Arfken Solutions 2

7/30/2019 Arfken Solutions 2

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Physics 451 Fall 2004

Homework Assignment #1 Solutions

Textbook problems: Ch. 1: 1.1.5, 1.3.3, 1.4.7, 1.5.5, 1.5.6

Ch. 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 vbldenote 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

wherevbw = 4 km/hr @ 50

= (2.57x + 3.06y) km/hr

vbl = 6.12 km/hr @ 30 = (5.3x + 3.06y) km/hr

Thusvwl = 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 ra 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

ra

a

r


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(b) (r a ) r = 0

This time the vector ra 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

ra

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 + A2

xB2

z + A2

yB2

x + A2

yB2

z + A2

zB2

x + A2

zB2

y

2AxBxAyBy 2AxBxAzBz 2AyByAzBz

= (A2x + A2

y + A2

z)(B2

x + B2

y + B2

z) (AxBx + AyBy + AzBz)2

where we had to add and subtract A2xB2x+A2yB2y+A2zB2z and do some factorizationto obtain the last line.

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

| A B |2 = (AB sin )2 = (AB)2(1 cos2 ) = (AB)2 (AB cos )2

= (AB)2 ( A B )2


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1.5.5 The orbital angular momentum L of a particle is given by L = rp = mrv where pis the linear momentum. With linear and angular velocity related by v = r, showthat

L = mr2[ r(r )]

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

Using L = mr v and v = r, we find

L = mr ( r )

Because of the double cross product, this is the perfect opportunity to use theBACCAB rule: A ( B C) = B( A C) C( A B)

L = m[(r r ) r(r )] = m[r2 r(r )]

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

L = mr2[ r(r )] (1)

1.5.6 The kinetic energy of a single particle is given by T = 12

mv2. For rotational motionthis becomes 1

2m( r )2. Show that

T = 12

m[r22 (r )2]

We can use the result of problem 1.4.7:

T = 12

m( r )2 = 12

m[(r)2 ( r )2] = 12

m[r22 (r )2]

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

T = 12

mr2[2 (r )2]

Comparing this with (1) above, we find that

T = 12

L

which is not a coincidence.


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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 bb 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 geta bb a

+

c dd 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 bb a

c dd 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)=

1

a2 + b2(a ib)

This corresponds to the 2 2 matrix

(a + ib)1 1

a2 + b2

a bb a


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

(a + ib)1

a bb 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 angular

momentum 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 be

expressed as [ P , Jx] = 0 and [ P , Jy] = 0 or equivalently as P Jx = Jx P andP Jy = Jy P. We also take the hint into account and note that Jx and Jy satisfy

the 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]] = P JxJy P JyJx JxJy P + JyJx P

= Jx P Jy Jy P Jx Jx P Jy + Jy P Jx

= 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.


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

Tr (A + B) = i

(aii + bii) = i

aii + i

bii = Tr (A) + 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(A) = n 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.


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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 A A = I and B B = I.We now denote the product ofA and B by C = AB. To show that Cis orthogonal,we compute CC and see what happens. Recalling that the transpose of a productis the reversed product of the transposes, we have

C

C = (AB)(

AB) = AB

B

A = A

A = 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 A A = I, or

a b

c da c

b d = a2 + b2 ac + bd

ac + bd c2

+ d2 =

1 0

0 1This 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


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

a = cos , b = sin

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

c = cos , d = sin

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

cos cos + sin sin = cos( ) = 0

This means that = /2 or = 3/2. We must consider both casesseparately.

= /2: This gives

c = cos( /2) = sin , d = sin( /2) = cos

or

A1 =

cos sin sin cos

(1)

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

= 3/2: This gives

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

or

A2 =

cos sin

sin cos

(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 det A2 = 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 sin

sin cos


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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 ofall 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

= SA S where is diagonalWe can invert this relation by multiplying both sides on the left by

S and on

the right by S. This yields A = SSTaking the transpose of A, we find

A = ( SS) = SSHowever, the transpose of a transpose is the original matrix,

S = S, and thetranspose of a diagonal matrix is the original matrix,

= . Hence

A = SS = ASince 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.


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3.5.6 A has eigenvalues i and corresponding eigenvectors |xi. Show that A1 has thesame eigenvectors but with eigenvalues 1i .

