Table Of ContentInstructor’s Manual
MATHEMATICAL
METHODS FOR
PHYSICISTS
A Comprehensive Guide
SEVENTH EDITION
George B. Arfken
Miami University
Oxford, OH
Hans J. Weber
University of Virginia
Charlottesville, VA
Frank E. Harris
University of Utah, Salt Lake City, UT;
University of Florida, Gainesville, FL
AMSTERDAM • BOSTON • HEIDELBERG • LONDON
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Contents
1
Introduction
1
2
Errata and Revision Status
3
3
Exercise Solutions
7
1.
Mathematical Preliminaries
. . . . . . . . . . . . . . . . . . .
7
2.
Determinants and Matrices . . . . . . . . . . . . . . . . . . . .
27
3.
Vector Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.
Tensors and Differential Forms . . . . . . . . . . . . . . . . . .
58
5.
Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
6.
Eigenvalue Problems
. . . . . . . . . . . . . . . . . . . . . . .
81
7.
Ordinary Differential Equations . . . . . . . . . . . . . . . . .
90
8.
Sturm-Liouville Theory . . . . . . . . . . . . . . . . . . . . . .
106
9.
Partial Differential Equations
. . . . . . . . . . . . . . . . . .
111
10.
Green’s Functions . . . . . . . . . . . . . . . . . . . . . . . . .
118
11.
Complex Variable Theory
. . . . . . . . . . . . . . . . . . . .
122
12.
Further Topics in Analysis . . . . . . . . . . . . . . . . . . . .
155
13.
Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . .
166
14.
Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
192
15.
Legendre Functions . . . . . . . . . . . . . . . . . . . . . . . .
231
16.
Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . .
256
17.
Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . .
268
18.
More Special Functions . . . . . . . . . . . . . . . . . . . . . .
286
19.
Fourier Series
. . . . . . . . . . . . . . . . . . . . . . . . . . .
323
20.
Integral Transforms . . . . . . . . . . . . . . . . . . . . . . . .
332
21.
Integral Equations . . . . . . . . . . . . . . . . . . . . . . . . .
364
22.
Calculus of Variations . . . . . . . . . . . . . . . . . . . . . . .
373
23.
Probability and Statistics . . . . . . . . . . . . . . . . . . . . .
387
4
Correlation, Exercise Placement
398
5
Unused Sixth Edition Exercises
425
iv
Chapter 1
Introduction
The seventh edition of Mathematical Methods for Physicists is a substantial and
detailed revision of its predecessor. The changes extend not only to the topics
and their presentation, but also to the exercises that are an important part
of the student experience. The new edition contains 271 exercises that were
not in previous editions, and there has been a wide-spread reorganization of the
previously existing exercises to optimize their placement relative to the material
in the text. Since many instructors who have used previous editions of this text
have favorite problems they wish to continue to use, we are providing detailed
tables showing where the old problems can be found in the new edition, and
conversely, where the problems in the new edition came from. We have included
the full text of every problem from the sixth edition that was not used in the
new seventh edition. Many of these unused exercises are excellent but had to
be left out to keep the book within its size limit. Some may be useful as test
questions or additional study material.
Complete methods of solution have been provided for all the problems that
are new to this seventh edition. This feature is useful to teachers who want to
determine, at a glance, features of the various exercises that may not be com-
pletely apparent from the problem statement. While many of the problems from
the earlier editions had full solutions, some did not, and we were unfortunately
not able to undertake the gargantuan task of generating full solutions to nearly
1400 problems.
Not part of this Instructor’s Manual but available from Elsevier’s on-line
web site are three chapters that were not included in the printed text but which
may be important to some instructors. These include
• A new chapter (designated 31) on Periodic Systems, dealing with mathe-
matical topics associated with lattice summations and band theory,
• A chapter (32) on Mathieu functions, built using material from two chap-
ters in the sixth edition, but expanded into a single coherent presentation,
and
1
CHAPTER 1. INTRODUCTION
2
• A chapter (33) on Chaos, modeled after Chapter 18 of the sixth edition
but carefully edited.
In addition, also on-line but external to this Manual, is a chapter (designated
1) on Infinite Series that was built by collection of suitable topics from various
places in the seventh edition text. This alternate Chapter 1 contains no material
not already in the seventh edition but its subject matter has been packaged into
a separate unit to meet the demands of instructors who wish to begin their
course with a detailed study of Infinite Series in place of the new Mathematical
Preliminaries chapter.
