Hints and answers

4.9 HINTS AND ANSWERS
4.35
4.36
find a closed-form expression for α, the Madelung constant for this (unrealistic)
lattice.
One of the factors contributing to the high relative permittivity of water to static
electric fields is the permanent electric dipole moment, p, of the water molecule.
In an external field E the dipoles tend to line up with the field, but they do not
do so completely because of thermal agitation corresponding to the temperature,
T , of the water. A classical (non-quantum) calculation using the Boltzmann
distribution shows that the average polarisability per molecule, α, is given by
p
α = (coth x − x−1 ),
E
where x = pE/(kT ) and k is the Boltzmann constant.
At ordinary temperatures, even with high field strengths (104 V m−1 or more),
x 1. By making suitable series expansions of the hyperbolic functions involved,
show that α = p2 /(3kT ) to an accuracy of about one part in 15x−2 .
In quantum theory, a certain method (the Born approximation) gives the (socalled) amplitude f(θ) for the scattering of a particle of mass m through an angle
θ by a uniform potential well of depth V0 and radius b (i.e. the potential energy
of the particle is −V0 within a sphere of radius b and zero elsewhere) as
f(θ) =
2mV0
(sin Kb − Kb cos Kb).
2 K 3
Here is the Planck constant divided by 2π, the energy of the particle is 2 k 2 /(2m)
and K is 2k sin(θ/2).
Use l’Hôpital’s rule to evaluate the amplitude at low energies, i.e. when k and
hence K tend to zero, and so determine the low-energy total cross-section.
[ Note: the differential cross-section is given by |f(θ)|2 and the total crossπ
section by the integral of this over all solid angles, i.e. 2π 0 |f(θ)|2 sin θ dθ. ]
4.9 Hints and answers
4.1
4.3
4.5
4.7
4.9
499
Write as 2( 1000
n=1 n −
n=1 n) = 751 500.
Divergent for r ≤ 1; convergent for r ≥ 2.
(a) ln(N + 1), divergent; (b) 13 [1 − (−2)n ], oscillates infinitely; (c) Add 13 SN to the
3
3
[1 − (−3)−N ] + 34 N(−3)−N−1 , convergent to 16
.
SN series; 16
Write the nth term as the difference between two consecutive values of a partialfraction function of n. The sum equals 12 (1 − N −2 ).
Sum the geometric series with rth term exp[i(θ + rα)]. Its real part is
{cos θ − cos [(n + 1)α + θ] − cos(θ − α) + cos(θ + nα)} /4 sin2 (α/2),
4.11
which can be reduced to the given answer.
(a) −1 ≤ x < 1; (b) all x except x = (2n ± 1)π/2; (c) x < −1; (d) x < 0; (e)
always divergent. Clearly divergent for x > −1. For −X = x < −1, consider
∞
Mk
k=1 n=Mk−1 +1
4.13
1
,
(ln Mk )X
where ln Mk = k and note that Mk − Mk−1 = e−1 (e − 1)Mk ; hence show that the
series diverges.
(a) Absolutely convergent, compare with exercise 4.10(b). (b) Oscillates finitely.
(c) Absolutely convergent for all x. (d) Absolutely convergent; use partial fractions. (e) Oscillates infinitely.
149
SERIES AND LIMITS
4.15
4.17
4.19
4.21
4.23
4.25
4.27
4.29
4.31
4.33
4.35
Divide the series into two series, n
odd and n even. For r = 2 both are absolutely
convergent, by comparison
n−2 . For r = 1 neither series is convergent,
−1 with
by comparison with
n . However, the sum of the two is convergent, by the
alternating sign test or by showing that the terms cancel in pairs.
The first term has value 0.833 and all other terms are positive.
|A|2 (1 − r)2 /(1 + r2 − 2r cos φ).
Use the binomial expansion and collect terms up to x4 . Integrate both sides of
the displayed equation. tan x = x + x3 /3 + 2x5 /15 + · · · .
For example, P5 (x) = 24x4 − 72x2 + 9. sinh−1 x = x − x3 /6 + 3x5 /40 − · · · .
Set a = D + δ and b = D − δ and use the expansion for ln(1 ± δ/D).
The limit is 0 for m > n, 1 for m = n, and ∞ for m < n.
(a) − 21 , 12 , ∞; (b) −4; (c) −1 + 2/π.
(a) First approximation 0.886 452; second approximation 0.886 287. (b) Set y =
sin x and re-express f(x) = 2 as a polynomial equation. y = sin(0.886 287) =
0.774 730. −nx
If S(x) = ∞
evaluate S(x) and consider dS(x)/dx.
n=0 e
E = hν[exp(hν/kT ) − 1]−1 . px 1
x2
The series expansion is
−
+ ··· .
E 3 45
150