Chapter 1
Review of Complex
Numbers
Complex numbers are defined in terms of the imaginary unit, i, having the property
i2
= −1. (1.1)
A general complex number has the form
z = x + iy, (1.2)
where x, y are real numbers. We also often write
z = Re z + iIm z, (1.3)
where Re z is the “real part of z,” and Im z is the “imaginary part of z.” Complex numbers are added and multiplied just like real numbers: If
z1 = x1+ iy1, (1.4a) z2 = x2+ iy2, (1.4b) then z1+ z2 = (x1+ x2) + i(y1+ y2), (1.5a) z1z2 = x1x2+ iy1x2+ ix1y2+ i 2 y1y2 = x1x2− y1y2+ i(x1y2+ x2y1). (1.5b)
The complex conjugate of a number is obtained by reversing the sign of i: If z = x + iy, we define the complex conjugate of z by
z∗= x − iy. (1.6)
2 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS -6 y axis Imaginary axis x axis Real axis z-plane x y r θ
Figure 1.1: Geometrical interpretation of a complex number z = x + iy.
(Sometimes the notation ¯z is used for the complex conjugate of z.) Note that
Re z = z + z∗
2 , (1.7a)
Im z = z − z∗
2i . (1.7b)
Note also that
zz∗= x2
+ y2
(1.8) is purely real and non-negative, so we define the modulus, or magnitude, or absolute value of z by
|z| =√zz∗=p(Re z)2+ (Im z)2, (1.9)
where the positive square root is implied.
We give a simple geometrical interpretation to complex numbers, by thinking of them as two-dimensional vectors, as sketched in Fig. 1.1. Here the length of the vector is the magnitude of the complex number,
r = |z|, (1.10)
and the angle the vector makes with the real axis is θ, where
tan θ = y/x; (1.11) the quadrant θ lies in is determined by the sign of x and y. We call
θ = arg z (1.12)
the argument or phase of z. The above geometrical picture is sometimes called an Argand diagram.
3 Version of August 22, 2011 -6 y axis x axis z-plane r θ z = x + iy @ @ @ @ @ R r −θ z∗= x − iy
Figure 1.2: Geometrical interpretation of complex conjugation.
There is an arbitrariness in the choice of the argument θ of a complex number z, for one can always add an arbitrary multiple of 2π to θ without changing z,
θ → θ + 2πn, n an integer, z → z. (1.13) It is often convenient to define a single-valued argument function arg z. By convention, the principal value of arg z is that phase angle which satisfies the inequality
−π < arg z ≤ π. (1.14) (Note that radian measure is always employed.) For every z there is a unique arg z lying in this range.
The geometrical significance of complex conjugation is shown in Fig. 1.2. Complex conjugation corresponds to reflection in the x-axis.
From the Argand diagram we can write down the “polar representation” of a complex number,
z = r cos θ + ir sin θ
= r(cos θ + i sin θ), (1.15)
so if we have two complex numbers,
z1= r1(cos θ1+ i sin θ1), (1.16a)
z2= r2(cos θ2+ i sin θ2), (1.16b)
the product is
z1z2 = r1r2{cos θ1cos θ2− sin θ1sin θ2
+ i [cos θ1sin θ2+ cos θ2sin θ1]}
4 Version of August 22, 2011CHAPTER 1. REVIEW OF COMPLEX NUMBERS
That is, the moduli of the complex numbers multiply,
|z1z2| = |z1||z2|, (1.18a)
while the arguments add,
arg(z1z2) = arg z1+ arg z2. (1.18b)
The latter statement is to be understood as modulo 2π, i.e., equality up to the addition of an arbitrary integer multiple of 2π. In particular, note that
1 zz = 1 z |z| = 1, (1.19a) while 0 = arg 1 zz = arg1 z+ arg z, (1.19b) implying that 1 z = 1 |z|, (1.20a) arg1 z = −arg z. (1.20b)
1.1
De Moivre’s Theorem
From the above, if we choose a unit vector,
z = cos θ + i sin θ, (1.21) successive powers follow a simple pattern:
z2
= cos 2θ + i sin 2θ, (1.22a) z3 = cos 3θ + i sin 3θ, (1.22b) . . . , zn = cos nθ + i sin nθ, (1.22c) or (cos θ + i sin θ)n = cos nθ + i sin nθ, (1.23) where n is a positive integer. This is called De Moivre’s theorem.
