2. Find Solution using Newton's Forward Difference formula

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

x = -1

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

x	0	1	2	3
y	1	0	1	10

Newton's forward difference interpolation method to find solution

Newton's forward difference table is

x	y	`Deltay`	`Delta^2y`	`Delta^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 = -1`

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

`p = (x - x_0)/h = (-1 - 0)/1 = -1`

Newton's forward difference interpolation formula is
`y(x) = y_0 + p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_0 + (p(p - 1)(p - 2))/(3!) * Delta^3y_0`

`y(-1) = 1 + (-1) xx -1 + (-1 (-1 - 1))/(2) xx 2 + (-1 (-1 - 1)(-1 - 2))/(6) xx 6`

`y(-1) = 1 +1 +2 -6`

`y(-1) = -2`


Solution of newton's forward interpolation method `y(-1) = -2`

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