Interpolation and Approximation

Transcription

1 Interpolation and Approximation The Basic Problem: Approximate a continuous function f(x), by a polynomial p(x), over [a, b]. f(x) may only be known in tabular form. f(x) may be expensive to compute. Definition: A polynomial p(x) interpolates f(x) at the nodes x 0, x 1, x n if p(x i ) = f(x i ) for i = 0, 1, n. (Intuitively if f(x) and p(x) agree at the x i then they should be close at nearby points.) CSCC51H- Numerical Approx, Int and ODEs p.17/177

2 Key Theorem Given distinct nodes x 0, x 1, x n and arbitrary f 0, f 1, f n, there is a unique polynomial p n (x) of degree at most n that interpolates f(x) at x 0, x 1, x n. Proof: Existence: (constructive) Let P n (x) = a 0 + a 1 (x x 0 ) + a n (x x 0 ) (x x n 1 ), For any choice of a 0, a 1, a n, P n (x) will be of degree at most n. We will choose the a i to ensure that P n (x i ) = f i. This results in a system of n + 1 linear equations in the n + 1 unknowns, a 0, a 1, a n. The i th equation is: a 0 + a 1 (x i x 0 ) + a n (x i x 0 ) (x i x n 1 ) = f i, CSCC51H- Numerical Approx, Int and ODEs p.18/177

3 Proof (cont) In matrix form then, Ba = f, where a T = [a 0, a 1, a n ] T, f = [f 0, f 1, f n ] T, and B is the matrix, 1 for j = 0;, b ij = (x i x 0 ) (x i x j 1 ) for j = 1, n;. B is lower triangular since b ij contains a factor (x i x i ) for j > i. b ii 0 since x 0, x 1, x n are distinct. This implies B is nonsingular and there exists a unique solution a. Solving this triangular linear system by forward substitution: a 0 = f 0, a 1 = (f 1 a 0 )/(x 1 x 0 ),. a n = [f n (a 0 + a n 1 (x n x 0 ) (x n x n 2 ))] /[(x n x 0 )(x n x 1 ) (x n x n 1 )]. CSCC51H- Numerical Approx, Int and ODEs p.19/177

4 Two Key Observations The first r + 1 terms, a 0, a 1, a r, determine a polynomial of degree at most r that interpolates f(x) at x 0, x 1, x r. The resulting P n (x) satisfies the interpolation conditions and the coefficients a 0, a 1, a n define the Newton Form of this polynomial. (The Newton Form depends on the nodes and their order.) CSCC51H- Numerical Approx, Int and ODEs p.20/177

5 Proof of Uniqueness Let q n (x) be a polynomial of degree at most n such that q n (x i ) = f i for i = 0, 1, n. Then with r(x) P n (x) q n (x) we observe that r(x) is a polynomial of degree at most n such that r(x i ) = 0 for i = 0, 1, n. Then, by Rolle s theorem, r (x) has n distinct zeros, r (x) has n 1 distinct zeros,... and r n (x) has one zero in the interval containing the nodes. But this is impossible unless r(x) 0, in which case q n (x) must equal P n (x). CSCC51H- Numerical Approx, Int and ODEs p.21/177

6 Representation of P n (x) The interpolating polynomial P n (x) is unique but it can be represented in different ways. Consider the following representations of P n (x): Monomials (powers of x or (x w) ): P n (x) = c 0 + c 1 x + c n x n, (In this case P n (x) is represented by the coefficients c 0, c 1, c n and we do not need to know the nodes or their order to evaluate or use this polynomial.) Newton Basis (or divided differences): P n (x) = a 0 + a 1 (x x 0 ) + a n (x x 0 ) (x x n 1 ), (In this case P n (x) is represented by the a i s and one must know the nodes and their order to use this polynomial.) CSCC51H- Numerical Approx, Int and ODEs p.22/177

7 Representation of P n (x) (cont) Lagrange Basis: We introduce the j th Lagrange basis function l j (x) (associated with the nodes [x i ] n i=0 ) by l j (x) = n n (x x i )/ (x j x i ) = i=0,i j i=0,i j n i=0,i j (x x i ) (x j x i ), for j = 0, 1, n. Then it is clear that l j (x) is a polynomial of degree n such that 0 for j = 0, 1, n; j i, l j (x i ) = 1 for j = i. CSCC51H- Numerical Approx, Int and ODEs p.23/177

8 Lagrange Basis (cont) It is also clear that any linear combination of the l j (x) will be a polynomial of degree at most n. In particular, P n (x) = n f j l j (x), j=0 since P n (x) is unique and the polynomial n j=o f jl j (x) satisfies, n f j l j (x i ) = f i, for i = 0, 1, n. j=0 (In this case P n (x) is represented by the function values, f 0, f 1, f n and we must know the nodes to use P n (x).) The particular choice of what representation to use will depend on the application. Often we will use the Lagrange form in our analysis but other forms in our implementations. CSCC51H- Numerical Approx, Int and ODEs p.24/177

9 An Example Consider the unique quadratic defined by the three interpolating conditions, P 2 ( 1) = 7, P 2 (0) = 2, and P 2 (1) = 1 (that is, x 0 = 1, x 1 = 0, x 2 = 1 and f 0 = 7, f 1 = 2, f 2 = 1). The Lagrange basis is, l 0 (x) = l 1 (x) = l 2 (x) = (x 0)(x 1) ( 1 0)( 1 1) = x2 x, 2 (x + 1)(x 1) (0 + 1)(0 1) = x2 1 1 (x + 1)x (1 + 1)(1 0) = x2 + x. 2 = 1 x 2, CSCC51H- Numerical Approx, Int and ODEs p.25/177