If A has eigenvalues i and eigenvectors |xi, that means

A|xi = i|xi

Multiplying both sides by A1 on the left, we find

A1A|xi = iA1|xi

or|xi = iA1|xi

Rewriting this as

A1

|xi = 1i |xi

it is now obvious that A1 has the same eigenvectors, but eigenvalues 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

= U AU, = V BV

where is the same diagonal matrix of eigenvalues. This means

U AU = V BV B = VU AUV

If we let W = VU, its Hermitian conjugate is W = (VU) = UV. This meansthat

B = W AW where W = VU

and W W = VU UV = I. Hence A and B are related by a unitary similaritytransformation.

3.5.30 a) Determine the eigenvalues and eigenvectors of

1 1

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


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We first find the eigenvalues through the secular equation 1 1 = (1 )2 2 = 0

This is easily solved

(1 )2 2 = 0 ( 1)2 = 2 ( 1) = (3)

Hence the two eigenvalues are + = 1 + and = 1 .For the eigenvectors, we start with + = 1 + . Substituting this into the eigen-value problem (A I)|x = 0, we find

a

b = 0 (a b) = 0 a = bSince the problem did not ask to normalize the eigenvectors, we can take simply

+ = 1 + : |x+ =

11

For = 1 , we obtain instead

a

b = 0 (a + b) = 0 a = bThis gives

= 1 : |x =

11

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

b) Determine the eigenvalues and eigenvectors of

1 12 1

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


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In this nonsymmetric case, the secular equation is 1 12 1 = (1 )2 2 = 0

Interestingly enough, this equation is the same as (3), even though the matrix is

different. Hence this matrix has the same eigenvalues + = 1 + and = 1 .For + = 1 + , the eigenvector equation is 1

2

ab

= 0 a + b = 0 b = a

Up to normalization, this gives

+ = 1 + : |x+ =

1

(4)

For the other eigenvalue, = 1 , we find 1

2 a

b = 0 a + b = 0 b = a

Hence, we obtain

= 1 : |x =

1

(5)

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

when = 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 for

0 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 or

cos =x+|x

x+|x+1/2x|x1/2

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

cos =1 2

1 + 2

1 + 2=

1 21 + 2

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


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

Ch. 2: 2.4.8, 2.4.11

Chapter 1

1.7.1 For a particle moving in a circular orbit r = x r cos t + y r sin t

(a) evaluate r r

Taking a time derivative of r, we obtain

r = x r sin t + y r cos t (1)

Hence

r r = (x r cos t + y r sin t) (x r sin t + y r cos t)

= (x y)r2 cos2 t (y x)r2 sin2 t

= z r2(sin2 t + cos2 t) = z r2

(b) Show that r + 2r = 0

The acceleration is the time derivative of (1)

r = x r2 cos t y r2 sin t = 2(x r cos t + y r sin t) = 2r

Hence r + 2r = 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 BACCAB 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)


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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 theBACCAB 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

(r ) =p r

4or3

Find the electric field, E = at r.

We first use the quotient rule to write

E = = 1

40

p r

r3

=

1

40

r3(p r ) (p r )(r3)

r6

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

E = 1

40

r3(p r ) 3r2(p r )(r)

r6

We now evaluate the two separate gradients

(p r ) = xi

xi(pjxj) = xipj

xjxi

= xipjij = xipi = p

and

r = xi

xi

x21

+ x22

+ x23

= xi1

2

x21

+ x22

+ x23

2xi =xixi

r=

r

r= r


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Hence

E = 1

40

r3p 3r2(p r )r

r6=

1

40

p 3(p r)r

r3

Note that we have used the fact that p is a constant, although this was never

stated 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 dividing

by the constant k)

A = 0.

We now return to the first equation and simplify the double curl using the BACCAB 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.


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1.10.4 Evaluate

r dr

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

2

1

r dr = 12

2

1

d(r2) = 12

r221 = 12r22 12r21However for a closed path, point 1 and point 2 are the same. Thus the integralalong a closed loop vanishes,

r dr = 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 dr =

S

r d

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

1.12.9 Prove that uv d =

v u d

This is an application of Stokes theorem. Let us write

(uv + v u) d =

S

(uv + v u) d (4)

We now expand the curl using

(uv) = (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

(uv + v u) d =

S

[(u) (v) + ( v) (u)] d = 0

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


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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

= (cos , sin , 0), = ( sin , cos , 0), z = (0, 0, 1)

their derivatives are

= ( sin , cos , 0) = ,

= ( cos , sin , 0) =

Using the chain rule, this indicates that

=

= , =

= (5)

