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A textbook of Computer Based Numerical and Statiscal Techniques part 33

A textbook of Computer Based Numerical and Statiscal Techniques part 33. 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. | 306 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Here x 2.04 h 0.02 u - 1at x 2.03 n h dx h _ 2.03 - 2.04 _ 0.01 u 0.02 - 0.02 1 2 Then by Newton s backward formula we have u u 1 - 2 u u 1 u 2 3 u u 1 u 2 u 3 3 y x yn uVy v w 2y ------f------V iyn -----------------V iyn 2 6 24 Differentiating w.r.t. x we have 1 L 2u 1 2 3u2 6u 2 3 4u3 18u2 22u 6 4 y x T Vyn V2yn ---V3yn ------------V4yn h 2 6 24 1 y 2.03 ----- 0.02 -0.0090 0 1Ï 2 2 22--- -0.0002 . 1 . 2 1 Y C 1A2 C 1A 4I- 18I- 22I- 6 I 2 I 2 I 2 X 2 2 YY -22------------------- -0.0004 50 -0.0090 0.000008 0.000017 -0.44875 Ans. Again differentiating equation 2 w.r.t. x 1 L2 6u 6 _3 12u2 36u 22 4 hi V yn V yn --------24-------V yn . 1 y x T 1 y 2.03 -------7 0.02 2 1 A2 C 1A z x 121- 361- 22 1 A I 2 I 2 -0.0002 1 - - 11 - 0.0002 X 2 2 24 X 2 ---- -0.0004 2500 - 0.0002 - 0.0001 - 0.00012 y 2.03 -1.05. Ans. Example 9. Find f 5 from the following table x 1 2 4 8 10 f x 0 1 5 21 27 Sol. Here the arguments are not equally spaced. So we use Newton s divide difference formula. NUMERICAL DIFFERENTIATION AND INTEGRATION 307 Difference table x f x Af x A2f x A3f x A4 f x 1 0 1 2 1 2 1 3 0 4 5 1 1 4 3 1 144 16 8 10 21 27 3 1 6 Newton s divided difference formula is given by f x f Xo X- Xo Af Xo x- Xo X- X1 A2f Xo x- Xg x- x1 x- x2 A3 f xo x - xo x - x1 x - x2 x - x3 A4 f xo . 1 Differentiating 1 w.r.t. x we get f x Af Xo 2x Xo Xi A f Xo x Xi x x2 x Xo x x2 x Xo x Xi A f Xo . At x 5 1 f 5 1 io -1 - 2 5 - 2 5 - 4 5 -1 5 - 4 5 -1 5 - 2 x o _.-1 5 - 2 5 - 4 5 - 8 5 -1 5 - 4 5 - 8 5 -1 5 - 2 5 - 8 5 -1 5 - 2 5 - 4 f 1 7 f1L 1 -9 -12 - 36 12 1 7 -45 3 144 3 144 Hence f 5 3.6458. Ans. Example 10. Find f 5 from the data given below x 2 4 9 13 16 21 29 f x 57 1345 66340 402052 1118209 4287844 21242820 Sol. Here the arguments are not equally spaced and therefore we shall apply Newton s divided difference formula. f x f xo x - xo Af Xg x - x0 x - X1 A2f x0 x - x0 x - X1 X - x2 I I A3 f xg x - xg x - x1 x - x2 x - x3 A4 f xo . 1 308