In a Taylor series expansion about a, what is the coefficient of (x-a)^2 in f(x) ≈ f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ...?

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Multiple Choice

In a Taylor series expansion about a, what is the coefficient of (x-a)^2 in f(x) ≈ f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ...?

Explanation:
In a Taylor series about a, f(x) is expanded as f(a) + f'(a)(x−a) + f''(a)/2! (x−a)^2 + f'''(a)/3! (x−a)^3 + … . The coefficient in front of (x−a)^2 is exactly f''(a)/2!, which matches the given term. Since 2! = 2, that coefficient is f''(a)/2. The other terms correspond to lower or higher powers: f(a) for the constant term, f'(a) for the linear term, and f'''(a)/3! for the cubic term.

In a Taylor series about a, f(x) is expanded as f(a) + f'(a)(x−a) + f''(a)/2! (x−a)^2 + f'''(a)/3! (x−a)^3 + … . The coefficient in front of (x−a)^2 is exactly f''(a)/2!, which matches the given term. Since 2! = 2, that coefficient is f''(a)/2. The other terms correspond to lower or higher powers: f(a) for the constant term, f'(a) for the linear term, and f'''(a)/3! for the cubic term.

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