Now, we note that the position vector is given by

r = + zz

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

v = r = + zz + + zz = + zz +

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

a = v = + zz + ( + ) + + zz +

= + zz + ( + ) + 2

= ( 2) + zz + ( + 2 )

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

(v ( v )) =

02( v )

Here is the viscosity and 0 the density of the fluid. For axial flow in a cylindricalpipe we take the velocity v to be

v = zv()


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From Example 2.4.1 (v ( v )) = 0

for this choice ofv. Show that2( v) = 0

leads to the differential equation

1

d

d

d2v

d2

1

2dv

d= 0

and that this is satisfied byv = v0 + a2

2

This problem is an exercise in applying the vector differential operators in cylin-

drical coordinates. Let us first compute V = v

V = v =1

z

z0 0 v()

=

dv

d V =

dv

d

Note that, since v() is a function of a single variable, partial derivatives of v arethe same as ordinary derivatives. Next we need to compute the vector Laplacian2( v ) = 2V. Using (2.35) in the textbook, and the fact that on the Vcomponent is non-vanishing, we find

(2V) = 2

2V

= 0

(2V) = 2(V)

1

2V =

2

dv

d

+

1

2dv

d

(2V)z = 0

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

(2V) = 1

d

d

d2v

d2

+

1

2dv

d(6)

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

To prove that v = v0 + a22 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


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

1

d

d

df

d

1

2f = 0

Expanding the derivatives in the first term yields

d2f

d2+

1

df

d

1

2f = 0

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

( 1) + 1 = 0 = 1

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

f = 2a + b1

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

v =

f d = v0 + a

2 + b log

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 = 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 + a

2.

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() = vmax

1

2

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).


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Chapter 2

2.5.11 A particle m moves in response to a central force according to Newtons second law

mr = r f(r )

Show that r r = c, a constant, and that the geometric interpretation of this leadsto Keplers second law.

Actually, r r is basically the angular momentum, L = r p = mr r. To show

that L is constant, we can take its time derivative

L =d

dt(mr r ) = mr r + mr r

The first cross-product vanishes. So, by using Newtons 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 is

conserved). 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 dr |

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

dA

dt= 1

2

r drdt = 12 |r r |

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


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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 four

dimensions, 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 44 antisymmetric

matrix, this means instead of having 42

= 16 independent elements, we actuallyonly have 12

(4)(3) = 6 independent choices for the last index pair (this is thenumber of elements in an antisymmetric 44 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


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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 = 112

d2(d2 1)

Putting in d = 4 yields the expected 20. However, it is fun to note that putting

in 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(r2ij 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 = jmxixj i = j


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We can enforce the condition i = j by using the Kronecker delta, ij . Similarly,the condition i = j can be enforced by the opposite expression 1 ij . Thismeans we can write

Iij = m(r2 x2i )ij mxixj(1 ij) (no sum)

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

end up with

Iij = mr2ij mx

2i ij mxixj + mx

2i ij = m(r

2ij xixj)

Note that there is a typo in the books version of the homework exercise!

b) Show thatIij = MilMlj = milkxkljmxm

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

We may calculate

MilMlj = milkxkljmxm = mlkiljmxkxm

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

lkiljm = kjim kmij (2)

This is actually the BACCAB rule in index notation. Hence

MilMlj = m(kjim kmij)xkxm = m(kjxkimxm kmxkxmij)

= m(xjxi xkxkij) = m(r2ij xixj) = Iij

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

22 + x

23 = r

2.

2.9.12 Given Ak =12

ijkBij with Bij = Bji , antisymmetric, show that

Bmn = mnkAk

Given Ak =12

ijkBij , we compute

mnkAk =12

mnkkijBij =12

kmnkijBij =12

(minj mjni)Bij

= 12

(Bmn Bnm) = Bmn

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


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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 = cos , y = sin , z = z (3)

to compute the differentials

dx = d cos sin d, dy = d sin + cos d, dz = dz

The line element is then

ds2 = dx2 + dy2 + dz2 = (d cos sin d)2 + (d sin + cos d)2 + dz2

= d2 + 2d2 + dz2

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

gij =

1 0 00 2 0

0 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

e = x cos + y sin

e = (x sin + y cos )

ez = z

Then

g = e e = (x cos + y sin ) (x cos + y sin ) = cos2 + sin2 = 1

g = e e = (x cos + y sin ) (x sin + y cos )

= ( cos sin + sin cos ) = 0

etc . . .

