9
$\begingroup$

Let $(q;\,q)_n$ denote the $q$-Pochhammer symbol: $$(q;\,q)_n = \prod_{k=1}^n (1 - q^k), \quad(q;\,q)_0 = 1.\tag1$$ Consider a formal power series in $z$: $$f(z) = \sum_{n=1}^\infty \frac{(-1)^{n+1}P_n(q)}{n!\,(q;\,q)_{n-1}}z^n,\tag2$$ where $P_n(q)$ are some (yet unknown) polynomials in $q$: $$P_n(q) = \sum_{k=0}^{m} c_{n,k} \, q^k,\tag3$$ where $m=\binom{n-1}2 = \frac{(n-1)(n-2)}2$ and $c_{n,k}$ are some integer coefficients.

Suppose the formal power series $f(z)$ satisfies the functional equation $$\exp(f(z)) = 1 + f(q\,z)/q.\tag4$$ Expanding the left-hand side of $(4)$ in powers of $z$ using the exponential partial Bell polynomials, and comparing coefficients at corresponding powers of $z$ at both sides, we can obtain a system of equations, by solving which we can find the coefficients of the polynomials $P_n(q)$: $$ \begin{align} P_1(q) &= 1\\ P_2(q) &= 1\\ P_3(q) &= 2 + q\\ P_4(q) &= 6 + 6 q + 5 q^2 + q^3\\ P_5(q) &= 24 + 36 q + 46 q^2 + 40 q^3 + 24 q^4 + 9 q^5 + q^6\\ \dots \end{align}\tag5 $$ This is a quite slow process, even when done on a computer. I computed the polynomials up to $n=27$ (they can be found here) using a Mathematica program that can be found here.

There are some patterns in the coefficients I computed (so far they are just conjectures): $$ \begin{align} c_{n,0} &= (n-1)!&\vphantom{\Huge|}\\ c_{n,1} &= \frac{(n-2)(n-1)!}2, &n\ge2\\ c_{n,2} &= \frac{(3n+8)(n-3)(n-1)!}{24}, &n\ge3\\ c_{n,3} &= \frac{(n^2 + 5 n - 34)\,n!}{48}, & n\ge4 \end{align} \tag6 $$ and $$ \begin{align} c_{n,m} &= 1&\vphantom{\Huge|}\\ c_{n,m-1} &= \frac{(n+1)(n-2)}2, &n\ge2\\ c_{n,m-2} &= \frac{(3 n^3 - 5 n^2 + 6 n + 8)(n-3)}{24}, &n\ge3\\ c_{n,m-3} &= \frac{(n^4 - 10 n^3 + 43 n^2 - 74 n + 16) (n - 1) \, n}{48}, &n\ge4 \end{align} \tag7 $$ where $m=\binom{n-1}2$. Other coefficients seem to follow more complicated patterns. We can also observe that $$ \begin{align} P_n(1) &= \frac{(n-1)!\,n!}{2^{n-1}}\\ P_{2n}(-1) &= \frac{(2n-1)!\,n!}{3^{n-1}}\\ P_{2n-1}(-1) &= \frac{(2n-1)!!\,(2n-2)!}{6^{n-1}}, \end{align}\tag8 $$ where $n!!$ denotes the double factorial.

I am trying to find a more direct formula for the polynomials $P_n(q)$ or their coefficients $c_{n,k}$ (possibly, containing finite products and sums, but not requiring to solve equations).

$\endgroup$
  • $\begingroup$ Vladimir, I have a initial discussion of this at pg 24 ff in go.helms-net.de/math/tetdocs/ContinuousfunctionalIteration.pdf and more to the point in go.helms-net.de/math/tetdocs/APT.htm where the coefficients are computed as terms of the Schröderfunction by diagonalization of the Carlemanmatrix for t^x-1 . The computation up to 32 is fast by that eigensystem-solver and even up to 64 is doable in seconds using Pari/GP. You can get the Pari/GP-code for the solver. $\endgroup$ – Gottfried Helms Nov 6 '16 at 8:10
  • 1
    $\begingroup$ A quation in MO asks for an elegant, nonrecursive formula for this coefficients: mathoverflow.net/questions/57627/… $\endgroup$ – Gottfried Helms Nov 6 '16 at 19:46
  • $\begingroup$ @GottfriedHelms Thanks. It looks that my question is an exact duplicate of that MO question. $\endgroup$ – Vladimir Reshetnikov Nov 6 '16 at 19:54
  • $\begingroup$ P.s. if you like you can get the $P_k(q)$ up to $k=63$ from a readable Pari/GP file (approx. 7MB) $\endgroup$ – Gottfried Helms Nov 6 '16 at 20:41
5
$\begingroup$

Using the first recurrence relation here, we can find a recurrence for the polynomials $P_n(q)$: $$P_1(q) = 1, \quad P_n(q) = \sum_{k=1}^{n-1} {{n-1} \choose {k-1}} {{n-2} \brack {k-1}}_q P_k(q) \, P_{n-k}(q) \, q^{n-k-1},$$ where $n \choose k$ is the binomial coefficient, and ${n \brack k}_q$ is the $q$-binomial coefficient (also known as the Gaussian binomial coefficient). A Mathematica program that computes them using this recurrence can be found here.

If we introduce a notation for the coefficients of the formal power series $f(z)$, that are rational functions of $q$: $$f(z) = \sum_{n=1}^\infty Q_n(q)\,z^n, \quad Q_n(q) = \frac{(-1)^{n+1}P_n(q)}{n!\,(q;\,q)_{n-1}},$$ then we can have a simpler recurrence for them: $$Q_1(q) = 1, \quad Q_n(q) = \frac1{n \, (1-q^{1-n})}\sum_{k=1}^{n-1} k \, q^{-k} \, Q_k(q) \, Q_{n-k}(q).$$ It would be nice to find a more direct, non-recurrent formula for them.

$\endgroup$

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.