Arfken Weber Math Selected Problem Solution

Physics 451 Fall 2004

Homework Assignment #1 — Solutions

Textbook problems: Ch. 1: 1.1.5, 1.3.3, 1.4.7, 1.5.5, 1.5.6Ch. 3: 3.2.4, 3.2.19, 3.2.27

Chapter 1

1.1.5 A sailboat sails for 1 hr at 4 km/hr (relative to the water) on a steady compass headingof 40◦ east of north. The saiboat is simultaneously carried along by a current. At theend of the hour the boat is 6.12 km from its starting point., The line from its startingpoint to its location lies 60◦ east of north. Find the x (easterly) and y (northerly)components of the water velocity.

This is a straightforward relative velocity (vector addition) problem. Let ~vbl

denote the velocity of the boat with respect to land, ~vbw the velocity of the boatwith respect to the water and ~vwl the velocity of the water with respect to land.Then

~vbl = ~vbw + ~vwl

where~vbw = 4 km/hr @ 50◦ = (2.57x + 3.06y) km/hr~vbl = 6.12 km/hr @ 30◦ = (5.3x + 3.06y) km/hr

Thus~vwl = ~vbl − ~vbw = 2.73x km/hr

1.3.3 The vector ~r, starting at the origin, terminates at and specifies the point in space(x, y, z). Find the surface swept out by the tip of ~r if

(a) (~r − ~a ) · ~a = 0

The vanishing of the dot product indicates that the vector ~r−~a is perpendicularto the constant vector ~a. As a result, ~r − ~a must lie in a plane perpendicularto ~a. This means ~r itself must lie in a plane passing through the tip of ~a andperpendicular to ~a

r−a

a

r

(b) (~r − ~a ) · ~r = 0

This time the vector ~r−~a has to be perpendicular to the position vector ~r itself.It is perhaps harder to see what this is in three dimensions. However, for twodimensions, we find

r−a

r

a

which gives a circle. In three dimensions, this is a sphere. Note that we can alsocomplete the square to obtain

(~r − ~a ) · ~r = |~r − 12~a |

2 − | 12~a |2

Hence we end up with the equation for a circle of radius |~a |/2 centered at thepoint ~a/2

|~r − 12~a |

2 = | 12~a |2

1.4.7 Prove that ( ~A× ~B) · ( ~A× ~B) = (AB)2 − ( ~A · ~B )2.

This can be shown just by a straightforward computation. Since

~A× ~B = (AyBz −AzBy)x + (AzBx −AxBz)y + (AxBy −AyBx)z

we find

| ~A× ~B |2 = (AyBz −AzBy)2 + (AzBx −AxBz)2 + (AxBy −AyBx)2

= A2xB2

y + A2xB2

z + A2yB2

x + A2yB2

z + A2zB

2x + A2

zB2y

− 2AxBxAyBy − 2AxBxAzBz − 2AyByAzBz

= (A2x + A2

y + A2z)(B

2x + B2

y + B2z)− (AxBx + AyBy + AzBz)2

where we had to add and subtract A2xB2

x+A2yB2

y+A2zB

2z and do some factorization

to obtain the last line.

However, there is a more elegant approach to this problem. Recall that crossproducts are related to sin θ and dot products are related to cos θ. Then

| ~A× ~B |2 = (AB sin θ)2 = (AB)2(1− cos2 θ) = (AB)2 − (AB cos θ)2

= (AB)2 − ( ~A · ~B )2

1.5.5 The orbital angular momentum ~L of a particle is given by ~L = ~r×~p = m~r×~v where ~pis the linear momentum. With linear and angular velocity related by ~v = ~ω×~r, showthat

~L = mr2[~ω − r(r · ~ω )]

Here, r is a unit vector in the ~r direction.

Using ~L = m~r × ~v and ~v = ~ω × ~r, we find

~L = m~r × (~ω × ~r )

Because of the double cross product, this is the perfect opportunity to use the“BAC–CAB” rule: ~A× ( ~B × ~C) = ~B( ~A · ~C)− ~C( ~A · ~B)

~L = m[~ω(~r · ~r )− ~r(~r · ~ω )] = m[~ωr2 − ~r(~r · ~ω )]

Using ~r = r r, and factoring out r2, we then obtain

~L = mr2[~ω − r(r · ~ω )] (1)

1.5.6 The kinetic energy of a single particle is given by T = 12mv2. For rotational motion

this becomes 12m(~ω × ~r )2. Show that

T = 12m[r2ω2 − (~r · ~ω )2]

We can use the result of problem 1.4.7:

T = 12m(~ω × ~r )2 = 1

2m[(ωr)2 − (~ω · ~r )2] = 12m[r2ω2 − (~r · ~ω )2]

Note that we could have written this in terms of unit vectors

T = 12mr2[ω2 − (r · ~ω)2]

Comparing this with (1) above, we find that

T = 12~L · ~ω

which is not a coincidence.

Chapter 3

3.2.4 (a) Complex numbers, a + ib, with a and b real, may be represented by (or areisomorphic with) 2× 2 matrices:

a + ib ↔(

a b−b a

)

Show that this matrix representation is valid for (i) addition and (ii) multiplica-tion.

Let us start with addition. For complex numbers, we have (straightforwardly)

(a + ib) + (c + id) = (a + c) + i(b + d)

whereas, if we used matrices we would get(a b−b a

)+

(c d−d c

)=

((a + c) (b + d)−(b + d) (a + c)

)which shows that the sum of matrices yields the proper representation of thecomplex number (a + c) + i(b + d).

We now handle multiplication in the same manner. First, we have

(a + ib)(c + id) = (ac− bd) + i(ad + bc)

while matrix multiplication gives(a b−b a

) (c d−d c

)=

((ac− bd) (ad + bc)−(ad + bc) (ac− bd)

)which is again the correct result.

(b) Find the matrix corresponding to (a + ib)−1.

We can find the matrix in two ways. We first do standard complex arithmetic

(a + ib)−1 =1

a + ib=

a− ib

(a + ib)(a− ib)=

1a2 + b2

(a− ib)

This corresponds to the 2× 2 matrix

(a + ib)−1 ↔ 1a2 + b2

(a −bb a

)

Alternatively, we first convert to a matrix representation, and then find the inversematrix

(a + ib)−1 ↔(

a b−b a

)−1

=1

a2 + b2

(a −bb a

)Either way, we obtain the same result.

3.2.19 An operator ~P commutes with Jx and Jy, the x and y components of an angularmomentum operator. Show that ~P commutes with the third component of angularmomentum; that is,

[~P , Jz] = 0

We begin with the statement that ~P commutes with Jx and Jy. This may beexpressed as [~P , Jx] = 0 and [~P , Jy] = 0 or equivalently as ~PJx = Jx

~P and~PJy = Jy

~P . We also take the hint into account and note that Jx and Jy satisfythe commutation relation

[Jx, Jy] = iJz

or equivalently Jz = −i[Jx, Jy]. Substituting this in for Jz, we find the doublecommutator

[~P , Jz] = [~P ,−i[Jx, Jy]] = −i[~P , [Jx, Jy]]

Note that we are able to pull the −i factor out of the commutator. From here,we may expand all the commutators to find

[~P , [Jx, Jy]] = ~PJxJy − ~PJyJx − JxJy~P + JyJx

~P

= Jx~PJy − Jy

~PJx − Jx~PJy + Jy

~PJx

= 0

To get from the first to the second line, we commuted ~P past either Jx or Jy asappropriate. Of course, a quicker way to do this problem is to use the Jacobiidentity [A, [B,C]] = [B, [A,C]]− [C, [A,B]] to obtain

[~P , [Jx, Jy]] = [Jx, [~P , Jy]]− [Jy, [~P , Jx]]

The right hand side clearly vanishes, since ~P commutes with both Jx and Jy.

3.2.27 (a) The operator Tr replaces a matrix A by its trace; that is

Tr (a) = trace(A) =∑

i

aii

Show that Tr is a linear operator.

Recall that to show that Tr is linear we may prove that Tr (αA+βB) = α Tr (A)+β Tr (B) where α and β are numbers. However, this is a simple property ofarithmetic

Tr (αA + βB) =∑

i

(αaii + βbii) = α∑

i

aii + β∑

i

bii = α Tr (A) + β Tr (B)

(b) The operator det replaces a matrix A by its determinant; that is

det(A) = determinant of A

Show that det is not a linear operator.

In this case all we need to do is to find a single counterexample. For example, foran n× n matrix, the properties of the determinant yields

det(αA) = αn det(A)

This is not linear unless n = 1 (in which case A is really a single number andnot a matrix). There are of course many other examples that one could come upwith to show that det is not a linear operator.



Textbook problems: Ch. 3: 3.3.1, 3.3.12, 3.3.13, 3.5.4, 3.5.6, 3.5.9, 3.5.30

Chapter 3

3.3.1 Show that the product of two orthogonal matrices is orthogonal.

Suppose matrices A and B are orthogonal. This means that AA = I and BB = I.We now denote the product of A and B by C = AB. To show that C is orthogonal,we compute CC and see what happens. Recalling that the transpose of a productis the reversed product of the transposes, we have

CC = (AB)(AB) = ABBA = AA = I

The statement that this is a key step in showing that the orthogonal matrices forma group is because one of the requirements of being a group is that the productof any two elements (ie A and B) in the group yields a result (ie C) that is alsoin the group. This is also known as closure. Along with closure, we also needto show associativity (okay for matrices), the existence of an identity element(also okay for matrices) and the existence of an inverse (okay for orthogonalmatrices). Since all four conditions are satisfied, the set of n × n orthogonalmatrices form the orthogonal group denoted O(n). While general orthogonalmatrices have determinants ±1, the subgroup of matrices with determinant +1form the “special orthogonal” group SO(n).

3.3.12 A is 2× 2 and orthogonal. Find the most general form of

A =(

a bc d

)Compare with two-dimensional rotation.

Since A is orthogonal, it must satisfy the condition AA = I, or(a bc d

) (a cb d

)=

(a2 + b2 ac + bdac + bd c2 + d2

)=

(1 00 1

)This gives three conditions

i) a2 + b2 = 1, ii) c2 + d2 = 1, iii) ac + bd = 0

These are three equations for four unknowns, so there will be a free parameterleft over. There are many ways to solve the equations. However, one nice way is

to notice that a2 + b2 = 1 is the equation for a unit circle in the a–b plane. Thismeans we can write a and b in terms of an angle θ

a = cos θ, b = sin θ

Similarly, c2 + d2 = 1 can be solved by setting

c = cos φ, d = sinφ

Of course, we have one more equation to solve, ac + bd = 0, which becomes

cos θ cos φ + sin θ sinφ = cos(θ − φ) = 0

This means that θ − φ = π/2 or θ − φ = 3π/2. We must consider both casesseparately.

φ = θ − π/2: This gives

c = cos(θ − π/2) = sin θ, d = sin(θ − π/2) = − cos θ

or

A1 =(

cos θ sin θsin θ − cos θ

)(1)

This looks almost like a rotation, but not quite (since the minus sign is in thewrong place).

φ = θ − 3π/2: This gives

c = cos(θ − 3π/2) = − sin θ, d = sin(theta− 3π/2) = cos θ

or

A2 =(

cos θ sin θ− sin θ cos θ

)(2)

which is exactly a rotation.

Note that we can tell the difference between matrices of type (1) and (2) bycomputing the determinant. We see that det A1 = −1 while detA2 = 1. In fact,the A2 type of matrices form the SO(2) group, which is exactly the group ofrotations in the plane. On the other hand, the A1 type of matrices representrotations followed by a mirror reflection y → −y. This can be seen by writing

A1 =(

1 00 −1

) (cos θ sin θ− sin θ cos θ

)

Note that the set of A1 matrices by themselves do not form a group (since theydo not contain the identity, and since they do not close under multiplication).However the set of all orthogonal matrices {A1, A2} forms the O(2) group, whichis the group of rotations and mirror reflections in two dimensions.

3.3.13 Here |~x 〉 and |~y 〉 are column vectors. Under an orthogonal transformation S, |~x ′〉 =S|~x 〉, |~y ′〉 = S|~y 〉. Show that the scalar product 〈~x |~y 〉 is invariant under this orthog-onal transformation.

To prove the invariance of the scalar product, we compute

〈~x ′|~y ′〉 = 〈~x |SS|~y 〉 = 〈~x |~y 〉

where we used SS = I for an orthogonal matrix S. This demonstrates that thescalar product is invariant (same in primed and unprimed frame).

3.5.4 Show that a real matrix that is not symmetric cannot be diagonalized by an orthogonalsimilarity transformation.

We take the hint, and start by denoting the real non-symmetric matrix by A.Assuming that A can be diagonalized by an orthogonal similarity transformation,that means there exists an orthogonal matrix S such that

Λ = SAS where Λ is diagonal

We can ‘invert’ this relation by multiplying both sides on the left by S and onthe right by S. This yields

A = SΛS

Taking the transpose of A, we find

A = ( ˜SΛS) = SΛ˜

S

However, the transpose of a transpose is the original matrix, ˜S = S, and the

transpose of a diagonal matrix is the original matrix, Λ = Λ. Hence

A = SΛS = A

Since the matrix A is equal to its transpose, A has to be a symmetric matrix.

However, recall that A is supposed to be non-symmetric. Hence we run into acontradiction. As a result, we must conclude that A cannot be diagonalized byan orthogonal similarity transformation.

3.5.6 A has eigenvalues λi and corresponding eigenvectors |~xi〉. Show that A−1 has thesame eigenvectors but with eigenvalues λ−1

i .

If A has eigenvalues λi and eigenvectors |~xi〉, that means

A|~xi〉 = λi|~xi〉

Multiplying both sides by A−1 on the left, we find

A−1A|~xi〉 = λiA−1|~xi〉

or|~xi〉 = λiA

−1|~xi〉

Rewriting this asA−1|~xi〉 = λ−1

i |~xi〉

it is now obvious that A−1 has the same eigenvectors, but eigenvalues λ−1i .

3.5.9 Two Hermitian matrices A and B have the same eigenvalues. Show that A and B arerelated by a unitary similarity transformation.

Since both A and B have the same eigenvalues, they can both be diagonalizedaccording to

Λ = UAU†, Λ = V BV †

where Λ is the same diagonal matrix of eigenvalues. This means

UAU† = V BV † ⇒ B = V †UAU†V

If we let W = V †U , its Hermitian conjugate is W † = (V †U)† = U†V . This meansthat

B = WAW † where W = V †U

and WW † = V †UU†V = I. Hence A and B are related by a unitary similaritytransformation.

3.5.30 a) Determine the eigenvalues and eigenvectors of(1 εε 1

)

Note that the eigenvalues are degenerate for ε = 0 but the eigenvectors are or-thogonal for all ε 6= 0 and ε → 0.

We first find the eigenvalues through the secular equation∣∣∣∣ 1− λ εε 1− ε

∣∣∣∣ = (1− λ)2 − ε2 = 0

This is easily solved

(1− λ)2 − ε2 = 0 ⇒ (λ− 1)2 = ε2 ⇒ (λ− 1) = ±ε (3)

Hence the two eigenvalues are λ+ = 1 + ε and λ− = 1− ε.

For the eigenvectors, we start with λ+ = 1 + ε. Substituting this into the eigen-value problem (A− λI)|x〉 = 0, we find(

−ε εε −ε

) (ab

)= 0 ⇒ ε(a− b) = 0 ⇒ a = b

Since the problem did not ask to normalize the eigenvectors, we can take simply

λ+ = 1 + ε : |x+〉 =(

11

)

For λ− = 1− ε, we obtain instead(ε εε ε

) (ab

)= 0 ⇒ ε(a + b) = 0 ⇒ a = −b

This gives

λ− = 1− ε : |x−〉 =(

1−1

)Note that the eigenvectors |x+〉 and |x−〉 are orthogonal and independent of ε. Ina way, we are just lucky that they are independent of ε (they did not have to turnout that way). However, orthogonality is guaranteed so long as the eigenvaluesare distinct (ie ε 6= 0). This was something we proved in class.

b) Determine the eigenvalues and eigenvectors of(1 1ε2 1

)

Note that the eigenvalues are degenerate for ε = 0 and for this (nonsymmetric)matrix the eigenvectors (ε = 0) do not span the space.

