It is precisely the somewhat random appearance of the Bernoulli numbers, coupled with the knowledge that they arise in fundamental circumstances, that make mathematicians feel they are looking at the manifestation of some deep and mysterious secret. ...
The Bernoulli numbers are among the most exotic creatures in all of mathematics. It is easy enough to define them--very briefly, they are the coefficients of $x$ in the expansion of the function $$ { x \over e^{x} - 1 } $$
in powers of $x$. I'll say something about the meaning of this definition in a moment. But my point here is that it raises more questions than it answers. Some major ones are: How on earth did anyone come up with that definition? Why did they come up with it? Why would you expect these numbers to have a significance beyond this single bare specification?
And yet, these numbers have fascinated mathematicians for several centuries. Here are a few of them: $$ \eqalign { \beta_{0} &= \phantom{-}1 \cr \beta_{1} &= -1/2 \cr \beta_{2} &= \phantom{-}1/6 \cr \beta_{4} &= -1/30 \cr \beta_{6} &= \phantom{-}1/42 \cr \beta_{8} &= -1/30 \cr \beta_{10} &= \phantom{-}5/66 \cr \beta_{12} &= -691/2730 \cr \beta_{14} &= \phantom{-}7/6 \cr \beta_{16} &= -3617/510 \cr \beta_{18} &= \phantom{-}43867/798 \cr \beta_{20} &= -174611/330 \cr \beta_{22} &= \phantom{-}854513/138 \cr \beta_{24} &= -236364091/2730 \cr \beta_{26} &= \phantom{-}8553103/6 \cr \beta_{28} &= -23749461029/870 \cr \beta_{28} &= \phantom{-}8615841276005/14322 \cr \beta_{30} &= -7709321041217/510 \cr \beta_{32} &= \phantom{-}2577687858367/6 \cr \beta_{34} &= -26315271553053477373/1919190 \cr \beta_{36} &= \phantom{-}2929993913841559/6 \cr } $$ All the $\beta_{n}=0$ for odd $n > 1$.
There is no apparent pattern to this sequence, is there? In fact, it is precisely the somewhat random appearance of these numbers, coupled with the knowledge that they arise in fundamental circumstances, that make mathematicians feel they are looking at the manifestation of some deep and mysterious secret.
In the next section I'll tell you how to do your own calculations. After that I'll answer all the questions asked above by recounting how these numbers were discovered by the Swiss mathematician Jakob Bernoulli, and then investigated later and more deeply by Leonhard Euler, another Swiss mathematician. This is a story told often and well, but then good stories are often worth retelling.
What does the definition mean? How can we compute a few of these coefficients? Well, if $f(x)$ is any well-behaved function defined for $x$ near $0$, it may be written as a series $$ f(x) = c_0 + c_1 x + c_2 x^{2} + \cdots $$ in which the coefficients may be evaluated in terms of derivatives of $f$: $$ \eqalign { c_{0} &= f(0) \cr c_{1} &= f'(0) \cr c_{2} &= { f''(0) \over 2 } \cr & \dots \cr c_{n} &= { f^{(n)}(0) \over n! } \, . } $$ It turns out that the function $x/(e^{x} -1)$ qualifies, although that might not be obvious, since setting $x=0$ we get $0/0$. However, you can write it as $$ f(x) = { 1 \over (e^{x}-1)/x } $$ and then apply the usual rules for derivatives of quotients to get the coefficients. This, you would quickly learn, leads to chaos.
