Formula
1. For `x=x_n`
`[(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 + ...)`
`[(d^2y)/(dx^2)]_(x=x_n) = 1/h^2 * (grad^2 Y_n + grad^3 Y_n + 11/12 * grad^4 Y_n + ...)`
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2. For any value of `x`
`[(dy)/(dx)] = 1/h * (grad Y_n + (2t+1)/2 * grad^2 Y_n + (3t^2+6t+2)/6 * grad^3 Y_n + (4t^3+18t^2+22t+6)/24 * grad^4 Y_n + ...)`
`[(d^2y)/(dx^2)] = 1/h^2 * (grad^2 Y_n + (t+1) * grad^3 Y_n + (12t^2+36t+22)/24 * grad^4 Y_n + ...)`
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Examples
1. Using Newton's Backward Difference formula to find solution
| x | f(x) |
| 1.4 | 4.0552 |
| 1.6 | 4.9530 |
| 1.8 | 6.0496 |
| 2.0 | 7.3891 |
| 2.2 | 9.0250 |
x = 2.2
Solution:
Numerical differentiation method to find solution.
The value of table for `x` and `y`
| x | 1.4 | 1.6 | 1.8 | 2 | 2.2 |
|---|
| y | 4.0552 | 4.953 | 6.0496 | 7.3891 | 9.025 |
|---|
Newton's backward differentiation table is
| x | y | `grady` | `grad^2y` | `grad^3y` | `grad^4y` |
| 1.4 | 4.0552 | | | | |
| | 0.8978 | | | |
| 1.6 | 4.953 | | 0.1988 | | |
| | 1.0966 | | 0.0441 | |
| 1.8 | 6.0496 | | 0.2429 | | 0.0094 |
| | 1.3395 | | 0.0535 | |
| 2 | 7.3891 | | 0.2964 | | |
| | 1.6359 | | | |
| 2.2 | 9.025 | | | | |
The value of x at you want to find `f(x) : x_n = 2.2`
`h = x_1 - x_0 = 1.6 - 1.4 = 0.2`
`[(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=2.2) = 1/0.2 xx (1.6359 + 1/2 xx 0.2964 + 1/3 xx 0.0535 + 1/4 xx 0.0094)`
`:.[(dy)/(dx)]_(x=2.2) = 9.02142`
`[(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=2.2) = 1/0.04 * (0.2964 + 0.0535 + 11/12 xx 0.0094)`
`:.[(d^2y)/(dx^2)]_(x=2.2) = 8.96292`
`:.` `Pn'(2.2) = 9.02142` and `Pn''(2.2) = 8.96292`
This material is intended as a summary. Use your textbook for detail explanation.
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