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Solution Manual for Mathematical Methods and Physical Insights An Integrated Approach (ISBN 9781107156418) [Updated December 2022]

Alec J. Schramm Occidental College

[Corrections? Please let me know: alec@oxy.edu]

©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

Contents

2 Coordinating Coordinates

page 1

3 Complex Numbers

4 Index Algebra

5 Brandishing Binomials

6 Series

7 Orbits in a Central Potential

8 Integration

9 Dirac Delta

10 Coda: Statistical Mechanics

11 Visualizing Vector Fields

12 Grad, Div & Curl

13 Interlude: Irrotational and Incompressible

14 Integrating Scalar and Vector Fields

15 The Theorems of Gauss and Stokes

16 Mostly Maxwell

109

17 Coda: Simply Connected Regions

113

18 Path Independence in the Complex Plane

115

19 Series, Singularities & Branches

122

20 Interlude: Conformal Mapping

130

21 The Calculus of Residues

136

22 Coda: Analyticity & Causality

152

23 Prelude: Superposition

156

24 Vector Space

157

25 The Inner Product

159

26 Interlude: Rotations

170

27 The Eigenvalue Problem

182

28 Coda: Normal Modes

201

29 Cartesian Tensors

212

30 Beyond Cartesian

221

31 Prelude: 1 2 3 . . . Infinity

237

32 Eponymous Polynomials

239

33 Fourier Series

253

34 Convergence & Completeness

260

35 Interlude: Beyond the Straight & Narrow

264

36 Fourier Transforms

275

37 Coda: Of Time Intervals and Frequency Bands

294

38 First-Order ODEs

298

39 Second-Order ODEs

302

40 The Sturm-Liouville Problem

318

41 Partial Differential Equations

332

42 Green’s Functions

352

43 Coda: Quantum Scattering

371

Appendix B

Rotations in R3

375

Appendix C

The Bessel Family of Functions

384

Coordinating Coordinates

2.1 Starting with ~ r = ρ cos φı̂ + ρ sin φ̂: ρ̂ =

∂~ r/∂ρ cos φı̂ + sin φ̂ = cos φı̂ + sin φ̂ X = p |∂~ r/∂ρ| cos2 φ + sin2 φ

and φ̂ =

∂~ r/∂φ −ρ sin φı̂ + ρ cos φ̂ = p = − sin φı̂ + cos φ̂ X |∂~ r/∂φ| ρ2 sin2 φ + ρ2 cos2 φ

Similar manipulations in spherical coordinates verify Eqn. (2.16).

2.2

⇢ˆ2

ˆ2

@ ⇢ˆ d @ @ˆ d @

⇢2

ˆ1

⇢ˆ1

⇢1

∂~ r ∂~ r ∂~ r dρ + ∂φ dφ + ∂z dz: 2.3 Cylindrical: d~ r = ∂ρ

∂~ r ∂~ r = ρ̂ = ρ̂ |cos φı̂ + sin φ̂| = ρ̂ ∂ρ ∂ρ ∂~ r ∂~ r = φ̂ = φ̂ |−ρ sin φı̂ + ρ cos φ̂| = φ̂ρ ∂φ ∂φ ∂~ r ∂~ r = k̂ = k̂ ∂z ∂z ∂~ r ∂~ r ∂~ r Spherical: d~ r = ∂~ dr + ∂θ dθ + ∂φ dφ: r

∂~ r ∂~ r = r̂ = r̂ sin θ cos φı̂ + sin θ sin φ̂ + cos θk̂ = r̂ ∂r ∂r ∂~ r ∂~ r = θ̂ = θ̂ r cos θ cos φı̂ + cos θ sin φ̂ − sin θk̂ = θ̂ r ∂θ ∂θ ∂~ r ∂~ r = φ̂ = φ̂ |r (− sin θ sin φı̂ + sin θ cos φ̂)| = φ̂ r sin θ ∂φ ∂φ ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.

cos φ sin φ 0 2.4 From Eqn. (2.18), the matrix mapping {ı̂, ̂, k̂} to {ρ̂, φ̂, k̂} is M = − sin φ cos φ 0 . 0 0 1 Similarly, Eqn. (2.19) gives the matrix N mapping {ı̂, ̂, k̂} to {r̂, θ̂, φ̂}. So the matrix mapping spherical coordinates into cylindrical coordinates is M N −1 . Since these are rotations, we can save a lot of work invoking N −1 = N T . Then that the mapping from {r̂, θ̂, φ̂} to {ρ̂, φ̂, k̂} multiplies out to be

sin θ 0 cos θ

cos θ 0 − sin θ

0 1 0

The inverse transformation is N M T — which is just the transpose of M N T .

2.5 Writing out the matrix equation r̂ θ̂ φ̂

! =

sin θ cos φ cos θ cos φ − sin φ

sin θ sin φ cos θ sin φ cos φ

cos θ − sin θ 0

ı̂ ̂ k̂

! ,

it’s straightforward to verify that r̂ · r̂ = θ̂ · θ̂ = φ̂ · φ̂ = 1 and that r̂ × θ̂ = φ̂, etc.

2.6 (a) cartesian: x2 + y 2 + z 2 = 1 cylindrical: ρ2 + z 2 = 1 spherical: r = 1 (b) cartesian: x2 + y 2 = 1 cylindrical: ρ = 1 spherical: r sin θ = 1

2.7 The direction cosines are the cartesian components of a unit vector from the origin making ~ and B, ~ we see A ~ ·B ~ = cos θ angles α, β, γ with the axes. Thus, given two different unit vectors A gives the identity in (b); part (a) is just a special case of this result.

2.8 Decomposing the vectors into cartesian components, but using spherical coordinates, ~ r = r sin θ cos φı̂ + r sin θ sin φ̂ + r cos θk̂ ~ r 0 = r 0 sin θ 0 cos φ 0 ı̂ + r 0 sin θ 0 sin φ 0 ̂ + r 0 cos θ 0 k̂ Then ~ r·~ r 0 ≡ rr 0 cos γ = rr 0 sin θ sin θ 0 cos φ cos φ 0 + rr 0 sin θ sin θ 0 sin φ sin φ 0 + rr 0 cos θ cos θ 0 . Solving:

cos γ = = sin θ sin θ 0 cos φ cos φ 0 + sin φ sin φ 0 + cos θ cos θ 0 = sin θ sin θ 0 cos(φ − φ 0 ) + cos θ cos θ 0 .

2.9 Executing the steps outlined in Example 2.2:

d ρ̇ρ̂ + ρφ̇φ̂ dt ˙ = ρ̈ρ̂ + ρ̇ρ̂˙ + ρ̇φ̇φ̂ + ρφ̈φ̂ + ρφ̇φ̂

~a =

= ρ̈ρ̂ + ρ̇φ̇φ̂ + ρ̇φ̇φ̂ + ρφ̈φ̂ − ρφ̇φ̇ρ̂ = ρ̈ρ̂ + 2ρ̇φ̇φ̂ + ρφ̈φ̂ − ρφ̇φ̇ρ̂

= ρ̈ − ρφ̇2 ρ̂ + ρφ̈ + 2ρ̇φ̇ φ̂

= ρ̈ − ρω 2 ρ̂ + (ρα + 2ρ̇ω) φ̂ .

2.10 First, |J|2 = |J| |J| = |J T | |J| = |J T J|. Then

|J| =

∂x/∂u ∂x/∂v

∂y/∂u ∂y/∂v

∂x/∂u ∂y/∂u

∂x/∂v ∂y/∂v

∂y 2 ∂x 2 ( ∂u ) + ( ∂u )

∂y ∂y ∂x ∂x + ∂u ∂u ∂v ∂v

∂y ∂x ∂x + ∂y ∂v ∂u ∂v ∂u

( ∂x )2 + ( ∂y )2 ∂v ∂v

Before trying to calculate this horrific determinant, note that since û ∼ ∂~ r/∂u and v̂ ∼ ∂~ r/∂v, the off-diagonal terms are just û · v̂ — which vanishes for an orthogonal system. Moreover, the diagonal terms are just the scale factors h2u and h2v . Thus |J|2 =

h2u 0

0 h2v

= h2u h2v .

