A product approximation to the Taylor series of the exponential - MathOverflow

A product approximation to the Taylor series of the exponential

2011-11-21T01:18:55Z

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 $n+1$?

Answer by Gjergji Zaimi for A product approximation to the Taylor series of the exponential

2011-11-22T11:41:02Z

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

Answer by Tom Copeland for A product approximation to the Taylor series of the exponential

2011-12-02T16:31:40Z

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 $exp(-x)$.