3.1 Interpolation and the Lagrange Polynomial

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1 MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall Consider a sample x x 0 x 1 x n y y 0 y 1 y n. Can we get a function out of discrete data above that gives a reasonable estimate of y at the points x for which data are not collected? This process is called interpolation. Most useful functions are polynomials and the following theorem gives us a theoretical back ground for polynomial interpolation. Theorem 1 Weierstrass Approximation Theorem Suppose that f is defined and continuous on [a, b]. For each ɛ > 0, there exists a polynomial P (x) such that f(x) P (x) < ɛ for all x [a, b] Also from 2.6 we know the following useful fact about uniqueness. Theorem 2 Let P (x) and Q(x) be polynomials of degree at most n. If x 1, x 2,..., x k, with k > n, are distinct numbers with P (x i ) = Q(x i ) for i = 1, 2,..., k, then P (x) = Q(x) for all values of x. 3.1 Interpolation and the Lagrange Polynomial One of the oldest problems in Mathematics is concerned with interpolations: constructing a smooth function from a discrete set of data, and constructing an approximation to a given function from simpler functions, typically polynomials. More precisely, 1. Given a set of nodes {x i ; 0 i n} and corresponding data values {y i ; 0 i n}, find the polynomial P n (x) of degree less than of equal to n, such that P n (x i ) = y i, 0 i n. 2. Given a set of nodes {x i ; 0 i n} and a continuous function f(x) find the polynomial P n (x) of degree less than of equal to n, such that P n (x i ) = f(x i ), 0 i n. Since the second one is a special case of the first (take y i = f(x i )), we will present material here in terms of second version of the problem. The Taylor polynomials approximate functions only locally, so alternative methods are needed. Example 1 Suppose (x 0, y 0 ) and (x 1, y 1 ) are given. Let L 0 (x) = x x 1 x 0 x 1 and L 1 (x) = x x 0 x 1 x 0, then is the first-degree interpolating polynomial. P (x) = f(x 0 )L 0 (x) + f(x 1 )L 1 (x) Similarly we first construct, for each k = 0, 1,..., n, a function L n,k (x) with the property that { 0 when i k L n,k (x i ) = 1 when i = k. (1)

2 MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall then P (x) = y 0 L n,0 (x) + + y n L n,n (x) = n y k L n,k (x) will do the job. Moreover this polynomial, called the nth Lagrange interpolating polynomial, is unique due to Theorem 2. We now define L n,k (x) as L n,k (x) = (x x 0) (x x k 1 )(x x k+1 ) (x x n ) n (x k x 0 ) (x k x k 1 )(x k x k+1 ) (x k x n ) = x x i x k x i and it is easy to see that L n,k (x) satisfies (1). In summary we have the following theorem that states the existence and uniqueness of the interpolating polynomial. Theorem 3 If x 0, x 1,..., x n are distinct real numbers, then for arbitrary values y 0, y 1,..., y n, there is a unique polynomial P n of degree at most n such that k=0 i k P n (x i ) = y i (0 i n). Example 2 Using the nodes x 0 = 2, x 1 = 2.5 and x 2 = 4 to find the second interpolating polynomial for f(x) = 1/x. We close this section by stating a remainder (error bound) formula for the truncation error. Theorem 4 If x 0, x 1,..., x n are distinct real numbers in the interval [a, b], then for each for x in [a, b], there exists a number ξ(x) such that f(x) = P n (x) + f n+1 (ξ(x)) (x x 0 )(x x 1 ) (x n ). (n + 1)! Remark This error estimate works for any interpolating polynomial due to the uniqueness. 3.2 Divided differences In Lagrange interpolation whenever we add a point to the set of nodes, we have to completely recompute all of the L n,k (x) functions; we cannot easily write P n+1 (x) in terms of P n (x) using the Lagrange construction. We introduce an alternative method due to Newton. Suppose that P n (x) is the n-th Lagrange polynomial that agrees with the function f at the distinct numbers x 0, x 1,..., x n. The divided differences of f with respect to x 0, x 1,..., x n are used to express P n (x) in the form P n (x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a n (x x 0 )(x x 1 ) (x n 1 ) (2) for appropriate constants a 0, a 1,..., a n. It can be easily shown that We define divided difference as follows a 0 = P n (x 0 ) = f(x 0 ), a 1 = f(x 1) f(x 0 ) x 1 x 0,... f[x i ] = f(x i ) - the zeroth divided difference of f, f[x i, x i+1 ] = f[x i+1] f[x i ] x i+1 x i - the first divided difference of f, f[x i, x i+1, x i+2 ] = f[x i+1, x i+2 ] f[x i, x i+1 ] x i+2 x i - the second divided difference of f.

