# Inverse relationship between Stirling numbers of the first and second kind via generating functions

In combinatorics, a well-known result is that the matrix formed by the Stirling numbers of the second kind $$\left(S(n,k)\right)_{n,k\geq 0}$$ and the matrix of the signed Stirling numbers of the first kind $$\left((-1)^{n-k}c(n,k)\right)_{n,k\geq 0}$$ are inverses of each other. This result has many proofs, including bijective and generating function approaches.

My question concerns the interpretation of this relationship in terms of bivariate generating functions. Specifically, we know the generating functions:

$$S(u,z) := \sum_{n,k\geq 0} S(n,k) u^k \frac{z^n}{n!} = \exp(u(e^z-1)),$$ and $$s(u,z) := \sum_{n,k\geq 0} (-1)^{n-k} c(n,k) u^k \frac{z^n}{n!} = (1+z)^u.$$

Is it possible to directly derive the inverse relationship between $$\left(S(n,k)\right)_{n,k\geq 0}$$ and $$\left((-1)^{n-k}c(n,k)\right)_{n,k\geq 0}$$ from the relationship between $$S(u,z)$$ and $$s(u,z)$$? For example, could this be achieved via a differential equation or some other analytical method?

• Maybe look up the keyword "exponential Riordan array". Also somewhat related: mathoverflow.net/questions/418266 Commented Aug 11 at 16:02
• The inverse relationship is obvious if you use the generating functions $$\sum_k (-1)^{n-k} s(n,k)x^k = x(x+1)\cdots (x+n-1)$$ and $$\sum_{n\geq k}S(n,k)x^n = \frac{x^k}{(1-x)(1-2x)\cdots (1-kx)}.$$ Commented Aug 11 at 18:23
• Also for what it's worth, $s(n,k)$ is normally the notation for the signed Stirling numbers of the 1st kind, whereas $c(n,k)=|s(n,k)|$ is used for their absolute values. For instance, this is what Wikipedia uses: en.wikipedia.org/wiki/Stirling_number Commented Aug 11 at 18:32
• If you write $s(u,z)$ as $\exp(u(\log (1+z))$ then the inverse relation becomes clear—$\log(1+z)$ is the compositional inverse of $e^z-1$. More generally, if $f(z)$ and $g(z)$ are compositional inverses then we have similar inverse matrices formed from the coefficients of $\exp(u f(z))$ and $\exp(u g(z))$. Commented Aug 18 at 2:27
• Rather I show almost explicitly. One has to make the last leap/connection between umbral composition and the multiplication of lower triangular coefficient matrices, which applies to all Sheffer polynomial sequences--binomial, Appell, and generic. Commented Aug 27 at 12:06

In umbral notation and maneuvers, it's quite simple. With an obvious change of notation, let the Stirling polynomials of the first kind (a binomial Sheffer sequence of polynomials in the hybrid umbral-finite operator calculus) be denoted by

$$(ST1.(x))^n = ST1_n(x) = \sum_{k=0}^n ST1_{n,k}x^k$$

and similarly for the binomial Sheffer polynomials $$ST2_n(x).$$

The e.g.f.s are

$$e^{x\ln(t+1)} = e^{ST1.(x)t}$$

and

$$e^{x(e^t-1)} = e^{ST2.(x)t}.$$

Then umbral composition gives

$$e^{ST2.(ST1.(x))t} = e^{ST1.(x)(e^t-1)} = e^{x\ln[1+(e^t-1)]} = e^{xt},$$

and, of course,

$$e^{ST1.(ST2.(x))t} = e^{ST2.(x)(\ln(1+t))} = e^{x[e^{\ln(1+t)}-1]} = e^{xt},$$

implying that the two polynomial sequences form an umbral compositional inverse pair, i.e.,

$$ST2_n(ST1.(x))= x^n = ST1_n(ST2.(x))$$

In terms of the pair of lower triangular matrices of coefficients of the two sequences, this implies the pair of matrices is an inverse pair.

The diff op reps of the two binomial Sheffer sequences are

$$ST2_n(:xD:) = (xD)^n$$

and

$$ST1_n(xD) = :xD:^n = x^nD^n,$$

where $$:xD:^k = x^kD^k$$ by definition with $$D = \partial_x$$. Then we have the diff op equivalent of the umbrally compositional inverse relation:

$$ST2_n(:xD:) = ST2_n(ST1.(xD)) = (xD)^n$$

and

$$ST1_n(xD) = ST1_n(ST2.(:xD:)) = :xD:^n.$$

The e.g.f.s follow from the conjugation $$Ad_{e^{-x}}$$ as

$$e^{ST2.(x)t} = e^{-x} e^{ST2(:xD:)t}e^x = e^{-x} e^{txD}e^x =e^{-x}\sum_{n\geq 0} e^{tn}\frac{x^n}{n!} = e^{x(e^t-1)}$$

and from $$Ad_{x^{-y}}$$ as

$$e^{ST1.(y)t} = x^{-y} e^{ST1.(xD)t}x^y = x^{-y} e^{t:xD:}x^y =\sum_{n\geq 0} \binom{y}{n}t^n = (1+t)^y = e^{y\ln(1+t)}.$$

In addition, the lowering ops defined for any Sheffer polynomial sequence $$P_n(x)$$ by

$$L_P P_n(x) = n \; P_{n-1}(x)$$

are given by the inverse functions in the arguments of the e.g.f.s as

$$L_{ST1} = e^{D}-1$$

and

$$L_{ST2} = \ln(1+D).$$

If we go one step further and define the finite difference operator

$$\delta_x = e^{\partial_x}-1 = L_{ST1},$$

then

$$\partial_x = \ln(1 + \delta_x),$$

and we have an entre into the finite difference calculus with reps of the (Heaviside fractional) differ-integral calculus and the Appell Sheffer Bernoull polynomials and function (essentially the Hurwitz zeta function) along with their umbral inverses with their connections to the Bernoulli-Todd-Hirzebruch operator and class, exact Euler-MacLaurin series, volumes of lattice polytopes, the BCH theorem, and more via the ops

$$\frac{\partial_x}{\delta_x} = \frac{\partial_x}{e^{\partial_x}-1} = \frac{\ln(1+\delta_x)}{\delta_x}$$

and their multiplicative inverse ops

$$\frac{\delta_x}{\partial_x} = \frac{e^{\partial_x}-1}{\partial_x} = \frac{\delta_x}{\ln(1+\delta_x)}.$$

• You can go the Riordan array route, but that's a baby only a pure algebraist or computer programmer can love. I prefer Riordan's and others' umbral and diff op approach. Commented Aug 12 at 0:06
• For the inverse relation using operational calculus and o.g.f.s along with other common notation, see mathoverflow.net/questions/172955/… where $ST2_n(x) = \phi_n(x)$ are called the Bell polynomials (also called the Touchard / exponential / Steffensen and probably should be called the Scherk polynomials as well) and $ST1_n(x) = (x)_n = x!/(x-n)!$ are called the falling factorials. Commented Aug 12 at 11:50