Question

The first non-zero term in the Taylor series expansion of \( (1 - x) - e^{-x \) about \( x = 0 \) is:}

• A. 1
• B. -1
• C. \( \frac{x^2}{2} \)
• D. \( -\frac{x^2}{2} \)

Hint:

For Taylor series problems, expand each term up to the required order, combine like terms, and identify the first non-zero coefficient.

Answer 1

The Correct Option is D

Step 1: Expand \( e^{-x} \) as a Taylor series about \( x = 0 \). The Taylor series expansion of \( e^{-x} \) is: \[ e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \] Step 2: Simplify \( (1 - x) - e^{-x \).} Substitute the expansion of \( e^{-x} \) into \( (1 - x) - e^{-x} \): \[ (1 - x) - e^{-x} = (1 - x) - \left( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \right). \] Simplify the terms: \[ (1 - x) - e^{-x} = 1 - x - 1 + x - \frac{x^2}{2} + \frac{x^3}{6} - \cdots. \] \[ (1 - x) - e^{-x} = -\frac{x^2}{2} + \frac{x^3}{6} - \cdots. \] Step 3: Identify the first non-zero term. The first non-zero term is \( -\frac{x^2}{2} \). Step 4: Conclusion. The first non-zero term in the Taylor series expansion is \( -\frac{x^2}{2} \).