Table Of ContentBernoulli Polynomials
Talks given at LSBU, October and November 2015
Tony Forbes
Bernoulli Polynomials
The Bernoulli polynomials B (x) are defined by
n
B (x) = 1,
0
(cid:90) 1
B(cid:48)(x) = nB (x) and B (x)dx = 0, n ≥ 1. (1)
n n−1 n
0
Thus
1 1
B (x) = x− , B (x) = x2 −x+ ,
1 2
2 6
3x2 x 1
B (x) = x3 − + , B (x) = x4 −2x3 +x2 − ,
3 4
2 2 30
5x4 5x3 x 5x4 x2 1
B (x) = x5 − + − , B (x) = x6 −3x5 + − + ,
5 6
2 3 6 2 2 42
7x6 7x5 7x3 x 14x6 7x4 2x2 1
B (x) = x7 − + − + , B (x) = x8 −4x7 + − + − ,
7 8
2 2 6 6 3 3 3 30
9x8 21x5 3x 15x8 3x2 5
B (x) = x9− +6x7− +2x3− , B (x) = x10−5x9+ −7x6+5x4− + ,
9 10
2 5 10 2 2 66
11x10 55x9 11x3 5x
B (x) = x11 − + −11x7 +11x5 − + ,
11
2 6 2 6
33x8 33x4 691
B (x) = x12 −6x11 +11x10 − +22x6 − +5x2 − , ....
12
2 2 2730
The constant term gives the nth Bernoulli number, B = B (0),
n n
1 1 1 1 1
B = 1, B = − , B = , B = 0, B = − , B = 0, B = , B = 0, B = − ,
0 1 2 3 4 5 6 7 8
2 6 30 42 30
5 691 7 3617
B = 0, B = , B = 0, B = − , B = 0, B = , B = 0, B = − ,
9 10 11 12 13 14 15 16
66 2730 6 510
43867 174611
B = 0, B = , B = 0, B = − , ....
17 18 19 20
798 330
There is some merit in defining what one might call the alternative Bernoulli polynomials:
B(cid:98) (x) = 1,
0
(cid:90) 1
B(cid:98)(cid:48)(x) = nB(cid:98) (x) and B(cid:98) (x)dx = 1, n ≥ 1.
n n−1 n
0
Then B(cid:98) (x) = B (x)+nxn−1 = (−1)nB (−x). The corresponding alternative Bernoulli
n n n
numbers are B(cid:98) = B(cid:98) (0) = B (1). The only difference occurs when n = 1:
n n n
1
B(cid:98) = = −B , B(cid:98) = B , n = 0, 2, 3, 4, ....
1 1 n n
2
We won’t investigate the alternative polynomials, but watch out for B(cid:98) in what follows.
n
1
The graphs show that B (1−x) = (−1)nB (x). From the definition, the coefficient of x
n n
is B(cid:48)(0) = nB . More generally, the coefficient of xm in B (x) is (cid:0)n(cid:1)B ; hence
n n−1 n m n−m
n (cid:18) (cid:19)
(cid:88) n
B (x) = B xm.
n n−m
m
m=0
Moreover, for x ∈ [0,1] we have these approximations for large n:
B (x) ≈ B cos2πx, n even, B (x) ≈ B (1/4)sin2πx, n odd,
n n n n
and
(cid:90) 1
B (x)B (x)dx = 0 when n+m is odd.
n m
0
(cid:82)1
Exercise for reader: compute B (x)B (x)dx when n+m is even.
0 n m
0.15
0.10
0.05
0.2 0.4 0.6 0.8 1.0
(cid:45)0.05
Bernoulli polynomials B (x) for n = 2, 3, ..., 8
n
The generating functions for the Bernoulli polynomials and the Bernoulli numbers are
respectively
text (cid:88)∞ tn t (cid:88)∞ tn
= B (x) and = B . (2)
et −1 n n! et −1 n n!
n=0 n=0
These could be used as definitions; then the left-hand equality in (1) is recovered by
differentiating the left-hand equality in (2) with respect to x:
t2ext (cid:88)∞ tn text (cid:88)∞ tn−1 (cid:88)∞ B(cid:48) (x) tn
= B(cid:48)(x) ⇒ = B(cid:48)(x) = n+1 · .
ex −1 n n! ex −1 n n! n+1 n!
n=1 n=1 n=0
From the right-hand equality in (2) we obtain two facts: (i) B = 0 for n ≥ 1 since the
2n+1
function t/(et −1)+t/2 is even, and (ii) multiplying the series by the Taylor expansion
of (et −1)/t gives
n (cid:18) (cid:19)
(cid:88) n+1
B = 0n.