2.11 Since θ̂ and φ̂ span the tangent plane to the sphere, then using little more than ı̂ × ̂ = k̂ does the trick:

n̂ ≡ θ̂ × φ̂ = ı̂ cos θ cos φ + ̂ cos θ sin φ − k̂ sin θ × (−ı̂ sin φ + ̂ cos φ) = k̂ cos θ cos2 φ + k̂ cos θ sin2 φ + ̂ sin θ sin φ + ı̂ sin θ cos φ = ı̂ sin θ cos φ + ̂ sin θ sin φ + k̂ cos θ ≡ r̂

2.12 Area elements (a) In cylindrical coordinates, the scale factors are hρ = 1, hφ = ρ, hz = 1. Then i. on surface of constant ρ, d~a = ρ̂ ρ dφ dz ii. on surface of constant φ, d~a = φ̂ dρ dz iii. on surface of constant z, d~a = k̂ ρ dρ dφ (b) In spherical coordinates, the scale factors are hr = 1, hθ = r, hφ = r sin θ. Then i. on surface of constant r, d~a = r̂ r2 sin θ dθ dφ = r̂ r2 dΩ ii. on surface of constant θ, d~a = θ̂ r sin θ dr dφ iii. on surface of constant φ, d~a = φ̂ r dr dθ

2.13 (a) Directly leveraging Eqn. (2.12) immediately yields radial equation: mr̈ − mrφ̇2 = − rk2 angular equation: rφ̈ + 2ṙφ̇ = 0 d 1 d (b) Simplifying the angular equation as r1 dt r2 φ̇ = mr mr2 φ̇ = 0 reveals angular modt mentum conservation: ` ≡ mr2 φ̇ = constant

d~ r dt

1 m ṙ2 + r2 φ̇2 . 2

Using angular momentum conservation in part (b), the kinetic energy is KE =

1 `2 mṙ2 + . 2 2mr2

For a circular orbit ṙ = 0, so the entire kinetic energy is due to angular motion.

~ = q 2.14 (a) A point charge q has electric field E 4π

r̂

0 r2

~ · n̂ dA = dΦ = E

∆A

n /

, so that

q q r̂ · n̂ dA = cos θ dA 4π 0 r2 4π 0 r2

r n r θ

∆Acosθ

∆A

∆Ω q

∆Ω

(b) ~ ≡ dA n̂ perpendicular to the radial line is (c) The component of the vector-valued area dA dA cos θ, and subtends the same solid angle as dA; using the definition of solid angle (which requires r perpendicular to the subtended area), dΩ =

dA cos θ r2

dΦ =

q dΩ . 4π 0

we have

Since the total solid angle is 4π, the net flux is

I Φ=

~ · n̂ dA = E

q 4π 0

I dΩ = S

q , 0

where no explicit integration was necessary!

2.15 Starting with x = r sin θ cos φ ,

y = r sin θ sin φ ,

z = r cos θ

we get

|J| ≡

∂x/∂r ∂y/∂r ∂z/∂r

∂x/∂θ ∂y/∂θ ∂z/∂θ

sin θ cos φ sin θ sin φ cos θ

∂x/∂φ ∂y/∂φ ∂z/∂φ

r cos θ cos φ r cos θ sin φ −r sin θ

−r sin θ sin φ r sin θ cos φ 0

= . . . = r2 sin θ .

2.16 (a)

y2 x2 2 2 + a2 sinh 2 u = cos φ+sin φ = 1. For constant u, this describes an ellipse with semia2 cosh2 u 2 2 2 2

major and minor axes a cosh u, a sinh u and foci at c = ±

a cosh u − a sinh u = ±a.

elliptic[u_, φ_] := {a Cosh[u]Cos[φ], a Sinh[u]Sin[φ]}; a = 1; uPlots = ParametricPlot[elliptic[u, φ], {u, 0, 1}, {φ, 0, 2π}, BoundaryStyle → Dashed, Mesh → 9, Frame → False, PlotStyle → Gray, Ticks → None, PlotRange → All]; zeroPlot = ParametricPlot[elliptic[0, φ], {φ, 0, 2π}, PlotStyle → {Red, Thick}, Ticks → None]; Show[uPlots,zeroPlot]

(b) The easiest approach is via Eqn. (2.27b):

h2u =

∂x ∂u

∂y ∂u

= (a sinh u cos φ)2 + (a cosh u sin φ)2

= a2 sinh2 u + sin2 φ

= a2 sinh2 u 1 − sin2 φ + 1 + sinh2 u sin2 φ = a2 sinh2 u + sin2 φ Similarly,

h2φ =

∂x ∂φ

∂y ∂φ

= (−a cosh u sin φ)2 + (a sinh u cos φ)2 = a2

2.17 (a) –

–

1 + sinh2 u sin2 φ + sinh2 u 1 − sin2 φ

x2 +y 2 z2 + a2 sinh 2u a2 cosh2 u

= sin2 θ + cos2 θ = 1. For constant u, this is the equation of an ellipsoidal surface. Note that the surface intersects the xy-plane in circles of radius a cosh u, whereas ellipses in the xz- and yz-planes have semi-major axis a cosh u, and semi-minor axis a sinh u. x2 +y 2 z2 − a2 cos 2θ a2 sin2 θ

= cosh2 u − sinh2 u = 1. For constant θ, this is the equation of a hyperbolic surface. Once again, the curves in the xy-plane are circles (this time of radius a sin θ), but those in the xz- and yz-planes are hyperbolas.

(b) Starting with x = a cosh u sin θ cos φ, y = a cosh u sin θ sin φ, z = a sinh u cos θ, the easiest

approach is via Eqn. (2.27b): h2u =

∂x ∂u

∂y ∂u

∂z ∂u

= (a sinh u sin θ cos φ)2 + (a sinh u sin θ sin φ)2 + (a cosh u cos θ)2 = a2 sinh2 u sin2 θ + cosh2 u cos2 θ

= a2 sinh2 u 1 − cos2 θ + 1 + sinh2 u cos2 θ = a2 sinh2 u + cos2 θ

It’s easy to see that hθ gives the same. As for hφ : h2φ =

∂x ∂φ

∂y ∂φ

∂z ∂φ

= (−a cosh u sin θ sin φ)2 + (a cosh u sin θ cos φ)2 = a2 cosh2 u sin2 θ

= a3

∂x/∂u ∂y/∂u ∂z/∂u

∂x/∂θ ∂y/∂θ ∂z/∂θ

∂x/∂φ ∂y/∂φ ∂z/∂φ

sinh u sin θ cos φ sinh u sin θ sin φ cosh u cos θ

cosh u cos θ cos φ cosh u cos θ sin φ − sinh u sin θ

= . . . = a3 cosh u sin θ sinh2 u + cos2 θ

− cosh u sin θ sin φ cosh u sin θ cos φ 0

(d) Surfaces of constant u have da = hθ hφ dθ dφ = a2 cosh u sin θ

sinh2 u + cos2 θ dθ dφ .

Since hu = hθ , surfaces of constant θ have the almost-identical da = hu hφ du dφ = a2 cosh u sin θ

sinh2 u + cos2 θ du dφ .

2.18 Using the hi for Problem 2.17 in Eqns. (2.27), and with ~ r = a cosh u sin θ cos φ ı̂ + a cosh u sin θ sin φ ̂ + a sinh u cos θ k̂ , we find êu =

1 ∂~ r 1 = p sinh u sin θ cos φ ı̂ + sinh u sin θ sin φ ̂ + a cosh u cos θ k̂ hu ∂u 2 2 sinh u + cos θ

êθ =

1 ∂~ 1 r = p cosh u cos θ cos φ ı̂ + cosh u cos θ sin φ ̂ − sinh u sin θ k̂ hθ ∂θ sinh2 u + cos2 θ

and êφ =

1 ∂~ r 1 = (− cosh u sin θ sin φ ı̂ + cosh u sin θ cos φ ̂) . hφ ∂φ cosh u sin θ

Multiple use of the identities cos2 + sin2 = 1 and cosh2 − sinh2 = 1 readily demonstrates that these vectors form an orthonormal set.

