Notes on Bernoulli numbers and Euler’s summation formula [expository notes]

9 Pages·2014·0.232 MB·English

by  Mark Wildon

NOTES ON BERNOULLI NUMBERS AND EULER’S
SUMMATION FORMULA
MARK WILDON
1. Bernoulli numbers
1.1. Definition. We define the Bernoulli numbers B for m ≥ 0 by
m
m (cid:18) (cid:19)
(cid:88) m+1
(1) B = [m = 0]
r
r
r=0
Bernoulli numbers are named after Johann Bernoulli (the most prolific
Bernoulli, and the discoverer of the Bernoulli effect).
1.2. Exponential generating function. If f(z) = (cid:80)a zn/n! and g(z) =
n
(cid:80)b zn/n! then f(z)g(z) = (cid:80)c zn/n! where the coefficients c are given
n n n
by c = (cid:80)(cid:0)n(cid:1)a b . Thus if we put β(z) = (cid:80)B zn/n! then
n r r n−r n
(cid:18) (cid:19)
1 (cid:88) n
[zn]β(z)expz = B
r
n! n−r
n−1(cid:18) (cid:19)
1 (cid:88) n Bn
= B +
r
n! r n!
r=0
B
n
= [n = 1]+
n!
from which it follows that β(z)expz = z+β(z). Therefore
(cid:88) zn z
(2) β(z) = B = .
n
n! expz−1
This shows that the radius of convergence of β(z) is π. So although it is
tempting to make a substitution in
(cid:88) (cid:88)(cid:88) e−tmn (cid:88) 1 tm
logP(e−t) = − log(1−e−tn) = =
m tm2etm−1
n n m m
this will not work, as tm is eventually more than π.
1.3. Some values. If we add z/2 to both sides of (2) we obtain
z zexpz+1
(3) β(z)+ = = z/2cothz/2,
2 2expz−1
which is an even function of z. So B = 1, B = −1/2 and B = 0 if n ≥ 3
0 1 n
is odd. Some further values are given below.
n 0 1 2 3 4 5 6 7 8
B 1 −1/2 1/6 0 −1/30 0 1/42 0 −1/30
n
Date: November 6, 2014.
1
2 MARKWILDON
Note that the numerators need not be ±1. For instance B = 5/66.
10
1.4. Connection with FLT. Kummer proved that there are no positive
integralsolutionsofxp+yp = zp wheneverpisanoddprimenotdividingthe
numerators of any of the Bernoulli numbers B ,B ,...,B . Such primes
2 4 p−3
are said to be regular. For instance as B = 691/2730, the prime 691 is not
12
regular.
1.5. Estimate for B . In [1] it is noted that B = 854513 > 6192; indeed
n 22
138
the Bernoulli numbers are unbounded. The authors explain that using Eu-
ler’s formula for cotz (which has a nice elementary proof via the so-called
Herglotz trick),
(cid:88) z2
zcotz = 1−2
k2π2−z2
k≥1
one gets
β(2z)+z = zcothz
(cid:88)∞ z2
= 1+2
k2π2+z2
k=1
(cid:88)∞ (cid:88)∞ (cid:18)−z2 (cid:19)m
= 1−2
k2π2
k=1m=1
for |z| < π. Hence, comparing coefficients of z2n,
2(2n)!
B = (−1)n−1 ζ(2n).
2n (2π)2n
Thus
2 |B | 4
2n
≤ ≤ .
(2π)2n (2n)! (2π)2n
Note this shows that |B |/(2n)! is a decreasing sequence. As these are
2n
coefficients in the convergent generating function β(z), |B |/(2n)! tends
2n
to 0 as n tends to ∞.
1.6. Connection with uniform distribution. Let X be distributed uni-
formly on [0,1]. Then the moment generating function for X is
(cid:90) 1 1 1
EezX = eztdt = (ex−1) = .
x β(z)
0
The coefficients of 1/β(z) give us a recurrence satisfied by the Bernoulli
numbers – but as
(cid:88)∞ zn
1/β(z) =
(n+1)!
n=0
this is just the defining recurrence (1).
NOTES ON BERNOULLI NUMBERS AND EULER’S SUMMATION FORMULA 3
1.7. Sums of powers. We derive from first principles a special case of the
Euler summation formula. Let
n−1
(cid:88)
S (n) = 0m+1m+...+(n−1)m = km.
m
k=0
Consider the following generating function where m varies and n is fixed:
Sˆ(z) = (cid:80)∞ S (n)zm. We have
m=0 m m!
