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Faulhaber's formula was generalized by V. Guo and J. Zeng to a ] {{harv|Guo|Zeng|2005}}. Faulhaber's formula was generalized by V. Guo and J. Zeng to a ] {{harv|Guo|Zeng|2005}}.


==Definition== == Definitions ==
===Generating functions===
The Bernoulli numbers may also be defined using ]s. Their ] is ''x''/(''e<sup>x</sup>''&nbsp;−&nbsp;1), so that:


Many characterizations of the Bernoulli numbers have been found in the last
:<math>
300 years, and each could be used to introduce theses
\frac{x}{e^x-1} = \sum_{n=0}^{\infin} B_n \frac{x^n}{n!}
numbers. Here only four of the most useful ones are mentioned:
</math>


* a recursive equation,
for all values of ''x'' of ] less than 2π (the ] of this ]).
* an explicit formula,
* a generating function,
* an algorithmic description.


For the proof of the ] of the four approaches the reader is
These definitions can be shown to be equivalent using ]. The initial condition <math>B_0 = 1</math> is immediate from ]. To obtain the recurrence, multiply both sides of the equation by <math>e^x-1</math>. Then, using the ] for the ],
referred to mathematical expositions like {{Harv|Ireland|Rosen|1990}} or {{Harv|Conway|Guy|1996}}.


Unfortunately in the literature the definition is given in two variants:
: <math>x = \left( \sum_{j=1}^{\infty} \frac{x^j}{j!} \right) \left( \sum_{k=0}^{\infty} \frac{B_k x^k}{k!} \right). </math>
Despite the fact, that Bernoulli defined ''B''<sub>1</sub>&nbsp;=&nbsp;1/2
some authors set ''B''<sub>1</sub>&nbsp;=&nbsp;—1/2
(more on different conventions below).&nbsp;In order to prevent potential confusions
both variants will be described here, side by side.


=== Recursive definition ===
By expanding this as a ] and rearranging slightly, one obtains


The recursive equation is best introduced in a slightly more general form
: <math> x = \sum_{m=0}^{\infty} \left( \sum_{j=0}^{m} {m+1 \choose j} B_j \right) \frac{x^{m+1}}{(m+1)!}. </math>


: <math> B_m(n)=n^m-\sum_{k=0}^{m-1}\binom mk\frac{B_k(n)}{m-k+1} \ . </math>
It is clear from this last equality that the coefficients in this power series satisfy the same recurrence as the Bernoulli numbers.


This equation defines integers ''B''<sub>''m''</sub>(''n'') for all integers ''n'' ≥ 0, ''m'' ≥ 0.
=== Other forms and conventions ===
0<sup>0</sup> has to be interpreted as 1. The recursion has its
base in ''B''<sub>0</sub>(''n'') = 1 for all ''n''.
The two variants now follow by setting ''n'' = 0 respectively ''n'' = 1.
Additionally the notation is simplified by erasing the reference
to the parameter ''n''.
{| class="wikitable" style="text-align:center; width:600px; height:120px; border=1"
|-
! <i>n</i> = 0
! <i>n</i> = 1
|-
| <math> B_m = \left -\sum_{k=0}^{m-1}\binom mk\frac{B_k}{m-k+1} </math>
| <math> B_m= 1 - \sum_{k=0}^{m-1}\binom mk\frac{B_k}{m-k+1} </math>
|}


Here the expression has the value 1 if ''m'' = 0 and 0 otherwise (Iverson bracket).
One may also write
Whenever a confusion between the two kinds of definitions might arise
it can be avoided by refering to the more general definition and by reintroducing
the erased parameter: writing ''B''<sub>m</sub>(0) in the first case and
''B''<sub>m</sub>(1) in the second will unambiguously denote the value in question.


=== Explicit definition ===
:<math>\sum_{k=0}^{m-1} k^n = \frac{B_{n+1}(m)-B_{n+1}(0)}{n+1},</math>


Starting again with a slightly more general formula
where ''B''<sub>''n''&nbsp;+&nbsp;1</sub>(''m'') is the (''n''&nbsp;+&nbsp;1)th-degree ].


: <math> B_m(n)=\sum_{k=0}^m\sum_{v=0}^k(-1)^v\binom kv\frac{\left( n+v\right) ^m}{k+1} , </math>
Bernoulli numbers may be calculated by using the following ] formula:


the choices ''n'' = 0 and ''n'' = 1 lead to
:<math>\sum_{j=0}^m{m+1\choose{j}}B_j = 0</math>
{| class="wikitable" style="text-align:center; width:640px; height:120px; border=1"
|-
! <i>n</i> = 0
! <i>n</i> = 1
|-
| <math> B_m=\sum_{k=0}^m\sum_{v=0}^k(-1)^v\binom kv\frac{v^m}{k+1} \ ,</math>
| <math> B_m=\sum_{k=1}^{m+1}\sum_{v=1}^{k+1}(-1)^{v+1}\binom{k-1}{v-1}\frac{v^m}k \ .</math>
|}


There is a widespread misinformation that no simple closed formulas
for ''m'' > 0, and ''B''<sub>0</sub> = 1.
for the Bernoulli numbers exist. The last two equations show that
this is not true. Moreover, already in 1893
listed a total of 38 explicit formulas for the Bernoulli numbers {{Harv|Saalschütz|1893}},
usually giving some reference in the older literature.


=== Generating function ===
An alternative convention for the Bernoulli numbers is to set ''B''<sub>1</sub> = 1/2 rather than −1/2. If this convention is used, then all the Bernoulli numbers may be calculated by a different ] formula without further qualification:


The general formula for the generating function is
:<math>\sum_{j=0}^m{m+1\choose{j}}B_j = m+1.</math>


: <math> \frac{te^{nt}}{e^t-1}=\sum_{m=0}^\infty B_m(n)\frac{t^m}{m!} \ . </math>
Dividing both sides by ''m'' + 1 then gives a form suggestive of the connection with the Riemann zeta function if the ''j''=0 case is understood as a limit to deal with the pole at ζ(1):


The choices ''n'' = 0 and ''n'' = 1 lead to
:<math>\sum_{j=0}^m{m\choose{j-1}}\frac{B_j}{j} = 1</math>
{| class="wikitable" style="text-align:center; width:500px; height:120px; border=1"
|-
! <i>n</i> = 0
! <i>n</i> = 1
|-
| <math> \frac t{e^t-1}=\sum_{m=0}^\infty B_m\frac{t^m}{m!} \ , </math>
| <math> \frac t{1-e^{-t}}=\sum_{m=0}^\infty B_m\frac{t^m}{m!} \ . </math>
|}


=== Algorithmic description ===
The terms of this sum are the coefficients given by ] for the closed form of <math>\sum_{x=1}^n x^m</math> and so this recursive definition is merely reflecting the fact that these sums evaluate to 1 when ''n''=1 for any ''m''. In the alternative convention, the generating function is


Although the above recursive formula can be used for computation it is
:<math>\frac{xe^x}{e^x-1}.</math>
mainly used to establish the connection with the sum of powers because it
is ].
However, both simple and high-end algorithms
for computing Bernoulli numbers exist. A simple one is given in pseudo code
below in the text box 'Akiyama-Tanigawa algorithm' and pointers to high-end
algorithms are given in another section below.


== Values of the Bernoulli numbers == == Values of the Bernoulli numbers ==

Revision as of 10:09, 17 July 2009

In mathematics, the Bernoulli numbers are a sequence of rational numbers with deep connections to number theory. They are closely related to the values of the Riemann zeta function at negative integers.

In Europe, they were first studied by Jakob Bernoulli. They appear in the Taylor series expansions of the tangent and hyperbolic tangent functions, in the Euler–Maclaurin formula, and in expressions for certain values of the Riemann zeta function.

In note G of Ada Lovelace's notes on the Analytical engine from 1842, Lovelace describes an algorithm for generating Bernoulli numbers with Babbage's machine . As a result, the Bernoulli numbers have the distinction of being the subject of the first computer program.

Introduction

The Bernoulli numbers are rooted in the early history of the computation of sums of integer powers, which have been of interest to mathematicians since antiquity.

