Find Solution of an Equation 2x^3-4x+1
x1 = 2 and x2 = 4
x = 2.25
Step value (h) = 0.25 using Newton's Forward Difference formula
Solution:
Equation is `f(x) = 2x^3-4x+1`.
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 | 9 | 14.7812 | 22.25 | 31.5938 | 43 | 56.6562 | 72.75 | 91.4688 | 113 |
|---|
Newton's forward differentiation 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 `f(x) : x_1 = 2.25`
`h = x_1 - x_0 = 2.25 - 2 = 0.25`
`[(dy)/(dx)]_(x=x_1) = 1/h * (Delta y_1 - 1/2 * Delta^2 y_1 + 1/3 * Delta^3 y_1 - 1/4 * Delta^4 y_1)`
`:.[(dy)/(dx)]_(x=2.25) = 1/0.25 * (7.4688 - 1/2 xx 1.875 + 1/3 xx 0.1875 - 1/4 xx 0)`
`:.[(dy)/(dx)]_(x=2.25) = 26.375`
`[(d^2y)/(dx^2)]_(x=x_1) = 1/h^2 * (Delta^2 y_1 - Delta^3 y_1 + 11/12 * Delta^4 y_1)`
`:.[(d^2y)/(dx^2)]_(x=2.25) = 1/0.0625 * (1.875 - 0.1875 + 11/12 xx 0)`
`:.[(d^2y)/(dx^2)]_(x=2.25) = 27`
`:.` `Pn'(2.25) = 26.375` and `Pn''(2.25) = 27`
This material is intended as a summary. Use your textbook for detail explanation.
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