tailieunhanh - A textbook of Computer Based Numerical and Statiscal Techniques part 35

A textbook of Computer Based Numerical and Statiscal Techniques part 35. 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. | 326 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 2 12 7 0 32 4 12 7 32 9 7 12 32 15 12 14 32 8 7 3 45 708 Hence the required area of the cross-section of the river 708 sq. m. Ans. ii By Simpson s one-third rule 80 h I ydx - y0 y8 4 y1 y3 y5 y7 2 y2 y4 y6 0 3 10 y 0 3 4 4 9 15 8 2 7 12 14 710 Hence the required area of the cross-section of the river 710 sq. m. Ans. Example 10. Evaluate 1 dx 1 - by dividing the interval of integration into 8 equal parts. Hence find loge 2 approximately. Sol. Since the interval of integration is divided into an even number of subintervals we shall use Simpson s one-third rule. 1 Here y- 1 f f x f ß 1_ 1 _ 8 yo_f _ 1 0_ 1 yi I8J 1 1 9 y2_fI i--5 8 _ f 31_ 8 _ 41_ 2 _ r 51_ 8 y3 8 J ii y4 ß 8J 3 y5 ß 8J 13 r6 l 4 r7 1_ 8 1 y6 - J I 8 I-7 y7 - f I 8 I-15 and y8 - f i - 0- Hence x 0 1 8 2 8 3 8 4 8 5 8 6 8 7 8 1 y 1 8 4 8 2 8 4 8 1 9 5 11 3 13 7 15 2 yo y1 y2 y3 y4 y5 y6 y7 ys By Simpson s one-third rule I n h y0 ys 4 y 1 y3 y5 y 7 2 y2 y4 y 0 1 x 3 IT1 11 418 ß 214 2 4 Y Hence h 1 8 24 _ 2 J 9 11 13 15 J 5 3 7 J_ . Ans. NUMERICAL DIFFERENTIATION AND INTEGRATION 327 Since f 1 ÈL 10g l X J log 2 0 1 x 0 1oge 2 . Ans. EULER-MACLAURIN S FORMULA This formula is based on the expansion of operators. Suppose AF x f x then an operator A 1 called inverse operator is defined as F x A-1 f x Again we have AF x f x0 F x1 - F xo f x0 F x2 - F x1 f x1 F xn - F xn-1 f xn-1 Adding all these we get n-1 F xn - F x0 X f xi i 0 . 1 where x0 x1 xn are the n 1 equidistant values of x with interval h. Now F x A-1f x E -1 -1 f x ehD -1 -1 f x h3 D3 ------ x I-1 3 f x 2 3 f x fhD h2D2 hD 1 - 1 I ---- I 2 3 - u 1D-1 LfhD d 1 -1 -2 fhD h2D2 h I 2 3 . I 2 I 2 3 -4 f x _1 L hD h2D2 h4 D4 3 D I 1------------------------ . f x h I 2 12 720 f h jf x dx -1 f x 12f x - 72 f x . 2 328 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Between limits x x0 and x xn from equation 2 we have F xn -F xo 1 f nf x dx-1 f xn -f xo - f xn -f x0 h xg 2