Question

Function \( f(x) \) by Maclaurin's series (as an infinite series) can be expressed as:

• A. \( f(x) = f(1) + x f'(1) + \frac{x^2}{2!} f''(1) + \frac{x^3}{3!} f^{(3)}(1) + \dots + \infty \)
• B. \( f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f^{(3)}(0) + \dots + \infty \)
• C. \( f(x) = f(1) - x f'(1) + \frac{x^2}{2!} f''(1) - \frac{x^3}{3!} f^{(3)}(1) + \dots + \infty \)
• D. \( f(x) = f(0) - x f'(0) + \frac{x^2}{2!} f''(0) - \frac{x^3}{3!} f^{(3)}(0) + \dots + \infty \)

Hint:

Maclaurin’s series is simply the Taylor series expansion of a function at \( x = 0 \).

Answer 1

The Correct Option is B

Step 1: Understanding Maclaurin’s series.
Maclaurin's series is a special case of Taylor's series expansion where the expansion is done around \( x = 0 \). The general form of the Maclaurin series for a function \( f(x) \) is: \[ f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f^{(3)}(0) + \dots \] Step 2: Conclusion.
Hence, the correct answer is option (B), where the series is expanded around \( x = 0 \). Final Answer: \[ \boxed{(B)} \]