In this nonsymmetric case, the secular equation is∣∣∣∣ 1− λ 1ε2 1− λ

∣∣∣∣ = (1− λ)2 − ε2 = 0

Interestingly enough, this equation is the same as (3), even though the matrix isdifferent. Hence this matrix has the same eigenvalues λ+ = 1+ ε and λ− = 1− ε.

For λ+ = 1 + ε, the eigenvector equation is(−ε 1ε2 −ε

) (ab

)= 0 ⇒ −εa + b = 0 ⇒ b = εa

Up to normalization, this gives

λ+ = 1 + ε : |x+〉 =(

1ε

)(4)

For the other eigenvalue, λ− = 1− ε, we find(ε 1ε2 ε

) (ab

)= 0 ⇒ εa + b = 0 ⇒ b = −εa

Hence, we obtain

λ− = 1− ε : |x−〉 =(

1−ε

)(5)

In this nonsymmetric case, the eigenvectors do depend on ε. And furthermore,

when ε = 0 it is easy to see that both eigenvectors degenerate into the same(

10

).

c) Find the cosine of the angle between the two eigenvectors as a function of ε for0 ≤ ε ≤ 1.

For the eigenvectors of part a), they are orthogonal, so the angle is 90◦. Thusthis part really refers to the eigenvectors of part b). Recalling that the angle canbe defined through the inner product, we have

〈x+|x−〉 = |x+| |x−| cos θ

or

cos θ =〈x+|x−〉

〈x+|x+〉1/2〈x−|x−〉1/2

Using the eigenvectors of (4) and (5), we find

cos θ =1− ε2√

1 + ε2√

1 + ε2=

1− ε2

1 + ε2

Recall that the Cauchy-Schwarz inequality guarantees that cos θ lies between −1and +1. When ε = 0 we find cos θ = 1, so the eigenvectors are collinear (anddegenerate), while for ε = 1, we find instead cos θ = 0, so the eigenvectors areorthogonal.



Textbook problems: Ch. 1: 1.7.1, 1.8.11, 1.8.16, 1.9.12, 1.10.4, 1.12.9Ch. 2: 2.4.8, 2.4.11

Chapter 1

1.7.1 For a particle moving in a circular orbit ~r = x r cosωt+ y r sinωt

(a) evaluate ~r × ~r

Taking a time derivative of ~r, we obtain

~r = −x rω sinωt+ y rω cosωt (1)

Hence

~r × ~r = (x r cosωt+ y r sinωt)× (−x rω sinωt+ y rω cosωt)

= (x× y)r2ω cos2 ωt− (y × x)r2ω sin2 ωt

= z r2ω(sin2 ωt+ cos2 ωt) = z r2ω

(b) Show that ~r + ω2~r = 0

The acceleration is the time derivative of (1)

~r = −x rω2 cosωt− y rω2 sinωt = −ω2(x r cosωt+ y r sinωt) = −ω2~r

Hence ~r+ ω2~r = 0. This is of course the standard kinematics of uniform circularmotion.

1.8.11 Verify the vector identity

~∇× ( ~A× ~B) = ( ~B · ~∇) ~A− ( ~A · ~∇) ~B − ~B(~∇ · ~A) + ~A(~∇ · ~B)

This looks like a good time for the BAC–CAB rule. However, we have to becareful since ~∇ has both derivative and vector properties. As a derivative, itoperates on both ~A and ~B. Therefore, by the product rule of differentiation, wecan write

~∇× ( ~A× ~B) =↓

~∇× ( ~A× ~B) +↓

~∇× ( ~A× ~B)

where the arrows indicate where the derivative is acting. Now that we havespecified exactly where the derivative goes, we can treat ~∇ as a vector. Using theBAC–CAB rule (once for each term) gives

~∇× ( ~A× ~B) =↓~A(~∇ · ~B)−

↓~B(~∇ · ~A) +

↓~A(~∇ · ~B)−

↓~B(~∇ · ~A) (2)

The first and last terms on the right hand side are ‘backwards’. However, we canturn them around. For example

↓~A(~∇ · ~B) =

↓~A( ~B · ∇) =

↓( ~B · ~∇) ~A

With all the arrows in the right place [after flipping the first and last terms in(2)], we find simply

~∇× ( ~A× ~B) = ( ~B · ~∇) ~A− ~B(~∇ · ~A) + ~A(~∇ · ~B)− ( ~A · ~∇) ~B

which is what we set out to prove.

1.8.16 An electric dipole of moment ~p is located at the origin. The dipole creates an electricpotential at ~r given by

ψ(~r ) =~p · ~r

4πεor3

Find the electric field, ~E = −~∇ψ at ~r.

We first use the quotient rule to write

~E = −~∇ψ = − 14πε0

~∇(~p · ~rr3

)= − 1

4πε0r3~∇(~p · ~r )− (~p · ~r )~∇(r3)

r6

Applying the chain rule to the second term in the numerator, we obtain

~E = − 14πε0

r3~∇(~p · ~r )− 3r2(~p · ~r )~∇(r)r6

We now evaluate the two separate gradients

~∇(~p · ~r ) = xi∂

∂xi(pjxj) = ~xipj

∂xj

∂xi= xipjδij = xipi = ~p

and

~∇r = xi∂

∂xi

√x2

1 + x22 + x2

3 = xi1

2√x2

1 + x22 + x2

3

2xi =xixi

r=~r

r= r

Hence~E = − 1

4πε0r3~p− 3r2(~p · ~r )r

r6= − 1

4πε0~p− 3(~p · r)r

r3

Note that we have used the fact that ~p is a constant, although this was neverstated in the problem.

1.9.12 Show that any solution of the equation

~∇× (~∇× ~A)− k2 ~A = 0

automatically satisfies the vector Helmholtz equation

∇2 ~A+ k2 ~A = 0

and the solenoidal condition~∇ · ~A = 0

We actually follow the hint and demonstrate the solenoidal condition first. Takingthe divergence of the first equation, we find

~∇ · ~∇× (~∇× ~A)− k2~∇ · ~A = 0

However, the divergence of a curl vanishes identically. Hence the first term isautomatically equal to zero, and we are left with k2~∇ · ~A = 0 or (upon dividingby the constant k) ~∇ · ~A = 0.

We now return to the first equation and simplify the double curl using the BAC–CAB rule (taking into account the fact that all derivatives must act on ~A)

~∇× (~∇×A) = ~∇(~∇ · ~A)−∇2 ~A (3)

As a result, the first equation becomes

~∇(~∇ · ~A)−∇2 ~A− k2 ~A = 0

However, we have shown above that ~∇· ~A = 0 for this problem. Thus (3) reducesto

∇2 ~A+ k2 ~A = 0

which is what we wanted to show.

1.10.4 Evaluate∮~r · d~r

We have evaluated this integral in class. For a line integral from point 1 to point2, we have ∫ 2

1

~r · d~r = 12

∫ 2

1

d(r2) = 12r

2∣∣21

= 12r

22 − 1

2r21

However for a closed path, point 1 and point 2 are the same. Thus the integralalong a closed loop vanishes,

∮~r · d~r = 0. Note that this vanishing of the line

integral around a closed loop is the sign of a conservative force.

Alternatively, we can apply Stokes’ theorem∮~r · d~r =

∫S

~∇× ~r · d~σ

It is easy to see that ~r is curl-free. Hence the surface integral on the right handside vanishes.

1.12.9 Prove that ∮u~∇v · d~λ = −

∮v~∇u · d~λ

This is an application of Stokes’ theorem. Let us write∮(u~∇v + v~∇u) · d~λ =

∫S

~∇× (u~∇v + v~∇u) · d~σ (4)

We now expand the curl using

~∇× (u~∇v) = (~∇u)× (~∇v) + u~∇× ~∇v = (~∇u)× (~∇v)

where we have also used the fact that the curl of a gradient vanishes. Returningto (4), this indicates that∮

(u~∇v + v~∇u) · d~λ =∫

S

[(~∇u)× (~∇v) + (~∇v)× (~∇u)] · d~σ = 0

where the vanishing of the right hand side is guaranteed by the antisymmetry ofthe cross-product, ~A× ~B = − ~B × ~A.

Chapter 2

2.4.8 Find the circular cylindrical components of the velocity and acceleration of a movingparticle

We first explore the time derivatives of the cylindrical coordinate basis vectors.Since

ρ = (cosϕ, sinϕ, 0), ϕ = (− sinϕ, cosϕ, 0), z = (0, 0, 1)

their derivatives are

∂ρ

∂ϕ= (− sinϕ, cosϕ, 0) = ϕ,

∂ϕ

∂ϕ= (− cosϕ,− sinϕ, 0) = −ρ

Using the chain rule, this indicates that

˙ρ =∂ρ

∂ϕϕ = ϕϕ, ˙ϕ =

∂ϕ

∂ϕϕ = −ρϕ (5)

Now, we note that the position vector is given by

~r = ρρ+ zz

So all we have to do to find the velocity is to take a time derivative

~v = ~r = ρρ+ zz + ˙ρρ+ ˙zz = ρρ+ zz + ϕρϕ

Note that we have used the expression for ˙ρ in (5). Taking one more time deriva-tive yields the acceleration

~a = ~v = ρρ+ zz + ϕ(ρϕ+ ρϕ) + ˙ρρ+ ˙zz + ˙ϕρϕ

= ρρ+ zz + ϕ(ρϕ+ ρϕ) + ϕρϕ− ρρϕ2

= ρ(ρ− ρϕ2) + zz + ϕ(ρϕ+ 2ρϕ)

2.4.11 For the flow of an incompressible viscous fluid the Navier-Stokes equations lead to

−~∇× (~v × (~∇× ~v )) =η

ρ0∇2(~∇× ~v )

Here η is the viscosity and ρ0 the density of the fluid. For axial flow in a cylindricalpipe we take the velocity ~v to be

~v = zv(ρ)

From Example 2.4.1~∇× (~v × (~∇× ~v )) = 0

for this choice of ~v. Show that∇2(~∇× ~v) = 0

leads to the differential equation

1ρ

d

dρ

(ρd2v

dρ2

)− 1ρ2

dv

dρ= 0

and that this is satisfied byv = v0 + a2ρ

2

This problem is an exercise in applying the vector differential operators in cylin-drical coordinates. Let us first compute ~V = ~∇× ~v

~V = ~∇× ~v =1ρ

∣∣∣∣∣∣∣ρ ρϕ z∂

∂ρ

∂

∂ϕ

∂

∂z0 0 v(ρ)

∣∣∣∣∣∣∣ = −ϕdvdρ

⇒ Vϕ = −dvdρ

Note that, since v(ρ) is a function of a single variable, partial derivatives of v arethe same as ordinary derivatives. Next we need to compute the vector Laplacian∇2(~∇ × ~v ) = ∇2~V . Using (2.35) in the textbook, and the fact that on the Vϕ

component is non-vanishing, we find

(∇2~V )ρ = − 2ρ2

∂Vϕ

∂ϕ= 0

(∇2~V )ϕ = ∇2(Vϕ)− 1ρ2Vϕ = −∇2

(dv

dρ

)+

1ρ2

dv

dρ

(∇2~V )z = 0

This indicates that only the ϕ component of the vector Laplacian gives a non-trivial equation. Finally, we evaluate the scalar Laplacian ∇2(dv/dρ) to obtain

(∇2~V )ϕ = −1ρ

d

dρ

(ρd2v

dρ2

)+

1ρ2

dv

dρ(6)

Setting this equal to zero gives the equation that we were asked to prove.

To prove that v = v0 + a2ρ2 satisfies the (third order!) differential equation, all

we have to do is substitute it in. However, it is more fun to go ahead and solve

the equation. First we notice that v only enters through its derivative f = dv/dρ.Substituting this into (6), we find

1ρ

d

dρ

(ρdf

dρ

)− 1ρ2f = 0

Expanding the derivatives in the first term yields

d2f

dρ2+

1ρ

df

dρ− 1ρ2f = 0

Since this is a homogeneous equation, we may substitute in f = ρα to obtain thealgebraic equation

α(α− 1) + α− 1 = 0 ⇒ α = ±1

This indicates that the general solution for f(ρ) is of the form

f = 2aρ+ bρ−1

where the factor of 2 is chosen for later convenience. Integrating f once to obtainv, we find

v =∫f dρ = v0 + aρ2 + b log ρ

which agrees with the given solution, except for the log term. However, now wecan appeal to physical boundary conditions for fluid flow in the cylindrical pipe.The point ρ = 0 corresponds to the central axis of the pipe. At this point, thefluid velocity should not be infinite. Hence we must throw away the log, or inother words we must set b = 0, so that v = v0 + aρ2.

Incidentally, the fluid flow boundary conditions should be that the velocity van-ishes at the wall of the pipe. If we let R be the radius of the pipe, this meansthat we can write the solution as

v(ρ) = vmax

(1− ρ2

R2

)

where the maximum velocity vmax is for the fluid along the central axis (with thevelocity going to zero quadratically as a function of the radius).




Chapter 2

2.5.11 A particle m moves in response to a central force according to Newton’s second law

m~r = r f(~r )

Show that ~r × ~r = ~c, a constant, and that the geometric interpretation of this leadsto Kepler’s second law.

Actually, ~r× ~r is basically the angular momentum, ~L = ~r× ~p = m~r× ~r. To showthat ~L is constant, we can take its time derivative

~L =d

dt(m~r × ~r ) = m~r × ~r + m~r × ~r

The first cross-product vanishes. So, by using Newton’s second law, we end upwith

~L = ~r × r f(~r ) = (~r × ~r )f(~r )

r= 0

This indicates that the angular momentum ~L is a constant in time (ie that it isconserved). The constant vector ~c of the problem is just ~L/m. Note that thisproof works for any central force, not just the inverse square force law.

For the geometric interpretation, consider the orbit of the particle m

dr

r

The amount of area swept out by the particle is given by the area of the triangle

dA = 12 |~r × d~r |

So the area swept out in a given time dt is simply

dA

dt= 1

2

∣∣∣∣~r × d~r

dt

∣∣∣∣ = 12 |~r × ~r |

Since this is a constant, we find that equal areas are swept out in equal times.This is just Kepler’s second law (which is also the law of conservation of angularmomentum).

2.6.5 The four-dimensional, fourth-rank Riemann-Christoffel curvature tensor of generalrelativity Riklm satisfies the symmetry relations

Riklm = −Rikml = −Rkilm

With the indices running from 0 to 3, show that the number of independent compo-nents is reduced from 256 to 36 and that the condition

Riklm = Rlmik

further reduces the number of independent components to 21. Finally, if the com-ponents satisfy an identity Riklm + Rilmk + Rimkl = 0, show that the number ofindependent components is reduced to 20.

