E ULER –M ACLAURIN F ORMULA
The Bernoulli numbers were introduced by Jacques (James, Jacob) Bernoulli. There are several equivalent definitions, but extreme care must be taken, for some authors introduce
18H. B. Dwight, Tables of Integrals and Other Mathematical Data. New York: Macmillan (1947).
variations in numbering or in algebraic signs. One relatively simple approach is to define the Bernoulli numbers by the series19
x ex−1=
∞
n=0
Bnxn
n! , (5.144)
which converges for |x|<2π by the ratio test substitut Eq. (5.153) (see also Exam- ple 7.1.7). By differentiating this power series repeatedly and then settingx=0, we obtain
Bn= dn
dxn x
ex−1
x=0
. (5.145)
Specifically,
B1= d dx
x ex−1
x=0
= 1
ex−1 − xex (ex−1)2
x=0
= −1
2, (5.146)
as may be seen by series expansion of the denominators. UsingB0=1 andB1= −12, it is easy to verify that the function
x
ex−1 −1+x 2 =
∞
n=2
Bnxn
n! = − x
e−x−1−1−x
2 (5.147)
is even inx, so allB2n+1=0.
To derive a recursion relation for the Bernoulli numbers, we multiply ex−1
x x
ex−1=1= ∞
m=0
xm (m+1)!
1−x
2 + ∞
n=1
B2n x2n (2n)!
=1+ ∞
m=1
xm 1
(m+1)!− 1 2m!
+ ∞
N=2
xN
1≤n≤N/2
B2n
(2n)!(N−2n+1)!. (5.148) ForN >0 the coefficient ofxN is zero, so Eq. (5.148) yields
1
2(N+1)−1=
1≤n≤N/2
B2n N+1
2n
=1
2(N−1), (5.149)
19The functionx/(ex−1)may be considered a generating function since it generates the Bernoulli numbers. Generating functions of the special functions of mathematical physics appear in Chapters 11, 12, and 13.
Table 5.1 Bernoulli Numbers
n Bn Bn
0 1 1.0000 00000
1 −12 −0.5000 00000
2 16 0.1666 66667
4 −301 −0.0333 33333
6 421 0.0238 09524
8 −301 −0.0333 33333
10 665 0.0757 57576
Note. Further values are given in National Bureau of Stan- dards, Handbook of Mathematical Functions (AMS-55).
See footnote 4 for the reference.
which is equivalent to
N−1 2 =
N n=1
B2n
2N+1 2n
,
N−1=
N−1 n=1
B2n 2N
2n
.
(5.150)
From Eq. (5.150) the Bernoulli numbers in Table 5.1 are readily obtained. If the variablex in Eq. (5.144) is replaced by 2ixwe obtain an alternate (and equivalent) definition ofB2n (B1is set equal to−12by Eq. (5.146)) by the expression
xcotx= ∞
n=0
(−1)nB2n
(2x)2n
(2n)! , −π < x < π. (5.151) Using the method of residues (Section 7.1) or working from the infinite product represen- tation of sinx(Section 5.11), we find that
B2n=(−1)n−12(2n)! (2π )2n
∞
p=1
p−2n, n=1,2,3, . . . . (5.152) This representation of the Bernoulli numbers was discovered by Euler. It is readily seen from Eq. (5.152) that|B2n|increases without limit asn→ ∞. Numerical values have been calculated by Glaisher.20 Illustrating the divergent behavior of the Bernoulli numbers, we have
B20= −5.291×102 B200= −3.647×10215.
20J. W. L. Glaisher, table of the first 250 Bernoulli’s numbers (to nine figures) and their logarithms (to ten figures). Trans.
Cambridge Philos. Soc. 12: 390 (1871–1879).
Some authors prefer to define the Bernoulli numbers with a modified version of Eq. (5.152) by using
Bn= 2(2n)! (2π )2n
∞
p=1
p−2n, (5.153)
the subscript being just half of our subscript and all signs positive. Again, when using other texts or references, you must check to see exactly how the Bernoulli numbers are defined.
