Question

Expansion of $f(x) = \log (1+x)$ by Taylor series is :

• A. $x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
• B. $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
• C. $1 - x + \frac{x^2}{2} - \frac{x^3}{3} + \dots$
• D. $x - \frac{x^2}{2!} + \frac{x^3}{3!} - \dots$

Hint:

Unlike the exponential series $e^x$, the logarithmic series expansion for $\log(1+x)$ does not contain factorials in its denominators.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Taylor series expansion of a function $f(x)$ about $x = 0$ is known as its Maclaurin series expansion.
Key Formula or Approach:
The Maclaurin series of a function $f(x)$ is given by:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]

Step 2: Detailed Explanation:
Let us find the derivatives of $f(x) = \log(1+x)$ and evaluate them at $x = 0$:
- $f(x) = \log(1+x) \implies f(0) = \log(1) = 0$
- $f'(x) = \frac{1}{1+x} \implies f'(0) = \frac{1}{1} = 1$
- $f''(x) = -\frac{1}{(1+x)^2} \implies f''(0) = -1$
- $f'''(x) = \frac{2}{(1+x)^3} \implies f'''(0) = 2$
- $f^{(iv)}(x) = -\frac{6}{(1+x)^4} \implies f^{(iv)}(0) = -6$
Substitute these derivative values into the Maclaurin expansion:
\[ \log(1+x) = 0 + 1 \cdot x + \frac{-1}{2!} x^2 + \frac{2}{3!} x^3 + \frac{-6}{4!} x^4 + \dots \]
Simplify the fractional coefficients:
- $\frac{1}{2!} = \frac{1}{2}$
- $\frac{2}{3!} = \frac{2}{6} = \frac{1}{3}$
- $\frac{6}{4!} = \frac{6}{24} = \frac{1}{4}$
This yields:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \]
Therefore, the correct series expansion is $x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$.

Step 3: Final Answer
The correct option is (A).