Here we just have to do some counting. For a general rank-4 tensor in fourdimensions, since each index can take any of four possible values, the number ofindependent components is simply

independent components = 44 = 256

Taking into account the first symmetry relation, the first part

Riklm = −Rikml

indicates that the Riemann tensor is antisymmetric when the last pair of indicesis switched. Thinking of the last pair of indices as specifying a 4×4 antisymmetricmatrix, this means instead of having 42 = 16 independent elements, we actuallyonly have 1

2 (4)(3) = 6 independent choices for the last index pair (this is thenumber of elements in an antisymmetric 4×4 matrix). Similarly, the second partof the first symmetry relation

Riklm = −Rkilm

indicates that the Riemann tensor is antisymmetric in the first pair of indices. Asa result, the same argument gives only 6 independent choices for the first indexpair. This accounts for

independent components = 6 · 6 = 36

We are now able to handle the second condition

Riklm = Rlmik

By now, it should be obvious that this statement indicates that the Riemanntensor is symmetric when the first index pair is interchanged with the second

index pair. The counting of independent components is then the same as that fora 6 × 6 symmetric matrix. This gives

independent components = 12 (6)(7) = 21

Finally, the last identity is perhaps the trickiest to deal with. As indicated in thenote, this only gives additional information when all four indices are different.Setting iklm to be 0123, this gives

R0123 + R0231 + R0312 = 0 (1)

As a result, this can be used to remove one more component, leading to

independent components = 21 − 1 = 20

We can, of course, worry that a different combination of iklm (say 1302 or some-thing like that) will give further relations that can be used to remove additionalcomponents. However, this is not the case, as can be seen by applying the firstto relations.

Note that it is an interesting exercise to count the number of independent com-ponents in the Riemann tensor in d dimensions. The result is

independent components for d dimensions = 112d2(d2 − 1)

Putting in d = 4 yields the expected 20. However, it is fun to note that puttingin d = 1 gives 0 (you cannot have curvature in only one dimension) and puttingin d = 2 gives 1 (there is exactly one independent measure of curvature in twodimensions).

2.9.6 a) Show that the inertia tensor (matrix) of Section 3.5 may be written

Iij = m(r2δij − xixj) [typo corrected!]

for a particle of mass m at (x1, x2, x3).

Note that, for a single particle, the inertia tensor of Section 3.5 is specified as

Ixx = m(r2 − x2), Ixy = −mxy, etc

Using i = 1, 2, 3 notation, this is the same as indicating

Iij ={

m(r2 − x2i ) i = j

−mxixj i 6= j

We can enforce the condition i = j by using the Kronecker delta, δij . Similarly,the condition i 6= j can be enforced by the ‘opposite’ expression 1 − δij . Thismeans we can write

Iij = m(r2 − x2i )δij − mxixj(1 − δij) (no sum)

distributing the factors out, and noting that it is safe to set xixjδij = x2i δij , we

end up with

Iij = mr2δij − mx2i δij − mxixj + mx2

i δij = m(r2δij − xixj)

Note that there is a typo in the book’s version of the homework exercise!

b) Show thatIij = −MilMlj = −mεilkxkεljmxm

where Mil = m1/2εilkxk. This is the contraction of two second-rank tensors andis identical with the matrix product of Section 3.2.

We may calculate

−MilMlj = −mεilkxkεljmxm = −mεlkiεljmxkxm

Note that the product of two epsilons can be re-expressed as

εlkiεljm = δkjδim − δkmδij (2)

This is actually the BAC–CAB rule in index notation. Hence

−MilMlj = −m(δkjδim − δkmδij)xkxm = −m(δkjxkδimxm − δkmxkxmδij)

= −m(xjxi − xkxkδij) = m(r2δij − xixj) = Iij

Note that we have used the fact that xkxk = x21 + x2

2 + x23 = r2.

2.9.12 Given Ak = 12εijkBij with Bij = −Bji, antisymmetric, show that

Bmn = εmnkAk

Given Ak = 12εijkBij , we compute

εmnkAk = 12εmnkεkijBij = 1

2εkmnεkijBij = 12 (δmiδnj − δmjδni)Bij

= 12 (Bmn − Bnm) = Bmn

We have used the antisymmetric nature of Bij in the last step.

2.10.6 Derive the covariant and contravariant metric tensors for circular cylindrical coordi-nates.

There are several ways to derive the metric. For example, we may use the relationbetween Cartesian and cylindrical coordinates

x = ρ cos ϕ, y = ρ sinϕ, z = z (3)

to compute the differentials

dx = dρ cos ϕ − ρ sinϕ dϕ, dy = dρ sinϕ + ρ cos ϕ dϕ, dz = dz

The line element is then

ds2 = dx2 + dy2 + dz2 = (dρ cos ϕ − ρ sinϕ dϕ)2 + (dρ sinϕ + ρ cos ϕ dϕ)2 + dz2

= dρ2 + ρ2dϕ2 + dz2

Since ds2 = gijdxidxj [where (x1, x2, x3) = (ρ, ϕ, z)] we may write the covariantmetric tensor (matrix) as

gij =

1 0 00 ρ2 00 0 1

(4)

Alternatively, the metric is given by gij = ~ei · ~ej where the basis vectors are

~ei =∂~r

∂xi

Taking partial derivatives of (3), we obtain

~eρ = x cos ϕ + y sinϕ

~eϕ = ρ(−x sinϕ + y cos ϕ)~ez = z

Then

gρρ = ~eρ · ~eρ = (x cos ϕ + y sinϕ) · (x cos ϕ + y sinϕ) = cos2 ϕ + sin2 ϕ = 1gρϕ = ~eρ · ~eϕ = (x cos ϕ + y sinϕ) · ρ(−x sinϕ + y cos ϕ)

= ρ(− cos ϕ sinϕ + sinϕ cos ϕ) = 0etc . . .

The result is the same as (4).

The contravariant components of the metric is given by the matrix inverse of (4)

gij =

1 0 00 ρ−2 00 0 1

(5)

2.10.11 From the circular cylindrical metric tensor gij calculate the Γkij for circular cylindrical

coordinates.

We may compute the Christoffel components using the expression

Γijk = 12 (∂kgij + ∂jgik − ∂igjk)

However, instead of working out all the components one at a time, it is more effi-cient to examine the metric (4) and to note that the only non-vanishing derivativeis

∂ρgϕϕ = 2ρ

This indicates that the only non-vanishing Christoffel symbols Γijk are the oneswhere the three indices ijk are some permutation of ρϕϕ. It is then easy to seethat

Γρϕϕ = −ρ, Γϕρϕ = Γϕϕρ = ρ

Finally, raising the first index using the inverse metric (5) yields

Γρϕϕ = −ρ, Γϕ

ρϕ = Γϕϕρ =

1ρ

(6)

Note that the Christoffel symbols are symmetric in the last two indices.

2.10.12 Using the Γkij from Exercise 2.10.11, write out the covariant derivatives V i

;j of avector ~V in circular cylindrical coordinates.

Recall that the covariant derivative of a contravariant vector is given by

V i;j = V i

,j + ΓijkV k

where the semi-colon indicates covariant differentiation and the comma indicatesordinary partial differentiation. To work out the covariant derivative, we justhave to use (6) for the non-vanishing Christoffel connections. The result is

V ρ;ρ = V ρ

,ρ + ΓρρkV k = V ρ

,ρ

V ϕ;ρ = V ϕ

,ρ + ΓϕρkV k = V ϕ

,ρ + ΓϕρϕV ϕ = V ϕ

,ρ +1ρV ϕ

V z;ρ = V z

,ρ + ΓzρkV k = V z

,ρ

V ρ;ϕ = V ρ

,ϕ + ΓρϕkV k = V ρ

,ϕ + ΓρϕϕV ϕ = V ρ

,ϕ − ρV ϕ

V ϕ;ϕ = V ϕ

,ϕ + ΓϕϕkV k = V ϕ

,ϕ + ΓϕϕρV

ρ = V ϕ,ϕ +

1ρV ρ

V z;ϕ = V z

,ϕ + ΓzϕkV k = V z

,ϕ

V ρ;z = V ρ

,z + ΓρzkV k = V ρ

,z

V ϕ;z = V ϕ

,z + ΓϕzkV k = V ϕ

,z

V z;z = V z

,z + ΓzzkV k = V z

,z

Note that, corresponding to the three non-vanishing Christoffel symbols, onlythree of the expressions are modified in the covariant derivative.



Textbook problems: Ch. 5: 5.1.1, 5.1.2

Chapter 5

5.1.1 Show that∞∑

n=1

1(2n− 1)(2n + 1)

=12

We take the hint and use mathematical induction. First, we assume that

sm =m

2m + 1(1)

In this case, the next partial sum becomes

sm+1 = sm + am+1 =m

2m + 1+

1(2(m + 1)− 1)(2(m + 1) + 1)

=m

2m + 1+

1(2m + 1)(2m + 3)

=m(2m + 3) + 1

(2m + 1)(2m + 3)

=2m2 + 3m + 1

(2m + 1)(2m + 3)=

(m + 1)(2m + 1)(2m + 1)(2m + 3)

=(m + 1)

2(m + 1) + 1

which is of the correct form (1). Finally, by explicit computation, we see thats1 = 1/(1 · 3) = 1/3 = 1/(2 · 1 + 1), so that (1) is correct for s1. Therefore, byinduction, we conclude that the mth partial sum is exactly sm = m/(2m + 1).

It is now simple to take the limit to obtain

S = limm→∞

sm = limm→∞

m

2m + 1=

12

Note that we could also have evaluated this sum by partial fraction expansion

∞∑n=1

1(2n− 1)(2n + 1)

=∞∑

n=1

(1

2(2n− 1)− 1

2(2n + 1)

)Since this is a telescoping series, we have

sm =1

2(2 · 1− 1)− 1

2(2m + 1)=

m

2m + 1

which agrees with (1).

5.1.2 Show that∞∑

n=1

1n(n + 1)

= 1

This problem may be solved in a similar manner. While there is no hint of forthe partial sum, we may try a few terms

s1 =12, s2 = s1 +

12 · 3

=23, s3 = s2 +

13 · 4

=34

This suggests thatsm =

m

m + 1(2)

We now prove this statement by induction. Starting from sm, we compute

sm+1 = sm + am+1 =m

m + 1+

1(m + 1)(m + 2)

=m(m + 2) + 1

(m + 1)(m + 2)

=(m + 1)2

(m + 1)(m + 2)=

m + 1m + 2

=(m + 1)

(m + 1) + 1

Therefore if (2) holds for m, it also holds for m + 1. Finally, since (2) is correctfor s1 = 1/2, it must be true for all m by induction.

Taking the limit yields

S = limm→∞

sm = limm→∞

m

m + 1= 1

The partial fraction approach to this problem is to note that

∞∑n=1

1n(n + 1)

=∞∑

n=1

(1n− 1

n + 1

)Hence

sm =11− 1

m + 1=

m

m + 1

which reproduces (2).

Additional Problems

1. The metric for a three-dimensional hyperbolic (non-Euclidean) space can be writtenas

ds2 = L2 dx2 + dy2 + dz2

z2

where L is a constant with dimensions of length. Calculate the non-vanishing Christof-fel coefficients for this metric.

We first note that the metric is given in matrix form as

gij =

L2/z2 0 00 L2/z2 00 0 L2/z2

so the non-zero components are

gxx =L2

z2, gyy =

L2

z2, gzz =

L2

z2(3)

The covariant components of the Christoffel connection are obtained from themetric by

Γijk = 12 (gij,k + gik,j − gjk,i)

where the comma denotes partial differentiation. According to (3), the only non-zero metric components have repeated indices. In addition, only the z-derivativeis non-vanishing. Hence we conclude that the only non-vanishing Christoffel sym-bols must have two repeated indices combined with a z index. Recalling that Γijk

is symmetric in the last two indices, we compute

Γzxx = − 12gxx,z =

L2

z3, Γxzx = Γxxz = 1

2gxx,z = −L2

z3

Γzyy = − 12gyy,z =

L2

z3, Γyzy = Γyyz = 1

2gyy,z = −L2

z3

Γzzz = 12gzz,z = −L2

z3

Raising the first index using the inverse metric gij = (z2/L2)δij finally yields

Γzxx =

1z, Γx

zx = Γxxz = −1

z

Γzyy =

1z, Γy

zy = Γyyz = −1

z

Γzzz = −1

z

(4)

2. The motion of a free particle moving along a path xi(t) in hyperbolic space is governedby the geodesic equation

xi(t) + Γijkxj(t)xk(t) = 0

Taking (x1, x2, x3) to be (x, y, z), and using the Christoffel coefficients calculatedabove, show that the geodesic equation is given explicitly by

x− 2zxz = 0

y − 2zyz = 0

z +1z(x2 + y2 − z2) = 0

Using the Christoffel coefficients in (4), we compute the three components of thegeodesic equation

x + Γxxzxz + Γx

zxzx = 0 ⇒ x− 2zxz = 0 (5)

y + Γyyz yz + Γy

zy zy = 0 ⇒ y − 2zyz = 0 (6)

z + Γzxxxx + Γz

yy yy + Γzzz zz = 0 ⇒ z +

1z(x2 + y2 − z2) = 0 (7)

The geodesic equation is important because it describes the motion of free parti-cles in curved space. However, for this problem, all that is necessary is to showthat it gives a system of coupled ordinary differential equations (5), (6), (7).

3. Show that a solution to the geodesic equation of Problem 2 is given by

x = x0 + R cos ϕ tanh(v0t)y = y0 + R sinϕ tanh(v0t)z = R sech(v0t)

where x0, y0, R, ϕ and v0 are constants. Show that the path of the particle lies on asphere of radius R centered at (x0, y0, 0) in the Cartesian coordinate space given by(x, y, z). Note that this demonstrates the non-Euclidean nature of hyperbolic space;in reality the ‘sphere’ is flat, while the space is curved.

It should be a straightforward exercise to insert the x, y and z equations into (5),(6) and (7) to show that it is a solution. However it is actually more interestingto solve the equations directly. We start with the x equation, (5). If we aresomewhat clever, we could rewrite (5) as

x

x= 2

z

z⇒ d

dtlog x = 2

d

dtlog z

Both sides of this may be integrated in time to get

x = axz2 (8)

where ax is a constant. It should be clear that the y equation, (6) can be workedon in similar manner to get

y = ayz2 (9)

Of course, we have not yet completely solved for x and y. But we are a step closerto the solution. Now, inserting (8) and (9) into the z equation, (7), we obtain

z + (a2x + a2

y)z3 − z2

z= 0

This non-linear differential equation can be simplified by performing the substi-tution z(t) = 1/u(t). Noting that

z = − u

u2, z = − u

u2+ 2

u2

u3

the z equation may be rewritten as

uu− u2 = (a2x + a2

y)

While this equation is still non-linear, it is possible to obtain a general solution

u(t) =1v0

√a2

x + a2y cosh(v0(t− t0))

where v0 and t0 are constants.

Given the solution for z = 1/u, we now insert this back into (8) to obtain

x =ax

u2=

v20ax

a2x + a2

y

sech2(v0(t− t0))

which may be integrated to yield

x(t) = x0 +v0ax

a2x + a2

y

tanh(v0(t− t0))

Similarly, for y, we integrate (9) to find

y(t) = y0 +v0ay

a2x + a2

y

tanh(v0(t− t0))

Note that the three (coupled) second order differential equations give rise to sixconstants of integration, (x0, y0, ax, ay, v0, t0). The expressions may be simplifiedby defining

ax =v0

Rcos ϕ, ay =

v0

Rsinϕ

in which case we see that

x = x0 + R cos ϕ tanh(v0(t− t0))y = y0 + R sinϕ tanh(v0(t− t0))z = R sech(v0(t− t0))

which is the answer we wanted to show, except that here we have retained anextra constant t0 related to the time translation invariance of the system.

Finally, to show that the path of the particle lies in a sphere, all we need todemonstrate is that

(x− x0)2 + (y − y0)2 + z2

= R2 cos2 ϕ tanh2(v0t) + R2 sin2 ϕ tanh2(v0t) + R2 sech2(v0t)

= R2(tanh2(v0t) + sech2(v0t)) = R2

This is indeed the equation for a sphere, (x− x0)2 + (y − y0)2 + z2 = R2.



