TAILIEUCHUNG - A textbook of Computer Based Numerical and Statiscal Techniques part 15

A textbook of Computer Based Numerical and Statiscal Techniques part 15. 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. | 126 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Similarly 2 Similarly An 1 -1 -2 -3 . -n x x x 1 x 2 . x n -1 nn x x 1 x 2 . x n A abcx a A bcx a bc x 1 - bcx a bcx bc - bcx a bc-1 bcx. A2 abcx AA abx A a bc - 1 bcx a bc - 1 Abcx a bc -1 2 bcx Proceeding in the same manner we get An abcx a bc - 1 nbcx. Example 29. If p q r and s be the successive entries corresponding to equidistant arguments in a table show that when third differences are taken into account the entry corresponding to the argument half way between the arguments of q and r is A -1B where A is the arithmetic mean of q r and B is the arithmetic mean of 3q - 2p - s and 3r - 2s - p. Sol. On taking h being the interval of differencing the difference table is as x ux Aux A2 ux A3 ux a p q - p a h q r - q r - 2q p s - 3r 3q - p a 2h r s - r s - 2r q a 3h s The argument half way between the arguments of q and r is a h a 2h . a 3 h- Hence the required entry is given by U 3 2 h X 1 A 3 2 Ua 3 3 3 1 1A 1 -A . A2 . I A3 2 222 22 I 2 3 ua Higher order differences being neglected . 2 1 3. Therefore u u Au A u-----------A u. tl uit ua 3 2 h a 2 a 8 a 16 a 3. 3 A x 1 p g q - p -g r - 2q p - s -3r 3q- p 2 8 16 CALCULUS OF FINITE DIFFERENCES 127 m 3 3 1 PI 1 2 8 16 3 3 3 2 4 16 I- s 16 19 9 1 --p -q -r-s 16 16 16 16 1 11A 1 p q r --s 16-------------- 16 2 2 16 1 x 1 - q r q r - p- s 2 16 1 Again A arithmetic mean of q and r 2 q r B Arithmetic mean of 3q - 2p - s and 3r - 2s - p is 1 3 3q - 2p - s 3r -2s - p q r - s - p . A B q r - s - p . 24 2 16 Substituting this value in 1 we get ua 3 2 h A 2 B. Example 30. Given u0 u1 u2 u3 u4 and u5. Assuming that fifth order differences to be constant. 1 c 2 Show that u 1 2 2 25 c - b 3 a - c -------7777------. where a u0 u5 b u1 u4 c u2 u3 256 U i 21 2 Sol. . E5 2u0 1 A 5 2u0 5 i 5 -1 5 . .21 2 . 2 A i------A2 2 2 5 I 5 Y 5 Y 5 Y 5 A 5I3 5 -1 I 5 - 2 I 5 - 3 I 5 - 4 212 X 2 X 2 X 2 a5 5 u0 5 A uo Auo -A uo ---A uo-----A uo .

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