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The Eulerian-polynomial exponential generating function in $Q (t) ⟦ x ⟧$

Statement

In the formal power series ring $Q (t) ⟦ x ⟧$ , where $t$ is an indeterminate,

$\sum_{n \geq 0} A_{n} (t) \frac{x^{n}}{n!} = \frac{t - 1}{t - e^{(t - 1) x}} .$

Facts & Assumptions

Given: The Eulerian recurrence, the Eulerian polynomials $A_{n} (t) = \sum_{k = 0}^{n - 1} A (n, k) t^{k}$ , and the formal series ring $Q (t) ⟦ x ⟧$ .

[L1]

The Eulerian numbers satisfy $A (n, k) = (k + 1) A (n - 1, k) + (n - k) A (n - 1, k - 1)$ (The Eulerian numbers satisfy $A (n, k) = (k + 1) A (n - 1, k) + (n - k) A (n - 1, k - 1)$ ).

[L2]

Formal $\exp$ and $\log$ are inverse homomorphisms and obey the expected derivative rules (Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws, Formal differentiation is linear and satisfies product, power, quotient, chain, and coefficient-recovery laws).

Proof

technique · direct

1.1

Summing [L1] against $t^{k}$ gives the polynomial recurrence $A_{n} (t) = (1 + (n - 1) t) A_{n - 1} (t) + t (1 - t) A_{n - 1}^{'} (t)$ .

L1algebra

2.1

Let $F (x, t) : = \sum_{n \geq 0} A_{n} (t) x^{n} / n! \in Q (t) ⟦ x ⟧$ . Differentiating termwise in $x$ and using step 1.1 yields $(1 - t x) \partial_{x} F = F + t (1 - t) \partial_{t} F$ , with initial condition $F (0, t) = 1$ .

step 1.1algebra

3.1

Put $G (x, t) : = (t - 1) / (t - e^{(t - 1) x}) \in Q (t) ⟦ x ⟧$ ; this is well defined because the denominator has nonzero constant term $t - 1$ in the coefficient field $Q (t)$ . Direct formal differentiation shows that $G$ satisfies the same differential equation and the same initial condition as $F$ . Since the recurrence of step 1.1 determines the coefficient of $x^{n}$ from lower degrees uniquely, the formal solution is unique, so $F = G$ .

step 2.1L2algebra

4.1

Therefore $\sum_{n \geq 0} A_{n} (t) x^{n} / n! = (t - 1) / (t - e^{(t - 1) x})$ .

step 3.1∎

Depends on

Used by

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Sources

Richard P. Stanley, Enumerative Combinatorics, Volume 1, second edition (standard reference, not scraped)

The Eulerian-polynomial exponential generating function in Q(t)⟦x⟧Q(t)[[x]]