3
$\begingroup$

I am think about this expression : $e^{\lambda x \frac{d}{dx}}f(x)$. Let us look at each term in the expansion of the exponential operator $e^{\lambda x \frac{d}{dx}}$, $$\left(x\frac{d}{dx}\right)^n f(x) $$

For example, (denote $\partial = \frac{d}{dx}$) \begin{align*} n&=1 : \quad x \partial \\ n&=2 : \quad x^2 \partial^2 + x\partial \\ n&=3 : \quad x^3 \partial^3 + 3x^2\partial^2 + x\partial \\ n&=4 : \quad x^4 \partial^4+6x^3 \partial^3 + 7x^2\partial^2 + x\partial \\ n&=5 : \quad x^5 \partial^5+ 10x^4 \partial^4+25\ x^3 \partial^3 + 15x^2\partial^2 + x\partial \\ n&=6 : \quad x^6 \partial^6+15x^5 \partial^5+ 65x^4 \partial^4+90\ x^3 \partial^3 + 31x^2\partial^2 + x\partial \\ \end{align*}

I wonder if there is a general expression for the coefficient. I have expanded it out and it seems that they are related to binomial coefficients though its kind of tedious. If there is a general expression of these coefficients, what's its properties? Thank you very much.

$\endgroup$
  • $\begingroup$ Either here in MSE or in mathoverflow there was recently a discussion about a direct (=nonrecursive) expression for the Stirling numbers 2nd kind. Perhaps this is interesting for you... $\endgroup$ – Gottfried Helms Sep 26 '15 at 6:03
5
$\begingroup$

You have :

$$(x\frac{d}{dx})^nf(x)=\sum_{k=1}^ns(k,n)x^k\partial^k f(x) $$

Where $s(k,n)$ is the Stirling number of the second kind. See : http://oeis.org/A008277/table.

Whereas $s(k,n)$ is the number of partitions into $k$ sets of a set of cardinal $n$, I don't think there is a combinatorial way to prove the identity. What you can do is to show that the Stirling numbers of the second kind and the coefficients of $(x\frac{d}{dx})^nf(x)$ follow the same induction formula.

One can use this to make a link between Newton sums ($1^k+...+n^k$), the Stirling numbers and some binomials (using generating functions machinery).

$\endgroup$
  • $\begingroup$ Thank you very much! Indeed it is a nice answer. Is there any explanation (instead of direct calculation) why the expansion of the operator gives these coefficients? $\endgroup$ – mastrok Sep 25 '15 at 10:15
  • $\begingroup$ @mastrok, unlike what I wrote in the answer, there may be a purely combinatorial way to prove this (i.e. without the induction) but I have not been able to find it at the time. It might be worth trying it as a new question on the site (or wait for a better answer to this question). $\endgroup$ – Clément Guérin Sep 25 '15 at 10:25
  • $\begingroup$ Thank you very much. You answer helps a lot! Its my first time learning this Stirling number of the second kind. I noticed that it means the number of ways to partition a set of $n$ objects into $k$ non-empty subsets. It's quite interesting, I will try to figure it out. $\endgroup$ – mastrok Sep 25 '15 at 10:33
  • $\begingroup$ @ClémentGuérin: I've added a combinatorial explanation. $\endgroup$ – David Bevan Sep 25 '15 at 17:30
2
$\begingroup$

There’s a purely combinatorial explanation of Clément Guérin's answer. Suppose $f(x)$ is the generating function for some combinatorial class $\mathcal{F}$, where the exponent of $x$ records the size of (i.e. the number of atoms in) an object in $\mathcal{F}$, and the coefficient of $x^m$ is the number of elements of size $m$ in $\mathcal{F}$.

Then $x\partial_x f(x)$ is the generating function for the class consisting of elements of $\mathcal{F}$ with a single distinguished atom. Flajolet and Sedgewick (p.102) call this pointing, i.e. “pointing at a distinguished atom”.

Moreover, the operator $x^k\partial_x^k$ corresponds to pointing at $k$ distinct atoms, whereas $(x\partial_x)^n$ corresponds to pointing at $n$ atoms that are not necessarily distinct.

Thus $(x\partial_x)^n f(x) = \sum_{k=1}^n s(n,k) x^k\partial_x^k f(x)$, where the sum distinguishes between the number of distinct distinguished atoms and $s(n,k)$ is the number of ways of assigning $n$ pointers to $k$ distinct atoms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.