Find Solution of an equation x^3-x+1 using Newton's Forward Difference formula
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.5
Finding f(2)

Solution:
Equation is `f(x)=x^3-x+1`.

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

x	2	2.5	3	3.5	4
y	7	14.125	25	40.375	61

Newton's forward difference interpolation method to find solution

Newton's forward difference table is

x	y	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
2	7
		7.125
2.5	14.125		3.75
		10.875		0.75
3	25		4.5		0
		15.375		0.75
3.5	40.375		5.25
		20.625
4	61

The value of `x` at you want to find the `f(x) : x = 2.25`

`h = x_1 - x_0 = 2.5 - 2 = 0.5`

`p = (x - x_0)/h = (2.25 - 2)/0.5 = 0.5`

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 + (p(p - 1)(p - 2)(p - 3))/(4!) * Delta^4y_0`

`y(2.25) = 7 + 0.5 xx 7.125 + (0.5 (0.5 - 1))/(2) xx 3.75 + (0.5 (0.5 - 1)(0.5 - 2))/(6) xx 0.75 + (0.5 (0.5 - 1)(0.5 - 2)(0.5 - 3))/(24) xx 0`

`y(2.25) = 7 +3.5625 -0.4688 +0.0469 +0`

`y(2.25) = 10.1406`


Solution of newton's forward interpolation method `y(2.25) = 10.1406`

---

---