INTERPOLATION Background Polynomial Approximation Problem:

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1 INTERPOLATION Background Polynomial Approximation Problem: given f(x) C[a, b], find P n (x) = a 0 + a 1 x + a 2 x a n x n with P n (x) close to f(x) for x [a, b]. Motivations: f(x) might be difficult to compute, but P n (x) is easy to compute, integrate, differentiate,... Weierstrass Approximation Theorem: if f C[a, b], ɛ > 0, there exists a polynomial P (x) with f(x) P (x) < ɛ x [a, b]. Problem: how to find a good P (x)? Possible solution: Taylor polynomials? P n (x) = f(x 0 )+(x x 0 )f (x 0 )+ +(x x 0 ) n f (n) (x 0 ). But Taylor P n (x) is often bad unless x is close to x 0 ; Taylor polynomial needs derivatives. Another possible solution: interpolating polynomial. Given points x 0, x 1,..., x n, and function values f(x 0 ), f(x 1 ),..., f(x n ), an interpolating polynomial P (x) for f(x) satisifies P (x i ) = f(x i ), i = 0, 1,..., n. Issues: how to find, represent, compute P (x); errors?

2 LAGRANGE INTERPOLATION Lagrange Interpolating Polynomial Linear Interpolation: given x 0, f(x 0 ), x 1, f(x 1 ), point-slope form for interpolating P 1 (x) is P 1 (x) = f(x 0 ) + (x x 0 ) f(x 1) f(x 0 ) x 1 x 0. Lagrange form for interpolating P 1 (x) is P 1 (x) = x x 1 x 0 x 1 f(x 0 ) + x x 0 x 1 x 0 f(x 1 ). Quadratic Interpolation: given (x 0, f(x 0 )), (x 1, f(x 1 )), (x 2, f(x 2 )), point-slope form for interpolating P 2 (x) (see 3.3) is P 2 (x) = f(x 0 ) + (x x 0 ) f(x 1) f(x 0 ) x 1 x 0 +??? Lagrange form for interpolating P 2 (x) is P 2 (x) = (x x 1 )(x x 2 ) (x 0 x 1 )(x 0 x 2 ) f(x 0) + (x x 0)(x x 2 ) (x 1 x 0 )(x 1 x 2 ) f(x 1) + (x x 0)(x x 1 ) (x 2 x 0 )(x 2 x 1 ) f(x 2). 2

3 LAGRANGE INTERPOLATION CONT. Example: f(x) = cos(πx), for x [0, 2] Maxiumum error for x [0, 2]: 2(for P 1 (x)),.61(p 2 (x)),.089(p 4 (x)). 1.5 cos(π x) and Lagrange degree 1, 2, 4 approximations

4 LAGRANGE INTERPOLATION CONT. General Lagrange Interpolation Lagrange Basis functions: given n + 1 distinct x i s L n,k (x) = (x x 0) (x x k 1 )(x x k+1 ) (x x n ) (x k x 0 ) (x k x k 1 )(x k x k+1 ) (x k x n ) n (x x i ) = (x k x i ). i=0, k Notice: L n,k (x j ) = 0, if j k, L n,k (x k ) = 1, and degree of L n,k is n. Lagrange Interpolating Polynomial: given data set D = {(x i, f(x i ))} n i=0, with distinct x i s, the Lagrange interpolating polynomial for D is P (x) = f(x 0 )L n,0 (x) + f(x 1 )L n,1 (x) + + f(x n )L n,n (x) n = f(x k )L n,k (x). k=0 Efficient evaluation: use n n f(x i ) P (x) = (x x i ) w i (x x i ), w i = i=0 i=0 n j=0, i (x i x j ). With precomputed w i s (O(n 2 )), P (x) needs O(n) flops. 4

5 LAGRANGE INTERPOLATION CONT. Theorem: given {f(x i )} n i=0, with distinct x i s, there is a unique polynomial P (x) of degree n which satisfies P (x i ) = f(x i ) for i = 0, 1,..., n. How to find power form for P (x)? Find a i s so that P n (x) = a 0 + a 1 x + a 2 x a n x n. Solve the linear system determined by P n (x i ) = a 0 + a 1 x i + a 2 x 2 i + + a n x n i = f(x i ), i = 0, 1,..., n, for the a i s. In system form V a = f, with Vandermonde matrix 1 x 0 x x n 0 a 0 f(x 0 ) V = 1 x 1 x x n , a = a 1., f = f(x 1 )., 1 x n x 2 n... x n n a n f(x n ) Problems: Cost to find a i s? V is often illconditioned; a i s are often not explicitly needed. 5

6 NEWTON INTERPOLATION Newton Divided Difference Background Motivation: to provide Taylor -like general formula where terms are added successively as degree increases: 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 n 1 ). Iterative construction of P n : Notice P n (x 0 ) = f(x 0 ) = a 0 ; P n (x 1 ) = f(x 1 ) = f(x 0 ) + a 1 (x 1 x 0 ), so Then a 1 = f(x 1) f(x 0 ) x 1 x 0. P n (x 2 ) = f(x 2 ) = f(x 0 )+a 1 (x 2 x 0 )+a 2 (x 2 x 0 )(x 2 x 1 ), so f(x 2 ) f(x 1 ) x a 2 = 2 x 1 f(x 1) f(x 0 ) x 1 x 0. x 2 x 0 For general case, use divided differences for all a k s. 6

