Introduction on Bernoulli's numbers

 

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1  Introduction

Bernoulli's numbers play an important and quite mysterious role in mathematics and in various places like analysis, number theory and differential topology. They first appeared in Ars Conjectandi, page 97, a famous (and posthumous) treatise published in 1713, by Jakob Bernoulli (1654-1705) when he studied the sums of powers of consecutive integers

sp(n)= n-1
å
k=1 
 kp,
(1)
where p and n are two given positive integers.

Bernoulli's numbers also appear in the computation of the numbers

z(2p)= ¥
å
k=1 
  1
k2p
and in the expansion of many usual functions as tan(x), tanh(x), 1/sin(x), ¼

Perhaps one of the most important result is Euler-Maclaurin summation formula, where Bernoulli's numbers are contained and which allows to accelerate the computation of slow converging series (see the essay on Euler's constant at [9]). They also appear in numbers theory (Fermat's theorem) and in many other domains and have caused the creation of a huge literature (see the 2700 and more entries enumerated in [6]).

According to Louis Saalschültz [17], the term Bernoulli's numbers was used for the first time by Abraham De Moivre (1667-1754) and also by Leonhard Euler (1707-1783) in 1755.

2  Bernoulli's approach

During the first years of the Calculus period and in its first integral computations of the function x® xp, Pierre de Fermat (1601-1665) in 1636 had to evaluate the sums sp(n) defined by (1). You can see this by replacing the area under the curve x® xp by it's rectangular approximations and naturally comes the need to compute sp(n).

Also in 1631, Johann Faulhaber (1580-1635) developed explicit formulas for these sums up to p=17 (read the excellent [12] for the beginnings of integration and [18] for some excerpts of Bernoulli's work). Thus, it was already known to Jakob Bernoulli that

s0(n)
=
n
s1(n)
=
 1
2
 n2-  1
2
 n=  n(n-1)
2
s2(n)
=
 1
3
 n3-  1
2
 n2+  1
6
 n=  n(n-1)(2n -1)
6
s3(n)
=
 1
4
 n4-  1
2
 n3+  1
4
 n2=  n2(n-1)2
4
s4(n)
=
 1
5
 n5-  1
2
 n4+  1
3
 n3-  1
30
 n=  n(n-1)(2n -1)(3n2 -3n -1)
30
s5(n)
=
 1
6
 n6-  1
2
 n5+  5
12
 n4-  1
12
 n2=  n2(2n2-2n -1)(n-1) 2
12
...

Jakob Bernoulli then, empirically, noticed that the polynomials sp(n) have the form

sp(n)=  1
p+1
 np+1-  1
2
 np+  p
12
 np-1+0×np -2+...

In this expression, the numbers (1,-1/2,1/12,0,...) are appearing and do not depend on p. More generally, the sums sp(n) can be written in the form


sp(n)
=
p
å
k=0 
  Bk
k!
  p!
(p+1-k)!
 np+1-k
=
 B0
0!
  np+1
p+1
 +  B1
1!
 np+  B2
2!
 pnp-1+  B3
3!
 p(p-1)np -2+...+  Bp
1!
 n
where the Bkare numbers which are independent of p and called Bernoulli's numbers.

We find by identification

B0=1,B1=-  1
2
 ,B2=  1
6
 ,B3=0,...

To illustrate the usefulness of his formula, Bernoulli computed the astonishing value of s10(1000) with little effort (in less than ''half a quarter of an hour'' he says ... [18])

91409924241424243424241924242500
(you can check it !). To achieve this he needed to find B0up to B10.

There is good evidence that the famous Japanese mathematician, Seki Takakazu (1642-1708) also discovered Bernoulli's numbers at the same time. The famous Indian mathematician Srinivasa Ramanujan (1887-1920) independently studied and rediscovered those numbers in 1904. He wrote one of his first article on this subject in 1911 [15].

3  A more modern definition

An equivalent definition of the Bernoulli's numbers is obtained from the series expansion of the function z/(ez-1):

G(z)=  z
ez-1
 = ¥
å
k=0 
 Bk  zk
k!
       | z| < 2p
(2)

In other words, the generating function of the Bernoulli's numbers Bk is z/(ez-1). The first terms of the expansion of this function are

G(z)= æ
è 		1+  z
2!
 +  z2
3!
 +  z3
4!
 +... ö
ø 		-1

 
 =1-  z
2
 +  z2
12
 -  z4
720
 +  z6
30240
 -  z8
1209600
 +...
which permit to obtain the first value of the Bernoulli's numbers:
B0=1
,
    B1=-  1
2
 ,    B2=  1
6
 ,    B3=0,
B4=-  1
30
,
    B5=0,    B6=  1
42
 ,   B7=0,    B8=-  1
30
 .