Because this Instructor’s Manual exists only on-line, there is an opportunity
for its continuing updating and improvement, and for communication, through
it, of errors in the text that will surely come to light as the book is used. The
authors invite users of the text to call attention to errors or ambiguities, and
it is intended that corrections be listed in the chapter of this Manual entitled
Errata and Revision Status. Errata and comments may be directed to the au-
thors at harris⟨at⟩qtp.ufl.edu or to the publisher. If users choose to forward
additional materials that are of general use to instructors who are teaching from
the text, they will be considered for inclusion when this Manual is updated.
Preparation of this Instructor’s Manual has been greatly facilitated by the
efforts of personnel at Elsevier. We particularly want to acknowledge the assis-
tance of our Editorial Project Manager, Kathryn Morrissey, whose attention to
this project has been extremely valuable and is much appreciated.
It is our hope that this Instructor’s Manual will have value to those who
teach from Mathematical Methods for Physicists and thereby to their students.
Chapter 2
Errata and Revision Status
Last changed: 06 April 2012
Errata and Comments re Seventh Edition text
Page 522
Exercise 11.7.12(a)
This is not a principal-value integral.
Page 535
Figure 11.26
The two arrowheads in the lower part of the
circular arc should be reversed in direction.
Page 539
Exercise 11.8.9
The answer is incorrect; it should be π/2.
Page 585
Exercise 12.6.7
Change the integral for which a series is sought
to
� ∞
0
e−xv
1 + v2 dv. The answer is then correct.
Page 610
Exercise 13.1.23
Replace (−t)ν by e−πiνtν.
Page 615
Exercise 13.2.6
In the Hint, change Eq. (13.35) to Eq. (13.44).
Page 618
Eq. (13.51)
Change l.h.s. to B(p + 1, q + 1).
Page 624
After Eq. (13.58)
C1 can be determined by requiring consistency
with the recurrence formula zΓ(z) = Γ(z + 1).
Consistency with the duplication formula then
determines C2.
Page 625
Exercise 13.4.3
Replace “(see Fig. 3.4)” by “and that of the
recurrence formula”.
Page 660
Exercise 14.1.25
Note that α2 = ω2/c2, where ω is the angular
frequency, and that the height of the cavity is l.
3
CHAPTER 2. ERRATA AND REVISION STATUS
4
Page 665
Exercise 14.2.4
Change Eq. (11.49) to Eq. (14.44).
Page 686
Exercise 14.5.5
In part (b), change l to h in the formulas for
amn and bmn (denominator and integration
limit).
Page 687
Exercise 14.5.14
The index n is assumed to be an integer.
Page 695
Exercise 14.6.3
The index n is assumed to be an integer.
Page 696
Exercise 14.6.7(b)
Change N to Y (two occurrences).
Page 709
Exercise 14.7.3
In the summation preceded by the cosine
function, change (2z)2s to (2z)2s+1.
Page 710
Exercise 14.7.7
Replace nn(x) by yn(x).
Page 723
Exercise 15.1.12
The last formula of the answer should read
P2s(0)/(2s + 2) = (−1)s(2s − 1)!!/(2s + 2)!!.
Page 754
Exercise 15.4.10
Insert minus sign before P 1
n(cos θ).
Page 877
Exercise 18.1.6
In both (a) and (b), change 2π to
√
2π.
Page 888
Exercise 18.2.7
Change the second of the four members of the
first display equation to
�x + ip
√
2
�
ψn(x), and
change the corresponding member of the
second display equation to
�x − ip
√
2
�
ψn(x).
Page 888
Exercise 18.2.8
Change x + ip to x − ip.
Page 909
Exercise 18.4.14
All instances of x should be primed.
Page 910
Exercise 18.4.24
The text does not state that the T0 term (if
present) has an additional factor 1/2.
Page 911
Exercise 18.4.26(b)
The ratio approaches (πs)−1/2, not (πs)−1.
Page 915
Exercise 18.5.5
The hypergeometric function should read
2F1
� ν
2 + 1
2, ν
2 + 1; ν + 3
2; z−2�
.
Page 916
Exercise 18.5.10
Change (n − 1
2)! to Γ(n + 1
2).
Page 916
Exercise 18.5.12
Here n must be an integer.
Page 917
Eq. (18.142)
In the last term change Γ(−c) to Γ(2 − c).
Page 921
Exercise 18.6.9
Change b to c (two occurrences).
Page 931
Exercise 18.8.3
The arguments of K and E are m.
Page 932
Exercise 18.8.6
All arguments of K and E are k2; In the
integrand of the hint, change k to k2.