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 2 0

0 0 1

(5)


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

coordinates.

We may compute the Christoffel components using the expression

ijk =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

g = 2

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

= , = =

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

= , = = 1

(6)

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

2.10.12 Using the kij from Exercise 2.10.11, write out the covariant derivatives Vi;j of a

vector V in circular cylindrical coordinates.

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

Vi;j = Vi,j +

ijkV

k

where the semi-colon indicates covariant differentiation and the comma indicates

ordinary partial differentiation. To work out the covariant derivative, we justhave to use (6) for the non-vanishing Christoffel connections. The result is

V; = V, +

kV

k = V,

V; = V, +

kV

k = V, + V

= V, +1

V

Vz ; = Vz, +

zkV

k = Vz,

V; = V, +

kV

k = V, + V

= V, V

V; = V, +

kV

k = V, + V

= V, +1

V

Vz ; = Vz, + zkVk = Vz,

V;z = V,z +

zkV

k = V,z

V;z = V,z +

zkV

k = V,z

Vz ;z = Vz,z +

zzkV

k = Vz,z

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


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Textbook problems: Ch. 5: 5.1.1, 5.1.2

Chapter 5

5.1.1 Show thatn=1

1

(2n 1)(2n + 1)=

1

2

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=

1

2

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


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which agrees with (1).

5.1.2 Show thatn=1

1

n(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 =1

2, s2 = s1 +

1

2 3=

2

3, s3 = s2 +

1

3 4=

3

4

This suggests that

sm =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 + 1

m + 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

1

n(n + 1)=

n=1

1

n

1

n + 1

Hence

sm =1

1

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 = L2dx2 + dy2 + dz2

z2


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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 0

0 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 the

metric by ijk =1

2(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 ijkis symmetric in the last two indices, we compute

zxx = 1

2gxx,z =

L2

z3, xzx = xxz =

1

2gxx,z =

L2

z3

zyy = 1

2gyy,z = L

2

z3, yzy = yyz =

1

2gyy,z = L

2

z3

zzz =1

2gzz,z =

L2

z3

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

zxx =1

z, xzx =

xxz =

1

z

zyy =1

z

, yzy = yyz =

1

zzzz =

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) + ijk xj(t)xk(t) = 0


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

x2

zxz = 0

y 2z yz = 0

z +1

z(x2 + y2 z2) = 0

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

x + xxzxz + xzx zx = 0 x

2

zxz = 0 (5)

y + yyz yz + yzy zy = 0 y 2z yz = 0 (6)

z + zxxxx + zyy yy +

zzz zz = 0 z +

1

z(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 tanh(v0t)

y = y0 + R sin tanh(v0t)

z = R sech(v0t)

where x0, y0, R, 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 interesting

to 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)


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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 closer

to the solution. Now, inserting (8) and (9) into the z equation, (7), we obtain

z + (a2x + a2y)z

3

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 + a2y)

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

u(t) =1

v0

a2x + a

2y 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 =axu2

=v20ax

a2x + a2y

sech2(v0(t t0))

which may be integrated to yield

x(t) = x0 +v0ax

a2x + a2y

tanh(v0(t t0))

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

y(t) = y0 +v0ay

a2x + a2y

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 =v0R

cos , ay =v0R

sin


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in which case we see that

x = x0 + R cos tanh(v0(t t0))

y = y0 + R sin 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 tanh2(v0t) + R2 sin2 tanh2(v0t) + R

2 sech2(v0t)

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

2

This is indeed the equation for a sphere, (x x0)2 + (y y0)

2 + z2 = R2.


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Chapter 5

5.2.6 Test for convergence

a)n=2

(ln n)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 ln n grows very slowly, so that its

inverse goes to zero very slowly too slowly, in fact, to converge. To prove this,we can perform a simple comparison test. Since ln n < n for n 2, we see that

an = (ln n)1 > n1

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

anan+1

=10

n + 1

Taking the limit, we obtain

limn

anan+1

= 0

hence the series diverges by the ratio test.


