A Note on a Theorem of Guo, Mező, and Qi

Antônio Francisco Neto^{1} DEPRO, Escola de Minas Campus Morro do Cruzeiro, UFOP 35400-000 Ouro Preto MG Brazil

                 antfrannet@gmail.com

Abstract

In a recent paper, Guo, Mező, and Qi proved an identity representing the Bernoulli polynomials at non-negative integer points m in terms of the m-Stirling numbers of the second kind. In this note, using a new representation of the Bernoulli polynomials in the context of the Zeon algebra, we give an alternative proof of the aforementioned identity.

1 Introduction

In an interesting recent paper [9], Guo, Mező, and Qi found the following identity

B_{n}(m) = ∑_{l=0}^{n}(−1)^{l} \frac{l!}{l + 1}S_{m}(n + m, l + m)  (1)

relating the Bernoulli polynomials B_{n}(m) at non-negative integer points m with m-Stirling numbers of the second kind S_{m}(n + m, l + m). Eq. (1) is a generalization of the identity [8, p. 560]

B_{n} = ∑_{m=0}^{n}(−1)^{m} \frac{m!}{m + 1}S(n, m).

1This work was supported by Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq-Brazil) under grant 307617/2012-2.

Note that B_{n} ≡ B_{n}(0) and S ≡ S_{0} are the usual Bernoulli numbers [16, Chap. 2] and Stirling numbers of the second kind [16, Chap. 1], respectively.

In this work, we will give another proof of (1) by showing that (1) is a straightforward consequence of a new Zeon representation [7, 11], [10, Chap. 5] of the Bernoulli polynomials.

We believe the approach here is of interest because it gives a straightforward demonstration of Guo, Mező, and Qi result, and, as a consequence, it provides another instance where computations involving Zeons and/or Grassmann variables provide direct and interesting results [1, 2, 5, 11, 12, 13, 15]. For more on Grassmann variables we refer the reader to the books of Berezin [3, Chap. 1], DeWitt [6, Chap. 1], and Rogers [14, Chap. 3].

For completeness, we recall some basic definitions and results already stated in previous work [11]. Throughout this work we let ℝ denote the real numbers, ℕ the positive integers, and ℕ_{0} = {0} ∪ ℕ the non-negative integers.

2 Zeon Algebra and Grassmann-Berezin Integral

Definition 1. The Zeon algebra Z_{n} ⊃ ℝ is defined as the associative algebra generated by the collection {ε_{i}}_{i=1}^{n} (n < ∞) and the scalar 1 ∈ ℝ, such that 1ε_{i} = ε_{i} = ε_{i}1, ε_{i}ε_{j} = ε_{j}ε_{i} ∀ i, j and ε_{i}^{2} = 0 ∀ i.

For {i, j, …, k} ⊂ {1, 2, …, n} and ε_{ij···k} ≡ ε_{i}ε_{j} ⋯ ε_{k} the most general element with n generators ε_{i} can be written as (with the convention of sum over repeated indices implicit)

ϕ_{n} = a + a_{i}ε_{i} + a_{ij}ε_{ij} + ⋯ + a_{12···n}ε_{12···n} = ∑_{i∈2^{[n]}} a_{i}ε_{i},  (2)

with a, a_{i}, a_{ij}, …, a_{12···n} ∈ ℝ, 2^{[n]} being the power set of [n] := {1, 2, …, n}, and 1 ≤ i < j < ⋯ ≤ n. We define the soul of ϕ_{n} by s (ϕ_{n}) := ϕ_{n} − a [6, Chap. 1].

Definition 2. The Grassmann-Berezin integral on Z_{n}, denoted by ∫, is the linear functional ∫ : Z_{n} → ℝ such that (we use throughout this work the compact notation dν_{n} := dε_{n} ⋯ dε_{1})

dε_{i}dε_{j} = dε_{j}dε_{i}, ∫ ϕ_{n}(ε̂_{i})dε_{i} = 0 and ∫ ϕ_{n}(ε̂_{i})ε_{i}dε_{i} = ϕ_{n}(ε̂_{i}),

where ϕ_{n}(ε̂_{i}) means any element of Z_{n} with no dependence on ε_{i}. Multiple integrals are iterated integrals, i.e.,

∫ f(ϕ_{n})dν_{n} = ∫ ⋯ (∫ (∫ f(ϕ_{n})dε_{n})dε_{n−1}) ⋯ dε_{1}.

