# Сlosed formula for $(g\partial)^n$

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{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} 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?

• 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. – Dan Fox Jul 31 '19 at 11:27
• 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/… – Abdelmalek Abdesselam Aug 1 '19 at 11:47
• @AbdelmalekAbdesselam many thanks for adding further insight – Wakabaloola Aug 1 '19 at 11:54
• 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. – Abdelmalek Abdesselam Aug 1 '19 at 12:13
• Related history mathoverflow.net/questions/287742/… – 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 $$n>0$$: $$F_n = z_0 F_{n-1} + z_0 \sum_{i=0}^{n-2} z_{i+1}\partial_{z_i} F_{n-1}.$$

Examples.

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

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.

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

Example:

$$(g(x)D_x)^3$$

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

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

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

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