Using Newton's Backward Difference formula to find solution
| x | f(x) |
| 0.0 | 1.0000 |
| 0.1 | 0.9975 |
| 0.2 | 0.9900 |
| 0.3 | 0.9776 |
| 0.4 | 0.8604 |
x = 0.3
Solution:
Numerical differentiation method to find solution.
The value of table for `x` and `y`
| x | 0 | 0.1 | 0.2 | 0.3 | 0.4 |
|---|
| y | 1 | 0.9975 | 0.99 | 0.9776 | 0.8604 |
|---|
Newton's backward differentiation table is
| x | y | `grady` | `grad^2y` | `grad^3y` | `grad^4y` |
| 0 | 1 | | | | |
| | -0.0025 | | | |
| 0.1 | 0.9975 | | -0.005 | | |
| | -0.0075 | | 0.0001 | |
| 0.2 | 0.99 | | -0.0049 | | -0.1 |
| | -0.0124 | | -0.0999 | |
| 0.3 | 0.9776 | | -0.1048 | | |
| | -0.1172 | | | |
| 0.4 | 0.8604 | | | | |
The value of x at you want to find `f(x) : x_n = 0.3`
`h = x_1 - x_0 = 0.1 - 0 = 0.1`
`[(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=0.3) = 1/0.1 xx (-0.0124 + 1/2 xx -0.0049 + 1/3 xx 0.0001 + 1/4 xx 0)`
`:.[(dy)/(dx)]_(x=0.3) = -0.1482`
`[(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=0.3) = 1/0.01 * (-0.0049 + 0.0001 + 11/12 xx 0)`
`:.[(d^2y)/(dx^2)]_(x=0.3) = -0.48`
`:.` `Pn'(0.3) = -0.1482` and `Pn''(0.3) = -0.48`
This material is intended as a summary. Use your textbook for detail explanation.
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