Formula

Everett's formula

`p = (x - x_0)/h` `q = 1-p` `y_p=qy_0 + (q(q^2 - 1^2))/(3!) * Delta^2y_(-1) + (q(q^2 - 1^2)(q^2 - 2^2))/(5!) * Delta^4y_(-2)+...+py_1 + (p(p^2 - 1^2))/(3!) * Delta^2y_(0) + (p(p^2 - 1^2)(p^2 - 2^2))/(5!) * Delta^4y_(-1)+...`

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Examples
1. Find Solution using Everett's formula

x	f(x)
1	1
1.1	1.049
1.2	1.096
1.3	1.140

x = 1.15

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

x	1	1.1	1.2	1.3
y	1	1.049	1.096	1.14

Everett method to find solution

`h=1.1-1=0.1`

Taking `x_0=1.1` then `p=(x-x_0)/h=(x-1.1)/0.1`

Now the central difference table is

`x`	`p=(x-1.1)/0.1`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`
1	-1	1
			0.049
1.1	0	1.049		-0.002
			0.047		-0.001
1.2	1	1.096		-0.003
			0.044
1.3	2	1.14

`x = 1.15`

`p = (x - x_0)/h = (1.15 - 1.1)/0.1 = 0.5`

`y_0=1.049, Delta y_0=0.047,Delta^2y_(-1)=-0.002`

Everett interpolation formula is
`y_p=qy_0 + (q(q^2 - 1^2))/(3!) * Delta^2y_(-1)+...+py_1 + (p(p^2 - 1^2))/(3!) * Delta^2y_(0)+...`

`y_(0.5) = (0.5)*(1.049) + ((0.5)(0.25 - 1))/(6) * (-0.002)+...+(0.5)*(1.096) + ((0.5)(0.25 - 1))/(6) * (-0.003)+...`

`y_(0.5)=0.5245 +0.000125 + 0.548 +0.0001875`

`y_(0.5)=1.07281`


Solution of Everett interpolation is `y(1.15) = 1.07281`

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