The Bernoulli numbers occur frequently in number theory. The von Staudt–Clausen the- orem states that
B2n=An− 1 p1− 1
p2− 1
p3− ã ã ã − 1 pk
, (5.154)
in whichAnis an integer andp1, p2, . . . , pkare prime numbers so thatpi−1 is a divisor of 2n. It may readily be verified that this holds for
B6(A3=1, p=2,3,7),
B8(A4=1, p=2,3,5), (5.155)
B10(A5=1, p=2,3,11), and other special cases.
The Bernoulli numbers appear in the summation of integral powers of the integers, N
j=1
jp, pintegral,
and in numerous series expansions of the transcendental functions, including tanx, cotx, ln|sinx|,(sinx)−1, ln|cosx|, ln|tanx|,(coshx)−1, tanhx, and cothx. For example,
tanx=x+x3 3 + 2
15x5+ ã ã ã +(−1)n−122n(22n−1)B2n
(2n)! x2n−1+ ã ã ã. (5.156) The Bernoulli numbers are likely to come in such series expansions because of the defining equations (5.144), (5.150), and (5.151) and because of their relation to the Riemann zeta function,
ζ (2n)= ∞
p=1
p−2n. (5.157)
Bernoulli Polynomials
If Eq. (5.144) is generalized slightly, we have xexs ex−1 =
∞
n=0
Bn(s)xn
n! (5.158)
Table 5.2 Bernoulli Polynomials
B0=1 B1=x−12 B2=x2−x+16 B3=x3−32x2+12x B4=x4−2x3+x2−301 B5=x5−52x4+53x3−16x B6=x6−3x5+52x4−12x2+421
Bn(0)=Bn, Bernoulli number
defining the Bernoulli polynomials, Bn(s). The first seven Bernoulli polynomials are given in Table 5.2.
From the generating function, Eq. (5.158),
Bn(0)=Bn, n=0,1,2, . . . , (5.159) the Bernoulli polynomial evaluated at zero equals the corresponding Bernoulli number.
Two particularly important properties of the Bernoulli polynomials follow from the defin- ing relation, Eq, (5.158): a differentiation relation
d
dsBn(s)=nBn−1(s), n=1,2,3, . . . , (5.160) and a symmetry relation (replacex→ −x in Eq. (5.158) and then sets=1)
Bn(1)=(−1)nBn(0), n=1,2,3, . . . . (5.161) These relations are used in the development of the Euler–Maclaurin integration formula.
Euler–Maclaurin Integration Formula
One use of the Bernoulli functions is in the derivation of the Euler–Maclaurin integration formula. This formula is used in Section 8.3 for the development of an asymptotic expres- sion for the factorial function — Stirling’s series.
The technique is repeated integration by parts, using Eq. (5.160) to create new deriva- tives. We start with
1 0
f (x) dx= 1
0
f (x)B0(x) dx. (5.162)
From Eq. (5.160) and Exercise 5.9.2,
B1′(x)=B0(x)=1. (5.163)
SubstitutingB1′(x)into Eq. (5.162) and integrating by parts, we obtain 1
0
f (x) dx=f (1)B1(1)−f (0)B1(0)− 1
0
f′(x)B1(x) dx
=1 2
f (1)+f (0) − 1
0
f′(x)B1(x) dx. (5.164) Again using Eq. (5.160), we have
B1(x)=1
2B2′(x), (5.165)
and integrating by parts we get 1
0
f (x) dx=1 2
f (1)+f (0) − 1 2!
f′(1)B2(1)−f′(0)B2(0)
+ 1 2!
1 0
f(2)(x)B2(x) dx. (5.166)
Using the relations
B2n(1)=B2n(0)=B2n, n=0,1,2, . . .
(5.167) B2n+1(1)=B2n+1(0)=0, n=1,2,3, . . .
and continuing this process, we have 1
0
f (x) dx=1 2
f (1)+f (0) − q p=1
1 (2p)!B2p
f(2p−1)(1)−f(2p−1)(0)
+ 1 (2q)!
1 0
f(2q)(x)B2q(x) dx. (5.168a)
This is the Euler–Maclaurin integration formula. It assumes that the functionf (x)has the required derivatives.
