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Official PYQ · 1996

GATE CS 1996 Q1.6 — Calculus: Limits, Continuity, Differentiability

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The formula used to compute an approximation for the second derivative of a function ff at a point x0x_0 is

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Answer: D

To find the approximation for the second derivative f′′(x0)f''(x_0), we use the Taylor series expansions of f(x0+h)f(x_0 + h) and f(x0−h)f(x_0 - h) around the point x0x_0:

f(x0+h)=f(x0)+hf′(x0)+h22f′′(x0)+h36f′′′(x0)+O(h4)— (1)f(x_0 + h) = f(x_0) + h f'(x_0) + \frac{h^2}{2} f''(x_0) + \frac{h^3}{6} f'''(x_0) + \mathcal{O}(h^4) \quad \text{--- (1)}

f(x0−h)=f(x0)−hf′(x0)+h22f′′(x0)−h36f′′′(x0)+O(h4)— (2)f(x_0 - h) = f(x_0) - h f'(x_0) + \frac{h^2}{2} f''(x_0) - \frac{h^3}{6} f'''(x_0) + \mathcal{O}(h^4) \quad \text{--- (2)}

Adding equations (1) and (2) to eliminate the first derivative term f′(x0)f'(x_0) gives:

f(x0+h)+f(x0−h)=2f(x0)+h2f′′(x0)+O(h4)f(x_0 + h) + f(x_0 - h) = 2f(x_0) + h^2 f''(x_0) + \mathcal{O}(h^4)

Rearranging the terms to solve for f′′(x0)f''(x_0):

h2f′′(x0)=f(x0+h)−2f(x0)+f(x0−h)−O(h4)h^2 f''(x_0) = f(x_0 + h) - 2f(x_0) + f(x_0 - h) - \mathcal{O}(h^4)

f′′(x0)≈f(x0+h)−2f(x0)+f(x0−h)h2f''(x_0) \approx \frac{f(x_0 + h) - 2f(x_0) + f(x_0 - h)}{h^2}

This is the standard central difference formula for the second derivative.

Therefore, the correct option is D.

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