Methods to calculate the sum of the first n positive integers, the sum of the squares and of the cubes of the first n positive integers were known, but there were no real 'formulas', only descriptions given entirely in words. Among the great mathematicians of antiquity which considered this problem were: Pythagoras (c. 572-497 BCE, Greece), Archimedes ((287-212 BCE, Italy), Aryabhata (b. 476, India), Abu Bakr al-Karaji (d. 1019, Persia) and Abu Ali al-Hasan ibn al-Hasan ibn al-Haytham (965-1039, Egypt).

Only during the late sixteenth and early seventeenth centuries mathematicians made progress. In the West Thomas Harriot (1560-1621) of England, Johann Faulhaber (1580-1635) of Germany, Pierre de Fermat (1601-1665) and fellow French mathematician Blaise Pascal (1623-1662) played an important role in this development.

Thomas Harriot seems to have been the first to derive and write formulas for sums of powers using symbolic notation, but even he calculated only up to the sum of the fourth powers. Johann Faulhaber gave formulas for sums of powers up to the 17th power in his 1631 Academia Algebrae, far higher than anyone before him, but he did not give a general formula.

The Swiss mathematician Jakob Bernoulli (1654-1705) was the first to realize the existence of a single sequence of constants B0, B1, B2,... which provide a uniform formula for all sums of powers (Knuth 1993). At approximately the same time in Japan an equivalent method for calculating sums of powers was discovered by Seki Kowa. However, Seki did not present his method as a formula based on a sequence of constants.

The joy Bernoulli experienced when he hit upon the pattern needed to compute quickly and easily the coefficients of his formula for the sum of the c-th powers for any positive integer c can be seen from his comment. He wrote:

“With the help of this table, it took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum
91,409,924,241,424,243,424,241,924,242,500.”

Bernoulli's formula for sums of powers is the most useful and generalizable formulation to date. The coefficients in Bernoulli's formula are nowadays called Bernoulli numbers, following a suggestion of Abraham de Moivre.

Reconstruction of 'Summae Potestatum'

Jakob Bernoulli's Summae Potestatum, 1713

The Bernoulli numbers were introduced by Jakob Bernoulli in the book 'Ars Conjectandi' published posthumously in 1713. The main formula can be seen in the second half of the facsimile given above. The constant coefficients denoted A, B, C and D by Bernoulli are mapped to the notation which is now prevalent as A = B2, B = B4, C = B6, D = B8. In the expression c·c−1·c−2·c−3 the small dots are used as grouping symbols, not as signs for multiplication. Using today's terminology these expressions are falling factorial powers c k _ {\displaystyle c^{\underline {k}}} . The factorial notation k! as a shortcut for 1×2×...×k was not introduced until 100 years later. The integral symbol on the left hand side goes back to Gottfried Wilhelm Leibniz in 1675 who used it as a long letter S for "summa" (sum). (The Mathematics Genealogy Project shows Leibniz as the doctoral adviser of Jakob Bernoulli. See also the Earliest Uses of Symbols of Calculus .) The letter n on the left hand side is not an index of summation but gives the upper limit of the range of summation which is to be understood as 1,2,...,n. Putting things together, for positive c, today a mathematician is likely to write Bernoulli's formula as

0 < k n k c = n c + 1 c + 1 + 1 2 n c + k 2 B k k ! c k 1 _ n c k + 1   . {\displaystyle \sum _{0<k\leq n}k^{c}={\frac {n^{c+1}}{c+1}}+{\frac {1}{2}}n^{c}+\sum _{k\geq 2}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}\ .}

In fact this formula imperatively suggests to set B1 = 1/2 when switching from the so called 'archaic' enumeration which uses only the even indices 2,4,.. to the modern form (more on different conventions in the next paragraph). Most striking in this context is the fact that the falling factorial c k 1 _ {\displaystyle c^{\underline {k-1}}} has for k = 0 the value 1/(c + 1). (See the explanation in .) Thus Bernoulli's formula can and has to be written

0 < k n k c = k 0 B k k ! c k 1 _ n c k + 1   . {\displaystyle \sum _{0<k\leq n}k^{c}=\sum _{k\geq 0}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}\ .}

if B1 stands for the value Bernoulli himself has given to the coefficient at that position.

Sum of powers

Bernoulli's Formula

Closed forms of the sum of powers for fixed values of m

    S m ( n ) = k = 1 n k m = 1 m + 2 m + + n m     {\displaystyle \ \ S_{m}(n)=\sum _{k=1}^{n}k^{m}=1^{m}+2^{m}+\cdots +{n}^{m}\ \ }

are always polynomials in n of degree m + 1. Note that Sm(0) = 0 for all m ≥ 0 because in this case the sum is the empty sum. The coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's formula:

S m ( n ) = 1 m + 1 k = 0 m ( m + 1 k ) B k n m + 1 k {\displaystyle S_{m}(n)={1 \over {m+1}}\sum _{k=0}^{m}{m+1 \choose {k}}B_{k}n^{m+1-k}}

Let n ≥ 0. Taking m to be 0 and B0 = 1 gives the natural number 0,1,2,3,… (sequence OEISA001477).

0 + 1 + 1 + + 1 = 1 1 ( B 0 n ) = n . {\displaystyle 0+1+1+\cdots +1={\frac {1}{1}}\left(B_{0}n\right)=n.}

Taking m to be 1 and B1 = 1/2 gives the triangular number 0,1,3,6,… (sequence OEISA000217).

0 + 1 + 2 + + n = 1 2 ( B 0 n 2 + 2 B 1 n 1 ) = 1 2 ( n 2 + n ) . {\displaystyle 0+1+2+\cdots +n={\frac {1}{2}}\left(B_{0}n^{2}+2B_{1}n^{1}\right)={\frac {1}{2}}\left(n^{2}+n\right).}

Taking m to be 2 and B2 = 1/6 gives the square pyramidal number 0,1,5,14,… (sequence OEISA000330).

0 2 + 1 2 + 2 2 + + n 2 = 1 3 ( B 0 n 3 + 3 B 1 n 2 + 3 B 2 n 1 ) = 1 3 ( n 3 + 3 2 n 2 + 1 2 n ) . {\displaystyle 0^{2}+1^{2}+2^{2}+\cdots +n^{2}={\frac {1}{3}}\left(B_{0}n^{3}+3B_{1}n^{2}+3B_{2}n^{1}\right)={\frac {1}{3}}\left(n^{3}+{\frac {3}{2}}n^{2}+{\frac {1}{2}}n\right).}

Although Bernoulli's formula reflects faithfully what Bernoulli has written some authors state Bernoulli's formula in a different way which is neither in accordance with Bernoulli's statement nor has any obvious advantage compared to it. They write:

S m ( n ) = 1 m + 1 k = 0 m ( 1 ) k ( m + 1 k ) B k n m + 1 k {\displaystyle S_{m}(n)={1 \over {m+1}}\sum _{k=0}^{m}(-1)^{k}{m+1 \choose {k}}B_{k}n^{m+1-k}}

To avoid a contradiction to the valid formula above these authors have to set B1 = -1/2. In the next section consequences of the resulting differences will be commented on as they are likely to produce some confusion.

Faulhaber's Formula

Bernoulli's formula is sometimes called Faulhaber's formula. There is no evidence which justifies this nomenclature. Johann Faulhaber found remarkable ways to calculate sum of powers but he never stated Bernoulli's formula.

