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

Question

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.

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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?

Answer 1

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

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