Unit 6: Meet the Bernoulli numbers

Reading time: about 14 minutes. Activity: about 20 minutes.

1. What you should be able to do afterwards

1. Say, in plain terms, what a Bernoulli number is and where the sequence comes from.
2. Derive B4 from the recurrence, by hand, and show your working.
3. Explain why two correct-looking programs can disagree about B1, and say which convention this course uses.
4. Translate between Lovelace's Note G labels and modern indices.

2. Why this unit comes before the AI unit

In Unit 7 you will ask an AI system to produce a routine that returns Bernoulli numbers. It will produce something, and what it produces can look competent whether or not it is right.

The question that decides whether that output is useful to you is not "does this look right" but "what should the answer be". If you have nothing independent of the output to answer the second question with — a value you derived, a published table, a separate method — you have no way of answering the first. So this unit gives you the independent answer first, deliberately, before anything is generated.

This ordering is the whole point of the course's approach to critical thinking: build enough understanding to judge, then use the tool.

3. What a Bernoulli number is

The Bernoulli numbers are a fixed sequence of exact rational numbers — ordinary fractions — with one value for each whole-number index from 0 upwards. They are written B0, B1, B2, and so on.

They are not approximations, not estimates, and not a matter of taste. Given the index and the convention (a fixed choice, explained under Trap 1 below), the value is completely determined. That property is what makes them a good subject for this course: there is a right answer, it can be written down exactly, and disagreement can always be resolved.

Here is the sequence this course works with.

All odd B_n for n > 1 are exactly zero. That is not a rounding artefact or a convenient approximation; those values are zero, precisely.

Two things are worth noticing before you read on. First, the values do not shrink tidily — B14 is larger than B12, and B30 is enormous. Second, the denominators are already awkward by B12. Both facts matter later: a routine that returns decimals can look right at every index and be exact at almost none of them, so only an exact comparison catches it.

4. Where the sequence comes from

You do not need this section to complete the unit. It is here so the numbers do not feel as though they were invented for the exercise.

Ask a simple question: what is 1² + 2² + … + 9²? The answer is 285, and you can get it by adding nine numbers. But there is also a closed formula that gives the answer for any upper limit, without adding anything up. There is another for cubes, another for fourth powers, and so on.

Write those formulas in a single general form covering every power at once, and the same sequence of fractions appears in the coefficients each time. That sequence is the Bernoulli numbers. They were established mathematics well before Note G took them as its example: they appear in Jacob Bernoulli's Ars Conjectandi, published in 1713 [S9], and Note G already calls them "the Numbers of Bernoulli" [P2a]. For sources on Note G itself, see References and source policy.

One detail of that general formula is a preview of Trap 1 below. The reference form writes it through Bernoulli polynomials [M6, sec. 24.4(iii)]; expanded into coefficients with this course's B1 = -1/2, it sums up to one less than the limit you give it. For squares and the limit m it reads

0^2 + 1^2 + ... + (m-1)^2 = (m^3 + 3*B1*m^2 + 3*B2*m) / 3

and with B1 = -1/2, B2 = 1/6 and the limit 9 it gives (729 - 243/2 + 9/2) / 3 = 204, which is 0^2 + 1^2 + ... + 8^2; it returns 285 only for the limit 10. (The form that sums from 1 to the limit needs B1 = +1/2.) Every step of that calculation is correct, and it still answers a different question from the one asked: checking that a procedure was carried out correctly is not checking that it was the right procedure for your question.

5. The rule that generates them

Every Bernoulli number is fixed by the ones before it, through one identity, which holds for every n from 1 on (B0 = 1 is where it starts):

sum_{j=0}^{n} C(n+1, j) * B_j = 0      for n >= 1

Read sum_{j=0}^{n} as "add up one term for each j from 0 to n": the expression after it is written out once with j = 0, once with j = 1, and so on up to j = n, and the terms are added. B_j is the Bernoulli number with index j.

C(n+1, j) is the binomial coefficient — "n plus one choose j" — the numbers from row n+1 of Pascal's triangle. Each row of the triangle begins and ends with 1, and every other number is the sum of the two above it. Its rows are counted from 0, and so are the positions within a row, so C(n+1, j) is the number at position j in row n+1:

row 0:       1
row 1:      1 1
row 2:     1 2 1
row 3:    1 3 3 1

What you need before you start works through this notation more slowly. Row n+1 has n+2 numbers and j stops at n, so its last number is not used: it would multiply B_{n+1}, which the identity for n+1 fixes. For a given n, the identity gives you a sum of terms, all of them known except the last, which contains B_n. Substitute what you know, and solve.

The smallest case is n = 1. Row 2 of Pascal's triangle is 1, 2, 1, and j runs from 0 to 1, so the identity has two terms: 1*B0 + 2*B1 = 0. With B0 = 1, that gives 2*B1 = -1, so B1 = -1/2, the value in the table.

Starting from B0 = 1, this generates the whole sequence, one value at a time, in a fixed order, with no judgement required at any step. It is an ordered procedure in exactly the sense of Unit 2.

The same identity also works as a check, and checking is the cheaper direction: given a list of claimed values, substitute them all and see whether each row sums to exactly zero. Nothing has to be solved. Unit 8 uses this.

Worked example: B2

Take n = 2. The identity gives three terms:

C(3,0)·B0 + C(3,1)·B1 + C(3,2)·B2 = 0

Row n + 1 = 3 of Pascal's triangle is 1, 3, 3, 1; j runs from 0 to 2, so the coefficients are 1, 3, 3. Substituting B0 = 1 and B1 = -1/2:

(1)(1) + (3)(-1/2) + (3)(B2) = 0

Work the known part first:

1 - 3/2 = -1/2

So:

-1/2 + 3·B2 = 0
        3·B2 = 1/2
          B2 = 1/6

Which is what the table says. Notice what you did not need: a calculator, a computer, or anyone's assurance. You needed the rule, the earlier values, and care with fractions.

In the activity you will do the same for B4, where n = 4 gives five terms instead of three. That is the derivation checkpoint cp06-b4-by-hand, and it is worth doing before you read further into this page. Stop here and attempt Task 1 from your own working. On the course site, the rest of this reading is hidden in a labelled section you open yourself, because it contains the result.

from fractions import Fraction

b0 = Fraction(1)
b1 = Fraction(-1, 2)
b2 = Fraction(1, 6)

print(b0 + 3 * b1)        # the known part of the n = 2 identity

python3 selfcheck.py run cp06-b4-by-hand  # or in the browser