TITLE: A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?
QUESTION [16 upvotes]: Let $f(x)$ be sufficiently regular (e.g. a smooth function or a formal power series in characteristic 0 etc.). In my research the following recursion made a surprising entrance
$$
f_1(x) = f(x),\ f_{n+1}(x) = f(x) f'_{n}(x)
$$
Thus I would like to understand the sequence
$$
\left(f(x) \frac{d}{dx} \right)^n f(x)
$$
This looks like a classical question that must have been already studied in dynamical systems, or Weyl algebras (say for $f(x) \in R[x]$, $R$ a commutative ring of characteristic $0$, and the derivation $\partial := f(x) \frac{d}{dx} \in A_1(R)$), or generating functions in combinatorics. But I have been unable to pinpoint it. My question is this:

Is there a known formula in the spirit of the general Leibniz formula which expresses
$$
\left(f(x) \frac{d}{dx} \right)^n f(x)
$$
in terms of $f$ and its derivatives $f',f'',\dots, f^{(n)}$?

Any references would be also very much appreciated!

REPLY [3 votes]: This, too, is not an answer, just a preview of a paper I probably won't be writing for a while (one of only 9 papers on my immediate to-do list).
It so happens I derived a related formula a few weeks ago when proving an
apocryphal identity of Hochschild. The formula needs some preliminary
definitions, and unfortunately my notations are different from the OP's, but I
hope it is of some use. Note that, while I am able to prove everything I
claim, I am not fluent in its combinatorial background (umbral calculus, in particular).
The general setup will be as follows:
We fix a commutative ring $R$. Let $d:R\rightarrow R$ be a derivation (i.e., a
$\mathbb{Z}$-linear map satisfying $d\left(  ab\right)  =d\left(  a\right)
\cdot b+a\cdot d\left(  b\right)  $ for all $a,b\in R$). Let $a\in R$ be an
element. Let $L_{a}$ be the map $R\rightarrow R,\ r\mapsto ar$ (known as "left
multiplication by $a$", but can just as well be called "right multiplication
by $a$" since $R$ is commutative). Note that $L_{a}\circ d$ is a derivation
(commonly denoted by $ad$), but $d\circ L_{a}$ is not (in general).
For each nonnegative integer $n$, we define the following:

Let $\left[  n\right]  $ be the set $\left\{  1,2,\ldots,n\right\}  $.

A set partition of $\left[  n\right]  $ means a set $P$ of subsets of
$\left[  n\right]  $ (called the blocks of $P$) such that each element of
$\left[  n\right]  $ belongs to exactly one block of $P$.

Let $\operatorname*{SP}\left(  n\right)  $ denote the set of all set
partitions of $\left[  n\right]  $. For instance,
\begin{align*}
\operatorname*{SP}\left(  3\right)    & ={\Large \{}\left\{  \left\{
1,2,3\right\}  \right\}  ,\ \ \ \left\{  \left\{  1\right\}  ,\left\{
2,3\right\}  \right\}  ,\ \ \ \left\{  \left\{  2\right\}  ,\left\{
1,3\right\}  \right\}  ,\\
& \ \ \ \ \ \ \left\{  \left\{  3\right\}  ,\left\{  1,2\right\}  \right\}
,\ \ \ \left\{  \left\{  1\right\}  ,\left\{  2\right\}  ,\left\{  3\right\}
\right\}  {\Large \}}.
\end{align*}
Note that $\left\vert \operatorname*{SP}\left(  n\right)  \right\vert $ is the
$n$-th Bell number $B_{n}$.

If $P=\left\{  P_{1},P_{2},\ldots,P_{k}\right\}  $ is a set partition of
$\left[  n\right]  $ (with $P_{1},P_{2},\ldots,P_{k}$ being distinct), then we
set $\ell\left(  P\right)  :=k$ and
\begin{align*}
b_{P}:=\prod_{i=1}^{k}\left(  d\circ L_{a}\right)  ^{\left\vert P_{i}
\right\vert -1}\left(  1\right)  .
\end{align*}


