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Question
Using Newton’s forward interpolation formula find the cubic polynomial.
| x | 0 | 1 | 2 | 3 |
| f(x) | 1 | 2 | 1 | 10 |
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Solution
Since we use the Newton’s forward interpolation formula.
`y_((x = x_0 + "h")) = y_0 + "n"/(1!) Deltay_0 + ("n"("n" - 1))/(2!) Delta^2y_0 ("n"("n" - 1)("n" - 2))/(3!) Delta^3y_0 + .........`
To find y at x
∴ x0 + nh = x
0 + n(1) = x
∴ n = x
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` |
| 0 | 1 | |||
| 1 | ||||
| 1 | 2 | – 2 | ||
| – 1 | 12 | |||
| 2 | 1 | 10 | ||
| 9 | ||||
| 3 | 10 |
y = `1 + x/(1!) + (x(x -1))/(2!) (-2) + (x(x - 1)(x - 2))/(3!) (12)`
= `1 + x/1 + (x(x - 1))/2 (-2) + (x(x - 1)(x - 2)(12))/6`
= 1 + x + (x2 – x)(– 1) + 2x(x2 – 3x + 2)
y = 1 + x – x2 + x + 2x3 – 6x2 + 4x
y = 2x3 – 7x2 + 6x + 1
∴ f(x) = 2x3 – 7x2 + 6x + 1
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