If `S = sum_(n =0)^(oo) ((logx)^(2n))/((2n)!)` these S is equal to

Options

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

Answer

To solve the problem, we need to evaluate the series

\[ S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} \]

### Step 1: Recognize the Series
The series resembles the Taylor series expansion for the exponential function. Specifically, the Taylor series for \( e^x \) is given by:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

### Step 2: Relate to the Cosine Hyperbolic Function
Notice that our series only includes even powers of \( \log x \). The series for the hyperbolic cosine function \( \cosh(x) \) is:

\[ \cosh(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \]

Thus, we can rewrite our series as:

\[ S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} = \cosh(\log x) \]

### Step 3: Simplify \( \cosh(\log x) \)
Using the definition of the hyperbolic cosine function, we have:

\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]

Substituting \( x = \log x \):

\[ S = \cosh(\log x) = \frac{e^{\log x} + e^{-\log x}}{2} \]

### Step 4: Simplify Further
We know that \( e^{\log x} = x \) and \( e^{-\log x} = \frac{1}{x} \). Therefore:

\[ S = \frac{x + \frac{1}{x}}{2} \]

### Final Result
Thus, the value of \( S \) is:

\[ S = \frac{x + \frac{1}{x}}{2} \]