Formula

1. For `x=x_n` `[(dy)/(dx)]_(x=x_n) = 1/h * (grad Y_n + 1/2 * grad^2 Y_n + 1/3 * grad^3 Y_n + 1/4 * grad^4 Y_n + ...)` `[(d^2y)/(dx^2)]_(x=x_n) = 1/h^2 * (grad^2 Y_n + grad^3 Y_n + 11/12 * grad^4 Y_n + ...)`

2. For any value of `x` `[(dy)/(dx)] = 1/h * (grad Y_n + (2t+1)/2 * grad^2 Y_n + (3t^2+6t+2)/6 * grad^3 Y_n + (4t^3+18t^2+22t+6)/24 * grad^4 Y_n + ...)` `[(d^2y)/(dx^2)] = 1/h^2 * (grad^2 Y_n + (t+1) * grad^3 Y_n + (12t^2+36t+22)/24 * grad^4 Y_n + ...)`

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Examples
1. Using Newton's Backward Difference formula to find solution

x	f(x)
1.4	4.0552
1.6	4.9530
1.8	6.0496
2.0	7.3891
2.2	9.0250

x = 2.2

Solution:
Numerical differentiation method to find solution.
The value of table for `x` and `y`

x	1.4	1.6	1.8	2	2.2
y	4.0552	4.953	6.0496	7.3891	9.025

Newton's backward differentiation table is

x	y	`grady`	`grad^2y`	`grad^3y`	`grad^4y`
1.4	4.0552
		0.8978
1.6	4.953		0.1988
		1.0966		0.0441
1.8	6.0496		0.2429		0.0094
		1.3395		0.0535
2	7.3891		0.2964
		1.6359
2.2	9.025

The value of x at you want to find `f(x) : x_n = 2.2`

`h = x_1 - x_0 = 1.6 - 1.4 = 0.2`


`[(dy)/(dx)]_(x=x_n) = 1/h * (grad y_n + 1/2 * grad^2 y_n + 1/3 * grad^3 y_n + 1/4 * grad^4 y_n)`

`:.[(dy)/(dx)]_(x=2.2) = 1/0.2 xx (1.6359 + 1/2 xx 0.2964 + 1/3 xx 0.0535 + 1/4 xx 0.0094)`

`:.[(dy)/(dx)]_(x=2.2) = 9.02142`


`[(d^2y)/(dx^2)]_(x=x_n) = 1/h^2 * (grad^2 y_n + grad^3 y_n + 11/12 * grad^4 y_n)`

`:.[(d^2y)/(dx^2)]_(x=2.2) = 1/0.04 * (0.2964 + 0.0535 + 11/12 xx 0.0094)`

`:.[(d^2y)/(dx^2)]_(x=2.2) = 8.96292`


`:.` `Pn'(2.2) = 9.02142` and `Pn''(2.2) = 8.96292`