Question

Use lagrange interpolation to find a polynomial that passes through the points. (a) (0, 1), (2, 3), (3, 0) (b) (−1, 0), (2, 1), (3, 1), (5, 2)

Answer 1

The Lagrange interpolation polynomial is a sum of products.

(a)
$f(x)=1\times \frac{(x-2)(x-3)}{(0-2)(0-3)}+3\times \frac{(x-0)(x-3)}{(2-0)(2-3)}+0\times \frac{(x-0)(x-2)}{(3-0)(3-2)}$
This can be simplified to
$f(x)=\frac{1}{3}(-4x^{2}+11x+3)$

(b)
$g(x)=0+\frac{1(x+1)(x-3)(x-5)}{(2+1)(2-3)(2-5)}+\frac{1(x+1)(x-2)(x-5)}{(3+1)(3-2)(3-5)}+\frac{2(x+1)(x-2)(x-3)}{(5+1)(5-2)(5-3)}$
This can be simplified to
$g(x)=\frac{1}{24}(x^{3}-6x^{2}+11x+18)$