4.3 Lagrange Approximation

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1 206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average of known function values at neighboring points. Linear interpolation uses a line segment that passes through two points. The slope between (x 0, y 0 ) and (x 1, y 1 ) is m = (y 1 y 0 )/(x 1 x 0 ), and the point-slope formula for the line y = m(x x 0 ) + y 0 can be rearranged as (1) y = P(x) = y 0 + (y 1 y 0 ) x x 0 x 1 x 0.

2 SEC. 4.3 LAGRANGE APPROXIMATION 207 When formula (1) is expanded, the result is a polynomial of degree 1. Evaluation of P(x) at x 0 and x 1 produces y 0 and y 1, respectively: (2) P(x 0 ) = y 0 + (y 1 y 0 )(0) = y 0, P(x 1 ) = y 0 + (y 1 y 0 )(1) = y 1. The French mathematician Joseph Louis Lagrange used a slightly different method to find this polynomial. He noticed that it could be written as x x 1 x x 0 (3) y = P 1 (x) = y 0 + y 1. x 0 x 1 x 1 x 0 Each term on the right side of (3) involves a linear factor; hence the sum is a polynomial of degree 1. The quotients in (3) are denoted by (4) L 1,0 (x) = x x 1 and L 1,1 (x) = x x 0. x 0 x 1 x 1 x 0 Computation reveals that L 1,0 (x 0 ) = 1, L 1,0 (x 1 ) = 0, L 1,1 (x 0 ) = 0, and L 1,1 (x 1 ) = 1 so that the polynomial P 1 (x) in (3) also passes through the two given points: (5) P 1 (x 0 ) = y 0 + y 1 (0) = y 0 and P 1 (x 1 ) = y 0 (0) + y 1 = y 1. The terms L 1,0 (x) and L 1,1 (x) in (4) are called Lagrange coefficient polynomials based on the nodes x 0 and x 1. Using this notation, (3) can be written in summation form (6) P 1 (x) = 1 y k L 1,k (x). k=0 Suppose that the ordinates y k are computed with the formula y k = f (x k ).IfP 1 (x) is used to approximate f (x) over the interval [x 0, x 1 ], we call the process interpolation. If x < x 0 (or x 1 < x), then using P 1 (x) is called extrapolation. The next example illustrates these concepts. Example 4.6. Consider the graph y = f (x) = cos(x) over [0.0, 1.2]. (a) Use the nodes x 0 = 0.0 and x 1 = 1.2 to construct a linear interpolation polynomial P 1 (x). (b) Use the nodes x 0 = 0.2 and x 1 = 1.0 to construct a linear approximating polynomial Q 1 (x). (a) Using (3) with the abscissas x 0 = 0.0 and x 1 = 1.2 and the ordinates y 0 = cos(0.0) = and y 1 = cos(1.2) = produces P 1 (x) = x 1.2 x = (x 1.2) (x 0.0).

3 208 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION y y y = f(x) y = f(x) 0.6 y = P 1 (x) 0.6 y = Q 1 (x) x x (a) Figure 4.11 (a) The linear approximation y = P 1 (x) where the nodes x 0 = 0.0 and x 1 = 1.2 are the endpoints of the interval [a, b]. (b) The linear approximation y = Q 1 (x) where the nodes x 0 = 0.2 andx 1 = 1.0 lie inside the interval [a, b]. (b) (b) When the nodes x 0 = 0.2 and x 1 = 1.0 with y 0 = cos(0.2) = and y 1 = cos(1.0) = are used, the result is Q 1 (x) = x 1.0 x = (x 1.0) (x 0.2). Figure 4.11(a) and (b) show the graph of y = cos(x) and compare it with y = P 1 (x) and y = Q 1 (x), respectively. Numerical computations are given in Table 4.6 and reveal that Q 1 (x) has less error at the points x k that satisfy 0.1 x k 1.1. The largest tabulated error, f (0.6) P 1 (0.6) = , is reduced to f (0.6) Q 1 (0.6) = by using Q 1 (x). The generalization of (6) is the construction of a polynomial P N (x) of degree at most N that passes through the N + 1 points (x 0, y 0 ), (x 1, y 1 ),...,(x N, y N ) and has the form (7) P N (x) = N y k L N,k (x), where L N,k is the Lagrange coefficient polynomial based on these nodes: (8) 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 ). k=0 It is understood that the terms (x x k ) and (x k x k ) do not appear on the right side of

