2. Find Solution using Newton's Forward Difference formula
x = -1
Solution:
The value of table for `x` and `y`
Newton's forward difference interpolation method to find solution
Newton's forward difference table is
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` |
| 0 | 1 | | | |
| | -1 | | |
| 1 | 0 | | 2 | |
| | 1 | | 6 |
| 2 | 1 | | 8 | |
| | 9 | | |
| 3 | 10 | | | |
The value of `x` at you want to find the `f(x) : x = -1`
`h = x_1 - x_0 = 1 - 0 = 1`
`p = (x - x_0)/h = (-1 - 0)/1 = -1`
Newton's forward difference interpolation formula is
`y(x) = y_0 + p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_0 + (p(p - 1)(p - 2))/(3!) * Delta^3y_0`
`y(-1) = 1 + (-1) xx -1 + (-1 (-1 - 1))/(2) xx 2 + (-1 (-1 - 1)(-1 - 2))/(6) xx 6`
`y(-1) = 1 +1 +2 -6`
`y(-1) = -2`
Solution of newton's forward interpolation method `y(-1) = -2`
This material is intended as a summary. Use your textbook for detail explanation.
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