7 DIVIDED DIFF. INTERP. CONT. Divided Differences Notation: the k th divided difference for x i, x i+1,..., x i+k, f(x i ), f(x i+1 ),..., f(x i+k ), is a number denoted by f[x i, x i+1,..., x i+k ]. Divided Difference Formula: using f[x i ] f(x i ) f[x i, x i+1,..., x i+k ] = f[x i+1, x i+2,..., x i+k ] f[x i, x i+1,..., x i+k 1 ] x i+k x i. Divided Difference Table computed row at a time x 0 f[x 0 ] x 1 f[x 1 ] f[x 0, x 1 ] x 2 f[x 2 ] f[x 1, x 2 ] f[x 0, x 1, x 2 ].... x n f[x n ] f[x n 1, x n ] f[x n 2, x n 1, x n ]... f[x 0, x 1,..., x n ] with a k = f[x 0, x 1,..., x k ], k = 0, 1,..., n. Example: find the divided diff. table for the given data x f(x)

8 DIVIDED DIFF. INTERP. CONT. Simple Matlab algorithm for divided diff. table. n = 4; x=[0:n]/n; f=cos(pi*x); % Data Values dd = zeros(1,n+1); for i = 1 : n+1, t = dd(1); dd(1) = f(i); for j = 1 : i-1, dm = t; t = dd(j+1); dd(j+1) = (dd(j)-dm)/(x(i)-x(i-j)); end, d(i) = dd(i); disp([x(i) dd(1:i)]) end % d contains DD coefficients %x s f s DD s e-15 Newton Interpolation Formula: define n P n (x) = f[x 0 ]+ f[x 0, x 1,..., x k ](x x 0 ) (x x k 1 ) k=1 Example: using diagonal from computed DD table: P 4 (x) = Note: actual degree of P 4 is 3. 8

9 DIVIDED DIFF. INTERP. CONT. Computation with DD polynomial: setup cost is O(n 2 ), but nested evaluation costs only O(n) for P (x). Nested evaluation algorithm for P(x) a) set P = f[x 0, x 1,..., x n ], b) for k = n 1 : 1 : 0, set P = f[x 0, x 1,..., x k ] + (x x k )P. Example: find f(0.45)(.15643) given previous data. P 4 (x) = P 4 (.45) = Algorithm steps: k = 4; P = 0; k = 3; P = (.3)0 = ; k = 2; P = (.05) = ; k = 1; P = (.2)( ) = ; k = 0; P = 1 + (.45)( ) = Example: given the DD table below, what is f(x)? % x f(x) 1st 2nd 3rd 4th 5th 6th

10 DIVIDED DIFF. INTERP. CONT. Example: how many degree d polynomials through n points? d = 3, with points (1,-1), (3,2), (4,5)? d = 4? Application Example: efficient function representation. Find a good interpolation approximation for sin(x). a) pick fundamental domain [0, π/2]; for other x s use sin(x) = sin(π x), x [π/2, π], sin(x) = sin(2π x), x [π, 2π], sin(x + 2kπ) = sin(x), x > 2π, with sin(x) = sin( x). b) given ɛ, find smallest d so that P d (x) sin(x) < ɛ. E.g. with evenly spaced x i s d = 3, has P 3 (x) sin(x).0024, d = 4, has P 4 (x) sin(x).00022, d = 5, has P 5 (x) sin(x) , d = 6, has P 6 (x) sin(x) , d = 7, has P 7 (x) sin(x) A single precision interpolating polynomial for sin(x) needs only 8 DD coefficients. 10

11 8 x 10 8 P 7 (x) sin(x) n = 7; x = pi*[0:n]/(2*n); f = sin(x); xt = pi*[0:128]/256; [fv,d] = divdif(x,f,xt); disp(max(abs(fv-sin(xt)))) plot( xt, fv-sin(xt), x, zeros(1,n+1) ); grid on disp(d) % DD diagonal entries

12 DD table computation and interp. poly value function function [ f, d, fp ] = divdif( xs, fs, x, ip ) if nargin < 4, ip = 0; end % Computes DD table for data xs, fs, and % evaluates interp. poly at points in x, % returning poly values in f, derivatives in fp, % and DD table diagonal entries in d. % n = length(xs); dd = zeros(1,n); for i = 1 : n, db = dd(1); dd(1) = fs(i); if ip>0, fprintf( %8.1f%11.2f,xs(i),fs(i)),end % compute and display row i of DD table for j = 1 : i-1, da = db; db = dd(j+1); dd(j+1) = ( dd(j) - da )/( xs(i) - xs(i-j) ); if ip>0, fprintf( %10.4f,dd(j+1)), end end, d(i) = dd(i); if ip>0, fprintf( \n ), end end, f = d(n); fp = 0; % nested poly and derivative values at x points for i = n-1 : -1 : 1, fp = f + (x-xs(i)).*fp; f = d(i) + (x-xs(i)).*f; end % end divdif 12