Find Solution using Stirling's formula

x	f(x)
0	0
5	0.0875
10	0.1763
15	0.2679
20	0.3640
25	0.4663
30	0.5774

x = 16
Finding f(2)

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

x	0	5	10	15	20	25	30
y	0	0.0875	0.1763	0.2679	0.364	0.4663	0.5774

Stirling's method to find solution

`h=5-0=5`

Taking `x_0=15` then `p=(x-x_0)/h=(x-15)/5`

The difference table is

`x`	`p=(x-15)/5`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`	`Delta^5y`	`Delta^6y`
0	-3	0
			0.0875
5	-2	0.0875		0.0013
			0.0888		0.0015
10	-1	0.1763		0.0028		0.0002
			0.0916		0.0017		-0.0002
15	0	0.2679		0.0045		0		0.0011
			0.0961		0.0017		0.0009
20	1	0.364		0.0062		0.0009
			0.1023		0.0026
25	2	0.4663		0.0088
			0.1111
30	3	0.5774

`x = 16`

`p = (x - x_0)/h = (16 - 15)/5 = 0.2`

`y_0=0.2679, Delta y_0=0.0961,Delta^2y_(-1)=0.0045,Delta^3y_(-1)=0.0017,Delta^4y_(-2)=0,Delta^5y_(-2)=0.0009,Delta^6y_(-3)=0.0011`

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) + (p(p^2 - 1^2)(p^2 - 2^2))/(5!) * (Delta^5y_(-2)+Delta^5y_(-3))/2 + (p^2(p^2 - 1^2)(p^2 - 2^2))/(6!) * Delta^6y_(-3)`

`y_(0.2) = 0.2679 + (0.2)*((0.0961+0.0916))/2 + ((0.04))/(2)*(0.0045) + ((0.2)(0.04 - 1))/(6)*((0.0017+0.0017))/2 + ((0.04)(0.04 - 1))/(24)*(0) + ((0.2)(0.04 - 1)(0.04 - 4))/(120)*((0.0009))/2 + ((0.04)(0.04 - 1)(0.04 - 4))/(720)*(0.0011)`

`y_(0.2)=0.2679+0.01877 +0.00009 -0.0000544 +0 +0.0000022176 +0.0000002323`

`y_(0.2)=0.2867`


Solution of Stirling's interpolation is `y(16) = 0.2867`

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