Find Solution using Everett's formula

x	f(x)
310	2.49136
320	2.50515
330	2.51851
340	2.53148
350	2.54407
360	2.55630

x = 337.5
Finding f(2)

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

x	310	320	330	340	350	360
y	2.4914	2.5052	2.5185	2.5315	2.5441	2.5563

Everett method to find solution

`h=320-310=10`

Taking `x_0=330` then `p=(x-x_0)/h=(x-330)/10`

Now the central difference table is

`x`	`p=(x-330)/10`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`
310	-2	2.4914
			0.0138
320	-1	2.5052		-0.0004
			0.0134		0
330	0	2.5185		-0.0004
			0.013		0
340	1	2.5315		-0.0004
			0.0126		0
350	2	2.5441		-0.0004
			0.0122
360	3	2.5563

`x = 337.5`

`p = (x - x_0)/h = (337.5 - 330)/10 = 0.75`

`q=1-p=1-0.75=0.25`

`y_0=2.5185, Delta y_0=0.013,Delta^2y_(-1)=-0.0004`

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_(0.75) = (0.25)*(2.5185) + ((0.25)(0.0625 - 1))/(6) * (-0.0004)+...+(0.75)*(2.5315) + ((0.75)(0.5625 - 1))/(6) * (-0.0004)+...`

`y_(0.75)=0.6296 +0.0000152344 + 1.8986 +0.0000207813`

`y_(0.75)=2.5283`


Solution of Everett interpolation is `y(337.5) = 2.5283`

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