Consider the function $\;\Phi(A)=\phi A\phi^{-1},\;$ where $\phi\::\:x^n\:\mapsto\:x(x-1)\cdots(x-n+1)$ and $A$ is an arbitrary linear operator over $\mathbb{C}[x]$.

It turns out that applying this to the derivative operator gives the forward difference operator: $\Phi(\mathcal{D})=\Delta$.

Furthermore, noticing that $\Phi(A^k)=(\Phi(A))^k$ for integer $k$, we see that $\Phi$ also maps the integral $\:\mathcal{D}^{-1}\:$ to the sum $\:\Delta^{-1}$.

Curious, I computed some values for $\Phi^{-1}(\mathcal{D})$ and ended up with what appears to be the sequence A238363, indicating that the operator $\:\Phi^{-1}(\mathcal{D})\:$ is in fact the commutator $[\ln\mathcal{D},\:\text{x}\mathcal{D}]$, where $\text{x}$ is the operator $\:\text{x}\::\:f(x)\:\mapsto\:xf(x)$.

So, since $\mathcal{D}=\Phi^{-1}(\Delta),$ I was wondering if $\mathcal{D}$ can be similarly realised as a commutator involving $\Delta$, e.g. $[\ln\Delta,\:\text{x}\Delta]$.

I also noticed that $\exp(\Phi^n(\mathcal{D}))\equiv\Phi^n(\mathcal{T})$, where $\mathcal{T}$ is the shift operator. So, I'm also wondering if this pattern continues in any meaningful way --- what do $e^\mathcal{T}$ and $\ln\mathcal{D}$ represent?

By multiplying $e^\Delta$ by $e$ we get the equation $e^\mathcal{T}=e\Phi(\mathcal{T})$, and the appearance of $\ln\mathcal{D}$ in the (hypothesised) commutator definition of $\Phi^{-1}(\mathcal{D})$ above surely means something.

I don't have a specific question and know next to nothing about this stuff; I'd just love to learn more about $\Phi^n$ - any insights or references would be more than appreciated.

So far the most relevant documentation I've found is Tom Copeland's works on the subject, namely his notebook "Goin' with the Flow", but it's quite dense and I'm struggling to wade through it.

Thanks in advance!

EDIT: $\phi$ and its inverse can be interpreted as matrices, $\phi$ populated by signed Stirling numbers of the first kind while $\phi^{-1}$ contains unsigned Stirling numbers of the second kind; i.e.

$$\phi=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 0 & 0 & 0 & \cdots\\ 0 & -1 & 1 & 0 & 0 & \cdots\\ 0 & 2 & -3 & 1 & 0 & \cdots\\ 0 & -6 & 11 & -6 & 1 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}\;\;\;\;\text{and}\;\;\;\; \phi^{-1}=\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 1 & 0 & 0 & \cdots\\ 0 & 1 & 3 & 1 & 0 & \cdots\\ 0 & 1 & 7 & 6 & 1 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$

For example, $\phi^{-1}(x^3)=x^3+3x^2+x$. Differentiating, $\mathcal{D}\phi^{-1}(x^3)=3x^2+6x+1$, and then applying $\phi$ to this gives $\Phi(\mathcal{D})(x^3)=3\phi(x^2)+6\phi(x)+1=3x^2+3x+1=\Delta(x^3)$.

If we had integrated instead of differentiating there, we would end up with a summation formula for $x^3$ instead.

I also made an introductory video on this topic.

EDIT 2: For what it's worth, I also just noticed that $\mathcal{D}\Phi(\mathcal{D}^{-1})(x^n)$ is the $n$th Bernoulli polynomial, which results in the equation $\mathcal{D}\Phi(\mathcal{D}^{-1})=\ln(\Delta+1)\Delta^{-1}$.

