Carlitz’s Identity for the Bernoulli Numbers and Zeon Algebra

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

                     antfrannet@gmail.com

Abstract In this work we provide a new short proof of Carlitz’s identity for the Bernoulli numbers. Our approach is based on the ordinary generating function for the Bernoulli numbers and a Grassmann-Berezin integral representation of the Bernoulli numbers in the context of the Zeon algebra, which comprises an associative and commutative algebra with nilpotent generators.

1 Introduction

In this work we will give a new, simple and short proof of Carlitz’s identity for the Bernoulli numbers [6]

∑_{i=0}^{m} \binom{m}{i}B_{n+i} = (−1)^{m+n} ∑_{j=0}^{n} \binom{n}{j}B_{m+j},  (1)

using the Zeon algebra [16, 17]. The identity in Eq. (1) has been re-obtained many times [7, 8, 12, 23, 25] and also very recently [13, 18, 24]. The proof given here is of independent interest, because of the simplicity of the arguments involved and, as it occurred in other contexts [1, 2, 5, 10, 11, 15, 16, 17, 20, 21, 22], the proof comprises another example of the usefulness of using the Zeon algebra and/or the Grassmann algebra towards obtaining combinatorial identities.

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

Before we continue, we establish the basic underlying algebraic setup needed to give the proof of Eq. (1). Throughout this work we let ℚ and ℝ denote the rational and real numbers, respectively.

2 Basic definitions: Zeon algebra and the 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.

Note that only linear elements in Z_{n} contribute to the calculations.

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 refer to a as the body of ϕ_{n} and write b(ϕ_{n}) = a and to ϕ_{n} − a as the soul such that s(ϕ_{n}) = ϕ_{n} − a.

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

For example, if we define φ_{n} := ε_{1} + ⋯ + ε_{n} it follows directly from Definition 2 and the multinomial theorem that

∫ φ_{n}^{i}dμ_{n} = i!δ_{i,n},  (3)

where δ_{i,n} is the Kronecker delta. For more details on Grassmann-Berezin integration, we refer the reader to the books of Berezin [3, Chapter 1] and [4, Chapter 2] or the books of DeWitt [9, Chapter 1] and Rogers [19, Chapter 11].

We will now recall some basic facts about the Zeon algebra. First, a + ϕ_{n} with s(a) = 0 = b(ϕ_{n}) is invertible iff b(a) ≠ 0. More precisely, we have

\frac{1}{a + ϕ_{n}} = \frac{1}{a} (1 − \frac{ϕ_{n}}{a} + \frac{ϕ_{n}^{2}}{a^{2}} + ⋯ + (−1)^{n} \frac{ϕ_{n}^{n}}{a^{n}} ) .  (4)

Second, the following expression holds

e^{φ_{n}} := ∑_{i=0}^{∞} \frac{φ_{n}^{i}}{i!} = ∑_{i=0}^{n} \frac{φ_{n}^{i}}{i!} = 1 + ∑_{1≤i≤n} ε_{i} + ∑_{1≤i<j≤n} ε_{ij} + ⋯ + ε_{12···n}.  (5)

To obtain Eq. (5) we have used the multinomial theorem and φ_{n}^{n+1} = 0 ∀ n ≥ 1. Third, let ϕ_{n} (ε̂_{i}, ε̂_{j}, …, ε̂_{k}) and dμ_{n} (ε̂_{i}, ε̂_{j}, …, ε̂_{k}) mean ϕ_{n} with ε_{i} = ε_{j} = ⋯ = ε_{k} = 0 and dμ_{n} with dε_{i}, dε_{j}, …, dε_{k} omitted, respectively. We have

∫ ϕ_{n}ε_{ij···k}dμ_{n} = ∫ ϕ_{n} (ε̂_{i}, ε̂_{j}, …, ε̂_{k}) ε_{ij···k}dμ_{n} = ∫ ϕ_{n} (ε̂_{i}, ε̂_{j}, …, ε̂_{k}) dμ_{n} (ε̂_{i}, ε̂_{j}, …, ε̂_{k}) . (6)

Eq. (6) follows directly from the general expression in Eq. (2) and Definition 2. Finally, from Definition 2, we conclude that the order of integration is irrelevant, i.e., a Fubini-like theorem holds in the setting of Grassmann-Berezin integration.