Faulhaber realized that for odd m, Sm(n) is not just a polynomial in n but a polynomial in the triangular number N = n(n + 1)/2. For example Faulhaber's formulas read as follows:

1 + 2 + + n = N ; {\displaystyle 1+2+\cdots +n=N;}
1 2 + 2 2 + + n 2 = N ( 2 n + 1 ) / 3 ; {\displaystyle 1^{2}+2^{2}+\cdots +n^{2}=N\left(2n+1\right)/3;}
1 3 + 2 3 + + n 3 = N 2 . {\displaystyle 1^{3}+2^{3}+\cdots +n^{3}=N^{2}.}

To call Bernoulli's formula Faulhaber's formula does injustice to Bernoulli and simultaneously hides the genius of Faulhaber as Faulhaber's formula is in fact more efficient than Bernoulli's formula. According to Knuth (Knuth 1993) a rigorous proof of Faulhaber’s formula was first published by Carl Jacobi in 1834 (Jacobi 1834). Donald E. Knuth's in-depth study of Faulhaber's Formula concludes:

“Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a single sequence of constants B0, B1, B2, … would provide a uniform

n m = 1 m + 1 ( B 0 n m + 1 ( m + 1 1 ) B 1 n m + ( m + 1 2 ) B 2 n m 1 + ( 1 ) m ( m + 1 m ) B m n ) {\displaystyle \quad \sum n^{m}={\frac {1}{m+1}}\left(B_{0}n^{m+1}-{\binom {m+1}{1}}B_{1}n^{m}+{\binom {m+1}{2}}B_{2}n^{m-1}-\cdots +(-1)^{m}{\binom {m+1}{m}}B_{m}n\right)}

for all sums of powers. He never mentioned, for example, the fact that almost half of the coefficients turned out to be zero after he had converted his formulas for n m {\displaystyle \sum n^{m}} from polynomials in N to polynomials in n.” (Knuth 1993, p. 14)

Faulhaber's formula was generalized by V. Guo and J. Zeng to a q-analog (Guo & Zeng 2005).

Definitions

Many characterizations of the Bernoulli numbers have been found in the last 300 years, and each could be used to introduce theses numbers. Here only four of the most useful ones are mentioned:

  • a recursive equation,
  • an explicit formula,
  • a generating function,
  • an algorithmic description.

For the proof of the equivalence of the four approaches the reader is referred to mathematical expositions like (Ireland & Rosen 1990) or (Conway & Guy 1996).

Unfortunately in the literature the definition is given in two variants: Despite the fact, that Bernoulli defined B1 = 1/2 some authors set B1 = —1/2 (more on different conventions below). In order to prevent potential confusions both variants will be described here, side by side.

Recursive definition

The recursive equation is best introduced in a slightly more general form

B m ( n ) = n m k = 0 m 1 ( m k ) B k ( n ) m k + 1   . {\displaystyle B_{m}(n)=n^{m}-\sum _{k=0}^{m-1}{\binom {m}{k}}{\frac {B_{k}(n)}{m-k+1}}\ .}

This equation defines integers Bm(n) for all integers n ≥ 0, m ≥ 0. 0 has to be interpreted as 1. The recursion has its base in B0(n) = 1 for all n. The two variants now follow by setting n = 0 respectively n = 1. Additionally the notation is simplified by erasing the reference to the parameter n.

n = 0 n = 1
B m = [ m = 0 ] k = 0 m 1 ( m k ) B k m k + 1 {\displaystyle B_{m}=\left-\sum _{k=0}^{m-1}{\binom {m}{k}}{\frac {B_{k}}{m-k+1}}} B m = 1 k = 0 m 1 ( m k ) B k m k + 1 {\displaystyle B_{m}=1-\sum _{k=0}^{m-1}{\binom {m}{k}}{\frac {B_{k}}{m-k+1}}}

Here the expression has the value 1 if m = 0 and 0 otherwise (Iverson bracket). Whenever a confusion between the two kinds of definitions might arise it can be avoided by refering to the more general definition and by reintroducing the erased parameter: writing Bm(0) in the first case and Bm(1) in the second will unambiguously denote the value in question.

Explicit definition

Starting again with a slightly more general formula

B m ( n ) = k = 0 m v = 0 k ( 1 ) v ( k v ) ( n + v ) m k + 1 , {\displaystyle B_{m}(n)=\sum _{k=0}^{m}\sum _{v=0}^{k}(-1)^{v}{\binom {k}{v}}{\frac {\left(n+v\right)^{m}}{k+1}},}

the choices n = 0 and n = 1 lead to

n = 0 n = 1
B m = k = 0 m v = 0 k ( 1 ) v ( k v ) v m k + 1   , {\displaystyle B_{m}=\sum _{k=0}^{m}\sum _{v=0}^{k}(-1)^{v}{\binom {k}{v}}{\frac {v^{m}}{k+1}}\ ,} B m = k = 1 m + 1 v = 1 k + 1 ( 1 ) v + 1 ( k 1 v 1 ) v m k   . {\displaystyle B_{m}=\sum _{k=1}^{m+1}\sum _{v=1}^{k+1}(-1)^{v+1}{\binom {k-1}{v-1}}{\frac {v^{m}}{k}}\ .}

There is a widespread misinformation that no simple closed formulas for the Bernoulli numbers exist. The last two equations show that this is not true. Moreover, already in 1893 Louis Saalschütz listed a total of 38 explicit formulas for the Bernoulli numbers (Saalschütz 1893), usually giving some reference in the older literature.

Generating function

The general formula for the generating function is

t e n t e t 1 = m = 0 B m ( n ) t m m !   . {\displaystyle {\frac {te^{nt}}{e^{t}-1}}=\sum _{m=0}^{\infty }B_{m}(n){\frac {t^{m}}{m!}}\ .}

The choices n = 0 and n = 1 lead to

n = 0 n = 1
t e t 1 = m = 0 B m t m m !   , {\displaystyle {\frac {t}{e^{t}-1}}=\sum _{m=0}^{\infty }B_{m}{\frac {t^{m}}{m!}}\ ,} t 1 e t = m = 0 B m t m m !   . {\displaystyle {\frac {t}{1-e^{-t}}}=\sum _{m=0}^{\infty }B_{m}{\frac {t^{m}}{m!}}\ .}

Algorithmic description

Although the above recursive formula can be used for computation it is mainly used to establish the connection with the sum of powers because it is computationally expensive. However, both simple and high-end algorithms for computing Bernoulli numbers exist. A simple one is given in pseudo code below in the text box 'Akiyama-Tanigawa algorithm' and pointers to high-end algorithms are given in another section below.

Values of the Bernoulli numbers

Bn = 0 for all odd n other than 1. B1 = 1/2 or −1/2 depending on the convention adopted (see below). The first few non-zero Bernoulli numbers (sequences OEISA027641 and OEISA027642 in OEIS) are listed below.

n Numerator Denominator Decimal approximation (N/D)
0 1 1 +1.00000000000
1 -1 2 -0.50000000000
2 1 6 +0.16666666667
4 -1 30 -0.03333333333
6 1 42 +0.02380952381
8 -1 30 -0.03333333333
10 5 66 +0.07575757576
12 -691 2730 -0.25311355311
14 7 6 +1.16666666667
16 -3617 510 -7.09215686275
18 43867 798 +54.9711779448

Efficient computation of Bernoulli numbers

In some applications it is useful to be able to compute the Bernoulli numbers B0 through Bp − 3 modulo p, where p is a prime; for example to test whether Vandiver's conjecture holds for p, or even just to determine whether p is an irregular prime. It is not feasible to carry out such a computation using the above recursive formulae, since at least (a constant multiple of) p arithmetic operations would be required. Fortunately, faster methods have been developed (Buhler et al. 2001) which require only O(p (log p)) operations (see big-O notation).

David Harvey (Harvey 2008) describes an algorithm for computing Bernoulli numbers by computing Bn modulo p for many small primes p, and then reconstructing Bn via the Chinese Remainder Theorem. Harvey writes that the asymptotic time complexity of this algorithm is O(n log(n)) and claims that this implementation is significantly faster than implementations based on other methods. Harvey's implementation is included in Sage since version 3.1. Using this implementation Harvey computed Bn for n = 10, which is a new record (October 2008). Prior to that Bernd Kellner (Kellner 2002) computed Bn to full precision for n = 10 on December 2002 and Oleksandr Pavlyk (Pavlyk 2008) for n = 10 with 'Mathematica' on April 2008.

