A textbook of Computer Based Numerical and Statiscal Techniques part 18

A textbook of Computer Based Numerical and Statiscal Techniques part 18. 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. | 156 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES ERRORS IN POLYNOMIAL INTERPOLATION Let the function y x having n 1 points x y for i 0 1 2 n be continuous and differentiable. Let this function be approximated by a polynomial pn x of degree not exceeding n such that pn xi yi i 0 1 2 n . a Now to obtain approximate values of y x at some points other than those given by a let y x - Pn x vanishes for x x0 x1 x2 . xn and y x - pn x Lnn 1 x . b Where nn i x x - xo x - xi x-xn and L is to be determined such that equation b holds for any intermediate values of x say x x x0 x xn Therefore L . c n. 1 x Now construct a function F x such that F x y x - Pn x - Lnn 1 x . d Where L is given by equation c . it is clear that F xo F xi F x2 . F xn F x 0 . F x vanishes n 2 times in the interval x0 x xn According to Rolle s Theorem F x must vanish n 1 times F x must vanish n times etc in the interval x0 x xn. In particular Fn 1 x must vanish once in the interval. Let this part be x q x0 q xn. On differentiating equation d n 1 times with respect to x and substituting x q. We get 0 yn 1 n - L n 1 L yn 1 n n 1 . e On comparing c and e we get y x - Pn x n In nn 1 x Dropping the prime on x y x - Pn x Vyn In nn 1 x X0 x xn This is required expression for polynomial interpolation. CALCULUS OF FINITE DIFFERENCES 157 . DIFFERENCES OF ZEROS Let y xm be a function of x where m be a positive integer. If x 0 1 2 3. we get 0m 1m 2m 3m . respectively. If we constructed difference table for this data we get leading differences is as A0m A20m A30m and these differences are called differences of zeros. To find differences of zeros Anxm E - 1 xm En - nC1En-11 nC2En-2I2 -. -1 In xm En xm - nC1En-1 I xm nC2En-2 I2 xm - -1 I xm x n m - nC1 x n - 1 m nC2 x n - 2 m - -1 xm Taking interval of differencing h 1 This expression for x 0 becomes Anxm x 0 nm - C1 n - 1 m C2 n - 2 m -. nCn-1 -1 n-1 . a The expression a for Anxm x 0 is written as a 0m and is known as differences of zero. If we .

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