Physics 451
Fall 2004 Homework Assignment #8 — Solutions
Textbook problems: Ch. 5: 5.6.2, 5.6.19, 5.7.4, 5.7.15, 5.9.11, 5.10.1, 5.10.7 Chapter 5 5.6.2 Derive a series expansion of cot x by dividing cos x by sin x. Since cos x = 1 − 12 x2 + to obtain
1 4 24 x
− · · · and sin x = x − 16 x3 +
1 5 120 x
− · · ·, we divide
1 4 1 4 1 − 12 x2 + 24 1 − 21 x2 + 24 x − ··· x − ··· cot x = = 1 1 x − 16 x3 + 120 x5 − · · · x(1 − 61 x2 + 120 x4 − · · ·)
We now run into an issue of dividing one series by another. However, instead of division, we may change this into a multiplication problem by using (1 − r)−1 = 1 + r + r2 + r3 + · · · to rewrite the denominator (1 − 16 x2 +
4 1 120 x
− · · ·)−1 = 1 + ( 16 x2 − = 1 + 61 x2 + = 1 + 16 x2 +
4 1 1 2 120 x + · · ·) + ( 6 x 1 1 (− 120 + 36 )x4 + · · · 4 7 360 x + · · ·
−
4 1 120 x
+ · · ·)2 + · · ·
where we have only kept terms up to O(x4 ). Returning to cot x, we now find cot x = x−1 (1 − 12 x2 + = x−1 (1 + = x−1 (1 −
4 1 4 1 2 7 24 x − · · ·)(1 + 36 x + 360 x + · · ·) 1 1 7 (− 12 + 16 )x2 + ( 24 − 12 + 360 )x4 + · · ·) 1 4 1 2 3 x − 45 x + · · ·)
In principle, we could work this out to higher orders by keeping more powers of x in the series expansions. Note that there is a nice expression for cot x in terms of the Bernoulli numbers. This may be obtained by noting that the generating function definition of Bn is ∞ ∞ X X x Bn n 1 B2p 2p = x =− x+ x x e − 1 n=1 n! 2 (2p)! p=0
where we have used the fact that all odd Bernoulli numbers vanish except for B1 = − 12 . Moving the − 12 x to the left hand side, and using the identity x 1 x ex + 1 x x + x= · x = coth x e −1 2 2 e −1 2 2
we obtain
∞
x x X B2p 2p coth = x 2 2 p=0 (2p)! or, by substituting x → 2x and dividing through by x ∞ X 2B2p
coth x =
p=0
(2p)!
(2x)2p−1
Finally, to change coth into cot, we may work in the complex domain and note that coth iz = −i cot z. Therefore we make the substitution x → ix to yield ∞ X 2B2p
−i cot x =
p=0
(2p)!
(2ix)2p−1
Multiplying through by i and simplifying then gives the expression cot x =
∞ X (−1)p 22p B2p
(2p)!
p=0
5.6.19
x2p−1
a) Planck’s theory of quandized oscillators leads to an average energy ∞ X
hεi =
nε0 exp(−nε0 /kT )
n=1 ∞ X
exp(−nε0 /kT )
n=0
where ε0 is a fixed energy. Identify the numerator and denominator as binomial expansions and show that the ratio is hεi =
ε0 exp(ε0 /kT ) − 1
To simplify the expressions, we begin with the substitution r = exp(−ε0 /kT ). This yields hεi = N/D where the numerator and denominator are N=
∞ X n=1
n
nε0 r ,
D=
∞ X
rn
n=0
We now see that the denominator is a simple geometric series. Hence D = d 1/(1 − r). For the numerator, we note that nrn = r dr (rn ). Hence we may write ! ∞ d X n d r ε0 r N = ε0 r r = ε0 r = dr n=1 dr 1 − r (1 − r)2
Dividing the numerator and denominator finally yields hεi =
ε0 ε0 ε0 r = −1 = 1−r r −1 exp(ε0 /kT ) − 1
b) Show that the hεi of part (a) reduces to kT , the classical result, for kT ε0 . In this limit, ε0 /KT 1, we may expand the exponential in the denominator exp(ε0 /kT ) ≈ 1 + As a result hεi ≈
