- A Note on the Discoverer of the Bernoulli Numbers
- The Power Series As a Basis for the Bernoulli Numbers
- Summing the Series of Natural Numbers Using Bernoulli's Formula
- A List of Some Bernoulli Numbers
- Examples of Using Bernoulli's Formula to Find Sums of Powers
- Generating Bernoulli Numbers

Jacques had a brother, who was also a distinguished mathematician, Jean Bernoulli (1677-1748), also known in English as John Bernoulli and in German as Johann. There were at least four Jean Bernoullis! According to Carl Boyer, "they were as quick to offend as to be offended".

Power | m+1 | m | m-1 | m-2 | m-3 | m-4 | m-5 | Formula |

1 | 1/2 | 1/2 | ||||||

2 | 1/3 | 1/2 | 1/6 | |||||

3 | 1/4 |
1/2 |
1/4 |
|||||

4 | 1/5 |
1/2 |
1/3 |
-1/30 | ||||

5 | 1/6 | 1/2 | 5/12 | -1/12 | ||||

6 | 1/7 | 1/2 | 1/2 | -1/6 | 1/42 | |||

7 | 1/2 | 7/12 | -7/24 | 1/12 | ||||

8 | 1/2 | 2/3 | -7/15 | 2/9 | -1/30 |

Some of the things that are apparent by studying the table are:

- The coefficient of the first term is always 1/(m+1) (Terms begin at 0!)
- The coefficient of the second term is always 1/2 (Term 1)
- There seems to be a pattern to these formula. And the coefficient of the third term is m/12, for instance.
- The coefficients add up to 1
- Counting the first term as 0, all the odd terms, except 1, are 0.
- The number of terms in each sum seems to be (m+1), with any odd terms, except 1, vanishing. So for the 7th Power, we expect 8 terms (0..7) with the 7, 5 and 3 terms vanishing. This leaves 8-3=5. The guess, or floor(m/2)+2 seems to work for the formulae in the table (but doesn't work for zero).

On this page so far, we do not yet know what these Bernoulli Numbers are. Bernoulli developed the formula by studying the series of the various powers and realised there were certain constants that appeared (Which he wrote as A, B, C, etc)

The series has the following pattern: that is, the formula for the sum to n terms of the m powers of the natural numbers is:

[3.1]

While Bernoulli used slightly different notation, he effectively deduced the above series by observing the sums of the powers.

Where the B's are the Bernoulli Numbers, r is the term (0, 1, 2...). Because the odd terms (except 1) vanish, I have not written odd number terms, such as term 3 with B

There are some observations about this formula:

- The formulae produced by 3.1 end with an n term or higher power of n (no constants!).

We can immediately find some Bernoulli Numbers by comparing formula 3.1 with series above.

- Except for 1, all the other odd number Bernoulli Numbers are 0.
- B
_{0}=1, because all the series have 1/(m+1) as the coefficient of term0. - B
_{1}=-1/2, because in the series above, the term 1 is always 1/2. All the odd terms in the series are negative, but only the term 1 appears (the others vanish), so we could, for summing series make all the signs in the formula positive, and make B_{1}1/2. However, we need B_{1}to equal -1/2 in, for instance, trigonometry. - B
_{2}=1/6 - B5=-1/30
- B
_{7}=1/42

B_{0} |
B_{1} |
B_{2} |
B_{4} |
B_{6} |
B_{8} |
B_{10} |
B_{12} |

1 | -1/2 | 1/6 | -1/30 | 1/42 | -1/30 | 5/66 | -691/2730 |

Perhaps there are two things to notice immediately. The numbers alternate between positive and negative. And B

[3.1, repeated]

We obtain the formula with B

We expected and found (0+1) terms.

There is no shame in starting off simple!

Substituting the values for the Bernoulli Numbers: B

We note the sum of the coefficients is 1

Substituting values for the Bernoulli Numbers:

And computing the values:

We expected 9 terms less term 3, 5 and 7, leaving 6 terms. We expected the sum of the coefficients to equal 1. Both these expectations were met, giving us confidence in our algebra and arithmetic!

For instance, set n=1, to find B

Writing the powers of B in the following way, we have:

Or B

Setting n=2, for B

Substituting the value for B

Giving B

We can set n= (-1) to find B

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