The objective is to obtain a closed formula for: $$ \boxed{A(n)=\big(g(z)\,\partial_z\big)^n,\qquad n=1,2,\dots} $$ where $g(z)$ is smooth in $z$ and $\partial_z$ is a derivative with respect to $z$. I think the first few terms are, \begin{equation} \begin{aligned} A(1) &= g\,\partial\\ A(2)&= g\,(\partial g)\,\partial+g^2\,\partial^2\\ A(3)&= \big[(\partial^2g)g^2+(\partial g)^2g\big]\partial+3(\partial g)g^2\,\partial^2+g^3\partial^3\\ A(4) &= \big[(\partial^3g)g^3+4(\partial^2g)(\partial g)g^2+(\partial g)^3g\big]\partial\\ &\quad +\big[4(\partial^2g)g^3+7(\partial g)^2g^2\big]\partial^2+6(\partial g)g^3\partial^3+g^4\partial^4\\ &\,\,\vdots \end{aligned} \end{equation} and perhaps there is a simple pattern that I'm failing to see.

The partitionings of the $\partial$ and $g$ are reminiscent of Bell polynomials but the coefficients are more complicated. Perhaps it is useful to make explicit that the general expansion is of the form: $$ (g\,\partial)^n=g^n\sum_{p=0}^{n-1}a_{n,p}(g)\,\partial^{\,n-p} $$ with, $$ a_{n,p}(g)=\sum_{m_1+2m_2+\dots+pm_{p}=p} C_{n,p}(m_1,\dots,m_{p})\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{p} g}{g}\Big)^{m_{p}}\qquad (*) $$ and the latter sum is over all non-negative integers, $\{m_1,\dots,m_{p}\}$, subject to: $$ m_1+2m_2+\dots+pm_{p}=p $$

From this viewpoint the objective is to determine the coefficients $C_{n,p}(m_1,\dots,m_{p})$, which in turn depend on all integers, $n$, $p$ and $\{m_1,\dots,m_p\}$.

Any ideas?

Many thanks in advance.

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    $\begingroup$ The paper arxiv.org/abs/1010.0354 of Blasiak and Flajolet treats this sort of question in a more general setting. In particular, see the appendix. $\endgroup$ – Dan Fox Jul 31 '19 at 11:27
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    $\begingroup$ The natural objects for indexing the terms of the expansion are trees. This goes back to Cayley who called expressions like $g(z)\partial_z$ "operandators" because they are at the same time operators and operands. See mathoverflow.net/questions/168888/… $\endgroup$ – Abdelmalek Abdesselam Aug 1 '19 at 11:47
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    $\begingroup$ @AbdelmalekAbdesselam many thanks for adding further insight $\endgroup$ – Wakabaloola Aug 1 '19 at 11:54
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    $\begingroup$ I did not know about the work of Scherk in the reference given by Dan. It is certainly relevant. I looked at it and it does not have explicit drawings of trees. I also looked at Cayley's paper and he does not mention Scherk's thesis which definitely preceded him in this investigation. $\endgroup$ – Abdelmalek Abdesselam Aug 1 '19 at 12:13
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    $\begingroup$ Related history mathoverflow.net/questions/287742/… $\endgroup$ – Tom Copeland Aug 20 '19 at 20:00

In OEIS A124796 I considered a similar problem of computing the coefficients of $(\partial_z\circ M_g)^n$, where $M_g$ is the operator of multiplying by $g(z)$.

It turns out that the coefficients represent generalized Stirling numbers indexed by infinite vectors of nonnegative integers ${\cal S}([k_0,k_1,k_2,\dots])$ with a finite number of nonzero components, where ${\cal S}([k_0,k_1,0,0,\dots]) = S(k_0+k_1+1,k_0+1)$ are conventional Stirling number of the 2nd kind.

The expansion for $(\partial_z\circ M_g)^n$ is given by $$(\partial_z\circ M_g)^n = \sum_{k_0+k_1+\dots=n\atop k_1+2k_2+\dots\leq n} {\cal S}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)}.$$

The coefficients satisfy a recurrence relation: $${\cal S}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k_1,\dots]) + (k_0+1){\cal S}([k_0,k_1-1,k_2,\dots]) + \sum_{i\geq 1} (k_i+1) {\cal S}([k_0-1,k_1,...,k_{i-1},k_i+1,k_{i+1}-1,k_{i+2},\dots])$$ with ${\cal S}([0,0,\dots])=1$, and ${\cal S}([k_0,k_1,\dots])=0$ when any $k_i<0$ or when $k_1+2k_2+\dots>k_0+k_1+k_2+\dots$ (in other words, $k_2+2k_3+\dots > k_0$). In particular, the sum in the r.h.s. of the recurrence relation consists of just a finite number of nonzero terms.

