1. Formula & Example-1
Formula
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Newton's Forward Difference formula
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`p = (x - x_0)/h`
`y(x) = y_0 + p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_0 + (p(p - 1)(p - 2))/(3!) * Delta^3y_0 + (p(p - 1)(p - 2)(p - 3))/(4!) * Delta^4y_0 + ...`
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Examples
1. Find Solution using Newton's Forward Difference formula
| x | f(x) | | 1891 | 46 | | 1901 | 66 | | 1911 | 81 | | 1921 | 93 | | 1931 | 101 |
x = 1895
Finding option 1. Value f(2)
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 forward difference interpolation method to find solution
Newton's forward difference table is
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` | | 1891 | 46 | | | | | | | `66-46=20` | | | | | 1901 | 66 | | `15-20=-5` | | | | | `81-66=15` | | `-3--5=2` | | | 1911 | 81 | | `12-15=-3` | | `-1-2=-3` | | | `93-81=12` | | `-4--3=-1` | | | 1921 | 93 | | `8-12=-4` | | | | | `101-93=8` | | | | | 1931 | 101 | | | | |
The value of `x` at you want to find the `f(x) : x = 1895`
`h = x_1 - x_0 = 1901 - 1891 = 10`
`p = (x - x_0)/h = (1895 - 1891)/10 = 0.4`
Newton's forward difference interpolation formula is
`y(x) = y_0 + p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_0 + (p(p - 1)(p - 2))/(3!) * Delta^3y_0 + (p(p - 1)(p - 2)(p - 3))/(4!) * Delta^4y_0`
`y(1895) = 46 + 0.4 xx 20 + (0.4 (0.4 - 1))/(2) xx -5 + (0.4 (0.4 - 1)(0.4 - 2))/(6) xx 2 + (0.4 (0.4 - 1)(0.4 - 2)(0.4 - 3))/(24) xx -3`
`y(1895) = 46 +8 +0.6 +0.128 +0.1248`
`y(1895) = 54.8528`
Solution of newton's forward interpolation method `y(1895) = 54.8528`
This material is intended as a summary. Use your textbook for detail explanation.
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