Is there a systematic relation between the generating functions for the rows vs that for columns of infinite sized Carleman-matrices?

Question

(I've asked this in MSE but nobody had an idea since dec 14...)

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(Roughly related, but generalizing, of this earlier MSE question)

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Background: The first part of the following(the column-wise-focus) is also described in Eri Jabotinski's 1953-treatize Representation of functions by matrices (at jstor)

Consider the matrix of Stirling numbers 2nd kind, factorially rescaled in columns and rows; let's call it $S$. I show here only the top left edge; but it is actually meant as of infinite size:

$\qquad $

It is well known (see for instance Abramowitz&Stegun) that the generating function for the $c$'th column is $f_c(x)=(\exp(x)-1)^c $ , and for instance the leftmost column (index $c=0$) is related to $f_0(x)=(\exp(x)-1)^0 = 1 $ and the second column (index $c=1$ is related to the well known function $f_1(x)=\exp(x)-1$. So this matrix is also an example for (and in the form of) the (transposed) "Carleman"-matrices, and in this question I'm interested in a general property of such Carleman-matrices.

If I extend now that matrix by columns, for which the generating functions are accordingly $f_{-1}(x)=(\exp(x)-1)^{-1} $,$f_{-2}(x)=(\exp(x)-1)^{-2} $ and so on then I have not only to left-prepend new columns but also I must extend the matrix with prepended rows as well. The central segement of this now two-way infinite-indexed matrix, let's call it $S^*$ looks like this

$ \qquad $

Well, I'm having that the gf columnwise are $f_c(x)=(\exp(x)-1)^c$ with the column-index now from $-\infty$ to $ \infty$ . Matrices in this two-way-infinite form have been discussed by Eri Jabotinsky but I've not seen a discussion by him of my question so far.

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Now let's change our view to focus the rows instead.

I have found by pattern analyzing, that the rowwise generating functions (in this practical example) are $$g_r(t) = t/(1+t)/\log(1+t)^{r+1} $$ where I have now to replace $t =1/x$ to match the column-index for the exponents at $x$, so actually it is $$ h_r(x)= 1/(1+x)/\log(1+1/x)^{r+1} $$ The index $r=c=0$ is at the single $1$ in the center of the image, and $r=1$ indicates the row below, which reads, form right to left, $g_1(t)=1 -1/2t+5/12t^2-3/8t^3 ...$ and is also $h_1(x)=1 -1/2/x+5/12/x^2-3/8/x^3 ...$

I've also checked the similarly extended starred version of the matrix of Stirling numbers 1st kind, whose entries column-wise are generated by the functions $f_c(x)=\log(1+x)^c $.
Here for the row-wise generating functions I've guessed $$g_r(t)= t \exp(t) / (\exp(t)-1)^{r+1} $$ and $$h_r(x) = g_r(1/x) $$ (Correct me if my guess is wrong here)
The relation of the guessed generating functions of the rows and of the columns is somehow striking, and even might come out simple and possibly trivial. So my question:

Q: Is there any simple/memorizable rule for the relations of generating functions of the transposed Carleman matrices in comparable / general cases?
(Possibly this applies only to triangular Carlemanmatrices, but I don't know that)

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[update] A reference to a discussion of this might be sufficient; I think I've seen something like this several years ago but could not remember, where...

Answer 1

Assume that $f_c(x) = x^c q_c(x)$, where $q_c(0)\ne 0$. Then the generating function for the row $r$ in the left matrix is $$g_r(t)=\sum_{c=0}^{\infty} [x^{r+c}]\ q_{-c}(x)\cdot t^c,$$ where the operator $[x^d]$ extracts the coefficient of $x^d$ from the series.

Example 1. For $f_c(x)=(\exp(x)-1)^c$, we have $q_c(x) = \left( \frac{\exp(x)-1}{x} \right)^c$ with $q_c(0)=1$. So we need to find $$g_r(t)=\sum_{c=0}^{\infty} [x^{r+c}]\ \left( \frac{\exp(x)-1}{x} \right)^{-c}\cdot t^c,$$ which is done by Lagrange inversion (Lagrange–Bürmann formula).

To do so, denote $n=r+c$, $\Phi(x) = \frac{x}{\exp(x)-1}$, and $\Psi(x)=\left( \frac{\exp(x)-1}{x} \right)^r$. Then $$g_r(t)=\frac{1}{t^r}\sum_{n=0}^\infty [x^n]\ \Psi(x)\cdot \Phi(x)^n\cdot t^n.$$ By Lagrange inversion, we solve $w=t\cdot \Phi(w)$ to get $w=\log(1+t)$ and derive $$g_r(t) = \frac{1}{t^r}\cdot \Psi(w)\cdot \frac{\partial w}{\partial t}\cdot \frac{t}{w} = \frac{t}{(1+t)\cdot \log(1+t)^{r+1}}$$ as expected.

Example 2. For $f_c(x)=\log(1+x)^c$, we have $q_c(x) = \left( \frac{\log(1+x)}{x} \right)^c$ with $q_c(0)=1$. So we need to find $$g_r(t)=\sum_{c=0}^{\infty} [x^{r+c}]\ \left( \frac{\log(1+x)}{x} \right)^{-c}\cdot t^c,$$ which is again done by Lagrange inversion.

To do so, denote $n=r+c$, $\Phi(x) = \frac{x}{\log(1+x)}$, and $\Psi(x)=\left( \frac{\log(1+x)}{x} \right)^r$. Then $$g_r(t)=\frac{1}{t^r}\sum_{n=0}^\infty [x^n]\ \Psi(x)\cdot \Phi(x)^n\cdot t^n.$$ By Lagrange inversion, we solve $w=t\cdot \Phi(w)$ to get $w=\exp(t)-1$ and derive $$g_r(t) = \frac{1}{t^r}\cdot \Psi(w)\cdot \frac{\partial w}{\partial t}\cdot \frac{t}{w} = \frac{t\cdot \exp(t)}{(\exp(t)-1)^{r+1}}$$ as expected.

Example 3 (generalizing Examples 1 and 2).

Let $p(x)=x+O(x^2)$ and $f_c(x)=p(x)^c$ and thus $q_c(x)=\left(\frac{p(x)}{x}\right)^c$, $\Phi(x)=\frac{x}{p(x)}$, and $\Psi(x)=\left(\frac{p(x)}{x}\right)^r$.

By Lagrange inversion, we solve $w=t\cdot \Phi(w)$ to get $w=p^{-1}(t)$ and derive $$g_r(t) = \frac{1}{t^r}\cdot \Psi(w)\cdot \frac{\partial w}{\partial t}\cdot \frac{t}{w} = \frac{1}{t^r} \cdot \left(\frac{t}{p^{-1}(t)}\right)^r\cdot \frac{1}{p'(p^{-1}(t))}\cdot \frac{t}{p^{-1}(t)}=\frac{t}{p'(p^{-1}(t))\cdot p^{-1}(t)^{r+1}}.$$ E.g., in Example 1, we have $p(x)=\exp(x)-1$ and thus $p'(x)=\exp(x)$ and $p^{-1}(t)=\log(1+t)$. Plugging in the above formula, we get $$g_r(t) = \frac{t}{(1+t)\cdot \log(1+t)^{r+1}}$$ as expected.