Question

Using Newton’s forward interpolation formula find the cubic polynomial.

x	0	1	2	3
f(x)	1	2	1	10

Answer 1

Since we use the Newton’s forward interpolation formula.

`y_((x  =  x_0  +  "h")) = y_0 + "n"/(1!) Deltay_0 + ("n"("n" - 1))/(2!) Delta^2y_0  ("n"("n" - 1)("n" - 2))/(3!) Delta^3y_0 +  .........`

To find y at x

∴ x₀ + nh = x

0 + n(1) = x

∴ n = x

x	y	`Deltay`	`Delta^2y`	`Delta^3y`
0	1
		1
1	2		– 2
		– 1		12
2	1		10
		9
3	10

y = `1 + x/(1!) + (x(x -1))/(2!) (-2) + (x(x - 1)(x - 2))/(3!) (12)`

= `1 + x/1 + (x(x - 1))/2 (-2) + (x(x - 1)(x - 2)(12))/6`

= 1 + x + (x² – x)(– 1) + 2x(x² – 3x + 2)

y = 1 + x – x² + x + 2x³ – 6x² + 4x

y = 2x³ – 7x² + 6x + 1

∴ f(x) = 2x³ – 7x² + 6x + 1