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A textbook of Computer Based Numerical and Statiscal Techniques part 32

A textbook of Computer Based Numerical and Statiscal Techniques part 32. 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. | 296 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Aliter We know that I A E ehD D dx hD log 1 A or D2 also D3 D A A2 2 A3 3 A4 4 A5 5 A6 6 1 r a2 A3 A4 A5 A6 A - - h 2 3 4 5 6 1 1 h2 . A2 A3 A4 A5 A6 A 4 P h2 L 2 a 3 11 a 4 2 A3 A4 12 -12 2 1 h3 A2 A3 A4 - 3 4 R h3 L 3 3 4 3 -A4 2 3 A 2 3 4 5 2 3 4 Applying equations 1 2 and 3 for y0 we get 6 1 2 . 1 3 1 4 - 2 A yo 3 A yo - 4 A y0 and d 1 dx J x x0 1 r. h K d2 y dx2 Jx xo 11 1 r .2 .3 11.2 72 A yo -A yo A yo h L 12 12 d3y dx3 -lx xo 1 Fa 3 3.4 h3 A y0 - 2A y0 6.2.2 Derivatives Using Newton s Backward Difference Formula Newton s backward interpolation formula is given by u u 1 2 u u 1 u 2 3 y yn u Vyn 2 V2 yn - - V3 yn 3 1 where u X n Differentiating both sides of equation 1 with respect to x we get dy 1 V7 T T Vyn dx h 2 2u 1 2 3u 6u 2 3 ------V2y ------------V3y 2 n ry f 7 n 3 2 NUMERICAL DIFFERENTIATION AND INTEGRATION 297 At x xn u 0. Therefore putting u 0 in 2 we get or and and and and dy 1 n 1 vz2 1V73 Tx h pn 2 Vyn 5 x xn Again differentiating both sides of equation 2 w.r.t. x we get d2 y 1 dx2 h2 2 6u 6 3 12u2 36u 22 4 VWn V3 yn ------------V4 yn 6 24 At x xnu 0 . Therefore putting u 0 in 3 we get Similarly d2 y dx2 x xn 1 V7 2 Vz3 11 74 7T V yn V yn TTV yn h2 L 12 d3 y dx3 x xn -4 v 3 yn -h3 L 2 3 V4 yn and so on. . 3 . 4 . 5 Aliter We know that 1 -V E 1 e hD -hD log 1 -V V2 V3 V4 V5 - V 2 3 4 5 1 D - V h V2 V3 V4 V5 -----1---- 2 3 4 5 D2 V2 1 V3 11V4 h2 L 2 12 d3 23 V3 3 V4 . Applying these identities to yn we get dy dx d2 y dx2 n d3 y dx3 n _1 L 1 h LVyn 2 2 V2 yn 1V3 yn 1V4 yn -4 v 2 yn v 3 yn h2 L Jn Jn 12 -4 v 3 yn -h L ÿn 2 11 4 12 V yn 3 V 4 yn 298 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.2.3 Derivatives Using Stirling s Formula If we want to determine the values of the derivatives of the function near the middle of the given set of arguments. We may apply any central difference formula. Therefore using Stirling s formula we get. y yo u Ayo Ay-i T u2 .2 -1 A2y_ 1 2 2 1 3 u u2 -1 f