Formula
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Newton's Divided Difference Interpolation formula
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`y(x) = y_0 + (x - x_0) f[x_0, x_1] + (x - x_0)(x - x_1) f[x_0, x_1, x_2] + ...`
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Examples
1. Find Solution using Newton's Divided Difference Interpolation formula
| x | f(x) |
| 300 | 2.4771 |
| 304 | 2.4829 |
| 305 | 2.4843 |
| 307 | 2.4871 |
x = 301
Solution:
The value of table for `x` and `y`
| x | 300 | 304 | 305 | 307 |
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| y | 2.4771 | 2.4829 | 2.4843 | 2.4871 |
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Numerical divided differences method to find solution
Newton's divided difference table is
| x | y | `1^(st)` order | `2^(nd)` order |
| 300 | 2.4771 | | |
| | `(2.4829-2.4771)/(304-300)=0.0014` | |
| 304 | 2.4829 | | `(0.0014-0.0014)/(305-300)=0` |
| | `(2.4843-2.4829)/(305-304)=0.0014` | |
| 305 | 2.4843 | | `(0.0014-0.0014)/(307-304)=0` |
| | `(2.4871-2.4843)/(307-305)=0.0014` | |
| 307 | 2.4871 | | |
The value of `x` at you want to find the `f(x) : x = 301`
Newton's divided difference interpolation formula is
`f(x)=y_0 +(x-x_0) f[x_0, x_1]+(x-x_0)(x-x_1) f[x_0, x_1, x_2]`
`y(301) = 2.4771 + (301 -300) xx 0.0014 + (301 -300)(301 -304) xx 0`
`y(301) = 2.4771 + (1) xx 0.0014 + (1)(-3) xx 0`
`y(301) = 2.4771 +0.0014 +0`
`y(301) = 2.4785`
Solution of divided difference interpolation method `y(301) = 2.4785`
This material is intended as a summary. Use your textbook for detail explanation.
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