Find Solution using Everett's formula
| x | f(x) |
| 310 | 2.49136 |
| 320 | 2.50515 |
| 330 | 2.51851 |
| 340 | 2.53148 |
| 350 | 2.54407 |
| 360 | 2.55630 |
x = 337.5
Finding f(2)
Solution:
The value of table for `x` and `y`
| x | 310 | 320 | 330 | 340 | 350 | 360 |
|---|
| y | 2.4914 | 2.5052 | 2.5185 | 2.5315 | 2.5441 | 2.5563 |
|---|
Everett method to find solution
`h=320-310=10`
Taking `x_0=330` then `p=(x-x_0)/h=(x-330)/10`
Now the central difference table is
| `x` | `p=(x-330)/10` | `y` | `Deltay` | `Delta^2y` | `Delta^3y` |
| 310 | -2 | 2.4914 | | | |
| | | 0.0138 | | |
| 320 | -1 | 2.5052 | | -0.0004 | |
| | | 0.0134 | | 0 |
| 330 | 0 | 2.5185 | | -0.0004 | |
| | | 0.013 | | 0 |
| 340 | 1 | 2.5315 | | -0.0004 | |
| | | 0.0126 | | 0 |
| 350 | 2 | 2.5441 | | -0.0004 | |
| | | 0.0122 | | |
| 360 | 3 | 2.5563 | | | |
`x = 337.5`
`p = (x - x_0)/h = (337.5 - 330)/10 = 0.75`
`q=1-p=1-0.75=0.25`
`y_0=2.5185, Delta y_0=0.013,Delta^2y_(-1)=-0.0004`
Everett interpolation formula is
`y_p=qy_0 + (q(q^2 - 1^2))/(3!) * Delta^2y_(-1)+...+py_1 + (p(p^2 - 1^2))/(3!) * Delta^2y_(0)+...`
`y_(0.75) = (0.25)*(2.5185) + ((0.25)(0.0625 - 1))/(6) * (-0.0004)+...+(0.75)*(2.5315) + ((0.75)(0.5625 - 1))/(6) * (-0.0004)+...`
`y_(0.75)=0.6296 +0.0000152344 + 1.8986 +0.0000207813`
`y_(0.75)=2.5283`
Solution of Everett interpolation is `y(337.5) = 2.5283`
This material is intended as a summary. Use your textbook for detail explanation.
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