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c)n=1

1

2n(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)

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

1

2n + 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=

1

2ln(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(ln n)q]

Since the ln n 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


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more precise, we can use Raabes test

an =1

np(ln n)q

anan+1

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

np(ln n)q

=

1 +

1

np

1 +

ln(1 + 1n )

ln n

q

=

1 +

1

n

p1 +

1

n ln n+

q

=

1 +p

n+

1 +

q

n ln n+

=

1 +p

n+

q

n ln n+

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

limn

n

an

an+1 1

= lim

n

p +

q

ln n+

= p

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

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

p = 1 an =1

n(ln n)q

we evaluate 2

dxx(ln x)q =ln 2

duuq

where we have used the substitution u = ln x. 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 Gausss hypergeometric series

F(,,; x) = 1 +

1!x +

( + 1)(+ 1)

2!(+ 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 ,, are well behaved, this


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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 =( + 1) ( + n 1)(+ 1) (+ n 1)

n!(+ 1) (+ n 1)xn

an

an+1= (n + 1)(+ n)

( + n)(+ n)x1

This allows us to begin with the ratio test

limn

anan+1

= limn

(n + 1)(+ n)

( + n)(+ n)x1 = x1

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

anan+1

=(n + 1)(+ n)( + n)(+ n)

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

anan+1

= 1 +

(n + 1)(+ n)

( + n)(+ n) 1

= 1 +

( + 1)n +

( + n)(+ n)

We can now see that the fraction approaches (+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 = + 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 = + 1, we see that the hypergeometric series F(,,; 1)converges for > + (h > 1) and diverges otherwise.

To summarize, we have proven that for non-negative x, the hypergeometric seriesconverges for x < 1 (any ,,) and x = 1 if > + , 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 > + . Based on

the 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 + 1 < + (this is harder toshow).


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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 Raabes test

as =(2s 1)!!

(2s)!!(2s + 1)

asas+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

asas+1

= 1 +

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

(2s + 1)2 1

= 1 +

6s + 5

(2s + 1)2

Then

lims

s

as

as+1 1

= lim

ss

6s + 5

(2s + 1)2

=

3

2

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 + 12

8s2 + 18s + 7= 1 +

4s + 5

8s2 + 18s + 7> 1

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


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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 ismissing 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 + 6

8s2 + 18s + 7= 1

4s + 1

8s2 + 18s + 7< 1

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

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


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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 + x1 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+1xn

nn=1

(1)n+1(x)n

n

=

n=1((1)n+1 + 1)

xn

n= 2

n oddxn

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 n4. Find the resulting series.

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

(1 + a1x + a2x2

+a3x3

) 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


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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 + 13x

3) + a1(x2 12x

3) + a2x3

+n=4

(1)n+1

xn

n

a1xn

n 1+

a2xn

n 2

a3xn

n 3

= x + (a1 12)x

2 + (a2 12a1 +

13)x

3

+n=4

(1)n+1

1

n

a1n 1

+a2

n 2

a3n 3

xn

(1)

Combining the terms over a common denominator yields

1n

a1n 1 +

a2n 2

a3n 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)n

2

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

n(n 1)(n 2)(n 3)

To make this converge as n4, we need to cancel the coefficients of n3, n2 and n

in 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 + 116 x

3+6

n=4(1)n

xn

n(n 1)(n 2)(n 3)

or

ln(1 + x) =x + 52x

2 + 116 x3 + 6

n=4(1)n x

n

n(n1)(n2)(n3)

(1 + x)3


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5.4.3 Show that

a)

n=2

[(n) 1] = 1

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

n=2

[(n) 1] =n=2

p=1

1

pn

1

=n=2

p=2

1

pn

=

p=2

n=2

1

pn

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

pn =p2

1 p1=

1

p(p 1)

In this case

n=2

[(n) 1] =

p=2

1

p(p 1)=

p=2

1

p 1

1

p

= 1

since this is a telescoping series.

b)n=2

(1)n[(n) 1] =1

2

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

n=2

(1)n[(n) 1] =n=2

(1)n

p=2

1

pn

=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=

1

p(p + 1)

In this case

n=2

(1)n[(n) 1] =

p=2

1

p(p + 1)=

p=2

1

p

1

p + 1

=

1

2


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5.5.2 For what range ofx 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 is

convergent for 0 s < 1. While this is all very obvious, the introduction of

this 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), the

geometric 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 sin nx)

is uniformly convergent for < x < .

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

|(x)| = |an cos nx + bn sin nx| |an cos nx| + |bn sin nx| |an| + |bn|

for the entire domain x (, ). Since the problem states that

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

(x) is uniformly

convergent for x (, ).


