Discussion
Given the data:
| x | 1 | 2 | 3 | 4 | 5 |
| y | -1 | 2 | -3 | 4 | -5 |
If the derivative of \(y(x)\) is approximated as:
\[ \begin{aligned} y'(x_{k}) &\approx \frac{1}{h}\Big(\Delta y_{k} + \frac{1}{2}\Delta^{2} y_{k} \\ &\qquad - \frac{1}{4}\Delta^{3} y_{k}\Big), \end{aligned} \]
then the value of \(y'(2)\) is
Answer: B
The formula gives \(y'(2) = 8\).
Here \(h = 1\) and \(x_{k} = 2\), so the forward differences start from \(y = 2\). From the values \(2, -3, 4, -5\):
\[ \begin{aligned} \Delta y &: -5,\ 7,\ -9 \\ \Delta^{2} y &: 12,\ -16 \\ \Delta^{3} y &: -28 \end{aligned} \]
Substituting the leading differences into the given formula:
\[ \begin{aligned} y'(2) &\approx -5 + \tfrac{1}{2}(12) - \tfrac{1}{4}(-28) \\ &= -5 + 6 + 7 = 8 \end{aligned} \]
The paper prints its own coefficients (\(+\frac{1}{2}, -\frac{1}{4}\)), and the answer follows them as given. Using the differences that start at \(x = 1\) instead, or dropping the third-difference term (which gives 1), leads away from 8; 4, 12 and 16 come from such slips.
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