Question

Find by lagrange’s formula, the value of

U5 if U0 = 1 , U3 = 19 , U4 = 49 , U6 = 181

Answer 1

Here the intervals are unequal. By Lagrange’s interpolation formula we have

$$\begin{matrix} x_0=0, & x_1=3, & x_2=4, & x_3=6 \\ y_0=1, & y_1=19, & y_2=49, & y_3=181 \end{matrix}$$

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

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

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

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

Put $x=5$

$$f(5)=\dfrac{(5-3)(5-4)(5-6)}{(0-3)(0-4)(0-6)}\times 1$$

$$+\dfrac{(5-0)(5-4)(5-6)}{(3-0)(3-4)(3-6)}\times 19$$

$$+\dfrac{(5-0)(5-3)(5-6)}{(4-0)(4-3)(4-6)}\times 49$$

$$+\dfrac{(5-0)(5-3)(5-4)}{(6-0)(6-3)(6-4)}\times 181$$

$$=101$$

$U5=101$