2.19 From Eqns. (2.27), we have hi êi = ∂~ r/dui : hr r̂ = (sin ψ sin θ cos φ, sin ψ sin θ sin φ, sin ψ cos θ, cos ψ) hψ ψ̂ = r (cos ψ sin θ cos φ, cos ψ sin θ sin φ, cos ψ cos θ, − sin ψ) hθ θ̂ = r (sin ψ cos θ cos φ, sin ψ cos θ sin φ, − sin ψ sin θ, 0) hφ φ̂ = r (− sin ψ sin θ sin φ, sin ψ sin θ cos φ, 0, 0)

Each is indeed orthogonal to the other three, and the magnitude of each gives the scale factor: hr = 1

hψ = r

hθ = r sin ψ

hφ = r sin ψ sin θ ,

which together produce the required line element in R4 . Finally, the product of the four gives the Jacobian, |J| = r3 sin2 ψ sin θ.

2.20 For hyperspherical coordinates x4 = r cos ψ x3 = r sin ψ cos θ x2 = r sin ψ sin θ cos φ x1 = r sin ψ sin θ sin φ , the Jacobian matrix is (in reverse order, as suggested) |J| =

∂(x4 , x3 , x2 , x1 ) ∂(r, ψ, θ, φ)

cos ψ sin ψ cos θ = sin ψ sin θ cos φ sin ψ sin θ sin φ

−r sin ψ r cos ψ cos θ r cos ψ sin θ cos φ r cos ψ sin θ sin φ

0 −r sin ψ sin θ r sin ψ cos θ cos φ r sin ψ cos θ sin φ

0 0 −r sin ψ sin θ sin φ r sin ψ sin θ cos φ

r cos ψ cos θ = cos ψ r cos ψ sin θ cos φ r cos ψ sin θ sin φ

−r sin ψ sin θ r sin ψ cos θ cos φ r sin ψ cos θ sin φ

0 −r sin ψ sin θ sin φ r sin ψ sin θ cos φ

sin ψ cos θ +r sin ψ sin ψ sin θ cos φ sin ψ sin θ sin φ

−r sin ψ sin θ r sin ψ cos θ cos φ r sin ψ cos θ sin φ

0 −r sin ψ sin θ sin φ . r sin ψ sin θ cos φ

Now each additive element of a determinant has exactly one contribution from each row and column. So a common multiplicative factor of a column or row can be moved outside the determinant. We can do just this with all the factors of sin ψ and cos ψ, giving |J| = r cos2 ψ sin2 ψ + r sin4 ψ

cos θ sin θ cos φ sin θ sin φ

−r sin θ r cos θ cos φ r cos θ sin φ

0 −r sin θ sin φ . r sin θ cos φ

Of course, we could also have taken out two more factors of r, but leaving them helps us recognize the remaining 3 × 3 determinant as the Jacobian r2 sin θ of standard spherical coordinates in R3 . Thus |J| = (r cos2 ψ sin2 ψ + r sin4 ψ) r2 sin θ = r3 sin2 ψ sin θ .

Complex Numbers

3.1 (a) |e−iπ |2 = eiπ e−iπ = 1 X (b) π ie = eie ln π X (c) (−π)ie = (eiπ eln π )ie = e−eπ eie ln π × eπ 2 (d) ieπ = eiπ/2 = eieπ /2 X i

(e) (−i)eπ = (ei3π/2 )e

(1+i ln π)

= ei3eπ/2 e−3π ln(π)/2 ×

3.2 In the cartesian representation z = a + ib, |z|2 = zz ∗ = (a + ib)(a − ib) = a2 + b2 |z 2 | = |(a + ib)2 | = |(a2 − b2 ) + i(2ab)| =

(a2 − b2 )2 + (2ab)2 =

(a2 + b2 )2 = a2 + b2

In the polar representation z = reiϕ it’s even easier: |z|2 = zz ∗ = reiϕ re−iϕ = r2 |z 2 | = |(reiϕ )2 | = |r2 e2iϕ | = r2 In general, |z n | = |z|n = rn .

3.3 Starting with z = cos θ + i sin θ, we have dz = (− sin θ + i cos θ)dθ = i(cos θ + i sin θ)dθ = izdθ . Thus

dz =i z

Z dθ

z = eiθ ,

=⇒

where the constant of integration is determined from z(θ = 0) = 1.

M N = 325 = (2 · 3 − 3 · 4)2 + (3 · 3 − 2 · 4)2 = 62 + 172

√ arg(z1 ) = arctan 1/ 3 = π6 =⇒ z1 = 2eiπ/6 √ =⇒ z2 = 2ei5π/6 = 2e−iπ/6 z2 = i − 3 : |z2 | = 2, arg(z1 ) = arctan −1/ 3 = 5π 6 √ √ √ (b) 2i = 2eiπ/2 2i = ± 2eiπ/4 = ± 2 √1 + √i = ±(1 + i)

3.5 (a) z1 = i +

√

3 : |z1 | =

√

1 + 3 = 2,

2 + 2i

√

3 = 4eiπ/3

2 + 2i

√

3 = ±2eiπ/6 = ±2

√

3 + 2i 2

= ±(

√

3 + i)

3.6 Real: z = z ∗ ; Imaginary z = −z ∗ (a) [(−1)1/i ]∗ = [(−1)−i ]∗ = (−1)i = [1/(−1)]−i = [1/(−1)]1/i = (−1)1/i = z Using −1 = eiπ , (−1)1/i = e+π ≈ 23.1 (b)

=⇒ REAL

∗

(z/z ∗ )i = (z ∗ /z)−i = (z/z ∗ )i =⇒ REAL Using z = reiθ , z/z ∗ = e2iθ so (z/z ∗ )i = e−2θ

(c) [z1 z2∗ − z1∗ z2 ]∗ = z1∗ z2 − z1 z2∗ = −[z1 z2∗ − z1∗ z2 ] =⇒ IMAGINARY z1 z2∗ − z1∗ z2 = r1 r2 ei(θ1 −θ2 ) − e−i(θ1 −θ2 ) = 2ir1 r2 sin(θ1 − θ2 ) = 2iIm[z1 z2∗ ] (d)

N P

∗ einθ

P∞ n=0

e−inθ

=⇒ COMPLEX

n=0

(e)

N P

∗ einθ

P∞ n=−∞

e−inθ =

P∞ n=−∞

einθ

=⇒ REAL

n=−N

The imaginary parts for positive n cancel in pairs with the imaginary contributions for negative n, so

N P

einθ = 2

n=−N

3.7 (a) (1 + i)4 = (b) (c) (d)

√

2eiπ/4

N P

cos (nθ)

n=0

= 4eiπ = −4

1+2i 2+i · 2+i = 5i =i 2−i 5 2 (1+4i)(1−4i) 1+4i 17 = (4+i)(4−i) = 17 =1 4+i (3+4i)4 (3−4i)3

z4 z ∗3

= |z| = 5

√ 3.8 – a ≡ (2 + i)(3 + i) = 5 + 5i = 5(1 + i) = 5 2eiπ/4 √ −1 – b ≡ (2 + i) = 5ei tan (1/2) √ −1 – c ≡ (3 + i) = √10ei tan (1/3) √ iπ/4 a = b · c =⇒ 5 2e = 5 2 exp{i[tan−1 ( 12 ) + tan−1 ( 13 )]} π = tan−1 ( 12 ) + tan−1 ( 13 ) 4

3.9 (a)

1 i(a+b) e + e−i(a+b) + ei(a−b) + e−i(a−b) 2 1 ia ib = e e + e−ib + e−ia eib + e−ib 2 1 ia = e · 2 cos b + e−ia · 2 cos b 2 = cos b eia + e−ia = 2 cos a cos b .

cos (a + b) + cos (a − b) =

(b) Same procedure works here, but for variety, let’s do it a little differently:

sin (a + b) + sin (a − b) = =m ei(a+b) + ei(a−b) = =m eia eib + e−ib

= =m eia (2 cos b)

= 2 sin a cos b .

