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I am trying to find a series expansion of the following function: $$\left(\frac{\log x}{x}\right)^n$$ I need hints or methods for going about doing this. Is it even possible?

I am on to something with the general case without the exponent (With some help form robjohn): $$D^n\left(\frac{\log(x)}{x}\right)=(-1)^nn!\frac{\log(x)-H_n}{x^{n+1}}$$

Ali Caglayan
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  • The first derivative is $n\left(\frac {\log x}x\right)^{n-1}\left(\frac {1-\log x}{x^2}\right)$, correct? – abiessu Oct 28 '13 at 20:37
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    What have you tried? Do you know how Taylor series are constructed? – dfeuer Oct 28 '13 at 20:38
  • Where did this problem come from? – Mhenni Benghorbal Oct 28 '13 at 21:05
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    My best advice for right now: see if you can find a Taylor series for $n=1$ and $n=2$ before you attempt the (complicated) general case. You will likely want to calculate it around $x=1$. – dfeuer Oct 28 '13 at 21:08
  • Related problems. Nice problem by the way? – Mhenni Benghorbal Oct 28 '13 at 21:22
  • @dfeuer I have a vague idea on how they are constructed but this one poses a different level of difficulty. – Ali Caglayan Oct 28 '13 at 22:18
  • I'm confident the best way to start is by calculating the first few terms of the series for $n=1$. There's not much chance that the general case will be simpler than the simplest case. – dfeuer Oct 28 '13 at 22:24
  • @dfeuer yeh it gives you a feel for the pattern. I'd expect trying to pull out a $-1$ at each step and trying to incorporate a factorial are usually good ideas. Or even what i found with another taylor series was keeping the sign of the highest power the "same" – snulty Apr 22 '14 at 16:22

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If you're curious, I was doing the same over Christmas recently. Let $D^n=\frac{d^n}{dx^n}$.

Define the recursive Harmonic Numbers as follows (with extra conditions for our purposes) for $n,k \in \mathbb{N}_0$ as

$\text{H}_{0,n}=1 \quad\forall \;n\in \mathbb{N}_0$

$\text{H}_{k,n}=0 \quad\text{if } n<k$

$\text{H}_{k,n}:=\text{H}_{k,n-1}+1/n\cdot\text{H}_{k-1,n-1} \quad\text{for } n,m\geq1$.

Note that $\text{H}_{1,n}$ are the ordinary harmonic numbers.

Then let $n,m \geq 1$

$D^n \left(\log x\right)^m=\frac{m\left(-1\right)^{n-1}\left(n-1\right)!}{x^n}\left(\sum\limits_{k=1}^m (-1)^{k+1}H_{ k-1,n-1}\; {}^{m-1}P_{m-k} \;\left(\log x\right)^{m-k}\right)$

with ${}^{m-1}P_{m-k}:=\frac{(m-1)!}{(m-k)!}$

Providing there isn't a typo you can verify this by induction. It involves pulling out the first term in a sum, relabeling, combining the two sums back in to one and using the above relation for $\text{H}_{m,n}$.

Maybe there's a modification of this that'll help you. Also not a chance this would fit as a comment. Also also, I under the impression steve roman might have derived this in one of his papers on Harmonic Logarithms but I haven't the chance to read the paper fully.

snulty
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