The range of integration in Eq. (5.168a) may be shifted from[0,1]to[1,2]by replacing f (x)byf (x+1). Adding such results up to[n−1, n], we obtain
n 0
f (x) dx=1
2f (0)+f (1)+f (2)+ ã ã ã +f (n−1)+1 2f (n)
− q p=1
1 (2p)!B2p
f(2p−1)(n)−f(2p−1)(0)
+ 1 (2q)!
1 0
B2q(x)
n−1
ν=0
f(2q)(x+ν) dx. (5.168b) The terms 12f (0)+f (1)+ ã ã ã +12f (n)appear exactly as in trapezoidal integration, or quadrature. The summation overp may be interpreted as a correction to the trapezoidal approximation. Equation (5.168b) may be seen as a generalization of Eq. (5.22); it is the
Table 5.3 Riemann Zeta Function
s ζ (s)
2 1.6449340668
3 1.2020569032
4 1.0823232337
5 1.0369277551
6 1.0173430620
7 1.0083492774
8 1.0040773562
9 1.0020083928
10 1.0009945751
form used in Exercise 5.9.5 for summing positive powers of integers and in Section 8.3 for the derivation of Stirling’s formula.
The Euler–Maclaurin formula is often useful in summing series by converting them to integrals.21
Riemann Zeta Function
This series, ∞
p=1p−2n, was used as a comparison series for testing convergence (Sec- tion 5.2) and in Eq. (5.152) as one definition of the Bernoulli numbers,B2n. It also serves to define the Riemann zeta function by
ζ (s)≡ ∞
n=1
n−s, s >1. (5.169)
Table 5.3 lists the values ofζ (s)for integrals,s=2,3, . . . ,10. Closed forms for evens appear in Exercise 5.9.6. Figure 5.11 is a plot ofζ (s)−1. An integral expression for this Riemann zeta function appears in Exercise 8.2.21 as part of the development of the gamma function, and the functional relation is given in Section 14.3.
The celebrated Euler prime number product for the Riemann zeta function may be de- rived as
ζ (s)
1−2−s
=1+ 1 2s + 1
3s + ã ã ã − 1
2s + 1 4s + 1
6s + ã ã ã
; (5.170)
eliminating all then−s, wherenis a multiple of 2. Then ζ (s)
1−2−s
1−3−s
=1+ 1 3s + 1
5s + 1 7s + 1
9s + ã ã ã
− 1
3s + 1 9s + 1
15s + ã ã ã
; (5.171)
21See R. P. Boas and C. Stutz, Estimating sums with integrals. Am. J. Phys. 39: 745 (1971), for a number of examples.
FIGURE5.11 Riemann zeta function,ζ (s)−1 versuss.
eliminating all the remaining terms in whichn is a multiple of 3. Continuing, we have ζ (s)(1−2−s)(1−3−s)(1−5−s)ã ã ã(1−P−s), whereP is a prime number, and all terms n−s, in whichnis a multiple of any integer up throughP, are canceled out. AsP → ∞,
ζ (s)
1−2−s
1−3−s
ã ã ã
1−P−s
→ζ (s)
#∞
P (prime)=2
1−P−s
=1. (5.172) Therefore
ζ (s)=
#∞
P (prime)=2
1−P−s−1
, (5.173)
givingζ (s)as an infinite product.22
This cancellation procedure has a clear application in numerical computation. Equa- tion (5.170) will giveζ (s)(1−2−s)to the same accuracy as Eq. (5.169) givesζ (s), but
22This is the starting point for the extensive applications of the Riemann zeta function to analytic number theory. See H. M. Ed- wards, Riemann’s Zeta Function. New York: Academic Press (1974); A. Ivi´c, The Riemann Zeta Function. New York: Wiley (1985); S. J. Patterson, Introduction to the Theory of the Riemann Zeta Function. Cambridge, UK: Cambridge University Press (1988).
with only half as many terms. (In either case, a correction would be made for the neglected tail of the series by the Maclaurin integral test technique — replacing the series by an inte- gral, Section 5.2.)
Along with the Riemann zeta function, AMS-55 (Chapter 23. See Exercise 5.2.22 for the reference.) defines three other Dirichlet series related toζ (s):
η(s)= ∞
n=1
(−1)n−1n−s=
1−21−s ζ (s),
λ(s)= ∞
n=0
(2n+1)−s=
1−2−s ζ (s), and
β(s)= ∞
n=0
(−1)n(2n+1)−s.
