2. Find Solution using Everett's formula
| x | f(x) |
| 25 | 4 |
| 26 | 3.846 |
| 27 | 3.704 |
| 28 | 3.571 |
| 29 | 3.448 |
| 30 | 3.333 |
x = 27.4
Solution:
The value of table for `x` and `y`
| x | 25 | 26 | 27 | 28 | 29 | 30 |
|---|
| y | 4 | 3.846 | 3.704 | 3.571 | 3.448 | 3.333 |
|---|
Everett method to find solution
`h=26-25=1`
Taking `x_0=27` then `p=(x-x_0)/h=(x-27)/1`
Now the central difference table is
| `x` | `p=(x-27)/1` | `y` | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` | `Delta^5y` |
| 25 | -2 | 4 | | | | | |
| | | -0.154 | | | | |
| 26 | -1 | 3.846 | | 0.012 | | | |
| | | -0.142 | | -0.003 | | |
| 27 | 0 | 3.704 | | 0.009 | | 0.004 | |
| | | -0.133 | | 0.001 | | -0.007 |
| 28 | 1 | 3.571 | | 0.01 | | -0.003 | |
| | | -0.123 | | -0.002 | | |
| 29 | 2 | 3.448 | | 0.008 | | | |
| | | -0.115 | | | | |
| 30 | 3 | 3.333 | | | | | |
`x = 27.4`
`p = (x - x_0)/h = (27.4 - 27)/1 = 0.4`
`y_0=3.704, Delta y_0=-0.133,Delta^2y_(-1)=0.009,Delta^4y_(-2)=0.004`
Everett interpolation formula is
`y_p=qy_0 + (q(q^2 - 1^2))/(3!) * Delta^2y_(-1) + (q(q^2 - 1^2)(q^2 - 2^2))/(5!) * Delta^4y_(-2)+...+py_1 + (p(p^2 - 1^2))/(3!) * Delta^2y_(0) + (p(p^2 - 1^2)(p^2 - 2^2))/(5!) * Delta^4y_(-1)+...`
`y_(0.4) = (0.6)*(3.704) + ((0.6)(0.36 - 1))/(6) * (0.009) + ((0.6)(0.36 - 1)(0.36 - 4))/(120) * (0.004)+...+(0.4)*(3.571) + ((0.4)(0.16 - 1))/(6) * (0.01) + ((0.4)(0.16 - 1)(0.16 - 4))/(120) * (-0.003)+...`
`y_(0.4)=2.2224 -0.000576 +0.000046592 + 1.4284 -0.00056 -0.000032256`
`y_(0.4)=3.64968`
Solution of Everett interpolation is `y(27.4) = 3.64968`
This material is intended as a summary. Use your textbook for detail explanation.
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