EDIT 3: I derived $\:\Phi(\mathcal{T})=\exp(\exp(\mathcal{D})-1)\:$ and $\:\mathcal{T}=\exp(\exp(\Phi^{-1}(\mathcal{D}))-1)$, so conjecture: $$\Phi^n(\mathcal{T})=\exp(\exp(\Phi^{n-1}(\mathcal{D}))-1).$$

EDIT 4: proven edit 3 and would love to know what I just did.

  • $\begingroup$ Very interesting, but I am unsure to understand what you mean by $\;\Phi(A)=\phi A\phi^{-1}\;$. In particular, what is $\phi^{-1}$ ? Are you thinking in terms of (infinite) matrices? Could you provide an example ? $\endgroup$
    – Jean Marie
    Sep 14, 2022 at 5:27
  • $\begingroup$ I updated the post, please tell me if I should add anything else! $\endgroup$
    – Supware
    Sep 14, 2022 at 8:46
  • $\begingroup$ Very satisfying explanations ! Thanks ! $\endgroup$
    – Jean Marie
    Sep 14, 2022 at 10:47

2 Answers 2


I've been playing with this thing a lot since posting this question, and have found enough to consider it solved: $\Phi$ is a linear operator over the space of linear operators over $\mathbb{C}[x]$, which is closely related to exponentiation when applied to differential operators.

$\Phi$ inherits linearity from $\phi$, and we also see that for $k\in\mathbb{N}$ $$(\Phi^n(A))^k=\underbrace{(\phi^nA\phi^{-n})(\phi^nA\phi^{-n})\cdots(\phi^nA\phi^{-n})}_{k\;\;\text{terms}}=\phi^nA^k\phi^{-n}=\Phi^n(A^k).$$ With this we also see that for $$f(A)=\sum_{k=0}^\infty a_kA^k$$ we have $$\Phi^n(f(A))=\sum_{k=0}^\infty a_k\Phi^n(A^k)=\sum_{k=0}^\infty a_k(\Phi^n(A))^k=f(\Phi^n(A))$$ e.g. $\Phi(\sin\mathcal{D})=\sin\Delta$.

Therefore, since $\Phi(\mathcal{D})=\Delta=\exp(\mathcal{D})-1$, we have $$\Phi^{n+1}(\mathcal{D})=\Phi^n(\exp(\mathcal{D}))-1=\exp(\Phi^n(\mathcal{D}))-1$$ hence for any differential operator $D$ of the form $\Phi^n(\mathcal{D})$ we get the equation $$\exp(D)=\Phi(D)+1.$$ Furthermore, by exponentiating both sides we also get $$\exp(T)=e\Phi(T)$$ where $T=\exp(D)$.

I think this is interesting enough to continue exploring, and I'd still really appreciate any relevant references if you know of any, but these properties are enough to describe what $\Phi$ "is" (which is all the question was asking for).


Just came across this question. I'm the author of both the OEIS entry A238363 and the pdf Goin’ with the Flow: Logarithm of the Derivative, cited by the OP, posted at my blog Shadows of Simplicity along with an erratum. Numerous posts on the Stirling polynomials of the first, $ST1_n(x) = x(x-1)\cdots(x-n+1$), and second kinds and the Bernoulli polynomials and several associated diff ops are available at my blog, began in 2008, as well as contributions in the OEIS, began in 2007.

Reserve the symbol $D_u$ for the partial derivative $\frac{\partial}{\partial u}$ and $i,j,k,n = 0,1,2,3,\cdots \; .$

First, the OP asks what the meaning of $\ln(D)$ is in my interpretation related to the commutator. This is briefly noted in the cited pdf which links to the MO-Q "Riemann zeta function at positive integers and an Appell sequence of polynomials related to fractional calculus" which in turn links to the MSE-Q "[Lie group heuristics for a raising operator for $(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}][4]$". The pdf and the questions clearly state that $\ln(D)$ is to be interpreted as a raising operator for a specific set of Appell Sheffer polynomials related to fractional integro-derivatives (FIDs), and, therefore, it can be interpreted as the infinitesimal generator of the (FIDs). Explicit integral and diff op reps of $R = \ln(D)$ are given in the questions along with their action. These are re-iterated and more examples of their action given in my answer to the two-year-old MO question "What's the matrix of logarithm of derivative operator, ln(D)? What is the role of this operator in various math fields?". (This log of the derivative operative should not be conflated with another interpretation by other authors, such as Khesin, based on pseudo-differential operators and their symbols.) So, with a little bit of diligence, it isn't so hard to find other lit on this topic, especially if searching with my name attached.