Computer Year n Digits
J. Bernoulli ~1689 10 1
L. Euler 1748 30 8
J.C. Adams 1878 62 36
D.E. Knuth, T.J. Buckholtz 1967 360 478
G. Fee, S. Plouffe 1996 10000 27677
G. Fee, S. Plouffe 1996 100000 376755
B.C. Kellner 2002 1000000 4767529
O. Pavlyk 2008 10000000 57675260
D. Harvey 2008 100000000 676752569
History of the computation of Bernoulli numbers

Digits is to be understood as the exponent of 10 when B(n) is written as a real in normalized scientific notation.

Different viewpoints and conventions

The Bernoulli numbers can be regarded from four main viewpoints:

  • as standalone arithmetical objects,
  • as combinatorial objects,
  • as values of a sequence of certain polynomials,
  • as values of the Riemann zeta function.

Each of these viewpoints leads to a set of more or less different conventions.

  • Bernoulli numbers as standalone arithmetical objects.
    Associated sequence: 1/6, −1/30, 1/42, −1/30,...
    This is the viewpoint of Jakob Bernoulli. (See the cutout from his Ars Conjectandi, first edition, 1713). The Bernoulli numbers are understood as numbers, recursive in nature, invented to solve a certain arithmetical problem, the summation of powers, which is the paradigmatic application of the Bernoulli numbers. It is misleading to call this viewpoint 'archaic'. For example Jean-Pierre Serre uses it in his highly acclaimed book A Course in Arithmetic which is a standard textbook used at many universities today.
  • Bernoulli numbers as combinatorial objects.
    Associated sequence: 1, +1/2, 1/6, 0,....
    This view focuses on the connection between Stirling numbers and Bernoulli numbers and arises naturally in the calculus of finite differences. In its most general and compact form this connection is summarized by the definition of the Stirling polynomials σn(x), formula (6.52) in Concrete Mathematics by Graham, Knuth and Patashnik.
    ( z e z e z 1 ) x = x n 0 σ n ( x ) z n {\displaystyle \left({\frac {ze^{z}}{e^{z}-1}}\right)^{x}=x\sum _{n\geq 0}\sigma _{n}(x)z^{n}}
    In consequence Bn = n! σn(1) for n ≥ 0.
  • Bernoulli numbers as values of a sequence of certain polynomials.
    Assuming the Bernoulli polynomials as already introduced the Bernoulli numbers can be defined in two different ways:
    Bn = Bn(0). Associated sequence: 1, −1/2, 1/6, 0,....
    Bn = Bn(1). Associated sequence: 1, +1/2, 1/6, 0,....
    The two definitions differ only in the sign of B1. The choice Bn = Bn(0) is the convention used in the Handbook of Mathematical Functions.
  • Bernoulli numbers as values of the Riemann zeta function.
    Associated sequence: 1, +1/2, 1/6, 0,....
    The Bernoulli numbers as given by the Riemann zeta function.

    This convention agrees with the convention Bn = Bn(1) (for example J. Neukirch and M. Kaneko). The sign '+' for B1 matches the representation of the Bernoulli numbers by the Riemann zeta function. In fact the identity nζ(1−n) = (−1)Bn valid for all n > 0 is then replaced by the simpler nζ(1−n) = -Bn. (See the paper of S. C. Woon.)
    (Note that in the foregoing equation for n = 0 and n = 1 the expression −nζ(1 − n) is to be understood as limx → n −xζ(1 − x).)

  B n = n ! σ n ( 1 ) = B n ( 1 ) = n ζ ( 1 n ) ( n 0 )   {\displaystyle \ B_{n}=n!\sigma _{n}(1)=B_{n}(1)=-n\zeta (1-n)\quad (n\geq 0)\ }

Application of the Bernoulli numbers

Arguably the most important application of the Bernoulli number in mathematics is their use in the Euler–MacLaurin formula. Assuming that ƒ is a sufficiently often differentiable function the Euler–MacLaurin formula can be written as

a k < b f ( k ) = a b f ( x ) d x   + k = 1 m B k k ! ( f ( k 1 ) ( b ) f ( k 1 ) ( a ) ) + R ( f , m )   . {\displaystyle \sum \limits _{a\leq k<b}f(k)=\int _{a}^{b}f(x)\,dx\ +\sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .}

This formulation assumes the convention B1 = −1/2. However, if one sets B1 = 1/2 then this formula can also be written as

a < k b f ( k ) = a b f ( x ) d x + k = 1 m B k k ! ( f ( k 1 ) ( b ) f ( k 1 ) ( a ) ) + R ( f , m ) .   {\displaystyle \sum \limits _{a<k\leq b}f(k)=\int _{a}^{b}f(x)\,dx+\sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m).\ }

Here ƒ = ƒ which is a commonly used notation identifying the zero-th derivative of ƒ with ƒ. Moreover, let ƒ denote an antiderivative of ƒ. By the fundamental theorem of calculus,

a b f ( x ) d x   = f ( 1 ) ( b ) f ( 1 ) ( a ) . {\displaystyle \int _{a}^{b}f(x)\,dx\ =f^{(-1)}(b)-f^{(-1)}(a).}

Thus the last formula can be further simplified to the following succinct form of the Euler–Maclaurin formula

a < k b f ( k ) = k = 0 m B k k ! ( f ( k 1 ) ( b ) f ( k 1 ) ( a ) ) + R ( f , m ) .   {\displaystyle \sum \limits _{a<k\leq b}f(k)=\sum \limits _{k=0}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m).\ }

This form is for example the source for the important Euler–MacLaurin expansion of the zeta function (B1 = 1/2)

ζ ( s ) = k = 0 m B k k ! s k 1 ¯ + R ( s , m ) = B 0 0 ! s 1 ¯ + B 1 1 ! s 0 ¯ + B 2 2 ! s 1 ¯ + + R ( s , m ) = 1 s 1 + 1 2 + 1 12 s + + R ( s , m ) . {\displaystyle {\begin{aligned}\zeta (s)&=\sum _{k=0}^{m}{\frac {B_{k}}{k!}}s^{\overline {k-1}}+R(s,m)\\&={\frac {B_{0}}{0!}}s^{\overline {-1}}+{\frac {B_{1}}{1!}}s^{\overline {0}}+{\frac {B_{2}}{2!}}s^{\overline {1}}+\cdots +R(s,m)\\&={\frac {1}{s-1}}+{\frac {1}{2}}+{\frac {1}{12}}s+\cdots +R(s,m).\end{aligned}}}

Here s k ¯ {\displaystyle s^{\overline {k}}} denotes the rising factorial power.

Bernoulli numbers are also frequently used in other kinds of asymptotic expansions. The following example is the classical Poincaré-type asymptotic expansion of the digamma function (again B1 = 1/2).

ψ ( z ) ln z k = 1 B k k z k {\displaystyle \psi (z)\sim \ln z-\sum _{k=1}^{\infty }{\frac {B_{k}}{kz^{k}}}}

Combinatorial definitions

The connection of the Bernoulli number to various kinds of combinatorial numbers is based on the classical theory of finite differences and on the combinatorial interpretation of the Bernoulli numbers as an instance of a fundamental combinatorial principle, the inclusion-exclusion principle.

Connection with the Worpitzky number

The definition to proceed with was developed by Julius Worpitzky in 1883. Besides elementary arithmetic only the factorial function n! and the power function k is employed. The signless Worpitzky numbers are defined as

W n , k = v = 0 k ( 1 ) v + k ( v + 1 ) n k ! v ! ( k v ) !   . {\displaystyle W_{n,k}=\sum _{v=0}^{k}(-1)^{v+k}\left(v+1\right)^{n}{\frac {k!}{v!(k-v)!}}\ .}

They can also be expressed through the Stirling set number

W n , k = k ! { n + 1 k + 1 } . {\displaystyle W_{n,k}=k!\left\{{\begin{matrix}n+1\\k+1\end{matrix}}\right\}.}

A Bernoulli number is then introduced as an inclusion-exclusion sum of Worpitzky numbers weighted by the sequence 1, 1/2, 1/3,...

B n = k = 0 n ( 1 ) k W n , k k + 1   =   k = 0 n 1 k + 1 v = 0 k ( 1 ) v ( k v ) ( v + 1 ) n   . {\displaystyle B_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {W_{n,k}}{k+1}}\ =\ \sum _{k=0}^{n}{\frac {1}{k+1}}\sum _{v=0}^{k}(-1)^{v}{k \choose v}\left(v+1\right)^{n}\ .}

This representation has B1 = 1/2.

