Formula
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Everett's formula
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`p = (x - x_0)/h`
`q = 1-p`
`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)+...`
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Examples
1. Find Solution using Everett's formula
| x | f(x) |
| 1 | 1 |
| 1.1 | 1.049 |
| 1.2 | 1.096 |
| 1.3 | 1.140 |
x = 1.15
Solution:
The value of table for `x` and `y`
| x | 1 | 1.1 | 1.2 | 1.3 |
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| y | 1 | 1.049 | 1.096 | 1.14 |
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Everett method to find solution
`h=1.1-1=0.1`
Taking `x_0=1.1` then `p=(x-x_0)/h=(x-1.1)/0.1`
Now the central difference table is
| `x` | `p=(x-1.1)/0.1` | `y` | `Deltay` | `Delta^2y` | `Delta^3y` |
| 1 | -1 | 1 | | | |
| | | 0.049 | | |
| 1.1 | 0 | 1.049 | | -0.002 | |
| | | 0.047 | | -0.001 |
| 1.2 | 1 | 1.096 | | -0.003 | |
| | | 0.044 | | |
| 1.3 | 2 | 1.14 | | | |
`x = 1.15`
`p = (x - x_0)/h = (1.15 - 1.1)/0.1 = 0.5`
`y_0=1.049, Delta y_0=0.047,Delta^2y_(-1)=-0.002`
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_(0.5) = (0.5)*(1.049) + ((0.5)(0.25 - 1))/(6) * (-0.002)+...+(0.5)*(1.096) + ((0.5)(0.25 - 1))/(6) * (-0.003)+...`
`y_(0.5)=0.5245 +0.000125 + 0.548 +0.0001875`
`y_(0.5)=1.07281`
Solution of Everett interpolation is `y(1.15) = 1.07281`
This material is intended as a summary. Use your textbook for detail explanation.
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