Newton-Gregory Backward Difference Interpolation polynomial: ... falls towards the end or say in the second half of the data set then it may be better to start ...
f0x1 - f1x · f(x) @ P1(x) = x + ; x - x · f0 +. f ; x - x · (. x - x · ) · f1 -. f ; x - x · f1 -. (f1- f0) ; = f(xn + sh) = Es f(xn), s = ( x - xn )/ h ...
NEWTON'S GREGORY FORWARD INTERPOLATION FORMULA : This formula is particularly useful for interpolating the values of f(x) near the beginning of the set of ...
17.10.2017 · Newton Forward And Backward Interpolation. Interpolation is the technique of estimating the value of a function for any intermediate value of the …
10.06.2015 · Now, Newton’s interpolation or polynomial can be expressed as: N (x) = [y 0] + [y 0, y 1] (x –x 0) + . . . + [y 0, . . .y k ] ( x – x 0) ( x – x 1) . . . ( x – x k-1) This form of Newton’s polynomial can be simplified by arranging x o, x 1, x 2, … x k in consecutively equal space. For simplicity, use the notation h = x i+1 – x i
07.11.2014 · Thus, in order to calculate the value of X from the Newton formula of interpolation, we can either take X o = 1997 and a = 1990 or we can take X o = 1998 and a = 1991. Both will provide the same value of X. Thus: X = X o – a h = 1997 – 1990 2 = 7 2 = 3.5 X = 7 2, f ( a) = 355, Δ f ( a) = 73 Δ 2 f ( a) = – 43, Δ 3 f ( a) = 20, Δ 4 f ( a) = 34
(Next method) 1. Formula & Examples Formula Newton's Backward Difference formula p = x - xn h y(x) = yn + p ∇ yn + p(p + 1) 2! ⋅ ∇2yn + p(p + 1)(p + 2) 3! ⋅ ∇3yn + p(p + 1)(p + 2)(p + 3) 4! ⋅ ∇4yn + ... Examples 1. Find Solution using Newton's Backward Difference formula x = 1925 Solution: The value of table for x and y
Thus the first backward differences are : NEWTON’S GREGORY BACKWARD INTERPOLATION FORMULA: This formula is useful when the value of f(x) is required near the end of the table. h is called the interval of difference and u = ( x – an ) / h, Here an is last term. Example: Input : Population in 1925