Physics 451
Fall 2004 Homework Assignment #10 — Solutions
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 m
(1 + z)
∞ X m(m − 1) 2 m zn = 1 + mz + z + ··· = n 1·2 n=0
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 0 (z) = m(1 + z)m−1 , f 00 (z) = m(m − 1)(1 + z)m−2 , f 000 (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! (1 + z)m−n (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 the Taylor series has the form ∞ ∞ ∞ X X X 1 (n) m! m n n f (0)z = z = zn f (z) = n n! n!(m − n)! n=0 n=0 n=0 For non-integer m (but integer n), the binomial coefficient may be defined by the Gamma function, or alternately by
m n
=
n Y m(m − 1)(m − 2) · · · (m − n + 1) m−k+1 = 1 · 2 · 3···n k k=1
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| < 1 is 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 radius of convergence of |z| < 1. On the other hand, for m a non-negative integer, the series terminates (giving a traditional binomial expansion for (1 + z) raised to an integer power), and the radius of convergence is infinite. This is consistent with the 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 any way we wish. What we can do here is to start with a Taylor expansion of the denominator ez − 1 = z + 12 z 2 + 16 z 3 + · · · = z(1 + 21 z + 16 z + · · ·) Hence f (z) = (ez − 1)−1 = z −1 (1 + 12 z + 61 z 2 + · · ·)−1 For small z, we invert the series using (1 + r)−1 = 1 − r + r2 − · · · where r = 1 1 2 2 z + 6 z + · · ·. This gives f (z) = z −1 1 − ( 12 z + 16 z 2 + · · ·) + ( 12 z + 16 z 2 + · · ·)2 − · · · 1 2 = z −1 1 − 12 z + 12 z + ··· 1 1 1 = − + z + ··· z 2 12
(1)
Of course, we could also take the hint and use the generating function of Bernoulli numbers to write 1 f (z) = z = z −1 e −1
z ez − 1
=z
−1
∞ X Bn n B0 1 1 z = +B1 + B2 z + B3 z 2 +· · · n! z 2 6 n=0
Inserting B0 = 1, B1 = − 21 and B2 = 61 then immediately yields the last line of (1). However, this method requires us to either remember or look up the values of the Bernoulli numbers. 6.6.2 What part of the z-plane corresponds to the interior of the unit circle in the w-plane if z−1 a) w = z+1
Note that, by trying a few numbers, we can see that z = 0 gets mapped to w = −1 and z = 1 gets mapped to w = 0 z
1 z+
w
z z −1
w=
z −1 z +1
In fact, the unit circle in the w-plane is given by the equation |w| = 1, which maps to |z − 1| = |z + 1| in the z-plane. Geometrically, this is saying that the point z is equidistant to both +1 and −1. This can only happen on the imaginary axis (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 may conclude that the right half (first and fourth quadrants) of the complex z-plane gets 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 to the points +i and −i instead. Thus in this case the real axis (y = 0) gets mapped to the circle. The upper half plane (first and second quadrants) gets mapped to the interior of the unit circle. 6.6.7 For noninteger m, show that the binomial expansion of Exercise 6.5.2 holds only for a 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 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 a suitable branch must be defined. We take the branch cut to run from w = 0 along the negative real axis in the w-plane. However, the simple transformation z = w − 1 allows us to return to the original z-plane. In this case, w = 0 is the same 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 z
z+ α
cut
1
z
−1
where the principle value is taken to be −π < α ≤ π. In this case, f (z) =
|1 + z|m eimα . Since the Taylor series is expanded about z = 0, the radius of convergence is |z| < 1, which is the distance to the nearest singularity (the branch point at z = −1). This is why it is desired to take the branch cut running along the left (otherwise, if it goes inside the unit circle, it will reduce or eliminate the radius 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, f20 (z0 ) 6= 0. Show that a−1 , the coefficient of (z − z0 )−1 in a Laurent expansion of f (z) at z = z0 , is given by a−1 =
f1 (z0 ) f20 (z0 )
Since f1 (z) and f2 (z) are both analytic, they may be expanded as Taylor series f1 (z) = f1 (z0 ) + f10 (z0 )(z − z0 ) + · · · , f2 (z) = f20 (z0 )(z − z0 ) + 12 f200 (z0 )(z − z0 )2 + · · · Here we have already set f2 (z0 ) = 0 since the function vanishes at z = z0 . As a result, we have f1 (z0 ) + f10 (z0 )(z − z0 ) + · · · f1 (z) = 0 f2 (z) f2 (z0 )(z − z0 ) + 21 f200 (z0 )(z − z0 )2 + · · · f1 (z0 )/f20 (z0 ) 1 + (f10 /f1 )(z − z0 ) + · · · = z − z0 1 + 12 (f200 /f20 )(z − z0 ) + · · ·
f (z) =
For z ≈ z0 , the denominator 1 + 21 (f200 /f20 )(z − z0 ) + · · · may be inverted using the geometric series relation 1/(1 + r) = 1 − r + r2 − · · ·. The result is a Laurent series of the form f1 (z0 )/f20 (z0 ) f10 f200 f (z) = 1 + ( − 0 )(z − z0 ) + · · · z − z0 f1 2f2 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. The branch points are joined by a cut line along the real (x) axis. a) Show that Q0 (z) = 21 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 at z = ±1 as promised. Joining the branch points by a cut line along the real axis gives the picture
z
z
z−
1
z +1
β
α −1
1
Of course, to make this picture well defined, we provide a principle value for the arguments z + 1 = |z + 1|eiα , −π < α ≤ π, z − 1 = |z − 1|eiβ ,
−π < β ≤ π
Thus Q0 (z) =
1 2
ln(z + 1) −
1 2
ln(z − 1) =
1 2
z + 1 i + (α − β) ln z − 1 2
(2)
It is the manner in which the arguments α and β show up in (2) that indicate the branch 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 there is no discontinuity in Q0 (x > 1) and thus no branch cut. For −1 < x < 1, on the other 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 below the axis. This shows that the value of Q0 changes by ±iπ when crossing the real axis. For x < −1, the situation is more interesting, as both α and β jump when crossing the axis. However the difference (α − β) is unchanged. In this sense, the two branch cuts cancel each other out, so that the function Q0 (x < −1) is well defined 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 both points. 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) =
1 2
ln
1+x 1−x
Show that Q0 (x) = 12 [Q0 (x + i0) + Q0 (x − i0)] The branch cut prescription described in part a) is somewhat unfortunate for real arguments |x| < 1, since those values of x sit right on top of the cut. To make this 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 is doing for us. Noting that (for |x| < 1) the arguments have the following values x + i0 (above the cut) : x − i0 (below the cut) :
α ≈ 0, α ≈ 0,
β ≈ π, β ≈ −π
The expression of (2) yields x + 1 − Q0 (x + i0) = ln x − 1 x + 1 1 + Q0 (x − i0) = 2 ln x − 1 1 2
iπ , 2 iπ 2
(3)
Taking the average gives Q0 (x) =
1 2 [Q0 (x
+ i0) + Q0 (x − i0)] =
1 2
x + 1 = ln x − 1
1 2
ln
1+x 1−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, ±iπ/2 in (3).