Further, we observe that

G(z)+  z
2
 =  z
2
 æ
è 		 2
ez-1
 +1 ö
ø 		=  z
2
  ez/2+e-z/2
ez/2-e -z/2
 =  z
2
 coth  z
2
 ,
where coth is the hyperbolic tangent, hence G(z)+z/2 is an even function and consequently every Bernoulli's numbers of the form B2k+1(k > 0) is null.

3.1  Bernoulli's polynomials

With a little modification it's possible to define Bernoulli's polynomials Bk(x) by

G(z,x)=  zezx
ez-1
 = ¥
å
k=0 
 Bk(x)  zk
k!
and G(z,0)=G(z) hence the value of the function Bk(x) at x=0 is Bkand because
 G
x
 (z,x)=zG(z,x)= ¥
å
k=0 
  dBk
dx
 (x)  zk
k!
it follows the important relation
 dBk
dx
 (x)=kBk-1(x).

Then, it's easy to deduce that Bk(x) are polynomials of degree k, and the first one are

B0(x)
=
1
B1(x)
=
x-  1
2
B2(x)
=
x2-x+  1
6
B3(x)
=
x3-  3
2
 x2+  1
2
 x
B4(x)
=
x4-2x3+x 2-  1
30
B5(x)
=
x5-  5
2
 x4+  5
3
 x3-  1
6
 x
...

Thanks to Bernoulli's polynomials, it's possible to rewrite the expression of the sums sp(n) as


sp(n)= n-1
å
k=0 
 kp=  1
p+1
 ( Bp+1(n)-Bp+1).

There are many relations with these polynomials, for example

Bk(1-x)
=
(-1)kBk(x),
(-1)kBk( -x)
=
Bk(x)+kxk-1,
| B2k(x)|
<
| B2k|        k=1,2,... and 0 < x < 1,
Bk æ
è 		 1
2
 ö
ø
=
-(1-2 1-k)Bk       k=0,1,...
...

Consult [1] for other formulas.

4  Properties

4.1  Recurrence relation

In the expression

sp(n)= p
å
k=0 
  Bk
k!
  p!
(p+1-k)!
 np+1-k
we let n=1, giving
0= p
å
k=0 
  Bk
k!
  1
(p+1-k)!
or equivalently
Bp=-  1
p+1
 p-1
å
k=0 
 \binomp+1kBk.
(3)

The recurrence relation (3) allows an easy generation of Bernoulli's numbers and shows that the numbers Bp are all rational numbers. It's convenient to rewrite this relation as the symbolic equation

(B+1)p+1-Bp+1=0,
and expand the binomial (B+1)p+1where each power Bkmust be replaced by Bk.

Example 1 With p=4 we have

5B4+10B3+10B2+5B1+B 0=0
thus
5B4+ æ
è 		0+10.  1
6
 -5.  1
2
 +1 ö
ø 		=0    therefore    B4=-  1
30
 .

4.2  Bernoulli's numbers and the zeta function

In 1735, the solution of the Basel problem, expressed by Jakob Bernoulli some years before, was one of Euler's most sensational discovery. The problem was to find the limit of

z(2)= ¥
å
n=1 
  1
n2
 ,
he found it to be p2/6. He also discovered the values of the sums
z(2k)= ¥
å
n=1 
  1
n2k
for k up to 13 ([8] and for more details [7]).

It's an extraordinary result that z(2k) can be expressed with Bernoulli's numbers ; the values of these sums are given by

z(2k)= ¥
å
n=1 
  1
n2k
 =  4k|B2k| p2k
2(2k)!
        k > 0,
(4)
(a proof based on two different expansions of z cot(z) is given in [4] p. 383). No similar expression is known for the odd values of the Zeta function.

On the extension of this function to negative values, we also have

z(1-2k)= -  B2k
2k
        k > 0,
which may also be used to compute B2k(see [5]).

4.3  Asymptotic expansion of Bernoulli's numbers

From the previous relation (4) with the Zeta function, it's clear that


| B2k| =  2(2k)!
(2p)2k
 z(2k)
(5)
and because, when k becomes large, thanks to Stirling's formula
z(2k)
~
1,
(2k)!
~
(2k)2ke-2k
Ö
 
4pk
 
,
we have
| B2k| ~ 4 æ
è 		 k
pe
 ö
ø 		2k

 
Ö
 
pk
 
.