3 MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall Thus, after the (k 1)st divided differences, f[x i, x i+1, x i+2,..., x i+k 1 ] and f[x i+1, x i+2,..., x i+k 1, x i+k ] have been determined, the k-th divided difference relative to x i, x i+1, x i+2,..., x i+k is given by f[x i, x i+1, x i+2,..., x i+k ] = f[x i+1, x i+2,..., x i+k 1,, x i+k ] f[x i, x i+1, x i+2,..., x i+k 1 ] x i+k x i. As might be expected the coefficients of P n (x) are given by and P n (x) can be written as P n (x) = f[x 0 ] + a k = f[x 0, x 1, x 2,..., x k ]. n f[x 0, x 1, x 2,..., x k ](x x 0 ) (x x k 1 ). (3) k=1 Example 1 Let f(x) = cos x. Approximate f(0.13) using the Newton s divided difference formula with the given data f(0.1) = , f(0.2) = , f(0.3) = , and f(0.4) = Compare the actual error with an upper bound of the truncation error (Theorem 4). Divided differences table i x i f[x i ] f[x i, x i+1 ] f[x i, x i+1, x i+2 ] f[x i, x i+1, x i+2, x i+2 ] For detailed solution refer the handout distributed in class. For the algorithms to find Newton s Divided-Difference polynomials and evaluation using Horner s algorithm, refer pp and the handout on Horner s Algorithm for Newton s Interpolation (# 2 of Computer Problems, HW #4). 3.4 Cubic Splines As seen in Problem (II)#3 of Homework 4, if some part(s) of a function behaves very differently, an interpolating polynomial would not give reasonable approximation and can cause large errors. An alternative approach is to divide the interval into a collection of subintervals and construct a different approximating polynomial on each subinterval. Approximation of this type is called piecewise-polynomial approximation. The simplest piecewise-polynomial approximation is piecewise-linear approximation that connects data points by lines. But the resulting interpolating function may not be smooth. The most common differential piecewise-polynomial approximation uses cubic polynomial between each successive pair of nodes and is called cubic spline interpolation. Definition 1 Let a = x 0 < x 1 < < x n = b be a partition of [a, b] and Y = {y 0, y 1,, y n } be given data. A cubic spline interpolant S for Y is a function C 2 [a, b] that satisfies the following conditions:

4 MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall (a) S(Y ; x) is a cubic polynomial, denoted S j (x), on the subinterval [x j, x j+1 ] for each j = 0, 1,..., n 1, i.e., S(Y ; x) [x j, x j+1 ] = S j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ) 3 such that S(Y ; x j ) = y j for each j = 0, 1, 2,..., n. (4) (b) S j+1 (x j+1 ) = S j (x j+1 ) for each j = 0, 1, 2,..., n 2. (c) S j+1 (x j+1) = S j (x j+1) for each j = 0, 1, 2,..., n 2. (d) S j+1 (x j+1) = S j (x j+1) for each j = 0, 1, 2,..., n 2. (f) Boundary Conditions: (a) S (a) = S (b) = 0, i.e., S 0 (x 0 ) = S n(x 0 ) = 0 (free or natural boundary); (b) S (a) = y 0 and S (b) = y n for some y 0 and y n. In case we are interpolating a function f C 2 [a, b] this condition becomes S (a) = f (a) and S (b) = f (b) (clamped boundary). How to get the coefficients for the cubic polynomial S j (x) = a j + b j (x x j ) + c j (x x j ) 2 + d j (x x j ) 2 for each j = 0, 1, 2,..., n 1? We use the boundary conditions and the recursive formulas obtained applying the conditions (a) (d) to the cubic polynomials S j s. Note that a j s are the given data Y, i.e., a j = S j (x j ) = y j for j = 0, 1, 2,..., n. We first find c j s by solving the linear system of equations: h j 1 c j 1 + 2(h j 1 + h j )c j + h j c j+1 = 3 h j (a j+1 a j ) 3 h j 1 (a j a j 1 ) for j = 1, 2,..., n 1, (5) and for c 0 and c n use the boundary conditions (free or clamped). Then get b j s and d js for j = 0, 1,..., n 1 from the following equations: b j = 1 h j (a j+1 a j ) h j 3 (2c j + c j+1 ) (6) c j+1 = c j + 3d j h j (7) Theorem 5 If f is defined at a = x 0 < x 1 < < x n = b, then f has a unique spline interpolant S on the nodes x 0, x 1, x n ; that satisfies the boundary conditions S (a) = 0 and S (b) = 0. (Natural Spline) proof Use boundary conditions to get c 0 = 0 and c n = 0. Equations (5), (6) and (7) produce a linear system described by the matrix equation Ax = b: c 0 0 2( + h 1 ) h c 3(a 2 a 1) h 1 2(h 1 + h 2 ) h (a1 a0) = 0 h n 2 2(h n 2 + ) c n 1 3(a n a n 1) 3(an 1 an 2) h n c n 0 The matrix A in the above equation is strictly diagonally dominant. Hence the linear system has a unique solution for c 0, c 1, c n by Theorem 6.19 from Chapter 6. A similar result holds in the case of clamped boundary condition.

5 MATH 4073 Chapter 3 Interpolation and Polynomial Approximation Fall Theorem 6 If f is defined at a = x 0 < x 1 < < x n = b, then f has a unique spline interpolant S on the nodes x 0, x 1, x n ; that satisfies the boundary conditions S (a) = f (a) and S (b) = f (b). (Clamped Spline) The corresponding matrix equation is given by ( + h 1 ) h h 1 2(h 1 + h 2 ) h h n 2 2(h n 2 + ) c 0 c 1 c n 1 c n = 3(a 1 a 0) 3(a 2 a 1) h 1 3(a n a n 1) 3f (a) 3(a1 a0) 3(an 1 an 2) h n 2 3f (b) 3(an an 1) Homework: Do the following problems by hand. pp : 1, 2, 3 [(a),(c)], 4 [(a),(c)], 5 [(a),(c)] [note there is a typo in the data values for 5 (a); use f (8.3) = ], 6 [(a),(c)], 7, 8;