Leg `g(x)=ln(f(x)),` whre `f(x)` is a twice differentiable positive function on `(0,oo)` such that `f(x+1)=xf(x)dot` 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 step-by-step, we need to analyze the given functions and apply differentiation techniques.

### Step 1: Define the functions
We have:
- \( g(x) = \ln(f(x)) \)
- The functional equation: \( f(x+1) = x f(x) \)

### Step 2: Find \( g(x+1) \)
Using the functional equation, we can express \( g(x+1) \):
\[ g(x+1) = \ln(f(x+1)) = \ln(x f(x)) = \ln(x) + \ln(f(x)) = \ln(x) + g(x) \]

### Step 3: Find the difference \( g(x+1) - g(x) \)
From the expression derived in Step 2:
\[ g(x+1) - g(x) = \ln(x) \]

### Step 4: Differentiate \( g(x) \)
Now, we differentiate both sides with respect to \( x \):
\[ g'(x+1) - g'(x) = \frac{1}{x} \]

### Step 5: Differentiate again to find \( g''(x) \)
Differentiating again:
\[ g''(x+1) - g''(x) = -\frac{1}{x^2} \]

### Step 6: Evaluate \( g^{(N+1/2)} - g^{(1/2)} \) for \( N = 1, 2, 3 \)
We can evaluate this for \( N = 1, 2, 3 \) using the results from Steps 4 and 5.

1. For \( N = 1 \):
\[ g'(1/2 + 1) - g'(1/2) = g'(3/2) - g'(1/2) = \frac{1}{1/2} = 2 \]

2. For \( N = 2 \):
\[ g''(1/2 + 1) - g''(1/2) = g''(3/2) - g''(1/2) = -\frac{1}{(1/2)^2} = -4 \]

3. For \( N = 3 \):
\[ g'''(1/2 + 1) - g'''(1/2) = g'''(3/2) - g'''(1/2) = \text{(higher order terms)} \]

### Final Result
The final expressions for \( g^{(N+1/2)} - g^{(1/2)} \) for \( N = 1, 2, 3 \) can be summarized as follows:
- For \( N = 1 \): \( g'(3/2) - g'(1/2) = 2 \)
- For \( N = 2 \): \( g''(3/2) - g''(1/2) = -4 \)
- For \( N = 3 \): (requires further evaluation)