Find Solution of an Equation x^3+x+2
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.25 using Newton's Forward Difference formula
Solution:
Equation is `f(x) = x^3+x+2`.
Numerical differentiation method to find solution.
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 | 12 | 15.6406 | 20.125 | 25.5469 | 32 | 39.5781 | 48.375 | 58.4844 | 70 |
|---|
Newton's forward differentiation table is
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` |
| 2 | 12 | | | | |
| | 3.6406 | | | |
| 2.25 | 15.6406 | | 0.8438 | | |
| | 4.4844 | | 0.0938 | |
| 2.5 | 20.125 | | 0.9375 | | 0 |
| | 5.4219 | | 0.0938 | |
| 2.75 | 25.5469 | | 1.0312 | | 0 |
| | 6.4531 | | 0.0938 | |
| 3 | 32 | | 1.125 | | 0 |
| | 7.5781 | | 0.0938 | |
| 3.25 | 39.5781 | | 1.2188 | | 0 |
| | 8.7969 | | 0.0938 | |
| 3.5 | 48.375 | | 1.3125 | | 0 |
| | 10.1094 | | 0.0938 | |
| 3.75 | 58.4844 | | 1.4062 | | |
| | 11.5156 | | | |
| 4 | 70 | | | | |
The value of `x` at you want to find `f(x) : x_0 = 2.25`
`h = x_1 - x_0 = 2.25 - 2 = 0.25`
`[(dy)/(dx)]_(x=x_0) = 1/h * (Delta y_0 - 1/2 * Delta^2 y_0 + 1/3 * Delta^3 y_0 - 1/4 * Delta^4 y_0)`
`:.[(dy)/(dx)]_(x=2.25) = 1/0.25 * (4.4844 - 1/2 xx 0.9375 + 1/3 xx 0.0938 - 1/4 xx 0)`
`:.[(dy)/(dx)]_(x=2.25) = 16.1875`
`[(d^2y)/(dx^2)]_(x=x_0) = 1/h^2 * (Delta^2 y_0 - Delta^3 y_0 + 11/12 * Delta^4 y_0)`
`:.[(d^2y)/(dx^2)]_(x=2.25) = 1/0.0625 * (0.9375 - 0.0938 + 11/12 xx 0)`
`:.[(d^2y)/(dx^2)]_(x=2.25) = 13.5`
`:.` `Pn'(2.25) = 16.1875` and `Pn''(2.25) = 13.5`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