ε0 + ··· kT
ε0 ≈ kT ε0 /kT + · · ·
5.7.4 The analysis of the diffraction pattern of a circular opening involves Z
2π
cos(c cos ϕ) dϕ 0
Expand the integrand in a series and integrate by using 2π
Z
cos 0
2n
(2n)! 2π, ϕ dϕ = 2n 2 (n!)2
Z
2π
cos2n+1 ϕ dϕ = 0
0
Setting x = c cos ϕ, we expand ∞ X (−1)n 2n x cos x = (2n)! n=0
so that Z
2π
cos(c cos ϕ) dϕ = 0
∞ 2π X
(−1)n 2n c cos2n ϕ dϕ (2n)! 0 n=0 ∞ X (−1)n c2n Z 2π = cos2n ϕ dϕ (2n)! 0 n=0 Z
∞ ∞ X X 2π(−1)n c 2n (−1)n c2n 2π(2n)! = = (2n)! 22n (n!)2 (n!)2 2 n=0 n=0
5.7.15 The Klein-Nishina formula for the scattering of photons by electrons contains a term of the form (1 + ε) 2 + 2ε ln(1 + 2ε) f (ε) = − ε2 1 + 2ε ε Here ε = hν/mc2 , the ratio of the photon energy to the electron rest mass energy. Find lim f (ε) ε→0
This problem is an exercise in taking Taylor series. Note that, if we simply set ε = 0 in f (ε), the first term (1 + ε)/ε2 would diverge as ε−2 . Hence this provides a hint that we should keep at least two powers of ε in any series expansion we perform. Keeping this in mind, we first work on the fraction 2 + 2ε = 2(1+ε)(1+2ε)−1 = 2(1+ε)(1−2ε+4ε2 −· · ·) = 2(1−ε+2ε2 +· · ·) (1) 1 + 2ε next we turn to the log ln(1 + 2ε) = ε−1 ln(1+2ε) = ε−1 (2ε− 12 (2ε)2 + 31 (2ε)3 +· · ·) = (2−2ε+ 38 ε2 −· · ·) ε (2) 2 Subtracting (2) from (1), and combining with the prefactor (1 + ε)/ε , we find (1 + ε) [2(1 − ε + 2ε2 + · · ·) − (2 − 2ε + 38 ε2 − · · ·)] ε2 (1 + ε) 2 8 2 (1 + ε) 4 2 = [4ε − ε + · · ·] = [ 3 ε + · · ·] = 43 (1 + ε)[1 + · · ·] 3 ε2 ε2
f (ε) =
We are now in a position to take the limit ε → 0 to obtain lim f (ε) =
ε→0
4 3
5.9.11 The integral Z
1
dx x
[ln(1 − x)]2
0
appears in the fourth-order correction to the magnetic moment of the electron. Show that it equals 2ζ(3). We begin with the variable substitution 1 − x = e−t , to obtain Z
1
2 dx
[ln(1 − x)] 0
x
dx = e−t dt Z = 0
∞
t2
e−t dt 1 − e−t
This integral involves powers and exponentials, and is not so easy to do. Thus we expand the fraction as a series ∞ X e−t −t −t −1 −t −t −2t −3t = e (1 − e ) = e (1 + e + e +e + · · ·) = e−nt 1 − e−t n=1
This gives Z
1
2 dx
[ln(1 − x)] 0
x
Z
∞ X
∞
=
t
2
0
! −nt
e
dt =
n=1
∞ Z X n=1
∞
e−nt t2 dt
0
This integral may be evaluated by integration by parts (twice). Alternatively, we make the substitution s = nt to arrive at Z 1 Z ∞ ∞ ∞ X X −3 2 dx −s 2 n n−3 Γ(3) = 2ζ(3) [ln(1 − x)] = e s ds = x 0 0 n=1 n=1 Here we have used the definition of the Gamma function Z ∞ Γ(z) = e−s sz−1 dz 0
as well as the zeta function ζ(z) =
∞ X
n−z
n=1
5.10.1 Stirling’s formula for the logarithm of the factorial function is N X 1 B2n 1 ln x − x − x1−2n ln(x!) = ln 2π + x + 2 2 2n(2n − 1) n=1 The B2n are the Bernoulli numbers. Show that Stirling’s formula is an asymptotic expansion. P Instead of using the textbook definition of an asymptotic series an (x), we aim to demonstrate the two principle facts; i) that the series diverges for fixed x when N → ∞, and ii) that the remainder vanishes for fixed N when x → ∞. To do so, we first examine the form of an (x) an (x) = −
