If `S=Sigma_(n=0)^(oo) (logx)^(2n)/(2n!)` , then S equals

Options

A. `x+x^(-1)`
B. `x-x^(-1)`
C. `1/2(x+x^(-1))`
D. none of these

Answer

To solve the problem \( S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} \), we can recognize that this series resembles the Taylor series expansion of the exponential function.

### Step-by-Step Solution:

1. **Identify the Series**:
The given series can be rewritten as:
\[ S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} \]
This series includes only the even powers of \( \log x \).

2. **Use the Exponential Series**:
The Taylor series expansion for \( e^x \) is:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
To get only the even terms, we can consider \( e^{\log x} \) and \( e^{-\log x} \).

3. **Combine the Exponential Functions**:
The series for \( e^{\log x} \) and \( e^{-\log x} \) can be combined:
\[ e^{\log x} + e^{-\log x} = x + \frac{1}{x} \]
The even terms of the series expansion can be derived from:
\[ e^x + e^{-x} = 2 \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]
Hence, we have:
\[ S = \frac{1}{2} \left( e^{\log x} + e^{-\log x} \right) = \frac{1}{2} \left( x + \frac{1}{x} \right) \]

4. **Final Result**:
Therefore, we conclude that:
\[ S = \frac{1}{2} \left( x + \frac{1}{x} \right) \]

### Conclusion:
The final expression for \( S \) is:
\[ S = \frac{1}{2} \left( x + \frac{1}{x} \right) \]