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`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
