Interestingly, Bernoulli numbers may have been discovered earlier by Seki Kowa and were rediscovered independently by a 17-year-old Srinivasa Ramanujan.

Bernoulli was analyzing formulas for the sums of powers, that have the following form:

S

_{m}(n) = ∑ (i=1, n) i

^{m}

He looked at, for example:

S

_{0}(n) = n

S

_{1}(n) = (1/2)n

^{2}- (1/2)n

S

_{2}(n) = (1/3)n

^{3}- (1/2)n

^{2}+ (1/6)n

S

_{3}(n) = (1/4)n

^{4}- (1/2)n

^{3}+ (1/4)n

^{2}

S

_{4}(n) = (1/5)n

^{5}- (1/2)n

^{4}+ (1/3)n

^{3}- (1/30)n

S

_{5}(n) = (1/6)n

^{6}- (1/2)n

^{5}+ (5/12)n

^{4}- (1/12)n

^{2}

S

_{6}(n) = (1/7)n

^{7}- (1/2)n

^{6}+ (1/2)n

^{5}- (1/6)n

^{3}+ (1/42)n.

S

_{7}(n) = (1/8)n

^{8}- (1/2)n

^{7}+ (7/12)n

^{6}- (7/24)n

^{4}+ (1/2)n

^{2}

S

_{8}(n) = (1/9)n

^{9}- (1/2)n

^{8}+ (2/3)n

^{7}- (7/15)n

^{5}+ (2/9)n

^{3}- (1/30)n

S

_{9}(n) = (1/10)n

^{10}- (1/2)n

^{9}+ (3/4)n

^{8}- (7/10)n

^{6}+ (1/2)n

^{4}- (3/20)n

^{2}

S

_{10}(n) = (1/11)n

^{11}- (1/2)n

^{10}+ (5/6)n

^{9}- n

^{7}+ n

^{5}- (1/2)n

^{3}+ (5/66)n.

By looking at these formulas, he noticed a pattern and in analyzing this pattern, he came up with what were termed Bernoulli numbers by Abraham de Moivre.

Well, for each S

_{m}(n), we can see that:

1) the first term is 1/(m+1)*n

^{m+1}

2) the second term, when it exists, is (-1/2)n

^{m}

3) the third term, when it exists, is (m/12)*n

^{m-1}

4) the fourth term, when it exists, is (-m)(m-1)(m-2)/720*n

^{m-3}

Bernoulli had discovered that if he defined a sequence of the numbers, he got a simple formula for defining the sum of powers.

The formula he discovered was the following:

In the above formula, B

_{k}refers to the sequence of Bernoulli numbers where Bernoulli numbers are defined in the following way:

Definition 1: Bernoulli Numbers

with B

_{0}= 1.

As a convention, I will refer to Bernoulli numbers as B

_{i}.

So, to figure out, B

_{1}, we have:

This gives us:

2*B

_{0}+ 2*B

_{1}= 0

B

_{1}= (1/2)[-2*B

_{0}] = (-1/2)(1) = (-1/2)

We determine B

_{2}based on solving:

In this way, we find that:

B

_{0}= 1

B

_{1}= -(1/2)

B

_{2}= (1/6)

B

_{3}= 0

B

_{4}= -(1/30)

B

_{5}= 0

B

_{6}= (1/42)

B

_{7}= 0

B

_{8}= -(1/30)

B

_{9}= 0

B

_{10}= (5/66)

B

_{11}= 0

B

_{12}= -(691/2730)

Bernoulli wrote that he was able to use his formula to compute S

_{10}(1000) in less than 10 minutes (see Smith, A Source Book in Mathematics, below).

Using his formula, this comes down to:

Evaluating this gives us:

91,409,924,241,424,243,424,241,924,242,500

I will now show a proof for Bernoulli's formula for sums. This is taken straight from Graham et al. Concrete Mathematics.

Definition 2: S

_{m}(n)

Let S

_{m}(n) = 0

^{m}+ 1

^{m}+ ... + (n-1)

^{m}

I will use this shorthand in the lemma and theorem below.

Postulate 1: 0

^{0}= 1

(This is taken from Graham et al.)

Graham et al argue in Concrete Mathematics that if we let 0

^{0}= 1, it provides a more elegant notation for the Binomial Theorem. Their argument is that the Binomial Theorem is very important while 0

^{0}is almost irrelevant. (see Concrete Mathematics, page 162).

Lemma 1:

Proof:

(1) S

_{m+1}(n) + n

^{m+1}=

= ∑(k=0,n-1) k

^{m+1}+ n

^{m+1}=

= ∑(k=0,n) k

^{m+1}=

= 0

^{m+1}+ ∑(k=1,n) k

^{m+1}= ∑(k=1,n) k

^{m+1}=

= ∑ (k=0,n-1) (k+1)

^{m+1}

(2) Using the Binomial Theorem, we get:

(3) Putting this together with step #1 gives us:

This gives us:

(4) Since S

_{j}(n) = o

^{j}+ 1

^{j}+ ... + (n-1)

^{j}, we have shown that:

(5) Now, since

The conclusion follows from simple subtraction since [(m+1)!]/[(m+1)!(0)!] = 1.

QED

Theorem:

Proof:

(1) We can see that this is true for m=0 since:

S

_{0}= n = (1/[1+1])∑ (k=0,0) ([0+1]!/k![1-k]!)B

_{k}n

^{0+1-k}=

= (1)*(1!/[0!][1!])B

_{0}n

^{1}= (1)*(1)*(1)*n = n [Since B

_{0}= 1, see above]

(2) Let's assume that this is true up to m-1.

(3) Let T

_{m}(n) =

(4) Let Δ = S

_{m}(n) - T

_{m}(n)

(5) Using Lemma 1 above, we know that:

(6) Using our definition of Δ, we can also see that:

(7) The goal in this proof will be to show that Δ = 0.

(8) Adding the definition of T

_{j}(n) [see step#3] gives us:

I also simplified [(m+1)!/(m!)(1!)] to (m+1).

(9) We can move all values that are independent of k before the ∑ (k=0,j):

(10) Since we can change the order of the terms of a finite sum and still get the same result, we can switch each value of k with (j-k) to get:

(11) Now, since (j+1) - (j-k) = k+1, we have:

(12) Putting this result in step #11 gives us:

(13) We can see that in step #12, 0 ≤ k ≤ j ≤ m. The sum is in fact all combinations of j,k where this is the case. This means that we can rearrange the sum to the following:

(14) Since n

^{k+1}is independent of j, we can move it over to get the following:

(15) Since,

we can also rearrange the result in step #14 to give us:

(16) Since:

We can rearrange terms to get:

(17) Since:

We can rearrange terms to get:

(18) Now, we know from the definition of Bernoulli numbers (see Definition 1 above) that

in all cases except where m = 0. This occurs only when k = m.

This allows us to simplify the result in step #17 to the following:

(19) Now since [(m+1)!/(m!)(1!)] = m+1, we get:

= n

^{m+1}+ (m+1)Δ

(20) But in step #5, we started out with n

^{m+1}so we are left with:

n

^{m+1}= n

^{m+1}+ (m+1)Δ.

(21) The only way that this is true is if Δ = 0.

QED

In another blog, I show how z/[e

^{z}- 1] can be used as a generating function for the Bernoulli numbers.

References

- Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989
- David Eugene Smith, A Source Book in Mathematics, Dover, 1959
- "Bernoulli Numbers", Wikipedia.Com