Many functions of ordinary calculus admit extensions to the realm of Zeon algebra [11]. For instance, if ϕ_{n} = a + s (ϕ_{n}) in Eq. (2), we have

e^{ϕ_{n}} := e^{a} ∑_{m=0}^{n} \frac{s^{m} (ϕ_{n})}{m!}  (3)

and only a finite number of terms is present in the sum on the right-hand side of Eq. (3), since s^{m} (ϕ_{n}) = 0 for m > n.

Likewise, we have

\frac{1}{ϕ_{n}} = \frac{1}{a + s (ϕ_{n})} := \frac{1}{a} ∑_{m=0}^{n} (−\frac{1}{a})^{m} s^{m} (ϕ_{n}) .  (4)

3 Proof of Eq. (1)

We are now ready to prove Eq. (1). We take φ_{n} := ε_{1} + ⋯ + ε_{n} ∈ Z_{n} from now on. We start with

B_{n}(x) = ∑_{m=0}^{n} \frac{(−1)^{m}}{m + 1} ∫ e^{xφ_{n}} (e^{φ_{n}} − 1)^{m} dν_{n}.  (5)

We will proceed by showing that

∑_{m=0}^{n} \binom{n + 1}{m}B_{m} = 0  (6)

with n ∈ ℕ, B_{0} ≡ 1, \binom{n}{m} := n!/(m!(n − m)!), and

B_{n}(x) = ∑_{m=0}^{n} \binom{n}{m}B_{m}x^{n−m}.  (7)

Eqs. (6) and (7) can be regarded as the definitions of the Bernoulli numbers [8, Eq. (6.79)] and the Bernoulli polynomials [8, Eq. (7.80)], respectively.

We will first show by induction on n that

φ_{n} = ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m + 1} (e^{φ_{n}} − 1)^{m+1} .  (8)

Indeed, it is easy to see that both sides of Eq. (8) give φ_{1} ≡ ε_{1} for n = 1. Next, using Eqs. (3) and (4), we have

φ_{n+1} = φ_{n} + ε_{n+1}
= φ_{n} + ε_{n+1} \frac{e^{φ_{n}}}{1 + (e^{φ_{n}} − 1)}
= φ_{n} + ε_{n+1} ∑_{m=0}^{n}(−1)^{m}e^{φ_{n}} (e^{φ_{n}} − 1)^{m}

= ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m + 1} (e^{φ_{n}} − 1)^{m+1} + ε_{n+1} ∑_{m=0}^{n}(−1)^{m}e^{φ_{n}} (e^{φ_{n}} − 1)^{m}

= ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m + 1} \binom{m + 1}{0} (e^{φ_{n}} − 1)^{m+1} + ε_{n+1} ∑_{m=0}^{n} \frac{(−1)^{m}}{m + 1} \binom{m + 1}{1}e^{φ_{n}} (e^{φ_{n}} − 1)^{m}
= ∑_{m=0}^{n} \frac{(−1)^{m}}{m + 1} (e^{φ_{n}} + ε_{n+1}e^{φ_{n}} − 1)^{m+1}
= ∑_{m=0}^{n} \frac{(−1)^{m}}{m + 1} (e^{φ_{n+1}} − 1)^{m+1} ,

and the result follows, i.e., Eq. (8) is true for all n ≥ 1. Now we can prove Eq. (6). Starting with Eq. (8) and using Eq. (3) we have

φ_{n} = ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m + 1} ∑_{l=1}^{n} \frac{φ_{n}^{l}}{l!} (e^{φ_{n}} − 1)^{m} .

