Computing Bernoulli Numbers Quickly

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1 Computing Bernoulli Numbers Quickly Kevin J. McGown December 8, 2005 The Bernoulli numbers are defined via the coefficients of the power series expansion of t/(e t ). Namely, for integers m 0 we define B m so that t e t = m=0 B m m! tm. Multiplying both sides above by e t and equating coefficients of t m+ yields: B 0 =, (m + )B m = m k=0 ( m + k ) B k Some authors take the above recurrence to be the definition of the Bernoulli numbers. This recurrence provides a straightforward method for calculating B m and is especially convenient for computing B m for all m up to some bound. The first few Bernoulli numbers are: B 0 =, B = 2 B 2 = 6, B 3 = 0, B 4 = 30, B 5 = 0, B 6 = 42, B 7 = 0, B 8 = 30, B 9 = 0, B 0 = 55 66, B = 0, B 2 = B 3 = 0, B 4 = 7 6 The values above provide evidence for two basic results regarding Bernoulli numbers. First, B m = 0 for odd m 3, and secondly, for even m 2, B m = ( ) m/2+ B m. Henceforth we will denote B m = a d

2 where a, d Z, d, and (a, d) =. From the properties mentioned above, it is clear that a = ( ) m/2+ a for even m 2. The goal of this paper is to compute B m rapidly, when m is potentially very large. Computing B m via the recurrence is slow; it requires us to sum over m(m + )/2 terms. In addition, this method requires storing the numbers B 0,..., B m in memory. In order to speed up this computation, we will describe an important connection between the Bernoulli numbers and the Riemann Zeta Function. Much of what we will describe was gleaned from the PARI-2.2..alpha source code. The algorithm this version of PARI uses to compute Bernoulli numbers was written by Henri Cohen and later refined by Karim Belabas; it was originally designed to speed up the computation of zeta values. For real s >, Euler defined the function ζ(s) = n s, n= for which he proved the product formula ζ(s) = p ( p s ). (Riemann showed that ζ(s) has an analytic continuation to the entire complex plane with a simple pole at s =, and hence the function bears his name.) In addition to giving the product formula, Euler was able to evaluate the zeta function at the even positive integers. [] For any integer m, 2ζ(2m) = ( )m+ (2π) 2m (2m)! It follows that for any even integer m 4, B m = 2m! (2π) m ζ(m). B 2m. It is now clear how to compute decimal approximations to B m ; we merely approximate ζ(m) using the Euler product and plug the result into the above equation. A priori, this is not enough to compute B m as a ratio of integers, but fortunately a theorem of Clausen and von Staudt precisely describes the denominator of B m in terms of the divisors of m. [] For even m 2, d := denom(b m ) = p. 2 p m

3 Now we show how to compute a. First define K := 2m! (2π) m so that B m = Kζ(m). Using the Euler product, we may approximate ζ(m) from below with arbitrary precision. Suppose that we have computed a number z such that then we have and hence 0 ζ(m) z < (Kd) ; 0 B m zk < d 0 a zkd <. It follows that a = zkd and hence a = ( ) m/2+ zkd. It remains to explicitly compute z. In order to accomplish this, we consider the following problem: given s > and ε > 0, find N Z + so that when we set z := p N( p s ), we are guaranteed that 0 ζ(s) z < ε. We always have 0 ζ(s) z. Further, it is not hard to see that ( p s ) and therefore n N n s p N ζ(s) z = n=n+ N n s x s dx (s )N s. If we choose N > ε /(s ), then we have (s )N s N < ε, s which implies ζ(s) z < ε, as required. For our purposes, we have s = m and ε = (Kd) and therefore it suffices to choose N > (Kd) /(m ). 3

4 The Algorithm: Suppose m 2 is even K := 2m! (2π) m d := p m p N := (Kd) /(m ) z := p N ( p m ) 5. a := ( ) m/2+ dkz 6. B m = a d Some remarks are in order. In step (), we must be careful to compute K to sufficient precision so that the calculation in (5) gives the desired result. In order to compute (4), it is useful to first compute all primes p N; this may be done quickly using the Sieve of Eratosthenes. One may also compute the product in (2) via a sieving process. Finally, for the value of N we may choose any integer greater than or equal to the one specified in (3), so we need not worry about computing (Kd) /(m ) to much precision. It is interesting to note that the algorithm above also gives a way of approximating ζ(m) quickly for even m. Namely, compute B m as a rational number using this algorithm and plug it into Euler s formula for ζ(m) along with an approximation of π sufficiently many decimal places. 4

5 Example: We use the modest size example of m = 50 for the sake of readability. Using 50 digits of precision, we compute K = The divisors of m are, 2, 5, 0, 25, 50 and hence d = (2)(3)() = 66. We find N = 4 and compute z = Finally we compute dkz = and therefore B 50 = References [] Ireland, Kenneth. Rosen, Michael. A Classical Introduction to Modern Number Theory. New York: Springer-Verlag,

Computing Bernoulli Numbers

Computing Bernoulli Numbers (joint work with Kevin McGown of UCSD) February 16, 2006 Bernoulli Numbers Defined by Jacques Bernoulli in posthumous work Ars conjectandi Bale, 1713. x e x 1 = n=0 B n n! x n B 0 = 1, B 1 = 1 2 B 2 =