(cid:88)∞ (cid:32)n(cid:88)−1 (cid:33) zm n(cid:88)−1 (cid:88)∞ (kz)m expnz−1
Sˆ(z) = km = =
m! m! expz−1
m=0 k=0 k=0m=0 .
= β(z)expnz−1 = (cid:32)(cid:88)∞ Bnzn(cid:33)(cid:32)(cid:88)∞ nm+1 zm(cid:33)
z n! (m+1)!
n=0 m=0
Comparing coefficients of zm/m! gives
(4) S (n) = m!(cid:88)m Br nm−r+1 = (cid:88)m Br (cid:18)m+1(cid:19)nm+1−r.
m
r! (m−r+1)! m+1 r
r=0 r=0
1.8. Bernoulli polynomials. Define
(cid:18) (cid:19)
(cid:88) m
B (z) = B zm−k.
m k
k
k
The first few Bernoulli polynomials are B (z) = 1, B (z) = z− 1, B (z) =
0 1 2
2
z2−z+ 1, B (z) = z3− 3z+ 1, on so on. Note that B (0) = B and that
3 m m
6 2 2
(cid:18) (cid:19)
(cid:88) m+1
B (n) = B nm+1−k = (m+1)S (n)+B
m+1 k m m+1
k
k
so
B (n)−B (0)
m+1 m+1
S (n) =
m
m+1
which is the obvious definition for S (z).
m
Some useful properties:
(cid:18) (cid:19) m−1(cid:18) (cid:19)
(cid:88) m (cid:88) m
B (1) = B = B +B = [m = 1]+B
m k k m m
k k
k k=1
so B (1) = B (0) = B unless m = 1 in which case B (1) = −1 = −B .
m m m 1 1
2
For m ≥ 1 we have
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:88) m (cid:88) m−1
B (z)(cid:48) = B (m−k)zm−k−1 = m B zm−k−1
m k k
k k
k k
= mB (z).
m−1
2. The Euler summation formula
From now on let f : R → R be a function with as many derivatives as
needed. Euler’s summation formula is:
4 MARKWILDON
Theorem 2.1 (Euler). Let m ≥ 1 and let a < b be integers. Then
(5) (cid:88) f(k) = (cid:90) bf(x)dx+(cid:88)m Bkf(k−1)(x)(cid:12)(cid:12)b +R
a k! (cid:12)a m
a≤k<b k=1
where the remainder R is given by
m
(cid:90) b B ({x})
R = (−1)m+1 m f(m)(x)dx.
m
m!
a
2.1. Application to sums of powers. Set f(x) = xr and take m > r.
Then f(m) = 0 so the remainder term vanishes. Putting a = 0 and b = n we
obtain
n(cid:88)−1kr = nr+1 +(cid:88)r+1 Bkrk−1nr+1−k = (cid:88)r+1 Bk (cid:18)r+1(cid:19)nr+1−k
r+1 k! r+1 k
k=1 k=1 k=0
which agrees with (4).
2.2. First proof. As all the terms telescope nicely it is sufficient to prove
the formula when a = 0 and b = 1. This has the advantage that we can
use B (x) rather than B ({x}). We proceed by induction on m. The case
m m
m = 1 states that
(cid:90) 1 (cid:90) 1
f(0) = f(x)dx− 1(f(1)−f(0))+ (x− 1)f(cid:48)(x)dx.
2 2
0 0
This follows from a simple integration by parts:
(cid:90) 1 (cid:90) 1
(x− 1)f(cid:48)(x)dx = f(1)− f(x)dx− 1(f(1)−f(0)).
2 2
0 0
For m > 1 we can write the right-hand-side as
(cid:90) 1f(x)dx+m(cid:88)−1 Bk(f(k−1)(1)−f(k−1)(0))
k!
0
k=1
B (cid:90) 1 B (x)
+ m(f(m−1)(1)−fm−1(0))+(−1)m+1 m f(m)(x)dx
m! m!
0
Applying the inductive hypothesis gives:
(cid:90) 1 B (x)
f(0)+(−1)m+1 m−1 f(m−1)(x)
(m−1)!
0
B (cid:90) 1 B (x)
+ m(f(m−1)(1)−f(m−1)(0))+(−1)m+1 m f(m)(x)dx
m! m!
