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Journal of Pure and Applied Mathematics

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Abdelhay Benmoussa*
 
Independent Researcher, Morocco, Email: bibo93035@gmail.com
 
*Correspondence: Abdelhay Benmoussa, Independent Researcher, Morocco, Email: bibo93035@gmail.com

Received: 04-May-2024, Manuscript No. puljpam-24-7045; Editor assigned: 05-May-2024, Pre QC No. puljpam-24-7045(PQ); Accepted Date: May 27, 2024; Reviewed: 06-May-2024 QC No. puljpam-24-7045(Q); Revised: 08-May-2024, Manuscript No. puljpam-24-7045(R); Published: 31-May-2024, DOI: 10.37532/2752- 8081.24.8(3).01-04

Citation: Benmoussa A. A very elementary proof of an explicit formula of Bernoulli Numbers. J Pure Appl Math. 2024; 8(3):01-04.

This open-access article is distributed under the terms of the Creative Commons Attribution Non-Commercial License (CC BY-NC) (http://creativecommons.org/licenses/by-nc/4.0/), which permits reuse, distribution and reproduction of the article, provided that the original work is properly cited and the reuse is restricted to noncommercial purposes. For commercial reuse, contact reprints@pulsus.com

Abstract

The aim of this paper is to give an easy and very elementary proof of a well-known explicit formula for Bernoulli numbers.

Key Words

Stirling numbers of the second kind; Bernoulli numbers; Bernoulli polynomials

Introduction

The numbers :

equation

are called the Bernoulli numbers. They can be defined by the following exponential generating function:

equation

When this sequence of numbers appeared in 1713, mathematicians didn’t know a formula to compute bn directly, so they used recursive formulas like this one [1]:

equation

In 1883, Worpitzky gave the following formula for bn [2]:

equation

We can also find other mathematicians from the 19th century who proved formula (1), such as Cesàro in 1885 and d’Ocagne in 1889 [3, 4].

For our part, we present here an elementary proof of the formula (1).

Stirling Numbers of the Second Kind

Let Y be an arbitrary function and set:

equation

If we develop the first four functions, we find:

equation

We conjecture that:

equation

The coefficients equation are called Stirling numbers of the second kind.They can be represented in a triangle similar to Pascal’s triangle. The triangle of the numbers equation is the following:

TABLE 1 The law for forming the numbers

  k=0 k=1 k=2 k=3 k=4
n=0 1          
n=1 0 1        
n=2 0 1 1      
n=3 0 1 3 1    
n=4 0 1 7 6 1  

We observe that:

equation

The law for forming the numbers in the above table is given by:

equation

The Explicit Formula for Stirling Numbers of the second kind

If we put Y = ex in the formula (2), we obtain:

equation

One can easily prove that Dnxi = inxi, so:

equation

If we develop the left-hand side we obtain:

equation

Comparing coefficients in both summations, we conclude that:

equation

Relation between Bernoulli Numbers and Stirling Numbers of the Second Kind

Putting Y = xy in the formula (2), we get:

equation

We know that (xy)(k) = y(y-1)...(y-k+1)xy-k and Dnxy = ynxy so we get:

equation

The polynomialy (y-1)...(y-k+1) is called the falling factorial of y of order k. Pochhammer used the symbol (yk) to denote it, so the formula (4) becomes using Pochhammer symbol:

equation

One interesting property of the falling factorial function is the following:

Proposition

Let y,n be non-negative integers, then:

equation

Proof

equation

We are going to use this property in the proof of the following proposition.

Proposition

Let (n,m) ∈ℕ2. We have:

equation

Proof

If we sum for y in the formula (4’) we find:

equation

Therefore:

equation

Definition

Let ∈ℕ

The Bernoulli polynomials Bn(x) are defined by the following generating function:

equation

One interesting observation to make about Bernoulli polynomials is that if we put x=0 we get:

equation

This generating function corresponds to the generating function of Bernoulli numbers bn. Hence for all n∈ℕ, we have:

Bn(0) =bn

Another interesting property of the Bernoulli polynomials is the following:

Proposition

Let n∈ℕ

equation

Proof

On the one hand:

equation

On the other hand:

equation

Comparing coefficients of both summations we conclude that:

equation

Proposition

Let n∈ℕ

equation

Proof

equation

Therefore:

equation

If we sum for y in the relation B(n+1)(y+1) - B(n+1)(y) = (n+1)yn, we obtain :

equation

Thus:

equation

Comparing formula (5) with formula (6). We conclude that:

equation

If we develop the expression of (X)k+1 in terms of the powers of X we find:

equation

Therefore:

equation

If we apply the above formula for (m+1)k+1 in the formula (7) we find:

equation

Substituting also Bn+1(m+1) by its explicit expression, we finally get:

equation

We have equality between two polynomials in m+1, both of degree n, so the coefficients of the terms of the same degree are equal. In particular for j = 0 we have:

equation

To get the explicit expression of bn in terms of n, we substitute equation in the above identity by its explicit expression, and after simplification we obtain the remarkable formula (1) for the Bernoulli numbers.

References

 
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