Interpolation and Polynomial Approximation

Chapter 3 Interpolation and Polynomial Approximation3.1 Interpolation and the Lagrange Polynomial3.1.1 Lagrangian FormConsider a polynomial of degree (n  1): n-1 n-2 P(x) = a1x + a2x +... + an-1x + an where the ai are constants. The polynomial can be written in Lagrangian form:P(x) = c1(x  2) (x  3)... (x  n) + c2(x  1) (x  3)... (x  n) + ... ci(x  1) (x  2) ... (x  i-1) (x  i+1) ... (x  n) + ... cn(x  1) (x  2)... (x  n-1) where i, i = 1, 2, ..., n are arbitrary scalars, while the constants ci are related to the constants ai.Example 3.1-1 ______Write the polynomial P(x) = x 2  4x + 3 in the Lagrangian form.SolutionThe Lagrangian form for P(x) = x 2  4x + 3 is P(x) = c1(x  2) (x  3) + c2(x  1) (x  3) + c3(x  1) (x  2) where i, i = 1, 2, 3 are arbitrary scalars. Let 1 = 1, 2 = 2, 3 = 3, then P(x) = c1(x  2) (x  3) + c2(x  1) (x  3) + c3(x  1) (x  2)The constants c1 can be evaluated from the above relation by substituting x = 1 = 1P(x = 1) = 1  4 + 3 = c1(1  2) (1  3)  c1 = 0For x = 2 = 2P(x = 2) = 4  8 + 3 = c2(2  1) (2  3)  c2 = 1For x = 3 = 3P(x = 3) = 9  12 + 3 = c3(3  1) (3  2)  c3 = 01 The Lagrangian form for the polynomial isP(x) = (x  1)(x  3)Let 1 =  2, 2 =  1, 3 = 2, then P(x) = c1(x + 1) (x  2) + c2(x + 2) (x  2) + c3(x + 2) (x + 1)The constants ci can be evaluated to obtain: c1 = 3.7500, c2 = -2.6667, and c3 = -0.0833. The Lagrangian form for the polynomial isP(x) = 3.7500 (x + 1) (x  2)  2.6667 (x + 2) (x  2)  0.0833 (x + 2) (x + 1)A short form notation for P(x) is n n (x ) P(x) = ci   k i1 k1,#i n where  (x  k ) denotes product of all terms (x  k), for k varying from 1 to n except i. k1,#iLet x = i thenP(i) = ci(i  1) (i  2) ... (i  i-1) (i  i+1) ... (i  n)The constant ci can be expressed as P(i ) n ci =  (i   k ) k1,#i3.1.2 Polynomial ApproximationConsider a function f(x) that passes through the two distinct points (x0, f(x0)) and (x1, f(x1)) as shown in Figure 3.1-1. The first order polynomial that approximates the function between these two points can be expressed asP(x) = a + bx where a and b are constants. P(x) can also be written in Lagrangian form asP(x) = c0(x  x1) + c1(x  x0) 2 f(x1 ) f(x2 ) f(x) f(x) f(x1 ) f(x0 ) f(x0 ) x0 x1 x x0 x1 x2 x Figure 3.1-1 First and second order polynomial approximation. where P(xi ) n ci =  (xi  xk ) k0,#i orP(x0 ) f (x0 ) P(x1 ) f (x1 ) c0 = = , and c1 = = (x0  x1 ) (x0  x1 ) (x1  x0 ) (x1  x0 )The approximating polynomial is finally(x  x1 ) (x  x0 ) P(x) = f(x0) + f(x1) (x0  x1 ) (x1  x0 )The first order polynomial basis function L0(x) is defined as(x  x1 ) 1 at x  x0 L0(x) = =  (x0  x1 ) 0 at x  x1Similarly, the first order polynomial basis function L1(x) is defined as(x  x0 ) 1 at x  x1 L1(x) = =  (x1  x0 ) 0 at x  x0In terms of the basis function, P(x) can be written as P(x) = L0(x) f(x0) + L1(x) f(x1)If a second order polynomial is used to approximate the function using three points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)) then3 (x  x1 )(x  x2 ) (x  x0 )(x  x2 ) (x  x0 )(x  x1 ) P(x) = f(x0) + f(x1) + f(x2) (x0  x1 )(x0  x2 ) (x1  x0 )(x1  x2 ) (x2  x0 )(x2  x1 )P(x) can also be written in terms of the second order polynomial basis function L2,k(x)P(x) = L2,0(x) f(x0) + L2,1(x)f(x1) + L2,2(x)f(x2)(x  x1 )(x  x2 )  1 at x  x0 where L2,0(x) = =  (x0  x1 )(x0  x2 ) 0 at x  x1 and at x  x2In general: L2,k(xk) = 1 at node k, L2,k(xi) = 0 at other nodes. We now seek a polynomial P(x) of degree n that interpolates a given function f(x) between the node xi of the grid for which there are n+1 nodes x0, x1, , xn and P(xk) = f(xk) for each k = 1, 2, , nThe polynomial is given by n P(x) = Ln,0(x) f(x0) + Ln,1(x) f(x1) +  + Ln,n(x)f(xn) =  Ln,k (x) f(xk) k0 n (x  xi ) where Ln,k(x) =  ; Ln,k(xi) = 0 and Ln,k(xk) = 1 i0,k (xk  xi )Polynomial approximation constitutes the foundation upon which we shall build the various numerical methods. The approximation P(x) to f(x) is known as a Lagrange interpolation polynomial, and the function Ln,k(x) is called a Lagrange basis polynomial.Example 3.1-2 ______Find the Lagrange interpolation polynomial that takes the values prescribed below xk 0 1 2 4 f(xk) 1 1 2 5Solution 3 P(x) =  L3,k (x) f(xk) k0(x 1)(x  2)(x  4) (x  0)(x  2)(x  4) P(x) = (1) + (1) (0 1)(0  2)(0  4) (1 0)(1 2)(1 4)(x  0)(x 1)(x  4) (x  0)(x 1)(x  2) + (2) + (5) (2  0)(2 1)(2  4) (4  0)(4 1)(4  2)4 When working with grids having large numbers of intervals one typically assigns a set of low degree (n = 1, 2, or 3) basis functions to each adjacent set of n+1 = 2, 3, or 4 nodes. Example 3.1-3 ______Use global interpolation by one polynomial and piecewise polynomial interpolation with quadratic for the following nodes. xk 0 1 2 4 5 f(xk) 0 16 48 88 0Solution4 Global interpolation by one polynomial: P(x) =  L4,k (x) f(xk) k0(x 1)(x  2)(x  4)(x  5) (x  0)(x  2)(x  4)(x  5) P(x) = (0) + (16) (0 1)(0  2)(0  4)(0  5) (1 0)(1 2)(1 4)(1 5)(x  0)(x 1)(x  4)(x  5) (x  0)(x 1)(x  2)(x  5) + (48) + (88) + 0 (2  0)(2 1)(2  4)(2  5) (4  0)(4 1)(4  3)(4  5)Piecewise polynomial interpolation with quadratic(x 1)(x  2) (x  0)(x  2) (x  0)(x 1) P(x) = (0) + (16) + (48); 0  x  2 (0 1)(0  2) (1  0)(1  2) (2  0)(2 1)(x  4)(x  5) (x  2)(x  5) (x  2)(x  4) P(x) = (48) + (88) + (0); 2  x  5 (2  4)(2  5) (4  2)(4  5) (5  2)(5  4)The error En(x) associated with the interpolation of f(x) by Pn(x) over the interval [x0, xn] can be estimated as n1 Wn (x) d f En(x) = f(x)  Pn(x) = () (n  1)! dx n1 where  is some number lying in the open interval (x0, xn) and Wn(x) = (x  x0)(x  x1)  (x  xn)When the spacial increments are uniform xk+1  xk = h, k = 0, 1, 2, , n-1Let x = x0 + h, since 5 x1 = x0 + h  x  x1 = (  1)h xn = x0 + nh  x  xn = (  n)hWn(x) = (x  x0)(x  x1)  (x  xn) = (h)[(  1)h] [(  n)h]The error associated with interpolation is then n1 n1 Wn (x) d f 1 d f En(x) = () = (h)[(  1)h] [(  n)h] () (n  1)! dx n1 (n  1)! dx n1The only variable in the above expression is h the spacing of the nodes, therefore n+1 En(x) = Ch , x0 <  < xn where C is a coefficient independent of h. n+1 n+1 We can therefore write En(x) = O(h ) meaning that the ratio En(x)/ h is bounded by a constant as h  0. As the increment h decreases, so also will the interpolation error En.Example 3.1-4 ______For the function f(x) = ln(x + 1), construct interpolation polynomials of degree one and two to approximate f(0.45) from the given nodes. Find the error bound and the actual error. xk 0 0.6 0.9 ln(x + 1) 1 0.47000 0.64185SolutionFirst degree polynomial x  0.6 x  0 P (x) = (0) + (0.47) = 0.78334x 1 0  0.6 0.6  0P1(0.45) = 0.35251 d n1 f Error bound: En(x) = (x  x0)(x  x1)  (x  xn) () (n  1)! dx n1 f "( ) E (x) = | (x  x )(x  x )| 1 2! 0 11 1 1 f(x) = ln(x + 1)  f’(x) =  f”(x) =  f””(x) = x  1 (x  1)2 (x 1)36 1 1 E (x) = | (0.45  0)(0.45  0.6)| = 3.37510-2 1 (0 1)2 2-2 Actual error = |ln(1 + 0.45)  P1(0.45)| = 1.90610Second degree polynomial(x  0.6)(x  0.9) (x  0)(x  0.9) P (x) = (0) + (0.47) 2 (0  0.6)(0  0.9) (0.6  0)(0.6  0.9) (x  0)(x  0.6) + (0.64185) (0.9  0.6)(0.9  0.6)P2(0.45) = 0.36829 f "'( ) Error bound: E (x) = | (x  x )(x  x )(x  x )| 2 3! 0 1 21 1 E (x) = | (0.45  0)(0.45  0.6)(0.45  0.9)| = 1.012510-2 2 (0 1)2 6-3 Actual error = |ln(1 + 0.45)  P2(0.45)| = 3.2729107

Interpolation and Polynomial Approximation

Details

Download

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

Support