Question

If y(1) = – 3, y(3) = 9, y(4) = 30, and y(6) = 132, find the four-point

Lagrange interpolation polynomial that takes the same values as the function y at the given points.

Answer 1

$$\begin{matrix} x_0=1, & x_1=3, & x_2=4, & x_3=6 \\ y_0=-3, & y_1=9, & y_2=30, & y_3=132 \end{matrix}$$

The interpolating polynomial is:

$$L(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3$$

$$L(x)=\dfrac{(x-3)(x-4)(x-6)}{(1-3)(1-4)(1-6)}\times (-3)$$

$$+\dfrac{(x-1)(x-4)(x-6)}{(3-1)(3-4)(3-6)}\times 9$$

$$+\dfrac{(x-1)(x-3)(x-6)}{(4-1)(4-3)(4-6)}\times30$$

$$+\dfrac{(x-1)(x-3)(x-4)}{(6-1)(6-3)(6-4)}\times 132$$

$$=x^3-3x^2+5x-6$$

$$L(x)=x^3-3x^2+5x-6$$