# Given data y i at abscissae x i as before the unique

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Given data y i at abscissae x i as before, the unique polynomial interpolant of degree at most n can now be written as p ( x ) = n j = 0 y j L j ( x ). Indeed, p is of degree at most n (being the linear combination of polynomials of degree n ), and it satis fi es the interpolation conditions because p ( x i ) = n j = 0 y j L j ( x i ) = 0 +···+ 0 + y i L i ( x i ) + 0 +···+ 0 = y i . Downloaded 10/19/18 to 132.174.255.3. Redistribution subject to SIAM license or copyright; see
10.3. Lagrange interpolation 303 Example 10.2. Let us use the same three data pairs of Example 10.1, namely, (1,1), (2,3), and (4,3), to demonstrate the construction of Lagrange polynomials. To make L 0 ( x ) vanish at x = 2 and x = 4 we write L 0 ( x ) = a ( x 2)( x 4). Requiring L 0 (1) = 1 then determines a (1 2)(1 4) = 1, i.e., a = 1 3 , and thus L 0 ( x ) = 1 3 ( x 2)( x 4). Similarly we determine that L 1 ( x ) = 1 2 ( x 1)( x 4), L 2 ( x ) = 1 6 ( x 1)( x 2). These Lagrange polynomials are depicted in Figure 10.3. We thus obtain the interpolant p 2 ( x ) = y 0 3 ( x 2)( x 4) y 1 2 ( x 1)( x 4) + y 2 6 ( x 1)( x 2) = 1 3 ( x 2)( x 4) 3 2 ( x 1)( x 4) + 3 6 ( x 1)( x 2). Despite the different form, this is precisely the same quadratic interpolant as the one in Exam- ple 10.1, so in fact, p 2 ( x ) = ( 2 x 2 + 12 x 7) / 3. It is also easy to verify that here, too, p 2 (3) = 11 3 . All this should not come as a surprise it s an illustration of the uniqueness of polynomial interpo- lation; see the theorem on page 301. 0 1 2 3 4 5 6 3 2 1 0 1 2 3 x Quadratic Lagrange polynomials L 0 (x) L 1 (x) L 2 (x) Figure 10.3. The quadratic Lagrange polynomials L 0 ( x ), L 1 ( x ), and L 2 ( x ) based on points x 0 = 1, x 1 = 2, x 2 = 4 , used in Example 10.2 . Downloaded 10/19/18 to 132.174.255.3. Redistribution subject to SIAM license or copyright; see
304 Chapter 10. Polynomial Interpolation Properties of Lagrange polynomials What properties do Lagrange polynomials have? What do they look like? In the general case where n + 1 data abscissae x i are speci fi ed, the Lagrange polynomials uniquely exist, because they are really nothing other than polynomial interpolants for special data. 41 This is also straightforward to verify directly, by explicitly specifying L j ( x ) = ( x x 0 ) ··· ( x x j 1 )( x x j + 1 ) ··· ( x x n ) ( x j x 0 ) ··· ( x j x j 1 )( x j x j + 1 ) ··· ( x j x n ) = n i = 0 i ̸ = j ( x x i ) ( x j x i ) . Indeed, the polynomial of degree n , written in terms of its roots as ( x x 0 ) ··· ( x x j 1 )( x x j + 1 ) ··· ( x x n ), clearly interpolates the 0-values at all data abscissae other than x j , and dividing by its value at x j normalizes the expression to yield L j ( x j ) = 1. Another picture of a Lagrange polynomial is provided in Figure 10.4. 0 1 2 3 4 5 6 7 1 0.5 0 0.5 1 1.5 x L 2 Figure 10.4. The Lagrange polynomial L 2 ( x ) for n = 5 . Guess what the data abscissae x i are.