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Chapter 5

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

Since cos x = 1 12x2 + 124x

4 and sin x = x 16x3 + 1120x

5 , we divideto obtain

cot x =1 12x

2 + 124x4

x 16x3 + 1120x

5 =

1 12x2 + 124x

4

x(1 16x2 + 1120x

4 )

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 + 1120x

4 )1 = 1 + (16x2 1120x

4 + ) + ( 16x2 1120x

4 + )2+

= 1 + 16x2 + ( 1120 +

136 )x

4 +

= 1 + 16x2 + 7360x

4 +

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

cot x = x1(1 12x2 + 124x4 )(1 + 136x2 + 7360x4 + )

= x1(1 + (12 +16)x

2 + ( 124 112 +

7360 )x

4 + )

= x1(1 13x2 145x

4 + )

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

xex 1

=

n=1

Bnn!

xn = 12

x +

p=0

B2p(2p)!

x2p

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

12 . Moving the

12x to the left hand side, and using the identity

x

ex 1+

1

2x =

x

2

ex + 1

ex 1=

x

2coth

x

2


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we obtainx

2coth

x

2=

p=0

B2p(2p)!

x2p

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

coth x =

p=0

2B2p(2p)!

(2x)2p1

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)2p1

Multiplying through by i and simplifying then gives the expression

cot x =

p=0

(1)p22pB2p(2p)!

x2p1

5.6.19 a) Plancks theory of quandized oscillators leads to an average energy

=

n=1

n0 exp(n0/kT)

n=0

exp(n0/kT)

where 0 is a fixed energy. Identify the numerator and denominator as binomialexpansions and show that the ratio is

=0

exp(0/kT) 1

To simplify the expressions, we begin with the substitution r = exp(0/kT).This yields = N/D where the numerator and denominator are

N =

n=1

n0rn, 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 = 0rd

dr

n=1

rn

= 0r

d

dr

r

1 r=

0r

(1 r)2


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Dividing the numerator and denominator finally yields

=0r

1 r=

0r1 1

=0

exp(0/kT) 1

b) Show that the of part (a) reduces to kT, the classical result, for kT 0.

In this limit, 0/KT 1, we may expand the exponential in the denominator

exp(0/kT) 1 +0kT

+

As a result

00/kT +

kT

5.7.4 The analysis of the diffraction pattern of a circular opening involves20

cos(c cos ) d

Expand the integrand in a series and integrate by using

20

cos2n d =(2n)!

22n(n!)22,

20

cos2n+1 d = 0

Setting x = c cos , we expand

cos x =n=0

(1)n

(2n)!x2n

so that

20

cos(c cos ) d =

20

n=0

(1)n

(2n)!c2n cos2n d

=

n=0

(1)nc2n

(2n)!20

cos2n d

=n=0

(1)nc2n

(2n)!

2(2n)!

22n(n!)2=n=0

2(1)n

(n!)2

c2

2n


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

f() =(1 + )

2

2 + 2

1 + 2

ln(1 + 2)

Here = h/mc2, the ratio of the photon energy to the electron rest mass energy.

Findlim0

f()

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

2 + 2

1 + 2= 2(1+)(1+2)1 = 2(1+)(12+42 ) = 2(1+22+ ) (1)

next we turn to the log

ln(1 + 2)

= 1 ln(1+2) = 1(2 12 (2)

2+ 13(2)3+ ) = (22+ 83

2 )

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

f() =(1 + )

2[2(1 + 22 + ) (2 2 + 83

2 )]

=(1 + )

2[42 83

2 + ] =(1 + )

2[43

2 + ] = 43 (1 + )[1 + ]

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

lim0

f() =4

3

5.9.11 The integral 10

[ln(1 x)]2dx

x

appears in the fourth-order correction to the magnetic moment of the electron. Show

that it equals 2(3).We begin with the variable substitution

1 x = et, dx = etdt

to obtain 10

[ln(1 x)]2dx

x=

0

t2et

1 etdt


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

et

1 et= et(1 et)1 = et(1 + et + e2t + e3t + ) =

n=1ent

This gives

10

[ln(1 x)]2dx

x=

0

t2

n=1

ent

dt =

n=1

0

entt2dt

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

n3

0

ess2ds =

n=1

n3(3) = 2(3)

Here we have used the definition of the Gamma function

(z) =

0

essz1dz

as well as the zeta function

(z) =n=1

nz

5.10.1 Stirlings formula for the logarithm of the factorial function is

ln(x!) =1

2ln 2 +

x +

1

2

ln x x

Nn=1

B2n2n(2n 1)

x12n

The B2n are the Bernoulli numbers. Show that Stirlings 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 do

so, we first examine the form of an(x)

an(x) = B2n

2n(2n 1)x12n

Using the relation

B2n =(1)n+12(2n)!

