Interpolation & Polynomial Approximation. Hermite Interpolation I

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1 Interpolation & Polynomial Approximation Hermite Interpolation I Numerical Analysis (th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011 Brooks/Cole, Cengage Learning

2 Outline 1 Osculating & Hermite Polynomials 2 The Precise Form of the Hermite Polynomials 3 Example: Constructing the Hermite Polynomial using Lagrange Polynomials Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 2 / 23

3 Outline 1 Osculating & Hermite Polynomials 2 The Precise Form of the Hermite Polynomials 3 Example: Constructing the Hermite Polynomial using Lagrange Polynomials Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 3 / 23

4 Hermite Interpolation: Osculating Polynomials Osculating Polynomials Osculating polynomials generalize both the Taylor polynomials and the Lagrange polynomials. Suppose that we are given n + 1 distinct numbers x 0, x 1,...,x n in [a, b] and nonnegative integers m 0, m 1,...,m n, and m = max{m 0, m 1,...,m n }. The osculating polynomial approximating a function f C m [a, b] at x i, for each i = 0,...,n, is the polynomial of least degree that has the same values as the function f and all its derivatives of order less than or equal to m i at each x i. Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 4 / 23

5 Hermite Interpolation: Osculating Polynomials Osculating Polynomials (Cont d) The degree of this osculating polynomial is at most M = n m i + n i=0 because the number of conditions to be satisfied is n i=0 m i + (n + 1), and a polynomial of degree M has M + 1 coefficients that can be used to satisfy these conditions. Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 5 / 23

6 Hermite Interpolation: Osculating Polynomials Definition: Osculating Polynomial Let x 0, x 1,...,x n be n + 1 distinct numbers in [a, b] and let m i be a nonnegative integer for i = 0, 1,...,n. Suppose that f C m [a, b], where m = max 0 i n m i. The osculating polynomial approximating f is the polynomial P(x) of least degree such that d k P(x i ) dx k = dk f(x i ) dx k for each i = 0, 1,...,n and k = 0, 1,...,m i. The osculating polynomial approximating f is the m 0 th Taylor polynomial for f at x 0 when n = 0 and the nth Lagrange polynomial interpolating f on x 0, x 1,...,x n when m i = 0 for each i. Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 6 / 23

7 Hermite Polynomials Osculating Polynomials d k P(x i ) dx k = dk f(x i ) dx k for each i = 0, 1,...,n and k = 0, 1,...,m i. Hermite Polynomials The case when m i = 1, for each i = 0, 1,...,n, gives the Hermite polynomials. For a given function f, these polynomials agree with f at x 0, x 1,...,x n. In addition, since their first derivatives agree with those of f, they have the same shape as the function at (x i, f(x i )) in the sense that the tangent lines to the polynomial and the function agree. Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 7 / 23

8 Outline 1 Osculating & Hermite Polynomials 2 The Precise Form of the Hermite Polynomials 3 Example: Constructing the Hermite Polynomial using Lagrange Polynomials Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 8 / 23

9 Precise Form of the Hermite Polynomials Theorem If f C 1 [a, b] and x 0,...,x n [a, b] are distinct, the unique polynomial of least degree agreeing with f and f at x 0,...,x n is the Hermite polynomial of degree at most 2n + 1 given by H 2n+1 (x) = n n f(x j )H n,j (x) + f (x j )Ĥn,j(x) j=0 j=0 where, for L n,j (x) denoting the jth Lagrange coefficient polynomial of degree n, we have H n,j (x) = [1 2(x x j )L n,j (x j)]l 2 n,j (x) and Ĥ n,j (x) = (x x j )L 2 n,j (x) Continued on the next slide... Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires / 23

10 Precise Form of the Hermite Polynomials H 2n+1 (x) = n n f(x j )H n,j (x) + f (x j )Ĥn,j(x) j=0 j=0 Theorem (Cont d) Moreover, if f C 2n+2 [a, b], then f(x) = H 2n+1 (x) + (x x 0) 2...(x x n ) 2 f (2n+2) (ξ(x)) (2n + 2)! for some (generally unknown) ξ(x) in the interval (a, b). Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 10 / 23

11 Precise Form of the Hermite Polynomials Proof (1/4) First recall that L n,j (x i ) = { 0, if i j 1, if i = j Hence when i j, H n,j (x i ) = 0 and Ĥ n,j (x i ) = 0 whereas, for each i, H n,i (x i ) = [1 2(x i x i )L n,i (x i)] 1 = 1 and Ĥ n,i (x i ) = (x i x i ) 1 2 = 0 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 11 / 23