We are now ready to prove Eq. (1).

3 Proof of Eq. (1)

Let us write ℚ[[z]] for the ring of formal power series in the variable z over ℚ. We recall the generating function for the Bernoulli numbers B_{j} in ℚ[[z]] [26], i.e.,

\frac{1}{∑{∞ \atop i=0} \frac{z^{i}}{(i+1)!}} = \frac{z}{e^{z} − 1} = ∑_{j=0}^{∞} B_{j} \frac{z^{j}}{j!}  (7)

and, making the change z → −z in Eq. (7), we get

\frac{e^{z}}{∑{∞ \atop i=0} \frac{z^{i}}{(i+1)!}} = \frac{ze^{z}}{e^{z} − 1} = ∑_{j=0}^{∞} B_{j} \frac{(−z)^{j}}{j!} .  (8)

Following the strategy of our previous work [16, 17], we consider Eqs. (7) and (8) in the context of the Zeon algebra with the replacement z → ϕ_{k} ≡ φ_{k}. Therefore, we get

\frac{1}{∑{k \atop i=0} \frac{φ_{k}^{i}}{(i+1)!}} = ∑_{j=0}^{k} B_{j} \frac{φ_{k}^{j}}{j!}  (9)

and

\frac{e^{φ_{k}}}{∑{k \atop i=0} \frac{φ_{k}^{i}}{(i+1)!}} = ∑_{j=0}^{k} B_{j} \frac{(−φ_{k})^{j}}{j!} ,  (10)

using that φ_{k}^{k+1} = 0 ∀ k ≥ 1. We observe that b(∑_{i=0}^{k} \frac{φ_{k}^{i}}{(i+1)!}) = 1 ≠ 0 and, hence, ∑_{i=0}^{k} \frac{φ_{k}^{i}}{(i+1)!} is invertible in Z_{k}.

Now, integrating Eq. (9) in the Zeon algebra and using Eq. (3) we get

∫ \frac{1}{∑{j \atop i=0} \frac{φ_{j}^{i}}{(i+1)!}} dμ_{j} = ∑_{k=0}^{j} \frac{B_{k}}{k!} ∫ φ_{j}^{k}dμ_{j} = B_{j}  (11)

∀ j ≥ 1. It is straightforward to verify that the representation in Eq. (11) is equivalent to a well-known representation of the Bernoulli numbers [14, Theorem 3.1], i.e.,

B_{n} = n! ∑_{i=1}^{n}(−1)^{i} ∑_{\begin{matrix}i_{1},i_{2},...,i_{n}≥0 \\ i_{1}+i_{2}+···+i_{n}=i \\ i_{1}+2i_{2}+···+ni_{n}=n \end{matrix}} \frac{i!}{i_{1}!i_{2}! ⋯ i_{n}!} \frac{1}{2!^{i_{1}}3!^{i_{2}} ⋯ (n + 1)!^{i_{n}}} .

Indeed, we have

B_{n} = ∑_{i=1}^{n}(−1)^{i} ∫ (\frac{φ_{n}}{2!} + \frac{φ_{n}^{2}}{3!} + ⋯ + \frac{φ_{n}^{n}}{(n + 1)!})^{i} dμ_{n}

= ∑_{i=1}^{n}(−1)^{i} ∑_{\begin{matrix}i_{1},i_{2},...,i_{n}≥0 \\ i_{1}+i_{2}+···+i_{n}=i \end{matrix}} \frac{i!}{i_{1}!i_{2}! ⋯ i_{n}!} ∫ \frac{φ_{n}^{i_{1}}φ_{n}^{2i_{2}} ⋯ φ_{n}^{ni_{n}}}{2!^{i_{1}}3!^{i_{2}} ⋯ (n + 1)!^{i_{n}}} dμ_{n}

= n! ∑_{i=1}^{n}(−1)^{i} ∑_{\begin{matrix}i_{1},i_{2},...,i_{n}≥0 \\ i_{1}+i_{2}+···+i_{n}=i \end{matrix}} \frac{i!}{i_{1}!i_{2}! ⋯ i_{n}!} \frac{δ_{n,i_{1}+2i_{2}+···+ni_{n}}}{2!^{i_{1}}3!^{i_{2}} ⋯ (n + 1)!^{i_{n}}}

