Formula
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Newton's Divided Difference formula
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1. Find equation using Newton's Divided Difference Interpolation formula
`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] + (x - x_0)(x - x_1)(x - x_2) f[x_0, x_1, x_2, x_3] + ...`
2. Now, differentiate f(x) with respect to x to get f'(x) and f''(x)
3. Now, substitute value of `x` in f'(x) and f''(x)
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Examples
1. Using Newton's Divided Difference formula to find solution
x = 5
Solution:
The value of table for `x` and `y`
Numerical divided differences method to find solution
Newton's divided difference table is
| x | y | `1^(st)` order | `2^(nd)` order | `3^(rd)` order |
| 2 | 4 | | | |
| | 26 | | |
| 4 | 56 | | 15 | |
| | 131 | | 1 |
| 9 | 711 | | 23 | |
| | 269 | | |
| 10 | 980 | | | |
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] + (x - x_0)(x - x_1)(x - x_2) f[x_0, x_1, x_2, x_3]`
`f(x) = 4 + (x -2) xx 26 + (x -2)(x -4) xx 15 + (x -2)(x -4)(x -9) xx 1`
`f(x) = 4 + (x-2) xx 26 + (x^2-6x+8) xx 15 + (x^3-15x^2+62x-72) xx 1`
`f(x) = 4 + (26x-52) + (15x^2-90x+120) + (x^3-15x^2+62x-72) `
`f(x) = x^3-2x `
Now, differentiate with x
`f'(x)=3x^2-2`
`f''(x)=6x`
Now, substitute `x=5`
`f'(5)=3 xx 5^2-2=73`
`f''(5)=6 xx 5=30`
This material is intended as a summary. Use your textbook for detail explanation.
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