I recently came across the following in something I'm working on, and I'd never seen it before. Consider
\begin{align*}
f_1(x) &= (1+x)^{1/1} \\\
f_2(x) &= (1+x)^{2/1} (1+2x)^{-1/2} \\\
f_3(x) &= (1+x)^{3/1} (1+2x)^{-3/2} (1+3x)^{1/3} \\\
f_4(x) &= (1+x)^{4/1} (1+2x)^{-6/2} (1+3x)^{4/3} (1+4x)^{-1/4} \\\
& \cdots \\\
f_n(x) &= \prod_{k=1}^n (1+kx)^{(-1)^{k-1}\binom{n}{k}/k}.
\end{align*}
Think of these as formal power series with coefficients in $\mathbb{Q}$. It turns out that these series approximate the Taylor expansion for the exponential, in the sense that
$$ f_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} +O(x^{n+1}).$$
(This isn't too hard to prove, using a logarithmic derivative:
\begin{align*}
1-f_n'(x)/f_n(x) &= \sum_{k=0}^n (-1)^k\binom{n}{k}(1+kx)^{-1}
\\\
&= \sum_j \biggl(\sum_k (-1)^k\binom{n}{k}k^j\biggr) (-x)^j.
\end{align*}
The coefficient of $x^j$ calculates the number of surjections $\{1,\dots,j\}\to \{1,\dots,n\}$ (up to a sign), and so is $0$ if $j<n$.)
My questions include (but are not limited to): Do these things have a name? Is there some kind of combinatorial interpretation of the $f_n(x)$ (e.g., as a generating function of some kind)? Is there a more direct proof that $f_n(x)$ agrees with $e^x$ to order $n+1$?
Best Answer
I'm going to switch a couple of signs and focus on $g_n(x)=\frac{1}{f_n(-x)}$ for a moment. It is an exponential generating function for the number of properly $n$-colored rooted monotonic planar forests of depth at most $1$.
I should clarify that by "$n$-colored" I mean that each edge is assigned one of given $n$ colors, and by "properly $n$-colored" I mean $n$-colored with the property that or any pair $i,j$, any edge with color $i$ is adjacent to some edge of color $j$. A monotonic rooted tree is a labelled tree for which the label of a vertex is greater than the labels of its children. It is now clear that $g_n(x)$ agrees with $e^x$ up to order $n$, since the only $g$-structures of order up to $n$ are graphs with no edges (if there were any edges, "properly $n$ colored" needs at least $n+1$ vertices).
It remains to see why the generating function of these structures is what it is. Combinatorial species give a nice framework for interpreting combinatorially what different operations on the generating function mean at the level of objects being counted. Here is a sketch: