Is there a slick/quick/better way to get the $nth$ derivative of $x f(x) = \log x$? Or, to show $$f^{(n)}(1) = (-1)^{n+1} n! \left(1 + \frac12 + \ldots +\frac1n\right)$$

Question

kinggeorge · Answer

Perhaps induction?

anonymous · Answer

Hmm how do I use induction on this? I don't even know the pattern. Is differentiating and trying to make out some pattern the only way?

Here's the actual question.

kinggeorge · Answer

First, rearrange to get $f(x)=\log(x)/x$, and then take the first one or two or three or four derivatives. From this you might be able to find some formula for the $n$-th derivative, and from that find $$f^{(n)} (1)$$

kinggeorge · Answer

Looking at Wolfram, it seems like the $n$-th derivative would be $$f^{(n)} = (-1)^n \cdot {n! \log(x) - (2n! -1) \over x^{n+1}}$$And it looks possible to prove this with induction.

anonymous · Answer

Thanks, I will see what I can get. I did try differentiating earlier but wasn't able to make out the pattern.

asnaseer · Answer

I don't know if this helps, but you could try implicit differentiation:$$\begin{align}
xf(x)&=\log(x)\
xf^1(x)+f(x)&=\frac{1}{x}\
xf^2(x)+f^1(x)+f^1(x)&=-\frac{1}{x^2}\
xf^2(x)+2f^1(x)&=-\frac{1}{x^2}\
...
\end{align}$$
this eventually seems to lead to the following pattern:$$xf^n(x)+nf^{n-1}(x)=(-1)^{n+1)*\frac{n-1}{x^n}$$

asnaseer · Answer

the rendering of the last line hasn't come out right. it should be:$$xf^n(x)+nf^{n-1}(x)=(-1)^{n+1}*\frac{n-1}{x^n}$$

asnaseer · Answer

so setting x=1 leads to:$$f^n(1)+nf^{n-1}(1)=(-1)^{n+1}*(n-1)$$

kinggeorge · Answer

We would still need to solve for get rid of the recurrence relation though. And that may or may not be easier.

asnaseer · Answer

true...

kinggeorge · Answer

I didn't type that sentence very well :(

asnaseer · Answer

it's lucky this isn't the English or Writing group then or you would be real trouble ;-)

asnaseer · Answer

*be in

anonymous · Answer

I did try that too but I wasn't able to get it all together. :/

asnaseer · Answer

maybe you can use the recurrence result to prove KingGeorge's result above using induction?

asnaseer · Answer

hmmm... but even with that I can't see how it would lead to:$$f^{(n)}(1) = (-1)^{n+1} n! \left(1 + \frac12 + \ldots +\frac1n\right)$$

anonymous · Answer

I am trying it's only the algebra now, with enough trials I might get it.

asnaseer · Answer

ok - good luck! :)

kinggeorge · Answer

I went a few terms farther with my possible definition of $f^{(n)}$ and it doesn't seem to work anymore.

asnaseer · Answer

I worked with the recurrence relation and ended up with something close (but no cigar!):$$f^n(1)=(-1)^{n+1}(\frac{1}{n(n-2)(n-3)...}+\frac{1}{(n-1)(n-3)(n-4)...}+...+1)$$

asnaseer · Answer

sorry that should be:$$f^n(1)=(-1)^{n+1}n!(\frac{1}{n(n-2)(n-3)...}+\frac{1}{(n-1)(n-3)(n-4)...}+...+1)$$

asnaseer · Answer

I'm guessing I either made a mistake in the algebra somewhere or the derivation of the recurrence relation had an error in it.

anonymous · Answer

Here's what I am getting 
$$xf^n(x) = \small (-1)^n\cdot \frac 1{x^n}\cdot \left((n-1) + n(n-2) + n(n-1)(n-3) + n(n-1)(n-2)(n-4) + \ldots \right)$$

asnaseer · Answer

yes Ishaan94 - if you take an n! out that you will get the same as I did

asnaseer · Answer

you can set x=1 to simplify the algebra

kinggeorge · Answer

This may or may not be helpful, but the new possibility I have for the $n$-th derivative is $$\large f^{(n)}(x) = (-1)^n \cdot {n! \log(x) - n!({1 \over 1}+{1 \over 2}+{1 \over 3}+...+{1 \over n}) \over x^{n+1}} $$Which would get you what you need if you factor a $-1$ out of the top. Proving this, is still not easy.

asnaseer · Answer

does this lead directly from your previous result?

kinggeorge · Answer

My previous result was flawed. After finding some more patterns however, this seems correct. It works perfectly through the 5th derivative.

asnaseer · Answer

ok - then this looks a lot more promising - great work!

asnaseer · Answer

KingGeorge - why do we need to prove this - I thought you said it comes from repeated derivatives and then spotting the pattern? isn't that proof enough?

anonymous · Answer

I don't understand why didn't we get the answer even after solving the recurrence?

Nice work, KingGeorge :-)

kinggeorge · Answer

We still need to prove the pattern works for all $n$. And if I'm doing my derivatives correctly, I think this is straightforward to prove using induction.

kinggeorge · Answer

It uses a lot of paper, but it's straightforward.

asnaseer · Answer

ok - I guess you now have a way forward now Ishaan94 (thanks to the King!) :)

anonymous · Answer

Awesome, it's perfect. Thanks asnaseer and KingGeorge :-) 

#Sorry for replying late, I was afk.

kinggeorge · Answer

In the derivative, the only part that's a little hard is proving that $$n!+n!\left(1+{1 \over2}+...+{1 \over n}\right) = (n+1)!\left(1+{1 \over 2}+...+{1 \over n+1}\right)$$The rest is rather easy.

anonymous · Answer

I did prove it and it works perfectly.

anonymous · Answer

Thanks again. and great work. :D

kinggeorge · Answer

You're welcome.