10 An Example (cont) In Lagrange form P 2 (x) is P 2 (x) = f 0 l 0 (x) + f 1 l 1 (x) + f 2 l 2 (x), = 7 ( x2 x ) + 2(1 x 2 ) + x2 + x. 2 2 ( = 2x 2 3x + 2 (in monomial form)) CSCC51H- Numerical Approx, Int and ODEs p.26/177

11 Divided Differences Definition: The divided difference, f[x 0 x 1 x i ] is defined to be a i the (i + 1) s t coefficient of the interpolating polynomial P n (x) written in Newton form. P n (x) = a 0 + a 1 (x x 0 ) + a n (x x 0 ) (x x n 1 ). Recall that the first r + 1 coefficients of the Newton form of P n (x) determine a polynomial of degree at most r that interpolates f(x) at the nodes (x 0, x 1, x r ). CSCC51H- Numerical Approx, Int and ODEs p.27/177

12 Properties of Divided Differences f[x 0 x 1 x i ] = f[x j0 x j1 x ji ], where j 0, j 1, j i is a permutation of 0, 1, i. f[x 0 x 1 x i ] = f[x 0x 1 x i 1 ] f[x 1 x 2 x i ] x 0 x i. CSCC51H- Numerical Approx, Int and ODEs p.28/177

13 Proof of First Property The polynomial of degree at most i interpolating f(x) at (x 0, x 1, x i ) is P i (x) = f[x 0 ] + f[x 0 x 1 ](x x 0 ) + f[x 0 x 1 x i ](x x 0 ) (x x i 1 ), Note that the coefficient of x i in this polynomial is f[x 0 x 1 x i ]. Similarly the polynomial Pi (x) interpolating f(x) at (x j0, x j1, x ji ) is P i (x) =f[x j0 ]+f[x j0 x j1 ](x x j0 )+ f[x j0 x j1 x ji ](x x j0 ) (x x ji 1 ). Now the sets [x 0, x 1, x i ] and [x j0, x j1, x ji ] are identical (only the order may be different) so by uniqueness we have P i (x) = P i (x). In particular the respective coefficients of x i must agree and we have, f[x 0 x 1 x i ] = f[x j0 x j1 x ji ]. CSCC51H- Numerical Approx, Int and ODEs p.29/177

14 Proof of Second Property Consider the orderings (x 1, x 2, x i, x 0 ) and (x 0, x 1, x i ). Writing P i (x) using the two different Newton Forms (corresponding to these different orderings) we have, P i (x) = f[x 0 ] + f[x 0 x 1 ](x x 0 ) + f[x 0 x 1 x i ](x x 0 ) (x x i 1 ) = f[x 1 ] + f[x 1 x 2 ](x x 1 ) + f[x 1 x 2 x i x 0 ](x x 1 ) (x x i ). Multiplying the first equation by (x x i ), the second equation by (x x 0 ) and subtracting we obtain, or... (x x i )P i (x) (x x 0 )P i (x) = (x 0 x i )P i (x), CSCC51H- Numerical Approx, Int and ODEs p.30/177

15 Proof (cont) (x 0 x i )P i (x) = f[x 0 ](x x i ) f[x 1 ](x x 0 ). +f[x 0 x 1 x i 1 ](x x 0 ) (x x i 2 )(x x i ) f[x 1 x 2 x i ](x x 0 ) (x x i 1 ) +f[x 0 x 1 x i ](x x 0 ) (x x i ) f[x 1 x 2 x i x 0 ](x x 0 ) (x x i ). But the last term in this expansion vanishes and this implies the coefficient of x i in the polynomial (x 0 x i )P i (x) is f[x 0 x 1 x i 1 ] f[x 1 x 2 x i ]. We know this coefficient is (x 0 x i )f[x 0 x 1 x i ] so we can conclude, f[x 0 x 1 x i ] = f[x 0x 1 x i 1 ] f[x 1 x 2 x i ] x 0 x i. CSCC51H- Numerical Approx, Int and ODEs p.31/177

16 Error in Interpolation Let E n (x) = f(x) P n (x). To investigate the behaviour of E n (x) consider fixing x and determine the polynomial p n+1 (z) of degree at most n + 1 (in z), interpolating f(z) at the n + 2 nodes (x 0, x 1, x n, x). p n+1 (z) = f[x 0 ] + f[x 0 x 1 ](z x 0 ) + f[x 0 x 1 x n x](z x 0 ) (z x n ) = P n (z)+f[x 0 x 1 x n x](z x 0 ) (z x n ). Evaluating this expression at z = x and using the fact that p n+1 (x) = f(x) we have, E n (x) = f(x) P n (x) = p n+1 (x) P n (x), = f[x 0 x 1 x n x](x x 0 ) (x x n ). CSCC51H- Numerical Approx, Int and ODEs p.32/177

17 We have shown, Error Expression E n (x) = f[x 0 x 1 x n x](x x 0 ) (x x n ) Therefore if f[x 0 x 1 x n x] (as a function of x) is slowly varying, then we can estimate E n (x) by for some x n+ 1. est n (x) = n i= 0 (x x i)f[x 0 x 1 x n x n+ 1 ] CSCC51H- Numerical Approx, Int and ODEs p.33/177