Textbook problems: Ch. 5: 5.2.6, 5.2.8, 5.2.9, 5.2.19, 5.3.1

Chapter 5

5.2.6 Test for convergence

a)∞∑

n=2

(lnn)−1

As in all these convergence tests, it is good to first have a general idea of whetherwe expect this to converge or not, and then find an appropriate test to confirmour hunch. For this one, we can imagine that lnn grows very slowly, so that itsinverse goes to zero very slowly — too slowly, in fact, to converge. To prove this,we can perform a simple comparison test. Since lnn < n for n ≥ 2, we see that

an = (lnn)−1 > n−1

since the harmonic series diverges, and each term is larger than the correspondingharmonic series term, this series must diverge.

Note that in this and all subsequent tests, there may be more than one way toprove convergence/divergence. Your solution may be different than that givenhere. But any method is okay, so long as the calculations are valid.

b)∞∑

n=1

n!10n

In this case, when n gets large (which is the only limit we care about), the factorialin the numerator will start to dominate over the power in the denominator. Sowe expect this to diverge. As a proof, we can perform a simple ratio test.

an =n!

10n⇒ an

an+1=

10n + 1

Taking the limit, we obtainlim

n→∞

an

an+1= 0

hence the series diverges by the ratio test.

c)∞∑

n=1

12n(2n + 1)

We first note that this series behaves like 1/4n2 for large n. As a result, we expectit to converge. To see this, we may consider a simple comparison test

an =1

2n(2n + 1)<

12n · 2n

=14

(1n2

)

Since the series ζ(2) =∑∞

n=1(1/n2) converges, this series converges as well.

d)∞∑

n=1

[n(n + 1)]−1/2

This series behaves as 1/n for large n. Thus we expect it to diverge. Whilethe square root may be a bit awkward to manipulate, we can actually perform asimple comparison test with the harmonic series

an =1√

n(n + 1)>

1√(n + 1)(n + 1)

=1

n + 1

Because the harmonic series diverges (and we do not care that the comparisonstarts with the second term in the harmonic series, and not the first) this seriesalso diverges.

e)∞∑

n=0

12n + 1

Since this behaves as 1/2n for large n, the series ought to diverge. We may eithercompare this with the harmonic series or perform an integral test. Consider theintegral test ∫ ∞

0

dx

2x + 1=

12

ln(2x + 1)∣∣∣∣∞0

= ∞

Thus the series diverges

5.2.8 For what values of p and q will the following series converge?∞∑

n=2

1/ [np(lnn)q]

Since the lnn term is not as dominant as the power term np, we may have someidea that the series ought to converge or diverge as the 1/np series. To make this

more precise, we can use Raabe’s test

an =1

np(lnn)q⇒ an

an+1=

(n + 1)p(ln(n + 1))q

np(lnn)q

=(

1 +1n

)p (1 +

ln(1 + 1n )

lnn

)q

=(

1 +1n

)p (1 +

1n lnn

+ · · ·)q

=(1 +

p

n+ · · ·

) (1 +

q

n lnn+ · · ·

)=

(1 +

p

n+

q

n lnn+ · · ·

)Note that we have Taylor (or binomial) expanded the expressions several times.Raabe’s test then yields

limn→∞

n

(an

an+1− 1

)= lim

n→∞

(p +

q

lnn+ · · ·

)= p

This gives convergence for p > 1 and divergence for p < 1.

For p = 1, Raabe’s test is ambiguous. However, in this case we can perform anintegral test. Since

p = 1 ⇒ an =1

n(lnn)q

we evaluate ∫ ∞2

dx

x(lnx)q=

∫ ∞ln 2

du

uq

where we have used the substitution u = lnx. This converges for q > 1 anddiverges otherwise. Hence the final result is

p > 1, any q convergep = 1, q > 1 convergep = 1, q ≤ 1 divergep < 1, any q diverge

5.2.9 Determine the range of convergence for Gauss’s hypergeometric series

F (α, β, γ;x) = 1 +αβ

1!γx +

α(α + 1)β(β + 1)2!γ(γ + 1)

x2 + · · ·

We first consider non-negative values of x (so that this is a positive series). Moreor less, this is a power series in x. So as long as α, β, γ are well behaved, this

series ought to converge for x < 1 (just like an ordinary geometric series). To seethis (and to prepare for Gauss’ test), we compute the ratio

an =α(α + 1) · · · (α + n− 1)β(β + 1) · · · (β + n− 1)

n!γ(γ + 1) · · · (γ + n− 1)xn

⇒ an

an+1=

(n + 1)(γ + n)(α + n)(β + n)

x−1

This allows us to begin with the ratio test

limn→∞

an

an+1= lim

n→∞

(n + 1)(γ + n)(α + n)(β + n)

x−1 = x−1

Hence the series converges for x < 1 and diverges for x > 1. However, the ratiotest is indeterminate for x = 1. This is where we must appeal to Gauss’ test.Setting x = 1, we have

an

an+1=

(n + 1)(γ + n)(α + n)(β + n)

Since this approaches 1 as n → ∞, we may highlight this leading behavior byadding and subtracting 1

an

an+1= 1 +

((n + 1)(γ + n)(α + n)(β + n)

− 1)

= 1 +(γ − α− β + 1)n + γ − αβ

(α + n)(β + n)

We can now see that the fraction approaches (γ−α−β+1)/n as n gets large. Thisis the h/n behavior that we need to extract for Gauss’ test: an/an+1 = 1+h/n+B(n)/n2. In principle, we may add and subtract h/n where h = γ −α− β + 1 inorder to obtain an explicit expression for the remainder term B(n)/n2. However,it should be clear based on a power series expansion that this remainder willindeed behave as ∼ 1/n2, which is the requirement for applying Gauss’ test.Thus, with h = γ −α− β + 1, we see that the hypergeometric series F (α, β, γ; 1)converges for γ > α + β (h > 1) and diverges otherwise.

To summarize, we have proven that for non-negative x, the hypergeometric seriesconverges for x < 1 (any α, β, γ) and x = 1 if γ > α + β, and diverges otherwise.In fact, for negative values of x, we may consider the series for |x|. In this case,we have absolute convergence for |x| < 1 and |x| = 1 if γ > α + β. Based onthe ratio test, it is not hard to see that the series also diverges for |x| > 1 (fornegative x, each subsequent term gets larger than the previous one). However,there is also conditional convergence for α + β − 1 < γ ≤ α + β (this is harder toshow).

5.2.19 Show that the following series is convergent.

∞∑s=0

(2s− 1)!!(2s)!!(2s + 1)

It is somewhat hard to see what happens when s gets large. However, we canperform Raabe’s test

as =(2s− 1)!!

(2s)!!(2s + 1)⇒ as

as+1=

(2s− 1)!!(2s)!!(2s + 1)

× (2s + 2)!!(2s + 3)(2s + 1)!!

=(2s− 1)!!(2s + 2)!!(2s + 3)(2s + 1)!! (2s)!! (2s + 1)

=(2s + 2)(2s + 3)(2s + 1)(2s + 1)

By adding and subtracting 1, we obtain

as

as+1= 1 +

((2s + 2)(2s + 3)

(2s + 1)2− 1

)= 1 +

6s + 5(2s + 1)2

Then

lims→∞

s

(as

as+1− 1

)= lim

s→∞s

(6s + 5

(2s + 1)2

)=

32

Since this is greater than 1, the series converges.

5.3.1 a) From the electrostatic two hemisphere problem we obtain the series

∞∑s=0

(−1)s(4s + 3)(2s− 1)!!(2s + 2)!!

Test it for convergence.

Since this is an alternating series, we may check if it is monotonic decreasing.Taking the ratio, we see that

|as||as+1|

=(4s + 3)(2s− 1)!!(2s + 4)!!(4s + 7)(2s + 1)!!(2s + 2)!!

=(4s + 3)(2s + 4)(4s + 7)(2s + 1)

=8s2 + 22s + 128s2 + 18s + 7

= 1 +4s + 5

8s2 + 18s + 7> 1

As a result|as| > |as+1|

and hence the series converges based on the Leibniz criterion. (Actually, to becareful, we must also show that lims→∞ as = 0. However, I have ignored thissubtlety.)

b) The corresponding series for the surface charge density is

∞∑s=0

(−1)s(4s + 3)(2s− 1)!!

(2s)!!

Test it for convergence.

This series is rather similar to that of part a). However the denominator is‘missing’ a factor of (2s + 2). This makes the series larger (term by term) thanthe above. To see whether the terms get too large, we may take the ratio

|as||as+1|

=(4s + 3)(2s− 1)!!(2s + 2)!!(4s + 7)(2s + 1)!! (2s)!!

=(4s + 3)(2s + 2)(4s + 7)(2s + 1)

=8s2 + 14s + 68s2 + 18s + 7

= 1− 4s + 18s2 + 18s + 7

< 1

In this case|as| < |as+1|

and the series diverges since the terms get larger as s →∞.




Chapter 5

5.4.1 Given the series

ln(1 + x) = x− x2

2+

x3

3− x4

4+ · · · , −1 < x ≤ 1

show that

ln(

1 + x

1− x

)= 2

(x +

x3

3+

x5

5+ · · ·

), −1 < x < 1

We use the property ln(a/b) = ln a− ln b to write

ln(

1 + x

1− x

)= ln(1 + x)− ln(1− x) =

∞∑n=1

(−1)n+1 xn

n−∞∑

n=1

(−1)n+1 (−x)n

n

=∞∑

n=1

((−1)n+1 + 1)xn

n= 2

∑n odd

xn

n

Note that, since we use the ln(1 + x) series for both +x and −x, the commonrange of convergence is the intersection of −1 < x ≤ 1 and −1 < −x ≤ 1, namely|x| < 1.

5.4.2 Determine the values of the coefficients a1, a2, and a3 that will make (1+a1x+a2x2 +

a3x3) ln(1 + x) converge as n−4. Find the resulting series.

Using the expansion for ln(1 + x), we write

(1 + a1x + a2x2+a3x

3) ln(1 + x)

=∞∑

n=1

(−1)n+1

(xn

n+

a1xn+1

n+

a2xn+2

n+

a3xn+3

n

)

We want to collect identical powers of x on the right hand side. To do this, wemust shift the index n according to n → n − 1, n → n − 2 and n → n − 3 forthe second, third and last terms on the right hand side, respectively. After doing

so, we may combine terms with powers x4 and higher. The first few terms (x, x2

and x3) may be treated as exceptions. The result is

(1 + a1x + a2x2+a3x

3) ln(1 + x)

= (x− 12x2 + 1

3x3) + a1(x2 − 12x3) + a2x

3

+∞∑

n=4

(−1)n+1

(xn

n− a1x

n

n− 1+

a2xn

n− 2− a3x

n

n− 3

)= x + (a1 − 1

2 )x2 + (a2 − 12a1 + 1

3 )x3

+∞∑

n=4

(−1)n+1

(1n− a1

n− 1+

a2

n− 2− a3

n− 3

)xn

(1)

Combining the terms over a common denominator yields(1n− a1

n− 1+

a2

n− 2− a3

n− 3

)

=

(n− 1)(n− 2)(n− 3)− a1n(n− 2)(n− 3) + a2n(n− 1)(n− 3)−a3n(n− 1)(n− 2)

n(n− 1)(n− 2)(n− 3)

=

(1− a1 + a2 − a3)n3 + (−6 + 5a1 − 4a2 + 3a3)n2

+(11− 6a1 + 3a2 − 2a3)n− 6n(n− 1)(n− 2)(n− 3)

To make this converge as n−4, we need to cancel the coefficients of n3, n2 and nin the numerator. Solving for

1− a1 + a2− a3 = 0, −6 + 5a1− 4a2 + 3a3 = 0, 11− 6a1 + 3a2− 2a3 = 0

yields the solutiona1 = 3, a2 = 3, a3 = 1

Finally, inserting this back into (1), we obtain

(1+3x+3x2 +x3) ln(1+x) = x+ 52x2 + 11

6 x3 +6∞∑

n=4

(−1)n xn

n(n− 1)(n− 2)(n− 3)

or

ln(1 + x) =x + 5

2x2 + 116 x3 + 6

∑∞n=4(−1)n xn

n(n−1)(n−2)(n−3)

(1 + x)3

5.4.3 Show that

a)∞∑

n=2

[ζ(n)− 1] = 1

Using the sum formula for the Riemann zeta function, we have

∞∑n=2

[ζ(n)− 1] =∞∑

n=2

[( ∞∑p=1

1pn

)− 1

]=∞∑

n=2

[ ∞∑p=2

1pn

]=∞∑

p=2

[ ∞∑n=2

1pn

]

where in the last step we have rearranged the order of summation. In doing so,we have now changed this to a geometric series, with sum

∞∑n=2

p−n =p−2

1− p−1=

1p(p− 1)

In this case

∞∑n=2

[ζ(n)− 1] =∞∑

p=2

1p(p− 1)

=∞∑

p=2

(1

p− 1− 1

p

)= 1

since this is a telescoping series.

b)∞∑

n=2

(−1)n[ζ(n)− 1] =12

The solution to this is similar to that of part a). The addition of (−1)n yields

∞∑n=2

(−1)n[ζ(n)− 1] =∞∑

n=2

(−1)n

[ ∞∑p=2

1pn

]=∞∑

n=2

[ ∞∑p=2

1(−p)n

]=∞∑

p=2

[ ∞∑n=2

1(−p)n

]

The sum over n is still a geometric series, this time with

∞∑n=2

(−p)−n =(−p)−2

1− (−p)−1=

1p(p + 1)

In this case

∞∑n=2

(−1)n[ζ(n)− 1] =∞∑

p=2

1p(p + 1)

=∞∑

p=2

(1p− 1

p + 1

)=

12

5.5.2 For what range of x is the geometric series∑∞

n=0 xn uniformly convergent?

We use the Weierstrass M test. We first note that the geometric series∑∞

n=0 xn

is absolutely convergent for |x| < 1. This means that the series∑∞

n=0 sn isconvergent for 0 ≤ s < 1. While this is all very obvious, the introduction ofthis convergent series in s allows us to bound the x series by an x-independentconvergent one. This is precisely the setup of the Weierstrass M test.

We simply choose Mn = sn. Then, so long as |x|n ≤ Mn (ie |x| ≤ s), thegeometric series is uniformly convergent. Therefore we have shown that

∑∞n=0 xn

is uniformly convergent provided |x| ≤ s < 1.

5.5.4 If the series of the coefficients∑

an and∑

bn are absolutely convergent, show thatthe Fourier series ∑

(an cos nx + bn sinnx)

is uniformly convergent for −∞ < x < ∞.

This is also a case for the Weierstrass M test. Note that, if we let α(x) =an cos nx + bn sinnx denote the n-th element of the series, then

|α(x)| = |an cos nx + bn sinnx| ≤ |an cos nx|+ |bn sinnx| ≤ |an|+ |bn|

for the entire domain x ∈ (−∞,∞). Since the problem states that∑

an and∑bn are absolutely convergent, we now take simply Mn = |an| + |bn|. Clearly,∑Mn converges, and since |α(x)| ≤ Mn, we conclude that

∑α(x) is uniformly

convergent for x ∈ (−∞,∞).




Chapter 5

5.6.2 Derive a series expansion of cot x by dividing cos x by sinx.

Since cos x = 1− 12x2 + 1

24x4 − · · · and sin x = x− 16x3 + 1

120x5 − · · ·, we divideto obtain

cot x =1− 1

2x2 + 124x4 − · · ·

x− 16x3 + 1

120x5 − · · ·=

1− 12x2 + 1

24x4 − · · ·x(1− 1

6x2 + 1120x4 − · · ·)

We now run into an issue of dividing one series by another. However, instead ofdivision, we may change this into a multiplication problem by using (1− r)−1 =1 + r + r2 + r3 + · · · to rewrite the denominator

(1− 16x2 + 1

120x4 − · · ·)−1 = 1 + ( 16x2 − 1

120x4 + · · ·) + ( 16x2 − 1

120x4 + · · ·)2+ · · ·= 1 + 1

6x2 + (− 1120 + 1

36 )x4 + · · ·= 1 + 1

6x2 + 7360x4 + · · ·

where we have only kept terms up to O(x4). Returning to cot x, we now find

cot x = x−1(1− 12x2 + 1

24x4 − · · ·)(1 + 136x2 + 7

360x4 + · · ·)= x−1(1 + (− 1

2 + 16 )x2 + ( 1

24 −112 + 7

360 )x4 + · · ·)= x−1(1− 1

3x2 − 145x4 + · · ·)

In principle, we could work this out to higher orders by keeping more powers ofx in the series expansions.