Worpitzky's representation of the Bernoulli number
B0  =  1/1
B1  =  1/1 − 1/2
B2  =  1/1 − 3/2 + 2/3
B3  =  1/1 − 7/2 + 12/3 − 6/4
B4  =  1/1 − 15/2 + 50/3 − 60/4 + 24/5
B5  =  1/1 − 31/2 + 180/3 − 390/4 + 360/5 − 120/6
B6  =  1/1 − 63/2 + 602/3 − 2100/4 + 3360/5 − 2520/6 + 720/7

A second formula representing the Bernoulli numbers by the Worpitzky numbers is for n ≥ 1

B n = n 2 n + 1 2 k = 0 n 1 ( 2 ) k W n 1 , k   . {\displaystyle B_{n}={\frac {n}{2^{n+1}-2}}\sum _{k=0}^{n-1}(-2)^{-k}\;W_{n-1,k}\ .}

Connection with the Stirling set number

A similar combinatorial representation derives from

B n = k = 0 n ( 1 ) k + n k ! k + 1 { n k }   . {\displaystyle B_{n}=\sum _{k=0}^{n}(-1)^{k+n}{\frac {k!}{k+1}}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}\ .}

Here the Bernoulli numbers are an inclusion-exclusion over the set of length-n words, where the sum is taken over all words of length n with k distinct letters, and normalized by k + 1. The combinatorics of this representation can be seen from:

B 0 = + |   { }   | {\displaystyle B_{0}=+\left\vert \ \left\{\varnothing \right\}\ \right\vert }
B 1 = 1 1 |     | + 1 2 | { a } | {\displaystyle B_{1}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{a\right\}\right\vert }
B 2 = + 1 1 |     | 1 2 | { a a } | + 1 3 | { a b , b a } | {\displaystyle B_{2}=+{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert -{\frac {1}{2}}\left\vert \left\{aa\right\}\right\vert +{\frac {1}{3}}\left\vert \left\{ab,ba\right\}\right\vert }
B 3 = 1 1 |     | + 1 2 | { a a a } | 1 3 | { a a b , a b a , b a a , a b b , b a b , b b a } | + 1 4 | { a b c , a c b , b a c , b c a , c a b , c b a } | {\displaystyle B_{3}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{aaa\right\}\right\vert -{\frac {1}{3}}\left\vert \left\{aab,aba,baa,abb,bab,bba\right\}\right\vert +{\frac {1}{4}}\left\vert \left\{abc,acb,bac,bca,cab,cba\right\}\right\vert }

Connection with the Stirling cycle number

Let [ n m ] {\displaystyle \left} denote the signless Stirling cycle number. The two main formulas relating these number to the Bernoulli number (B1 = 1/2) are

1 m ! k = 0 m ( 1 ) k [ m + 1 k + 1 ] B k = 1 m + 1   , {\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\leftB_{k}={\frac {1}{m+1}}\ ,}

and the inversion of this sum (for n ≥ 0, m ≥ 0)

1 m ! k = 0 m ( 1 ) k [ m + 1 k + 1 ] B n + k = A n , m   . {\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\leftB_{n+k}=A_{n,m}\ .}

Here the number An,m are the rational Akiyama-Tanigawa number, the first few of which are displayed in the following table.

Akiyama-Tanigawa number
n \ m 0 1 2 3 4
0 11/21/31/41/5
1 1/21/31/41/5...
2 1/61/63/20......
3 01/30.........
4 −1/30............

These relations lead to a simple algorithm to compute the Bernoulli number. The input is the first row, A0,m = 1/(m + 1) and the output are the Bernoulli number in the first column An,0 = Bn . This transformation is shown in pseudo-code below.

Akiyama-Tanigawa algorithm for Bn
Enter integer n.

For m from 0 by 1 to n do

    A ← 1/(m + 1)
    For j from m by −1 to 1 do
   
    A ← j × (A − A)
Return A  (which is Bn).

Connection with the Eulerian number

The Eulerian number are the number of permutations of {1,2,...,n} with m ascents. They were introduced by Leonhard Euler in 1755 and in the notation of D. E. Knuth written as n m . {\displaystyle \left\langle {n \atop m}\right\rangle .} The two main formulas connecting the Eulerian number to the Bernoulli number are:

m = 0 n ( 1 ) m n m = 2 n + 1 ( 2 n + 1 1 ) B n + 1 n + 1   , {\displaystyle \sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }=2^{n+1}(2^{n+1}-1){\frac {B_{n+1}}{n+1}}\ ,}
m = 0 n ( 1 ) m n m ( n m ) 1 = ( n + 1 ) B n   . {\displaystyle \sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }{\binom {n}{m}}^{-1}=(n+1)B_{n}\ .}

Both formulas are valid for n ≥ 0 if B1 is set to 1/2. If B1 is set to −1/2 they are valid only for n ≥ 1 and n ≥ 2 respectively.

A binary tree representation

The Stirling polynomials σn(x) are related to the Bernoulli numbers by Bn = nn(1). S. C. Woon (Woon 1997) described an algorithm to compute σn(1) as a binary tree.

Woon's tree for σn(1)

Woon's recursive algorithm (for n ≥ 1) starts by assigning to the root node N = . Given a node N = [a1,a2,..., ak] of the tree, the left child of the node is L(N) =  and the right child R(N) = .

Given a node N the factorial of N is defined as

N ! = a 1 k = 2... length ( N ) a k !   . {\displaystyle N!=a_{1}\prod _{k=2...{\text{length}}(N)}a_{k}!\ .}

Restricted to the nodes N of a fixed tree-level n the sum of 1/N! is σn(1), thus

B n = N   node of tree-level   n n ! N !   . {\displaystyle B_{n}=\sum _{N\ {\text{node of tree-level}}\ n}{\frac {n!}{N!}}\ .}

For example B1 = 1!(1/2!), B2 = 2!(-1/3!+1/(2!2!)), B3 = 3!(1/4!-1/(2!3!)-1/(3!2!)+1/(2!2!2!)).

Asymptotic approximation

Leonhard Euler expressed the Bernoulli numbers in terms of the Riemann zeta function as

B 2 n = ( 1 ) n + 1 2 ( 2 n ) ! ( 2 π ) 2 n [ 1 + 1 2 2 n + 1 3 2 n + 1 4 2 n + ] . {\displaystyle B_{2n}=(-1)^{n+1}{\frac {2(2n)!}{(2\pi )^{2n}}}\left.}

It then follows from the Stirling formula that, as n goes to infinity,

| B 2 n | 4 π n ( n π e ) 2 n . {\displaystyle |B_{2n}|\sim 4{\sqrt {\pi n}}\left({\frac {n}{\pi e}}\right)^{2n}.}

Integral representation and continuation

The integral

b ( s ) = 2 e s i π / 2 0 s t s 1 e 2 π t d t t {\displaystyle b(s)=2e^{si\pi /2}\int _{0}^{\infty }{\frac {st^{s}}{1-e^{2\pi t}}}{\frac {dt}{t}}}

has as special values b(2n) = B2n for n > 0. The integral might be considered as a continuation of the Bernoulli numbers to the complex plane and this was indeed suggested by Peter Luschny in 2004.

For example b(3) = (3/2)ζ(3)ΠΙ and b(5) = −(15/2) ζ(5) ΠΙ. Here ζ(n) denotes the Riemann zeta function and Ι the imaginary unit. It is remarkable that already Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

p = 3 2 π 3 ( 1 + 1 2 3 + 1 3 3 + etc.   ) = 0.0581522   , {\displaystyle p={\frac {3}{2\pi ^{3}}}\left(1+{\frac {1}{2^{3}}}+{\frac {1}{3^{3}}}+{\text{etc.}}\ \right)=0.0581522\ldots \ ,}
q = 15 2 π 5 ( 1 + 1 2 5 + 1 3 5 + etc.   ) = 0.0254132   . {\displaystyle q={\frac {15}{2\pi ^{5}}}\left(1+{\frac {1}{2^{5}}}+{\frac {1}{3^{5}}}+{\text{etc.}}\ \right)=0.0254132\ldots \ .}

Euler's values are unsigned and real, but obviously his aim was to find a meaningful way to define the Bernoulli numbers at the odd integers n > 1.

