Therefore Newton-Raphson method gives x n 1 _ x _ f xn n f n x4 - 5x33 20x2 - 40xn 60 x n 1 Xn 4x2 - 15x2 40xn - 40 3x4 - 10x3 20xn - 60 x n 1 4x3 - 15x2 40xn - 40 Put n 0 take x0 2 1 i by trial we get x1 1 i x2 ALGEBRAIC AND TRANSCENDENTAL EQUATION IT Since imaginary roots occur in conjugate pairs roots are upto 3 places of decimal. Assuming other pair of roots to be a p then a i ß a - iß Sum 2a 5 - a Also products of roots are a2 p2 2 2 60 p Hence other two roots are . Example 11. Apply Newton s formula to prove that the recurrence formula for finding the nth roots of a is xi 1 n -1 x a n-nxn Hence evaluate 240 1 5. Sol. Let x a1 n xn a or xn - a 0 Let f x xn - a 0 f x nxn1. Now by Newtons s-Raphson method we have x i x - fN or xi 1 Now to find the value of 240 1 5 n -1 x a nxf 1 . 1 We know that 243 1 5 35 1 5 3 Take a 240 and n 5 we get xi 1 4x5 240 5x4 . 2 First approximation Let i 0 x x0 say then