3.10 (a) As for any complex number, we can write eiα + eiβ ≡ Reiθ , where (using the cartesian

representation) R2 = [cos α + cos β]2 + [sin α + sin β]2 = 2 + 2 (cos α cos β + sin α sin β)

= 2 [1 + cos (α − β)] = 2 2 cos2

α−β 2

= 4 cos2

α−β . 2

Alternatively, one can stay in the polar representation: R2 = eiα + eiβ

= eiα + eiβ

eiα + eiβ

∗

= 2 + 2 cos(α − β) .

As for the phase (using the results of Problem 3.9 ): θ = tan−1

= tan

−1

sin α + sin β cos α + cos β

cos

α−β 2 α−β 2

α+β 2

2 cos = tan−1 tan

α+β 2 α+β 2

2 sin

cos =

i α+β

2 (b) Same procedure establishes eiα −eiβ = 2i sin α−β e . (It helps to note that the cot x 2 is odd, and to recall that the cot of an angle is the tan of its complement.)

These results can also be worked out geometrically by drawing eiα and eiβ in the complex plane and adding them with the usual rules of vector addition.

3.11 (a) z = z1 z2 = r1 r2 ei(φ1 +φ2 ) . Thus |z| = r1 r2 , and arg(z) = φ1 + φ2 . So geometrically, multiplication of a complex number by reiθ rotates it by θ and magnifies it by r. (b) z = z1 + z2 : |z|2 = |z1 + z2 |2 = (z1 + z2 )(z1 + z2 )∗ = (z1 + z2 )(z1∗ + z2∗ ) = |z1 |2 + |z2 |2 + z1∗ z2 + z1 z2∗ = |z1 |2 + |z2 |2 + 2<e[z1∗ z2 ] = |r1 |2 + |r2 |2 + 2r1 r2 cos(φ1 − φ2 ) arg(z) = tan−1

−i(z−z ∗ ) z+z ∗

= tan−1

r1 sin φ1 +r2 sin φ2 r1 cos φ1 +r2 cos φ2

3.12 This is a ultimately a relationship between the cartesian and polar representations of complex number z = a + ib = reiφ , where φ = tan−1 (b/a): ln z = ln(reiφ ) = ln r + iφ . Substituting in the cartesian representation, φ = −i ln(z/r) = −i ln

3.13

dn dtn

at e

sin bt = =m

dn dtn

a + ib √ a2 + b2

(a+ib)t e

= =m (a2 + b2 )n/2 ein tan

a + ib 1 = + ln a − ib 2i

= =m (a + ib)n e(a+ib)t

−1

(b/a) e(a+ib)t −1

= (a2 +b2 )n/2 eat =m ei(bt+n tan

r = −i ln

(b/a))

a + ib a − ib

= (a2 +b2 )n/2 eat sin bt + n tan−1 (b/a)

3.14 (a) cos (3θ) = <e ei3θ = <e (cos θ + i sin θ)3 = cos3 θ − 3 cos θ sin2 θ = cos3 θ − 3 cos θ(1 − cos2 θ) = 4 cos3 θ − 3 cos θ (b) Similarly, sin (3θ) = =m (cos θ + i sin θ)3 = 3 cos2 θ sin θ − sin3 θ = 3 sin θ − 4 sin3 θ

3.15 z 6 = Reiθ

R = 1, θ = 2πn , 6

n = 0, 1, 2, 3, 4, 5.

z ei0 = 1 √ eiπ/3 = 12 (1 + i 3) √ ei2π/3 = 12 (−1 + i 3) eiπ = −1 √ ei4π/3 = − 12 (1 +√ i 3) ei5π/3 = 12 (1 − i 3)

0 1 2 3 4 5

3.16

√ i

i = (i)1/i = (eiπ/2 )−i = eπ/2 . But in fact, i = eiπ/2+i2nπ for integer n. Thus there is an infinity of solutions, (i)1/i = eπ/2+2nπ = eπ/2 , e5π/2 , e7π/2 , · · ·

1 eiz − e−iz = 0 3.17 (a) sin z = 2i eiz = e−iz e2iz = 1 ≡ ei2πn 2z = 2πn =⇒ z = πn (Note: no zeros off the <e axis) (b) cosh z = 12 ez + e−z = 0 ez = −e−z e2z = −1 ≡ eiπ(2n+1) 2z = iπ(2n + 1)

=⇒

z = iπ (2n + 1) (Note: all zeros on the =m axis) 2

3.18 Parts (a) and (b) are very similar and can be solved the same way—but for the sake of variety, we’ll use a different approach for each. (a) sinh z = sinh (x + iy) = 12 e(x+iy) − e−(x+iy)

= 12 ex eiy − e−x e−iy + e−x eiy − e−x eiy (adding and subtracting e−x eiy )

iy 1

−x

−iy

= 2 e ex − e + e−x eiy − e iy −x = e sinh x + e i sin y = (cos y + i sin y) sinh x + i (cosh x − sinh x) sin y = cos y sinh x + i cosh x sin y (b) Since cos ix = cosh x and sin ix = i sinh x, cosh z = cosh (x + iy) = cos (ix − y) = cos (ix) cos y + sin (ix) sin y = cosh x cos y + i sinh x sin y (c) Using the result of (a): | sinh z|2 = sinh2 x cos2 y + cosh2 x sin2 y

= sinh2 x 1 − sin2 y + 1 + sinh2 x sin2 y = sinh2 x + sin2 y

(d) Using the result of (b): | cosh z|2 = cos2 y cosh2 x + sinh2 x sin2 y

= cos2 y 1 + sinh2 x + sinh2 x 1 − cos2 y

= cos2 y + sinh2 x (e) tanh z = =

sinh z cosh z

ex eiy − e−x e−iy · ex eiy + e−x e−iy

ex e−iy + e−x e+iy ex e−iy + e−x e+iy

(e2x − e−2x ) + (e2iy − e−2iy ) (e2x + e−2x ) + (e2iy + e−2iy ) sinh (2x) + i sin (2y) = cosh (2x) + cos (2y)

3.19 It’s the triangle inequality — again: |z1 + z2 |2 = |z1 |2 + |z2 |2 + z1∗ z2 + z1 z2∗ = |z1 |2 + |z2 |2 + 2<e [z1∗ z2 ] (Note: <e[z1∗ z2 ] ∼ cos )

≤ |z1 |2 + |z2 |2 + 2|z1∗ z2 | = (|z1 | + |z2 |)2 ≥ |z1 |2 + |z2 |2 − 2|z1∗ z2 | = (|z1 | − |z2 |)2

|z1 | − |z2 | ≤ |z1 + z2 | ≤ |z1 | + |z2 |

3.20 A cos (kx + α) = A [cos kx cos α − sin kx sin α] = (A cos α) cos kx + (−A sin α) sin kx ≡ C1 cos kx + C2 sin kx ,

C1 = A cos α, C2 = −A sin α

B sin (kx + β) = B [sin kx cos β + cos kx sin β] = (B cos β) sin kx + (B sin β) cos kx ≡ C2 sin kx + C1 cos kx ,

C1 = B sin β, C2 = B cos β

D1 eikx + D2 e−ikx = D1 [cos (kx) + i sin (kx)] + D2 [cos (kx) − i sin (kx)] = (D1 + D2 ) cos (kx) + i (D1 − D2 ) sin (kx) ≡ C1 cos (kx) + C2 sin (kx) for real f (x), D1 = D2∗ Re(F eikx ) = Re [(a + ib)(cos kx + i sin kx)] = a cos kx − b sin kx ≡ C1 cos kx + C2 sin kx

