Given datayiat abscissaexias before, the unique polynomial interpolant of degree at mostncannow be written asp(x)=nj=0yjLj(x).Indeed,pis of degree at mostn(being the linear combination of polynomials of degreen), and itsatisfies the interpolation conditions becausep(xi)=nj=0yjLj(xi)=0+···+0+yiLi(xi)+0+···+0=yi.Downloaded 10/19/18 to 126.96.36.199. Redistribution subject to SIAM license or copyright; see
10.3. Lagrange interpolation303Example 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 makeL0(x) vanish atx=2 andx=4 we writeL0(x)=a(x−2)(x−4).RequiringL0(1)=1 then determinesa(1−2)(1−4)=1, i.e.,a=13, and thusL0(x)=13(x−2)(x−4).Similarly we determine thatL1(x)=−12(x−1)(x−4),L2(x)=16(x−1)(x−2).These Lagrange polynomials are depicted in Figure 10.3. We thus obtain the interpolantp2(x)=y03(x−2)(x−4)−y12(x−1)(x−4)+y26(x−1)(x−2)=13(x−2)(x−4)−32(x−1)(x−4)+36(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,p2(x)=(−2x2+12x−7)/3. It is also easy to verify that here, too,p2(3)=113.All this should not come as a surprise—it’s an illustration of the uniqueness of polynomial interpo-lation; see the theorem on page 301.0123456−3−2−10123xQuadratic Lagrange polynomialsL0(x)L1(x)L2(x)Figure 10.3.The quadratic Lagrange polynomials L0(x),L1(x),and L2(x)based onpoints x0=1,x1=2,x2=4, used in Example10.2.Downloaded 10/19/18 to 188.8.131.52. Redistribution subject to SIAM license or copyright; see
304Chapter 10. Polynomial InterpolationProperties of Lagrange polynomialsWhat properties do Lagrange polynomials have? What do they look like?In the general case wheren+1 data abscissaexiare specified, the Lagrange polynomialsuniquely exist, because they are really nothing other than polynomial interpolants for special data.41This is also straightforward to verify directly, by explicitly specifyingLj(x)=(x−x0)···(x−xj−1)(x−xj+1)···(x−xn)(xj−x0)···(xj−xj−1)(xj−xj+1)···(xj−xn)=ni=0i̸=j(x−xi)(xj−xi).Indeed, the polynomial of degreen, written in terms of its roots as(x−x0)···(x−xj−1)(x−xj+1)···(x−xn),clearly interpolates the 0-values at all data abscissae other thanxj, and dividing by its value atxjnormalizes the expression to yieldLj(xj)=1. Another picture of a Lagrange polynomial isprovided in Figure 10.4.01234567−1−0.500.511.5xL2Figure 10.4.The Lagrange polynomial L2(x)for n=5. Guess what the data abscissae xiare.