The relation to the Euler numbers and π

The Euler numbers are a sequence of integers intimately connected with the Bernoulli numbers. Comparing the asymptotic expansions of the Bernoulli and the Euler numbers shows that the Euler numbers E2n are in magnitude approximately (2/π)(4 − 2) times larger than the Bernoulli numbers B2n. In consequence:

π     2 ( 2 2 n 4 2 n ) B 2 n E 2 n   . {\displaystyle \pi \ \sim \ 2\left(2^{2n}-4^{2n}\right){\frac {B_{2n}}{E_{2n}}}\ .}

This asymptotic equation reveals that π lies in the common root of both the Bernoulli and the Euler numbers. In fact π could be computed from these rational approximations.

Bernoulli numbers can be expressed through the Euler numbers and vice versa. Since for n odd BnEn = 0 (with the exception B1), it suffices to regard the case when n is even.

B n = k = 0 n 1 ( n 1 k ) n 4 n 2 n E k ( n = 2 , 4 , 6 , ) {\displaystyle B_{n}=\sum _{k=0}^{n-1}{\binom {n-1}{k}}{\frac {n}{4^{n}-2^{n}}}E_{k}\quad (n=2,4,6,\ldots )}
E n = k = 1 n ( n k 1 ) 2 k 4 k k B k ( n = 2 , 4 , 6 , ) . {\displaystyle E_{n}=\sum _{k=1}^{n}{\binom {n}{k-1}}{\frac {2^{k}-4^{k}}{k}}B_{k}\quad (n=2,4,6,\ldots ).}

These conversion formulas express an inverse relation between the Bernoulli and the Euler numbers. But more important, there is a deep arithmetic root common to both kinds of numbers, which can be expressed through a more fundamental sequence of numbers, also closely tied to π. These numbers are defined for n > 1 as

S n = 2 ( 2 π ) n k = ( 4 k + 1 ) n ( k = 0 , 1 , 1 , 2 , 2 , ) {\displaystyle S_{n}=2\left({\frac {2}{\pi }}\right)^{n}\sum _{k=-\infty }^{\infty }\left(4k+1\right)^{-n}\quad (k=0,-1,1,-2,2,\ldots )}

and S1 = 1 by convention (Elkies 2003). The magic of these numbers lies in the fact that they turn out to be rational numbers. This was first proved by Leonhard Euler 1734 in a landmark paper `De summis serierum reciprocarum' (On the sums of series of reciprocals) and fascinated mathematicians ever since. The first few of these numbers are

1 , 1 , 1 2 , 1 3 , 5 24 , 2 15 , 61 720 , 17 315 , 277 8064 , 62 2835 , {\displaystyle 1,1,{\frac {1}{2}},{\frac {1}{3}},{\frac {5}{24}},{\frac {2}{15}},{\frac {61}{720}},{\frac {17}{315}},{\frac {277}{8064}},{\frac {62}{2835}},\ldots }

The Bernoulli numbers and Euler numbers are best understood as special views of these numbers, selected from the sequence Sn and scaled for use in special applications.

B n = ( 1 ) n / 2 [ n   even ] n ! 2 n 4 n S n   , ( n = 2 , 3 , )   , {\displaystyle B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left{\frac {n!}{2^{n}-4^{n}}}\,S_{n}\ ,\quad (n=2,3,\ldots )\ ,}
E n = ( 1 ) n / 2 [ n   even ] n ! S n + 1 ( n = 0 , 1 , )   . {\displaystyle E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\leftn!\,S_{n+1}\quad \qquad (n=0,1,\ldots )\ .}

The expression has the value 1 if n is even and 0 otherwise (Iverson bracket).

These identities show that the quotient of Bernoulli and Euler numbers at the beginning of this section is just the special case of Rn = 2 Sn / Sn+1 when n is even. The Rn are rational approximations to π and two successive terms always enclose the true value of π. Beginning with n = 1 the sequence starts

2 , 4 , 3 , 16 5 , 25 8 , 192 61 , 427 136 , 4352 1385 , π   . {\displaystyle 2,4,3,{\frac {16}{5}},{\frac {25}{8}},{\frac {192}{61}},{\frac {427}{136}},{\frac {4352}{1385}},\ldots \quad \longrightarrow \pi \ .}

These rational numbers also appear in the last paragraph of Euler's paper cited above. But it was only in September 2007 that this classical sequence found its way into the Encyclopedia of Integer Sequences (A132049).

An algorithmic view: the Seidel triangle

The sequence Sn has another unexpected yet important property: The denominators of Sn divide the factorial (n − 1)!. In other words: the numbers Tn = Sn(n − 1)! are integers.

T n = 1 , 1 , 1 , 2 , 5 , 16 , 61 , 272 , 1385 , 7936 , 50521 , 353792 , ( n = 1 , 2 , ) {\displaystyle T_{n}=1,1,1,2,5,16,61,272,1385,7936,50521,353792,\ldots \quad (n=1,2,\ldots )}

Thus the above representations of the Bernoulli and Euler numbers can be rewritten in terms of this sequence as

B n = ( 1 ) n / 2 [ n  even ] n 2 n 4 n T n   , ( n = 2 , 3 , )   , {\displaystyle B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left{\frac {n}{2^{n}-4^{n}}}\,T_{n}\ ,\quad (n=2,3,\ldots )\ ,}
E n = ( 1 ) n / 2 [ n  even ] T n + 1 ( n = 0 , 1 , )   . {\displaystyle E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\leftT_{n+1}\quad \quad \qquad (n=0,1,\ldots )\ .}

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers En are given immediately by T2n + 1 and the Bernoulli numbers B2n are obtained from T2n by some easy shifting, avoiding rational arithmetic.

What remains is to find a convenient way to compute the numbers Tn. However, already in 1877 Philipp Ludwig von Seidel published an ingenious algorithm which makes it extremely simple to calculate Tn.

Seidel's algorithm for Tn

Start by putting 1 in row 0 and let k denote the number of the row currently being filled. If k is odd, then put the number on the left end of the row k − 1 in the first position of the row k, and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper. At the end of the row duplicate the last number. If k is even, proceed similar in the other direction.

Seidel's algorithm is in fact much more general (see the exposition of Dominique Dumont (1981)) and was rediscovered several times thereafter.

Similar to Seidel's approach D. E. Knuth and T. J. Buckholtz (1967) gave a recurrence equation for the numbers T2n and recommended this method for computing B2n and E2n ‘on electronic computers using only simple operations on integers’.

V. I. Arnold rediscovered Seidel's algorithm in 1991 and later Millar, Sloane and Young popularized Seidel's algorithm under the name boustrophedon transform.

A combinatorial view: alternating permutations

Main article: Alternating permutations

Around 1880, three years after the publication of Seidel's algorithm, Désiré André proved a now classic result of combinatorial analysis. Looking at the first terms of the Taylor expansion of the trigonometric functions tan x and sec x André made a startling discovery.

tan x = 1 x 1 ! + 2 x 3 3 ! + 16 x 5 5 ! + 272 x 7 7 ! + 7936 x 9 9 ! + {\displaystyle \tan x=1{\frac {x}{1!}}+2{\frac {x^{3}}{3!}}+16{\frac {x^{5}}{5!}}+272{\frac {x^{7}}{7!}}+7936{\frac {x^{9}}{9!}}+\cdots }
sec x = 1 + 1 x 2 2 ! + 5 x 4 4 ! + 61 x 6 6 ! + 1385 x 8 8 ! + 50521 x 10 10 ! + {\displaystyle \sec x=1+1{\frac {x^{2}}{2!}}+5{\frac {x^{4}}{4!}}+61{\frac {x^{6}}{6!}}+1385{\frac {x^{8}}{8!}}+50521{\frac {x^{10}}{10!}}+\cdots }

The coefficients are the Euler numbers of odd and even index, respectively. In consequence the ordinary expansion of tan x + sec x has as coefficients the rational numbers Sn.

tan x + sec x = 1 + 1 x + 1 2 x 2 + 1 3 x 3 + 5 24 x 4 + 2 15 x 5 + 61 720 x 6 + {\displaystyle \tan x+\sec x=1+1x+{\frac {1}{2}}x^{2}+{\frac {1}{3}}x^{3}+{\frac {5}{24}}x^{4}+{\frac {2}{15}}x^{5}+{\frac {61}{720}}x^{6}+\cdots }

André then succeeded by means of a recurrence argument to show that the alternating permutations of odd size are enumerated by the Euler numbers of odd index (also called tangent numbers) and the alternating permutations of even size by the Euler numbers of even index (also called secant numbers).

