ALL 11 CHAPTERS COVERED
,ii
Contents
Preface iii
1 Introduction and Review 1
2 Free Fall and Harmonic Oscillators 25
3 Linear Algebra 69
4 Nonlinear Dynamics 111
5 The Harmonics of Vibrating Strings 141
6 Non-sinusoidal Harmonics 173
7 Complex Representations of Functions 219
8 Transform Techniques in Physics 245
9 Vector Analysis and EM Waves 291
10 Extrema and Variational Calculus 327
11 Problems in Higher Dimensions 357
A Review of Sequences and Infinite Series 379
B Quick Answers 395
Ch.1 Introduction . 395
Ch.2 Free Fall and Harmonic Oscillators 398
Ch.3 Linear Algebra 403
Ch.4 Nonlinear Dynamics 408
Ch.5 The Harmonics of Vibrating Strings 412
Ch.6 Non-sinusoidal Harmonics 415
Ch.7 Complex Representations of Functions 419
Ch.8 Transform Techniques in Physics 422
Ch.9 Vector Analysis and EM Waves 427
Ch.10 Extrema and Variational Calculus 431
Ch.11 Problems in Higher Dimensions 434
Ch. A Review of Sequences and Infinite Series 436
,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
cosines of sums of angles.
a. cos 2x = 2 cos2 x — 1.
cos 2x = cos2 x — sin2 x
= cos2 x — (1 — cos2 x)
= 2 cos2 x — 1.
b. sin 3x = A sin3 x + B sin x, for what values of A and B?
sin 3x = sin x cos 2x + sin 2x cos x
= sin x(cos2 x — sin2 x) + 2 sin x cos2 x
= 3 sin x(1 — sin 2 x) — sin 3 x
= 3 sin x — 4 sin 3 x.
So, A = — 4, B = 3.
θ π
c. sec θ + tan θ = tan + .
2 4
θ π
θ π sin 2
+ 4
tan + = θ π
2 4 cos 2 + 4
sin θ cos π + sin π cos θ
= 2 4 4 2
cos 2θ cos 4π — sin 4π sin 2θ
sin θ + cos θ
= 2 2
cos θ2 — sin 2θ
!
sin 2θ + cos 2θ cos θ2 + sin 2θ
=
cos θ2 — sin 2θ cos 2θ + sin 2θ
2
cos 2θ + sin θ2
=
cos2 θ2 — sin2 2θ
cos2 θ + sin2 θ + 2 sin θ cos θ
= 2 2 2 2
cos θ
, 2 MATHEMATICAL METHODS FOR PHYSICISTS
1 + sin θ
=
cos θ
= sec θ + tan θ.
2. Determine the exact values of
π
a. sin .
8
π 1 π
sin2 = 1 — cos
8 2 4
√ !
1 2
= 1—
2 2
1 √
= 2— 2 .
q 4
π 1 √
Therefore, sin = 2 — 2.
8 2
b. tan 15o.
tan 15o = tan(60o — 45o)
= tan 60 —otan 45 o
o o
1 + tan 60 tan 45
√
3 —1
= √
1+ 3
√ √ !
3 —1 1 — √ 3
= √
1+ 3 1— 3
√
= 2 — 3.
One can get the same answer using
√
2 o sin2 15o 2— 3 √ 2
1 — cos 30o
tan 15 = = = √ = (2 — 3) .
cos2 15o 1 + cos 30o 2+ 3
c. cos 105o.
1
cos2 105o =
(1 + cos 210o)
2
1
= (1 — cos 30o)
2
1 √
= (2 — 3).
4
√ √
Therefore, cos 105o = —21 2 — 3. One could also use
1
cos2 105o = sin2 15o = (1 — cos 30o) .
2
Note that this answer can be denested:
q q
1 √ 1 √
— 2— 3 = — 8—4 3
2 4
,ii
Contents
Preface iii
1 Introduction and Review 1
2 Free Fall and Harmonic Oscillators 25
3 Linear Algebra 69
4 Nonlinear Dynamics 111
5 The Harmonics of Vibrating Strings 141
6 Non-sinusoidal Harmonics 173
7 Complex Representations of Functions 219
8 Transform Techniques in Physics 245
9 Vector Analysis and EM Waves 291
10 Extrema and Variational Calculus 327
11 Problems in Higher Dimensions 357
A Review of Sequences and Infinite Series 379
B Quick Answers 395
Ch.1 Introduction . 395
Ch.2 Free Fall and Harmonic Oscillators 398
Ch.3 Linear Algebra 403
Ch.4 Nonlinear Dynamics 408
Ch.5 The Harmonics of Vibrating Strings 412
Ch.6 Non-sinusoidal Harmonics 415
Ch.7 Complex Representations of Functions 419
Ch.8 Transform Techniques in Physics 422
Ch.9 Vector Analysis and EM Waves 427
Ch.10 Extrema and Variational Calculus 431
Ch.11 Problems in Higher Dimensions 434
Ch. A Review of Sequences and Infinite Series 436
,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
cosines of sums of angles.
a. cos 2x = 2 cos2 x — 1.
cos 2x = cos2 x — sin2 x
= cos2 x — (1 — cos2 x)
= 2 cos2 x — 1.
b. sin 3x = A sin3 x + B sin x, for what values of A and B?
sin 3x = sin x cos 2x + sin 2x cos x
= sin x(cos2 x — sin2 x) + 2 sin x cos2 x
= 3 sin x(1 — sin 2 x) — sin 3 x
= 3 sin x — 4 sin 3 x.
So, A = — 4, B = 3.
θ π
c. sec θ + tan θ = tan + .
2 4
θ π
θ π sin 2
+ 4
tan + = θ π
2 4 cos 2 + 4
sin θ cos π + sin π cos θ
= 2 4 4 2
cos 2θ cos 4π — sin 4π sin 2θ
sin θ + cos θ
= 2 2
cos θ2 — sin 2θ
!
sin 2θ + cos 2θ cos θ2 + sin 2θ
=
cos θ2 — sin 2θ cos 2θ + sin 2θ
2
cos 2θ + sin θ2
=
cos2 θ2 — sin2 2θ
cos2 θ + sin2 θ + 2 sin θ cos θ
= 2 2 2 2
cos θ
, 2 MATHEMATICAL METHODS FOR PHYSICISTS
1 + sin θ
=
cos θ
= sec θ + tan θ.
2. Determine the exact values of
π
a. sin .
8
π 1 π
sin2 = 1 — cos
8 2 4
√ !
1 2
= 1—
2 2
1 √
= 2— 2 .
q 4
π 1 √
Therefore, sin = 2 — 2.
8 2
b. tan 15o.
tan 15o = tan(60o — 45o)
= tan 60 —otan 45 o
o o
1 + tan 60 tan 45
√
3 —1
= √
1+ 3
√ √ !
3 —1 1 — √ 3
= √
1+ 3 1— 3
√
= 2 — 3.
One can get the same answer using
√
2 o sin2 15o 2— 3 √ 2
1 — cos 30o
tan 15 = = = √ = (2 — 3) .
cos2 15o 1 + cos 30o 2+ 3
c. cos 105o.
1
cos2 105o =
(1 + cos 210o)
2
1
= (1 — cos 30o)
2
1 √
= (2 — 3).
4
√ √
Therefore, cos 105o = —21 2 — 3. One could also use
1
cos2 105o = sin2 15o = (1 — cos 30o) .
2
Note that this answer can be denested:
q q
1 √ 1 √
— 2— 3 = — 8—4 3
2 4