0
Integrating the final term by parts using the results in §1.8 gives
(cid:90) 1 B (x) B
f(0)+(−1)m+1 m−1 f(m−1)(x)+ m(f(m−1)(1)−f(m−1)(0))
(m−1)! m!
0
B (x) (cid:12)1 (cid:90) 1 B (x)
+(−1)m+1 m f(m−1)(x)(cid:12) +(−1)m m−1 f(m−1)(x)dx.
m! (cid:12)0 0 (m−1)!
The outermost two terms obviously cancel. And so do the inner two, as if
m is odd, then as m > 1, B = 0. (cid:3)
m
NOTES ON BERNOULLI NUMBERS AND EULER’S SUMMATION FORMULA 5
2.3. An alternative proof. There used to be a section here claiming an
alternating proof by taking a sequence of polynomials converging uniformly
to f on the interval [a,b] and using uniform convergence to interchange the
integral with the limit of the polynomials: this reduces to the case where
f(x) = xr. Howeveritisnolongerpossibletoreducetothecasewherea = 0
and b = 1, and the main inductive step in the proof is not substantially
simplified from the previous section.
It is however worth noting that if m > r and we take a = 0 and b = N
then the result comes out very easily. In this case Euler’s formula says that
(cid:88) kr = Nr+1 +(cid:88)r Bkrk−1Nr−(k−1)
r+1 k!
0≤k<N k=1
Using the identity 1(cid:0) r (cid:1) = 1 (cid:0)r+1(cid:1), the right-hand side can be rewritten
k k−1 r+1 k
as
Nr+1 +(cid:88)r Bk (cid:18)r+1(cid:19)Nr+1−k
r+1 r+1 k
k=1
which is equal to (cid:80) kr by (4).
0≤k<N
2.4. Estimates for error term. We can state Euler’s summation formula
in the following form
(6) (cid:88) f(k) = (cid:90) bf(x)dx−1(f(b)−f(a))+(cid:88)m B2k f(2k−1)(x)(cid:12)(cid:12)b+R .
a 2 (2k)! (cid:12)a 2m
a≤k<b k=1
It can be shown that |B ({x})| ≤ B so a rough estimate for the error
2m 2m
term is given by §1.5:
B (cid:90) b
|R | ≤ 2m |f(2m)(x)|dx.
2m
(2m)!
a
If f(2m) is positive then this shows that the magnitude of the error term is
at most the magnitude of the final term in the sum.
Note that very often the remainder term R will not tend to 0 as the
2m
upper limit in the summation, b tends to ∞. However it will often have
some non-zero limit. In this case we have
(cid:90) ∞ B ({x})
R = R (∞)− 2m f(2m)(x)dx
2m 2m
(2m)!
b
where the right-hand-side tends to R (∞) as b → ∞.
2m
Itisprovedin[1]p475thatiff(2m+2)andf(2m+4)arepositiveforx ∈ [a,b]
then
B (cid:12)b
R = θ 2m+2 f2m+1(x)(cid:12)
2m m(2m+2)! (cid:12)a
where 0 ≤ θ ≤ 1. So the remainder term lies somewhere between 0 and
m
what would have been the next term in the sum.
6 MARKWILDON
3. Examples of Euler summation
3.1. First example. We shall attempt to use Euler summation to find
n−1
(cid:88)
S = tk
n
k=0
where0 ≤ t < 1. Ofcoursewealreadyknowtheanswer,S = (1−tn)/(1−t),
n
andinfactEulersummationturnsouttobeaveryineffectivewayoffinding
it! Still there are some points of interest.
If f(x) = tx then f(r)(x) = (logt)rtx so by equation (6) we have
n(cid:88)−1tk = (cid:90) ntxdx− 1(tn−1)+(cid:88)m B2k (logt)2k−1tx(cid:12)(cid:12)n+R
0 2 (2k)! (cid:12)0 2m
k=0 k=1
(cid:32) m (cid:33)
= (1−tn) 1 − 1 −(cid:88) B2k (logt)2k−1 +R
2m
2 logt (2k)!
k=1
for n ≥ 1 and m ≥ 1. Now something a little suprising happens: if t > e−π
then |logt| < π and so we can use the exponential generating function for
the Bernoulli numbers to obtain
∞
(cid:88) B2k (logt)2k−1 = 1 − 1 + 1
(2k)! t−1 logt 2
k=1
and so
n(cid:88)−1tk = 1−tn +(1−tn) (cid:88)∞ B2k (logt)2k−1+R
2m
1−t (2k)!
k=0 k=m+1
As we know the answer, the last equation implies that R → 0 as m → ∞.
