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.

  • $\begingroup$ 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

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).

  • $\begingroup$ 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

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.


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