4 SEC. 4.3 LAGRANGE APPROXIMATION 209 Table 4.6 Comparison of f (x) = cos(x) and the Linear Approximations P 1 (x) and Q 1 (x) x k f (x k ) = cos(x k ) P 1 (x k ) f (x k ) P 1 (x k ) Q 1 (x k ) f (x k ) Q 1 (x k ) equation (8). It is appropriate to introduce the product notation for (8), and we write (9) L N,k (x) = Nj=0 j =k Nj=0 j =k (x x j ) (x k x j ). Here the notation in (9) indicates that in the numerator the product of the linear factors (x x j ) is to be formed, but the factor (x x k ) is to be left out (or skipped). A similar construction occurs in the denominator. A straightforward calculation shows that for each fixed k, the Lagrange coefficient polynomial L N,k (x) has the property (10) L N,k (x j ) = 1 when j = k and L N,k (x j ) = 0 when j = k. Then direct substitution of these values into (7) is used to show that the polynomial curve y = P N (x) goes through (x j, y j ): (11) P N (x j ) = y 0 L N,0 (x j ) + +y j L N, j (x j ) + +y N L N,N (x j ) = y 0 (0) + +y j (1) + +y N (0) = y j. To show that P N (x) is unique, we invoke the fundamental theorem of algebra, which states that a nonzero polynomial T (x) of degree N has at most N roots. In other words, if T (x) is zero at N + 1 distinct abscissas, it is identically zero. Suppose that P N (x) is not unique and that there exists another polynomial Q N (x) of degree N that also passes through the N +1 points. Form the difference polynomial T (x) =

5 210 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION y 1.0 y x y = f(x) y = P 2 (x) 0.5 x y = f(x) y = P 3 (x) (a) Figure 4.12 (a) The quadratic approximation polynomial y = P 2 (x) based on the nodes x 0 = 0.0, x 1 = 0.6, and x 2 = 1.2. (b) The cubic approximation polynomial y = P 3 (x) based on the nodes x 0 = 0.0, x 1 = 0.4, x 2 = 0.8, and x 3 = 1.2. (b) P N (x) Q N (x). Observe that the polynomial T (x) has degree N and that T (x j ) = P N (x j ) Q N (x j ) = y j y j = 0, for j = 0, 1,..., N. Therefore, T (x) 0 and it follows that Q N (x) = P N (x). When (7) is expanded, the result is similar to (3). The Lagrange quadratic interpolating polynomial through the three points (x 0, y 0 ), (x 1, y 1 ), and (x 2, y 2 ) is (x x (12) P 2 (x) = y 1 )(x x 2 ) 0 (x 0 x 1 )(x 0 x 2 ) + y (x x 0 )(x x 2 ) 1 (x 1 x 0 )(x 1 x 2 ) + y (x x 0 )(x x 1 ) 2 (x 2 x 0 )(x 2 x 1 ). The Lagrange cubic interpolating polynomial through the four points (x 0, y 0 ), (x 1, y 1 ), (x 2, y 2 ), and (x 3, y 3 ) is (13) (x x P 3 (x) = y 1 )(x x 2 )(x x 3 ) 0 (x 0 x 1 )(x 0 x 2 )(x 0 x 3 ) + y (x x 0 )(x x 2 )(x x 3 ) 1 (x 1 x 0 )(x 1 x 2 )(x 1 x 3 ) (x x + y 0 )(x x 1 )(x x 3 ) 2 (x 2 x 0 )(x 2 x 1 )(x 2 x 3 ) + y (x x 0 )(x x 1 )(x x 2 ) 3 (x 3 x 0 )(x 3 x 1 )(x 3 x 2 ). Example 4.7. Consider y = f (x) = cos(x) over [0.0, 1.2]. (a) Use the three nodes x 0 = 0.0, x 1 = 0.6, and x 2 = 1.2 to construct a quadratic interpolation polynomial P 2 (x). (b) Use the four nodes x 0 = 0.0, x 1 = 0.4, x 2 = 0.8, and x 3 = 1.2 to construct a cubic interpolation polynomial P 3 (x). (a) Using x 0 = 0.0, x 1 = 0.6, x 2 = 1.2 and y 0 = cos(0.0) = 1, y 1 = cos(0.6) =