For example, if $n=5$ and $P=\left\{  \left\{  1,4\right\}  ,\left\{
2,3,5\right\}  \right\}  $, then
\begin{align*}
b_{P}=\underbrace{\left(  d\circ L_{a}\right)  ^{2-1}\left(  1\right)
}_{\substack{=\left(  d\circ L_{a}\right)  \left(  1\right)  \\=d\left(
a\right)  }}\cdot\underbrace{\left(  d\circ L_{a}\right)  ^{3-1}\left(
1\right)  }_{\substack{=\left(  d\circ L_{a}\right)  ^{2}\left(  1\right)
\\=d\left(  ad\left(  a\right)  \right)  }}=d\left(  a\right)  \cdot d\left(
ad\left(  a\right)  \right)  .
\end{align*}
Of course, $b_{P}$ depends only on the sizes of the blocks of $P$.
Now, my formulas claim that each integer $m\geq0$ and each $w\in R$ satisfy
\begin{equation}
\left(  d\circ L_{a}\right)  ^{m}\left(  w\right)  =\sum_{P\in
\operatorname*{SP}\left(  m+1\right)  }b_{P}\cdot a^{\ell\left(  P\right)
-1}\cdot d^{\ell\left(  P\right)  -1}\left(  w\right)
\label{darij1.eq.damw}
\tag{1}
\end{equation}
and
\begin{equation}
\left(  L_{a}\circ d\right)  ^{m}\left(  w\right)  =\sum_{P\in
\operatorname*{SP}\left(  m\right)  }b_{P}\cdot a^{\ell\left(  P\right)
}\cdot d^{\ell\left(  P\right)  }\left(  w\right)  .
\label{darij1.eq.admw}
\tag{2}
\end{equation}
For instance, for $m=2$, these two formulas become
\begin{align*}
\left(  d\circ L_{a}\right)  ^{2}\left(  w\right)    &
=\underbrace{b_{\left\{  \left\{  1,2,3\right\}  \right\}  }}_{=\left(  d\circ
L_{a}\right)  ^{2}\left(  1\right)  }\cdot\underbrace{a^{1-1}}_{=1}
\cdot\underbrace{d^{1-1}\left(  w\right)  }_{=w}\\
& \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{  \left\{  1\right\}  ,\left\{
2,3\right\}  \right\}  }}_{\substack{=\left(  d\circ L_{a}\right)  ^{0}\left(
1\right)  \cdot\left(  d\circ L_{a}\right)  ^{1}\left(  1\right)  \\=\left(
d\circ L_{a}\right)  ^{1}\left(  1\right)  }}\cdot\underbrace{a^{2-1}}
_{=a}\cdot\underbrace{d^{2-1}}_{=d}\left(  w\right)  \\
& \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{  \left\{  2\right\}  ,\left\{
1,3\right\}  \right\}  }}_{\substack{=\left(  d\circ L_{a}\right)  ^{0}\left(
1\right)  \cdot\left(  d\circ L_{a}\right)  ^{1}\left(  1\right)  \\=\left(
d\circ L_{a}\right)  ^{1}\left(  1\right)  }}\cdot\underbrace{a^{2-1}}
_{=a}\cdot\underbrace{d^{2-1}}_{=d}\left(  w\right)  \\
& \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{  \left\{  3\right\}  ,\left\{
1,2\right\}  \right\}  }}_{\substack{=\left(  d\circ L_{a}\right)  ^{0}\left(
1\right)  \cdot\left(  d\circ L_{a}\right)  ^{1}\left(  1\right)  \\=\left(
d\circ L_{a}\right)  ^{1}\left(  1\right)  }}\cdot\underbrace{a^{2-1}}
_{=a}\cdot\underbrace{d^{2-1}}_{=d}\left(  w\right)  \\
& \ \ \ \ \ \ \ \ \ \ +\underbrace{b_{\left\{  \left\{  1\right\}  ,\left\{
2\right\}  ,\left\{  3\right\}  \right\}  }}_{\substack{=\left(  d\circ
L_{a}\right)  ^{0}\left(  1\right)  \cdot\left(  d\circ L_{a}\right)
^{0}\left(  1\right)  \cdot\left(  d\circ L_{a}\right)  ^{0}\left(  1\right)
\\=1}}\cdot\underbrace{a^{3-1}}_{=a^{2}}\cdot\underbrace{d^{3-1}}_{=d^{2}
}\left(  w\right)  \\
& =\left(  d\circ L_{a}\right)  ^{2}\left(  1\right)  \cdot w+3\cdot\left(
