2. Find Solution using Newton's Backward Difference formula
x = 4
Solution:
The value of table for `x` and `y`
Newton's backward difference interpolation method to find solution
Newton's backward difference table is
| x | y | `grady` | `grad^2y` | `grad^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 = 4`
`h = x_1 - x_0 = 1 - 0 = 1`
`p = (x - x_n) / h = (4 - 3) / 1 = 1`
Newton's backward difference interpolation formula is
`y(x) = y_n + p grad y_n + (p(p + 1))/(2!) * grad^2y_n + (p(p + 1)(p + 2))/(3!) * grad^3y_n`
`y(4) = 10 + 1 xx 9 + (1 (1 + 1))/(2) xx 8 + (1 (1 + 1)(1 + 2))/(6) xx 6`
`y(4) = 10 +9 +8 +6`
`y(4) = 33`
Solution of newton's backward interpolation method `y(4) = 33`
This material is intended as a summary. Use your textbook for detail explanation.
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