Find Solution of an equation 2x^3-4x+1 using Newton's Forward Difference formula
x1 = 2 and x2 = 4
x = 2.1
Step value (h) = 0.25
Finding f(2)

Solution:
Equation is `f(x)=2x^3-4x+1`.

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

x	2	2.25	2.5	2.75	3	3.25	3.5	3.75	4
y	9	14.7812	22.25	31.5938	43	56.6562	72.75	91.4688	113

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	9
		5.7812
2.25	14.7812		1.6875
		7.4688		0.1875
2.5	22.25		1.875		0
		9.3438		0.1875
2.75	31.5938		2.0625		0
		11.4062		0.1875
3	43		2.25		0
		13.6562		0.1875
3.25	56.6562		2.4375		0
		16.0938		0.1875
3.5	72.75		2.625		0
		18.7188		0.1875
3.75	91.4688		2.8125
		21.5312
4	113

The value of `x` at you want to find the `f(x) : x = 2.1`

`h = x_1 - x_0 = 2.25 - 2 = 0.25`

`p = (x - x_0)/h = (2.1 - 2)/0.25 = 0.4`

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(2.1) = 9 + 0.4 xx 5.7812 + (0.4 (0.4 - 1))/(2) xx 1.6875 + (0.4 (0.4 - 1)(0.4 - 2))/(6) xx 0.1875 + (0.4 (0.4 - 1)(0.4 - 2)(0.4 - 3))/(24) xx 0`

`y(2.1) = 9 +2.3125 -0.2025 +0.012 +0`

`y(2.1) = 11.122`


Solution of newton's forward interpolation method `y(2.1) = 11.122`

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