Formula

Newton's Divided Difference Interpolation formula

`y(x) = y_0 + (x - x_0) f[x_0, x_1] + (x - x_0)(x - x_1) f[x_0, x_1, x_2] + ...`

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Examples
1. Find Solution using Newton's Divided Difference Interpolation formula

x	f(x)
300	2.4771
304	2.4829
305	2.4843
307	2.4871

x = 301

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

x	300	304	305	307
y	2.4771	2.4829	2.4843	2.4871

Numerical divided differences method to find solution

Newton's divided difference table is

x	y	`1^(st)` order	`2^(nd)` order
300	2.4771
		`(2.4829-2.4771)/(304-300)=0.0014`
304	2.4829		`(0.0014-0.0014)/(305-300)=0`
		`(2.4843-2.4829)/(305-304)=0.0014`
305	2.4843		`(0.0014-0.0014)/(307-304)=0`
		`(2.4871-2.4843)/(307-305)=0.0014`
307	2.4871

The value of `x` at you want to find the `f(x) : x = 301`

Newton's divided difference interpolation formula is
`f(x)=y_0 +(x-x_0) f[x_0, x_1]+(x-x_0)(x-x_1) f[x_0, x_1, x_2]`

`y(301) = 2.4771 + (301 -300) xx 0.0014 + (301 -300)(301 -304) xx 0`

`y(301) = 2.4771 + (1) xx 0.0014 + (1)(-3) xx 0`

`y(301) = 2.4771 +0.0014 +0`

`y(301) = 2.4785`


Solution of divided difference interpolation method `y(301) = 2.4785`

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