A textbook of Computer Based Numerical and Statiscal Techniques part 28

A textbook of Computer Based Numerical and Statiscal Techniques part 28. 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. | 256 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 9. Using Newton s divided difference formula prove that f x f 0 xAf -1 X 1 X A2 f -1 X 1 3 X 1 A3f -2 . Sol. Taking the arguments 0 -1 1 -2 . the Newton s divided difference formula is xA A2 A3 f X f 0 -1 f 0 X X 1 -1 1 f 0 X X 1 X 1 -1 1-2 f 0 . . 1 23 f x f 0 x A f -1 x x 1 A1 f -1 x x 1 x - 1 0A-1 1 f -2 . Now A f -1 f 0 - -ft-1 Af -1 0 - -1 ft f -1 i-pj Af 0 -Af -1 1 Af 0 -Af -1 1 A2f -1 A3 f -2 TTE A2 f -1 - A2 f -2 -1 0 1 f 2 1 - -2 _0 1 -1 02 v f 1 Ta2 f -1 A2 f -2 32 2 A3 f -2 A3 f -2 Q 9 ------ ---- and so on. 3 x 2 3 Substituting these values in 1 f x f 0 x Af -1 X 1 X A2f -1 X 1 X X - 1 A3f -2 . 2 3 Example 10. Using Newton s divided difference formula calculate the value of f 6 from the following data x 1 2 7 8 f x 1 5 5 4 INTERPOLATION WITH UNEQUAL INTERVAL 257 Sol. The divided difference table is x f x Af x A2f x A3f x 1 1 4 2 5 0 2 - 3 1 14 7 5 -1 1 - 6 8 4 Applying Newton s divided difference formula 2 A 1 A f x 1 x - 1 4 x - 1 x - 2 I - 3 x - 1 x - 2 x - 7 I I 2 A 1 f 6 1 20 5 4 I - 3 I 5 4 -1 I 14 Example 11. Find the value of log10 656 using Newton s divided difference formula from the data given below x 654 658 659 661 log10x Sol. Divided difference table for the given data is as x 105 f x 105 Af x 105 A2 f x 105 A3 f x 654 658 659 661 281560 281820 281890 282020 260 65 4 70 70 1 130 65 2 70_65 1 5 65 - 70 3 -1 7 258 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES For the given argument the divided difference formula is f x y0 x - x0 Ayo x - Xo x - x1 A2y0 x - x0 x - x1 x - x2 A3y0 105 f x 281560 x - 654 65 x - 654 x - 658 .1 x - 654 x - 658 x - 659 On Substituting x 656 we get 105 f x 281560 2 x 65 -4 -4 -3 x 281560 130 - 4 105 f x 281690. 56 f x log10 656 PROBLEM SET 1. By means of Newton s divided difference formula Find the value of f 8 and f 15 from the following table x 4 5 7 10 11 .