2m
Less artificially, we can attempt to prove this directly. We find
(cid:90) n B ({x})
R = (logt)2m−1 2m txdx
2m
(2m)!
0
B
≤ |(logt)|2m−2 2m (1−tn)
(2m)!
4
≤ |(logt)|2m−2 (1−tn)
(2π)2m
So the error term R → 0 as m → ∞, provided e−2π < t < e2π, which
2m
holds as we have already assumed that e−π < t < 1.
The limiting behaviour with respect to n is also of interest. Knowing the
limit of S allows us to see that
n
∞
lim R = − (cid:88) B2k (logt)2k−1
2m
n→∞ (2k)!
k=m+1
This is an example of the point made earlier, that while the remainder term
will not usually tend to 0 as n tends to ∞, it may well have some limiting
value.
NOTES ON BERNOULLI NUMBERS AND EULER’S SUMMATION FORMULA 7
√
(cid:82)n
3.2. Summing square roots. Let f(x) = x. Note that f(x)dx =
0
3
2(n2 −1) and that
3 (cid:18) (cid:19)
f(r)(x) = 12 r!x12−r.
r
Euler’s summation formula gives
(cid:88)n √k = 2(n32 −1)+ 1√n+ 1 +(cid:88)m B2k(cid:18) 12 (cid:19)(n32−2k −1)+R2m.
3 2 2 2k 2k−1
k=1 k=1
Setting m = 1 we get
(cid:88)n √k = 2(n32 −1)+ 1√n+ 1 + 1 (n−12 −1)+R2.
3 2 2 12
k=1
We can estimate R by using the bound |B ({x})| < B = 1:
2 2 2
6
(cid:90) n
|R2| ≤ 1 x−32 = 1 (1−n−12)
6.8 24
1
So we have
(cid:88)n √k = 2(n32 −1)+ 1√n+C +O(n−12)
3 2
k=1
for some constant C. In general the remark on the previous page shows that
the error is given by
(cid:18) (cid:19)
R2m = θmB2m+2 12 (n12−2m−1) = C(m)+O(n12−2m)
2m+2 2m−1
where θ ∈ [0,1] and C(m) does not depend on n, This gives
m
(cid:88)n √k = 2(n23 −1)+ 1√n+C(m)+(cid:88)m B2k(cid:18) 12 (cid:19)n32−2k +O(n12−2m).
3 2 2k 2k−1
k=1 k=1
There is a small simplification: C(m) can be determined by taking a limit
with respect to n, so
(cid:32)(cid:88)n √ 3 √ (cid:33)
C = lim k− 2(n2 −1)− 1 n
n→∞ 3 2
k=1
and so does not depend on m. (Exercise 9.27 in [1] reveals that C = ζ(−1);
2
in fact the definition of ζ(α) for α > −1.)
3.3. An estimate for P(x).
∞
(cid:88)
logP(e−t) = −log(1−e−kt)
k=1
(cid:88)∞ (cid:88)∞ e−tmk
=
m
k=1m=1
∞
(cid:88) 1
= .
emt−1
m=1
This suggests 2 ways we might apply Euler summation. (But they turn out
to be equivalent.)
8 MARKWILDON
3.3.1. Taken from Knuth fascicle Exercise 25. Let
(cid:88)∞ e−mtx
f(x) = −log(1−e−tx) = .
m
m=1
Then we have logP(e−t) = (cid:80)∞ f(k). The integral of f is given by the Li
k=1 2
function:
(cid:90) xf(u)du = (cid:88)∞ e−mtu(cid:12)(cid:12)1 = (cid:88)∞ e−mt−(cid:88)∞ e−mtx = Li2(e−t)−Li2(e−tx).
1 tm2 (cid:12)x tm2 tm2 t t
m=1 m=1 m=1
The derivatives of f are connected with the Eulerian numbers: let (cid:10)n(cid:11) de-
m
notethenumberofpermutationsinS withexactlymascents(ordescents).
n
Then we claim
Lemma 3.1. For n ≥ 1 we have
e−tx(−t)n (cid:88)(cid:28)n−1(cid:29)
f(n)(x) = e−ktx.
