Find Solution using Gauss Backward formula
| x | f(x) |
| 1939 | 12 |
| 1949 | 15 |
| 1959 | 20 |
| 1969 | 27 |
| 1979 | 39 |
| 1989 | 52 |
x = 1974
Finding f(2)
Solution:
The value of table for `x` and `y`
| x | 1939 | 1949 | 1959 | 1969 | 1979 | 1989 |
|---|
| y | 12 | 15 | 20 | 27 | 39 | 52 |
|---|
Gauss's backward difference interpolation method to find solution
`h=1949-1939=10`
Taking `x_0=1969` then `p=(x-x_0)/h=(x-1969)/10`
Now the central difference table is
| `x` | `p=(x-1969)/10` | `y` | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` | `Delta^5y` |
| 1939 | -3 | 12 | | | | | |
| | | 3 | | | | |
| 1949 | -2 | 15 | | 2 | | | |
| | | 5 | | 0 | | |
| 1959 | -1 | 20 | | 2 | | 3 | |
| | | 7 | | 3 | | -10 |
| 1969 | 0 | 27 | | 5 | | -7 | |
| | | 12 | | -4 | | |
| 1979 | 1 | 39 | | 1 | | | |
| | | 13 | | | | |
| 1989 | 2 | 52 | | | | | |
`x = 1974`
`p = (x - x_0)/h = (1974 - 1969)/10 = 0.5`
`y_0=27, Delta y_(-1)=7,Delta^2y_(-1)=5,Delta^3y_(-2)=3,Delta^4y_(-2)=-7,Delta^5y_(-3)=-10`
Gauss's backward interpolation formula is
`y_p=y_0+p Delta y_(-1) + ((p + 1)p)/(2!) * Delta^2y_(-1) + ((p + 1)p(p - 1))/(3!) * Delta^3y_(-2) + ((p + 2)(p + 1)p(p - 1))/(4!) * Delta^4y_(-2) + ((p + 2)(p + 1)p(p - 1)(p - 2))/(5!) * Delta^5y_(-3)`
`y_(0.5) = 27 + (0.5)(7) + ((0.5 + 1)(0.5))/(2) * (5) + ((0.5 + 1)(0.5)(0.5 - 1))/(6) * (3) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1))/(24) * (-7) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1)(0.5 - 2))/(120) * (-10)`
`y_(0.5)=27 +3.5 +1.875 -0.1875 +0.2734375 -0.1171875`
`y_(0.5)=32.3438`
Solution of Gauss's backward interpolation is `y(1974) = 32.3438`
This material is intended as a summary. Use your textbook for detail explanation.
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