But there is another way to find the coefficients. First of all, we can write $$ \eqalign { { x \over e^{x} - 1 } &= { x \over (1 + x + x^{2}/2 + x^{3}/3! + \cdots ) - 1 } \cr &= { x \over x + x^{2}/2 + x^{3}/3! + \cdots } \cr &= { 1 \over 1 + x/2 + x^{2}/3! + \cdots \ } \, . \cr } $$ Suppose $c_{0} + c_{1}x + c_{2}x^{2} + \cdots $ to be the series we are looking for. In other words $$ { 1\over 1 + x/2 + x^{2}/3! + \cdots \ } = c_{0} + c_{1}x + c_{2}x^{2} + \cdots $$ But this is equivalent to the equation $$ 1 = ( 1 + x/2 + x^{2}/3! + x^{3}/4! + \cdots \ ) (c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + \cdots \ ) \, . $$ We can multiply the two series and match the powers of $x$ on the left and right to get $$ \eqalign { 1 &= c_{0} \cr c_{0} &= 1 \cr & \cr 0 &= c_{0}/2 + c_{1} \cr c_{1} &= -c_{0}/2 \cr &= - 1/2 \cr & \cr 0 &= c_{0}/3! + c_{1}/2! + c_{2} \cr c_{2}&= -c_{0}/3! - c_{1}/2! \cr &= 1/12 \cr & \cr 0 &= c_{0}/4! + c_{1}/3! + c_{2}/2! + c_{3} \cr c_{3} &= 0 \cr & \cr & \dots \cr & \cr 0 &= c_{0}/(n+1)! + c_{1}/n! + \cdots + c_{n-1}/2! + c_{n} \cr } $$ and then the recursive formula $$ c_{n} = -\big ( c_{0}/(n+1)! + c_{1}/n! + \cdots + c_{n-1}/2!\big ) \, . $$ In other words, once we know all the $c_{i}$ for $i \lt n$ we can calculate $c_{n}$. This works reasonably well on a computer, but there are certainly limits to what can be done. For one thing, as the examples show, the complexity of the fractions increases rapidly with $n$, and for another, all the $c_{i}$ have to be stored for later use.
The $n$-th Bernoulli number is $\beta_{n} = n! c_{n}$, so that $$ c_{n} = {\beta_{n} \over n! } \, . $$
The progress of mathematics is almost always the history of problems that were solved. And when interesting problems are solved, almost invariably other problems become apparent. I repeat: The history of mathematics is the history of problems. It does happen occasionally that great mathematical discoveries come out of a cloudless blue sky, but this happens very, very rarely. What problem led to the discovery of Bernoulli numbers? What problems led to the recognition of their importance?
We are lucky. Bernoulli numbers were discovered by Jakob Bernoulli in about 1700 and he has left us, in a few pages of his famous if posthumous book Ars conjectandi, a good--and in fact an extraordinarily lucid--account of how he happened to come across them.
The problem he was looking at was to find a formula for the sum of $k$-th powers of integers. For example, it was known in many cultures that $$ 1 + 2 + \cdots + n = { n(n+1) \over 2 } \, . $$ But what about $$ \eqalign { & 1 + 4 + 9 + \cdots + n^{2} = \sum_{m=1}^{n} m^{2} \, ? \cr & 1 + 8 + 27 + \cdots + n^{3} = \sum_{m=1}^{n} m^{3} \, ? \cr & 1 + 16 + 81 + \cdots + n^{4} = \sum_{m=1}^{n} m^{4} \, ? \cr } $$ By $1700$ the problem of summing powers of integers had been around for quite a while, and had been investigated by many European mathematicians. One can find formulas for the sums above more or less by a kind of guess-work. For example, one might guess that $$ \sum_{m=1}^{n} m^{2} $$ is a polynomial of degree $3$ in $n$. Computing examples you can guess that the leading term is $n^{3}/3$. Subtracting this off and multiplying by $3$ to make all numbers integral, you get the sequence $$ \matrix { n & n^2 & \sum m^{2} & 3\, \sum m^{2} - n^{3} \cr 1 & 1 & 1 & 2 \cr 2 & 4 & 5 & 7 \cr 3 & 9 & 14 & 15 \cr 4 & 16 & 30 & 26 \cr 5 & 25 & 55 & 40 \cr 6 & 36 & 91 & 57 \cr 7 & 49 & 140 & 77 \cr 8 & 64 & 204 & 100 \cr } $$ You can guess again that this last sequence has dominant term $3n^{2}/2$, and finally arrive at the formula $$ \sum_{1}^{n} m^{2} = n^{3}/3 + n^{2}/2 + n/6 \, . $$ This procedure does not constitute a proof, of course, but with a little work you can check that this formula is correct for many values of $n$.
Many such formulas were found before Bernoulli. The most systematic and successful of these predecessors was the remarkable Johann Faulhaber, who found formulas for the sum of powers $n^{k}$ for $k$ up to $17$.