12 Precise Form of the Hermite Polynomials Proof (2/4) As a consequence H 2n+1 (x i ) = n f(x j ) 0 + f(x i ) 1 + j=0 j i so H 2n+1 agrees with f at x 0, x 1,...,x n. n f (x j ) 0 = f(x i ) To show the agreement of H 2n+1 with f at the nodes, first note that L n,j (x) is a factor of H n,j (x), so H n,j (x i) = 0 when i j. j=0 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 12 / 23

13 Precise Form of the Hermite Polynomials Proof (3/4) In addition, when i = j we have L n,i (x i ) = 1, so H n,i (x i) = 2L n,i (x i) L 2 n,i (x i) + [1 2(x i x i )L n,i (x i)]2l n,i (x i )L n,i (x i) = 2L n,i (x i) + 2L n,i (x i) = 0 Hence, H n,j (x i) = 0 for all i and j. Finally, Ĥ n,j (x i) = L 2 n,j (x i) + (x i x j )2L n,j (x i )L n,j (x i) so Ĥ n,j (x i) = 0 if i j and Ĥ n,i (x i) = 1. = L n,j (x i )[L n,j (x i ) + 2(x i x j )L n,j (x i)] Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 13 / 23

14 Precise Form of the Hermite Polynomials Proof (4/4) Combining these facts, we have H 2n+1 (x i) = n f(x j ) 0 + j=0 n f (x j ) 0 + f (x i ) 1 = f (x i ) j=0 j i Therefore, H 2n+1 agrees with f and H 2n+1 with f at x 0, x 1,...,x n. Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 14 / 23

15 Outline 1 Osculating & Hermite Polynomials 2 The Precise Form of the Hermite Polynomials 3 Example: Constructing the Hermite Polynomial using Lagrange Polynomials Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 15 / 23

16 Constructing the Hermite Polynomial Example: Constructing H 5 (x) Use the Hermite polynomial that agrees with the data listed in the following table to find an approximation to f(1.5). k x k f(x k ) f (x k ) Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 16 / 23

17 Constructing the Hermite Polynomial H 5 (x) Solution (1/5) We first compute the Lagrange polynomials and their derivatives. This gives L 2,0 (x) = (x x 1)(x x 2 ) (x 0 x 1 )(x 0 x 2 ) = 50 x2 175 x L 100 2,0 (x) = x 175 L 2,1 (x) = (x x 0)(x x 2 ) (x 1 x 0 )(x 1 x 2 ) = 100 x x 247 L 2,1 (x) = 200 x Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 17 / 23

18 Constructing the Hermite Polynomial H 5 (x) Solution (2/5) and, finally L 2,2 = (x x 0)(x x 1 ) (x 2 x 0 )(x 2 x 1 ) = 50 x2 145 x L 100 2,2 (x) = x 145 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 18 / 23

19 Constructing the Hermite Polynomial H 5 (x) Solution (3/5) The polynomials H 2,j (x) and Ĥ2,j(x) are then ( 50 H 2,0 (x) = [1 2(x 1.3)( 5)] ( 50 = (10x 12) x2 175 x x2 175 x ) 2 ( 100 H 2,1 (x) = 1 x x 247 ( 50 H 2,2 (x) = 10(2 x) x2 145 x ) 2 ) 2 ) 2 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 1 / 23

20 Constructing the Hermite Polynomial H 5 (x) Solution (4/5) ( 50 Ĥ 2,0 (x) = (x 1.3) ( 100 Ĥ 2,1 (x) = (x 1.6) x2 175 x ) 2 x x 247 ( 50 Ĥ 2,2 (x) = (x 1.) x2 145 x ) 2 ) 2 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 20 / 23

21 Constructing the Hermite Polynomial H 5 (x) Solution (5/5): and finally... H 5 (x) = H 2,0 (x) H 2,1 (x) H 2,2 (x) Ĥ2,0(x) Ĥ2,1(x) Ĥ2,2(x) so that ( ) ( ) 4 64 H 5 (1.5) = ( ) ( ) = a result that is accurate to the places listed. ( ) ( Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 21 / 23 )

22 Constructing the Hermite Polynomial Observation Although the theorem provides a complete description of the Hermite polynomials, it is clear from this example that the need to determine and evaluate the Lagrange polynomials and their derivatives makes the procedure tedious even for small values of n. Remedy We will turn to an alternative method for generating Hermite approximations that has as its basis the Newton interpolatory divided-difference formula at x 0, x 1,...,x n, that is, P n (x) = f[x 0 ] + n f[x 0, x 1,...,x k ](x x 0 ) (x x k 1 ) k=1 Numerical Analysis (Chapter 3) Hermite Interpolation I R L Burden & J D Faires 22 / 23

23 Questions?