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Solutions for A Course in Mathematical Methods for Physicists, 1st Edition by Herman (All Chapters included) $29.49   Add to cart

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Solutions for A Course in Mathematical Methods for Physicists, 1st Edition by Herman (All Chapters included)

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  • A Course in Mathematical Methods for Physicists 1e
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  • A Course In Mathematical Methods For Physicists 1e

Complete Solutions Manual for A Course in Mathematical Methods for Physicists, 1st Edition by Russell L. Herman ; ISBN13: 9781466584679....(Full Chapters included from Chapter 1 to 11)...1 Introduction and Review 2 Free Fall and Harmonic Oscillators 3 Linear Algebra 4 Nonlinear Dynamics 5 The H...

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  • October 28, 2024
  • 441
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  • A Course in Mathematical Methods for Physicists 1e
  • A Course in Mathematical Methods for Physicists 1e
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SOLUTIONS MANUAL FOR




R . L . H E R M A N - V E R S I O N D AT E : N O V E M B E R 1 4 , 2 0 1 3

,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 − sin2 x ) − sin3 x
= 3 sin x − 4 sin3 x.

So, A = −4, B = 3.
 
θ π
c. sec θ + tan θ = tan + .
2 4
 
θ π

θ π
 sin 2 + 4
tan + =  
2 4 cos θ
+ π
2 4

sin 2θ cos π4 + sin π4 cos 2θ
=
cos 2θ cos π4 − sin π4 sin 2θ
sin 2θ + cos 2θ
=
cos 2θ − sin 2θ
!
sin 2θ + cos 2θ cos 2θ + sin 2θ
=
cos 2θ − sin 2θ cos 2θ + sin 2θ
 2
cos 2θ + sin 2θ
=
cos2 θ
2 − sin2 θ
2
cos2 θ
2 + sin2 2θ + 2 sin 2θ cos 2θ
=
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 .
4
π 1
q √
Therefore, sin = 2− 2.
8 2
b. tan 15o .


tan 15o = tan(60o − 45o )
tan 60o − tan 45o
=
1 + tan 60o tan 45o

3−1
= √
1+ 3
√ √ !
3−1 1− 3
= √ √
1+ 3 1− 3

= 2 − 3.

One can get the same answer using

2 osin2 15o 1 − cos 30o 2− 3 √ 2
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
p √
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:
1
q √ 1
q √
− 2− 3 = − 8−4 3
2 4

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