(2)2n(2n)


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we find

|an(x)| =2(2n 2)!(2n)

(2)2nx12n

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)!(2n)

(2)2n(2)2n+2

2(2n)!(2n + 2)x2 =

(2)2

2n(2n 1)

(2n)

(2n + 2)x2 (3)

Since limn (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 term

n = 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 Stirlings formula is an asymptoticexpansion.

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

=n

s=1

s1 ln n 12n

+Nk=1

B2k(2k)n2k

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

= limn

n

s=1

s1 ln n

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

into the Euler-Maclauren sum formulan

x=1

f(x) =

n1

f(x)dx +1

2f(1) +

1

2f(n) +

Np=1

1

(2p)!B2p[f

(2p1)(n) f(2p1)(1)]

1

(2N)!

10

B2N(x)n1=1

f(2N)(x + )dx

(4)


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However, we first note that, for f(x) = 1/x we haven1

f(x)dx =

n1

dx

x= ln n

as well as

f(k)(x) = (1)k k!xk+1

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

ns=1

s1 = ln n +1

2+

1

2n

Np=1

B2p2p

[n2p 1]

10

B2N(x)n1=1

(x + )2N+1dx

orn

s=1

s1 ln n =1

2+

1

2n

Np=1

B2p2p

[n2p 1] + RN(n) (5)

where the remainder RN(n) is given by

RN(n) =

10

B2N(x)n1=1

(x + )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

= limn

n

s=1

s1 ln n

= 12

+N

p=1

B2p2p

+ RN() (7)

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

RN(n) =

10

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)

ns=1

s1 + ln n = 12n

+

Np=1

B2p2p

1n2p

+ [RN() RN(n)]

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

=n

s=1

s1 ln n 1

2n+

Np=1

B2p2p

1

n2p


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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 complexquantity , we have ||2 = ()2 + ()2 ()2. Taking a square root gives|| || or || ||. For the present case (where = 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, z2and 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|).


50/79

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

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

for z > 0. This implies that |z + 1| > |z 1| for z > 0. The picture here is

that 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)N1n=0

cos nx = sin(N x/2)sin x/2

cos(N 1) x2

b)

N1n=0

sin nx =sin(N x/2)

sin x/2sin(N 1)

x

2

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

S =N1n=0

cos nx + iN1n=0

sin nx =N1n=0

(cos nx + i sin nx) =N1n=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 =

N1n=0

rn =1 rN

1 r=

1 eNix

1 eix=

e1

2Nix(e

1

2Nix e

1

2Nix)

e1

2ix(e

1

2ix e

1

2ix)


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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 sin = (ei ei)/2i. The result is

S = e1

2(N1)ix sin(N x/2)

sin x/2= cos 1

2

(N 1)x + i sin 1

2

(N 1)x sin(Nx/2)sin x/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) = ey sin x As above, we have

u

x=

v

y= ey sin x u = ey cos x + C(y)

u

y=

v

x= ey cos x u = ey 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 + ey cos x + iey sin x = c + ey(cos x + i sin x) = c + eixy = c + eiz


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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) andu

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+i0

zdz

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

zdz =

(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+i0

zdz =

10

(x iy)y=0

dx +

10

(x iy)x=1

idy

=

10

xdx +

10

(i + y)dy = 12x210

+ (iy + 12y2)10

= 1 + i


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Similarly, for path 2, we find

1+i0

zdz =

10

(x iy)x=0

idy +

10

(x iy)y=1

dx

=

1

0 ydy +

1

0 (x i)dx =

1

2y

210 + (

1

2x

2

ix)10 = 1 i

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

6.4.3 Solve Exercise 6.3.4 [C

dz/(z2+z) where C is a circle defined by |z| > 1] by separatingthe integrand into partial fractions and then applying Cauchys integral theorem formultiply connected regions.

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

C

dz

z z0 = 2i (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

1

z

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 = 2i 2i = 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 2i 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

=1

2

C

dz

z 1

1

2

C

dz

z + 1= i i = 0

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


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Textbook problems: Ch. 6: 6.5.2, 6.5.8, 6.6.2, 6.6.7

Ch. 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)m1,

f(z) = m(m 1)(1 + z)m2,

f(z) = m(m 1)(m 2)(1 + z)m3,

etc.

