# 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. Jul 31, 2019 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/… Aug 1, 2019 at 11:47
• @AbdelmalekAbdesselam many thanks for adding further insight Aug 1, 2019 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. Aug 1, 2019 at 12:13
• Related history mathoverflow.net/questions/287742/… Aug 20, 2019 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 Jul 31, 2019 at 15:00
• @Wakabaloola: I've added some more details. Aug 1, 2019 at 10:38
• fantastic, that's very helpful Aug 1, 2019 at 10:43

The Ihara reference "Derivations and automorphisms on non-commutative power series" (open archive now) in OEIS A139605 contains an explicit formula for the coefficients you are looking for, obtained from the Comtet ref. "Une formule explicite pour les puissances successives de l'opérateur de dérivation de Lie."

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

• great! it would be very helpful (also for future readers) if you would extract the relevant discussion and formula(s) and write it as an answer here Aug 1, 2019 at 21:15
• Aug 7, 2019 at 15:38
• Aug 20, 2019 at 20:03
• Related history mathoverflow.net/questions/97512/in-splendid-isolation/… Aug 20, 2019 at 20:07