Bernoulli numbers Ezra Getzler Department of Mathematics Northwestern University Tsinghua Math Camp 2017.7.20 Bernoulli numbers 1 / 43 Triangular numbers Everyone is familiar with the sum n(n + 1) 1 + 2 + ··· + n = : 2 These numbers are called the triangular numbers. For example, 1(1 + 1) 1 = = 1 2 2(2 + 1) 1 + 2 = = 3 2 3(3 + 1) 1 + 2 + 3 = = 6 2 4(4 + 1) 1 + 2 + 3 + 4 = = 10 2 5(5 + 1) 1 + 2 + 3 + 4 + 5 = = 15 2 Bernoulli numbers 2 / 43 Pyramidal numbers Many of you will also be familiar with the pyramidal numbers: these are the sums of squares: n(n + 1)(2n + 1) 12 + 22 + ··· + n2 = 6 For example, 1(1 + 1)(2 + 1) 12 = = 1 6 2(2 + 1)(4 + 1) 12 + 22 = = 5 6 3(3 + 1)(6 + 1) 12 + 22 + 32 = = 14 6 4(4 + 1)(8 + 1) 12 + 22 + 32 + 42 = = 30 6 5(5 + 1)(10 + 1) 12 + 22 + 32 + 42 + 52 = = 55 6 Bernoulli numbers 3 / 43 Generalizations? Is there a general story here? If k is a natural number larger than 2, is there a formula for the sum of the kth powers of the first n natural numbers? 1k + ··· + nk =??? Bernoulli numbers 4 / 43 Nicomachus Nicomachus (Niko'maqoc) was a Hellenistic mathematician who lived between around 60 CE and 120 CE in the Roman city of Gerasa, east of the Jordan River. He was a Neopythagorean, who believed in religious doctrines based on the Greek philosophers Pythagorus and Plato. Among their beliefs, they associated God with the number One. Bernoulli numbers 5 / 43 Nicomachus's Theorem Nicomachus gave a formula for the sum of cubes: 13 + 23 + ··· + n3 = 1 + 2 + ··· + n2 This is a beautiful formula but it doesn't really give an idea of how to understand the general sums 1k + ··· + nk =??? The proof uses mathematical induction. For n = 1, both sides of the equation 13 + 23 + ··· + n3 = 1 + 2 + ··· + n2 equal 1. Bernoulli numbers 6 / 43 Suppose we have proved that 13 + 23 + ··· + n3 = 1 + 2 + ··· + n2 We want to prove that replacing n by n + 1, both sides increase by the same amount, namely (n + 1)3. Consider the square with sides (n + 1)(n + 2) 1 + 2 + ··· + n + (n + 1) = ; 2 and subdivide the sides into two segments, of length n(n + 1) 1 + 2 + ··· + n = 2 and n + 1 respectively. Bernoulli numbers 7 / 43 n(n + 1)=2 n + 1 n(n + 1)=2 AB n + 1 CD We see that (1 + 2 + ··· + n + (n + 1))2 = A + B + C + D n(n + 1)2 n(n + 1)2 = (1 + 2 + ··· + n)2 + + + (n + 1)2 2 2 = (13 + 23 + ··· + n3) + n(n + 1)2 + (n + 1)2 = (13 + 23 + ··· + n3) + (n + 1)3: Bernoulli numbers 8 / 43 Johannes Faulhaber The next steps in this subject were taken by Johannes Faulhaber (1585-1635), a mathematician who was born and lived in the German city of Ulm. Bernoulli numbers 9 / 43 Faulhaber's Theorem Faulhaber showed that for any fixed natural number k, the sum of powers k k k Σk (n) = 1 + 2 + ··· + n ; thought of as a function of n, is a polynomial of degree k + 1. We have already seen this explicitly for k = 0, 1, 2 and 3, where the polynomials are respectively n, n(n + 1)=2, n(n + 1)(2n + 1)=6, and n2(n + 1)2=4. In other words, there are numbers bk;i for i = 1;:::; k + 1, such that k k k k+1 k 2 1 + 2 + ··· + n = bk;k+1n + bk;k n + ··· + bk;2n + bk;1n: It turns out that all of the coefficients of these polynomials are rational numbers. (The constant term bk;0 is absent because both sides vanish when n = 0.) Bernoulli numbers 10 / 43 When we take k = 4, something interesting happens | the coefficient of the linear term n has quite a large denominator. n5 n4 n3 n Σ (n) = + + − : 4 5 2 3 30 Let's prove this special case of Faulhaber's Theorem: the general case is actually proved in exactly the same way. Suppose that the left-hand side is a polynomial in n of degree 5: 5 4 3 2 Σ4(n) = b4;5n + b4;4n + b4;3n + b4;2n + b4;1n If we can find values of b4;k such that both sides are equal for n = 0, and the right-hand side jumps by n4 as we go from n − 1 to n, then the formula will be