Remark: I give much background because it might significantly help to find an idea how to generate a solution

I'm analyzing properties of a certain infinite matrix $U$, for whose columns we have the generating functions $$f_c(x) = (\exp(x)-1)^c \qquad \qquad c=-\infty...+\infty $$, assuming $c$ from $-\infty$ to $+\infty$ (This is the type of Carleman-matrices extended to two-way-infinite size which Eri Jabotinsky has discussed in one article I've found in some archive...)
Clearly the Stirling numbers 2nd kind are involved in the part of nonnegative column-indexes, but in the columns with negative index we have Zeta-values (or: Bernoulli-numbers) due to the generating functions having $\exp(x)-1$ to the negative power.

I succeded with finding that the generating functions for the rows are $$ g_r(x) = {1 \over 1+x}{1 \over \log(1+1/x)^{1+r}} \qquad \qquad c=-\infty...+\infty$$ for rowindex r from $-\infty$ to $+\infty$ .
(We can find the coefficients easier when we write $$ g^*_r(t)={t \over 1+t}{1 \over \log(1+t)^{1+r}} $$ get the Laurent-series for this and replace the powers of t by the negative powers of x .)

Now, because I'm looking at iterations of the exponential function I've the matrix $U$ to the second power, $U^2$ which has the following entries around its zero-indexes: matrix

Because the 2nd power of the matrix gives the second iterate of the function in the columns, I know that the generating functions of the columns are now $ f_{2,c}(x) = (\exp(\exp(x)-1)-1)^c$

Q: My problem is now to find the generating functions of the rows. Simply writing the iterate of the rows-functions of the original version of $U$ seems to lead to nothing when I let Pari/GP do the expression. Does someone see how I could express them given the beginning of the sequence of coefficients? For instance, at row $r=0$ whe have $$g_{2,0}(x) = 1 -1/x + 4/3/x^2-15/8/x^3+122/45/x^4 + \cdots = ??? $$ or $$g^*_{2,0}(t) = 1 -1t + 4/3t^2-15/8t^3+122/45t^4 + \cdots = ??? $$
A meaningful suggestion for only one of the rows might already be helpful because the overall system should have a simple systematic derived from one key-idea as it is also for the non-iterated case.


Ahh, this looks very good...

By some trial & error I got the following scheme, which extends also nicely to higher powers of $U$ (or: iterates of $\exp(x)-1$). I got the following four families of generating functions for iteration-counts $h=0,1,2,3$ depending on rowindex $r$: $$ \small \begin{array} {} U^0:& g^*_{0,r}(t)&=& t \cdot{1 \over t^{1+r} } \qquad \qquad \text{note: $U^0$ is the identity-matrix}\\ U^1:& g^*_{1,r}(t)&=& t \cdot {1\over 1+t}\cdot{1 \over \log(1+t)^{1+r} } \\ U^2:& g^*_{2,r}(t)&=&t \cdot {1\over 1+t}{1 \over 1+\log(1+t)}\cdot{1 \over \log(1+\log(1+t))^{1+r}} \\ U^3:& g^*_{3,r}(t)&=& t \cdot {1\over 1+t}{1 \over 1+\log(1+t)}{1 \over 1+\log(1+\log(1+t))}\cdot {1 \over \log(1+\log(1+\log(1+t)))^{1+r}} \\ \end{array} $$ and I assume that this is the general pattern also for the higher families.

Denoting $\Lambda(t)=\log(1+t)$ and for the iterations $\Lambda^{°h}(t)$ where $\Lambda^{°0}(t)=t$ we seem to have $$ g^*_{h,r}(t) = \left(\prod_{k=0}^{h-1} {1 \over 1+\Lambda^{°k}(t)} \right) \cdot {t \over \Lambda^{°h}(t)^{1+r}} $$

Still open: an analytical justification/confirmation for this heuristic...
Update 10. Jun 2015: In this answer there is an analytical derivation valid for more general cases (functions of the type $f(x) = x + O(x^2)$ )


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.