Key Concepts and Formulas

• Logarithm Properties: $\ln(ab) = \ln a + \ln b$, $\ln(a^b) = b \ln a$.
• Differentiation Rules:
  • Chain Rule: $\frac{d}{dx} [f(h(x))] = f'(h(x)) \cdot h'(x)$.
  • Derivative of $\ln x$: $\frac{d}{dx} (\ln x) = \frac{1}{x}$.
  • Derivative of $\frac{1}{x^n}$: $\frac{d}{dx} (x^{-n}) = -nx^{-n-1}$.
• Telescoping Sum: A sum where intermediate terms cancel out, e.g., $\sum_{i=1}^{n} (a_{i+1} - a_i) = a_{n+1} - a_1$.

Step-by-Step Solution

Step 1: Define g(x) and manipulate the given functional equation. We are given the function $f: S \to S$ where $S = (0, \infty)$ is twice differentiable, such that $f(x+1) = xf(x)$. We are also given $g(x) = \ln(f(x))$. Taking the natural logarithm of both sides of the functional equation $f(x+1) = xf(x)$, we get: $\ln(f(x+1)) = \ln(xf(x))$ Using the logarithm property $\ln(ab) = \ln a + \ln b$, we have: $\ln(f(x+1)) = \ln x + \ln(f(x))$

Step 2: Express the functional equation in terms of g(x). Since $g(x) = \ln(f(x))$, we can substitute this into the equation from Step 1: $g(x+1) = \ln x + g(x)$ Rearranging this equation, we get: $g(x+1) - g(x) = \ln x$ ..... (i)

Step 3: Differentiate the functional equation for g(x) to find g'(x). We differentiate both sides of the equation $g(x+1) = \ln x + g(x)$ with respect to $x$. Using the chain rule for $g(x+1)$ (where the outer function is $g$ and the inner function is $x+1$, whose derivative is 1), we get $\frac{d}{dx}[g(x+1)] = g'(x+1) \cdot 1 = g'(x+1)$. The derivative of $\ln x$ is $\frac{1}{x}$. The derivative of $g(x)$ is $g'(x)$. So, differentiating the equation $g(x+1) = \ln x + g(x)$ with respect to $x$ gives: $g'(x+1) = \frac{1}{x} + g'(x)$ Rearranging this, we get: $g'(x+1) - g'(x) = \frac{1}{x}$ ..... (ii)

Step 4: Differentiate the equation for g'(x) to find g''(x). Now, we differentiate both sides of the equation $g'(x+1) - g'(x) = \frac{1}{x}$ with respect to $x$. Differentiating $g'(x+1)$ with respect to $x$ using the chain rule gives $g''(x+1) \cdot 1 = g''(x+1)$. Differentiating $g'(x)$ with respect to $x$ gives $g''(x)$. Differentiating $\frac{1}{x}$ with respect to $x$ gives $-\frac{1}{x^2}$. So, differentiating the equation $g'(x+1) - g'(x) = \frac{1}{x}$ with respect to $x$ gives: $g''(x+1) - g''(x) = -\frac{1}{x^2}$ ..... (iii)

Step 5: Use the derived relation for g''(x) to find g''(5) - g''(1). We need to find the value of $|g''(5) - g''(1)|$. We can use the relation $g''(x+1) - g''(x) = -\frac{1}{x^2}$ and apply it for consecutive integer values of $x$ starting from $x=1$ up to $x=4$.

For $x=1$: $g''(1+1) - g''(1) = -\frac{1}{1^2}$ $g''(2) - g''(1) = -1$ ..... (iv)

For $x=2$: $g''(2+1) - g''(2) = -\frac{1}{2^2}$ $g''(3) - g''(2) = -\frac{1}{4}$ ..... (v)

For $x=3$: $g''(3+1) - g''(3) = -\frac{1}{3^2}$ $g''(4) - g''(3) = -\frac{1}{9}$ ..... (vi)

For $x=4$: $g''(4+1) - g''(4) = -\frac{1}{4^2}$ $g''(5) - g''(4) = -\frac{1}{16}$ ..... (vii)

Step 6: Sum the equations from Step 5 to obtain g''(5) - g''(1). We add equations (iv), (v), (vi), and (vii) together: $(g''(2) - g''(1)) + (g''(3) - g''(2)) + (g''(4) - g''(3)) + (g''(5) - g''(4)) = -1 - \frac{1}{4} - \frac{1}{9} - \frac{1}{16}$

This is a telescoping sum on the left side. The intermediate terms $g''(2)$, $g''(3)$, and $g''(4)$ cancel out: $g''(5) - g''(1) = - \left( 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} \right)$

Step 7: Calculate the sum and find the absolute value. We need to sum the fractions: $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}$ The least common multiple of 1, 4, 9, and 16 is 144. $1 = \frac{144}{144}$ $\frac{1}{4} = \frac{36}{144}$ $\frac{1}{9} = \frac{16}{144}$ $\frac{1}{16} = \frac{9}{144}$

So, the sum is: $\frac{144}{144} + \frac{36}{144} + \frac{16}{144} + \frac{9}{144} = \frac{144 + 36 + 16 + 9}{144} = \frac{205}{144}$

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

The question asks for the value of $|g''(5) - g''(1)|$: $|g''(5) - g''(1)| = \left| - \frac{205}{144} \right| = \frac{205}{144}$

Common Mistakes & Tips

• Incorrectly applying the chain rule: Remember that when differentiating $g(x+1)$ with respect to $x$, the derivative of the inner function $(x+1)$ is 1, so the chain rule simplifies nicely.
• Errors in fraction addition: Carefully find the least common multiple and add the fractions to avoid arithmetic mistakes.
• Forgetting the absolute value: The question specifically asks for the absolute value, so ensure you take the magnitude of the final result.

Summary

The problem involves a functional equation for a function $f(x)$ and its logarithmic transformation $g(x)$. By taking logarithms and differentiating twice, we derived a relation for the second derivative of $g(x)$. Using this relation, we expressed the difference $g''(x+1) - g''(x)$ in terms of $x$. By substituting consecutive integer values for $x$ and summing the resulting equations, we obtained $g''(5) - g''(1)$ using a telescoping sum. Finally, we calculated the numerical value of this difference and took its absolute value.

The final answer is $\boxed{\frac{205}{144}}$.