Second, note that the OP's statement in the third sentence that

$$\Phi(D_x^{-1})= \:\Delta^{-1} \; $$

is ill-defined. For example,

$$\Phi(D_x^{-1})(1) = \phi \; D_x^{-1} \; \phi^{-1} \; (1) = \phi \; D_x^{-1} \; 1 $$

$$= \phi \; \int^x_c dt = \phi \; (x-c) = ST1_1(x) - c = x - c ,$$

where $ST1_n(x) = n! \binom{x}{n} = x(x-1)\cdots(x-n+1)$ are the Stirling polynomials of the first kind. Depending on the definition of $D^{-1}$, $c$ can be any number, finite or infinite. Later the OP takes the derivative $D_x\Phi(D_x^{-1})$ which eliminates this ambiguity in this instance.

In the following, I'll be give both a more accurate and general analysis. I'll show that not only for the two Stirling sequences but for any pair of umbral inverse Appell Sheffer sequences, the operator

$$\phi\; D_x \; \phi^{-1}$$

is a lowering / annihilation / destruction operator for the member of the pair associated with the substitution op $\phi$.

In fact, this generalizes to any pair of umbral inverse Sheffer polynomial sequences. See my pdf at "The umbral compositional inverse of a Sheffer polynomial sequence and its lowering and raising operators".

I've explored these and related topics over the last sixteen years and posted notes at my blog, in MO and MSE Q&As, and in the OEIS. The general subject of the umbral Sheffer operational calculus is broad, deep, and elegant, touching areas in all facets of mathematics and physics and I feel I've barely scratched the surface.

Umbral operational analysis:

A binomial Sheffer sequence of polynomials has an e.g.f. of the form

$$e^{B.(x)t} = e^{h(t)x}$$

with $h(z)$ a function analytic at the origin or formal Taylor series with $h(0)=0$ and $D_{z=0}h(z)\neq 0$. The period as a subscript flags an umbral quantity, which, once an expression is reduced to a power / Taylor series in the umbral character, can be evaluated via lowering the exponent to the subscript position; e.g.,

$$(B.(x))^n = B_n(x) = \sum_{k=0}^n b_{n,k} x^k \; .$$

(See this MSE answer for more on umbral machinations.)

Due to the conditions above, $h(z)$ has a (at least a formal) compositional inverse $h^{(-1)}(z)$ about the origin from which a dual binomial Sheffer sequence $\bar{B}_n(x)$ can be formed with the e.g.f.

$$e^{\bar{B}.(x)t}= e^{xh^{(-1)}(t)}.$$

The two are an umbral inverse pair of binomial Sheffer sequences under umbral composition; i.e.,

$$B_n(\bar{B}.(x)) = x^n = \bar{B}_n(B.(x)),$$

where umbral composition is defined by, e.g.,

$$B_n(\bar{B}.(x)) = \sum_{n=0}b_{n,k} (\bar{B}.(x))^k =\sum_{n=0} b_{n,k} \bar{B}_k(x) .$$

This identity follows from simple umbral manipulations of the e.g.f.s;

$$e^{B.(\bar{B}.(x)) t} = e^{\bar{B}.(x)h(t)} =e^{xh^{(-1)}(h(t))} = e^{xt}.$$

The inverse of the substitution operator $\phi\::\:x^n\:\mapsto\:B_n(x)$ is then the substitution operator $\phi^{-1}\::\:x^n\:\mapsto\:\bar{B}_n(x)$.

