Find Solution of an equation x^3-x+1 using Newton's Backward Difference formula
x1 = 2 and x2 = 4
x = 3.75
Step value (h) = 0.5
Finding f(2)
Solution:
Equation is `f(x)=x^3-x+1`.
The value of table for `x` and `y`
| x | 2 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | 7 | 14.125 | 25 | 40.375 | 61 |
|---|
Newton's backward difference interpolation method to find solution
Newton's backward difference table is
| x | y | `grady` | `grad^2y` | `grad^3y` | `grad^4y` |
| 2 | `7` | | | | |
| | `7.125` | | | |
| 2.5 | `14.125` | | `3.75` | | |
| | `10.875` | | `0.75` | |
| 3 | `25` | | `4.5` | | `0` |
| | `15.375` | | `0.75` | |
| 3.5 | `40.375` | | `5.25` | | |
| | `20.625` | | | |
| 4 | `61` | | | | |
The value of x at you want to find the `f(x) : x = 3.75`
`h = x_1 - x_0 = 2.5 - 2 = 0.5`
`p = (x - x_n) / h = (3.75 - 4) / 0.5 = -0.5`
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 + (p(p + 1)(p + 2)(p + 3))/(4!) * grad^4y_n`
`y(3.75) = 61 + (-0.5) xx 20.625 + (-0.5 (-0.5 + 1))/(2) xx 5.25 + (-0.5 (-0.5 + 1)(-0.5 + 2))/(6) xx 0.75 + (-0.5 (-0.5 + 1)(-0.5 + 2)(-0.5 + 3))/(24) xx 0`
`y(3.75) = 61 -10.3125 -0.6562 -0.0469 +0`
`y(3.75) = 49.9844`
Solution of newton's backward interpolation method `y(3.75) = 49.9844`
This material is intended as a summary. Use your textbook for detail explanation.
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