1. Find Solution using Stirling's formula

x	f(x)
20	49225
25	48316
30	47236
35	45926
40	44306

x = 28

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

x	20	25	30	35	40
y	49225	48316	47236	45926	44306

Stirling's method to find solution

`h=25-20=5`

Taking `x_0=30` then `p=(x-x_0)/h=(x-30)/5`

The difference table is

`x`	`p=(x-30)/5`	`y`	`Deltay`	`Delta^2y`	`Delta^3y`	`Delta^4y`
20	-2	49225
			-909
25	-1	48316		-171
			-1080		-59
30	0	47236		-230		-21
			-1310		-80
35	1	45926		-310
			-1620
40	2	44306

`x = 28`

`p = (x - x_0)/h = (28 - 30)/5 = -0.4`

`y_0=47236, Delta y_0=-1310,Delta^2y_(-1)=-230,Delta^3y_(-1)=-80,Delta^4y_(-2)=-21`

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_(-0.4) = 47236 + (-0.4)*((-1310-1080))/2 + ((0.16))/(2)*(-230) + ((-0.4)(0.16 - 1))/(6)*((-80-59))/2 + ((0.16)(0.16 - 1))/(24)*(-21)`

`y_(-0.4)=47236+478 -18.4 -3.892 +0.1176`

`y_(-0.4)=47691.8256`


Solution of Stirling's interpolation is `y(28) = 47691.8256`