k
k
k=0
ThislastequalitycanbeusedtocomputetheB recursively; seeLovelace[3], forexample.
n
2
Stirling numbers of the second kind
The Stirling number of the second kind S(n,k) is the number of ways to partition a set
of n objects into k non-empty subsets:
1 (cid:88)k (cid:18)k(cid:19) (−1)k (cid:88)k (cid:18)k(cid:19)
S(n,k) = (−1)k−j jn = (−1)j jn,
k! j k! j
j=0 j=0
or they can be defined recursively
S(0,0) = 1, S(n,0) = 0 for n > 0, S(0,k) = 0 for k > 0,
S(n+1,k) = kS(n,k)+S(n,k −1) for k > 0.
k
n 0 1 2 3 4 5 6 7 8 9 10
0 1 0 0 0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0
2 0 1 1 0 0 0 0 0 0 0 0
3 0 1 3 1 0 0 0 0 0 0 0
4 0 1 7 6 1 0 0 0 0 0 0
5 0 1 15 25 10 1 0 0 0 0 0
6 0 1 31 90 65 15 1 0 0 0 0
7 0 1 63 301 350 140 21 1 0 0 0
8 0 1 127 966 1701 1050 266 28 1 0 0
9 0 1 255 3025 7770 6951 2646 462 36 1 0
10 0 1 511 9330 34105 42525 22827 5880 750 45 1
It is clear from the recursive definition that the S(n,k) are integers, that S(n,n) = 1 and
that S(n,k) = 0 if k > n. Interesting problem: prove directly that (cid:80)k (−1)j(cid:0)k(cid:1)jn = 0
j=0 j
whenever k > n > 0.
1000
800
600
400
200
k
50 100 150 200 250 300
Plot of logS(n,k)+1 for k = 5,10,... and n = 1,2,...,300
3
Euler–Maclaurin summation
Let f(x) be a sufficiently well-behaved function. Suppose we want to sum f(x) over
integer values of x and that we know how to integrate f(x). Let m < n be integers. Then
(cid:88)n f(r) = (cid:90) nf(x)dx+(cid:88)r (cid:0)f(k−1)(n)−f(k−1)(m)(cid:1) B(cid:98)k −(cid:90) nf(r)(x)B(cid:98)r((cid:98)x(cid:99)−x) dx,
k! r!
m m
k=m+1 k=1
or, making use of the fact that B = 0 for odd n ≥ 3,
n
(cid:88)n f(k) = (cid:90) nf(x)dx+ f(n)−f(m) +(cid:88)s (cid:0)f(2k−1)(n)−f(2k−1)(m)(cid:1) B2k
2 (2k)!
m
k=m+1 k=1
(cid:90) n B (x−(cid:98)x(cid:99))
− f(2r)(x) 2r dx, (r ≥ 1).
(2r)!
m
If f(x) is a polynomial, the last term will vanish for sufficiently large r. For example, to
get the formula for the sum of the first n squares put m = 0 and f(x) = x2:
(cid:88)n (cid:90) n n2 B n3 n2 n n(2n+1)(n+1)
k2 = x2dx+ +2n 2 = + + = .
2 2! 3 2 6 6
0
k=1
Poly-Bernoulli numbers
If we define the more general poly-Bernoulli numbers B(k) by
n
(cid:88)∞ tm Li (1−e−t) (cid:88)∞ tn
Li (t) = , k = B(k) ,
k mk 1−e−t n n!
m=1 n=0
thenk = 1givesLi1(x) = −log(1−x)andthegeneratingfunctiont/(1−e−t)ofBn(1) = B(cid:98)n.
Properties of the Bernoulli numbers
Theorem 1 (Explicit formula for the Bernoulli numbers) We have
(cid:88)n 1 (cid:88)k (cid:18)k(cid:19) (cid:88)n (−1)k
B = (−1)j jn = k!S(n,k). (3)
n
k +1 j k +1
k=0 j=0 k=0
Proof We refer to the generating function:
x log(1+(ex −1)) 1 (cid:88)∞ (ex −1)k (cid:88)∞ (ex −1)k
= = (−1)k+1 = (−1)k .
ex −1 ex −1 ex −1 k k +1
k=1 k=0
Now expand the binomial (ex −1)k and then expand ex:
x (cid:88)∞ (−1)k (cid:88)k (cid:18)k(cid:19) (cid:88)∞ (−1)k (cid:88)k (cid:18)k(cid:19)(cid:88)∞ jnxn
= (−1)k−j ejx = (−1)k−j
ex −1 k +1 j k +1 j n!