CHAPTER 2. ERRATA AND REVISION STATUS
5
Page 978
Exercise 20.2.9
The formula as given assumes that Γ > 0.
Page 978
Exercise 20.2.10(a)
This exercise would have been easier if the
book had mentioned the integral
representation J0(x) = 2
π
� 1
0
cos xt
√
1 − t2 dt.
Page 978
Exercise 20.2.10(b)
Change the argument of the square root
to x2 − a2.
Page 978
Exercise 20.2.11
The l.h.s. quantities are the transforms of
their r.h.s. counterparts, but the r.h.s.
quantities are (−1)n times the transforms
of the l.h.s. expressions.
Page 978
Exercise 20.2.12
The properly scaled transform of f(µ) is
(2/π)1/2injn(ω), where ω is the transform
variable. The text assumes it to be kr.
Page 980
Exercise 20.2.16
Change d3x to d3r and remove the limits
from the first integral (it is assumed to be
over all space).
Page 980
Eq. (20.54)
Replace dk by d3k (occurs three times)
Page 997
Exercise 20.4.10
This exercise assumes that the units and
scaling of the momentum wave function
correspond to the formula
ϕ(p) =
1
(2πℏ)3/2
�
ψ(r) e−ir·p/ℏ d3r .
Page 1007
Exercise 20.6.1
The second and third orthogonality equa-
tions are incorrect. The right-hand side
of the second equation should read:
N,
p = q = (0 or N/2);
N/2, (p + q = N) or p = q but not both;
0,
otherwise.
The right-hand side of the third equation
should read:
N/2,
p = q and p + q ̸= (0 or N);
−N/2,
p ̸= q and p + q = N;
0,
otherwise.
Page 1007
Exercise 20.6.2
The exponentials should be e2πipk/N and
e−2πipk/N.
Page 1014
Exercise 20.7.2
This exercise is ill-defined. Disregard it.
Page 1015
Exercise 20.7.6
Replace (ν − 1)! by Γ(ν) (two occurrences).
Page 1015
Exercise 20.7.8
Change M(a, c; x) to M(a, c, x) (two
CHAPTER 2. ERRATA AND REVISION STATUS
6
occurrences).
Page 1028
Table 20.2
Most of the references to equation numbers
did not get updated from the 6th edition.
The column of references should, in its
entirety, read: (20.126), (20.147), (20.148),
Exercise 20.9.1, (20.156), (20.157), (20.166),
(20.174), (20.184), (20.186), (20.203).
Page 1034
Exercise 20.8.34
Note that u(t − k) is the unit step function.
Page 1159
Exercise 23.5.5
This problem should have identified m as the
mean value and M as the “random variable”
describing individual student scores.
Corrections and Additions to Exercise Solutions
None as of now.
Chapter 3
Exercise Solutions
1.
Mathematical Preliminaries
1.1
Infinite Series
1.1.1.
(a) If un < A/np the integral test shows �
n un converges for p > 1.
(b) If un > A/n, �
n un diverges because the harmonic series diverges.
1.1.2.
This is valid because a multiplicative constant does not affect the conver-
gence or divergence of a series.
1.1.3.
(a) The Raabe test P can be written 1 + (n + 1) ln(1 + n−1)
ln n
.
This expression approaches 1 in the limit of large n. But, applying the
Cauchy integral test,
�
dx
x ln x = ln ln x,
indicating divergence.
(b) Here the Raabe test P can be written
1 + n + 1
ln n
ln
�
1 + 1
n
�
+ ln2(1 + n−1)
ln2 n
,
which also approaches 1 as a large-n limit. But the Cauchy integral test
yields
�
dx
x ln2 x = − 1
ln x,
indicating convergence.
1.1.4.
Convergent for a1 − b1 > 1. Divergent for a1 − b1 ≤ 1.
1.1.5.
(a) Divergent, comparison with harmonic series.
7
CHAPTER 3. EXERCISE SOLUTIONS
8
(b) Divergent, by Cauchy ratio test.
(c) Convergent, comparison with ζ(2).
(d) Divergent, comparison with (n + 1)−1.
(e) Divergent, comparison with 1
2(n+1)−1 or by Maclaurin integral test.
1.1.6.
(a) Convergent, comparison with ζ(2).
(b) Divergent, by Maclaurin integral test.
(c) Convergent, by Cauchy ratio test.
(d) Divergent, by ln
�
1 + 1
n
�
∼ 1
n.
(e) Divergent, majorant is 1/(n ln n).
1.1.7.
The solution is given in the text.
1.1.8.