d\circ L_{a}\right)  ^{1}\left(  1\right)  \cdot ad\left(  w\right)
+a^{2}\cdot d^{2}\left(  w\right)
\end{align*}
and
\begin{align*}
\left(  L_{a}\circ d\right)  ^{2}\left(  w\right)    & =b_{\left\{  \left\{
1,2\right\}  \right\}  }\cdot a^{1-1}\cdot d^{1-1}\left(  w\right)
+b_{\left\{  \left\{  1\right\}  ,\left\{  2\right\}  \right\}  }\cdot
a^{2-1}\cdot d^{2-1}\left(  w\right)  \\
& =\left(  d\circ L_{a}\right)  \left(  1\right)  +ad\left(  w\right)  ;
\end{align*}
both are easily verified.
The right hand sides of the formulas
\eqref{darij1.eq.damw} and \eqref{darij1.eq.admw} can be rewritten in terms of
exponential Bell
polynomials.
As a result, the formulas take the forms (I hope I got them right)
\begin{equation}
\left(  d\circ L_{a}\right)  ^{m}\left(  w\right)  =\sum_{k=0}^{m}
B_{m+1,k+1}\left(  u_{0},u_{1},u_{2},\ldots\right)  \cdot a^{k}\cdot
d^{k}\left(  w\right)  \nonumber
\end{equation}
and
\begin{align*}
\left(  L_{a}\circ d\right)  ^{m}\left(  w\right)  =\sum_{k=0}^{m}
B_{m,k}\left(  u_{0},u_{1},u_{2},\ldots\right)  \cdot a^{k}\cdot d^{k}\left(
w\right)  ,
\end{align*}
where
\begin{align*}
u_{i}:=\left(  d\circ L_{a}\right)  ^{i}\left(  1\right)
\ \ \ \ \ \ \ \ \ \ \text{for each integer }i\geq0.
\end{align*}
Proving the formulas is fairly easy. You can show \eqref{darij1.eq.damw} by
induction on $m$ (using the Leibniz rule and the easy observation that a set
partition $P\in\operatorname*{SP}\left(  m\right)  $ with $\ell\left(
P\right)  =k$ can be obtained from exactly $k+1$ different set partitions
$P^{\prime}\in\operatorname*{SP}\left(  m+1\right)  $ by removing the element
$m+1$). Then, \eqref{darij1.eq.admw} follows by applying
\eqref{darij1.eq.damw} to $m-1$ and $d\left(  w\right)  $ instead of $m$ and
$w$ (and multiplying the result by $a$). There are probably some more
essential ways to prove the formulas -- in particular, the uncanny similarity
to Faa di Bruno's formula cries out for an explanation -- but I am happy
enough with the induction.
To see why this all is related to the OP, let $R$ be the ring of functions,
and let $a=f\left(  x\right)  $ and $d=\dfrac{d}{dx}$. Not sure if the results
are of much use, though.
Now, what is the connection to Hochschild? Let $p$ be a prime number, and let
$w\in R$ be such that $pw=0$. (Usually, one just considers the case when $p=0$
in $R$, but all we need is $pw=0$.) Applying \eqref{darij1.eq.damw} to
$m=p-1$, we obtain
\begin{equation}
\left(  d\circ L_{a}\right)  ^{p-1}\left(  w\right)  =\sum_{P\in
\operatorname*{SP}\left(  p\right)  }b_{P}\cdot a^{\ell\left(  P\right)
-1}\cdot d^{\ell\left(  P\right)  -1}\left(  w\right)  .
\label{darij1.eq.p-1}
\tag{3}
\end{equation}
However, the cyclic group $\mathbb{Z}/p$ acts on the set $\operatorname*{SP}
\left(  p\right)  $ (by cyclically rotating each number in each block -- i.e.,
the element $\overline{1}\in\mathbb{Z}/p$ sends a set partition $\left\{
\left\{  a_{1},a_{2},\ldots,a_{k}\right\}  ,\left\{  b_{1},b_{2}