k=0 j=0 k=0 j=0 n=0
(cid:88)∞ xn (cid:88)∞ (−1)k (cid:88)k (cid:18)k(cid:19)
= (−1)k−j jn
n! k +1 j
n=0 k=0 j=0
(cid:88)∞ xn (cid:88)∞ (−1)k
= k!S(n,k).
n! k +1
n=0 k=0
4
But S(n,k) = 0 when k > n. Hence
x (cid:88)∞ xn (cid:88)n (−1)k
= k!S(n,k),
ex −1 n! k +1
n=0 k=0
as required to get the right-hand expression in (3). The other expression follows by
substituting for S(n,k). (cid:3)
Theorem 2 (von Staudt – Clausen) If n is a positive even integer, then
(cid:88) 1
B + ≡ 0 (mod 1). (4)
n
p
p−1|n, p prime
Proof Suppose n ≥ 2 is even. We consider each of the terms in the sum over k in either
of the two expressions of (3).
If k + 1 is not prime and k > 3, then k!/(k + 1) is an integer, as also is S(n,k); so the
term corresponding to this k is an integer. Also one can verify directly that the terms
corresponding to k = 0 and k = 3 are integers.
Now suppose k +1 is prime and k does not divide n. Write n = qk +r with 0 < r < k.
Then for 0 < j ≤ k, jqk ≡ 1 (mod k +1). Hence modulo k +1 the sum over j becomes
(cid:80)k (−1)j(cid:0)k(cid:1)jr = ±k!S(r,k) = 0 since k > r.
j=0 j
Finally, suppose k+1 is prime and k divides n. Then for 0 < j ≤ k, jn ≡ 1 (mod k+1),
and 0n = 0. Therefore (cid:80)k (−1)j(cid:0)k(cid:1)jn ≡ −1 (mod k+1) and hence modulo 1 there is a
j=0 j
contribution of −1/(k +1) to the sum. (cid:3)
When n is odd the sum is zero modulo 1; the k = 1 term contributes 1/2, as before, but
now the k = 3 term also contributes 1/2. All other k terms are integers.
The Von Staudt–Clausen theorem allows you to compute the fractional part of B for
n
some quite large n. For example, if n = 28000000000, the divisors of n are powers of two and
the denominators of the fractions in (4) are just 2 and the Fermat primes, 3, 5, 17, 257 and
65537. On the other hand, the calculation is currently impossible for, say, n = 29000000000.
It is convenient to have an expression for the absolute value of B . So we rearrange (4)
n
to get, writing {x} for x−(cid:98)x(cid:99),
(cid:110) (cid:111)
(cid:80)
1/p , n > 0, n ≡ 0 (mod 4),
p−1|n, pprime
(cid:110) (cid:111)
1− (cid:80) 1/p , n ≡ 2 (mod 4),
p−1|n, pprime
|B | = (cid:98)|B |(cid:99)+ (5)
n n 0, n = 0,
1/2, n = 1,
0, n > 1, n odd.
Thus for n = 2,4,..., we get
1 1 1 1 5 691 1 47 775 41 17 691 1 59 12899
, , , , , , , , , , , , , , ,
6 30 42 30 66 2730 6 510 798 330 138 2730 6 870 14322
47 1 638653 1 2011 1 53 41 14477 5 83 775 59 53 22298681
, , , , , , , , , , , , , , ,
510 6 1919190 6 13530 1806 690 282 46410 66 1590 798 870 354 56786730
1 47 62483 1 289 48540859 1 1 37 47717
, , , , , , , , , ,
6 510 64722 30 4686 140100870 6 30 3318 230010
77 1058237 1 5407 230759 77 1 1450679 1 4471
, , , , , , , , ,
498 3404310 6 61410 272118 1410 6 4501770 6 33330
5
Theorem 3 The Bernoulli numbers are related to the Riemann zeta-function by
2n!
B(cid:98) = −nζ(1−n), n ≥ 0; B = − ζ(n), even n ≥ 0. (6)
n n (2πi)n
Proof (Titchmarsh [6, section 2.4]) Let s be a complex veriable with Res > 1. Using
(cid:90) ∞ Γ(s)
xs−1e−kxdx =
ks
0
we get
(cid:88)∞ (cid:90) ∞ (cid:90) ∞ (cid:88)∞ (cid:90) ∞ xs−1
Γ(s)ζ(s) = xs−1e−ksdx = xs−1 e−kxdx = dx.
