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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\,\partial^2+g^2\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 presumably 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 not.

  Perhaps it is useful to rephrase the question by noting that there is an expansion of the form: $$ (g\,\partial)^n=g^n\sum_{p=1}^na_{n-p}(g)\,\partial^{\,p} $$ and we can write down a general expression for $a_{n-p}(g)$ in terms of unknown coefficients, $C_{n-p}(m_1,m_2,\dots,m_{n-p})$, $$ a_{n-p}(g)=\sum_{\{m_\ell\}} C_{n-p}(\{m_\ell\})\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{n-p+1} g}{g}\Big)^{m_{n-p+1}} $$ where the sum is over all non-negative integers, $\{m_\ell\}$, subject to: $$ m_1+2m_2+\dots+(n-p+1)m_{n-p+1}=n-p $$ So from this viewpoint the objective is to determine the coefficients $C_{n-p}(m_1,m_2,\dots,m_{n-p+1})$.

Any ideas?

Many thanks in advance.

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.

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\,\partial^2+g^2\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 presumably 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 not.

Perhaps it is useful to rephrase the question as follows: find a closed formula for the $a_n(g)$ in: $$ \boxed{(g\,\partial)^n=\sum_{p=1}^na_{n-p}(g)\,\partial^{\,p}} $$ More precisely, maybe it is useful to write down the general expression in terms of unknown coefficients, $C_{p}(m_1,m_2,\dots,m_{n-p})$, $$ A(n) =\sum_{p=1}^n \Bigg[\sum_{\{m_\ell\}} C_{p}(\{m_\ell\})g^n\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{n-p} g}{g}\Big)^{m_{n-p}}\Bigg]\partial^{\,p}, $$ where the sum in the brackets is over all non-negative integers, $\{m_\ell\}$, subject to: $$ m_1+2m_2+\dots+(n-p)m_{n-p}=n-p $$

Any ideas?

Many thanks in advance.

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\,\partial^2+g^2\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 presumably 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 not.

Perhaps it is useful to rephrase the question by noting that there is an expansion of the form: $$ (g\,\partial)^n=g^n\sum_{p=1}^na_{n-p}(g)\,\partial^{\,p} $$ and we can write down a general expression for $a_{n-p}(g)$ in terms of unknown coefficients, $C_{n-p}(m_1,m_2,\dots,m_{n-p})$, $$ a_{n-p}(g)=\sum_{\{m_\ell\}} C_{n-p}(\{m_\ell\})\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{n-p+1} g}{g}\Big)^{m_{n-p+1}} $$ where the sum is over all non-negative integers, $\{m_\ell\}$, subject to: $$ m_1+2m_2+\dots+(n-p+1)m_{n-p+1}=n-p $$ So from this viewpoint the objective is to determine the coefficients $C_{n-p}(m_1,m_2,\dots,m_{n-p+1})$.

Any ideas?

Many thanks in advance.