6 SEC. 4.3 LAGRANGE APPROXIMATION , and y 2 = cos(1.2) = in equation (12) produces (x 0.6)(x 1.2) (x 0.0)(x 1.2) P 2 (x) = ( )( ) ( )( ) (x 0.0)(x 0.6) ( )( ) = (x 0.6)(x 1.2) (x 0.0)(x 1.2) (x 0.0)(x 0.6). (b) Using x 0 = 0.0, x 1 = 0.4, x 2 = 0.8, x 3 = 1.2 and y 0 = cos(0.0) = 1.0, y 1 = cos(0.4) = , y 2 = cos(0.8) = , and y 3 = cos(1.2) = in equation (13) produces (x 0.4)(x 0.8)(x 1.2) P 3 (x) = ( )( )( ) (x 0.0)(x 0.8)(x 1.2) ( )( )( ) (x 0.0)(x 0.4)(x 1.2) ( )( )( ) (x 0.0)(x 0.4)(x 0.8) ( )( )( ) = (x 0.4)(x 0.8)(x 1.2) (x 0.0)(x 0.8)(x 1.2) (x 0.0)(x 0.4)(x 1.2) (x 0.0)(x 0.4)(x 0.8). The graphs of y = cos(x) and the polynomials y = P 2 (x) and y = P 3 (x) are shown in Figure 4.12(a) and (b), respectively. Error Terms and Error Bounds It is important to understand the nature of the error term when the Lagrange polynomial is used to approximate a continuous function f (x). It is similar to the error term for the Taylor polynomial, except that the factor (x x 0 ) N+1 is replaced with the product (x x 0 )(x x 1 ) (x x N ). This is expected because interpolation is exact at each of the N + 1 nodes x k, where we have E N (x k ) = f (x k ) P N (x k ) = y k y k = 0for k = 0, 1, 2,...,N. Theorem 4.3 (Lagrange Polynomial Approximation). Assume that f C N+1 [a, b] and that x 0, x 1,...,x N [a, b] are N + 1 nodes. If x [a, b], then (14) f (x) = P N (x) + E N (x),

7 212 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION where P N (x) is a polynomial that can be used to approximate f (x): (15) f (x) P N (x) = N k=0 f (x k )L N,k (x). The error term E N (x) has the form (16) E N (x) = (x x 0)(x x 1 ) (x x N ) f (N+1) (c) (N + 1)! for some value c = c(x) that lies in the interval [a, b]. Proof. As an example of the general method, we establish (16) when N = 1. The general case is discussed in the exercises. Start by defining the special function g(t) as follows: (17) g(t) = f (t) P 1 (t) E 1 (x) (t x 0)(t x 1 ) (x x 0 )(x x 1 ). Notice that x, x 0, and x 1 are constants with respect to the variable t and that g(t) evaluates to be zero at these three values; that is, g(x) = f (x) P 1 (x) E 1 (x) (x x 0)(x x 1 ) (x x 0 )(x x 1 ) = f (x) P 1(x) E 1 (x) = 0, g(x 0 ) = f (x 0 ) P 1 (x 0 ) E 1 (x) (x 0 x 0 )(x 0 x 1 ) = f (x 0 ) P 1 (x 0 ) = 0, (x x 0 )(x x 1 ) g(x 1 ) = f (x 1 ) P 1 (x 1 ) E 1 (x) (x 1 x 0 )(x 1 x 1 ) = f (x 1 ) P 1 (x 1 ) = 0. (x x 0 )(x x 1 ) Suppose that x lies in the open interval (x 0, x 1 ). Applying Rolle s theorem to g(t) on the interval [x 0, x] produces a value d 0, with x 0 < d 0 < x, such that (18) g (d 0 ) = 0. A second application of Rolle s theorem to g(t) on [x, x 1 ] will produce a value d 1, with x < d 1 < x 1, such that (19) g (d 1 ) = 0. Equations (18) and (19) show that the function g (t) is zero at t = d 0 and t = d 1. A third use of Rolle s theorem, but this time applied to g (t) over [d 0, d 1 ], produces a value c for which (20) g (2) (c) = 0.

8 SEC. 4.3 LAGRANGE APPROXIMATION 213 Now go back to (17) and compute the derivatives g (t) and g (t): (21) g (t) = f (t) P 1 (t) E 1(x) (t x 0) + (t x 1 ) (x x 0 )(x x 1 ), (22) g (t) = f 2 (t) 0 E 1 (x) (x x 0 )(x x 1 ). In (22) we have used the fact the P 1 (t) is a polynomial of degree N = 1; hence its second derivative is P 1 (t) 0. Evaluation of (22) at the point t = c and using (20) yields (23) 0 = f 2 (c) E 1 (x) (x x 0 )(x x 1 ). Solving (23) for E 1 (x) results in the desired form (16) for the remainder: (24) E 1 (x) = (x x 0)(x x 1 ) f (2) (c), 2! and the proof is complete. The next result addresses the special case when the nodes for the Lagrange polynomial are equally spaced x k = x 0 + hk, fork = 0, 1,..., N, and the polynomial P N (x) is used only for interpolation inside the interval [x 0, x N ]. Theorem 4.4 (Error Bounds for Lagrange Interpolation, Equally Spaced Nodes). Assume that f (x) is defined on [a, b], which contains equally spaced nodes x k = x 0 + hk. Additionally, assume that f (x) and the derivatives of f (x), up to the order N + 1, are continuous and bounded on the special subintervals [x 0, x 1 ], [x 0, x 2 ], and [x 0, x 3 ], respectively; that is, (25) f (N+1) (x) M N+1 for x 0 x x N, for N = 1, 2, 3. The error terms (16) corresponding to the cases N = 1, 2, and 3 have the following useful bounds on their magnitude: (26) E 1 (x) h2 M 2 8 valid for x [x 0, x 1 ], (27) E 2 (x) h3 M valid for x [x 0, x 2 ], (28) E 3 (x) h4 M 4 24 valid for x [x 0, x 3 ]. Proof. We establish (26) and leave the others for the reader. Using the change of variables x x 0 = t and x x 1 = t h, the error term E 1 (x) can be written as (29) E 1 (x) = E 1 (x 0 + t) = (t2 ht) f (2) (c) 2! for 0 t h.