B2n x1−2n 2n(2n − 1)
Using the relation B2n =
(−1)n+1 2(2n)! ζ(2n) (2π)2n
we find |an (x)| =
2(2n − 2)!ζ(2n) 1−2n x (2π)2n
For condition i), in order to show that the series diverges for fixed x, we may perform the ratio test 2(2n − 2)!ζ(2n) (2π)2n+2 ζ(2n) (2π)2 |an | 2 = x = x2 2n |an+1 | (2π) 2(2n)!ζ(2n + 2) 2n(2n − 1) ζ(2n + 2)
(3)
Since limn→∞ ζ(n) = 1, and since there are factors of n in the denominator, we see that |an | =0 (for fixed x) lim n→∞ |an+1 | and hence the ratio test demonstrates that the series diverges. For showing condition ii), on the other hand, we suppose the series stops at term n = N . Then the error or remainder is related to the subsequent terms aN +1 , aN +2 , etc. However, according to (3), if we take the limit x → ∞ for fixed N we have |aN | |aN | lim =∞ ⇒ |aN +1 | → = 0 as x → ∞ x→∞ |aN +1 | ∞ Hence the remainder terms fall off sufficiently fast to satisfy the criteria for an asymptotic series. We thus conclude that Stirling’s formula is an asymptotic expansion. 5.10.7 Derive the following Bernoulli number asymptotic series for the Euler-Mascheroni constant n N X X 1 B2k −1 s − ln n − γ= + 2n (2k)n2k s=1 k=1
Let us start by recalling the useful definition of the Euler-Mascheroni constant ! n X γ = lim s−1 − ln n n→∞
s=1
Essentially, the constant γ is the difference between the sum and the integral approximation. This suggests that we begin by inserting the function f (x) = 1/x into the Euler-Maclauren sum formula Z n n N X X 1 1 1 f (x) = f (x)dx + f (1) + f (n) + B2p [f (2p−1) (n) − f (2p−1) (1)] 2 2 (2p)! 1 p=1 x=1 1 − (2N )!
Z
1
B2N (x) 0
n−1 X
f (2N ) (x + ν)dx
ν=1
(4)
However, we first note that, for f (x) = 1/x we have Z n Z n dx f (x)dx = = ln n x 1 1 as well as f (k) (x) = (−1)k
k!
xk+1 Using these results, and returning to (4), we find n X s=1
or
N
−1
s
X B2p 1 1 = ln n + + − [n−2p − 1] − 2 2n p=1 2p n X
Z
1
B2N (x) 0
n−1 X
(x + ν)2N +1 dx
ν=1
N
−1
s
s=1
X B2p 1 1 − ln n = + − [n−2p − 1] + RN (n) 2 2n p=1 2p
where the remainder RN (n) is given by Z 1 n−1 X RN (n) = − (x + ν)2N +1 dx B2N (x) 0
(5)
(6)
ν=1
At this point, may note that the left hand side of (5) is close to the expression we want for the Euler-Mascheroni constant. However, we must recall that the sum formula (4) generally yields an asymptotic expansion (since the Bernoulli numbers diverge). Thus we have to be careful about the remainder term. Of course, we can still imagine taking the limit n → ∞ in (5) to obtain ! n N X 1 X B2p −1 γ = lim s − ln n = + + RN (∞) n→∞ 2 p=1 2p s=1
(7)
Noting that the remainder (6) is a sum of terms Z 1 1 1 1 RN (n) = − B2N (x) + + ··· + dx (x + 1)2N +1 (x + 2)2N +1 (x + n − 1)2N +1 0 and that the first few terms in the sum dominate, we may eliminate most (but not all) of the remainder by subtracting (5) from (7) γ−
n X
N
s−1 + ln n = −
s=1
X B2p 1 1 + + [RN (∞) − RN (n)] 2n p=1 2p n2p
Finally, dropping the difference of remainders, we obtain the result γ=
n X s=1
N
s−1 − ln n −
X B2p 1 1 + 2n p=1 2p n2p