Note that there is a nice expression for cot x in terms of the Bernoulli numbers.This may be obtained by noting that the generating function definition of Bn is

x

ex − 1=∞∑

n=1

Bn

n!xn = −1

2x +

∞∑p=0

B2p

(2p)!x2p

where we have used the fact that all odd Bernoulli numbers vanish except forB1 = − 1

2 . Moving the − 12x to the left hand side, and using the identity

x

ex − 1+

12x =

x

2· ex + 1ex − 1

=x

2coth

x

2

we obtainx

2coth

x

2=∞∑

p=0

B2p

(2p)!x2p

or, by substituting x → 2x and dividing through by x

cothx =∞∑

p=0

2B2p

(2p)!(2x)2p−1

Finally, to change coth into cot, we may work in the complex domain and notethat coth iz = −i cot z. Therefore we make the substitution x → ix to yield

−i cot x =∞∑

p=0

2B2p

(2p)!(2ix)2p−1

Multiplying through by i and simplifying then gives the expression

cot x =∞∑

p=0

(−1)p22pB2p

(2p)!x2p−1

5.6.19 a) Planck’s theory of quandized oscillators leads to an average energy

〈ε〉 =

∞∑n=1

nε0 exp(−nε0/kT )

∞∑n=0

exp(−nε0/kT )

where ε0 is a fixed energy. Identify the numerator and denominator as binomialexpansions and show that the ratio is

〈ε〉 =ε0

exp(ε0/kT )− 1

To simplify the expressions, we begin with the substitution r = exp(−ε0/kT ).This yields 〈ε〉 = N/D where the numerator and denominator are

N =∞∑

n=1

nε0rn, D =

∞∑n=0

rn

We now see that the denominator is a simple geometric series. Hence D =1/(1− r). For the numerator, we note that nrn = r d

dr (rn). Hence we may write

N = ε0rd

dr

( ∞∑n=1

rn

)= ε0r

d

dr

r

1− r=

ε0r

(1− r)2

Dividing the numerator and denominator finally yields

〈ε〉 =ε0r

1− r=

ε0

r−1 − 1=

ε0

exp(ε0/kT )− 1

b) Show that the 〈ε〉 of part (a) reduces to kT , the classical result, for kT � ε0.

In this limit, ε0/KT � 1, we may expand the exponential in the denominator

exp(ε0/kT ) ≈ 1 +ε0

kT+ · · ·

As a result〈ε〉 ≈ ε0

ε0/kT + · · ·≈ kT

5.7.4 The analysis of the diffraction pattern of a circular opening involves∫ 2π

0

cos(c cos ϕ) dϕ

Expand the integrand in a series and integrate by using∫ 2π

0

cos2n ϕ dϕ =(2n)!

22n(n!)22π,

∫ 2π

0

cos2n+1 ϕ dϕ = 0

Setting x = c cos ϕ, we expand

cos x =∞∑

n=0

(−1)n

(2n)!x2n

so that∫ 2π

0

cos(c cos ϕ) dϕ =∫ 2π

0

∞∑n=0

(−1)n

(2n)!c2n cos2n ϕ dϕ

=∞∑

n=0

(−1)nc2n

(2n)!

∫ 2π

0

cos2n ϕ dϕ

=∞∑

n=0

(−1)nc2n

(2n)!2π(2n)!22n(n!)2

=∞∑

n=0

2π(−1)n

(n!)2( c

2

)2n

5.7.15 The Klein-Nishina formula for the scattering of photons by electrons contains a termof the form

f(ε) =(1 + ε)

ε2

[2 + 2ε

1 + 2ε− ln(1 + 2ε)

ε

]Here ε = hν/mc2, the ratio of the photon energy to the electron rest mass energy.Find

limε→0

f(ε)

This problem is an exercise in taking Taylor series. Note that, if we simply setε = 0 in f(ε), the first term (1 + ε)/ε2 would diverge as ε−2. Hence this providesa hint that we should keep at least two powers of ε in any series expansion weperform. Keeping this in mind, we first work on the fraction

2 + 2ε

1 + 2ε= 2(1+ε)(1+2ε)−1 = 2(1+ε)(1−2ε+4ε2−· · ·) = 2(1−ε+2ε2+· · ·) (1)

next we turn to the log

ln(1 + 2ε)ε

= ε−1 ln(1+2ε) = ε−1(2ε− 12 (2ε)2+ 1

3 (2ε)3+ · · ·) = (2−2ε+ 83ε2−· · ·)

(2)Subtracting (2) from (1), and combining with the prefactor (1 + ε)/ε2, we find

f(ε) =(1 + ε)

ε2[2(1− ε + 2ε2 + · · ·)− (2− 2ε + 8

3ε2 − · · ·)]

=(1 + ε)

ε2[4ε2 − 8

3ε2 + · · ·] =(1 + ε)

ε2[ 43ε2 + · · ·] = 4

3 (1 + ε)[1 + · · ·]

We are now in a position to take the limit ε → 0 to obtain

limε→0

f(ε) =43

5.9.11 The integral ∫ 1

0

[ln(1− x)]2dx

x

appears in the fourth-order correction to the magnetic moment of the electron. Showthat it equals 2ζ(3).

We begin with the variable substitution

1− x = e−t, dx = e−tdt

to obtain ∫ 1

0

[ln(1− x)]2dx

x=∫ ∞

0

t2e−t

1− e−tdt

This integral involves powers and exponentials, and is not so easy to do. Thuswe expand the fraction as a series

e−t

1− e−t= e−t(1− e−t)−1 = e−t(1 + e−t + e−2t + e−3t + · · ·) =

∞∑n=1

e−nt

This gives∫ 1

0

[ln(1− x)]2dx

x=∫ ∞

0

t2

( ∞∑n=1

e−nt

)dt =

∞∑n=1

∫ ∞0

e−ntt2dt

This integral may be evaluated by integration by parts (twice). Alternatively, wemake the substitution s = nt to arrive at∫ 1

0

[ln(1− x)]2dx

x=∞∑

n=1

n−3

∫ ∞0

e−ss2ds =∞∑

n=1

n−3Γ(3) = 2ζ(3)

Here we have used the definition of the Gamma function

Γ(z) =∫ ∞

0

e−ssz−1dz

as well as the zeta function

ζ(z) =∞∑

n=1

n−z

5.10.1 Stirling’s formula for the logarithm of the factorial function is

ln(x!) =12

ln 2π +(

x +12

)lnx− x−

N∑n=1

B2n

2n(2n− 1)x1−2n

The B2n are the Bernoulli numbers. Show that Stirling’s formula is an asymptoticexpansion.

Instead of using the textbook definition of an asymptotic series∑

an(x), we aimto demonstrate the two principle facts; i) that the series diverges for fixed x whenN → ∞, and ii) that the remainder vanishes for fixed N when x → ∞. To doso, we first examine the form of an(x)

an(x) = − B2n

2n(2n− 1)x1−2n

Using the relation

B2n =(−1)n+12(2n)!

(2π)2nζ(2n)

we find

|an(x)| = 2(2n− 2)!ζ(2n)(2π)2n

x1−2n

For condition i), in order to show that the series diverges for fixed x, we mayperform the ratio test

|an||an+1|

=2(2n− 2)!ζ(2n)

(2π)2n

(2π)2n+2

2(2n)!ζ(2n + 2)x2 =

(2π)2

2n(2n− 1)ζ(2n)

ζ(2n + 2)x2 (3)

Since limn→∞ ζ(n) = 1, and since there are factors of n in the denominator, wesee that

limn→∞

|an||an+1|

= 0 (for fixed x)

and hence the ratio test demonstrates that the series diverges.

For showing condition ii), on the other hand, we suppose the series stops at termn = N . Then the error or remainder is related to the subsequent terms aN+1,aN+2, etc. However, according to (3), if we take the limit x →∞ for fixed N wehave

limx→∞

|aN ||aN+1|

= ∞ ⇒ |aN+1| →|aN |∞

= 0 as x →∞

Hence the remainder terms fall off sufficiently fast to satisfy the criteria for anasymptotic series. We thus conclude that Stirling’s formula is an asymptoticexpansion.

5.10.7 Derive the following Bernoulli number asymptotic series for the Euler-Mascheroniconstant

γ =n∑

s=1

s−1 − lnn− 12n

+N∑

k=1

B2k

(2k)n2k

Let us start by recalling the useful definition of the Euler-Mascheroni constant

γ = limn→∞

(n∑

s=1

s−1 − lnn

)Essentially, the constant γ is the difference between the sum and the integralapproximation. This suggests that we begin by inserting the function f(x) = 1/xinto the Euler-Maclauren sum formula

n∑x=1

f(x) =∫ n

1

f(x)dx +12f(1) +

12f(n) +

N∑p=1

1(2p)!

B2p[f (2p−1)(n)− f (2p−1)(1)]

− 1(2N)!

∫ 1

0

B2N (x)n−1∑ν=1

f (2N)(x + ν)dx

(4)

However, we first note that, for f(x) = 1/x we have∫ n

1

f(x)dx =∫ n

1

dx

x= ln n

as well asf (k)(x) = (−1)k k!

xk+1

Using these results, and returning to (4), we find

n∑s=1

s−1 = ln n +12

+12n−

N∑p=1

B2p

2p[n−2p − 1]−

∫ 1

0

B2N (x)n−1∑ν=1

(x + ν)2N+1dx

orn∑

s=1

s−1 − lnn =12

+12n−

N∑p=1

B2p

2p[n−2p − 1] + RN (n) (5)

where the remainder RN (n) is given by

RN (n) = −∫ 1

0

B2N (x)n−1∑ν=1

(x + ν)2N+1dx (6)

At this point, may note that the left hand side of (5) is close to the expressionwe want for the Euler-Mascheroni constant. However, we must recall that thesum formula (4) generally yields an asymptotic expansion (since the Bernoullinumbers diverge). Thus we have to be careful about the remainder term.

Of course, we can still imagine taking the limit n →∞ in (5) to obtain

γ = limn→∞

(n∑

s=1

s−1 − lnn

)=

12

+N∑

p=1

B2p

2p+ RN (∞) (7)

Noting that the remainder (6) is a sum of terms

RN (n) = −∫ 1

0

B2N (x)(

1(x + 1)2N+1

+1

(x + 2)2N+1+ · · ·+ 1

(x + n− 1)2N+1

)dx

and that the first few terms in the sum dominate, we may eliminate most (butnot all) of the remainder by subtracting (5) from (7)

γ −n∑

s=1

s−1 + lnn = − 12n

+N∑

p=1

B2p

2p

1n2p

+ [RN (∞)−RN (n)]

Finally, dropping the difference of remainders, we obtain the result

γ =n∑

s=1

s−1 − lnn− 12n

+N∑

p=1

B2p

2p

1n2p




Chapter 6

6.1.3 Prove algebraically that

|z1| − |z2| ≤ |z1 + z2| ≤ |z1|+ |z2|

Interpret this result in terms of vectors. Prove that

|z − 1| < |√

z2 − 1| < |z + 1|, for <(z) > 0

We start by evaluating |z1 + z2|2

|z1 + z2|2 = (z1 + z2)(z∗1 + z∗2) = |z1|2 + |z2|2 + z1z∗2 + z∗1z2

= |z1|2 + |z2|2 + (z1z∗2) + (z1z

∗2)∗ = |z1|2 + |z2|2 + 2<(z1z

∗2)

(1)

We now put a bound on the real part of z1z∗2 . First note that, for any complex

quantity ζ, we have |ζ|2 = (<ζ)2 + (=ζ)2 ≥ (<ζ)2. Taking a square root gives|ζ| ≥ |<ζ| or −|ζ| ≤ <ζ ≤ |ζ|. For the present case (where ζ = z1z

∗2) this gives

−|z1||z2| ≤ <(z1z∗2) ≤ |z1||z2|. Using this inequality in (1), we obtain

|z1|2 + |z2|2 − 2|z1||z2| ≤ |z1 + z2|2 ≤ |z1|2 + |z2|2 + 2|z1||z2|

or(|z1| − |z2|)2 ≤ |z1 + z2|2 ≤ (|z1|+ |z2|)2

Taking the square root then proves the triangle inequality. The reason this iscalled the triangle inequality is that, in terms of vectors, we can think of z1, z2

and z1 + z2 as the three sides of a triangle

21z2+ z

z1

z

Then the third side (|z1 + z2|) of a triangle can be no longer than the sum ofthe lengths of the other two sides (|z1| + |z2|) nor shorter than the difference oflengths (|z1| − |z2|).

Finally, for the second inequality, we start by proving that

|z + 1|2 = |z|2 + 1 + 2<z = (|z|2 + 1− 2<z) + 4<z = |z − 1|2 + 4<z > |z − 1|2

for <z > 0. This implies that |z + 1| > |z − 1| for <z > 0. The picture here isthat if z is on the right half of the complex plane then it is closer to the point 1than the point −1.

z

1+

−z

1

z

z

Given this result, it is simple to see that

|z − 1|2 < |z − 1||z + 1| < |z + 1|2

or, by taking a square root

|z − 1| < |√

(z − 1)(z + 1)| < |z + 1|

which is what we set out to prove.

6.1.7 Prove that

a)N−1∑n=0

cos nx =sin(Nx/2)

sinx/2cos(N − 1)

x

2

b)N−1∑n=0

sinnx =sin(Nx/2)

sinx/2sin(N − 1)

x

2

We may solve parts a) and b) simultaneously by taking the complex combination

S =N−1∑n=0

cos nx + iN−1∑n=0

sinnx =N−1∑n=0

(cos nx + i sinnx) =N−1∑n=0

einx

The real part of S gives part a) and the imaginary part of S gives part b). Whenwritten in this fashion, we see that S is a terminating geometric series with ratior = eix. Thus

S =N−1∑n=0

rn =1− rN

1− r=

1− eNix

1− eix=

e12 Nix(e

12 Nix − e−

12 Nix)

e12 ix(e

12 ix − e−

12 ix)

We performed the last step in order to ‘balance’ positive and negative exponentialsinside the parentheses. This is so that we may relate both the numerator anddenominator to sinα = (eiα − e−iα)/2i. The result is

S = e12 (N−1)ix sin(Nx/2)

sinx/2=

(cos 1

2 (N − 1)x + i sin 12 (N − 1)x

) sin(Nx/2)sinx/2

It should now be apparent that the real and imaginary parts are indeed thesolutions to parts a) and b).

6.2.5 Find the analytic function

w(z) = u(x, y) + iv(x, y)

a) if u(x, y) = x3 − 3xy2

We use the Cauchy-Riemann relations

∂v

∂x= −∂u

∂y= 6xy ⇒ v = 3x2y + C(y)

∂v

∂y=

∂u

∂x= 3x2 − 3y2 ⇒ v = 3x2y − y3 + D(x)

In order for these two expressions to agree, the functions C(y) and D(x) musthave the form C(y) = −y3 + c and D(x) = c where c is an arbitrary constant. Asa result, we find that v(x, y) = 3x2y − y3 + c, or

w(z) = (x3 − 3xy2) + i(3x2y − y3) + ic = z3 + ic

The constant c is unimportant.

b) v(x, y) = e−y sinx As above, we have

∂u

∂x=

∂v

∂y= −e−y sinx ⇒ u = e−y cos x + C(y)

∂u

∂y= −∂v

∂x= −e−y cos x ⇒ u = e−y cos x + D(x)

Thus we must have C(y) = D(x) = c with c a constant. The complex functionw(z) is

w(z) = c + e−y cos x + ie−y sinx = c + e−y(cos x + i sinx) = c + eix−y = c + eiz

6.2.6 If there is some common region in which w1 = u(x, y) + iv(x, y) and w2 = w∗1 =

u(x, y)− iv(x, y) are both analytic, prove that u(x, y) and v(x, y) are constants.