UPDATED. The original question concerns $(M_g\circ\partial_z)^n = M_g\circ (\partial_z\circ M_g)^{n-1}\circ \partial_z$. Hence, \begin{split} (M_g\circ\partial_z)^n &= g(z)\cdot \sum_{k_0+k_1+\dots=n-1\atop k_1+2k_2+\dots\leq n-1} {\cal S}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)} \\ &= \sum_{k_0+k_1+\dots=n\atop k_1+2k_2+\dots\leq n} {\cal C}([k_0,k_1,\dots]) \prod_{i\geq 0} (\partial_z^i g(z))^{k_i}\cdot \partial_z^{n-(k_1+2k_2+\dots)}, \end{split} where ${\cal C}([k_0,k_1,\dots]) = {\cal S}([k_0-1,k_1,\dots])$ for $k_0\geq 1$, and ${\cal C}([0,k_1,k_2,\dots])=0$ except for ${\cal C}([0,0,0,\dots])=1$. In fact, the formula with coefficients ${\cal C}([k_0,k_1,\dots])$ holds even for $n=0$.

Correspondingly, we have a recurrence relation: $${\cal C}([k_0,k_1,\dots]) = {\cal C}([k_0-1,k_1,\dots]) + k_0{\cal C}([k_0,k_1-1,k_2,\dots]) + \sum_{i\geq 1} (k_i+1) {\cal C}([k_0-1,k_1,...,k_{i-1},k_i+1,k_{i+1}-1,k_{i+2},\dots]).$$ Then the generating function $$F(z_0,z_1,\dots) := \sum_{k_0,k_1,\dots\geq 0} {\cal C}([k_0,k_1,\dots]) \prod_{i\geq 0}z_i^{k_i}$$ satisfies the differential equation: $$F = 1 + z_0 F + z_0 \sum_{i\geq 0} z_{i+1}\partial_{z_i} F.$$ If $F_n$ is the restriction of $F$ to the terms of degree $n$, then $F_0=1$ and for $n>0$: $$F_n = z_0 F_{n-1} + z_0 \sum_{i=0}^{n-2} z_{i+1}\partial_{z_i} F_{n-1}.$$


  • $F_1 = z_0$
  • $F_2 = z_0^2 + z_0z_1$
  • $F_3 = z_0^3 + 3z_0^2z_1 + z_0z_1^2 + z_0^2z_2$
  • $F_4 = z_0^4 + 6 z_0^3 z_1 + 7z_0^2z_1^2 + z_0z_1^3 + 4z_0^3z_2 + 4z_0^2z_1z_2 + z_0^3z_3$

As expected, the coefficients in $F_n(z_0,z_1,0,0,\dots)$ are Stirling numbers of the 2nd kind.

It's worth to notice that for $g(z)=z$, we have $(M_g\circ\partial_z)^n = \sum_{k=0}^n S(n,k) z^k \partial_z^k$, which is essentially an umbral Touchard polynomial.

  • $\begingroup$ many thanks for your insight, I'm hoping a more explicit answer can be found $\endgroup$ – Wakabaloola Jul 31 '19 at 15:00
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    $\begingroup$ @Wakabaloola: I've added some more details. $\endgroup$ – Max Alekseyev Aug 1 '19 at 10:38
  • $\begingroup$ fantastic, that's very helpful $\endgroup$ – Wakabaloola Aug 1 '19 at 10:43

See OEIS A139605 (also related OEIS A145271) for matrix computations for these partition polynomials and numerous references.

The formula section of A139605 contains the matrix formula. Multiply the $n$-th diagonal (with $n=0$ the main diagonal) of the lower triangular Pascal matrix A007318 by $g_n = D_x^n g(x)$ to obtain the matrix $VP$ with $VP_{n,k} = \binom{n}{k}g_{n-k} $. Then $$(g(x)D_x)^n = (1, 0, 0,..) [VP \dot \; S]^n (1, D, D^2, ..)^T,$$ where S is the shift matrix A129185, representing differentiation in the divided powers basis $x^n/n!$.



$$= (1, 0, 0, 0) [VP \dot \; S]^3 (1, D, D^2, D^3)^T$$

$$= \begin{pmatrix} 1 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & g_0 & 0 & 0 \\ 0 & g_1 & g_0 & 0\\ 0 & g_2 & 2g_1 & g_0 \\ 0 & g_3 & 3g_2 & 3g_1 \end{pmatrix}^3 \begin{pmatrix} 1 \\ D \\ D^2 \\ D^3 \end{pmatrix} $$

$$ = [g_0g_1^2 + g_0^2 g_2] D + 3 g_0^2g_1 D^2 + g_0^3D^3 $$

And, the pdf Mathemagical Forests gives a diagrammatic method for creating forests of trees through "natural growth" that represent the partition polynomials.


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