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

[lfdahl]

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 $x$.

 

[MarkFL]

My solution:

Spoiler

We have:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Shown as desired. :D

 

[lfdahl]

My solution:

Spoiler

We have:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Shown as desired. :D

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