Let $$f : S \to S$$ where $$S = (0, \infty)$$ be a twice differentiable function such that $$f(x+1) = xf(x)$$. If $$g : S \to R$$ be defined as $$g(x) = \log_e f(x)$$, then the value of $$|g''(5) - g''(1)|$$ is equal to:

We have $$f(x+1) = xf(x)$$ and $$g(x) = \ln f(x)$$. Taking logarithms: $$g(x+1) = \ln(xf(x)) = \ln x + g(x)$$, so $$g(x+1) - g(x) = \ln x$$.

Differentiating twice with respect to $$x$$: $$g''(x+1) - g''(x) = -\frac{1}{x^2}$$.

Now we compute $$g''(5) - g''(1)$$ by telescoping:

$$g''(2) - g''(1) = -\frac{1}{1^2} = -1$$

$$g''(3) - g''(2) = -\frac{1}{2^2} = -\frac{1}{4}$$

$$g''(4) - g''(3) = -\frac{1}{3^2} = -\frac{1}{9}$$

$$g''(5) - g''(4) = -\frac{1}{4^2} = -\frac{1}{16}$$

Adding all these: $$g''(5) - g''(1) = -1 - \frac{1}{4} - \frac{1}{9} - \frac{1}{16} = -\frac{144 + 36 + 16 + 9}{144} = -\frac{205}{144}$$.

Therefore, $$|g''(5) - g''(1)| = \frac{205}{144}$$.