Let `g(x) = ln f(x)` where f(x) is a twice differentiable positive function on `(0, oo)` such that `f(x+1) = x f(x)`. Then for N = 1,2,3 `g''(N+1/2)- g''(1/2) =`

Options

A. `-4{1+(1)/(9) + (1)/(25)+....+(1)/((2N - 1)^(2))}`
B. `4{1+(1)/(9) + (1)/(25)+....+(1)/((2N - 1)^(2))}`
C. `-4{1+(1)/(9) + (1)/(25)+....+(1)/((2N - 1)^(2))}`
D. `4{1+(1)/(9) + (1)/(25)+....+(1)/((2N - 1)^(2))}`

Answer

To solve the problem, we start with the given function \( g(x) = \ln f(x) \) and the functional equation \( f(x+1) = x f(x) \). We need to find \( g''(N + \frac{1}{2}) - g''(\frac{1}{2}) \) for \( N = 1, 2, 3 \).

### Step-by-Step Solution:

1. **Understanding the relationship between \( f(x) \) and \( g(x) \)**:
\[ g(x) = \ln f(x) \]
Therefore, the first derivative \( g'(x) \) can be expressed using the chain rule:
\[ g'(x) = \frac{f'(x)}{f(x)} \]

2. **Finding \( g(x + 1) \)**:
Using the functional equation \( f(x + 1) = x f(x) \), we can write:
\[ g(x + 1) = \ln f(x + 1) = \ln(x f(x)) = \ln x + \ln f(x) = \ln x + g(x) \]

3. **Finding the first derivative of \( g \)**:
From the equation \( g(x + 1) - g(x) = \ln x \), we differentiate both sides:
\[ g'(x + 1) - g'(x) = \frac{1}{x} \]

4. **Finding the second derivative of \( g \)**:
Differentiating again gives:
\[ g''(x + 1) - g''(x) = -\frac{1}{x^2} \]

5. **Generalizing the second derivative**:
Continuing this process, we can express \( g''(x + n) - g''(x) \) as:
\[ g''(x + n) - g''(x) = -\left( \frac{1}{x^2} + \frac{1}{(x + 1)^2} + \ldots + \frac{1}{(x + n - 1)^2} \right) \]

6. **Substituting \( x = \frac{1}{2} \)**:
We want to find \( g''(N + \frac{1}{2}) - g''(\frac{1}{2}) \):
\[ g''(N + \frac{1}{2}) - g''(\frac{1}{2}) = -\left( \frac{1}{(\frac{1}{2})^2} + \frac{1}{(\frac{1}{2} + 1)^2} + \ldots + \frac{1}{(\frac{1}{2} + N - 1)^2} \right) \]

7. **Calculating each term**:
- For \( N = 1 \):
\[ g''(1.5) - g''(0.5) = -\left( 4 + \frac{1}{\left(\frac{3}{2}\right)^2} \right) = -\left( 4 + \frac{4}{9} \right) = -\left( \frac{36}{9} + \frac{4}{9} \right) = -\frac{40}{9} \]
- For \( N = 2 \):
\[ g''(2.5) - g''(0.5) = -\left( 4 + \frac{4}{9} + \frac{4}{25} \right) \]
- For \( N = 3 \):
\[ g''(3.5) - g''(0.5) = -\left( 4 + \frac{4}{9} + \frac{4}{25} + \frac{4}{49} \right) \]

8. **Final Result**:
The final expression for \( g''(N + \frac{1}{2}) - g''(\frac{1}{2}) \) can be simplified into a common form for any \( N \).