2. Find Solution using Stirling's formula

x	f(x)
10	0.23967
11	0.28060
12	0.31788
13	0.35209
14	0.38368

x = 12.2

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

x	10	11	12	13	14
y	0.23967	0.2806	0.31788	0.35209	0.38368

Stirling's method to find solution

`h=11-10=1`

Taking `x_0=12` then `p=(x-x_0)/h=(x-12)/1`

The difference table is

`x`	`p=(x-12)/1`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
10	-2	0.23967
			0.04093
11	-1	0.2806		-0.00365
			0.03728		0.00058
12	0	0.31788		-0.00307		-0.00013
			0.03421		0.00045
13	1	0.35209		-0.00262
			0.03159
14	2	0.38368

`x = 12.2`

`p = (x - x_0)/h = (12.2 - 12)/1 = 0.2`

`y_0=0.31788, Delta y_0=0.03421,Delta^2y_(-1)=-0.00307,Delta^3y_(-1)=0.00045,Delta^4y_(-2)=-0.00013`

Stirling's formula is
`y_p=y_0+p*(Delta y_0+Delta y_(-1))/2 + (p^2)/(2!) * Delta^2y_(-1) + (p(p^2 - 1^2))/(3!) * (Delta^3y_(-1)+Delta^3y_(-2))/2 + (p^2(p^2 - 1^2))/(4!) * Delta^4y_(-2)`

`y_(0.2) = 0.31788 + (0.2)*((0.03421+0.03728))/2 + ((0.04))/(2)*(-0.00307) + ((0.2)(0.04 - 1))/(6)*((0.00045+0.00058))/2 + ((0.04)(0.04 - 1))/(24)*(-0.00013)`

`y_(0.2)=0.31788+0.007149 -0.0000614 -0.00001648 +0.000000208`

`y_(0.2)=0.324951`


Solution of Stirling's interpolation is `y(12.2) = 0.324951`

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