3.21 With z(t) = C(ω)eiωt = |C|ei(ωt−ϕ) for z(t) =

f0 eiωt ω02 − ω 2

+ 2iβω

we get

+iϕ

ω02 − ω 2 + 2iβω 2βω |C| −→ tan ϕ = 2 = , = q 2 z(0) ω0 − ω 2 2 2 2 2 + 4β ω ω0 − ω

as advertised. Then x = <e(z) and cos(a ± b) = cos a cos b ∓ sin a sin b together yield x(t) = q

f0 cos(ωt − ϕ) ω02 − ω 2

+ 4β 2 ω 2

3.22 (a) A root z0 satisfies Pn (z0 ) =

n X

ak z0k =

k=0

n X

ak rk eikθ = 0 ,

k=0

where z0 = reiθ . The complex conjugate equation is ∗

[Pn (z0 )] = 0 =

" n X

#∗ ak r e

k=0

n X

k ikθ

n X

a∗k rk e−ikθ

k=0

ak rk e−ikθ , for real ak

k=0

= Pn (z0∗ ) Thus if z0 is a root, so is z0∗ = re−iθ . (b) Since roots must occur in complex conjugate pairs, for an odd number of roots at least one must be its own complex conjugate — that is, real. (c) (z − z0 )(z − z0∗ ) = z 2 − (z0 + z0∗ )z + z0 z0∗ = z 2 − 2<e[z0 ]z + |z0 |2 Im n=2 (−1/2, 3/2)

n=1 (1/2, 3/2) π/3

(−1,0) n=3

π/3

(1,0) n=0

π/3 (−1/2, − 3/2) n=4

(1/2, − 3/2 ) n=5

Index Algebra

4.1 Tij = 21 (Tij + Tji ) + 21 (Tij − Tji ) ≡ Sij + Aij , where Sij = Sji and Aij = −Aji . 4.2 (AB)T ij = (AB)ji =

P k

(Ajk Bki ) =

T AT kj Bik =

P k

T AT = (B T AT ) Bik ij kj

4.3 Compare M Λ with ΛM : (M Λ)ij =

Mik λ(j) δkj = λ(j)

Mik δkj = λ(j) Mij

whereas (ΛM )ij =

λ(i) δik Mkj = λ(i)

δik Mkj = λ(i) Mij

So the matrices commute only if all the diagonal elements of Λ are equal, λ(i) = λ for all i. Simply put, Λ must be proportional to the identity, Λ = λ 11.

4.4 Let’s check the x component: ~ × B) ~ 1= (A

1jk Aj Bk = A2 B3 − A3 B2 ,

since 123 = − 132 , and all others are zero.

4.5 Easily verified by explicitly writing out the sum.

4.6 (a) (êi × êj )` =

mn` (êi )m (êj )n =

mn` δim δjn = ij` =

ijk δk` =

X k

(b) (êi × êj ) · êk =

(êi × êj )` (êk )` =

ij` δk` = ijk

4.7 Cylindrical: hr = 1, hφ = r, hz = 1: ds2 =

hi dui = dr2 + r2 dφ2 + dz 2

dτ = h1 h2 h3 du1 du2 du3 = r dr dφ dz The area element d~a =

ê k ijk k

darφ = ẑ rdr dφ

hi hj dui duj has components daφz = r̂ rdφ dz

dazr = φ̂ dr dz

ijk (êk )`

Similarly for spherical: hr = 1, hθ = r, hφ = r sin θ: ds =

hi dui = dr2 + r2 dθ2 + r2 sin2 θ dφ2

dτ = h1 h2 h3 du1 du2 du3 = r2 sin θ dr dθ dφ and darθ = φ̂ rdr dθ

daθφ = r̂ r2 sin θ dθ dφ

daφr = θ̂ r sin θ dr dφ

4.8 Using Eqn. (4.46) h2i =

∂~ r ∂~ r · , ∂ui ∂ui

with cartesian ~ r = (x1 , x2 , x3 , x4 ) and hyperspherical coordinates gives h2r =

∂~ r 2 2 = (sin ψ sin θ sin φ, sin ψ sin θ cos φ, sin ψ cos θ, cos ψ) = 1 ∂r

h2ψ =

∂~ r 2 2 = (r cos ψ sin θ sin φ, r cos ψ sin θ cos φ, r cos ψ cos θ, −r sin ψ) = r2 ∂ψ

h2θ =

∂~ r 2 2 = (r sin ψ cos θ sin φ, r sin ψ cos θ cos φ, −r sin ψ sin θ, 0) = r2 sin2 ψ ∂θ

h2φ =

∂~ r 2 2 = (r sin ψ sin θ cos φ, −r sin ψ sin θ sin φ, 0, 0) = r2 sin2 ψ sin2 θ . ∂φ

So the line element is d~ r = êr dr + êψ rdψ + êθ r sin ψ dθ + êφ r sin ψ sin θ dφ , and the volume element is dτ = hr hψ hθ hφ d3 x = r3 sin2 ψ sin θ dr dψ dθ dφ .

4.9 Write out the sum:

= mn1 ij1 + mn2 ij2 + mn3 ij3 . ` mn` ij`

Due to the antisymmetry of ijk , for any values of i, j, m, n at most only one of terms in the sum will contribute. To be concrete, consider the first term; there are only three possibilities: (mn1), (ij1) both cyclic, (mn1), (ij1) both anti-cyclic, and one cyclic, the other anti-cyclic. If they’re both cyclic, then m = i, n = j and the product is +1; ditto if they’re both anti-cyclic. If one is cyclic the other anti-cyclic, then it must be that m = j, n = i, and the product is −1.

4.10 (a)

! X

mjk njk =

X X j

mjk njk

(δmn δjj − δmj δnj )

= δmn

X j

δjj −

δmj δnj

= 3δmn − δmn = 2δmn

(b) Beginning where part (a) left off:

! X

X X

ijk ijk =

ijk

ijk ijk

(2δii ) = 6

4.11 (a)

~ × B) ~ i= (A

ijk Aj Bk

(relabeling j ↔ k)

ikj Ak Bj

=−

ijk Ak Bj

(antisymmetry of ijk )

=−

~ × A) ~ i ijk Bj Ak = −(B

(b)

! ~ · (A ~ × B) ~ = A

ijk Aj Bk

ijk Ai Aj Bk =

ijk Aj Ai Bk

ijk

! =−

jik Aj Ai Bk = −

ijk

jik Ai Bk

~ · (A ~ × B) ~ = −A

(c)

! ~ · (B ~ × C) ~ = A

ijk Bj Ck

ijk Ai Bj Ck

ijk

! ijk Ck Ai

! =

! k

ijk Ai Bj

~ · (C ~ × A) ~ =B

jki Ck Ai

! k

kij Ai Bj

~ · (A ~ × B) ~ =C

(d)

~ × (B ~ × C) ~ A

~ × C) ~ k ijk Aj (B

! =

ijk Aj

k`m B` Cm

! =

ijk k`m Aj B` Cm =

jk`m

X X j`m

ijk `mk

Aj B` Cm

δi` δjm − δim δj` Aj B` Cm

j`m

~ · C) ~ − Ci (A ~ · B) ~ = Bi (A (e)

~×B ~ · C ~ ×D ~ = A

X X i

ijk Aj Bk

! X

i`m C` Dm

ijk i`m Aj Bk C` Dm

ijk`m

δj` δkm − δjm δk` Aj Bk C` Dm

jk`m

~ · C)( ~ B ~ · D) ~ − (A ~ · D)( ~ B ~ · C) ~ = (A

4.12 (a)

2 ∂ (r −1 ) ∂r 1 ∂ (r ) = − r12 ∂x = − r12 2r = − rx3i . ∂xi ∂xi i 2

(b) Since r is a scalar and ∂x∂∂x = ∂x∂ ∂x , it must be that ∂x∂∂x i

1 r

is a symmetric, two-

index object (symmetric 2nd rank tensor, to be precise). Now the only indexed objects we have at our disposal are δij , ijk , and xi . But there’s no way to combine and contract ijk with the others and leave only two symmetric indices. Thus the result can only be of the form a δij + b xi xj . (c) Taking the trace gives Tr (a δij + b xi xj ) =

a δii + b xi xi = 3a + br2 .

