Neville s Method. MATH 375 Numerical Analysis. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Neville s Method

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1 Neville s Method MATH 375 Numerical Analysis J. Robert Buchanan Department of Mathematics Fall 2013

2 Motivation We have learned how to approximate a function using Lagrange polynomials and how to estimate the error in such an approximation. Today we will learn how to interpolate between data values found in a table without knowing the function that generated the values. We will see that we can perform the interpolation without explicitly writing out the interpolating polynomial.

3 Interpolation from Data Suppose we had only the following table of data for f (x): x f (x) Approximate f (27.5) using polynomial interpolation.

4 Interpolation from Data Suppose we had only the following table of data for f (x): x f (x) Approximate f (27.5) using polynomial interpolation. Linear approximation: P 1 (27.5) = ( ) 22.2) f (22.2)+(27.5 f (32.0) ( ) ( )

5 Higher Order Approximations Quadratic approximation: (two choices) P 2 (27.5) = ˆP 2 (27.5) = ( )( ) ( )( ) f (10.1) ( )( ) + ( )( ) f (22.2) ( )( ) + f (32.0) ( )( ) ( )( ) ( )( ) f (22.2) ( )( ) + ( )( ) f (32.0) ( )( ) + f (41.6) ( )( ) There are also two potential cubic interpolating polynomials and a single quartic polynomial.

6 Remarks Without knowing f (x) we have no idea of the size of the errors in these approximations. The highest degree polynomial does not necessarily deliver the smallest error. Knowing the Lagrange Interpolating Polynomial of degree k does not help us find the one of degree k + 1.

7 Remarks Without knowing f (x) we have no idea of the size of the errors in these approximations. The highest degree polynomial does not necessarily deliver the smallest error. Knowing the Lagrange Interpolating Polynomial of degree k does not help us find the one of degree k + 1. There is a method for using previous calculations to derive higher order interpolations.

8 Lagrange Polynomials Definition If f is a function defined at x 0, x 1,..., x n distinct real numbers and m 1, m 2,..., m k are k distinct integers with 0 m i n, then the Lagrange polynomial that agrees with f at the k points x m1, x m2,..., x mk is denoted P m1,m 2,,m k.

9 Example (1 of 2) Let x 0 = 1, x 1 = 2, x 2 = 4, x 3 = 5, and x 4 = 7 and f (x) = J 0 (x) (Bessel function of the first kind of order zero). Determine P 1,3,4 (x).

10 Example (1 of 2) Let x 0 = 1, x 1 = 2, x 2 = 4, x 3 = 5, and x 4 = 7 and f (x) = J 0 (x) (Bessel function of the first kind of order zero). Determine P 1,3,4 (x). P 1,3,4 (x) = = (x 5)(x 7) (2 5)(2 7) J (x 2)(x 7) 0(2) + (5 2)(5 7) J 0(5) (x 2)(x 5) + (7 2)(7 5) J 0(7) [ 1 15 x x + 7 ] [ J 0 (2) x x + 7 ] J 0 (5) 3 [ x 2 7 ] 10 x + 1 J 0 (7)

11 Example (2 of 2) Let x 0 = 1, x 1 = 2, x 2 = 4, x 3 = 5, and x 4 = 7 and f (x) = J 0 (x). Use P 1,3,4 (3) to approximate J 0 (3).

12 Example (2 of 2) Let x 0 = 1, x 1 = 2, x 2 = 4, x 3 = 5, and x 4 = 7 and f (x) = J 0 (x). Use P 1,3,4 (3) to approximate J 0 (3). P 1,3,4 (x) = (x 5)(x 7) (2 5)(2 7) J (x 2)(x 7) 0(2) + (5 2)(5 7) J 0(5) (x 2)(x 5) + (7 2)(7 5) J 0(7) P 1,3,4 (3) = 8 15 J 0(2) J 0(5) 1 5 J 0(7)

13 Example (2 of 2) Let x 0 = 1, x 1 = 2, x 2 = 4, x 3 = 5, and x 4 = 7 and f (x) = J 0 (x). Use P 1,3,4 (3) to approximate J 0 (3). P 1,3,4 (x) = (x 5)(x 7) (2 5)(2 7) J (x 2)(x 7) 0(2) + (5 2)(5 7) J 0(5) (x 2)(x 5) + (7 2)(7 5) J 0(7) P 1,3,4 (3) = 8 15 J 0(2) J 0(5) 1 5 J 0(7) Actual value: J 0 (3)

14 Theoretical Result Theorem Let f be defined at x 0, x 1,..., x k and x i and x j be two distinct real numbers in this set. Then the kth degree Lagrange polynomial that interpolates f at x 0, x 1,..., x k is P(x) = (x x j) (x i x j ) P 0,1,...,j 1,j+1,...,k(x) (x x i) (x i x j ) P 0,1,...,i 1,i+1,...,k(x).

