Find Solution using Bessel's formula
| x | f(x) |
| 10 | 20.9848 |
| 11 | 22.9816 |
| 12 | 24.9781 |
| 13 | 26.9743 |
| 14 | 28.9702 |
x = 12.3
Finding f(2)
Solution:
The value of table for `x` and `y`
| x | 10 | 11 | 12 | 13 | 14 |
|---|
| y | 20.9848 | 22.9816 | 24.9781 | 26.9743 | 28.9702 |
|---|
Bessel'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` |
| 10 | -2 | 20.9848 | | | |
| | | 1.9968 | | |
| 11 | -1 | 22.9816 | | -0.0003 | |
| | | 1.9965 | | 0 |
| 12 | 0 | 24.9781 | | -0.0003 | |
| | | 1.9962 | | 0 |
| 13 | 1 | 26.9743 | | -0.0003 | |
| | | 1.9959 | | |
| 14 | 2 | 28.9702 | | | |
`x = 12.3`
`p = (x - x_0)/h = (12.3 - 12)/1 = 0.3`
`y_0=24.9781, Delta y_0=1.9962,Delta^2y_(-1)=-0.0003,Delta^3y_(-1)=0`
Bessel's formula is
`y_p=(y_0+y_1)/2+(p-1/2)*Delta y_0 + (p(p-1))/(2!) * (Delta^2y_(-1)+Delta^2y_(0))/2 + ((p-1/2)p(p-1))/(3!) * Delta^3y_(-1)`
`y_(0.3) = (24.9781+26.9743)/2 + (0.3-1/2)*(1.9962) + (0.3(0.3-1))/(2)*((-0.0003))/2 + ((0.3-1/2)0.3(0.3-1))/(6)*(0)`
`y_(0.3)=25.9762-0.39924 +0.0000315 +0`
`y_(0.3)=25.577`
Solution of Bessel's interpolation is `y(12.3) = 25.577`
This material is intended as a summary. Use your textbook for detail explanation.
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