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


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 (Carleman-)matrix of Stirling numbers 2nd kind, factorially rescaled; let's call it $S$. I show here only the top left edge; but it is actually meant as of infinite size:

$\qquad $ (picture)

It is well known 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 $f_1(x)=\exp(x)-1$.

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 $ picture

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$


Now the view along rows, where my question is originated from.

I have found by pattern analyzing, that the rowwise generating functions 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 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 $, and 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 I'm wrong here)

My question:

Q: Is there a 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)


[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...

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