,\ldots,b_{\ell}\right\}  ,\ldots,\left\{  g_{1},g_{2},\ldots,g_{r}\right\}
\right\}  $ to $\left\{  \left\{  a_{1}+1,a_{2}+1,\ldots,a_{k}+1\right\}
,\left\{  b_{1}+1,b_{2}+1,\ldots,b_{\ell}+1\right\}  ,\ldots,\left\{
g_{1}+1,g_{2}+1,\ldots,g_{r}+1\right\}  \right\}  $, where addition happens
modulo $p$). This action splits $\operatorname*{SP}\left(  p\right)  $ into
orbits of sizes $1$ and $p$. Each orbit of size $p$ contributes a total of $0$
to the right hand side of \eqref{darij1.eq.p-1}, because the $p$ addends
corresponding to the entries of this orbit are all equal and contain a $w$ (so
summing them $p$ times yields $0$ since $pw=0$). What remains are the addends
corresponding to the orbits of size $1$. These orbits are the ones of the set
partitions $\left\{  \left\{  1,2,\ldots,p\right\}  \right\}  $ and $\left\{
\left\{  1\right\}  ,\left\{  2\right\}  ,\ldots,\left\{  p\right\}  \right\}
$. The corresponding addends are $\left(  d\circ L_{a}\right)  ^{p-1}\left(
1\right)  \cdot a^{1-1}\cdot d^{1-1}\left(  w\right)  =\left(  d\circ
L_{a}\right)  ^{p-1}\left(  1\right)  \cdot w$ and $\underbrace{\left(
\left(  d\circ L_{a}\right)  ^{1-1}\right)  ^{p}}_{=1^{p}=1}\cdot a^{p-1}\cdot
d^{p-1}\left(  w\right)  =a^{p-1}\cdot d^{p-1}\left(  w\right)  $. Thus,
\eqref{darij1.eq.p-1} simplifies to
\begin{equation}
\left(  d\circ L_{a}\right)  ^{p-1}\left(  w\right)  =\left(  d\circ
L_{a}\right)  ^{p-1}\left(  1\right)  \cdot w+a^{p-1}\cdot d^{p-1}\left(
w\right)  .
\label{darij1.eq.hoch1}
\tag{4}
\end{equation}
If we apply this to $d\left(  w\right)  $ instead of $w$, and multiply the
result by $a$, we obtain
\begin{equation}
\left(  L_{a}\circ d\right)  ^{p}\left(  w\right)  =\left(  d\circ
L_{a}\right)  ^{p-1}\left(  1\right)  \cdot ad\left(  w\right)  +a^{p}\cdot
d^{p}\left(  w\right)  .
\label{darij1.eq.hoch2}
\tag{5}
\end{equation}
In particular, if $p=0$ in $R$, then we thus have
\begin{align*}
\left(  L_{a}\circ d\right)  ^{p}  & =\underbrace{\left(  d\circ L_{a}\right)
^{p-1}\left(  1\right)  \cdot a}_{=\left(  L_{a}\circ d\right)  ^{p-1}\left(
a\right)  }d+a^{p}\cdot d^{p}\\
& =\left(  L_{a}\circ d\right)  ^{p-1}\left(  a\right)  \cdot d+a^{p}\cdot
d^{p}
\end{align*}
as maps $R\rightarrow R$. This is Proposition 1.1 in Andrzej Nowicki,
Integral derivations, Journal of Algebra
110(1):262-276 (another scan at
https://doi.org/10.1016/0021-8693(87)90045-7 ), where it is credited to
Hochschild. Something similar does indeed appear as Lemma 1 in a 1955 paper
by
Hochschild,
with a vague sketch of a proof I never decided whether to trust; the modern
statement seems to have been first made by Cartier in Questions de
rationalité des diviseurs en géométrie
algébrique
(see (43) therein). I am not sure if a complete proof without holes has ever
appeared in the literature.
Note that this is not the same as that other Hochschild
identity, which is (42) in Cartier's op.cit. and
doesn't seem to appear in Hochschild's work at all. But the latter identity,
too, can be proved combinatorially using a $\mathbb{Z}/p$-action :)