If u + iv and u − iv are both analytic, then they must both satisfy the Cauchy-Riemann equations. This corresponds to

∂u

∂x=

∂v

∂y,

∂u

∂y= −∂v

∂x

(from u + iv) and∂u

∂x= −∂v

∂y,

∂u

∂y=

∂v

∂x

(from u− iv). Clearly this indicates that

∂u

∂x=

∂u

∂y= 0,

∂v

∂x=

∂v

∂y= 0

Since all partial derivatives vanish, u and v can only be constants.

6.3.3 Verify that ∫ 1+i

0

z∗dz

depends on the path by evaluating the integral for the two paths shown in Fig. 6.10.

2

x

+1 iy

z

2

1

1

We perform this integral as a two-dimensional line integral∫z∗dz =

∫(x− iy)(dx + idy)

For path 1, we first integrate along the x-axis (y = 0; dy = 0) and then along they-axis (x = 1; dx = 0)∫ 1+i

0

z∗dz =∫ 1

0

(x− iy)∣∣∣y=0

dx +∫ 1

0

(x− iy)∣∣∣x=1

idy

=∫ 1

0

xdx +∫ 1

0

(i + y)dy = 12x2

∣∣∣10

+ (iy + 12y2)

∣∣∣10

= 1 + i

Similarly, for path 2, we find∫ 1+i

0

z∗dz =∫ 1

0

(x− iy)∣∣∣x=0

idy +∫ 1

0

(x− iy)∣∣∣y=1

dx

=∫ 1

0

ydy +∫ 1

0

(x− i)dx = 12y2

∣∣∣10

+ ( 12x2 − ix)

∣∣∣10

= 1− i

So we see explicitly that the integral depends on the path taken (1 + i 6= 1− i).

6.4.3 Solve Exercise 6.3.4 [∮

Cdz/(z2+z) where C is a circle defined by |z| > 1] by separating

the integrand into partial fractions and then applying Cauchy’s integral theorem formultiply connected regions.

Note that, by applying Cauchy’s integral formula to a constant function f(z) = 1,we may derive the useful expression∮

C

dz

z − z0= 2πi (2)

provided point z0 is contained inside the contour C (it is zero otherwise). Then,using partial fractions, we see that∮

C

dz

z2 + z=

∮C

dz

z(z + 1)=

∮C

(1z− 1

z + 1

)dz =

∮C

dz

z−

∮dz

z + 1

Since C is a circle of radius greater than one, it encompasses both points z0 = 0and z0 = −1. Thus, using (2), we find∮

C

dz

z2 + z= 2πi− 2πi = 0

Note that, if the radius of C is less than one, we would have encircled only thepole at z0 = 0. The result would then have been 2πi instead of zero.

6.4.4 Evaluate ∮C

dz

z2 − 1

where C is the circle |z| = 2.

Again, we use partial fractions and (2)∮C

dz

z2 − 1=

∮C

dz

(z + 1)(z − 1)=

∮C

(1/2

z − 1− 1/2

z + 1

)dz

=12

∮C

dz

z − 1− 1

2

∮C

dz

z + 1= πi− πi = 0

Here it is important that the contour of radius 2 encircles both points z0 = −1and z0 = 1.



Textbook problems: Ch. 6: 6.5.2, 6.5.8, 6.6.2, 6.6.7Ch. 7: 7.1.2, 7.1.4

Chapter 6

6.5.2 Derive the binomial expansion

(1 + z)m = 1 + mz +m(m− 1)

1 · 2z2 + · · · =

∞∑n=0

(mn

)zn

for m any real number. The expansion is convergent for |z| < 1. Why?

To derive the binomial expansion, consider generating the Taylor series for f(z)around z = 0 where f(z) = (1 + z)m. Taking derivatives of f(z), we find

f ′(z) = m(1 + z)m−1,

f ′′(z) = m(m− 1)(1 + z)m−2,

f ′′′(z) = m(m− 1)(m− 2)(1 + z)m−3,

etc.

In general, the n-th derivative is given by

f (n)(z) = m(m− 1)(m− 2) · · · (m− n + 1)(1 + z)m−n =m!

(m− n)!(1 + z)m−n

where the factorial for non-inteter m may be defined by the Gamma function,or by the expression indicated. In particular, f (n)(0) = m!/(m− n)!. Hence theTaylor series has the form

f(z) =∞∑

n=0

1n!

f (n)(0)zn =∞∑

n=0

m!n!(m− n)!

zn =∞∑

n=0

(mn

)zn

For non-integer m (but integer n), the binomial coefficient may be defined by theGamma function, or alternately by(

mn

)=

m(m− 1)(m− 2) · · · (m− n + 1)1 · 2 · 3 · · ·n

=n∏

k=1

m− k + 1k

Note that, for non-integer m, the expression (1+z)m has a branch point at z = −1.(This is explored in problem 6.6.7 below.) Since the radius of convergence of the

Taylor series is the distance to the nearest singularity, this explains why |z| < 1is necessary for convergence. For negative integer m, there is no branch point,but there is still a pole (of order |m|) at z = −1. The pole also results in a radiusof convergence of |z| < 1. On the other hand, for m a non-negative integer, theseries terminates (giving a traditional binomial expansion for (1+ z) raised to aninteger power), and the radius of convergence is infinite. This is consistent withthe absence of any singularity in this case.

6.5.8 Develop the first three nonzero terms of the Laurent expansion of

f(z) = (ez − 1)−1

about the origin.

Since the Laurent expansion is a unique result, we may obtain the expansion anyway we wish. What we can do here is to start with a Taylor expansion of thedenominator

ez − 1 = z + 12z2 + 1

6z3 + · · · = z(1 + 12z + 1

6z+ · · ·)

Hencef(z) = (ez − 1)−1 = z−1(1 + 1

2z + 16z2 + · · ·)−1

For small z, we invert the series using (1 + r)−1 = 1 − r + r2 − · · · where r =12z + 1

6z2 + · · ·. This gives

f(z) = z−1(1− ( 1

2z + 16z2 + · · ·) + ( 1

2z + 16z2 + · · ·)2 − · · ·

)= z−1

(1− 1

2z + 112z2 + · · ·

)=

1z− 1

2+

112

z + · · ·(1)

Of course, we could also take the hint and use the generating function of Bernoullinumbers to write

f(z) =1

ez − 1= z−1

(z

ez − 1

)= z−1

∞∑n=0

Bn

n!zn =

B0

z+B1+

12B2z+

16B3z

2+· · ·

Inserting B0 = 1, B1 = − 12 and B2 = 1

6 then immediately yields the last line of(1). However, this method requires us to either remember or look up the valuesof the Bernoulli numbers.

6.6.2 What part of the z-plane corresponds to the interior of the unit circle in the w-planeif

a) w =z − 1z + 1

Note that, by trying a few numbers, we can see that z = 0 gets mapped to w = −1and z = 1 gets mapped to w = 0

1z

+1z

−1z

+1z

z

z w

w =−

In fact, the unit circle in the w-plane is given by the equation |w| = 1, whichmaps to |z − 1| = |z + 1| in the z-plane. Geometrically, this is saying that thepoint z is equidistant to both +1 and −1. This can only happen on the imaginaryaxis (x = 0). Hence the imaginary axis maps to the circumference of the circle.Furthermore, since z = 1 gets mapped into the interior of the circle, we mayconclude that the right half (first and fourth quadrants) of the complex z-planegets mapped to the interior of the unit circle.

b) w =z − i

z + i

This map is similar to that of part a), except that the distances are measured tothe points +i and −i instead. Thus in this case the real axis (y = 0) gets mappedto the circle. The upper half plane (first and second quadrants) gets mapped tothe interior of the unit circle.

6.6.7 For noninteger m, show that the binomial expansion of Exercise 6.5.2 holds only fora suitably defined branch of the function (1 + z)m. Show how the z-plane is cut.Explain why |z| < 1 may be taken as the circle of convergence for the expansion ofthis branch, in light of the cut you have chosen.

Returning to the binomial expansion of f(z) = (1+z)m, we note that if w = 1+z,we end up with a function f(w) = wm which is multi-valued under w → we2πi

whenever m is nonintegral. This indicates that w = 0 is a branch point, and asuitable branch must be defined. We take the branch cut to run from w = 0along the negative real axis in the w-plane. However, the simple transformationz = w − 1 allows us to return to the original z-plane. In this case, w = 0 is thesame as z = −1, so the branch point is at z = −1, with a cut running to the left.The picture of the cut z-plane is then as follows

1

+z

1

z

cut α

−

z

where the principle value is taken to be −π < α ≤ π. In this case, f(z) =

|1 + z|meimα. Since the Taylor series is expanded about z = 0, the radius ofconvergence is |z| < 1, which is the distance to the nearest singularity (the branchpoint at z = −1). This is why it is desired to take the branch cut running alongthe left (otherwise, if it goes inside the unit circle, it will reduce or eliminate theradius of convergence).

Chapter 7

7.1.2 A function f(z) can be represented by

f(z) =f1(z)f2(z)

in which f1(z) and f2(z) are analytic. The denominator f2(z) vanishes at z = z0,showing that f(z) has a pole at z = z0. However, f1(z0) 6= 0, f ′2(z0) 6= 0. Show thata−1, the coefficient of (z− z0)−1 in a Laurent expansion of f(z) at z = z0, is given by

a−1 =f1(z0)f ′2(z0)

Since f1(z) and f2(z) are both analytic, they may be expanded as Taylor series

f1(z) = f1(z0) + f ′1(z0)(z − z0) + · · · ,f2(z) = f ′2(z0)(z − z0) + 1

2f ′′2 (z0)(z − z0)2 + · · ·

Here we have already set f2(z0) = 0 since the function vanishes at z = z0. As aresult, we have

f(z) =f1(z)f2(z)

=f1(z0) + f ′1(z0)(z − z0) + · · ·

f ′2(z0)(z − z0) + 12f ′′2 (z0)(z − z0)2 + · · ·

=f1(z0)/f ′2(z0)

z − z0

1 + (f ′1/f1)(z − z0) + · · ·1 + 1

2 (f ′′2 /f ′2)(z − z0) + · · ·

For z ≈ z0, the denominator 1 + 12 (f ′′2 /f ′2)(z − z0) + · · · may be inverted using

the geometric series relation 1/(1 + r) = 1− r + r2 − · · ·. The result is a Laurentseries of the form

f(z) =f1(z0)/f ′2(z0)

z − z0

(1 + (

f ′1f1− f ′′2

2f ′2)(z − z0) + · · ·

)This expansion has a single pole, and its residue is simply

a−1 =f1(z0)f2(z0)

7.1.4 The Legendre function of the second kind Qν(z) has branch points at z = ±1. Thebranch points are joined by a cut line along the real (x) axis.

a) Show that Q0(z) = 12 ln((z + 1)/(z − 1)) is single-valued (with the real axis

−1 ≤ x ≤ 1 taken as a cut line).

Because lnw has a branch point at w = 0, this ratio of logs has branch points atz = ±1 as promised. Joining the branch points by a cut line along the real axisgives the picture

z

1

+z1 1−

z

α

1

z

β

−

Of course, to make this picture well defined, we provide a principle value for thearguments

z + 1 = |z + 1|eiα, −π < α ≤ π,

z − 1 = |z − 1|eiβ , −π < β ≤ π

Thus

Q0(z) = 12 ln(z + 1)− 1

2 ln(z − 1) = 12 ln

∣∣∣∣z + 1z − 1

∣∣∣∣ + i2 (α− β) (2)

It is the manner in which the arguments α and β show up in (2) that indicate thebranch cut is as written. For x > 1 on the real axis, both α and β are smooth,α ≈ 0 and β ≈ 0 for going either a little bit above or below the axis. Hence thereis no discontinuity in Q0(x > 1) and thus no branch cut. For −1 < x < 1, on theother hand, the argument α ≈ 0 is smooth infinitesimally above or below the axis,but the argument β is discontinuous: β ≈ π above the axis, but β ≈ −π belowthe axis. This shows that the value of Q0 changes by ±iπ when crossing the realaxis. For x < −1, the situation is more interesting, as both α and β jump whencrossing the axis. However the difference (α−β) is unchanged. In this sense, thetwo branch cuts cancel each other out, so that the function Q0(x < −1) is welldefined without a cut.

Essentially, the branch cut prevents us from going around either of the pointsz = 1 or z = −1 individually. However, we can take a big circle around bothpoints. In this case, α → α + 2π and β → β + 2π, but once again the difference(α− β) in (2) is single-valued. So this is an appropriate branch cut prescription.

b) For real argument x and |x| < 1 it is convenient to take

Q0(x) = 12 ln

1 + x

1− x

Show thatQ0(x) = 1

2 [Q0(x + i0) + Q0(x− i0)]

The branch cut prescription described in part a) is somewhat unfortunate for realarguments |x| < 1, since those values of x sit right on top of the cut. To makethis well defined for real x, we must provide a prescription for avoiding the cut.This is what the x + i0 (above the cut) and x− i0 (below the cut) prescription isdoing for us. Noting that (for |x| < 1) the arguments have the following values

x + i0 (above the cut) : α ≈ 0, β ≈ π,

x− i0 (below the cut) : α ≈ 0, β ≈ −π

The expression of (2) yields

Q0(x + i0) = 12 ln

∣∣∣∣x + 1x− 1

∣∣∣∣− iπ

2,

Q0(x− i0) = 12 ln

∣∣∣∣x + 1x− 1

∣∣∣∣ +iπ

2

(3)

Taking the average gives

Q0(x) = 12 [Q0(x + i0) + Q0(x− i0)] = 1

2 ln∣∣∣∣x + 1x− 1

∣∣∣∣ = 12 ln

1 + x

1− x

where we have used the fact that |x− 1| = 1− x for |x| < 1. In this case, we seethat averaging the function below and above the cut cancels the imaginary parts,±iπ/2 in (3).




Chapter 7

7.2.5 The unit step function is defined as

u(s− a) ={

0, s < a1, s > a

Show that u(s) has the integral representations

a) u(s) = limε→0+

12πi

∫ ∞

−∞

eixs

x− iεdx

Let us first suppose we may close the contour with a semi-circle in the upper halfplane

θε

z

u(s)

C

IR

Ri

Always assuming the limit ε → 0+, we see that the real integral for u(s) may bepromoted to a closed contour integral

12πi

∮C

eizs

z − iεdz = u(s) + IR (1)

where IR denotes the integral along the semi-circle at infinity. We now show that,at least for s > 0, the integral IR vanishes. To do so, we make an explicit variablesubstitution

z = Reiθ, dz = iReiθdθ

so that

IR =1

2πi

∫ π

0

eisReiθ

Reiθ − iεiReiθdθ =

12π

∫ π

0

eisR(cos θ+i sin θ)dθ

where we have safely taken the limit ε → 0+. Expanding out the exponent, wefind

IR =12π

∫ π

0

eisR cos θe−sR sin θdθ (2)

This vanishes by Jordan’s lemma provided sR sin θ > 0 so that the real exponen-tial is suppressed instead of blowing up (in fact, this is Jordan’s lemma). SinceR →∞ is positive and sin θ > 0 in the upper half plane, this corresponds to therequirement s > 0 (as alluded to above). In this case, since IR = 0, (1) simplifiesto

u(s) =1

2πi

∮C

eizs

z − iεdz = residue of

eizs

z − iεat z = iε (s > 0)

The residue at iε is simply limε→0+ e−sε = 1. Hence we have confirmed thatu(s) = 1 for s > 0.

