Question

Let f be a function such that \(f(x)>0\) Assume that f has derivatives of all orders and that \(\ln f(x)=f(x) \int_{0}^{x} f(t) d t\). Find (i) \(\mathrm{f}(0)\), (ii) \(\mathrm{f}^{(1)}(0)\), (iii) \(\mathrm{f}^{2)}(0)\).

Short answer

Answer: The values for the function \(f(x)\) are given by: (i) \(f(0) = 1\), (ii) \(f^{(1)}(0) = 1\), and (iii) \(f^{(2)}(0) = 0\).

Step-by-step solution

Differentiate both sides of the equation with respect to \(x\)

Apply the chain rule to differentiate the left side of the equation with respect to \(x\): \(\frac{1}{f(x)} \frac{d}{dx}f(x)\) Now, differentiate the right side of the equation using the product rule and the fundamental theorem of calculus: \(\frac{d}{dx}(f(x)\int_{0}^{x} f(t) dt) = f(x) \frac{d}{dx}(\int_{0}^{x} f(t) dt) + \int_{0}^{x} f(t) dt \frac{d}{dx}f(x)\) By the fundamental theorem of calculus, \(\frac{d}{dx}(\int_{0}^{x}f(t)dt)=f(x)\). Thus, the equation becomes: \(\frac{1}{f(x)} \frac{d}{dx}f(x) = f(x) + \int_{0}^{x} f(t) dt \frac{d}{dx}f(x)\) Now we have: (i) \(f(0)\), (ii) \(f^{(1)}(0)\), (iii) \(f^{(2)}(0)\)

Find \(f(0)\)

To find \(f(0)\), substitute \(x=0\) into the given equation: \(\ln f(0) = f(0) \int_{0}^{0} f(t) dt\) The integral goes from 0 to 0, which evaluates to 0. Since the natural logarithm of 1 is 0, we have: \(f(0) = 1\)

Find \(f^{(1)}(0)\)

Substitute \(x=0\) into the differentiated equation: \(\frac{1}{f(0)} f^{(1)}(0) = f(0) + \int_{0}^{0} f(t) dt f^{(1)}(0)\) Since \(f(0) = 1\), the equation simplifies to: \(f^{(1)}(0) = 1 + 0 \cdot f^{(1)}(0)\) So, \(f^{(1)}(0) = 1\).

Differentiate both sides of the equation again with respect to \(x\)

Differentiate both sides of the equation from Step 1: \(\frac{-1}{f^{2}(x)}(f^{(1)}(x))^2 + \frac{1}{f(x)} f^{(2)}(x) = (f^{(1)}(x))^2 + \int_{0}^{x} f(t) dt f^{(2)}(x) + f(x) f^{(1)}(x)\)

Find \(f^{(2)}(0)\)

Substitute \(x=0\) into the twice-differentiated equation: \(\frac{-1}{1^2}(1)^2 + \frac{1}{1} f^{(2)}(0) = (1)^2 + \int_{0}^{0} f(t) dt f^{(2)}(0) + 1 \cdot 1\) This equation simplifies to: \(f^{(2)}(0) = 1 - 1 + 0 \cdot f^{(2)}(0)\) So, \(f^{(2)}(0) = 0\). In conclusion, we have: (i) \(f(0) = 1\), (ii) \(f^{(1)}(0) = 1\), and (iii) \(f^{(2)}(0) = 0\).