MHB Can you find $f^{(n)}(1)$ with the given conditions?

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The function \( f(x) \) is defined by the equation \( xf(x) = \ln x \) for \( x > 0 \). The discussion focuses on deriving the \( n \)-th derivative of \( f \) at \( x = 1 \), specifically showing that \( f^{(n)}(1) = (-1)^{n+1}n!\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right) \). Participants express appreciation for the solutions provided, indicating a collaborative effort to understand the problem. The conversation highlights the mathematical intricacies involved in calculating derivatives of the function. Overall, the thread emphasizes the importance of rigorous mathematical reasoning in deriving results.
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Let $f(x)$ be a function satisfying
\[xf(x)=\ln x \ \ \ \ \ \ \ \ \text{for} \ \ x>0\]
Show that $f^{(n)}(1)=(-1)^{n+1}n!\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$ where $f^{(n)}(x)$ denotes the $n$-th derivative evaluated at $x$.
 
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My solution:

We have:

$$f(x)=\frac{\ln(x)}{x}$$

Now, let's look at a few base cases to see if a pattern emerges:

$$f^{(1)}(x)=\frac{1-\ln(x)}{x^2}$$

$$f^{(2)}(x)=\frac{2\ln(x)-3}{x^3}$$

$$f^{(3)}(x)=\frac{11-6\ln(x)}{x^4}$$

$$f^{(4)}(x)=\frac{24\ln(x)-50}{x^5}$$

I'm ready to posit the inductive statement $P_n$:

$$f^{(n)}(x)=\frac{(-1)^nn!\left(\ln(x)-H_n\right)}{x^{n+1}}$$

Note: $H_n$ is the $n$th harmonic number, defined as:

$$H_n\equiv\sum_{k=1}^{n}\left(\frac{1}{k}\right)$$

We already know the base case is true, so as our inductive step, let's differentiate both sides w.r.t $x$:

$$f^{(n+1)}(x)=\frac{x^{n+1}\left((-1)^nn!\frac{1}{x}\right)-(-1)^nn!\left(\ln(x)-H_n\right)(n+1)x^n}{\left(x^{n+1}\right)^2}=\frac{(-1)^nn!\left(1-(n+1)(\ln(x)-H_n\right)}{x^{n+2}}=\frac{(-1)^{n+1}(n+1)!\left(\ln(x)-H_{n+1}\right)}{x^{(n+1)+1}}$$

We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction. Hence:

$$f^{(n)}(1)=\frac{(-1)^nn!\left(\ln(1)-H_n\right)}{1^{n+1}}=(-1)^{n+1}n!H_n$$

Shown as desired. :D
 
MarkFL said:
My solution:

We have:

$$f(x)=\frac{\ln(x)}{x}$$

Now, let's look at a few base cases to see if a pattern emerges:

$$f^{(1)}(x)=\frac{1-\ln(x)}{x^2}$$

$$f^{(2)}(x)=\frac{2\ln(x)-3}{x^3}$$

$$f^{(3)}(x)=\frac{11-6\ln(x)}{x^4}$$

$$f^{(4)}(x)=\frac{24\ln(x)-50}{x^5}$$

I'm ready to posit the inductive statement $P_n$:

$$f^{(n)}(x)=\frac{(-1)^nn!\left(\ln(x)-H_n\right)}{x^{n+1}}$$

Note: $H_n$ is the $n$th harmonic number, defined as:

$$H_n\equiv\sum_{k=1}^{n}\left(\frac{1}{k}\right)$$

We already know the base case is true, so as our inductive step, let's differentiate both sides w.r.t $x$:

$$f^{(n+1)}(x)=\frac{x^{n+1}\left((-1)^nn!\frac{1}{x}\right)-(-1)^nn!\left(\ln(x)-H_n\right)(n+1)x^n}{\left(x^{n+1}\right)^2}=\frac{(-1)^nn!\left(1-(n+1)(\ln(x)-H_n\right)}{x^{n+2}}=\frac{(-1)^{n+1}(n+1)!\left(\ln(x)-H_{n+1}\right)}{x^{(n+1)+1}}$$

We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction. Hence:

$$f^{(n)}(1)=\frac{(-1)^nn!\left(\ln(1)-H_n\right)}{1^{n+1}}=(-1)^{n+1}n!H_n$$

Shown as desired. :D

What a nice solution, MarkFL, thankyou for your participation! :cool:
 
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