The solution is given in the text.
1.1.10.
In the limit of large n, un+1/un = 1 + 1
n + O(n−2).
Applying Gauss’ test, this indicates divergence.
1.1.11.
Let sn be the absolute value of the nth term of the series.
(a) Because ln n increases less rapidly than n, sn+1 < sn and limn→∞ sn =
0. Therefore this series converges. Because the sn are larger than corre-
sponding terms of the harmonic series, this series is not absolutely con-
vergent.
(b) Regarding this series as a new series with terms formed by combin-
ing adjacent terms of the same sign in the original series, we have an
alternating series of decreasing terms that approach zero as a limit, i.e.,
1
2n + 1 +
1
2n + 2 >
1
2n + 3 +
1
2n + 4,
this series converges. With all signs positive, this series is the harmonic
series, so it is not aboslutely convergent.
(c) Combining adjacent terms of the same sign, the terms of the new series
satisfy
2
�1
2
�
> 1
2 + 1
3 > 2
�1
3
�
,
3
�1
4
�
> 1
4 + 1
5 + 1
6 > 3
�1
6
�
,
etc.
The general form of these relations is
2n
n2 − n + 2 > sn >
2
n + 1.
CHAPTER 3. EXERCISE SOLUTIONS
9
An upper limit to the left-hand side member of this inequality is 2/(n−1).
We therefore see that the terms of the new series are decreasing, with
limit zero, so the original series converges. With all signs positive, the
original series becomes the harmonic series, and is therefore not absolutely
convergent.
1.1.12.
The solution is given in the text.
1.1.13.
Form the nth term of ζ(2)−c1α1−c2α2 and choose c1 and c2 so that when
placed over the common denominator n2(n + 1)(n + 2) the numerator will
be independent of n. The values of the ci satisfying this condition are
c1 = c2 = 1, and our resulting expansion is
ζ(2) = α1 + α2 +
∞
�
n=1
2
n2(n + 1(n + 2) = 5
4 +
∞
�
n=1
2
n2(n + 1(n + 2) .
Keeping terms through n = 10, this formula yields ζ(2) ≈ 1.6445; to this
precision the exact value is ζ(2) = 1.6449.
1.1.14.
Make the observation that
∞
�
n=0
1
(2n + 1)3 +
∞
�
n=1
1
(2n)3 = ζ(3)
and that the second term on the left-hand side is ζ(3)/8). Our summation
therefore has the value 7ζ(3)/8.
1.1.15.
(a) Write ζ(n) − 1 as �∞
p=2 p−n, so our summation is
∞
�
n=2
∞
�
p=2
1
pn =
∞
�
p=2
∞
�
n=2
1
pn .
The summation over n is a geometric series which evaluates to
p−2
1 − p−1 =
1
p2 − p .
Summing now over p, we get
∞
�
p=2
1
p(p − 1) =
∞
�
p=1
1
p(p + 1) = α1 = 1 .
(b) Proceed in a fashion similar to part (a), but now the geometric series
has sum 1/(p2 + p), and the sum over p is now lacking the initial term of
α1, so
∞
�
p=2
1
p(p + 1) = α1 −
1
(1)(2) = 1
2.
CHAPTER 3. EXERCISE SOLUTIONS
10
1.1.16.
(a) Write
ζ(3) = 1 +
∞
�
n=2
1
n3 −
∞
�
n=2
1
(n − 1)n(n + 1) + α′
2 = 1 +
∞
�
n=2
� 1
n3 −
1
n(n2 − 1)
�
+ 1
4
= 1 + 1
4 −
∞
�
n=2
1
n3(n2 − 1) .
(b) Now use α′
2 and α′
4 =
∞
�
n=3
1
n(n2 − 1)(n2 − 4) = 1
96:
ζ(3) = 1 + 1
23 +
∞
�
n=3
1
n3 −
∞
�
n=3
1
n(n2 − 1) +
�
α′
2 − 1
6
�
−
∞
�
n=3
B
n(n2 − 1)(n2 − 4) + Bα′
4
= 29
24 + B
96 +
∞
�
n=3
� 1
n3 −
1
n(n2 − 1) −
B
n(n2 − 1)(n2 − 4)
�
= 29
24 − B
96 +
∞
�
n=3
4 − (1 + B)n2
n(n2 − 1)(n2 − 4) .
The convergence of the series is optimized if we set B = −1, leading to
the final result
ζ(3) = 29
24 − 1
96 +
∞
�
n=3
4
n(n2 − 1)(n2 − 4) .
