Question

From the following data find y at x = 43 and x = 84.

x	40	50	60	70	80	90
y	184	204	226	250	276	304

Answer 1

To find y at x = 43

Since the value of y is required near the beginning of the table

We use the Newton’s forward interpolation formula.

`y_((x = 2.8)) = 34 + ((-0.2))/(1!) (23) + ((-0.2)(-0.2 + 1))/2 (14) + ((-0.2)(-0.2 + 1)(-0.2 + 2))/(3!) (6) + ......`

= `34 - 4.6 + ((-0.2)(0.8)(14))/2 + ......`

x	y	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
40	184
		20
50	204		2
		22		0
60	226		2		0
		24		0
70	250		2		0
		26		0
80	276		2
		28
90	304

`y_((43)) = 184 + 0.3/(1!) (0) + ((0.3)(0.3 - 1))/(2!) (2) + ((0.3)(0.3 - 1)(0.3 - 2))/(3!) (0)`

= 184 + (0.3)(20) + (0.3)(– 0.7)

= 184 + 6.0 – 0.21

= 190 + 0.21

`y_((x = 43))` = 189.79

To find y at x = 84

Since the value of y is required at the end of the table, we apply backward interpolation formula.

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

x	y	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
40	184
		20
50	204		2
		22		0
60	226		2		0
		24		0
70	250		2		0
		26		0
80	276		2
		28
90	304

x_n + nh = x

90 + n(10) = 84

10n = 84 – 90

10n = – 6

∴ n = – 0.6

`y_((x = 84)) = 304 + ((0.6))/(1!) (28) + ((0.6)(-0.6 + 1))/(2!) (2) + ....`

= `304 + (0.6)(28) + ((-0.06)(0.4))/2 + 2`

= 304 – 16.8 – 0.24

= 304 – 17.04

= 286.96