proved. The case n = 0 is easy: both sides vanish. Bernoulli numbers 11 / 43 It remains to check that 5 4 3 2 b4;5n + b4;4n + b4;3n + b4;2n + b4;1n 5 4 3 2 4 − b4;5(n − 1) + b4;4(n − 1) + b4;3(n − 1) + b4;2(n − 1) + b4;1(n − 1) = n : The left-hand side equals 5 5 4 4 3 3 2 2 b4;5(n − (n − 1) ) + b4;4(n − (n − 1) ) + b4;3(n − (n − 1) ) + b4;2(n − (n − 1) ) + b4;1(n − (n − 1)) 4 3 2 3 2 2 = b4;5(5n − 10n + 10n − 5n + 1) + b4;4(4n − 6n + 4n − 1) + b4;3(3n − 3n + 1) + b4;2(2n − 1) + b4;1 The induction step consists of showing that the numbers b4;i may be chosen in such a way that this equals n4. Bernoulli numbers 12 / 43 To do this, it suffices to solve the simultaneous linear equations 5b4;5 = 1 −10b4;5 + 4b4;4 = 0 10b4;5 − 6b4;4 + 3b4;3 = 0 −5b4;5 + 4b4;4 − 3b4;3 + 2b4;2 = 0 b4;5 − b4;4 + b4;3 − b4;2 + b4;1 = 0: It is easy to see that these have a unique solution, in which each coefficient b4;i is a rational number: (b4;5; b4;4; b4;3; b4;2; b4;1) = (1=5; 1=2; 1=3; 0; −1=30): Two things about this solution are striking: the quadratic term is missing (that is, b4;2 = 0), and the linear term has the peculiar coefficient −1=30. Bernoulli numbers 13 / 43 Let us list the coefficients bk;i in a table for k = 1;:::; 6: k bk;1 bk;2 bk;3 bk;4 bk;5 bk;6 bk;7 0 1 1 1=2 1=2 2 1=6 1=2 1=3 3 0 1=4 1=2 1=4 4 −1=30 0 1=3 1=2 1=5 5 0 −1=12 0 5=12 1=2 1=6 6 1=42 0 −1=6 0 1=2 1=2 1=7 The numbers in the left column have a name: they are the Bernoulli numbers Bk . Along the diagonals, this table shows some nice patterns. A little experimentation shows that the following conjectural formula holds explains all of the diagonal patterns in the table: k! B b = k−i+1 : k;i i! (k − i + 1)! Bernoulli numbers 14 / 43 Jakob Bernoulli Jakob Bernoulli (1645{1705) was a member of a famous family of mathematicians in the Swiss city of Basel. He wrote an important book Ars Conjectandi (The Art of Conjecture) on probability and games of chance, and discovered the number e. Bernoulli numbers 15 / 43 Jakob Bernoulli gave the first description of the numbers now named after him. In modern language, the numbers Bk are defined recursively by k+1 k+1 k+1 k+1 k = 1 B1 + ··· + k−2 Bk−2 + k−1 Bk−1 + k Bk : 1 = 2 B1 , so that B1 = 1=2 2 = 3 B1 + 3 B2 , so that B2 = 1=6 3 = 4 B1 + 6 B2 + 4 B3 , so that B3 = 0 4 = 5 B1 + 10 B2 + 10 B3 + 5 B4 , so that B4 = −1=30 It is not hard to calculate a few more of these numbers for yourself. Eventually, you reach the bizarre number B12 = −691=2730: this hints that the Bernoulli numbers present a richer structure than most sequences of numbers you will have met in mathematics. Bernoulli numbers 16 / 43 The Bernoulli numbers have many remarkable properties. Theorem If k > 1 is odd, then Bk = 0. To see this, we extend the function Σk (n) to negative values of n. Observe that Σk (n) is completely determined by the formulas Σk (0) = 0 and k Σk (n) − Σk (n − 1) = n . This makes it natural to set k k k Σk (−n) = (−n + 1) + (−n + 2) + ··· + (−1) : This gives the remarkable identity k Σk (−n) = (−1) Σk (n − 1): Bernoulli numbers 17 / 43 If k is odd, we can calculate Σk (n) + Σk (−n) in two different ways: if k is odd, the above identity shows that k Σk (n) + Σk (−n) = Σk (n) − Σk (n − 1) = n : On the other hand, Faulhaber's Theorem shows that k k−2 k−4 Σk (n) + Σk (−n) = 2 bk;k n + bk;k−2n + bk;k−4n + ··· + bk;1n : It follows that if k > 1, Bk = bk;1 = 0 (and B1 = 1=2, but we already knew this). Bernoulli numbers 18 / 43 Theorem The signs of the even Bernoulli numbers oscillate: B2, B6, . , are positive, and B4, B8, . , are negative. There are two ways to prove this result. The first is as a corollary of the following theorem. Theorem The derivative k−1 d tan(x) k−1 dx x=0 k=2−1 k k equals (−1) 2 (2 − 1)Bk =k. It vanishes if k is odd, and is a positive integer if k is even.