But it was only Bernoulli who managed to find a systematic way to attack the problem. His trick was to relate the problem to one which was even in his time well known, related to the the coefficients appearing in binomial expansions. Recall that these are the coefficients appearing in the expansion of the algebraic expression $(x + y)^{n}$. For example $$ \eqalign { (x+y)^{0} &= 1 \cr (x+y)^{1} &= x + y \cr (x + y)^{2} &= x^{2} + 2xy + y^{2} \cr (x + y)^{3} &= x^{3} + 3x^{2}y + 3xy^{2} + y^{3} \cr (x + y)^{2} &= x^{4} + 4x^{3}y + 6 x^{2}y^{2} + 4 xy^{3} + y^{4} \cr } $$ The coefficients go into Pascal's Triangle. This displays numbers $C_{n,i}$ which are, in spite of the name, most naturally displayed in an infinite rectangle. They are generated by a few inductive rules: (1) In the first row, all the numbers except the first vanish, and $C_{0,0}=1$; (2) In the first column all numbers $C_{n,0}= 1$; (3) Elsewhere in the table, the number $C_{n,i}$ in row $n$ and column $i$ is the sum of two of those in the previous row: $$ C_{n,i} = C_{n-1,i} + C_{n-1,i-1} \, . $$ Here is an intial part of the display: $$ \matrix { 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \cr 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \cr 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \cr 1 & 3 & 3 & 1 & 0 & 0 & 0 & 0 \cr 1 & 4 & 6 & 4 & 1 & 0 & 0 & 0 \cr 1 & 5 & 10 & 10 & 5 & 1 & 0 & 0 \cr 1 & 6 & 15 & 20 & 15 & 6 & 1 & 0 \cr 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 \cr } $$ This rule for computing $C_{n,i}$ leads immediately to another: the entry $C_{n,i}$ is equal to the sum of all the $C_{m,i-1}$ with $0 \le m \le n-1$. Why? Well, we keep applying the basic rule over and over: $$ \eqalign { C_{n,i} &= C_{n-1,i} + C_{n-1,i-1} \cr &= C_{n-2,i} + C_{n-2,i-1} + C_{n-1,i-1} \cr &= C_{n-3,i} + C_{n-3,i-1} + C_{n-2,i-1} + C_{n-1,i-1} \cr & \dots \cr &= C_{0,i-1} + C_{1,i-1} + \cdots + C_{n-3,i-1} + C_{n-2,i-1} + C_{n-1,i-1} \, . \cr } $$ This is all very elementary. What is not quite so elementary is that one doesn't have to build up Pascal's `Triangle' by such simple means. There is a direct formula for the entries (discovered, at least in Europe, by Pascal): $$ C_{n,i} = { n \choose i } = \cases { { n(n-1)\dots (n-(i-1)) \over i! } & if $0 \le i \le n$ \cr \phantom{.}& \cr \phantom{xxxxxxxxxx} 0 & otherwise. \cr } $$ Combining these two facts about Pascal's triangle, we have a sum formula $$ { 1 \choose i } + { 2 \choose i } + { 3 \choose i } + \cdots + { n \choose i } = { n + 1 \choose i+1 } \, . \qquad (*) $$ Bernoulli pointed out that this gives us a systematic way to find formulas for sums of powers. Not yet a formula, but a systematic way to find a formula.