This must equal the trace of the double derivatives,

∂ 2 (1/r) ∂xi ∂xj

=−

X ∂ xi i

∂xi

=−

X 1 r3

−

3x2i r5

=0,

where we’ve used the result in part (a). (d) For i , j the kronecker delta vanishes, leaving bxi xj =

∂2 ∂xi ∂xj

1 r

=−

∂ ∂xj

xi r3

= −xi

∂ r−3 ∂xj

3xi xj xi ∂r =3 4 = , r ∂xj r5

where again we’ve used the result of part (a). Altogether then, we find a = −1/r3 and b = 3/r5 — and hence the desired identity.

q q

i j 1 4.13 Each charge pair qi , qj contributes energy 4π . Summing over all i and j would doubleri −~ rj | 0 |~ count the pairs; but dividing by two — and explicit disallowing i = j — gives the correct sum over distinct pairs,

qi qj 1X 1 1X U = = qi 2 4π 0 |~ ri − ~ rj | 2 i,j,i

qj 4π 0 |~ ri − ~ rj |

X 1

j,i

! ≡

1X qi V (~ ri ) , 2 i

where V (~ ri ) is the voltage at ~ ri due to all the other charges in the configuration.

4.14 The extremal values of ρ(~v ) = 2

∂ρ = ∂v`

A vv ij ij i j

A v i `i i

vk vk

v v − k k k

are given by

A vv ij ij i j

vk vk

(2v` )

P i

A`i vi − 2ρv`

P k

vk vk

A`i vi = ρv`

=⇒

A~v = ρ~v ,

which is the eigenvalue equation for A.

4.15 I=

y2 + z2 −yx −zx

−xy z 2 + x2 −zy

−xz −yz x2 + y 2

! .

3 4.16 As a uniform cube,R the mass density and can come out of the integrals. ρ = M/a2 is constant 2 Then, using Iij ≡ r δij − ri rj ρ dτ with r = x2 + y 2 + z 2 —

(a) Origin at center of cube, all the integrals extend from −a/2 to a/2:

Z a/2 I11 = ρ

(y 2 + z 2 ) dx dy dz =

−a/2

1 M a2 . 6

By symmetry, the same result holds for the other diagonal elements I22 , I33 . As for the off-diagonal elements,

Z a/2 I12 = −ρ

xy dx dy dz = 0 . −a/2

All elements Iij for i , j must similarly vanish. Thus the moment of inertia is a diagonal ~ = I~ ~ and ω matrix, so L ω shows that L ~ will be parallel around any axis through the origin. (b) Origin at one corner of cube, all the integrals extend from 0 to a:

Z a I11 = ρ 0

(y 2 + z 2 ) dx dy dz =

2 M a2 . 3

By symmetry, the same holds for the other diagonal elements I22 , I33 . As for the off-diagonal

elements,

Z a I12 = −ρ 0

1 xy dx dy dz = − M a2 . 4

where again symmetry dictates the same value for all elements Iij , i , j. Solving for axes ~ and ω through the origin around which L ~ are parallel is an example of an eigenvalue problem (Chapter 27). But inspection of the symmetry both ! of the geometry generally and I specif! 1 1 ~ and ω ically shows that L ~ are parallel around 1 . They are also parallel around 0 1 −1 ! 1 ~ = I~ and −1 , as direct substitution in L ω verifies. 0

4.17 (a) Clearly, Iij = Iji , so it’s symmetric. So of the 9 elements, the three elements above the diagonal elements are the same as those in below it diagonal. And then there are the 3 diagonal elements. So altogether, the moment of inertia has 6 independent elements. In n dimensions, there are n(n − 1)/2 elements above the diagonal and n elements along the diagonal. So a real n × n symmetric matrix has n(n − 1)/2 + n = n(n + 1)/2 independent elements. (b) The diagonal elements have integrands of the form Iii = r2 − ri2 ≥ 0. Since mass is always positive, Iii ≥ 0. (c) Tr(I) =

X ij

Iij δij =

r2 δii − ri2 = 3r2 − r2 = 2r2 .

Z Tr(I) = 2

r2 ρ dτ .

4.18 Using spherical coordinates in Eqn. (4.60) yields the expected results.

Brandishing Binomials

d 5.1 Only need to show that dx xN

= N xN −1 :

df f (x + h) − f (x) (x + h)N − xN = lim = lim dx h h h→0 h→0 = lim

( " N 1 X N m

h→0

#) m N −m

x h

−x

m=0

= lim

( "N −1 1 X N m

h→0

xm hN −m +

N N

xN − xN

m=0

= lim

( "N −1 1 X N

h→0

#) m N −m

x h

m=0

But only the m = N − 1 contribution survives the h → 0 limit, giving df = dx

N N −1

xN −1 = N xN −1

5.2 (a) n! n n n! + + = m!(n − m)! (m + 1)!(n − m − 1)! m m+1 n!(m + 1) + n!(n − m) = (m + 1)! (n − m)! n!(n + 1) = (m + 1)! (n − m)! (n + 1)! n + 1 = = (m + 1)! [(n + 1) − (m + 1)]! m+1

(b) Using the recursion relation in (a), n n − 1 = + m m n−1 = + m n − 1 + = m n−1 = + m n−1 = + m 4=

n − 1 m−1 n − 2 + m−1 n − 2 + m−1 n−2 + m−1 n−2 + m−1

m X n − 1 − k

m−k

n − 2 m−2 n − 3 n − 3 + m−3 m−2 n − 3 n − 1 − m + ··· + m−2 0 n − 3 + ··· + 1 m−2

k=0

(1 + 1)n =

n X n

=⇒

m=0

Similarly, (1 − 1)n =

Pn m=0

n m

= 2n

m=0

n−m 1

(−1)m =

m=0

n (−1)m m = 0.

(d) The sum over odd binomial coefficients can be extracted by subtracting from

m=0

2n − 0 =

Pn m=0

n (−1)m m

and using the results of (c):

n X n

[1 − (−1)m ] = 2

m=0

n X

X n n =⇒ = 2n−1 . 2m + 1 2m + 1

m=0

The sum over even binomial coefficients can similarly be extracted by adding the two series.

5.3 To simplify the notation, we’ll use Dn ≡ (d/dx)n = dn /dxn (a) for n = 1:

D(uv) =

1 X 1

[Dm u] D1−m v

m=0

1 1 u(Dv) + (Du)v 0 1 = u(Dv) + (Du)v =

which is the usual product rule. (b)

n+1

(uv) = D

" n X n m

# m

(D u) D

n−m

m=0

n X n

(Dm u) Dn−m+1 v + Dm+1 u

Dn−m v

m=0

To combine these into a single expression (and use the suggested recursion relation), we need to get the powers of D to match; we do this in the 2nd sum by shifting the index m → m − 1: Dn+1 (uv) =

n X n

(Dm u) Dn−m+1 v +

m=0

n+1 X

n m (D u) Dn−m+1 v . m−1

m=1

Now we explicitly separate out the m = 0 term from the first sum and the m = n + 1 term from the second, combine what’s left, and shift n → n − 1 at the end: Dn+1 (uv) =

n n+1 n (u) Dn+1 v + D u (v) + 0 n n h X n

n (Dm u) Dn−m+1 v m−1 i

m=1 n

= u Dn+1 v +

X n + 1 m

(Dm u) Dn−m+1 v + Dn+1 u (v)

m=1 n+1

X n + 1 m m=0

(Dm u) Dn−m+1 v =

n X n

(Dm u) Dn−m v

m=0

5.4 Binomial Theorem: (a + b)n = sin

n `

`=0

n−` ` a

b . In our case, n = 2m, so let’s try

mπ = =m eimπ/2 = =m 2

eiπ/4

= =m (cos π/4 + i sin π/4)2m = =m

1+i √ 2

2m i

X 1 = m =m 2

2m `

12m−` i`

`=0

The =m picks out only odd `, so: 2m−1

2m sin

X mπ = 2

2m `

−(−1)(`+1)/2

`odd

2m 1

−

2m 3

2m 5

− · · · + (−1)

m+1

2m 2m − 1

5.5 De Moivre gives

sin(5ϕ) = =m (cos ϕ + i sin ϕ)5 = ... = sin ϕ 16 sin4 ϕ − 20 sin2 ϕ + 5

For ϕ = π/5, this gives 16 sin4 (π/5) − 20 sin2 (π/5) + 5 = 0 so that

1 √ √ 1 20 ± 400 − 320 = 5± 5 . 32 8

sin2 (π/5) =

Now we (should) know that sin2 (π/5) < sin2 (π/4) = 12 , so the lower sign is the correct choice. √ 5 . For ϕ = 2π/5, we still have sin(5ϕ) = 0 and so end up with the √ same quadratic — but this time, the upper sign is the correct choice: sin2 (2π/5) = 18 5 + 5 . Thus sin2 (π/5) = 81 5 −

