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The $q$-exponential function is given by the power series

$$ e_q(x) = \sum_{n=0}^\infty \frac{x^n}{[n]!} $$

using the $q$-integers $[k]:=q^{k-1}+\cdots+q+1=(q^k-1)/(q-1)$ and $q$-factorials $[n]!=[n][n-1]\cdots[2][1]$.

What is the inverse function $\ln_q$ satisfying $\ln_q(e_q(x))=x$ and $e_q(\ln_q(1+x))=1+x$?

I am treating $x$ and $q$ as either indeterminates for formal power series or complex variables in whatever domain yields an answer. By hand I calculated the first few terms

$$ \ln_q(1+x)=x-\frac{1}{[2]}x^2+ \left(\frac{2}{\,[2]^2}-\frac{1}{[3][2]}\right)x^3- \left(\frac{5}{\,[2]^3}-\frac{5}{[3][2]^2}+\frac{1}{[4][3][2]}\right)x^4+\cdots $$

A vast generalization might be to treat $[1],[2],[3],\cdots$ as independent formal variables and treat these coefficients above as rational functions of them. At worst, the series for $\ln_q$ is no simpler than this generalization.

Curiously, this seems to be in no literature at all. Like, are these coefficients a special kind of polynomial? Is $\ln_q$ the $q$-integral of something? Is there an analogous identity for the fact $e_q(x+y)=e_q(x)e_q(y)$ when $(xy)=q(yx)$? All mentions of $q$-logarithms in search results seem to be talking about Tsallis statistics, which is an unrelated situation with a unrelated notion of $q$-exponential.

anon
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  • Do you have any more terms? – Pedro Oct 04 '21 at 08:33
  • A way to approach this is as follows: consider combinatorial species but with structure groups $\mathrm{GL}(n,q)$ for fixed $q$ and varying in (i.e. the category of finite $q$-vector spaces and isomorphisms). There is an exponential species there, and there is a "character" functor that takes a $q$-species and sends it to a power series. Then the combinatorial interpretation of the $q$-logarithm species may be clear, and you can explain the coefficients you get combinatorially. – Pedro Oct 04 '21 at 08:34
  • In particular, I think you should be writing your series in $x^n/[n]!$ to see that you get an alternating sum of $q$-binomials, weighted by numbers appearing in https://oeis.org/A145882. – Pedro Oct 04 '21 at 08:47
  • Having said the above, I suspect an answer to your question may be found at https://www.sciencedirect.com/science/article/pii/0097316576900285. Essentially, your exponential corresponds to the function $f(x,y) = 1$ when $x\leqslant y$, whose inverse is the Mobius function of that poset, so what you want to do is Mobius inverstion, right? See Theorem 3.1 in that paper. – Pedro Oct 04 '21 at 08:51

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