Find Solution of an Equation x^3+x+2
x1 = 2 and x2 = 4
x = 3.5
Step value (h) = 0.5 using Newton's Backward 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 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | 12 | 20.125 | 32 | 48.375 | 70 |
|---|
Newton's backward differentiation table is
| x | y | `grady` | `grad^2y` | `grad^3y` | `grad^4y` |
| 2 | 12 | | | | |
| | 8.125 | | | |
| 2.5 | 20.125 | | 3.75 | | |
| | 11.875 | | 0.75 | |
| 3 | 32 | | 4.5 | | 0 |
| | 16.375 | | 0.75 | |
| 3.5 | 48.375 | | 5.25 | | |
| | 21.625 | | | |
| 4 | 70 | | | | |
The value of x at you want to find `f(x) : x_n = 3.5`
`h = x_1 - x_0 = 2.5 - 2 = 0.5`
`[(dy)/(dx)]_(x=x_n) = 1/h * (grad y_n + 1/2 * grad^2 y_n + 1/3 * grad^3 y_n + 1/4 * grad^4 y_n)`
`:.[(dy)/(dx)]_(x=3.5) = 1/0.5 xx (16.375 + 1/2 xx 4.5 + 1/3 xx 0.75 + 1/4 xx 0)`
`:.[(dy)/(dx)]_(x=3.5) = 37.75`
`[(d^2y)/(dx^2)]_(x=x_n) = 1/h^2 * (grad^2 y_n + grad^3 y_n + 11/12 * grad^4 y_n)`
`:.[(d^2y)/(dx^2)]_(x=3.5) = 1/0.25 * (4.5 + 0.75 + 11/12 xx 0)`
`:.[(d^2y)/(dx^2)]_(x=3.5) = 21`
`:.` `Pn'(3.5) = 37.75` and `Pn''(3.5) = 21`
This material is intended as a summary. Use your textbook for detail explanation.
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