5.6 Start with de Moivre’s theorem, and apply the binomial theorem to the right-hand side: cos(N ϕ) + i sin(N ϕ) = (cos ϕ + i sin ϕ)N =

N X N

cosk ϕ (i sin ϕ)N −k

k=0

N X N

eiπ/2

N −k

cosk ϕ sinN −k ϕ

k=0

N h X N

cos

(N − k)π 2

+ i sin

(N − k)π 2

cosk ϕ sinN −k ϕ

k=0

Equating real and imaginary parts gives the sought-after identities. For N = 2 and 3, we find: cos(2ϕ) =

2 X 2

cos

(2 − k)π 2

cosk ϕ sin2−k ϕ = − sin2 ϕ + cos2 ϕ = 2 cos2 ϕ − 1

(3 − k)π 2

cosk ϕ sin3−k ϕ = −3 cos ϕ sin2 ϕ + cos3 ϕ = 4 cos3 ϕ − 3 cos ϕ .

k=0

cos(3ϕ) =

3 X 3

cos

k=0

Similarly, sin(2ϕ) =

2 X 2

sin

(2 − k)π 2

cosk ϕ sin2−k ϕ = 2 cos ϕ sin ϕ

(3 − k)π 2

cosk ϕ sin3−k ϕ = − sin3 ϕ + 3 cos2 ϕ sin ϕ = 3 sin ϕ − 4 sin3 ϕ .

k=0

sin(3ϕ) =

3 X 3

sin

k=0

The sine and cosine of multiples of π/2 mean only every other k in the sum contributes. So even/odd N renders cos(N ϕ) as an even/odd power series in cos ϕ, whereas only odd power series in sin ϕ (times cosines for N even) contribute to sin(N ϕ).

5.7

cos x =

eix + e−ix 2

n =

1 X n ix k −ix n−k 1 X n i(2k−n)x e e = n e . 2n k 2 k k=0

k=0

For real x, the sum has to yield a real function of x — that is, ahead and take the real part to find

= <e

— so we can go

cosn x =

1 X n cos[(2k − n)x] . k 2n k=0

5.8 (a) Even and odd powers of cos ϕ:

cos

ϕ=

eiϕ + e−iϕ 2

2N X 2N

1 22N

eikϕ e−i(2N −k)ϕ

k=0

2N 1 X 2N i2(k−N )ϕ

22N

k=0

cos2N ϕ =

1 X 2N 2i(k−N )ϕ 1 2N e + e2i(N −k)ϕ + 2N N 2 N 2 k

n n Now since m = n−m , we can split the sum into complex-conjugate pairs for k , N — so all but the k = N term yield cosines, N −1

k=0 N −1

1 2N X 2N + 2 cos [(2k − N )ϕ] 2N N k k=0

1 2N 2N

N −1

X 2N

22N −1

cos [2(k − N )ϕ] ,

k=0

as expected. The procedure is the same for odd powers, 2N + 1, but now all the terms combine to give cosines, so the sum can go all the way to N , cos2N +1 ϕ =

N X 2N + 1

22N

cos[(2(N − k) + 1) ϕ] .

k=0

For N = 1, we get the identities 1 2 1 2 + cos(2ϕ) 22 1 2 0 1 = (1 + cos 2ϕ) 2

cos2 ϕ =

and 1

1 X 3 cos [(3 − 2k)ϕ] 22 k

cos3 ϕ =

k=0

1 = (cos 3ϕ + 3 cos ϕ) . 4 (b) Even and odd powers of sin ϕ:

sin

ϕ=

eiϕ − e−iϕ 2i

(−1)N X 2N ikϕ e (−1)2N −k e−i(2N −k)ϕ k (2)2N k=0

1 22N

N −1

2N 1 X 2N i2(k−N )ϕ + 2N (−1)N +k e + ei2(N −k)ϕ N 2 k k=0

1 22N

N −1

X 2N 1 2N + 2N −1 (−1)N +k cos [2(k − N )ϕ] , N k 2 k=0

as in part (a). For odd powers, i2N +1 = i(−1)N , and (−1)2N +1−k = (−1)k+1 . As a result, the complex conjugate pairs now subtract, yielding sines rather then cosines. The desired result then follows,

sin2N +1 ϕ =

N X

(−1)N +k

22N

2N + 1 sin[(2(N − k) + 1)ϕ] k

k=0

For N = 1, we get the identities 1 2 1 2 − cos(2ϕ) 22 1 2 0 1 = (1 − cos 2ϕ) 2

sin2 ϕ =

and 1

1 X 3 (−1)k sin [(3 − 2k)ϕ] sin3 ϕ = − 2 2 k k=0

1 = − (sin 3ϕ − 3 sin ϕ) . 4

N m

5.9 (a) First, it’s easy to show that

µ=

N X

m P (m) =

m=0

N X

N m

N X

N −1 m−1

; then

m P (m) =

m=1

N X

N m N −m p q , m

q =1−p

m=1 N

X N − 1 N − 1 m N −m p q =N pm q N −m m−1 m−1

m=1

X M r

pr+1 q M −r ,

r =m−1 , M =N −1

r=0

= Np

M X M

pr q M −r = N p X

r=0

(b)

σ2 ≡

N X

(m − µ)2 P (m) =

m=0

N X

(m2 − 2µm + µ2 ) P (m)

m=0

m2 P (m) − 2µ

m=0

N X

m P (m) + µ2

m=0

N X

P (m)

m=0

m2 P (m) − 2µ2 + µ2 = hm2 i − hmi2 X

m=0

(c) The procedure is essentially the same as in (a), applied to the expression for σ 2 derived in (b). Start with the average of the square: N X

m2 P (m) = =

m=0

N X

N m N −m p q m

N − 1 m N −m p q m−1

m=1

N X m=1

M X

(r + 1)

M r+1 M −r p q , r

r =m−1 , M =N −1

r=0

= Np

"M X

M r M −r X M r M −r r p q + p q r r

r=0

= N p [M p + 1] , where we’ve used both the result from (a) (why rederive it?) and the normalization condition. With M = N − 1 and subtracting µ2 = (N p)2 , we get σ 2 = N p [(N − 1)p + 1] − µ2 = (N p)2 + N p(1 − p) − (N p)2 = N p(1 − p)

5.10 (a) N m N −M N! p q = pm q N −M m m!(N − m)! √ 1 N N N e−N pm q N −M ≈ √ √ √ m −m mm e N − m (N − m)N −m e−(N −m) 2π

P (m) =

1 = √ 2π

r =

N NN pm q N −M m m(N − m) m (N − m)N −m

N 2π m(N − m)

Np m

Nq N −m

N −m

(b) With m = x + N p and N − m = N q − x, 1 P (m) ≈ √ 2π

N (x + N p)(N q − x)

Np x + Np

x+N p

Nq Nq − x

N q−x

1 N ln 2 2π(x + N p)(N q − x)

+ (x + N p) ln

Np x + Np

+ (N q − x) ln

Nq Nq − x

For small x and/or large N and N p, the denominator of the first log is approximately 2π N 2 pq. As for the other logs: (x + N p) ln

Np x + Np

Nq Nq − x x + Np Nq − x = −(x + N p) ln − (N q − x) ln Np Nq x x = −(x + N p) ln 1 + − (N q − x) ln 1 − Np Nq

+ (N q − x) ln

≈ −(x + N p)

