Formula

Newton's Backward Difference formula

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

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

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

x = 1925

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

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

Newton's backward difference interpolation method to find solution

Newton's backward difference table is

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

The value of x at you want to find the `f(x) : x = 1925`

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

`p = (x - x_n) / h = (1925 - 1931) / 10 = -0.6`

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(1925) = 101 + (-0.6) xx 8 + (-0.6 (-0.6 + 1))/(2) xx -4 + (-0.6 (-0.6 + 1)(-0.6 + 2))/(6) xx -1 + (-0.6 (-0.6 + 1)(-0.6 + 2)(-0.6 + 3))/(24) xx -3`

`y(1925) = 101 -4.8 +0.48 +0.056 +0.1008`

`y(1925) = 96.8368`


Solution of newton's backward interpolation method `y(1925) = 96.8368`

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