If `y = x - (x^(2))/(2) + (x^(3))/(3) -...` then x =

Options

A. `log (1 + y)`
B. `e^(y)`
C. `e^(-y) -1`
D. `y + (y^(2))/(2!) + (y^(3))/(3!) + ...`

Answer

To solve the equation given by the series \( y = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \), we will recognize that this series represents the Taylor series expansion for the natural logarithm function.

### Step-by-step Solution:

1. **Recognize the Series**: The series can be rewritten as:
\[ y = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} \]
This is the Taylor series expansion for \( \ln(1+x) \) around \( x = 0 \).

2. **Set the Series Equal to the Logarithm**: From the recognition of the series, we can equate:
\[ y = \ln(1+x) \]

3. **Exponentiate Both Sides**: To solve for \( x \), we exponentiate both sides:
\[ e^y = 1 + x \]

4. **Rearrange to Solve for \( x \)**: Now, we can isolate \( x \):
\[ x = e^y - 1 \]

5. **Final Expression**: Thus, we have expressed \( x \) in terms of \( y \):
\[ x = e^y - 1 \]

### Final Answer:
\[ x = e^y - 1 \]