Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series
$$ A(x) = \frac{x}{1 – e^{-x}} = \sum_{m=0}^\infty \left( -\sum_{n=1}^\infty \frac{(-x)^n}{(n+1)!} \right)^m = 1 + \frac12 x + \frac1{12}x^2 + 0x^3 – \frac1{720}x^4 + \dots $$
where I might have made an arithmetic error in expanding it out.
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Are all the coefficients egyptian, in the sense that they are given by $A^{(n)}(0)/n! = 1/N$ for $N$ an integer? The answer is no, unless I made an error, e.g. the third coefficient. But maybe every non-zero coefficient is egyptian?
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If all the coefficients were positive eqyptian, then the sequence of denominators might count something — one hopes that the $n$th element of any sequence of nonnegative integers counts the number of ways of putting some type of structure on an $n$-element set.
Of course, generating functions really come in two types: ordinary and exponential. The difference is whether you think of the coefficients as $\sum a_n x^n$ or as $\sum A^{(n)} x^n/n!$. If it makes more sense as an exponential generating function, that's cool too.
So my question really is: is there a way of computing the $n$th coefficient of $A(x)$, or equivalently of computing $A^{(n)}(0)/n!$, without expanding products of power series the long way?
Where you might have seen this series
Let $\xi,\psi$ be non-commuting variables over a field of characteristic $0$, and let $B(\xi,\psi) = \log(\exp \xi \exp \psi)$ be the Baker-Campbell-Hausdorff series. Fixing $\xi$ and thinking of this as a power series in $\psi$, it is given by
$$B(\xi,\psi) = \xi + A(\text{ad }\xi)(\psi) + O(\psi^2)$$
where $A$ is the series above, and $\text{ad }\xi$ is the linear operator given by the commutator: $(\text{ad }\xi)(\psi) = [\xi,\psi] = \xi\psi – \psi\xi$.
More generally, $B$ can be written entirely in terms of the commutator, and so makes sense as a $\mathfrak g$-valued power series on $\mathfrak g$ for any Lie algebra $\mathfrak g$. It converges in a neighborhood of $0$ when $\mathfrak g$ is finite-dimensional over $\mathbb R$, in which case $\mathfrak g$ is a (generally noncommutative) "partial group".
(More generally, you can consider the "formal group" of $\mathfrak g$. Namely, take the commutative ring $\mathcal P(\mathfrak g)$ of formal power series on $\mathfrak g$; then $B$ defines a non-cocommutative comultiplication, making $\mathcal P = \mathcal P(\mathfrak g)$ into a Hopf algebra. Or rather, $B(\mathcal P)$ does not land in the algebraic tensor product $\mathcal P \otimes \mathcal P$. Instead, $\mathcal P$ is cofiltered, in the sense that it is a limit $\dots \to \mathcal P_2 \to \mathcal P_1 \to \mathcal P_0 = 0$, where (over characteristic 0, anyway) $\mathcal P_n = \text{Poly}(\mathfrak g)/(\mathfrak g \text{Poly}(\mathfrak g))^n$, where $\text{Poly}(\mathfrak g)$ is the ring of polynomial functions on $\mathfrak g$, and $\mathfrak g \text{Poly}(\mathfrak g)$ is the ideal of functions vanishing at $0$. Then $B$ lands in the cofiltered tensor product, which is just what it sounds like. (In arbitrary characteristic, $\mathcal P$ is the cofiltered dual of the filtered Hopf algebra $\mathcal S \mathfrak g$, the symmetric algebra of $\mathfrak g$, filtered by degree.))
