A textbook of Computer Based Numerical and Statiscal Techniques part 14. 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. | 116 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 14. Evaluate the following I. A2 cos 2x II. A2 3ex III. A tan-1 x IV. A x cos x the interval of differencing being h. Sol. I. We have A2 cos 2x E - 1 2 cos 2x because A E - 1. E2 - 2 E 1 cos 2x E2 cos 2x - 2E cos 2x cos 2x cos 2x 4h - 2 cos 2x 2h cos 2x cos 2x 4h - cos 2x 2h - cos 2x 2h cos 2x 2 sin 2x 3h sin - h - 2 sin 2x h sin h - 2 sin h sin 2x 3h - sin 2x h - 2 sin h 2 cos 2x 2h sin h - 4 sin2 h cos 2x 2h . II. We have A 3ex 3 A ex 3 ex h - ex 3ex eh-1 A2 3ex A A3ex A 3ex eh-1 3 eM Aex 3 eh-1 ex h - ex 3 eM e eh-1 3ex eM 2. III. We have A tan -1 x tan-1 x h - tan-1 x tan-1 tan-1 IV. We have A x cos x A x A cos x x h - x cos x h - cos x h 2 sin 2-V s n -h h h h - 2 sin I x I sin 2 x h -x 1 x h x h_ 1 xh x2 Now sin x h E E sin x h Example 15. Evaluate sin x h ---------- where h being the interval of differencing. E E sin x h Sol. To evaluate the given problem we use the operator property that is A E - 1 A2 srn x h E - 1 2 sin X h E -1 2 sin x h sin x 2h E CALCULUS OF FINITE DIFFERENCES 117 E - 2 E-1 sin x h E2 - 2E 1 sin x h sin x 2h sin x 2h - 2 sin x h sin x sin x 3h - 2 sin x 2h sin x h sin x 2h 2 sin x 2h cos h -1 2 sin x h cos h - 1 ---------------------- sin x 2h 2 cos h - 1 sin x h - 1 . r i Example 16. Show that B m 1 n -1 m Am I I where m is a positive integer. Sol. We know that ï e nx dx - . o n Therefore ra Am J e nx dx o Am 1 n or ra J Am e nx o dx Am 1 n where for Am e nx n is to be regarded variable and x is to be regarded as constant. Now Ame-nx Am-1 e - e-nx Am-1 e-nx e-x - 1 e-x - 1 Am-1 e-nx e-x - 1 2 Am-2 e-nx . e-x - 1 m e-nx Therefore J0 e nx e x -1 m dx Am Put e-x z so that -e-x dx dz or dx - 1 z dz. or or or Then J0 zn z-1 m -1 z dz Am n 1 n-1 1 -1 m J z 1 - z m dz Am I - 1 1 zn-1 1 - z m 1 -1 dz -1 m Am I n o 1 B m 1 n -1 m Am I I n 118 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 17. Show that ex ex. Eex A 2ex the interval of differencing being h. Sol. Let f x
