# A product approximation to the Taylor series of the exponential

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 $n+1$?

• The vanishing of the inner sum $\sum_k (-1)^k\binom{n}{k}k^j$ for $0 \leq j < n$ can also be seen by interpreting it as the $n$-th finite difference of a polynomial of degree less than $n$. – Noam D. Elkies Nov 21 '11 at 2:29

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:

Exponential Generating function for labelled monotonic planar rooted trees of depth at most 1 is $$\log(1-x)$$.

Therefore the EGF for $$k$$-colored labelled monotonic planar rooted trees of depth at most 1 is $$\frac{1}{k}\log(1-kx)$$.

The binomial tranform of order $$n$$ changes "$$k$$-colored" to "properly $$n$$-colored".

The exponential formula changes "tree" to "forest".

So this all gives $$g_n(x)=\exp\left(\sum_{k=1}^n (-1)^k\binom{n}{k}\frac{1}{k}\log(1-kx)\right)$$

• Nice! "Forests of depth at most one" is a bit of an awkward idea for me; could we re-express this as counting some kind of partitions of something? – Charles Rezk Nov 22 '11 at 22:28

See A019538, A049019, and A133314 for extensive formula for the coefficients and relation to face vectors of permutahedra/permutohedra. Your proof is a nice one in that A133314 clearly shows why the lower order coefficients as given by the finite differences must vanish if $1/f_n(x)$ truly is an approximation of $exp(-x)$.