Formula

Newton's Forward Difference formula

`p = (x - x_0)/h` `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 + ...`

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Examples
1. Find Solution using Newton's Forward Difference formula

x	f(x)
1891	46
1901	66
1911	81
1921	93
1931	101

x = 1895
Finding option 1. Value f(2)

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

x	1891	1901	1911	1921	1931
y	46	66	81	93	101

Newton's forward difference interpolation method to find solution

Newton's forward difference table is

x	y	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
1891	46
		`66-46=20`
1901	66		`15-20=-5`
		`81-66=15`		`-3--5=2`
1911	81		`12-15=-3`		`-1-2=-3`
		`93-81=12`		`-4--3=-1`
1921	93		`8-12=-4`
		`101-93=8`
1931	101

The value of `x` at you want to find the `f(x) : x = 1895`

`h = x_1 - x_0 = 1901 - 1891 = 10`

`p = (x - x_0)/h = (1895 - 1891)/10 = 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(1895) = 46 + 0.4 xx 20 + (0.4 (0.4 - 1))/(2) xx -5 + (0.4 (0.4 - 1)(0.4 - 2))/(6) xx 2 + (0.4 (0.4 - 1)(0.4 - 2)(0.4 - 3))/(24) xx -3`

`y(1895) = 46 +8 +0.6 +0.128 +0.1248`

`y(1895) = 54.8528`


Solution of newton's forward interpolation method `y(1895) = 54.8528`

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