Then the op $\phi\; D_x \; \phi^{(-1)}$ can be identified as the lowering operator $L_{B}$ of $B_n(x)$ defined by

$$ L_{B} \; B_n(x) = n \; B_{n-1}(x)$$


$$\phi\; D_x \;\phi^{(-1)}\; B_n(x) = \phi\; D_x \; B_n(\bar{B}.(x)) = \phi\; D_x \; x^n $$

$$ = \phi \; n \;x^{n-1} = n \;(B.(x))^{n-1} = n \; B_{n-1}(x) = L_B \; B_n(x).$$

By the same line of reasoning, the lowering op for the umbral inverse sequence is

$$L_{\bar{B}} = \phi^{-1}\; D_x \; \phi \; .$$

The alternative rep of the lowering op as

$$L_B = h^{(-1)}(D_x)$$

can be obtained several ways. One simple, direct way is via the e.g.f. argument

$$\sum_{n \geq 0} L_{B}\; B_{n}(x)\frac{t^n}{n!} = L_B \; e^{tB.(x)} = L_{B}\; e^{xh(t)} = h^{(-1)}(D_x) \; e^{xh(t)} = h^{(-1)}(h(t)) \; e^{xh(t)}$$

$$ = t \; e^{xh(t)} = t \; e^{tB.(x)} = \sum_{n \geq 1} n\; B_{n-1}(x)\frac{t^n}{n!}.$$


$$\phi\; D_x \; \phi^{-1} \; = L_{B} = h^{(-1)}(D_x) $$


$$\phi^{-1}\; D_x \; \phi \; = L_{\bar{B}} = h(D_x) \;.$$

The Stirling polynomials of the second kind $ST2_n(x)$, a.k.a. the Bell polynomials, have the e.g.f.

$$e^{ST2.(x)t} = e^{x (e^t-1)}$$

and that for the Stirling polynomials of the first kind, a.k.a. the falling factorial polynomials, is

$$e^{ST1.(x)t} = e^{x \ln(1+t)} = (1+t)^x.$$

Since $h^{(-1)}(t) = e^t-1$ and $h(t) = \ln(1+t)$ comprise a compositional inverse pair, the Stirling polynomials of the first and second kinds are an umbral inverse pair of binomial Sheffer sequences, so the identities above apply in agreement with some in the question.

More generally for an analytic function or formal Taylor series $F(z) = e^{a.z}$, from the conjugation property,

$$\phi \; F(D_z) \; \phi^{-1} = \phi \; e^{a.D_z} \; \phi^{-1} = e^{a.\phi\;D_z \phi^{-1}} = F(h^{(-1)}(D_z)) $$

and, of course,

$$\phi^{-1} \; F(D_z) \; \phi = F(h(D_z)) \; .$$

In particular, with

$$F(D_z) = \frac{\ln(1+D_z)}{D_z}$$

and the specialization $h^{(-1)}(t) = e^t-1$, then

$$\phi \; \frac{\ln(1+D_z)}{D_z}\; \phi^{-1} = \frac{D_z}{e^z-1} = e^{ber.D_z},$$

the Todd operator, where $ber_n$ are the famous, classical Bernoulli numbers, the moments of the the Appell Sheffer sequence of Bernoulli polynomials defined by the e.g.f.

$$e^{Ber.(x)t} = e^{(ber.+x)t} = e^{ber.t}e^{xt} = \frac{t}{e^t-1} e^{xt}\; .$$

The Todd op is the substitution op for the Bernoulli polynomials; i.e.,

$$\frac{D_z}{e^{D_z}-1}\; z^n = e^{ber.D_z} \; z^n = (ber.+z)^n = Ber_n(z)$$

In the same way,

$$\frac{\ln(1+D_z)}{D_z}\; z^n = (r.+z)^n = R_n(z) $$

illustrates the sub op for the row polynomials of the Appell Sheffer integer reciprocal polynomials with the moments $r_n = (-1)^n \frac{1}{n+1}$ and e.g.f.

$$e^{R.(x)t} = \frac{\ln(1+t)}{t}e^{xt} = 1 + \frac{2x - 1}{2} t + \frac{3 x^2 - 3 x + 2}{3} \frac{t^2}{2!} + \frac{4 x^3 - 6 x^2 + 8 x - 6}{4} \frac{t^3}{3!} + \cdots,$$

which are the shifted row polynomials of A238363 divided by $(n+1)$.

