2. Find Solution using Everett's formula

x	f(x)
25	4
26	3.846
27	3.704
28	3.571
29	3.448
30	3.333

x = 27.4

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

x	25	26	27	28	29	30
y	4	3.846	3.704	3.571	3.448	3.333

Everett method to find solution

`h=26-25=1`

Taking `x_0=27` then `p=(x-x_0)/h=(x-27)/1`

Now the central difference table is

`x`	`p=(x-27)/1`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`	`Delta^5y`
25	-2	4
			-0.154
26	-1	3.846		0.012
			-0.142		-0.003
27	0	3.704		0.009		0.004
			-0.133		0.001		-0.007
28	1	3.571		0.01		-0.003
			-0.123		-0.002
29	2	3.448		0.008
			-0.115
30	3	3.333

`x = 27.4`

`p = (x - x_0)/h = (27.4 - 27)/1 = 0.4`

`y_0=3.704, Delta y_0=-0.133,Delta^2y_(-1)=0.009,Delta^4y_(-2)=0.004`

Everett interpolation formula is
`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)+...`

`y_(0.4) = (0.6)*(3.704) + ((0.6)(0.36 - 1))/(6) * (0.009) + ((0.6)(0.36 - 1)(0.36 - 4))/(120) * (0.004)+...+(0.4)*(3.571) + ((0.4)(0.16 - 1))/(6) * (0.01) + ((0.4)(0.16 - 1)(0.16 - 4))/(120) * (-0.003)+...`

`y_(0.4)=2.2224 -0.000576 +0.000046592 + 1.4284 -0.00056 -0.000032256`

`y_(0.4)=3.64968`


Solution of Everett interpolation is `y(27.4) = 3.64968`

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