Generalizations by polynomials

The Bernoulli polynomials can be regarded as generalizations of the Bernoulli numbers the same as the Euler polynomials are generalizations of the Euler numbers.

Arithmetical properties of the Bernoulli numbers

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as Bn = − nζ(1 − n) for integers n ≥ 0 provided for n = 0 and n = 1 the expression − nζ(1 − n) is understood as the limiting value and the convention B1 = 1/2 is used. This intimately relates them to the values of the zeta function at negative integers. As such, they could be expected to have and do have deep arithmetical properties, a fact discovered by Kummer in his work on Fermat's last theorem.

Divisibility properties of the Bernoulli numbers are related to the ideal class groups of cyclotomic fields by a theorem of Kummer and its strengthening in the Herbrand-Ribet theorem, and to class numbers of real quadratic fields by Ankeny-Artin-Chowla. We also have a relationship to algebraic K-theory; if cn is the numerator of Bn/2n, then the order of K 4 n 2 ( Z ) {\displaystyle K_{4n-2}({\mathbb {Z} })} is −c2n if n is even, and 2c2n if n is odd.

The Agoh-Giuga conjecture postulates that p is a prime number if and only if pBp−1 is congruent to −1 mod p.

Von Staudt–Clausen theorem

The von Staudt–Clausen theorem was given by Karl Georg Christian von Staudt and Thomas Clausen independently in 1840. It describes the arithmetical structure of the Bernoulli numbers.

The von Staudt–Clausen theorem has two parts. The first one describes how the denominators of the Bernoulli numbers can be computed. Paraphrasing the words of Clausen it can be stated as:

“The denominator of the 2nth Bernoulli number can be found as follows: Add to all divisors of 2n, 1, 2, aa', ..., 2n the unity, which gives the sequence 2, 3, a + 1, a' + 1, ..., 2n + 1. Select from this sequence only the prime numbers 2, 3, pp', etc. and build their product.”

Clausen's algorithm translates almost verbatim to a modern computer algebra program, which looks similar to the pseudocode on the left hand site of the following table. On the right hand side the computation is traced for the input n = 88. It shows that the denominator of B88 is 61410. 

Clausen's algorithm for the denominator of Bn
Clausen: function(integer n)   |   n = 88
S = divisors(n);   |   {1, 2, 4, 8, 11, 22, 44, 88}
S = map(kk + 1, S);   |   {2, 3, 5, 9, 12, 23, 45, 89}
S = select(isprime, S);   |   {2, 3, 5, 23, 89}
return product(S);   |   61410

The second part of the von Staudt–Clausen theorem is a very remarkable representation of the Bernoulli numbers. This representation is given for the first few nonzero Bernoulli numbers in the next table.

Von Staudt–Clausen representation of Bn
B0  =  1
B1  =  1 − 1/2
B2  =  1 − 1/2 − 1/3
B4  =  1 − 1/2 − 1/3 − 1/5
B6  =  1 − 1/2 − 1/3 − 1/7
B8  =  1 − 1/2 − 1/3 − 1/5
B10  =  1 − 1/2 − 1/3 − 1/11

The theorem affirms the existence of an integer In such that

B n = I n 2 ( p 1 ) | n 1 p ( n 0 )   . {\displaystyle B_{n}={\frac {I_{n}}{2}}-\sum _{(p-1)|n}{\frac {1}{p}}\quad (n\geq 0)\ .}

The sum is over the primes p for which p − 1 divides n. These are the same primes which are employed in the Clausen algorithm. The proposition holds true for all n ≥ 0, not only for even n. I1 = 2 and for odd n > 1, In = 1.

Consequences of the von Staudt–Clausen theorem are: the denominators of the Bernoulli numbers are square-free and for n ≥ 2 divisible by 6.

Why do the odd Bernoulli numbers vanish?

The sum

φ k ( n ) = i = 0 n i k n k 2 {\displaystyle \varphi _{k}(n)=\sum _{i=0}^{n}i^{k}-{\frac {n^{k}}{2}}}

can be evaluated for negative values of the index n. Doing so will show that it is an odd function for even values of k, which implies that the sum has only terms of odd index. This and the formula for the Bernoulli sum imply that B2k+1−m is 0 for m odd and greater than 1; and that the term for B1 is cancelled by the subtraction. The von Staudt-Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for n > 1).

From the von Staudt-Clausen theorem it is known that for odd n > 1 the number 2Bn is an integer. This seems trivial if one knows beforehand that in this case Bn = 0. However, by applying Worpitzky's representation one gets

2 B n = m = 0 n ( 1 ) m 2 m + 1 m ! { n + 1 m + 1 } = 0 ( n > 1   odd ) {\displaystyle 2B_{n}=\sum _{m=0}^{n}\left(-1\right)^{m}{\frac {2}{m+1}}m!\left\{{\begin{matrix}n+1\\m+1\end{matrix}}\right\}=0\quad \left(n>1\ {\text{odd}}\right)}

as a sum of integers, which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let Sn,m be the number of surjective maps from {1, 2, ..., n} to {1, 2, ..., m}, then S n , m = m ! { n m } {\displaystyle S_{n,m}=m!\left\{{\begin{matrix}n\\m\end{matrix}}\right\}} . The last equation can only hold if

m = 1 , 3 , 5 , n 2 m 2 S n , m = m = 2 , 4 , 6 , n 2 m 2 S n , m ( n > 2  even ) .   {\displaystyle \sum _{m=1,3,5,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}=\sum _{m=2,4,6,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}\quad \left(n>2{\text{ even}}\right).\ }

This equation can be proved by induction. The first two examples of this equation are

n = 4 :  2 + 8 = 7 + 3,  n = 6:  2 + 120 + 144 = 31 + 195 + 40.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

p-adic continuity

An especially important congruence property of the Bernoulli numbers can be characterized as a p-adic continuity property. If b, m and n are positive integers such that m and n are not divisible by p − 1 and m n mod p b 1 ( p 1 ) {\displaystyle m\equiv n\,{\bmod {\,}}p^{b-1}(p-1)} , then

( 1 p m 1 ) B m m ( 1 p n 1 ) B n n mod p b . {\displaystyle (1-p^{m-1}){B_{m} \over m}\equiv (1-p^{n-1}){B_{n} \over n}\,{\bmod {\,}}p^{b}.}

Since B n = n ζ ( 1 n ) {\displaystyle B_{n}=-n\zeta (1-n)} , this can also be written

( 1 p u ) ζ ( u ) ( 1 p v ) ζ ( v ) mod p b , {\displaystyle (1-p^{-u})\zeta (u)\equiv (1-p^{-v})\zeta (v)\,{\bmod {\,}}p^{b}\,,}

where u = 1 − m and v = 1 − n, so that u and v are nonpositive and not congruent to 1 mod p − 1. This tells us that the Riemann zeta function, with 1 − p taken out of the Euler product formula, is continuous in the p-adic numbers on odd negative integers congruent mod p − 1 to a particular a 1 mod p 1 {\displaystyle a\not \equiv 1\,{\bmod {\,}}p-1} , and so can be extended to a continuous function ζ p ( s ) {\displaystyle \zeta _{p}(s)} for all p-adic integers Z p , {\displaystyle {\mathbb {Z}}_{p},\,} the p-adic Zeta function.

