Formula
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Newton's Backward Difference formula
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`p = (x - x_n) / h`
`y(x) = y_n + p grad y_n + (p(p + 1))/(2!) * grad^2y_n + (p(p + 1)(p + 2))/(3!) * grad^3y_n + (p(p + 1)(p + 2)(p + 3))/(4!) * grad^4y_n + ...`
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Examples
1. Find Solution using Newton's Backward Difference formula
| x | f(x) |
| 1891 | 46 |
| 1901 | 66 |
| 1911 | 81 |
| 1921 | 93 |
| 1931 | 101 |
x = 1925
Solution:
The value of table for `x` and `y`
| x | 1891 | 1901 | 1911 | 1921 | 1931 |
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| y | 46 | 66 | 81 | 93 | 101 |
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Newton's backward difference interpolation method to find solution
Newton's backward difference table is
| x | y | `grady` | `grad^2y` | `grad^3y` | `grad^4y` |
| 1891 | `46` | | | | |
| | `20` | | | |
| 1901 | `66` | | `-5` | | |
| | `15` | | `2` | |
| 1911 | `81` | | `-3` | | `-3` |
| | `12` | | `-1` | |
| 1921 | `93` | | `-4` | | |
| | `8` | | | |
| 1931 | `101` | | | | |
The value of x at you want to find the `f(x) : x = 1925`
`h = x_1 - x_0 = 1901 - 1891 = 10`
`p = (x - x_n) / h = (1925 - 1931) / 10 = -0.6`
Newton's backward difference interpolation formula is
`y(x) = y_n + p grad y_n + (p(p + 1))/(2!) * grad^2y_n + (p(p + 1)(p + 2))/(3!) * grad^3y_n + (p(p + 1)(p + 2)(p + 3))/(4!) * grad^4y_n`
`y(1925) = 101 + (-0.6) xx 8 + (-0.6 (-0.6 + 1))/(2) xx -4 + (-0.6 (-0.6 + 1)(-0.6 + 2))/(6) xx -1 + (-0.6 (-0.6 + 1)(-0.6 + 2)(-0.6 + 3))/(24) xx -3`
`y(1925) = 101 -4.8 +0.48 +0.056 +0.1008`
`y(1925) = 96.8368`
Solution of newton's backward interpolation method `y(1925) = 96.8368`
This material is intended as a summary. Use your textbook for detail explanation.
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