Find Solution of an equation 2x^3-4x+1 using Newton's Backward Difference formula
x1 = 2 and x2 = 4
x = 3.75
Step value (h) = 0.5
Finding f(2)

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

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

x	2	2.5	3	3.5	4
y	9	22.25	43	72.75	113

Newton's backward difference interpolation method to find solution

Newton's backward difference table is

x	y	`grady`	`grad^2y`	`grad^3y`	`grad^4y`
2	`9`
		`13.25`
2.5	`22.25`		`7.5`
		`20.75`		`1.5`
3	`43`		`9`		`0`
		`29.75`		`1.5`
3.5	`72.75`		`10.5`
		`40.25`
4	`113`

The value of x at you want to find the `f(x) : x = 3.75`

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

`p = (x - x_n) / h = (3.75 - 4) / 0.5 = -0.5`

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 + (p(p + 1)(p + 2)(p + 3))/(4!) * grad^4y_n`

`y(3.75) = 113 + (-0.5) xx 40.25 + (-0.5 (-0.5 + 1))/(2) xx 10.5 + (-0.5 (-0.5 + 1)(-0.5 + 2))/(6) xx 1.5 + (-0.5 (-0.5 + 1)(-0.5 + 2)(-0.5 + 3))/(24) xx 0`

`y(3.75) = 113 -20.125 -1.3125 -0.0938 +0`

`y(3.75) = 91.4688`


Solution of newton's backward interpolation method `y(3.75) = 91.4688`

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