Question

Using Lagrange’s formula, find the values of

(i) y5 if y1 = 4, y3 = 120, y4 = 340, y5 = 2544

(ii) y0 if y–30 = 30, y–12 = 34, y3 = 38, y18 = 42.

Answer 1

(i) y5 if y1 = 4, y3 = 120, y4 = 340, y5 = 2544

Answer y5=2544

y5 if y1 = 4, y3 = 120, y4 = 340, y6 = 2544

Here the intervals are unequal.

$$\begin{matrix} x_0=1, & x_1=3, & x_2=4, & x_3=6 \\ y_0=4, & y_1=120, & y_2=340, & y_3=2544 \end{matrix}$$

By Lagrange’s interpolation formula we have

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

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

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

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

Put $x=5$

$$f(5)=\dfrac{(5-3)(5-4)(5-6)}{(1-3)(1-4)(1-6)}\times 4$$ $$+\dfrac{(5-1)(5-4)(5-6)}{(3-1)(3-4)(3-6)}\times 120$$

$$+\dfrac{(5-1)(5-3)(5-6)}{(4-1)(4-3)(4-6)}\times 340$$

$$+\dfrac{(5-1)(5-3)(5-4)}{(6-1)(6-3)(6-4)}\times 2544$$

$$=1052$$

$y5=1052$

(ii)

Here the intervals are unequal.

$$\begin{matrix} x_0=-30, & x_1=-12, & x_2=3, & x_3=18 \\ y_0=30, & y_1=34, & y_2=38, & y_3=42 \end{matrix}$$

By Lagrange’s interpolation formula we have

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

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

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

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

Put $x=0$

$$f(0)=\dfrac{(0+12)(0-3)(0-18)}{(-30+12)(-30-3)(-30-18)}\times 30$$

$$+\dfrac{(0+30)(0-3)(0-18)}{(-12+30)(-12-3)(-12-18)}\times 34$$

$$+\dfrac{(0+30)(0+12)(0-18)}{(3+30)(3+12)(3-18)}\times 38$$

$$+\dfrac{(0+30)(0+12)(0-3)}{(18+30)(18+12)(18-3)}\times 42$$

$$=\dfrac{409}{11}$$

$y_0=\dfrac{409}{11}$