For s < 0, on the other hand, Jordan’s lemma makes it clear that we shouldinstead close the contour with a semi-circle in the lower half plane

ε

IR

z

u(s)

i

C

Since there are no residues inside the contour, we simply obtain u(s) = 0 fors < 0. Although the problem does not discuss the case when s = 0, it is worthconsidering. In this case, we might as well close the contour in the upper halfplane. Then IR can be directly evaluated by inserting s = 0 into (2). The resultis simply IR = 1

2 . Since the contour integral still has the value of 1 (residue atthe pole at iε), inserting IR = 1

2 into (1) gives

1 = u(0) + 12 ⇒ u(0) = 1

2

which is a nice result indicating that the step function is given completely by

u(s− a) =

{ 0, s < a12 , s = a1, s > 1

at least using this definition.

b) u(s) =12

+1

2πiP

∫ ∞

−∞

eixs

xdx

The principal value integral can be evaluated by deforming the contour aboveand below the pole and then taking the average of the results. For s > 0, thiscorresponds to something like

z

u(s)

C

IR

or

z

u(s)

C

IR

As in part a), the residue of the pole at z = 0 is simply 1. So for the contour onthe left we have 1

2πi

∮eizs

z dz = 1, while for the one on the right (no poles inside)we have 0. The principal value then gives 1/2 (the average of 1 and 0). Thisindicates that

u(s) =12

+1 + 0

2= 1, (s > 0)

For s < 0, on the other hand, we close the contour on the lower half plane. Again,we average between the case when the pole is inside and when it is outside thecontour. However, it is important to realize that by closing the contour on thelower half plane, we are actually choosing a clockwise (‘wrong direction’) contour.This means the contour integral gives either −1 or 0 depending on whether thepole is inside or outside. The principal value prescription then yields

u(s) =12

+−1 + 0

2= 0, (s < 0)

If we wanted to be careful, we could also work this out for s = 0 to find the sameanswer u(0) = 1

2 .

7.2.7 Generalizing Example 7.2.1, show that∫ 2π

0

dθ

a± b cos θ=∫ 2π

0

dθ

a± b sin θ=

2π

(a2 − b2)1/2

for a > |b|. What happens if |b| > |a|?

Since this integral is over a complete period, we note that we would get thesame answer whether we integrate cos θ or sin θ. Furthermore, it does not matterwhether we integrate a + b cos θ or a − b cos θ. This can be proven directly byconsidering the substitutions θ → π

2 − θ or θ → θ + π into the integral. In anycase, this means we only need to consider

I =∫ 2π

0

dθ

a + b cos θ

where we assume a > b > 0. For these types of trig integrals, we make thesubstitutions

z = eiθ, dz = ieiθdθ = izdθ, cos θ =z + z−1

2

to change the real integral into a contour integral on the unit circle |z| = 1

I =∮

C

dz

iz(a + b2 (z + z−1))

=−2i

b

∮C

dz

z2 + 2ab z + 1

Since the contour is already closed, we do not need to worry about finding a wayto close the contour. All we need is to identify the poles inside the contour and

their residues. To do this, we solve the quadratic equation in the denominator toobtain

I =−2i

b

∮C

dz

(z − z+)(z − z−)

where

z± = −a

b±√

a2

b2− 1 (3)

Since we have assumed a > b > 0, the two zeros of the denominator, z− and z+

are located as followsz

C

z − z +

In particular, it is not hard to check that the pole at z+ lies inside the circle ofunit radius. As a result

I = (2πi)−2i

b

(residue of

1(z − z+)(z − z−)

at z = z+

)=

4π

b

1(z+ − z−)

=4π

2b√

a2/b2 − 1=

2π√a2 − b2

Note that, for a < 0, the integrand would always be negative. In this case, Iwould be negative. Thus the complete answer is

I =2π√

a2 − b2sign(a)

For |b| > |a|, the integrand would blow up when θ = − cos−1(a/b) so the integralis not defined. What happens in this case is that, on the complex plane, the twopoles z+ and z−, which still solve (3), move off the real axis but stay on the unitcircle contour itself.

z +

z −

C

z

So the complex integral is just as bad as the real integral. This is an examplewhere we could consider using a principal value prescription to make the integralwell defined.

7.2.14 Show that (a > 0)

a)∫ ∞

−∞

cos x

x2 + a2dx =

π

ae−a

How is the right-hand side modified if cos x is replaced by cos kx?

For these types of integrals with sin or cos in the numerator, it is best to considersinx or cos x as the imaginary or real part of the complex exponential eix. Inthis case, we write

I =∫ ∞

−∞

cos x

x2 + a2dx = <

∫ ∞

−∞

eix

x2 + a2dx

Using Jordan’s lemma, we may close the contour using a semi-circle in the upperhalf plane.

ia

z

C

IR

Since IR = 0 (by Jordan’s lemma), we have simply

I = <∮

eiz

z2 + a2dz = <

∮eiz

(z − ia)(z + ia)dz = <

(2πi

e−a

2ia

)=

π

ae−a

(for a positive).

If cos x is replaced by cos kx, we would write the numerator as <eikx. In thiscase, for k > 0 we would close the contour in the upper half plane as before.In addition, the exponential factor in the residue would be e−ka, so for cos kx,the integral would be (π/a)e−ka. For k < 0, on the other hand, we could closethe contour in the lower half plane. However, it is actually easier to see thatcos(−kx) = cos kx, so the answer should be independent of the sign of k. Hence∫ ∞

−∞

cos kx

x2 + a2dx =

π

|a|e−|ka|

is valid for any sign of k and a.

b)∫ ∞

−∞

x sinx

x2 + a2dx = πe−a

How is the right-hand side modified if sinx is replaced by sin kx?

As above, we write sinx = =eix. Closing the contour in the same manner, andusing Jordan’s lemma to argue that IR = 0, we obtain∫ ∞

−∞

x sinx

x2 + a2dx = =

∮zeiz

z2 + a2dz = =

∮zeiz

(z − ia)(z + ia)dz

= =(

2πiiae−a

2ia

)= πe−a

If sinx is replaced by sin kx, the residue would get modified so that e−a is replacedby e−ka. As a result ∫ ∞

−∞

x sin kx

x2 + a2dx = πe−|ka|

The reason for the absolute value is the same as for part a) above.

7.2.20 Show that ∫ ∞

0

dx

(x2 + a2)2=

π

4a3, a > 0

This problem involves a double pole at z = ia. Choosing to close the contour inthe upper half plane

ia

z

C

IR

we obtain

I =∫ ∞

0

dx

(x2 + a2)2=

12

∫ ∞

−∞

dx

(x2 + a2)2=

12

∮C

dz

(z2 + a2)2

= πi (residue at z = ia)(4)

Although this is a double pole, it may still have a residue. To see this, imagineexpanding the integrand in a power series near the pole at z = ia

1(z2 + a2)2

= (z − ia)−2(z + ia)−2 = (z − ia)−2[2ia + (z − ia)]−2

= (z − ia)−2(2ia)−2

(1 +

z − ia

2ia

)−2

=−14a2

(z − ia)−2

(1− 2

(z − ia

2ia

)+

2 · 32

(z − ia

2ia

)2

− · · ·

)

=−1/4a2

(z − ia)2+−i/4a3

(z − ia)+ (3/16a4) + · · ·

Here we have used the binomial expansion for (1 + r)−2. This shows that, inaddition to the obvious double pole, there is a single pole ‘hidden’ on top of itwith residue a−1 = −i/4a3. Alternatively, we could have computed the residuemuch more quickly by noting that for a double pole in f(z) = 1/(z2 + a2)2, weform the non-singular function g(z) = (z − ia)2f(z) = 1/(z + ia)2. The residueis then the derivative

a−1 = g′(ia) =d

dz

1(z + ia)2

∣∣∣∣z=ia

=−2

(z + ia)3

∣∣∣∣z=ia

=−2

(2ia)3=−i

4a3

In either case, using this residue in (4), we find

I = πi

(−i

4a3

)=

π

4a3

or more precisely I = π/4|a|3, which is valid for either sign of a.

It is worth noting that, for this problem, the integrand falls off sufficiently fast atinfinity that we could choose to close the contour either in the upper half planeor the lower half plane. Had we worked with the lower half plane, we wouldhave found a pole at −ia with opposite sign for the residue. On the other hand,the clockwise contour would have contributed another minus sign. So overall wewould have found the same result either way (which is a good thing).

7.2.22 Show that ∫ ∞

0

cos(t2)dt =∫ ∞

0

sin(t2)dt =√

π

2√

2

Again, when we see sin or cos, it is worth considering this as the imaginary or realparts of the complex exponential. Hence we first choose to evaluate the integral

I =∫ ∞

0

eit2dt

Taking the hint into account, we write down a (closed) contour integral using thecontour

I

/2π

z

C

I

I R2

Thus ∮C

eiz2dz = I + IR + I2

We start by evaluating the contour integral on the left. Although eiz2has an

essential singularity at infinity, that actually lies outside the contour (this iscertainly true for any fixed large radius R; it also remains outside the contourin the limit R → ∞). Since there are no poles and no singularities inside thecontour, the contour integral vanishes. As a result,

0 = I + IR + I2 ⇒ I = −IR − I2

We now show that the integral on IR vanishes as well by a simple modificationto Jordan’s lemma. For IR, we let z = Reiθ, so that

IR =∫ π/2

0

eiR2e2iθ

iReiθdθ = iR

∫ π/2

0

eiR2 cos 2θe−R2 sin 2θeiθdθ

Hence

|IR| = R

∫ π/2

0

e−R2 sin 2θdθ

Using the same argument as in Jordan’s lemma, we can show that the integral∫ π/2

0e−R2 sin 2θdθ may be bounded by 1/R2. Hence |IR| itself falls off as 1/R, and

vanishes when we take R →∞. As a result, we are left with the observation

I = −I2

To examine I2, we note that the path of integration is a line of constant slope inthe complex plane

z = `eiπ/2, dz = eiπ/2d`

Thus

I2 =∫ 0

∞ei(`eiπ/2)2eiπ/2d` = −eiπ/2

∫ ∞

0

e−`2d`

Note that the minus sign came from consideration of the direction of integrationalong the I2 contour. At this point, complex analysis does not really help us, andwe must recall (or look up) the gaussian integral∫ ∞

0

e−`2d` =12

∫ ∞

−∞e−`2d` =

√π

2

Thus

I = −I2 = eiπ/2

√π

2= (cos(π/2) + i sin(π/2))

√π

2= (1 + i)

√π

2√

2

Since I =∫∞0

eit2dt, we may now take the real (cos) and imaginary (sin) parts ofI to obtain ∫ ∞

0

cos(t2)dt =∫ ∞

0

sin(t2)dt =√

π

2√

2

This is a curious result, as it is not directly obvious why integrating cos(t2) andsin(t2) would give identical results.




Chapter 8

8.2.2 The Laplace transform of Bessel’s equation (n = 0) leads to

(s2 + 1)f ′(s) + sf(s) = 0

Solve for f(s)

This equation is amenable to separation of variables

df

f= − s

s2 + 1ds ⇒

∫df

f= −

∫s

s2 + 1ds

⇒ ln f = − 12 ln(s2 + 1) + c

Exponentiating this and redefining the consant, we obtain

f(x) =C√

s2 + 1

8.2.5 A boat, coasting through the water, experiences a resisting force proportional to vn,v being the instantaneous velocity of the boat. Newton’s second law leads to

mdv

dt= −kvn

With v(t = 0) = v0, x(t = 0) = 0, integrate to find v as a function of time and thenthe distance.

This equation is separabledv

vn= − k

mdt

For n 6= 1, this may be integrated to give∫ v

v0

dv′

v′n= − k

m

∫ t

0

dt′ ⇒ − 1n− 1

(1

vn−1− 1

vn−10

)= − k

mt

⇒ v(t) =(

v−(n−1)0 +

(n− 1)km

t

)−1/(n−1)(1)

This may be integrated once more to obtain x as a function of t

x(t) =∫ t

0

v(t′)dt′ =∫ t

0

(v−(n−1)0 +

(n− 1)km

t′)−1/(n−1)

dt′

Although this may look somewhat scary, it is in fact trivial to integrate, as it isessentially t′ (plus a constant) to some fractional power. The only difficulty isbookkeeping the various constants. For n 6= 2, the result is

x(t) =1

1− 1/(n− 1)m

(n− 1)k

(v−(n−1)0 +

(n− 1)km

t′)1−1/(n−1)

∣∣∣∣∣t

0

=m

(n− 2)k

[(v−(n−1)0 +

(n− 1)km

t

)(n−2)/(n−1)

− v−(n−2)0

] (2)

If desired, the position and velocity, (2) and (1) may be rewritten as

x(t) =m

(n− 2)kvn−20

[(1 +

(n− 1)kvn−10

mt

)(n−2)/(n−1)

− 1

]

v(t) = v0

(1 +

(n− 1)kvn−10

mt

)−1/(n−1)

As a result, it is possible to eliminate t and obtain the velocity as a function ofposition

v = v0

(1 +

(n− 2)kvn−20 x

m

)−1/(n−2)

(3)

Note that we may definexk =

m

(n− 2)kvn−20

which represents a length scale related to the resisting force and initial velocity.In terms of xk, the velocity and position relation may be given as

v

v0=

(1 +

x

xk

)−1/(n−2)

or(v0

v

)n−2

= 1 +x

xk

Note that, in fact, it is possible to obtain (3) directly from Newton’s second lawby rewriting it as

dv

vn−1= − k

mv dt = − k

mdx

and then integrating∫ v

v0

dv′

v′n−1= − k

m

∫ x

0

dx′ ⇒ − 1n− 2

(1

vn−2− 1

vn−20

)= − k

mx

⇒(v0

v

)n−2

= 1 +(n− 2)kvn−2

0

mx

So far, what we have done does not apply to the special cases n = 1 or n = 2.For n = 1, we have

dv

v= − k

mdt ⇒ ln

(v

v0

)= − k

mt ⇒ v(t) = v0e

−(k/m)t

Integrating once more yields

x(t) =mv0

k(1− e−(k/m)t) ⇒ v

v0= 1− kx

mv0

which is in fact consistent with setting n = 1 in (3).

For n = 2, we have

dv

v2= − k

mdt ⇒ −1

v+

1v0

= − k

mt ⇒ v(t) =

v0

1 + (kv0/m)t

Integrating this for position yields

x(t) =m

kln

(1 +

kv0

mt

)⇒ kx

m= ln

(v0

v

)

8.2.6 In the first-order differential equation dy/dx = f(x, y) the function f(x, y) is a functionof the ratio y/x:

dy

dx= g

(y

x

)Show that the substitution of u = y/x leads to a separable equation in u and x.

If we let u = y/x, this means that y = xu. So, by the product rule

dy

dx= x

du

dx+ u

The above differential equation now becomes

xdu

dx+ u(x) = g(u) ⇒ du

g(u)− u=

dx

x

which is separated in u and x.

8.2.10 A certain differential equation has the form

f(x)dx + g(x)h(y)dy = 0

with none of the functions f(x), g(x), h(y) identically zero. Show that a necessaryand sufficient condition for this equation to be exact is that g(x) = const.

The check for exactness is

∂

∂yf(x) =

∂

∂x(g(x)h(y))

or

0 =dg(x)dx

h(y)

Since h(y) is not identically zero, we may divide out by h(y) (at least in anydomain away from isolated zeros of h), leading to dg(x)/dx = 0, which indicatesthat g(x) must be constant.

