Question

How do I complete compute f(0.3) for the data x 0 1 3 4 7 f 1 3 49 129 813 using Lagrange's interpolation formula?

Answer 1

The given table is

x = 0, 1, 3, 4, 7

f(x)= 1, 3, 49 , 129, 813

$\Rightarrow f(x)=\frac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_o-x_1)(x_o-x_2)(x_o-x_3)(x_o-x_4)}f_o+...\frac{(x-x_o)(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)}f_4$

$\Rightarrow f(x)=\frac{(0.3-1)(0.3-3)(0.3-4)(0.3-7)}{(-1)(-3)(-4)(-7)}1+\frac{(0.3-0)(0.3-3)(0.3-4)(0.3-7)}{(1)(-2)(-3)(-6)}3$

$+\frac{(0.3-1)(0.3-3)(0.3-4)(0.3-7)}{(3)(2)(-1)(-4)}49+\frac{(0.3-1)(0.3-3)(0.3-3)(0.3-7)}{(4)(3)(1)(-3)}129$

$\frac{(0.3-0)(0.3-1)(0.3-3)(0.3-4)}{(7)(6)(4)(3)}813$

After solving, we will get 1.831