Integrating, we get (n ≥ 2)

0 = ∫ φ_{n}dν_{n} = ∑_{l=1}^{n} ∑_{1≤k_{1},k_{2},...,k_{l}≤n} ∑_{m=0}^{n−1} \frac{(−1)^{m}}{m + 1} ∫ ε_{k_{1}k_{2}···k_{l}} (e^{φ_{n}} − 1)^{m} dν_{n}

= ∑_{l=1}^{n} \binom{n}{l} ∑_{m=0}^{n−l} \frac{(−1)^{m}}{m + 1} ∫ (e^{φ_{n−l}} − 1)^{m} dν_{n−l}
= ∑_{l=1}^{n} \binom{n}{l}B_{n−l}

and making the change of variables n − l ↦ l we obtain Eq. (6).

We will now show Eq. (7). Indeed, from Eq. (5) we have

B_{n}(x) = ∑_{l=0}^{n} ∑_{m=0}^{n} \frac{(−1)^{l}}{l + 1} \frac{x^{m}}{m!} ∫ φ_{n}^{m} (e^{φ_{n}} − 1)^{l} dν_{n}
= ∑_{l=0}^{n} \frac{(−1)^{l}}{l + 1} ∫ (e^{φ_{n}} − 1)^{l} dν_{n}

+ ∑_{l=0}^{n} ∑_{m=1}^{n} ∑_{1≤l_{1},l_{2},...,l_{m}≤n} \frac{(−1)^{l}}{l + 1} x^{m} ∫ ε_{l_{1}l_{2}···l_{m}} (e^{φ_{n}} − 1)^{l} dν_{n}

= ∑_{m=0}^{n} \binom{n}{m}x^{m} ∑_{l=0}^{n−m} \frac{(−1)^{l}}{l + 1} ∫ (e^{φ_{n−m}} − 1)^{l} dν_{n−m}
= ∑_{m=0}^{n} \binom{n}{m}x^{m}B_{n−m}

and making the change of variables n − m ↦ m we obtain Eq. (7).

We recall the generating function for the m-Stirling numbers [4, Thm. 16]

∑_{n=l}^{∞} S_{m}(n + m, l + m)\frac{x^{n}}{n!} = \frac{1}{l!}e^{mx} (e^{x} − 1)^{l}  (9)

with m ∈ ℕ_{0}. The Zeon representation of S_{m}(n + m, l + m) comes from the generating function in Eq. (9) taking, as in previous work [11], x → φ_{n} ∈ Z_{n} and doing a Grassmann-Berezin integration over the Zeon algebra to get the representation

S_{m}(n + m, l + m) = ∑_{k=l}^{n} S_{m}(k + m, l + m) \underbrace{∫ \frac{φ_{n}^{k}}{k!} dν_{n}}_{δ_{k,n}} = \frac{1}{l!} ∫ e^{mφ_{n}} (e^{φ_{n}} − 1)^{l} dν_{n}  (10)

with δ_{k,n} meaning the Kronecker delta. We note that the representation in Eq. (10) is a generalization of the representation of the usual Stirling numbers of the second kind [15, Prop. 2.1] obtained by setting m = 0 in Eq. (10). Therefore, by setting x ≡ m in Eq. (5), we conclude that

B_{n}(m) = ∑_{l=0}^{n} \frac{(−1)^{l}}{l + 1} ∫ e^{mφ_{n}} (e^{φ_{n}} − 1)^{l} dν_{n},

which is equivalent to Eq. (1) using the Zeon representation of the numbers S_{m}(n+m, l +m) in Eq. (10).

4 Acknowledgments

The author thanks the anonymous referee for suggestions that improved the paper.

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2010 Mathematics Subject Classification: Primary 11B68; Secondary 11B73; 33B10; 05A15; 05A19. Keywords: Zeon algebra, Berezin integration, Bernoulli number, m-Stirling number, generating function.

(Concerned with sequences \underline{A027641}, \underline{A027642}, \underline{A143494}, \underline{A143495}, and \underline{A143496}.)

Received March 13 2016; revised version received April 13 2016. Published in Journal of Integer Sequences, May 11 2016.

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