8.2.16 Bernoulli’s equationdy

dx+ f(x)y = g(x)yn

is nonlinear for n 6= 0 or 1. Show that the substitution u = y1−n reduces Bernoulli’sequation to a linear equation.

For n 6= 1, the substitution u = y1−n is equivalent to y = u1/(1−n). Thus

dy

dx=

11− n

u1/(1−n)−1 du

dx=

11− n

un/(1−n) du

dx

Bernoulli’s equation then becomes

11− n

un/(1−n) du

dx+ f(x)u1/(1−n) = g(x)un/(1−n)

Multiplying by u−n/(1−n) gives

11− n

du

dx+ f(x)u = g(x)

ordu

dx+ (1− n)f(x)u = (1− n)g(x)



Textbook problems: Ch. 8: 8.4.1, 8.4.3, 8.5.6, 8.5.11, 8.5.14, 8.5.17

Chapter 8

8.4.1 Show that Legendre’s equation has regular singularities at x = −1, 1, and ∞.

Legendre’s equation may be written as

y′′ − 2x1− x2

y′ +l(l + 1)1− x2

y = 0

so that

P (x) = − 2x1− x2

=2x

(x− 1)(x+ 1), Q(x) =

l(l + 1)1− x2

= − l(l + 1)(x− 1)(x+ 1)

Written in this fashion, we see that both P (x) and Q(x) have simple poles atx = 1 and x = −1. This is sufficient to indicate that these two points are regularsingular points.

For the point at ∞, we make the substitution x = 1/z. As worked out in thetext, we end up with

P (z) =2z − P (z−1)

z2=

2z + 2z−1/(1− z−2)z2

=2z

+2

z(z2 − 1)=

2zz2 − 1

and

Q(z) =Q(z−1)z4

=l(l + 1)/(1− z−2)

z4=

l(l + 1)z2(z2 − 1)

Examining the behavior of P and Q as z → 0, we see that P is regular, while Qhas a double pole. Because of the double pole in Q, Legendre’s equation also hasa regular singularity at x = ∞.

8.4.3 Show that the substitution

x→ 1− x2

, a = −l, b = l + 1, c = 1

converts the hypergeometric equation into Legendre’s equation.

Making the above substitution (along with dx→ − 12dx which implies y′ → (−2)y′

and y′′ → (−2)2y′′) into the Hypergeometric equation, we find

x(x− 1)y′′ + [(1 + a+ b)x− c]y′ + aby = 0

⇒ 1− x2

(1− x

2− 1

)(−2)2y′′ +

[(1− l + (l + 1))

1− x2

− 1]

(−2)y′

− l(l + 1)y = 0

⇒ −(1− x2)y′′ + 2xy′ − l(l + 1)y = 0

Changing an overall sign yields Legendre’s equation

(1− x2)y′′ − 2xy′ + l(l + 1)y = 0

This indicates that Legendre’s equation is in fact a special case of the more generalHypergeometric equation.

8.5.6 Develop series solutions for Hermite’s differential equation

a) y′′ − 2xy′ + 2αy = 0

Since x = 0 is a regular point, we develop a simply Taylor series solution

y =∞∑

n=0

anxn, y′ =

∞∑n=0

nanxn−1, y′′ =

∞∑n=0

n(n− 1)anxn−2

Substituting this in to Hermite’s equation, we find

∞∑n=0

[n(n− 1)anxn−2 − 2nanx

n + 2αanxn] = 0

⇒∞∑

n=0

[(n+ 2)(n+ 1)an+2 + 2(α− n)an]xn = 0

To obtain the second line, we had made the substitution n → n + 2 in the firstterm of the series so that we could collect identical powers of xn. Since this seriesvanishes for all values of x, each coefficient must vanish. This yields the recursionrelation

an+2 =2(n− α)

(n+ 2)(n+ 1)an (1)

which determines all higher an’s, given a0 and a1 as a starting point.

In fact, we obtain two series, one for n even and one for n odd. For n even, weset a0 = 1 and find

a0 = 1, a2 =2(−α)

2!, a4 =

2(2− α)4 · 3

a2 =22(−α)(2− α)

4!, etc.

This gives the even solution

yeven = 1 + 2(−α)x2

2!+ 22(−α)(2− α)

x4

4!+ 23(−α)(2− α)(4− α)

x6

6!+ · · · (2)

For n odd, we set a1 = 1 and find

a1 = 1, a3 =2(1− α)

3!, a5 =

2(3− α)5 · 4

a3 =22(1− α)(3− α)

5!, etc.

This results in the odd solution

yodd = x+2(1−α)x3

3!+22(1−α)(3−α)

x5

5!+23(1−α)(3−α)(5−α)

x7

7!+ · · · (3)

Note that, and an ordinary point, we did not have to solve the indicial equation.However, if we had chosen to do so, we would have found k = 0 or k = 1, yieldingthe even and odd solutions, respectively.

b) Show that both series solutions are convergent for all x, the ratio of successivecoefficients behaving, for large index, like the corresponding ratio in the expansionof exp(2x2).

To test for convergence, all we need is to use the ratio test

limn→∞

anxn

an+2xn+2= lim

n→∞

(n+ 2)(n+ 1)2(n− α)x2

= limn→∞

n

2x2= ∞ (4)

Since this is larger than 1, the series converges for all values of x. Note thatthe ratio an/an+2 was directly obtained from the recursion relation (1), and thisresult is valid for both yeven and yodd. Furthermore, if we compared this withexp(2x2), we would see that the n-th term in the Taylor series of the exponentialis bn = (2x2)n/n!, which leads to a ratio

bn−1

bn=

n

2x2

in direct correspondence with that of (4). Hence the solutions to Hermite’s equa-tions are (generically) asymptotic to exp(2x2).

c) Show that by appropriate choice of α the series solutions may be cut off andconverted to finite polynomials.

Examination of the series solutions (2) and (3) indicates that yeven terminatesfor α = 0, 2, 4, . . . and yodd terminates for α = 1, 3, 5, . . .. This means the for αa non-negative integer either yeven or yodd (depending on α being even or odd)terminates, yielding a finite ‘Hermite polynomial’.

8.5.11 Obtain two series solutions of the confluent hypergeometric equation

xy′′ + (c− x)y′ − ay = 0

Test your solutions for convergence.

We first observe that this equation has a regular singular point at x = 0 and anirregular one at x = ∞. We would like to develop a series solution around theregular singular point at x = 0. Thus we start with the indicial equation

y′′ +c− xx

y′ − a

xy = 0 ⇒ p0 = c, q0 = 0

and

k(k− 1)+ p0k+ q0 = 0 ⇒ k(k− 1)+ ck = 0 ⇒ k(k+ c− 1) = 0

This shows that the indices at x = 0 are k1 = 0 and k2 = 1 − c. We start withk1 = 0. Since the index vanishes, we attempt an ordinary Taylor series solution

y =∞∑

n=0

anxn, y′ =

∞∑n=0

nanxn−1, y′′ =

∞∑n=0

n(n− 1)anxn−2

Substituting this into the confluent hypergeometric equation, we obtain

∞∑n=0

[n(n− 1)anxn−1 + ncanx

n−1 − nanxn − aanx

n] = 0

Making the substition n→ n+ 1 in the first two terms and simplifying gives

∞∑n=0

[(n+ 1)(c+ n)an+1 − (a+ n)an]xn = 0

Therefore we have a recursion relation of the form

an+1 =a+ n

(n+ 1)(c+ n)an (5)

Setting a0 = 1, the first few terms in the series becomes

a0 = 1, a1 =a

c, a2 =

a+ 12(c+ 1)

a1 =a(a+ 1)2!c(c+ 1)

,

a3 =a+ 2

3(c+ 2)a2 =

a(a+ 1)(a+ 2)3!c(c+ 1)(c+ 2)

This indicates that

y = 1 +a

cx+

a(a+ 1)c(c+ 1)

x2

2!+a(a+ 1)(a+ 2)c(c+ 1)(c+ 2)

x3

3!+ · · ·

=∞∑

n=0

(a)n

(c)n

xn

n!

(6)

where the notation (a)n is given by

(a)n = a(a+ 1)(a+ 2) · · · (a+ n− 2)(a+ n− 1) =Γ(a+ n)

Γ(a)(7)

This is the ‘regular’ solution of the confluent hypergeometric equation. We nowtest this series for convergence using the ratio test. Given the recursion relation(5), we find

limn→∞

anxn

an+1xn+1= lim

n→∞

(n+ 1)(c+ n)(a+ n)x

= limn→∞

n

x= ∞

Therefore this series converges for all values of x, unless c is a non-positive integer,in which case the denominators in (6) will eventually all blow up.

Turning next to k2 = 1− c, we seek a series solution of the form

y = x1−c∞∑

n=0

anxn, y′ = x−c

∞∑n=0

(n+ 1− c)anxn,

y′′ = x−1−c∞∑

n=0

(n+ 1− c)(n− c)anxn

Substituting this into the confluent hypergeometric equation, we find

x1−c∞∑

n=0

[(n+1−c)(n−c)anxn−1+c(n+1−c)anx

n−1−(n+1−c)anxn−aanx

n] = 0

Performing the shift n→ n+ 1 in the first two terms and simplifying, we obtain

x1−c∞∑

n=0

[(n+ 2− c)(n+ 1)an+1 − (n+ 1 + a− c)an]xn = 0

which yields the recursion relation

an+1 =n+ 1 + a− c

(n+ 2− c)(n+ 1)an

Supposing that a0 = 1, the first few terms in this series are given by

a0 = 1, a1 =1 + a− c

2− c, a2 =

2 + a− c2(3− c)

a1 =(1 + a− c)(2 + a− c)

2!(2− c)(3− c),

a3 =3 + a− c3(4− c)

a2 =(1 + a− c)(2 + a− c)(3 + a− c)

3!(2− c)(3− c)(4− c)

Following the notation of (7), we may write the series solution as

ynew = x1−c∞∑

n=0

(1 + a− c)n

(2− c)n

xn

n!(8)

This series is rather similar to the standard one (6). In fact, the solution of (6)may be converted into ynew by making the substitions a→ a+1−c and c→ 2−cand multiplying y by the prefactor x1−c. [Why this works may be seen by makingthe substitutions directly into the confluent hypergeometric equation itself.] Asa result, by the same ratio test argument as before, ynew converges for all valuesof x, except when c = 2, 3, 4, . . . where the denominators in (8) would eventuallyall blow up.

To summarize, for non-integer values of c, the two solutions (6) and (8) forma complete linearly independent set. For c = 1, both (6) and (8) are preciselythe same, and we have found only one solution. For other integer values of c,only one of (6) or (8) makes sense (and the other one blows up because of a baddenominator). So in fact for all integer c, we have only obtained one solutionby the series method, and the second solution would be of the ‘irregular’ form(which is not fun at all).

8.5.14 To a good approximation, the interaction of two nucleons may be described by amesonic potential

V =Ae−ax

x

attractive for A negative. Develop a series solution of the resultant Schrodinger waveequation

h2

2md2ψ

dx2+ (E − V )ψ = 0

We begin by substituting the explicit potential in the Schrodinger equation

d2ψ

dx2+

(2mEh2 − 2mAe−ax

h2x

)ψ = 0

As in the text, it would be convenient to define

E =2mEh2 , A =

2mAh2

In this case, we want to solve the second order equation

ψ′′ +(E − Ae

−ax

x

)ψ = 0 (9)

which has a regular singular point at x = 0 and an irregular one at x = ∞. Wenow develop a series solution around x = 0. Noting that

P (x) = 0, Q(x) =e−ax

x⇒ p0 = 0, q0 = 0

the indicial equation is trivial, k(k − 1) = 0. Since we have k1 = 1 and k2 = 0,we look for the k1 = 1 series (the larger index one always ‘works’). Here we haveto worry that e−ax is non-polynomial. As a result, we will not be able to obtaina simple recursion relation. We thus content ourselves with just working out afew terms in the series. Normalizing the first term in the series to be x, we take

y = x+a2x2+a3x

3+· · · , y′ = 1+2a2x+3a3x2+· · · , y′′ = 2a2+6a3x+· · ·

Substitution into (9) gives

2a2 + 6a3x+ · · ·+ (Ex−Ae−ax)(1 + a2x+ a3x2 + · · ·) = 0

Since we have used a series for the wavefunction ψ(x), we ought to also expandthe exponential as a series, e−ax = 1 − ax + 1

2a2x2 − · · ·. Keeping appropriate

powers of x, we find

0 = 2a2 + 6a3x+ · · ·+ (Ex−A(1− ax+ · · ·))(1 + a2x+ · · ·)= 2a2 + 6a3x+ · · ·+ (−A+ (aA+ E)x+ · · ·)(1 + a2x+ · · ·)= 2a2 + 6a3x+ · · ·+ (−A) + (aA+ E − a2A)x+ · · ·= (2a2 −A) + (6a3 + aA+ E − a2A)x+ · · ·

Setting the coefficients to zero gives

a2 = 12A, a3 = 1

6 (a2A− E − aA) = 16 ( 1

2A2 − E − aA)

The series solution is the of the form

ψ = x+ 12Ax

2 + 16 ( 1

2A2 − E − aA)x3 + · · ·

8.5.17 The modified Bessel function I0(x) satisfies the differential equation

x2 d2

dx2I0(x) + x

d

dxI0(x)− x2I0(x) = 0

From Exercise 7.4.4 the leading term in an asymptotic expansion is found to be

I0(x) ∼ex

√2πx

Assume a series of the form

I0(x) ∼ex

√2πx

(1 + b1x−1 + b2x

−2 + · · ·)

Determine the coefficients b1 and b2

The (modified) Bessel equation has a regular singular point at x = 0 and anirregular one at x = ∞. Here we are asked to develop an asymptotic expansionaround x = ∞. Although this is an irregular one (witness the essential singularityex), we are given the form of the series. As a result, all we have to do is to takederivatives and insert the expressions into the differential equation. To make iteasier to obtain the derivatives, we write

I0(x) ∼ex

√2π

(x−12 + b1x

− 32 + b2x

− 52 + b3x

− 72 + · · ·)

The derivative d/dx acts either on the ex factor or the series in the parentheses.The resulting first derivative is

I ′0(x) ∼ex

√2π

(x−12 + (b1 − 1

2 )x−32 + (b2 − 3

2b1)x− 5

2 + (b3 − 52b2)x

− 72 + · · ·)

Taking one more derivative yields

I ′′0 (x) ∼ ex

√2π

(x−12 +(b1−1)x−

32 +(b2−3b1 + 3

4 )x−52 +(b3−5b2 + 15

4 b1)x− 7

2 + · · ·)

Substituting the above into the modified Bessel equation and collecting like pow-ers of x, we find

0 ∼ ex

√2π

(x32 + (b1 − 1)x

12 + (b2 − 3b1 + 3

4 )x−12 + (b3 − 5b2 + 15

4 b1)x− 3

2 + · · ·

+ x12 + (b1 − 1

2 )x−12 + (b2 − 3

2b1)x− 3

2 + · · ·

− x 32 − b1x

12 − b2x−

12 − b3x−

32 − · · ·)

∼ ex

√2π

((−2b1 + 14 )x−

12 + (−4b2 + 9

4b1)x− 3

2 + · · ·)

Setting the coefficients to zero gives

b1 = 18 , b2 = 9

16b1 = 9128

so that the asymptotic series develops as

I0(x) ∼ex

√2πx

(1 + 18x−1 + 9

128x−2 + · · ·)

Note that, in order to find b1 and b2, we needed to keep track of the b3 coefficient,even though it dropped out in the end.

Arfken Weber Math Selected Problem Solution

Documents

collect identical

a2x a2y

ad bc

dt 0sin

x2 y2 z2

lower half

z1 z2

conuent hypergeometric