(1−e−tx)n k
k
Proof. By induction on m. If m = 1 then the formula is readily verified.
Suppose true for m. Then
−me−2tx(−t)mt (cid:88)(cid:28)n−1(cid:29)
f(m+1)(x) = e−ktx
(1−e−tx)m+1 k
k
e−tx(−t)n(−t) (cid:88) (cid:28)n−1(cid:29)
+ k e−ktx
(1−e−tx)n k
k
(cid:32) (cid:33)
e−tx(−t)n+1 (cid:88) (cid:28)n−1(cid:29) (cid:88) (cid:28)n−1(cid:29)
= n +(1−e−tx) k e−ktx
(1−e−tx)n+1 k−1 k
k k
e−tx(−t)n+1 (cid:18) (cid:28)n−1(cid:29) (cid:28)n−1(cid:29)(cid:19)
= (n−k) +k e−ktx
(1−e−tx)n+1 k−1 k
To finish we apply the identity1 (n−m)(cid:10)n−1(cid:11)+(m+1)(cid:10)n−1(cid:11) = (cid:10)n(cid:11). (cid:3)
m−1 m m
We are now ready to apply Euler summation. Taking m = 1 in (6) we
get
(cid:88)n f(k) = Li2(e−t) − Li2(e−nt) − 1(log(1−e−nt)+log(1−e−t))
t t 2
k=1
e−tx(−t)(cid:12)n
+ 1 (cid:12) +R .
12 1−e−tx(cid:12)1 2
1Thisidentitycanbeprovedasfollows. Letg∈S haveexactlymascents. Ifnappears
n
in a position ...bna... or ...n then removing n gives a permutation in S with m−1
n−1
ascents. Conversely,givensuchapermutation, therearen−m−1descents,andputting
n between any 2 numbers involved in a descent, or at the end, gives a permutation in S
n
with m ascents. Otherwise we have ...anb... or n... in which case removing n does not
change the number of ascents. Conversely given a permutation in S with m ascents,
n−1
putting n between the two 2 numbers involved in an ascent, or at the start, does not
change the number of ascents.
NOTES ON BERNOULLI NUMBERS AND EULER’S SUMMATION FORMULA 9
where as the even derivatives are positive
e−tx(−t)(cid:12)n te−t t
R ≤ 1 (cid:12) ≤ = ≤ 1.
2 12 1−e−tx(cid:12)1 1−e−t et−1
So letting n → ∞ we get
Li (e−t)
logP(e−t) = 2 − 1 log(1−e−t)+O(1)
t 2
where O(1) stands for a quantity always lying in [0,1]. If we use the identity
suggested by Knuth, namely
Li (x)+Li (1−x) = ζ(2)−logxlog(1−x)
2 2
or, equivalently,
Li (e−t) Li (1−e−t) ζ(2)
2 = − 2 + +tlog(1−e−t
t t t
we get
ζ(2) Li (1−e−t)
logP(e−t) = − 2 − 3 log(1−e−t)+O(1)
t t 2
ζ(2)
= − 1 log(1−e−t)+O(1) ; where O(1) ∈ [−1,1]
t 2
ζ(2) logt
= − +O(1) ; where O(1) ∈ [−1−log2,1−log2].
t 2
As Li (y) ≤ −log(1−y) gives Li (1−e−t) ≤ t and log(1−e−t) = logt+
2 2
log(1−t/2+...) = logt+O(1). In particular by being a little more careful
with the O(1) errors we can get a lower bound for logP(e−t):
π2 logt π2 logt
+ −1−log2 ≤ logP(e−t) ≤ + +1−log2.
6t 2 6t 2
√
In fact the ‘right’ constant is −log( 2π) ≈ −0.9189; this follows from
π2 logt √
logP(e−t) = + −log 2π+O(t)
6t 2
where the O(t) term is (by the functional equation), − t +logP(e−4π2/t).
24
Note that
−1.6931 < −0.9189 < .6931.
3.3.2. Alternative. We might also try to use Euler summation to sum
∞
(cid:88)
−log(1−e−kt).
k=1
Perhaps surprisingly this turns out to be equivalent to the previous ap-
proach, since
d te−xt t
log(1−e−xt) = =
dx 1−e−xt ext−1
which is essentially the function we summed earlier.
References
[1] Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. Concrete Mathematics.
Addison Wesley, 1994.