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`
This material is intended as a summary. Use your textbook for detail explanation.
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