From the Bernoulli numbers (Exercise 5.9.6) or Fourier series (Example 14.3.3 and Exer- cise 14.3.13) special values are
ζ (2)=1+ 1 22+ 1
32+ ã ã ã =π2 6 ζ (4)=1+ 1
24+ 1
34+ ã ã ã =π4 90 η(2)=1− 1
22+ 1
32+ ã ã ã =π2 12 η(4)=1− 1
24+ 1
34+ ã ã ã =7π4 720 λ(2)=1+ 1
32+ 1
52+ ã ã ã =π2 8 λ(4)=1+ 1
34+ 1
54+ ã ã ã =π4 96 β(1)=1−1
3+1
5− ã ã ã =π 4 β(3)=1− 1
33+ 1
53− ã ã ã =π3 32. Catalan’s constant,
β(2)=1− 1 32+ 1
52− ã ã ã =0.91596559. . . , is the topic of Exercise 5.2.22.
Improvement of Convergence
If we are required to sum a convergent series∞
n=1anwhose terms are rational functions of n, the convergence may be improved dramatically by introducing the Riemann zeta function.
Example 5.9.1 IMPROVEMENT OFCONVERGENCE
The problem is to evaluate the series ∞
n=11/(1 +n2). Expanding (1 +n2)−1 = n−2(1+n−2)−1by direct division, we have
1+n2−1
=n−2
1−n−2+n−4− n−6 1+n−2
= 1 n2− 1
n4+ 1
n6− 1 n8+n6. Therefore
∞
n=1
1
1+n2=ζ (2)−ζ (4)+ζ (6)− ∞
n=1
1 n8+n6.
Theζ values are tabulated and the remainder series converges asn−8. Clearly, the process can be continued as desired. You make a choice between how much algebra you will do and how much arithmetic the computer will do. Other methods for improving computational effectiveness are given at the end of Sections 5.2 and 5.4.
Exercises
5.9.1 Show that
tanx= ∞
n=1
(−1)n−122n(22n−1)B2n
(2n)! x2n−1, −π
2 < x <π 2. Hint. tanx=cotx−2 cot 2x.
5.9.2 Show that the first Bernoulli polynomials are B0(s)=1 B1(s)=s−12
B2(s)=s2−s+16. Note thatBn(0)=Bn, the Bernoulli number.
5.9.3 Show thatBn′(s)=nBn−1(s),n=1,2,3, . . .. Hint. Differentiate Eq. (5.158).
5.9.4 Show that
Bn(1)=(−1)nBn(0).
Hint. Go back to the generating function, Eq. (5.158), or Exercise 5.9.2.
5.9.5 The Euler–Maclaurin integration formula may be used for the evaluation of finite series:
n m=1
f (m)= n
0
f (x) dx+1
2f (1)+1
2f (n)+B2
2!
f′(n)−f′(1) + ã ã ã.
Show that (a)
n m=1
m=1
2n(n+1).
(b) n m=1
m2=1
6n(n+1)(2n+1).
(c) n m=1
m3=1
4n2(n+1)2. (d)
n m=1
m4= 1
30n(n+1)(2n+1)
3n2+3n−1 . 5.9.6 From
B2n=(−1)n−12(2n)! (2π )2nζ (2n), show that
(a)ζ (2)=π2
6 (d)ζ (8)= π8 9450 (b)ζ (4)=π4
90 (e)ζ (10)= π10 93,555. (c)ζ (6)= π6
945
5.9.7 Planck’s blackbody radiation law involves the integral ∞
0
x3dx ex−1. Show that this equals 6ζ (4). From Exercise 5.9.6, ζ (4)=π4
90. Hint. Make use of the gamma function, Chapter 8.
5.9.8 Prove that
∞
0
xnexdx
(ex−1)2 =n!ζ (n).
Assumingn to be real, show that each side of the equation diverges ifn=1. Hence the preceding equation carries the conditionn >1. Integrals such as this appear in the quantum theory of transport effects — thermal and electrical conductivity.
5.9.9 The Bloch–Gruneissen approximation for the resistance in a monovalent metal is ρ=CT5
6 /T
0
x5dx (ex−1)(1−e−x), whereis the Debye temperature characteristic of the metal.