15 Theoretical Result Theorem Let f be defined at x 0, x 1,..., x k and x i and x j be two distinct real numbers in this set. Then the kth degree Lagrange polynomial that interpolates f at x 0, x 1,..., x k is P(x) = (x x j) (x i x j ) P 0,1,...,j 1,j+1,...,k(x) (x x i) (x i x j ) P 0,1,...,i 1,i+1,...,k(x). This theorem allows us to build a Lagrange polynomial of degree k from two Lagrange polynomials each of degree k 1.

16 Proof Let Q(x) = P 0,1,...,j 1,j+1,...,k (x) and ˆQ(x) = P 0,1,...,i 1,i+1,...,k (x), then P(x) = (x x j)q(x) (x x i ) ˆQ(x). (x i x j )

17 Proof Let Q(x) = P 0,1,...,j 1,j+1,...,k (x) and ˆQ(x) = P 0,1,...,i 1,i+1,...,k (x), then P(x) = (x x j)q(x) (x x i ) ˆQ(x). (x i x j ) If x = x i, then P(x i ) = Q(x i ) = f (x i ). If x = x j, then P(x j ) = ˆQ(x j ) = f (x j ). If l {0, 1,..., k} with i l j then P(x l ) = (x l x j )Q(x l ) (x l x i ) ˆQ(x l ) (x i x j ) = (x l x j )f (x l ) (x l x i )f (x l ) = f (x l ). (x i x j )

18 Proof Let Q(x) = P 0,1,...,j 1,j+1,...,k (x) and ˆQ(x) = P 0,1,...,i 1,i+1,...,k (x), then P(x) = (x x j)q(x) (x x i ) ˆQ(x). (x i x j ) If x = x i, then P(x i ) = Q(x i ) = f (x i ). If x = x j, then P(x j ) = ˆQ(x j ) = f (x j ). If l {0, 1,..., k} with i l j then P(x l ) = (x l x j )Q(x l ) (x l x i ) ˆQ(x l ) (x i x j ) = (x l x j )f (x l ) (x l x i )f (x l ) = f (x l ). (x i x j ) Thus P(x) is a polynomial of degree at most k + 1 which interpolates f (x) at x i for i = 0, 1,..., k and thus P(x) = P 0,1,...,k (x).

19 Recursive Generation of Polynomials P 0,1 (x) = P 1,2 (x) = P 0,1,2 (x) =. 1 [(x x 0 )P 1 (x x 1 )P 0 ] x 1 x 0 1 [(x x 1 )P 2 (x x 2 )P 1 ] x 2 x 1 1 [ ] (x x0 )P 1,2 (x x 2 )P 0,1 x 2 x 0

20 Neville s Method Define polynomial Q i,j (x) = P i j,i j+1,...,i (x) where 0 j i. j represents the degree of the polynomial. i determines the consecutive values of x used. Q i,j (x) can be found recursively. Q i,j (x) = (x x i j)q i,j 1 (x) (x x i )Q i 1,j 1 (x) x i x i j

21 Tabular Calculation Given x 0, x 1,..., x n and f (x 0 ), f (x 1 ),... f (x n ) we can arrange the Lagrange interpolating polynomials in the following table. x 0 f (x 0 ) = P 0 = Q 0,0 x 1 f (x 1 ) = P 1 = Q 1,0 P 0,1 = Q 1,1 x 2 f (x 2 ) = P 2 = Q 2,0 P 1,2 = Q 2,1 P 0,1,2 = Q 2,2.... x n f (x n ) = P n = Q n,0 P n 1,n = Q n,1 P n 2,n 1,n = Q n,2 P 0,1,...,n = Q n,n

22 Example Approximate f (27.5) from the following data. x i f (x i )

23 Example Order points by increasing distance from x = x i Q i,

24 Example x i Q i,0 Q i, = = = = ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

25 Example x i Q i,0 Q i,1 Q i, = = = ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

26 Example x i Q i,0 Q i,1 Q i,2 Q i, = = ( )( ) ( )( ) ( )( ) ( )( )

27 Example x i Q i,0 Q i,1 Q i,2 Q i,3 Q i, = ( )( ) ( )( )

28 Neville s Iterated Interpolation Algorithm Goal: evaluate the interpolating polynomial P on n + 1 distinct numbers x 0,..., x n at x to approximate f (x). INPUT values x, x 0,..., x n, f (x 0 ),..., f (x n ) STEP 1 For i = 0, 1,..., n set Q i,0 = f (x i ). STEP 2 For i = 1, 2,..., n for j = 1, 2,..., i set Q i,j = 1 x i x j [ (x xi j )Q i,j 1 (x x i )Q i 1,j 1 ]. STEP 3 OUTPUT Q. STOP.

29 Homework Read Section 3.2. Exercises: 1a, 3, 5, 7