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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. 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.

A rural reader
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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

1 Answers1

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Let $$ f(z)=e_q(z)-1=\sum_{n\ge 1}\frac{z^n}{[n]!} $$ Then $f(0)=0$ and $f'(0)=1$, so $f$ has a unique formal inverse satisfying $$ f(g(x))=x $$ By definition this inverse is $g(x)=\ln_q(1+x)$, since $$ e_q(g(x))=1+f(g(x))=1+x $$

Write $f(z)=z\phi(z)$ where $$ \phi(z)=\frac{f(z)}{z}=\frac{e_q(z)-1}{z} $$ Since $f(z)=z+O(z^2)$, we have $\phi(0)=1$ and $\phi$ is invertible in $\Bbb C(q)[[z]]$. The identity $f(g(x))=x$ becomes $$ x=g(x)\phi(g(x)). $$ Dividing by $\phi(g(x))$ gives $$ g(x)=x\,\psi(g(x)) $$ where $$ \psi(z)=\frac{1}{\phi(z)}=\frac{z}{e_q(z)-1} $$ Note that $\psi(0)=1\neq 0$.

Now use the Lagrange inversion: if $g(x)$ satisfies $g(x)=x\,\psi(g(x))$ and $\psi(0)\neq 0$, then for every $n\ge 1$ the coefficient of $x^n$ in $g(x)$ equals $ \frac{1}{n}$ times the coefficient of $z^{n-1}$ in $\psi(z)^n$.

Applying this with $\psi(z)=z/(e_q(z)-1)$ gives, for $n\ge 1$,

$$ \ln_q(1+x)=\sum_{n=1}^\infty \frac{x^n}{n} \left(\text{coefficient of }z^{n-1}\text{ in } \left(\frac{z}{e_q(z)-1}\right)^n\right) $$

Update:

Explicitly, writing $$ \ln_q(1+x)=\sum_{n\ge1}c_n x^n, $$ then we have $$ \boxed{ c_n=\frac1n\sum_{k=0}^{n-1}(-1)^k\binom{n+k-1}{k}\, B_{n-1,k}\!\left(\frac1{[2]!},\frac1{[3]!},\dots,\frac1{[n]!}\right) } $$ for $n \ge 1$, where $B_{n,k}(a_1, a_2, ...,)$ is the ordinary Bell polynomial.

As far as I'm aware, this is not a $q$-integral. It is unlikely to expect it to be as it is an inversion of a series, but that does not rule out the possibility of course.

Mako
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  • How is this answering the questions in the body of the post? – Pedro Dec 29 '25 at 01:08
  • @Pedro this shows what the coefficients are, answering "Curiously, this seems to be in no literature at all. Are these coefficients a special kind of polynomial? Is $\ln_q$ the $q$-integral of something?..." The edit/update shows an explicit formula – Mako Dec 29 '25 at 10:19
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    +1 The formula for the coefficients is definitely something that I don't see in the original post. ;) – Pedro Dec 29 '25 at 23:22