Bernoulli numbers and the Riemann hypothesis

The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of the Riemann hypothesis (RH) which uses only the Bernoulli number. In fact in 1916 M. Riesz proved that the RH is equivalent to the following assertion:

For every ε > 1/4 there exists a constant Cε > 0 (depending on ε) such that |R(x)| < Cε x as x → ∞.

Here R(x) is the Riesz function

R ( x ) = 2 k = 1 k k ¯ x k ( 2 π ) 2 k ( B 2 k / ( 2 k ) ) = 2 k = 1 k k ¯ x k ( 2 π ) 2 k β 2 k .   {\displaystyle R(x)=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\left(B_{2k}/(2k)\right)}}=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\beta _{2k}}}.\ }

n k ¯ {\displaystyle n^{\overline {k}}} denotes the rising factorial power in the notation of D. E. Knuth. The number βn = Bn/n occur frequently in the study of the zeta function and are significant because βn is a p-integer for primes p where p − 1 does not divide n. The βn are called divided Bernoulli number.

Ramanujan's congruences

The following relations, due to Ramanujan, provide a more efficient method for calculating Bernoulli numbers:

m 0 mod 6 ( m + 3 m ) B m = m + 3 3 j = 1 m / 6 ( m + 3 m 6 j ) B m 6 j {\displaystyle m\equiv 0\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{m/6}{m+3 \choose {m-6j}}B_{m-6j}}
m 2 mod 6 ( m + 3 m ) B m = m + 3 3 j = 1 ( m 2 ) / 6 ( m + 3 m 6 j ) B m 6 j {\displaystyle m\equiv 2\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{(m-2)/6}{m+3 \choose {m-6j}}B_{m-6j}}
m 4 mod 6 ( m + 3 m ) B m = m + 3 6 j = 1 ( m 4 ) / 6 ( m + 3 m 6 j ) B m 6 j . {\displaystyle m\equiv 4\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}=-{{m+3} \over 6}-\sum _{j=1}^{(m-4)/6}{m+3 \choose {m-6j}}B_{m-6j}.}

Use of Bernoulli numbers in topology

The Kervaire–Milnor formula for the order of the cyclic group of diffeomorphism classes of exotic (4n − 1)-spheres which bound parallelizable manifolds for n ≥ 2 involves Bernoulli numbers; if B(n) is the numerator of B4n/n, then

2 2 n 2 ( 1 2 2 n 1 ) B ( n ) {\displaystyle 2^{2n-2}(1-2^{2n-1})B_{(n)}}

is the number of such exotic spheres. (The formula in the topological literature differs because topologists use a different convention for naming Bernoulli numbers; this article uses the number theorists' convention.)

Assorted identities

  • Let n be non-negative and even
    ζ ( n ) = ( 1 ) n 2 1 B n ( 2 π ) n 2 n ! {\displaystyle \zeta (n)={\frac {\left(-1\right)^{{\frac {n}{2}}-1}B_{n}\left(2\pi \right)^{n}}{2n!}}}
  • The nth cumulant of the uniform probability distribution on the interval is Bn/n.
  • Let n¡ = 1/n! and n ≥ 1. Then Bn is n! times the determinant of the following matrix.
    Bn = 
    1
    ───
    n¡
    0 0 0 ... 0
    0 0 ... 0
    0 ... 0
    ... ... ... ... ... ... ...
    (n − 2)¡ ... ... 0
    (n − 1)¡ (n − 2)¡ ... ...
    n¡ (n − 1)¡ (n − 2)¡ ... ...

    Thus the determinant is σn(1), the Stirling polynomial at x = 1.

  • Let n ≥ 1.
    1 n k = 1 n ( n k ) B k B n k + B n 1 = B n (L. Euler) {\displaystyle {\frac {1}{n}}\sum _{k=1}^{n}{\binom {n}{k}}B_{k}B_{n-k}+B_{n-1}=-B_{n}\quad {\text{(L. Euler)}}}
  • Let n ≥ 1. Then (von Ettingshausen 1827)
    k = 0 n ( n + 1 k ) ( n + k + 1 ) B n + k = 0 {\displaystyle \sum _{k=0}^{n}{\binom {n+1}{k}}(n+k+1)B_{n+k}=0}
  • Let n ≥ 0. Then (Leopold Kronecker 1883)
    B n = k = 1 n + 1 ( 1 ) k k ( n + 1 k ) j = 1 k j n {\displaystyle B_{n}=-\sum _{k=1}^{n+1}{\frac {(-1)^{k}}{k}}{\binom {n+1}{k}}\sum _{j=1}^{k}j^{n}}
  • Let n ≥ 1 and m ≥ 1. Then (Carlitz 1968)
    ( 1 ) m r = 0 m ( m r ) B n + r = ( 1 ) n s = 0 n ( n s ) B m + s {\displaystyle (-1)^{m}\sum _{r=0}^{m}{\binom {m}{r}}B_{n+r}=(-1)^{n}\sum _{s=0}^{n}{\binom {n}{s}}B_{m+s}}
  • Let n ≥ 4 and
    H n = 1 k n k 1 {\displaystyle H_{n}=\sum _{1\leq k\leq n}k^{-1}}
    the harmonic number. Then
    n 2 k = 2 n 2 B n k n k B k k k = 2 n 2 ( n k ) B n k n k B k = H n B n (H. Miki, 1978) {\displaystyle {\frac {n}{2}}\sum _{k=2}^{n-2}{\frac {B_{n-k}}{n-k}}{\frac {B_{k}}{k}}-\sum _{k=2}^{n-2}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}=H_{n}B_{n}\qquad {\text{(H. Miki, 1978)}}}
  • Let n ≥ 4. Yuri Matiyasevich found (1997)
    ( n + 2 ) k = 2 n 2 B k B n k 2 l = 2 n 2 ( n + 2 l ) B l B n l = n ( n + 1 ) B n {\displaystyle (n+2)\sum _{k=2}^{n-2}B_{k}B_{n-k}-2\sum _{l=2}^{n-2}{\binom {n+2}{l}}B_{l}B_{n-l}=n(n+1)B_{n}}
  • Let n ≥ 1
    n 2 ( B n 1 ( x ) + k = 1 n 1 B k ( x ) k B n k ( x ) n k ) k = 0 n 1 ( n k ) B n k n k B k ( x ) = H n 1 B n ( x ) {\displaystyle {\frac {n}{2}}\left(B_{n-1}(x)+\sum _{k=1}^{n-1}{\frac {B_{k}(x)}{k}}{\frac {B_{n-k}(x)}{n-k}}\right)-\sum _{k=0}^{n-1}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}(x)=H_{n-1}B_{n}(x)}
    This is an identity by Faber-Pandharipande-Zagier-Gessel. Choose x = 0 or x = 1 to get a Bernoulli number identity according to your favourite convention.
  • The next formula is true for n ≥ 0 if B1 = B1(1) = 1/2, but only for n ≥ 1 if B1 = B1(0) = -1/2.
    k = 0 n ( n k ) B k n k + 2 = B n + 1 n + 1 {\displaystyle \sum _{k=0}^{n}{\binom {n}{k}}{\frac {B_{k}}{n-k+2}}={\frac {B_{n+1}}{n+1}}}
  • Let n ≥ 0 and = 1 if b is true, 0 otherwise.
    1 + k = 0 n ( n k ) 2 n k + 1 n k + 1 B k ( 1 ) = 2 n {\displaystyle -1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(1)=2^{n}}
    and
    1 + k = 0 n ( n k ) 2 n k + 1 n k + 1 B k ( 0 ) = [ n = 0 ] {\displaystyle -1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(0)=}

See also

Notes

  1. Note G in the Menabrea reference
  2. Mathematics Genealogy Project
  3. Earliest Uses of Symbols of Calculus
  4. Concrete Mathematics, (2.51)
  5. Concrete Mathematics, (9.67).
  6. Concrete Mathematics, (2.44) and (2.52)
  7. Concrete Mathematics, Table 254 Euler's triangle.

References

External links

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