Question

Find the value of f(x) when x = 32 from the following table:

x	30	5	40	45	50
f(x)	15.9	14.9	14.1	13.3	12.5

Answer 1

Since the value of f(x) is required near the beginning of the table

We use the Newton’s forward interpolation formula.

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

To find y at x = 32

∴ x₀ + nh = 32;

30 + n(5) = 32

5n = 32 – 30

⇒ 5n = 2

n = `2/5`

∴ n = 0.4

x	y	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
30	15.9
		– 1
35	14.9		0.2
		– 0.8		– 0.2
40	14.1		0		0.2
		– 0.8		0
45	13.3		0
		– 0.8
50	12.5

`y_((x = 32)) = 15.9 + ((0.4))/(1!) (1) + ((0.40.4 - 1))/(2!) (0.2) + ((0.4)(0.4 - 1)(0.4 - 2))/(3) (- 0.2) + ((0.4)(.4 - 1)(0.4 - 2)(0.4 - 3))/(4!) (0.2) +`

= 15.9 – 0.4 – 0.024 – 0.0128 – 0.00832

= 15.9 – 0.44512

= 15.45488

= 15.45

∴ When x = 32, f(x) = 15.45