Find Solution of an equation 2x^3-4x+1 using Newton's Forward Difference formula
x1 = 2 and x2 = 4
x = 2.1
Step value (h) = 0.25
Finding f(2)
Solution:
Equation is `f(x)=2x^3-4x+1`.
The value of table for `x` and `y`
| x | 2 | 2.25 | 2.5 | 2.75 | 3 | 3.25 | 3.5 | 3.75 | 4 |
|---|
| y | 9 | 14.7812 | 22.25 | 31.5938 | 43 | 56.6562 | 72.75 | 91.4688 | 113 |
|---|
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 | 9 | | | | |
| | 5.7812 | | | |
| 2.25 | 14.7812 | | 1.6875 | | |
| | 7.4688 | | 0.1875 | |
| 2.5 | 22.25 | | 1.875 | | 0 |
| | 9.3438 | | 0.1875 | |
| 2.75 | 31.5938 | | 2.0625 | | 0 |
| | 11.4062 | | 0.1875 | |
| 3 | 43 | | 2.25 | | 0 |
| | 13.6562 | | 0.1875 | |
| 3.25 | 56.6562 | | 2.4375 | | 0 |
| | 16.0938 | | 0.1875 | |
| 3.5 | 72.75 | | 2.625 | | 0 |
| | 18.7188 | | 0.1875 | |
| 3.75 | 91.4688 | | 2.8125 | | |
| | 21.5312 | | | |
| 4 | 113 | | | | |
The value of `x` at you want to find the `f(x) : x = 2.1`
`h = x_1 - x_0 = 2.25 - 2 = 0.25`
`p = (x - x_0)/h = (2.1 - 2)/0.25 = 0.4`
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(2.1) = 9 + 0.4 xx 5.7812 + (0.4 (0.4 - 1))/(2) xx 1.6875 + (0.4 (0.4 - 1)(0.4 - 2))/(6) xx 0.1875 + (0.4 (0.4 - 1)(0.4 - 2)(0.4 - 3))/(24) xx 0`
`y(2.1) = 9 +2.3125 -0.2025 +0.012 +0`
`y(2.1) = 11.122`
Solution of newton's forward interpolation method `y(2.1) = 11.122`
This material is intended as a summary. Use your textbook for detail explanation.
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