Solutions Manual for A Course in Mathematical Methods for Physicists

445 Pages·4.836 MB·English

by  Russell L. Herman

S O L U T I O N S M A N U A L F O R
R. L. HERMAN - VERSION DATE: NOVEMBER 14, 2013
SOLUTIONS MANUAL FOR
A COURSE IN
MATHEMATICAL
METHODS FOR
PHYSICISTS
by
Russell L. Herman
Boca Raton London New York
CRC Press is an imprint of the
Taylor & Francis Group, an informa business
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Printed on acid-free paper
Version Date: 20140114
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ii
Contents
Preface iii
1 IntroductionandReview 1
2 FreeFallandHarmonicOscillators 25
3 LinearAlgebra 69
4 NonlinearDynamics 111
5 TheHarmonicsofVibratingStrings 141
6 Non-sinusoidalHarmonics 173
7 ComplexRepresentationsofFunctions 219
8 TransformTechniquesinPhysics 245
9 VectorAnalysisandEMWaves 291
10 ExtremaandVariationalCalculus 327
11 ProblemsinHigherDimensions 357
A ReviewofSequencesandInfiniteSeries 379
B QuickAnswers 395
Ch.1 Introduction . 395
Ch.2 FreeFallandHarmonicOscillators 398
Ch.3 LinearAlgebra 403
Ch.4 NonlinearDynamics 408
Ch.5 TheHarmonicsofVibratingStrings 412
Ch.6 Non-sinusoidalHarmonics 415
Ch.7 ComplexRepresentationsofFunctions 419
Ch.8 TransformTechniquesinPhysics 422
Ch.9 VectorAnalysisandEMWaves 427
Ch.10 ExtremaandVariationalCalculus 431
Ch.11 ProblemsinHigherDimensions 434
Ch. A ReviewofSequencesandInfiniteSeries 436
Preface
Thisisthesolutionmanualforthetextbook,ACourseinMathematicalMeth-
ods for Physicists, by CRC Press. It has been provided for use by instruc-
tors only. It only contains solutions to problems in the text. The solu-
tions are worked out in detail in most cases and a list of answers to many
of the problems are provided at the end of this manual. Hopefully, it is
free from error. However, any noted errors will be posted at the website
1
http://www.russherman.com/cmmp/alongwitherratafromthetextbook. 1Thereareacoupleoferrorsintextbook
Animportantuseofthismanualisinscanningproblemsolutionsbefore problemsthatweremissedbeforeprint-
ing:
assigning them to students. While many problems are relatively straight
• Problem 19 in Chapters 4 and 5:
forward, there are a few challenging problems, or problems for which in- During editing, Problem 19 on find-
structors might decide to add hints for students depending on the level of ing the circumference of an ellipse
in Chapter 4 was overwritten by an
thecourse. Ofcourse,instructorsarefreetoaddtheirownfavoriteproblems intended correction for Problem 19
toaccompanythetext. Chapter5.
The book was originally written to use in an intermediate course on The correct problems are listed in this
manual.
mathematical methods in physics, but due to the addition of a few chap-
ters and topics, this book can certainly be used in other courses such as
applied mathematics, engineering mathematics, senior and first year grad-
uate courses in mathematical methods. Amongst some standard topics the
bookalsooffers:
• Aquickreviewofmathematicalprerequisites,proceedingtoapplications
ofdifferentialequationsandlinearalgebra
• Classroom-testedexplanationsofcomplexandFourieranalysisfortrigono-
metricandspecialfunctions
• Coverageofvectoranalysisandcurvilinearcoordinatesforsolvinghigher
dimensionalproblems
• Sectionsonnonlineardynamics,variationalcalculus,numericalsolutions
ofdifferentialequations,andGreen’sfunctions

