Question

Find by lagrange’s formula, the value of
U4 if U3 = 16 , U5 = 36 , U7 = 64 , U8 = 81 and U9 = 100

Answer 1

Here the intervals are unequal.

$$\begin{matrix} x_0=3, & x_1=5, & x_2=7, & x_3=8, & x_4=9 \\ y_0=16, & y_1=36, & y_2=64, & y_3=81,& y_4=100 \end{matrix}$$

$$y=f(x)=(x+1)^2$$

Put $x=4$

$$f(4)=(4+1)^2=25$$

By Lagrange’s interpolation formula we have

$$y=f(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)}\times y_3$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)}\times y_4$$

Put $x=4$

$$f(4)=\dfrac{(4-5)(4-7)(4-8)(4-9)}{(3-5)(3-7)(3-8)(3-9)}\times 16$$

$$+\dfrac{(4-3)(4-7)(4-8)(4-9)}{(5-3)(5-7)(5-8)(5-9)}\times 36$$

$$+\dfrac{(4-3)(4-5)(4-8)(4-9)}{(7-3)(7-5)(7-8)(7-9)}\times 64$$

$$+\dfrac{(4-3)(4-5)(4-7)(4-9)}{(8-3)(8-5)(8-7)(8-9)}\times 81$$

$$+\dfrac{(4-3)(4-5)(4-7)(4-8)}{(9-3)(9-5)(9-7)(9-8)}\times 100$$

$$=25$$

$U4=25$