Question

From the following table obtain a polynomial of degree y in x.

x	1	2	3	4	5
y	1	– 1	1	– 1	1

Answer 1

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

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

x	y	`Deltay`	`Delta^2y`	`Delta^3y``	Delta^4y`
1	1
		– 2
2	– 1		4
		2		– 8
3	1		– 4		16
		– 2		8
4	– 1		4
		2
5	1

To find y in terms of x

x_n + nh = x

5 + n(1) = x

∴ n = x – 5

`"y"_((x)) = 1 + ((x - 5))/(1!) (2) + ((x - )(x - 5 + 1))/(2!) (4) + ((x - 5)(x -5 + 1)(x - 5 + 2))/(3!) (8) + ((x - 5)(x - 5 + 1)(x - 5 + 2)(x - 5 + 3))/(4!) (16)`

= `1 + 2(x - 5) + ((x - 5)(x - 4)(4))/2 + ((x - 5)(x - 4)(x - 2)(8))/6 + ((x - 5)(x - 4)(x - 2)(16))/24`

= `1 + 2x - 10 + 2(x^2 - 9x + 20) + 4/3 (x - 5) (x^2 - 7x + 12) + 2/3(x^2 - 9x + 20)(x^2 - 5x + 6)`

= `1 + 2x - 10 + 2x^2 - 18x + 40 + 4/3 [x^3 - 7x^2 + 12x - 5x^2 + 35x - 60] + 2/3 [x^4 - 5x^3 +  6x^2 - 9x^3 + 45x^2 - 54x + 20x^2 - 100x + 120]`

= `2x^2 - 16x + 31 + 4/3 [x^3 - 12x^2 + 47x - 60] + 2/3 [x^4 - 14x^3 + 71x^2 - 154x + 120]`

= `2x^2 - 16x + 31 + 4/3 x^3 - 16x^2 + 188/3 x - 80 + 2/3 x^4 - 28/3 x^3 + 142/3 x^2 - 308/3 x + 80`

= `2/3 x^4 + (4/3 - 28/3) x^3 + (2 - 16 + 142/3) x^2 + (- 16 + 188/3 - 308/3)x + (31 - 80 + 80)`

= `2/3 x^4 + ((-24)/3)x^3+ ((6 - 48 + 142)/3) x^2 + ((-48 + 188 - 308)/3) x + 31`

`y(x) = 2/3 x^4 - 8x^3 + 100/3 x^2 - 56x + 31`