Then the op identity above is manifested as the double umbral composition

$$Ber_n(z) = B_n(R.(\bar{B}.(x))) = ST2_n(R.(ST1.(x))) \;.$$

This double umbral composition for the Bernoulli polynomials was noted several years ago in A238363 and also since derived various other ways in several old sets of notes of mine archived at my math blog.

Conversely, naturally,

$$ST1_n(Ber.(ST2.(x))) = R_n(x).$$

See the last section below for a little more info on Appell Sheffer sequences.

The function $\frac{\ln(1+u)}{u}$ that occurs in the op $\frac{\ln(1+D_z)}{D_z}\; z^n = (r.+z)^n = R_n(z)$ is also related to the Bernoulli numbers in my response to the MO-Q "Eulerian number identity"; explicitly,

$$\frac{\ln(1+u)}{u} = \frac{1}{1+c.u} = (1+u)^{-ber.-1},$$


$$\frac{\ln(1+D_z)}{D_z} = \frac{1}{1+c.D_z} = (1+D_z)^{-ber.-1}$$

where $c_n = \frac{1}{1+n}$. This then leads to the relation between the reciprocals of the integers and the Bernoulli numbers via the Stirling polynomials as well. Other old sets of my notes show the relations between the Appell umbral inverse pair of integer reciprocal and Bernoulli polynomials and the binomial umbral inverse pair of Stirling polynomials of the first and second kinds in yet other ways.

Matrix Rep as lower triangular matrices (illustrated in my pdf):

Define the lower triangular matrix $[B]$ with the elements $b_{i,j}$ and similarly for $[\bar{B}]$. Then the umbral inverse identity manifests as

$$[B][\bar{B}]=[B][B]^{-1} = [\bar{B}]^{-1}[\bar{B}] = [I],$$

the identity matrix.

The umbral composition $w_m(x) = q_m(B.(x))$ of a generic polynomial $q_m(x) = \sum_{k=0}^m c_k x^k$ has the matrix rep

$$(w_0 \; w_1 \cdots w_m) = (c_0 \; c_1 \; \cdots c_m) [B] \;.$$

From OEIS A074909:

My Apr 25 2014 formulas note that

$[padded \; A074909] = [ST2]*[dP]*[ST1] = A048993*A132440*[padded \; A008275],$

where $[dP]$ is the matrix rep of the derivative operator in the power basis $x^n$. The matrix identity is the matrix rep (in right multiplication of a row vector ) of the op $L_B$. A relation to the derivatives of a function of an operator w.r.t. another operator is given in the OEIS entry as well as in my pdf.

My Nov 12 2014 formulas note a relation to the Bernoulli polynomials.

From my OEIS entry A238363:

My formulas dated Apr 26, 2014 (item A),

$[M] = padded \; A238363 = A238385-I = [ST1]*[dP]*[ST2] = [padded \; A008275]*A132440*A048993.$

A relation to the Bernoulli polynomials and, therefore, the Bernoulli numbers is given in my contribution dated Nov 6, 2016.

For exponentiation, item A) in my Apr 17 2014 formula is

$$P(x)= \exp(x*dP) = \exp[x*(e^M-I)]$$

where $P(x)$ is the lower triangular Pascal / binomial coefficient matrix A007318 with the n'th diagonal multiplied by $x^n$. With $x=1$, this reduces to

$$P(1)= \exp(dP) = \exp[(e^M-I)]\;.$$

Use the search line of the OEIS to find more notes related to A238363.

