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A textbook of Computer Based Numerical and Statiscal Techniques part 10

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 2.8233 n cos n Similarly successive iterations are x2 2.7986 x3 2.7984 x4 2.7984. Since x3 x4 hence the required root is 2.798 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 0.6839397 1 2e j 2 e j Now f x1 0.3553424 and f x1 3.337012 So that Second approximation 0.3553424 A x2 0.6839397 - I 3 337012 0.5774545 Third approximation x3 0.5672297 Similarly x4 0.5671433 Hence the required root is 0.5671 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 1.92 1 i x2 1.915 1.908z ALGEBRAIC AND TRANSCENDENTAL EQUATION IT Since imaginary roots occur in conjugate pairs roots are 1.915 1.908 upto 3 places of decimal. Assuming other pair of roots to be a p then a i ß a - iß Sum 1.915 1.908 2a 3.83 5 1.915 - 1.908 a 0.585 Also products of roots are a2 p2 1.915 2 1.908 2 60 p 2.805 Hence other two roots are 0.585 2.805 . 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 2.9 say then