A first order divided difference

Transcription

1 A first order divided difference For a given function f (x) and two distinct points x 0 and x 1, define f [x 0, x 1 ] = f (x 1) f (x 0 ) x 1 x 0 This is called the first order divided difference of f (x). By the Mean-value theorem, f (x 1 ) f (x 0 ) = f (c)(x 1 x 0 ) for some c between x 0 and x 1. Thus f [x 0, x 1 ] = f (c) and the divided difference is very much like the derivative, especially if x 0 and x 1 are quite close together. In fact, ( ) f x1 + x 0 f [x 0, x 1 ] 2 is quite an accurate approximation of the derivative.

2 Second order divided difference Given three distinct points x 0, x 1, and x 2, define f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0 This is called the second order divided difference of f (x). We can show that f [x 0, x 1, x 2 ] = 1 2 f (c) for some c intermediate to x 0, x 1, and 2. In fact, f (x 1 ) 2f [x 0, x 1, x 2 ] in the case when nodes are evenly spaced, x 1 x 0 = x 2 x 1.

3 Example Consider the table x cos x Let x 0 = 1, x 1 = 1.1, and x 2 = 1.2. Then f [x 0, x 1 ] = = f [x 1, x 2 ] = = f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x1] x 2 x ( ) = = ( ) For comparison, f x1 + x 0 = sin(1.05) =

4 General divided differences Given n + 1 distinct points x 0,..., x n, with n 2, define f [x 0,..., x n ] = f [x 1,..., x n ] f [x 0,..., x n 1 ] x n x 0 This is a recursive definition of the n-th order divided difference of f (x), using divided differences of order n. We can also relate this to the derivative as follows: f [x 0,..., x n ] = 1 n! f (n) (c) for some c intermediate to the points {x 0,..., x n }.

5 Order of nodes We see that f [x 0, x 1 ] = f (x 1) f (x 0 ) x 1 x 0 = f (x 0) f (x 1 ) x 0 x 1 = f [x 1, x 0 ] The order of x 0 and x 1 does not matter. We have for 2nd order divided difference: f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0 f (x 0 ) = (x 0 x 1 )(x 0 x 2 ) f (x 1 ) + (x 1 x 0 )(x 1 x 2 ) f (x 2 ) + (x 2 x 0 )(x 2 x 1 )

6 Order of nodes Using this formula, we can show that f [x 0, x 1, x 2 ] = f [x 0, x 2, x 1 ] Mathematically, f [x 0, x 1, x 2 ] = f [x i0, x i1, x i2 ] for any permutation (i 0, i 1, i 2 ) of (0, 1, 2).

7 Newton s interpolating polynomial Recall that the general interpolation problem: find a polynomial P n (x) of degree n for which P n (x i ) = y i, i = 0, 1,..., n with given data points (x 0, y 0 ), (x 1, y 1 ),..., (x n, y n ), where {x 0,..., x n } distinct points.

8 Newton s interpolating polynomial Let the data values be generated from a function f (x): y i = f (x i ), i = 0, 1,..., n Using divided differences f [x 0, x 1 ], f [x 0, x 1, x 2 ],..., f [x 0,..., x n ] we can write the interpolation polynomials P 1 (x), P 2 (x),..., P n (x) in a way that is simple to compute. P 1 (x) = f (x 0 ) + f [x 0, x 1 ](x x 0 ) P 2 (x) = f (x 0 ) + f [x 0, x 1 ](x x 0 ) +f [x 0, x 1, x 2 ](x x 0 )(x x 1 ) = P 1 (x) + f [x 0, x 1, x 2 ](x x 0 )(x x 1 )

9 Newton s interpolating polynomial For the case of general n-th degree polynomial, we have P n (x) = f (x 0 ) + f [x 0, x 1 ](x x 0 ) +f [x 0, x 1, x 2 ](x x 0 )(x x 1 ) f [x 0,..., x n ](x x 0 )... (x x n 1 ) Therefore P n (x) = P n 1 (x) + f [x 0,..., x n ](x x 0 )... (x x n 1 ) in which P n 1 (x) interpolates f (x) at points in {x 0,..., x n 1 }

10 Example Consider the table Define x cos x D k f (x i ) = f [x i,..., x i+k ] i x i f (x i ) D 1 f (x i ) D 2 f (x i ) D 3 f (x i ) D 4 f (x i ) These were computed using the recursive definition f [x 0,..., x n ] = f [x 1,..., x n ] f [x 0,..., x n 1 ] x n x 0

11 Then P 1 (x) = (x 1) P 2 (x) = P 1 (x) (x 1)(x 1.1) P 3 (x) = P 2 (x) (x 1)(x 1.1)(x 1.2) P 4 (x) = P 3 (x) (x 1)(x 1.1)(x 1.2)(x 1.3) We can now estimate cos(1.05) using various order polynomials and the results are tabulated below: n P n (1.05) Error

12 Error of interpolation Let P n (x) be a n-th degree interpolating polynomial that agree with f (x) on n + 1 distinct points. Let the error E(x) be defined as E(x) = f (x) P n (x) Since E(x i ) = 0 for i = 0,..., n, we can write E(x) = (x x 0 )(x x 1 )... (x x n )g(x), where the function g(x) has to be determined. We see that f (x) P n (x) (x x 0 )(x x 1 )... (x x n )g(x) = 0

13 Error of interpolation Let us introduce a new function W (t) = f (t) P n (t) (t x 0 )(t x 1 )... (t x n )g(x) Note that W (t) = 0 at t = x, x 0, x 1,..., x n. Therefore, W (t) has n + 2 zeros. We assume that W (t) is continuous and differentiable and use mean value theorem. Therefore, W (c) = 0 if c is a number intermediate in {x 0, x 1,..., x n, x}.

14 Error of interpolation Using mean value theorem repeatedly, we get W n+1 (c) = 0 = f (n+1) (c) (n + 1)!g(x) for c intermediate in {x 0, x n, x}. Thus Hence g(x) = f (n+1) (c) (n + 1)! E(x) = (x x 0 )(x x 1 )... (x x n ) f (n+1) (c) (n + 1)!

15 Error of interpolation Example: If the function f (x) = sin x is approximated by a polynomial of degree 9 that interpolates f (x) at ten points in the interval [0, 1], estimate the interpolation error.