Discussion
Using Newton's forward interpolation formula, the cubic polynomial which takes the following value
| \(x\) : | 0 | 1 | 2 | 3 |
| \(f(x)\) : | 1 | 2 | 1 | 10 |
is :
Answer: A
The polynomial is \(2x^{3} - 7x^{2} + 6x + 1\).
Forward differences of 1, 2, 1, 10: first differences 1, -1, 9; second -2, 10; third 12. Newton's formula with \(h = 1\):
\[ \begin{aligned} f(x) &= 1 + x - 2\frac{x(x-1)}{2} + 12\frac{x(x-1)(x-2)}{6} \\ &= 2x^{3} - 7x^{2} + 6x + 1 \end{aligned} \]
Check: \(f(1) = 2 - 7 + 6 + 1 = 2\), \(f(2) = 16 - 28 + 12 + 1 = 1\), \(f(3) = 54 - 63 + 18 + 1 = 10\). Option B gives \(f(1) = 4\), option C gives \(f(1) = 2\) but \(f(2) = 17\), and option D gives \(f(1) = 8\).
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