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

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} x & 1 & 3 & 4 & 6 \\ \hline y & -3 & 9 & 30 & 132 \end{array}$

The Lagrange interpolation polynomial is

$f(x)= \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}y_0+ \frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+ \frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}y_2+ \frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3=\\$

$=\frac{1}{10}(x-3)(x-4)(x-6)+\frac{3}{2}(x-1)(x-4)(x-6)-5(x-1)(x-3)(x-6)+\frac{22}{5}(x-1)(x-3)(x-4)=\\$

$\frac{1}{10}(x^3-13x^2+54x-72)+\frac{3}{2}(x^3-11x^2+34x-24)-5(x^3-10x^2+27x-18)+\frac{22}{5}(x^3-8x^2+19x-12)=\\$

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

Answer: $f(x)=x^3-3x^2+5x-6.$