= ··· = −

x 1 − Np 2

x2 +O 2N pq

x Np

x3 (N p)2

− (N q − x) −

x 1 − Nq 2

(d) Putting it all back together and exponentiating,

2 1 N e−x /2N pq P (m) ≈ √ N 2 pq 2π 2 2 1 1 e− 2 (m−µ) /σ , = √ 2πσ 2

which is a gaussian distribution with µ = N p, σ 2 = N pq.

x Nq

Series

6 6.1

0.9999 . . . = 0.9 + 0.09 + 0.009 + . . . ∞ X 9

∞

9 X 1 = n 10 10 10n

n=1

9 10

6.2

q p 2

√

n=1

2−n

9 10 · =1 10 9

= 21 = 2 [Can confirm algebraically: x =

2x =⇒ x = 2]

p n

6.3 (a) ratio: limn→∞

1 1 − 10

2... = 22 24 28 ... = 2

2... =

n=0

= 1 — inconclusive

P∞ 1n+1 P∞ 1 √ > comparison: → ∞ — diverges 1 n n 1 n n2

(b) ratio: limn→∞

= 12 < 1 — converges

n+1

P∞ (n+1)2 P∞ 1 comparison: 1 n21n < n = 1 (geometric, starting at n = 1) — converges π−2 1 2 n = 1 — inconclusive (c) ratio: limn→∞ n+1 P∞ 2n2 +1 comparison: < 2ζ(1.1) – converges nπ 1 ln (n) (d) ratio: limn→∞ ln (n+1) = 1 — inconclusive P∞ 1 P∞ 1 comparison:

ln n

→ ∞ — diverges

6.4 (a) Recall that the series 1 = 1 + x + x2 + x3 + . . . 1−x

P∞

converges for all |x| < 1. This can then be compared directly to any series a . Though n=0 n |an+1 /an | may be greater than 1 for a finite number of terms, as long as limn→∞ |an+1 /an | < 1, the series converges by the comparison test. (b) The ratio test applied to ζ(s) = 1 + 21s + 31s + 41s + . . . gives lim n→∞

n s =1, n+1

and so is inconclusive for all s. But we already know that this series converges for any s > 1. The limit above is too crude in that all s-dependence is lost. So let’s retain the next order term in 1/n:

n s = n+1

s 1 ≈ 1 + 1/n

1−

s n

, for large n.

Just as in part (a), we can use the comparison P test with this known convergent series to develop a “new & improved" ratio test for a : if a series has the form n n an+1 −→ an

1−

s n

, for large n,

then comparison with ζ, the series converges for all s > 1. Pby ∞ 1 √ : (c) i. 1 n an+1 = an

n 1/2 = n+1

1/2 1 −→ 1 + 1/n

1−

1 2n

for large n.

From this, we see that s = 12 < 1, so the series diverges. ii.

P∞ 2n2 +1 nπ

: the ratio test gives an+1 −→ an

2−π n

Since s = π − 2 > 1, the series is convergent.

6.5 (a)

P∞ x2n 1

2 n n2

: r(x) = lim n→∞

x2(n+1) 2n n2 1 x2 2 = x lim = 2n 2 n→∞ 2 2(n+1) (n + 1)2 x

This converges only when less than 1, so must have |x| < (b)

n→∞

P∞ n1/n 1

P∞ n 1

n→∞

P∞ (x−1)n 1

nxn

n→∞

P∞ 1

|x| < 1

y n+1 (n + 1)1/(n+1) = |y| y n n1/n

|y| < 1

|x| > 1

(x − 1)n+1 nxn x−1 = <1 (n + 1)xn+1 (x − 1)n x

x > 1/2

: r(x) = lim

(e)

x3(n+1) n = |x|3 (n + 1) x3n

n1/n , y = 1/x:

r(x) = lim

(d)

: n r(x) = lim

(c)

√

P∞ x3n

n!xn :

r(x) = lim n→∞

(n + 1)! xn+1 = |x| lim (n + 1) −→ ∞ , so doesn’t converge for any x n! xn n→∞

6.6 Using the ratio test, |x| lim n→∞

n ζ(n + 1) ζ(n) n + 1

= |x| ,

so the series is absolutely convergent for |x| < 1. Since ζ(n)/n goes to zero for large n, the alternating series at x = 1 also converges. Thus the range x for which the series converges is −1 < x ≤ 1.

6.7 The ratio test quickly shows that the radius of convergence for all three is R = 1. (a) At x = 1, the series clearly diverges; at x = −1, we get an alternating series that also diverges. In fact, S1 diverges everywhere on the (complex) unit circle. (b) At x = 1, the series reduces to the harmonic series, which is divergent; at x = −1, the series becomes the alternating harmonic series which is convergent. So S2 converges at only some points on the unit circle. (c) At x = 1, the series is equivalent to the Riemann ζ(2) = π 2 /6, which is convergent; the alternating series at x = −1 is also convergent. In fact, this series converges everywhere on the unit circle.

6.8 Convert to a complex geometric series. The real part gives the cos series, n+1 X

S1 =

n+2 θ 2

cos cos kθ =

n+1 θ 2

sin

sin (θ/2)

k=1

the imaginary part the sin series,

S2 =

n+1 X

n+2 θ 2

sin sin kθ =

n+1 θ 2

sin

sin (θ/2)

k=1 2π At θ = n+1 , both series vanish. This is easily understood by representing the eikθ as vectors in the plane: the elements of the sum are equally distributed around the unit circle, so that the sum gives zero net displacement.

6.9 Recognizing the sum as a geometric series, N X

einx = e−iN x

n=−N

2N X

einx = e−iN x

1 − ei(2N +1)x 1 − eix

n=0

= e−iN x =

ei(N +1/2)x e−i(N +1/2)x − ei(N +1/2)x eix/2 e−ix/2 − eix/2

sin[(N + 1/2)x] . sin(x/2)

6.10 (a) With z = eiπx/a ,

" S(x) =

sin(nπx/a) =

n odd

= =m

=m(z ) = =m

n odd

"∞ X

# z

2j+1

= =m = =m

h h

z 1 − z2

= =m z

∞ X

# z

j=0

= =m

eiπx/a 1 − e2iπx/a

e−iπx/a − eiπx/a

n odd

j=0

= =m

# X

i 2 sin(πx/a)

= =m

1 2 sin(πx/a)

1 −2i sin(πx/a)

(b) With z = eiθ : S(x) =

N X

cos(2k − 1)θ = <e

k=1

# ei(2k−1)θ

= <e

1 X 2k z z

− zN

k=1

1 = <e z =

"N X

k=1

1 − z 2(N +1) −1 1 − z2

= <e z N

sin 2N θ cos N θ sin N θ = sin θ 2 sin θ

1 −z z

6.11 Using cos(kx) = 12 eikx + e−ikx , we can rewrite our sum in terms of two geometric series ∞ X cos(kx)

1 = 2

"∞ X eix k

k=0

k ∞ X e−ix r

# .

k=0

The geometric series sum to 1/(1 − e±ix /r), so that ∞ X cos(kx)

1 2

1 1 + 1 − e+ix /r 1 − e−ix /r

k=0

r r + r − e+ix r − e−ix r(r − cos x) = 2 X r − 2r cos x + 1 =

1 2

6.12 The total time (in hours) the trip takes is given by the sum T =1+

1 1 1 1 + + + ··· + . 2 3 4 100

This is just the first 100 terms of the harmonic series. You can tediously work it out, or approximate it by T ≈ γ + ln 100 = 5.182 , where γ ≈ 0.577 is the Euler-Mascheroni constant. So the trip takes about 5 hrs 11 mins. Note, by the way, that 1000 miles would only take 7 hrs 29 minutes — but you’d likely have your license revoked while setting a new land speed record!

6.13 (a) Differentiating the series for arctan term by term, d arctan z = 1 − z 2 + z 4 − z 6 + · · · dz 1 = . 1 + z2 (b) The derivative series only converges for |z| < 1, so this is also the radius of convergence for the arctan series. (c) Applying the ratio test to the series of arctan: lim n→∞

z 2n+1 /(2n + 1) 2n − 1 = |z|2 lim z 2n−1 /(2n − 1) n→∞ 2n + 1 = |z|2 < 1 ,

so the radius of convergence is, indeed, |z| < 1.