Question

Use Lagrange polynomial estimate (2) for the given data.

x -2 -1 0 4

f(x) -2 4 1 8

Answer 1

Lagrange polynomial:

$P(x)=f_0L_0(x)+f_1L_1(x)+f_2L_2(x)+f_3L_3(x)$

$L_0(x)=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}=\frac{(x+1)x(x-4)}{(-2+1)(-2)(-2-4)}=-\frac{(x+1)x(x-4)}{12}$

$L_1(x)=\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}=\frac{(x+2)x(x-4)}{(-1+2)(-1)(-1-4)}=\frac{(x+2)x(x-4)}{5}$

$L_2(x)=\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}=\frac{(x+2)(x+1)(x-4)}{(0+2)(0+1)(0-4)}=-\frac{(x+2)(x+1)(x-4)}{8}$

$L_3(x)=\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}=\frac{(x+2)(x+1)x}{4(4+2)(4+1)}=\frac{(x+2)(x+1)x}{120}$

$f(2)=P(2)=2\frac{2(2+1)(2-4)}{12}+4\frac{2(2+2)(2-4)}{5}-\frac{(2+2)(2+1)(2-4)}{8}+8\frac{2(2+2)(2+1)}{120}=$

$=-2-12.8+3+1.6=-10.2$