Example. Equation (*) says that $$ \sum_{m=1}^{n} { m(m-1) \over 2 } = { (n+1)n(n-1) \over 3! } $$ But we can write the left hand side as $$ \sum_{m=1}^{n} \left( { m^{2} \over 2 } - { m \over 2 } \right) $$ and we deduce that $$ \eqalign { \sum_{m=1}^{n} { m^{2} \over 2 } &= { (n+1)n(n-1) \over 3! } + \sum_{m=1}^{n} { m \over 2 } \cr &= { (n+1)n(n-1) \over 3! } + { n(n+1) \over 4 } + \cr \sum_{m=1}^{n} m^{2} &= { n^{3} \over 3 } + { n^{2} \over 2 } + { n \over 6} \, . \cr } $$ Proceeding in this way, Bernoulli arrived at the following table: $$ \matrix { {\sum}_{m=1}^{n} \, 1 &=& n && & \phantom{ 1 \over 2 } \cr {\sum}_{m=1}^{n} \, m &=& { n^{2} \over 2 } &+& { n \over 2 } \cr {\sum}_{m=1}^{n} \, m^{2} &=& { n^{3} \over 3 } &+& { n^{2} \over 2 } &+& { n \over 6 } \cr {\sum}_{m=1}^{n} \, m^{3} &=& { n^{4} \over 4 } &+& { n^{3} \over 2 } &+& { n^{2} \over 4 } \cr {\sum}_{m=1}^{n} \, m^{4} &=& { n^{5} \over 5 } &+& { n^{4} \over 2 } &+& { n^{3} \over 3 } &-& { n \over 30 } \cr {\sum}_{m=1}^{n} \, m^{5} &=& { n^{6} \over 6 } &+& { n^{5} \over 2 } &+& { 5 n^{4} \over 12 } &-& { n^{2} \over 12 } \cr {\sum}_{m=1}^{n} \, m^{6} &=& { n^{7} \over 7 } &+& { n^{6} \over 2 } &+& { n^{5} \over 2 } &-& { n^{3} \over 6 } &+& { n \over 42 } \cr {\sum}_{m=1}^{n} \, m^{7} &=& { n^{8} \over 8 } &+& { n^{7} \over 2 } &+& { 7n^{6} \over 12 } &-& { 7n^{4} \over 24 } &+& { n^{2} \over 12 } \cr {\sum}_{m=1}^{n} \, m^{8} &=& { n^{9} \over 9 } &+& { n^{8} \over 2 } &+& { 2n^{7} \over 3 } &-& { 7n^{5} \over 15 } &+& { 2n^{3} \over 9 } &-& { n \over 30 } \cr {\sum}_{m=1}^{n} \, m^{9} &=& { n^{10} \over 10 } &+& { n^{9} \over 2 } &+& { 3n^{8} \over 4 } &-& { 7n^{6} \over 10 } &+& { n^{4} \over 2 } &-& { n^{2} \over 12 } \cr {\sum}_{m=1}^{n} \, m^{10} &=& { n^{11} \over 11 } &+& { n^{10} \over 2 } &+& { 5n^{9} \over 6 } &-& n^{7} &+& n^{5} &-& { n^{3} \over 2 } &+& { 5n \over 66 } \cr} $$ Some patterns in this table are easy to spot, but others are not. Among the easy ones: (1) The coefficient of $n^{k+1}$ in the first column is $1/(k+1)$, and (2) every coefficient of $n^{k}$ in the second is $1/2$. The third column already looks a bit mysterious.
One thing that is also evident is that some powers of $n$ are always skipped, which means in effect we are looking at a $0$. Putting those zeros in, and not bothering to include the powers of $n$, we get $$ \matrix { & k = 0: & 1 & & \phantom{ 1 \over 2 } \cr & k = 1: & { 1 \over 2 } \phantom{x} & { 1 \over 2 } \phantom{x} \cr & k = 2: & { 1 \over 3 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 1 \over 6 } \cr & k = 3: & { 1 \over 4 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 1 \over 4 } & 0 \cr &k = 4: & { 1 \over 5 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 1 \over 3 } & 0 & -{ 1 \over 30 } \cr &k = 5: & { 1 \over 6 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 5 \over 12 } & 0 & -{ 1 \over 12 } & 0 \cr &k=6: & { 1 \over 7 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 1 \over 2 } & 0 & -{ 1 \over 6 } & 0 & { 1 \over 42 } \cr & k = 7: & { 1 \over 8 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 7 \over 12 } & 0 & -{ 7 \over 24 } & 0 & { 1 \over 12 } & 0 \cr & k = 8: & { 1 \over 9 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 2 \over 3 } & 0 & -{ 7 \over 15 } & 0 & { 2 \over 9 } & 0 & -{ 1 \over 30 } \cr & k = 9: & { 1 \over 10 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 3 \over 4 } & 0 & -{ 7 \over 10 } & 0 & { 1 \over 2 } & 0 & -{ 1 \over 12 } & 0 \cr & k = 10: & { 1 \over 11 } \phantom{x} & { 1 \over 2 } \phantom{x} & { 5 \over 6 } & 0 & -1 & 0 & 1 & 0 & -{ 1 \over 2 } & 0 & { 5 \over 66 } \cr } $$
We can now try to understand the third column. The secret is given away by expressing all coefficients in the third column with denominator $12$: $2/12$, $3/12$, $4/12$, ... , $10/12$. All are linearly proportional to the first entry. What we have in the third column in row $k$ is $k/12 = (k/2)(1/6)$.