In general, the n-th derivative is given by

f(n)(z) = m(m 1)(m 2) (m n + 1)(1 + z)mn = m!(m n)!

(1 + z)mn

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

1

n!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=

nk=1

m k + 1

k

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


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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 an

integer 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 the

denominator

ez 1 = z + 12z2 + 16z

3 + = z(1 + 12z +16z

+ )

Hencef(z) = (ez 1)1 = z1(1 + 12z +

16z

2 + )1

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

16z

2 + . This gives

f(z) = z1 1 (12z + 16z2 + ) + ( 12z + 16z2 + )2 = z1

1 12z +

112z

2 +

=1

z

1

2+

1

12z +

(1)

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

f(z) =1

ez 1= z1

z

ez 1

= z1

n=0

Bnn!

zn =B0z

+B1+1

2B2z +

1

6B3z

2+

Inserting B0 = 1, B1 = 12 and B2 = 16 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 1

z + 1


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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

+1z1z

+1zz

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 iz + 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 of

this branch, in light of the cut you have chosen.

Returning to the binomial expansion off(z) = (1+z)m, we note that ifw = 1+z,we end up with a function f(w) = wm which is multi-valued under w we2i

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

+z1

z

cut

z

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


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|1 + z|meim. 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) = 0, f

2(z0) = 0. Show that

a1, the coefficient of (z z0)1

in a Laurent expansion of f(z) at z = z0, is given by

a1 =f1(z0)

f2(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) +12f

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) +

f2(z0)(z z0) +12f

2 (z0)(z z0)2 +

=f1(z0)/f

2(z0)

z z0

1 + (f1/f1)(z z0) +

1 + 12 (f

2 /f

2)(z z0) +

For z z0, the denominator 1 +12 (f

2 /f

2)(z z0) + may be inverted usingthe 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 + (

f1f1

f22f2

)(z z0) +

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

a1 =f1(z0)

f2(z0)


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7.1.4 The Legendre function of the second kind Q(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 ln w 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

1

z

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

z + 1 = |z + 1|ei, < ,

z 1 = |z 1|ei , <

Thus

Q0(z) =12 ln(z + 1)

12 ln(z 1) =

12 ln

z + 1z 1+ i2 ( ) (2)

It is the manner in which the arguments and show up in (2) that indicate thebranch cut is as written. For x > 1 on the real axis, both and are smooth,

0 and 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 0 is smooth infinitesimally above or below the axis,but the argument is discontinuous: above the axis, but belowthe axis. This shows that the value of Q0 changes by i when crossing the realaxis. For x < 1, the situation is more interesting, as both and jump whencrossing the axis. However the difference ( ) 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 points

z = 1 or z = 1 individually. However, we can take a big circle around bothpoints. In this case, + 2 and + 2, but once again the difference( ) 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


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Show thatQ0(x) =

12 [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) : 0, ,

x i0 (below the cut) : 0,

The expression of (2) yields

Q0(x + i0) =1

2ln x + 1

x 1 i

2,

Q0(x i0) =12 ln

x + 1x 1+ i2

(3)

Taking the average gives

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

12 ln

x + 1x 1 = 12 ln 1 + x1 x

where we have used the fact that |x 1| = 1 x for |x| < 1. In this case, we see

that averaging the function below and above the cut cancels the imaginary parts,i/2 in (3).


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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) = lim0+

1

2i

eixs

x idx

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

z

u(s)

C

IR

R

i

Always assuming the limit 0+

, we see that the real integral for u(s) may bepromoted to a closed contour integral

1

2i

C

eizs

z i dz = 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 = Rei, dz = iReid

so that

IR =1

2i

0

eisRei

Rei i iReid =

1

2

0

eisR(cos +i sin )d

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

IR =1

2

0

eisR cos esR sin d (2)


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This vanishes by Jordans lemma provided sR sin > 0 so that the real exponen-tial is suppressed instead of blowing up (in fact, this is Jordans lemma). SinceR is positive and sin > 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) = 12i

C

eizs

z i dz = residue of eizs

z i at z = i (s > 0)The residue at i is simply lim0+ e

s = 1. Hence we have confirmed thatu(s) = 1 for s > 0.

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

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 =

12 . Since the contour integral still has the value of 1 (residue at

the pole at i), inserting IR =12 into (1) gives

1 = u(0) + 12 u(0) = 12which is a nice result indicating that the step function is given completely

Arfken Solutions 2

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