Question

A second degree polynomial passes though the point (1, –1) (2, –1) (3, 1) (4, 5). Find the polynomial

Answer 1

Point (1, –1), (2, –1), (3, 1), (4, 5)

x	1	2	3	4
y	– 1	– 1	1	5

We will use Newton’s backward interpolation formula to find the polynomial.

x	y	`Deltay`	`Delta^2y`	`Delta^3y`
1	– 1
		0
2	– 1		2
		2		0
3	1		2
		4
4	5

`y_((x =  x + "nh")) = y_"n" + "n"/(1!) ∇y_"n" ("n"("n" + 1))/(2!) ∇^2y_"n" + ("n"("n" + 1)("n" + 2))/(3!) Delta^3y_"n" + ......`

To find y in terms of x

`x_"n" + "nh" = x`

Here `x_"n"` = 4, h = 1

4 + n(1) = x

n = x – 4

`y_((x)) = 5 + ((x - 4))/(1!) (4) + ((x - 4)(x - 4 + 1))/(2!) (2) + ((x - 4)(x - 4 + 1)(x - 4 + 2))/(3!) (0) + ......`

= 5 + (x – 4)(4) + (x – 4)(x – 3) + 0

= 5 + 4x – 16 + x² – 7x + 12

y(x) = x² – 3x + 1