Looking at the fourth column in a similar way doesn't tell much. First we divide by the first non-zero entry, which is $-1/30$, giving us $$ \eqalign { k\phantom{:\,} & \cr 4: & 1 \cr 5: & 5/2 \cr 6: & 5 \cr 7: & 35/4 \cr 8: & 14 \cr 9: & 21 \cr 10: & 30 \cr } $$ Not so promising. What makes things much more comprehensible is to multiply this by $k+1$, giving us $$ \eqalign { k\phantom{:\,} & \cr 4: & 5 \cr 5: & 15 \cr 6: & 35 \cr 7: & 70 \cr 8: & 126 \cr 9: & 210 \cr 10: & 330 \cr } $$ Much better! Because these are from the fifth column of Pascal's triangle! So we guess that if $\beta = -1/30$ then the entry in the fourth column of Bernoulli's table is $$ { 1 \over k+1 } \cdot { (k+1)k(k-1)k \over 4.3.2.1 } \cdot \beta \, . $$ Of course we don't know exactly how Bernoulli himself proceeded, but what we do know is that eventually he arrived at a plausible formula. If $\beta_{n}$ is the entry in the $n$-th row, $n$-th column of the table constructed by Bernoulli, then $$ {\sum}_{m=1}^{n} m^{k} = { 1 \over k+1 } \cdot \Big( n^{k+1} + (k+1) \beta_{1} n^{k} + { (k+1)k \over 2 } \beta_{2} n^{k-1} + { (k+1)k(k-1) \over 3! } \beta_{3} n^{k-2} + \cdots + \beta_{k} n \Big ) \, . $$ Of course the numbers $\beta_{i}$ are what were later called the Bernoulli numbers. In other words, they contribute the complicated factors (as opposed to the factors ${k \choose i}$) of the coefficients in an expansion of ${\sum}_{1}^{n} m^{k}$ in powers of $n$. Bernoulli's formula is a remarkable discovery, in effect assigning the Bernoulli numbers a universal role.
Bernoulli's formula tells you how to compute the $\beta_{i}$, too. The formula is to be valid for all $n$, and in particular for $n=1$. But $\sum_{1}^{1} m^{k} = 1$, so we can substitute $n=1$ in the formula. But if we know all the $\beta_{i}$ for $i \lt k$ we recover a formula for $\beta_{k}$ in terms of them. It was Euler who later realized that it was exactly this formula that appeared in the definition of the $\beta_{n}$ as coefficients in the power series expansion of $x/(e^{x}-1)$.
Bernoulli, as far as I know, had no idea of how to verify rigourously that his formula was correct. This fell to Euler, who deduced it as a consequence of what is now called the Euler-Maclaurin sum forrnula, in which the function $x/(e^{x}-1)$ and the $\beta_{n}$ play an important role. It was probably at this point that it became clear to many that the Bernoulli numbers were in fact the manifestation of some deep and as yet secret world.
An anthology compiled by David Eugene Smith. Includes Jacob Bernoulli's own account of his discovery.
An earlier Feature Column on the Euler-Maclaurin sum fomrlua.
Nowadays, the naive method I describe above for computing the Bernoulli numbers is not used. But by modern methods, they have been computed, say, for $n=100,000,000$, as is discussed in this article. Numbers with more than 650,000 digits occur.
In the early nineteenth centuy, Charles Babbage built a kind of computer he called the Difference Engine that performed numerical computations, but he also designed what he called the Analytical Engine, based on a more ambitious idea. Although it was never built, he did start a project of writing programs for it. He also enlisted the help of Ada Lovelace, Lady Byron. One of the items high on his list of things the machine should do was to compute Bernoulli numbers, and this she undertook. Her remarkable discussion begins "We will terminate these Notes by following up in detail the steps through which the engine could compute the Numbers of Bernoulli, ... "
The Internet will tell you mch about Ada Lovelace's program for computing Bernoulli numbers. Some of it is even true. There is some controversy about her contributions. What is indisputable is that Charles Babbage was a remarkably creative man, and appreciated Lady Byron's contribution. In any case, I find Wolfram's comments relatively thorough.
A contemporary mathematician's account.
Faulhaber did not discover the sequence of Bernoulli numbers, but he did discover something else that was curious and intriguing.
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