Why I care
When $\mathfrak g$ is finite-dimensional over $\mathbb R$, and $U$ is the open neighborhood of $0$ in which $B$ converges, then $\mathfrak g$ acts as left-invariant derivations on $U$, where by left-invariant I mean under the multiplication $B$. Hence there is a canonical identification of the universal enveloping algebra $\mathcal U\mathfrak g$ with the algebra of left-invariant differential operators on $U$. Since $\mathfrak g$ is in particular a vector space, the "symbol" map gives a canonical identification between the algebra of differential operators on $U$ and the algebra of functions on the cotangent bundle $T^{\ast} U$ that are polynomial (of uniformly bounded degree) in the cotangent directions. Left-invariance then means that the operators are uniquely determined by their restrictions to the fiber $T^{\ast}_0\mathfrak g = \mathfrak g^{\ast}$, and the space of polynomials on $\mathfrak g^{\ast}$ is canonically the symmetric algebra $\mathcal S \mathfrak g$. This gives a canonical PBW map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$, a fact I learned from J. Baez and J. Dolan.
(In the formal group language, the noncocommutative cofiltered Hopf algebra $\mathcal P(\mathfrak g)$ is precisely the cofiltered dual to the filtered algebra $\mathcal U\mathfrak g$, whereas with its cocommutative Hopf structure $\mathcal P(\mathfrak g)$ is dual to $\mathcal S \mathfrak g$. But as algebras these are the same, and unpacking the dualizations gives the PBW map $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$, and explains why it is actually an isomorphism of coalgebras.)
Anyway, in one direction, the isomorphism $\mathcal U\mathfrak g \cong \mathcal S \mathfrak g$ is easy. Namely, the map $\mathcal S \mathfrak g \to \mathcal U \mathfrak g$ is given on monomials by the "symmetrization map" $\xi_1\cdots \xi_n \mapsto \frac1{n!} \sum_{\sigma \in S_n} \prod_{k=1}^n \xi_{\sigma(k)}$, where $S_n$ is the symmetric group on $n$ letters, and the product is ordered. (In this direction, the isomorphism of coalgebras is obvious. In fact, the corresponding symmetrization map into the full tensor algebra is a coalgebra homomorphism.)
In the reverse direction, I can explain the map $\mathcal U \mathfrak g \to \mathcal S \mathfrak g$ as follows. On a monomial $\xi_1\cdots \xi_n$, it acts as follows. Draw $n$ dots on a line, and label them $\xi_1,\dots,\xi_n$. Draw arrows between the dots so that each arrow goes to the right (from a lower index to a higher index), and each dot has either 0 or 1 arrow out of it. At each dot, totally order the incoming arrows. Then for each such diagram, evaluate it as follows. What you want to do is collapse each arrow $\psi\to \phi$ into a dot labeled by $[\psi,\phi]$ at the spot that was $\phi$, but never collapse $\psi\to \phi$ unless $\psi$ has no incoming arrows, and if $\phi$ has multiple incoming arrows, collapse them following your chosen total ordering. So at the end of the day, you'll have some dots with no arrows left, each labeled by an element of $\mathfrak g$; multiply these elements together in $\mathcal S\mathfrak g$. Also, multiply each such element by a numerical coefficient as follows: for each dot in your original diagram, let $m$ be the number of incoming arrows, and multiply the final product by the $m$th coefficient of the power series $A(x)$. Sum over all diagrams.
Anyway, the previous paragraph is all well and cool, but it would be better if the numerical coefficient could be read more directly off the diagram somehow, without having to really think about the function $A(x)$.
Best Answer
Two people have pointed it out already, but somehow I can't resist: your formal power series is precisely the defining power series of the Bernoulli numbers:
http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function
Accordingly, they are far from Egyptian: as came up recently in response to the question
When does the zeta function take on integer values?
the odd-numbered terms (except the first) are all zero, whereas the even-numbered terms alternate in sign and grow rapidly in absolute value, so only finitely many are reciprocals of integers.
I find it curious that you are looking at this sequence from such a sophisticated perspective and didn't know its classical roots. I feel like there should be a lesson here, but I don't know exactly what it is. Here's a possibility: every young mathematician should learn some elementary number theory regardless of their primary interests. Comments?