Find Solution of an equation x^3-x+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)=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 backward difference interpolation method to find solution

Newton's backward difference table is

x	y	`grady`	`grad^2y`	`grad^3y`	`grad^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 = 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) = 61 + (-0.5) xx 20.625 + (-0.5 (-0.5 + 1))/(2) xx 5.25 + (-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(3.75) = 61 -10.3125 -0.6562 -0.0469 +0`

`y(3.75) = 49.9844`


Solution of newton's backward interpolation method `y(3.75) = 49.9844`

---

---