(a) ForT → ∞, show that
ρ≈C 4 ã T
2. (b) ForT →0, show that
ρ≈5!ζ (5)CT5 6. 5.9.10 Show that
(a) 1
0
ln(1+x) x dx=1
2ζ (2), (b) lim
a→1
a 0
ln(1−x)
x dx=ζ (2).
From Exercise 5.9.6,ζ (2)=π2/6. Note that the integrand in part (b) diverges fora=1 but that the integrated series is convergent.
5.9.11 The integral
1 0
ln(1−x) 2dx x
appears in the fourth-order correction to the magnetic moment of the electron. Show that it equals 2ζ (3).
Hint. Let 1−x=e−t.
5.9.12 Show that ∞
0
(lnz)2 1+z2dz=4
1− 1
33 + 1 53 − 1
73+ ã ã ã
. By contour integration (Exercise 7.1.17), this may be shown equal toπ3/8.
5.9.13 For “small” values ofx,
ln(x!)= −γ x+ ∞
n=2
(−1)nζ (n) n xn,
whereγ is the Euler–Mascheroni constant andζ (n)is the Riemann zeta function. For what values ofx does this series converge?
ANS.−1< x≤1.
Note that ifx=1, we obtain
γ= ∞
n=2
(−1)nζ (n) n ,
a series for the Euler–Mascheroni constant. The convergence of this series is exceed- ingly slow. For actual computation ofγ, other, indirect approaches are far superior (see Exercises 5.10.11, and 8.5.16).
5.9.14 Show that the series expansion of ln(x!)(Exercise 5.9.13) may be written as (a) ln(x!)=1
2ln π x
sinπ x
−γ x− ∞
n=1
ζ (2n+1) 2n+1 x2n+1, (b) ln(x!)=1
2ln π x
sinπ x
−1 2ln
1+x 1−x
+(1−γ )x
− ∞
n=1
ζ (2n+1)−1 x2n+1 2n+1.
Determine the range of convergence of each of these expressions.
5.9.15 Show that Catalan’s constant,β(2), may be written as β(2)=2
∞
k=1
(4k−3)−2−π2 8 . Hint.π2=6ζ (2).
5.9.16 Derive the following expansions of the Debye functions forn≥1:
x 0
tndt et−1 =xn
1
n− x
2(n+1)+ ∞
k=1
B2kx2k (2k+n)(2k)!
, |x|<2π; ∞
x
tndt et−1 =
∞
k=1
e−kx xn
k +nxn−1
k2 +n(n−1)xn−2
k3 + ã ã ã + n! kn+1
forx >0.The complete integral(0,∞)equalsn!ζ (n+1), Exercise 8.2.15.
5.9.17 (a) Show that the equation ln 2=∞
s=1(−1)s+1s−1(Exercise 5.4.1) may be rewritten as
ln 2= ∞
s=2
2−sζ (s)+ ∞
p=1
(2p)−n−1
1− 1 2p
−1
. Hint. Take the terms in pairs.
(b) Calculate ln 2 to six significant figures.
5.9.18 (a) Show that the equationπ/4=∞
n=1(−1)n+1(2n−1)−1(Exercise 5.7.6) may be rewritten as
π
4 =1−2 ∞
s=1
4−2sζ (2s)−2 ∞
p=1
(4p)−2n−2
1− 1 (4p)2
−1
. (b) Calculateπ/4 to six significant figures.
5.9.19 Write a function subprogram ZETA(N )that will calculate the Riemann zeta function for integer argument. Tabulateζ (s)fors=2,3,4, . . . ,20. Check your values against Table 5.3 and AMS-55, Chapter 23. (See Exercise 5.2.22 for the reference.).
Hint. If you supply the function subprogram with the known values ofζ (2), ζ (3), and ζ (4), you avoid the more slowly converging series. Calculation time may be further shortened by using Eq. (5.170).
5.9.20 Calculate the logarithm (base 10) of|B2n|, n=10,20, . . . ,100.
Hint. Programζ (n)as a function subprogram, Exercise 5.9.19.
Check values. log|B100| =78.45 log|B200| =215.56.