An alternative method of derivation of the lowering op:

The umbral substitution ops have diff op reps as

$$B_n(x) = \phi\; x^n = e^{B.(x) D_t}\; t^n \; |_{t=0} = e^{x \; h(D_t)}|_{t=0}\; t^n \;|_{t=0}$$

$$= e^{-(1-B.(x)):tD_t:}\; t^n \;|_{t=1} = (t-(1-B.(x))t)^n \; |_{t=1}= (1-(1-B.(x))^n = (B.(x)t)^n \;|_{t=1} = B_n(x) t^n \; |_{t=1} ,$$

where for any two ops $M$ and $N$ by definition $:MN:^k = M^kN^k$, a generalized normal-ordering operation. (The diff op reps evaluated at $t=1$ are not necessary for the following arguments, but they do often allow for analytic continuation of an umbral variable $q_n$ to $q_s$ by action on $x^s$, where $s$ is real, via a Newton series, the binomial expansion of $(1-(1-q.))^s$ or via a formally equivalent Mellin transform interpolation with a different domain of convergence that can be analytically continued to a larger domain.)

Introducing the commutator $[M,\;N] = MN - NM$,

$$ \phi \; D_x \; \phi^{-1} = \phi \; [D_x, \;\phi^{-1}] + D_x \; .$$

Using the diff op rep of the substitution op, the commutator acting on a function $f(x)$ gives

$$[D_x, \;\phi^{-1}] \; f(x) = D_x e^{xh^{(-1)}(D_t)} \; f(t) |_{t=0} - e^{xh^{(-1)}(D_t)}\; D_t \;f(t) |_{t=0}$$

$$ = e^{xh^{(-1)}(D_t)} \;(h^{(-1)}(D_t) - D_t)\; f(t)\; |_{t=0} = \phi^{-1}\; (h^{(-1)}(D_x)-D_x)\, f(x),$$


$$ \phi \; D_x \; \phi^{-1} = \phi \; [D_x, \phi^{-1}] + D_x = h^{(-1)}(D_x) = L_B.$$

Compare this with the conjugation rep of the raising / creation op

$$R_{AS} = AS(D_x) \; x \; (AS(D_x))^{-1}$$

$$= e^{a.D_x} \; x \; (e^{a. D_x})^{-1} = e^{a.D_x} \; x \; e^{\bar{a}. D_x} $$

$$ = x + e^{a.D_x}[x,\;e^{\bar{a}.D_x}] = x + D_{t=D_x}\ln(AS(t))$$

for the Appell Sheffer polynomial $AS_n(x)$, defined by

$$R_{AS}\; AS_n(x)= AS_{n+1}(x)$$

where, with $a_0 =1$, the 'substitution op', or binomial diff op generator, is such that

$$AS_n(x) =AS(D_x)\; x^n = e^{a.D_x}\; x^n = (x+a.)^n,$$

and that for the umbral inverse sequence is such that

$$\bar{AS}_n(x) = \frac{1}{AS(D_x)}\; x^n = (e^{a.D_x})^{-1} \; x^n = e^{\bar{a}.D_x}\; x^n = (x+\bar{a}.)^n.$$

The e.g.f. for any Appell Sheffer sequence has the form

$$AS(x,t)= e^{AS.(x)t} = e^{(a.+x)t}= e^{a.t}e^{xt} = AS(t) e^{xt} ,$$

so $R_{AS} = D_{t = D_x} \ln(AS(x,t))$.

  • $\begingroup$ Related: mathoverflow.net/questions/396533/… $\endgroup$ Jul 8, 2023 at 5:41
  • $\begingroup$ Wow! This is an incredible answer; I'll have to spend some time to get through it. Thank you so much! In fact I tried to contact you directly about this when I posted the question, but you're apparently quite elusive haha. And apologies for such a late response! $\endgroup$
    – Supware
    Jan 8 at 16:39
  • $\begingroup$ A comment to an MSE or MO contribution or to a post at my Wordpress blog (link on my user profile) gets my attention. (There is a lot to assimilate.) $\endgroup$ Jan 8 at 18:39

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