9 214 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION The bound for the derivative for this case is (30) f (2) (c) M 2 for x 0 c x 1. Now determine a bound for the expression (t 2 ht) in the numerator of (29); call this term (t) = t 2 ht. Since (t) = 2t h, there is one critical point t = h/2 that is the solution to (t) = 0. The extreme values of (t) over [0, h] occur either at an end point (0) = 0, (h) = 0 or at the critical point (h/2) = h 2 /4. Since the latter value is the largest, we have established the bound (31) (t) = t 2 ht h2 4 = h2 4 for 0 t h. Using (30) and (31) to estimate the magnitude of the product in the numerator in (29) results in (32) E 1 (x) = (t) f (2) (c) 2! h2 M 2, 8 and formula (26) is established. Comparison of Accuracy and O(h N+1 ) The significance of Theorem 4.4 is to understand a simple relationship between the size of the error terms for linear, quadratic, and cubic interpolation. In each case the error bound E N (x) depends on h in two ways. First, h N+1 is explicitly present so that E N (x) is proportional to h N+1. Second, the values M N+1 generally depend on h and tend to f (N+1) (x 0 ) as h goes to zero. Therefore, as h goes to zero, E N (x) converges to zero with the same rapidity that h N+1 converges to zero. The notation O(h N+1 ) is used when discussing this behavior. For example, the error bound (26) can be expressed as E 1 (x) =O(h 2 ) valid for x [x 0, x 1 ]. The notation O(h 2 ) stands in place of h 2 M 2 /8 in relation (26) and is meant to convey the idea that the bound for the error term is approximately a multiple of h 2 ; that is, E 1 (x) Ch 2 O(h 2 ). As a consequence, if the derivatives of f (x) are uniformly bounded on the interval [a, b] and h < 1, then choosing N large will make h N+1 small, and the higher-degree approximating polynomial will have less error.

10 SEC. 4.3 LAGRANGE APPROXIMATION 215 y y y = E 2 (x) y = E 3 (x) x x (a) (b) Figure 4.13 (a) The error function E 2 (x) = cos(x) P 2 (x). (b) The error function E 3 (x) = cos(x) P 3 (x). Example 4.8. Consider y = f (x) = cos(x) over [0.0, 1.2]. Use formulas (26) through (28) and determine the error bounds for the Lagrange polynomials P 1 (x), P 2 (x), and P 3 (x) that were constructed in Examples 4.6 and 4.7. First, determine the bounds M 2, M 3, and M 4 for the derivatives f (2) (x), f (3) (x), and f (4) (x), respectively, taken over the interval [0.0, 1.2]: f (2) (x) = cos(x) cos(0.0) = = M 2, f (3) (x) = sin(x) sin(1.2) = = M 3, f (4) (x) = cos(x) cos(0.0) = = M 4. For P 1 (x) the spacing of the nodes is h = 1.2, and its error bound is (33) E 1 (x) h2 M 2 (1.2)2 ( ) = For P 2 (x) the spacing of the nodes is h = 0.6, and its error bound is (34) E 2 (x) h3 M (0.6)3 ( ) 9 = For P 3 (x) the spacing of the nodes is h = 0.4, and its error bound is (35) E 3 (x) h4 M 4 24 (0.4)4 ( ) 24 = From Example 4.6 we saw that E 1 (0.6) = cos(0.6) P 1 (0.6) = , so the bound in (33) is reasonable. The graphs of the error functions E 2 (x) = cos(x) P 2 (x) and E 3 (x) = cos(x) P 3 (x) are shown in Figure 4.13(a) and (b),

11 216 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Table 4.7 Comparison of f (x) = cos(x) and the Quadratic and Cubic Polynomial Approximations P 2 (x) and P 3 (x) x k f (x k ) = cos(x k ) P 2 (x k ) E 2 (x k ) P 3 (x k ) E 3 (x k ) respectively, and numerical computations are given in Table 4.7. Using values in the table, we find that E 2 (1.0) = cos(1.0) P 2 (1.0) = and E 3 (0.2) = cos(0.2) P 3 (0.2) = , which is in reasonable agreement with the bounds and given in (34) and (35), respectively.

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