(c) Number of terms required for error less than 5×10−7: ζ(3) alone, 999;
combined as in part (a), 27; combined as in part (b), 11.
1.2
Series of Functions
1.2.1.
(a) Applying Leibniz’ test the series converges uniformly for ε ≤ x < ∞
no matter how small ε > 0 is.
(b) The Weierstrass M and the integral tests give uniform convergence for
1 + ε ≤ x < ∞ no matter how small ε > 0 is chosen.
1.2.2.
The solution is given in the text.
1.2.3.
(a) Convergent for 1 < x < ∞.
(b) Uniformly convergent for 1 < s ≤ x < ∞.
CHAPTER 3. EXERCISE SOLUTIONS
11
1.2.4.
From | cos nx| ≤ 1, | sin nx| ≤ 1 absolute and uniform convergence follow
for −s < x < s for any s > 0.
1.2.5.
Since |uj+2
uj
| ∼ |x|2, |x| < 1 is needed for convergence.
1.2.6.
The solution is given in the text.
1.2.7.
The solution is given in the text.
1.2.8.
(a) For n = 0, 1, 2, . . . we find
d4n+1 sin x
dx4n+1
����
0
= cos x|0 = 1,
d4n+2 sin x
dx4n+2
����
0
= − sin x|0 = 0,
d4n+3 sin x
dx4n+3
����
0
= − cos x|0 = −1,
d4n sin x
dx4n
����
0
= sin x|0 = 0.
Taylor’s theorem gives the absolutely convergent series
sin x =
∞
�
n=0
(−1)n
x2n+1
(2n + 1)!.
(b) Similar derivatives for cos x give the absolutely convergent series
cos x =
∞
�
n=0
(−1)n x2n
(2n)!.
1.2.9.
cot x = 1
x − x
3 − x3
45 − 2x5
945 − · · · , −π < x < π.
1.2.10.
From coth y = η0 = ey + e−y
ey − e−y = e2y + 1
e2y − 1 we extract y = 1
2 ln η0 + 1
η0 − 1.
To check this we substitute this into the first relation, giving
η0 + 1
η0 − 1 + 1
η0 + 1
η0 − 1 − 1
= η0.
The series coth−1 η0 =
∞
�
n=0
(η0)−2n−1
2n + 1
follows from Exercise 1.6.1.
CHAPTER 3. EXERCISE SOLUTIONS
12
1.2.11.
(a) Since d√x
dx
����
0
=
1
2√x
����
0
does not exist, there is no Maclaurin expan-
sion.
(b) |x − x0| < x0 because the origin must be excluded.
1.2.12.
lim
x→x0
f(x)
g(x) = f(x + (x0 − x))
g(x + (x0 − x) = lim
x→x0
f(x) + (x0 − x)f ′(x) + · · ·
g(x) + (x0 − x)g′(x) + · · ·
= lim
x→x0
f ′(x)
g′(x) , where the intermediate formal expression f(x + (x0 − x))
g(x + (x0 − x)
may be dropped.
1.2.13.
(a) − ln
n
n − 1 = ln
�
1 − 1
n
�
= −
∞
�
ν=1
1
νnν .
Hence 1
n − ln
n
n − 1 = −
∞
�
ν=2
1
νnν < 0.
(b) ln n + 1
n
= ln(1 + 1
n) =
∞
�
ν=2
(−1)ν−1
νnν
, 1
n − ln n + 1
n
=
∞
�
ν=2
(−1)ν
νnν
> 0.
Summing (a) yields
0 >
n
�
m=2
1
m − ln
2 · 3 · · · n
1 · 2 · · · (n − 1) =
n
�
m=2
1
m − ln n → γ − 1.
Thus, γ < 1. Summing (b) yields
0 <
n−1
�
m=2
1
m − ln
2 · 3 · · · n
1 · 2 · · · (n − 1) =
n−1
�
m=2
1
m − ln n → γ.
Hence 0 < γ < 1.
1.2.14.
The solution is given in the text.
1.2.15.
The solutions are given in the text.
1.2.16.
If
����
an+1
an
���� → 1
R then
����
(n + 2)an+1
(n + 1)an
���� → 1
R and
����
an+1/(n + 2)
an/(n + 1)
���� → 1
R.
1.3
Binomial Theorem
1.3.1.
P(x) = C
�x
3 − x3
45 + · · ·
�
.
1.3.2.
Integrating termwise tan−1 1 = π
4 =
∞
�
n=0
(−1)n
� 1
0
x2ndx =
∞
�
n=0
(−1)n
2n + 1.
