`e^((x-1)-(1)/(2) (x-1)^(2) + (1)/(3) (x-1)^(3) - (1)/(4) (x-1)^(4) +...`

Options

A. x
B. `x-1`
C. `log x `
D. `log (x-1)`

Answer

To solve the given series \( e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} \), we can follow these steps:

### Step 1: Define the Series
Let \( y = e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} \).

### Step 2: Recognize the Series as a Logarithmic Expansion
The series inside the exponent resembles the Taylor series expansion for \( \log(1 + u) \), where \( u = (x-1) \). The Taylor series expansion for \( \log(1 + u) \) is given by:
\[ \log(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \ldots \]
Thus, we can rewrite the exponent as:
\[ y = e^{\log(1 + (x-1))} = e^{\log(x)}. \]

### Step 3: Apply the Exponential and Logarithmic Properties
Using the property of exponentials and logarithms, we know that:
\[ e^{\log(x)} = x. \]

### Step 4: Conclusion
Therefore, we conclude that:
\[ y = x. \]

### Final Answer
The final result is:
\[ y = x. \]

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