Find Solution of an equation x^3-x+1 using Newton's Forward Difference formula
x1 = 2 and x2 = 4
x = 2.25
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 forward difference interpolation method to find solution
Newton's forward difference table is
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^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 = 2.25`
`h = x_1 - x_0 = 2.5 - 2 = 0.5`
`p = (x - x_0)/h = (2.25 - 2)/0.5 = 0.5`
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 + (p(p - 1)(p - 2)(p - 3))/(4!) * Delta^4y_0`
`y(2.25) = 7 + 0.5 xx 7.125 + (0.5 (0.5 - 1))/(2) xx 3.75 + (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(2.25) = 7 +3.5625 -0.4688 +0.0469 +0`
`y(2.25) = 10.1406`
Solution of newton's forward interpolation method `y(2.25) = 10.1406`
This material is intended as a summary. Use your textbook for detail explanation.
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