In [16], the following results describes how the numerator N2k of B2k grows with k:

log| N2k| = 2klog(k)+O(k).

4.4  Bounds

It may be useful to estimate bounds for B2k, to achieve this we use the following relation between the function z(s) and the alternating series za(s)

z(s)
=
 za(s)
1-21-s
za(s)
=
¥
å
n=1 
  (-1)n -1
ns
and since
1-  1
2s
 < za(s) < 1,
we have the bounds for z(s)
 1-2-s
1-21-s
 < z(s) <  1
1-21-s
 .

If we use this relation with (5), the bounds for B2k are therefore

 2(2k)!(1-4 -k)
(2p)2k(1 -2.4-k)
 < | B2k| <  2(2k)!
(2p)2k(1 -2.4-k)
 .

5  Clausen-von Staudt's theorem

The following famous and important theorem was published in 1840 by Karl von Staudt (1798-1867) and it allows to compute easily the fractional part of Bernoulli's numbers (thus it also permits to compute the denominator of those numbers). This theorem was discovered the same year, independently, by Thomas Clausen (1801-1885).

Theorem 2 The value B2k, added to the sum of the inverse of prime numbers p such that (p-1) divides 2k, is an integer. In other words,

-B2k º
å
(p-1) | 2k 
  1
p
 mod 1

Proof. A complete proof is given in [11], p. 91.  

When k > 1, we observe that the primes p= 2,3 are such as (p-1) divides 2k. Let's illustrate this theorem with a few examples. For k=1, it becomes

-B2
º

å
(p-1) |2 
  1
p
 =  1
2
 +  1
3
 =  5
6
 mod 1
B2
º
 1
6
 mod 1
for k=5
-B10
º

å
(p-1) |10 
  1
p
 =  1
2
 +  1
3
 +  1
11
 =  61
66
 mod 1
B10
º
 5
66
 mod 1
and for k=8
-B16
º

å
(p-1) |16 
  1
p
 =  1
2
 +  1
3
 +  1
5
 +  1
17
 =  47
510
 mod 1
B16
º
 463
510
 mod 1

Corollary 3 (due to Rado 1934) For every prime numbers k of the form 3n+1

B2k º  1
6
 mod 1.

Proof. It's an easy consequence of the Staudt's theorem since p-1 divides 2k=2(3n+1) only if p-1 is one of 1,2,3n+1,6n+2, that is p is one of 2,3,3n+2,6n+3. But 6n+3 is divisible by 3 and 3n+2 is divisible by 2 because 3n+1 is prime so the only primes p candidates are 2 and 3.  

Example 4 The first primes of the form 3n+1 are 7,13,19,31,37,43,61,67,... hence we have

B14 º B26 º B38 º B62 º B74 º B86 º B122 º B134 º  1
6
 mod 1.

Clausen-von Staudt's theorem also permits to compute exactly a Bernoulli's number as soon as a sufficiently good approximation of it is known.

6  Expansion of usual functions

In a previous section we gave the definition

 z
ez-1
 = ¥
å
k=0 
 Bk  zk
k!
        |z| < 2p
and the consequence
 z
2
 coth  z
2
 = ¥
å
k=0 
 B2k  z2k
(2k)!
 .

It obviously leads to the two following expansions

zcoth(z)
=
¥
å
k=0 
 4kB2k  z2k
(2k)!
 =1+  z2
3
 -  z4
45
 +  2z6
945
 -...       | z| < p
zcot(z)
=
¥
å
k=0 
 (-4)kB2k  z2k
(2k)!
 =1-  z2
3
 -  z4
45
 -  2z6
945
 -...       | z| < p
where cot(z)=cos(z)/sin(z)=icoth(iz) is the cotangent function.

Now it's possible to find the expansion for tanh(z) and tan(z), if we observe that

2coth(2z)-coth(z)=2  cosh(2z)
sinh(2z)
 -  cosh(z)
sinh(z)
 =  cosh2(z)+sinh2(z)
sinh(z)cosh(z)
 -  cosh(z)
sinh(z)
 =tanh(z)
so that
tanh(z)
=
¥
å
k=1 
 4k(4k-1)B 2k  z2k-1
(2k)!
 =z-  z3
3
 +  2z5
15
 -  17z7
315
 +...       |z| <  p
2
tan(z)
=
¥
å
k=1 
 (-4)k(1 -4k)B2k  z2k-1
(2k)!
 =z+  z3
3
 +  2z5