2. Find Solution using Newton's Backward Difference formula

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

x = 4

Solution:
The value of table for `x` and `y`

x	0	1	2	3
y	1	0	1	10

Newton's backward difference interpolation method to find solution

Newton's backward difference table is

x	y	`grady`	`grad^2y`	`grad^3y`
0	`1`
		`-1`
1	`0`		`2`
		`1`		`6`
2	`1`		`8`
		`9`
3	`10`

The value of x at you want to find the `f(x) : x = 4`

`h = x_1 - x_0 = 1 - 0 = 1`

`p = (x - x_n) / h = (4 - 3) / 1 = 1`

Newton's backward difference interpolation formula is
`y(x) = y_n + p grad y_n + (p(p + 1))/(2!) * grad^2y_n + (p(p + 1)(p + 2))/(3!) * grad^3y_n`

`y(4) = 10 + 1 xx 9 + (1 (1 + 1))/(2) xx 8 + (1 (1 + 1)(1 + 2))/(6) xx 6`

`y(4) = 10 +9 +8 +6`

`y(4) = 33`


Solution of newton's backward interpolation method `y(4) = 33`

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