1
Introduction and Review
1. Prove the following identities using only the definitions of the trigono-
metric functions, the Pythagorean identity, or the identities for sines and
cosinesofsumsofangles.
a. cos2x =2cos2x−1.
cos2x = cos2x−sin2x
= cos2x−(1−cos2x)
= 2cos2x−1.
b. sin3x = Asin3x+Bsinx,forwhatvaluesof A and B?
sin3x = sinxcos2x+sin2xcosx
= sinx(cos2x−sin2x)+2sinxcos2x
= 3sinx(1−sin2x)−sin3x
= 3sinx−4sin3x.
So, A = −4, B =3.
(cid:18) (cid:19)
θ π
c. secθ+tanθ =tan + .
2 4
(cid:16) (cid:17)
(cid:18) (cid:19) sin θ + π
θ π 2 4
tan + = (cid:16) (cid:17)
2 4 cos θ + π
2 4
sinθ cosπ +sinπ cosθ
= 2 4 4 2
cosθ cosπ −sinπ sinθ
2 4 4 2
sinθ +cosθ
= 2 2
cosθ −sinθ
2 2
(cid:32) (cid:33)
sinθ +cosθ cosθ +sinθ
= 2 2 2 2
cosθ −sinθ cosθ +sinθ
2 2 2 2
(cid:16) (cid:17)2
cosθ +sinθ
2 2
=
cos2 θ −sin2 θ
2 2
cos2 θ +sin2 θ +2sinθ cosθ
= 2 2 2 2
cosθ
2 mathematical methods for physicists
1+sinθ
=
cosθ
= secθ+tanθ.
2. Determinetheexactvaluesof
π
a. sin .
8
π 1(cid:16) π(cid:17)
sin2 = 1−cos
8 2 4
(cid:32) √ (cid:33)
1 2
= 1−
2 2
1(cid:16) √ (cid:17)
= 2− 2 .
4
(cid:113) √
π 1
Therefore,sin = 2− 2.
8 2
b. tan15o.
tan15o = tan(60o−45o)
tan60o−tan45o
=
1+tan60otan45o
√
3−1
= √
1+ 3
√ (cid:32) √ (cid:33)
3−1 1− 3
= √ √
1+ 3 1− 3
√
= 2− 3.
Onecangetthesameanswerusing
√
sin215o 1−cos30o 2− 3 √
tan215o = = = √ = (2− 3)2.
cos215o 1+cos30o 2+ 3
c. cos105o.
1
cos2105o = (1+cos210o)
2
1
= (1−cos30o)
2
√
1
= (2− 3).
4
(cid:112) √
Therefore,cos105o = −1 2− 3.Onecouldalsouse
2
1
cos2105o =sin215o = (1−cos30o).
2
Notethatthisanswercanbedenested:
(cid:113) √ (cid:113) √
1 1
− 2− 3 = − 8−4 3
2 4
introduction and review 3
(cid:113) √ √
1
= − 6−2 2 6+2
4
(cid:113) √ √
1
= − ( 6− 2)2
4
√ √
1
= ( 2− 6).
4
3. Denestthefollowingifpossible.
(cid:112) √
a. 3−2 2.
(cid:112) √ √
Assumethat 3−2 2= a+b 2.Then,
√ √
3−2 2 = (a+b 2)2
√
= a2+2b2+2ab 2.
Thiswouldholdifa2+2b2 =3and2ab = −2.Thesesimultaneous
equations have the solutions a = −b = ±1. Therefore, the positive
(cid:112) √ √
solutionis 3−2 2= 2−1.
(cid:112) √
b. 1+ 2.
Following the reasoning from the last problem, one tries to solve
the system a2+2b2 = 1 and 2ab = 1. Solving the second equation
for b =1/2a andinsertingthisintothefirstequation,wehave
2a4+a2−2=0.
Solving,weobtain
√
17−1
a2 = .
4
(cid:112) √
Therefore, 1+ 2cannotbedenested.
(cid:112) √
c. 5+2 6.
√ √ √ √ √
Note that 5+2 6 = 2+3+2 2 3 = ( 2+ 3)2. Therefore,
(cid:112) √ √ √
5+2 6= 2+ 3.
(cid:112)√ (cid:112)√
d. 3 5+2− 3 5−2.
(cid:112)√ √
Wefirstguesstheform 3 5±2= a+b 5.Cubingbothsidesof
thisequation,wefind
√ √ √
5±2= a3+3a2b 5+15ab2+5b3 5.
Thisleadstothesystemofequations
a3+15ab2 = ±2
3a2b+5b3 = 1.
Solvingthesecondequationfor a,wehave
(cid:112)
3b(1−5b3)
a = ± .
3b
Substitutingthisintothefirstequationofthesystem,weobtain
(cid:112)
3b(1−5b3)(40b3+1)
= ±2.
9b2
4 mathematical methods for physicists
Now,squarebothsidesoftheequation. Rearrangingandfactoring,
wehave
(8b3−1)(1000b6−25b3+1) =0.
Theonlyrealsolutionscomefrom b3 =1/8.Thus, b =1/2.
Insertingthisresultintotheexpressionfor a,
(cid:112)
3/2(1−5/8) 3/4 1
a = ± = ± = ± .
3/2 3/2 2
√ √
Therefore, 5±2= 1( 5±1).
2
Usingthisresult,wecanproceedtosolvetheproblem:
(cid:113)√ (cid:113)√ √ √
3 5+2− 3 5−2= 1( 5+1)− 1( 5−1) =1.
2 2
√
e. Findtherootsof x2+6x−4 5=0insimplifiedform.
(cid:112) √ (cid:112) √
Fromthequadraticformula,x = −3± 9+4 5.Setting 9+4 5=
√
a+b 5,weneedtosolvetheequations
a2+5b2 = 9
2ab = 4.
Solvingforb = 2,andsubstitutingintothefirstequation,wehave
a
a4−9a2+20=0.
Solvingfor a2,
9±1
a2 = =4,5.
2
√ √ (cid:112) √
So,a =2, 5and,therefore,b =2/ 5,1.Bothcasesgive 9+4 5=
√
2+ 5.
Using this result, we find the solutions of the problem, x = −1+
√ √
5, x = −5− 5.
5
4. Determinetheexactvaluesof
(cid:18) (cid:19)
4 3
a. sin cos−1 .
θ 5
3 3 3
This problem takes the form sinθ for θ = cos−1 , or cosθ = .
Figure1.1:TriangleforProblem4a. 5 5
One can draw a triangle with base of length 3 and hypotenuse
4
of length 5 as shown in Figure 1.1. Then, sinθ = . This is also
5
obtainablefromthePythagoreanidentity,
(cid:18) (cid:19)2
3
7 +sin2θ =1.
5
x (cid:16) x(cid:17)
b. tan sin−1 .
θ 7
√49−x2 Similar to the last problem one can use the triangle in Figure 1.2.
x
Figure1.2:TriangleforProblem4b. Letsinθ = .Then,tanθ = x .
7 49−x2