= n! ∑_{i=1}^{n}(−1)^{i} ∑_{\begin{matrix}i_{1},i_{2},...,i_{n}≥0 \\ i_{1}+i_{2}+···+i_{n}=i \end{matrix}} \frac{i!}{i_{1}!i_{2}! ⋯ i_{n}!} \frac{δ_{n,i_{1}+2i_{2}+···+ni_{n}}}{2!^{i_{1}}3!^{i_{2}} ⋯ (n + 1)!^{i_{n}}} ,

using Eqs. (3), (4) and the multinomial theorem.

By considering Eq. (10), we take k = m + n and write φ_{m+n} = φ_{m} + ϕ_{n} with φ_{m} := ε_{1} + ⋯ + ε_{m}, ϕ_{n} := ϵ_{1} + ⋯ + ϵ_{n}, and ϵ_{i} := ε_{i+m} ∀ 1 ≤ i ≤ n. Next, we multiply both sides of Eq. (10) by e^{−ϕ_{n}}. Finally, integrating the resulting equation with dμ_{m} := dε_{m} ⋯ dε_{1} and dν_{n} := dϵ_{n} ⋯ dϵ_{1} we get

∫ (∫ \frac{e^{φ_{m}}}{∑{m+n \atop i=0} \frac{(φ_{m}+ϕ_{n})^{i}}{(i+1)!}} dμ_{m}) dν_{n} = ∑_{j=0}^{m+n} \frac{B_{j}}{j!} ∫ (∫ (−φ_{m} − ϕ_{n})^{j}e^{−ϕ_{n}}dν_{n}) dμ_{m}.  (12)

In Eq. (12) we have used a Fubini-like argument to perform the integrations. We first consider the left-hand side of Eq. (12). By expanding e^{φ_{m}} as in Eq. (5) and integrating with respect to dμ_{m} we will need to analyze terms such as

∑_{1≤i_{1}<i_{2}<···<i_{j}≤m} ∫ (∫ \frac{ε_{i_{1}i_{2}···i_{j}}}{∑{m+n \atop i=0} \frac{(φ_{m}+ϕ_{n})^{i}}{(i+1)!}} dμ_{m}) dν_{n}

= \binom{m}{j} ∫ (∫ \frac{1}{∑{m−j+n \atop i=0} \frac{(φ_{m−j}+ϕ_{n})^{i}}{(i+1)!}} dμ_{m−j}) dν_{n} = \binom{m}{j}B_{n+m−j}.  (13)

Therefore, using Eq. (13), we get for the left-hand side of Eq. (12)

∑_{i=0}^{m} \binom{m}{i}B_{n+i}.  (14)

Similarly, we expand e^{−ϕ_{n}} as in Eq. (5) and integrate with respect to dν_{n} to obtain for the right-hand side of Eq. (12)

(−1)^{m+n} ∑_{j=0}^{n} \binom{n}{j}B_{m+j}.  (15)

By equating the expressions in (14) and (15) we obtain the desired result, i.e., Eq. (1).

Let B_{i}^{(j)} be the i-th Bernoulli number of order j with generating function in ℚ[[z]] given by

(\frac{z}{e^{z} − 1})^{j} = ∑_{i=0}^{∞} B_{i}^{(j)} \frac{z^{i}}{i!} .

Note that B_{n}^{(1)} ≡ B_{n}. Following the procedure just described, it is straightforward to prove an analogous identity for the Bernoulli numbers of higher order, i.e.,

∑_{i=0}^{m} k^{i}\binom{m}{i}B_{n+i}^{(k)} = (−1)^{m+n} ∑_{j=0}^{n} k^{j}\binom{n}{j}B_{m+j}^{(k)}.

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 33B10, 05A15, 05A19. Keywords: Zeon algebra, Berezin integration, Bernoulli number, generating function.

(Concerned with sequences \underline{A027641} and \underline{A027642}.)

Received January 29 2015; revised version received April 6 2015. Published in Journal of Integer Sequences, May 25 2015.

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