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.
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).