Find Solution of an equation 2x^3-4x+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)=2x^3-4x+1`.
The value of table for `x` and `y`
| x | 2 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | 9 | 22.25 | 43 | 72.75 | 113 |
|---|
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 | `9` | | | | |
| | `13.25` | | | |
| 2.5 | `22.25` | | `7.5` | | |
| | `20.75` | | `1.5` | |
| 3 | `43` | | `9` | | `0` |
| | `29.75` | | `1.5` | |
| 3.5 | `72.75` | | `10.5` | | |
| | `40.25` | | | |
| 4 | `113` | | | | |
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) = 113 + (-0.5) xx 40.25 + (-0.5 (-0.5 + 1))/(2) xx 10.5 + (-0.5 (-0.5 + 1)(-0.5 + 2))/(6) xx 1.5 + (-0.5 (-0.5 + 1)(-0.5 + 2)(-0.5 + 3))/(24) xx 0`
`y(3.75) = 113 -20.125 -1.3125 -0.0938 +0`
`y(3.75) = 91.4688`
Solution of newton's backward interpolation method `y(3.75) = 91.4688`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
