Question

Find f(2.8) from the following table:

x	0	1	2	3
f(x)	1	2	11	34

Answer 1

Since the required value is at the end of the table

Apply backward interpolation formula.

x	`(f(x))/y`	`Deltay`	`Delta^2y`	`Delta^3y`
0	1
		1
1	2		8
		9		6
2	11		14
		23
3	34

`y_((x = x_0 + "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 at x = 2.8

∴ x₀ + nh = 2.8

∴ 3 + n(1) = 2.8

n = 2.8 – 3

n = – 0.2

`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 + ((-0.2)(0.8)(4))/2 + ((-0.2)(-0.8)(1.8))/6 (6)`

= 34 – 4.6 – 1.12 – 0.288

= 34 – 6.008

= 27.992

∴ f(2.8) = 27.992