ex −1
0 0 0
k=1 k=1
Let C , ρ > 0, denote the contour that goes from +∞ to ρ on the real axis, circles the
ρ
origin once anticlockwise and then returns to +∞. Let ρ → 0. Since the integral along
the circle tends to zero as ρ tends to zero, we have
(cid:90) zs−1 (cid:90) 0 zs−1 (cid:90) ∞ (e2πiz)s−1
dz = dz + dz = (e2πis −1)Γ(s)ζ(s).
ez −1 ez −1 ez −1
C0 ∞ 0
Using the well-known formula Γ(1−z)Γ(z) = π/(sinπz), we obtain
1 (cid:90) zs−1 e−πisΓ(1−s) (cid:90) zs−1
ζ(s) = dz = dz,
(e2πis −1)Γ(s) ez −1 2πi ez −1
C0 C0
which extends the domain of ζ(s) to the whole complex plane except for s = 1. Now let
n ≥ 1 be an integer and put s = 1−n. Then
(−1)n−1(n−1)! (cid:90) z−n (−1)n−1(n−1)! (cid:90) (cid:88)∞ B
ζ(1−n) = dz = k zk−n−1dz.
2πi ez −1 2πi k!
C0 C0 k=0
But the value of integral on the right is 2πi times the residue of the pole at z = 0, and
this is just the coefficient of 1/z in the sum. Hence, recalling that B = −B(cid:98) and B = 0
1 1 n
for odd n ≥ 3,
(−1)n−1(n−1)! B (−1)n−1B B(cid:98)
n n n
ζ(1−n) = ·2πi = = − .
2πi n! n n
This is the left-hand equality in (6) for n ≥ 1. For the remaining case, recall that
1
ζ(s) = −γ +O(s−1) as s → 1.
s−1
The right-hand equality follows from the functional equation of ζ(s),
(cid:18) (cid:19)
(cid:16)s(cid:17) 1−s
Γ π−s/2ζ(s) = Γ π−(1−s)/2ζ(1−s).
2 2
(See Edwards [1], for example.) (cid:3)
Of course we can discard the sign if we are interested only in the absolute value:
(cid:18) (cid:19)
2n! 2n! 1 1
|B | = ζ(n) = 1+ + +... , even n ≥ 2. (7)
n (2π)n (2π)n 2n 3n
6
We can use (5) and (7) to determine |B | exactly. Write b(n) = (2n!)1/n/(2π). Then split
n
the sum into three parts,
|B | = I +E +R ,
n n n n
(cid:100)(cid:88)b(n)(cid:101)(cid:18)b(n)(cid:19)n (cid:100)(cid:88)b(n)(cid:101)(cid:18)b(n)(cid:19)n (cid:88)∞ (cid:18)b(n)(cid:19)n
In = , En = −In, Rn = .
k k k
k=1 k=1 k=(cid:100)b(n)+1(cid:101)
We compute I exactly and E approximately, we assume that R is small enough to be
n n n
ignored, and we get the exact value of F , the fractional part of |B |, using the von Staudt
n n
– Clausen formula, (5). Then for positive even n, we have |B | = I +(cid:15)+F , where (cid:15) = 0
n n n
if E < F , (cid:15) = 1 otherwise. Example:
n n
(cid:100)b(1000)(cid:101) = 59, I = 5318704469...6955251716 ≈ 5.318704469415522×101769,
1000
55743421
E ≈ 0.1625177182 and F = ≈ 0.1625177220 > E .
1000 1000 1000
342999030
There is an interesting application concerning prime generating functions, [4], and also
an interesting problem. Is there an n that requires (cid:15) = 1? By simple-minded direct
calculation there are no instances for even n up to 3200. But could it possibly happen
thatE isverynearlyequaltobutlessthan1,F isverysmall,andR ,whichisverysmall
n n n
anyway, is just large enough so that E +R > 1? Then we would have F = E +R −1
n n n n n
and hence (cid:98)|B |(cid:99) = I +1. A similar problem is presented in M500 [2, 5].
n n
References
[1] H. M. Edwards, Riemann’s Zeta-Function.
[2] TF, Problem 267.4 – Bernoulli numbers, M500 267 (December 2015).
[3] A.A.Lovelace, NoteGin: L.F.Menabrea, SketchofTheAnalyticalEngineInvented
by Charles Babbage With notes upon the Memoir by the Translator, Ada Augusta,
Countess of Lovelace, Biblioth`eque Universelle de Gen`eve 82 (October 1842).
[4] R. Thompson, Bernoulli numbers and prime generating functions, M500 267 (De-
cember 2015).
[5] R. Thompson, Solution 267.4 – Bernoulli numbers, M500, to appear.
[6] E. C. Titchmarsh, The Theory of the Riemann Zeta-Function.
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