1.2
Roots
Suppose we wish to find all the nth roots of unity, that is, all solutions to the equation
zn
1.2. ROOTS 5 Version of August 22, 2011 • • • • • • • • 1 i −1 −i 1+i √ 2 −1+i√ 2 1−i√ 2 −1−i√ 2
Figure 1.3: The eight 8th roots of unity.
where n is a positive integer. If we take the polar form,
z = ρ(cos φ + i sin φ), (1.25) this means ρn (cos nφ + i sin nφ) = 1, (1.26) which implies ρ = 1, (1.27a) nφ = 2πk, (1.27b)
where k is any integer. Thus the nth root of unity has the form
z = cos2πk n + i sin
2πk
n . (1.28)
These are distinct for
k = 0, 1, 2, . . . , n − 1; (1.29) outside of these values of k, the roots repeat. Thus there are n distinct nth roots of unity. For example, for n = 8, the roots are as shown in Fig. 1.3, in the complex plane.
Chapter 2
Infinite Series
2.1
Sequences
A sequence of complex numbers {zn}∞n=1 is a countably infinite set of numbers,
z1, z2, z3, . . . , zn, . . . . (2.1)
That is, for every positive integer k, there is a number, the kth term of the sequence, zk, in the set {zn}∞n=1. Mathematically, a sequence is a
complex-valued function defined on the positive integers. We say that the sequence possesses a limit l,
lim
n→∞zn = l, or zn→ l as n → ∞, (2.2)
if, for every ǫ > 0, no matter how small, there exists a number N for which |zn− l| < ǫ for all n > N. (2.3)
(The number N will depend on ǫ.) That is, {zn}∞n=N +1all lie within a circle of
radius ǫ centered on the point l in the complex plane. A necessary and sufficient condition for a sequence {zn}
∞
n=1 to converge to
a limit is Cauchy’s criterion: A sequence {zn}∞n=1possesses a limit if and only
if for every ǫ > 0, no matter how small, it is possible to find a number N such that
|zn− zm| < ǫ for all n, m > N. (2.4)
(Note that the difference |n − m| may be arbitrarily large.) Thus, all elements of the sequence {zn}∞n=N +1lie within a disk of radius ǫ. Briefly, we say that the
Cauchy condition is
|zn− zm| → 0 for all n, m sufficiently large. (2.5)
Sequences having this property are called Cauchy sequences. Every Cauchy sequence of complex numbers possesses a limit (which is, of course, a complex number)—this property means that the complex numbers form a complete space.
8 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES
2.2
Series
Suppose we have a sequence {ak}∞k=1 from which we construct the finite sums
sn = n
X
k=1
ak, n = 1, 2, 3, . . . . (2.6)
The set of all these sums, {sn}∞n=1, itself forms a sequence. If this latter sequence
has a limit S,
sn → S as n → ∞, (2.7)
then we say that the infinite series
∞ X k=1 ak= lim n→∞ n X k=1 ak (2.8)
possesses the limit S (or converges to S),
∞
X
k=1
ak= S. (2.9)
By the Cauchy criterion, this will be true if and only if n X k=m ak < ǫ (2.10)
for any fixed ǫ > 0, whenever n ≥ m > N, N a number depending on ǫ. Obviously, a necessary condition for
∞
X
k=1
ak (2.11)
to converge is for ak → 0 as k → ∞. However, this is not sufficient, as the
following example shows.
2.3
Examples
2.3.1
Harmonic series
Consider the sum of the reciprocals of the integers,
1 + 1 2+ 1 3 + . . . = ∞ X n=1 1 n. (2.12)
Note that if the nth term of the series is denoted an= 1/n, we have for the sum
of n adjacent terms an+1+ . . . + a2n= 1 n + 1+ . . . + 1 2n> n 1 2n =1 2, (2.13) no matter how large n is. This violates Cauchy’s criterion, so the harmonic series diverges.
2.4. ABSOLUTE AND CONDITIONAL CONVERGENCE9 Version of August 27, 2011
2.3.2
Geometric series
Consider the series
∞
X
m=0
arm, (2.14)
where a is a constant and r ≥ 0. For r 6= 1, the nth partial sum is sn = n X m=0 arm= a1 − rn+1 1 − r , (2.15) so S = lim n→∞sn= a 1 − r if r < 1, (2.16) while the series diverges if r ≥ 1.
2.4
Absolute and Conditional Convergence
Suppose we have a convergent seriesP∞
n=1an. If alsoP ∞
n=1|an| converges, we
say that the original series converges absolutely. Otherwise, the original series is conditionally convergent. (That is, it converges because of sign alternations.) A sufficient condition for (at least) conditional convergence is provided by the following theorem due to Leibnitz:
If the terms of a series are of alternating sign and in addition their absolute values tend to zero, |an| → 0, monotonically, i.e., |an| > |an+1| for sufficiently
largen, then
∞
X
n=1
an converges. (2.17)
In absolutely convergent series one can rearrange the terms without affecting the value of the sum. With conditionally convergent series, one cannot rearrange terms; in fact, such rearrangements can make a conditionally convergent series converge to any desired value, or to diverge!
2.4.1
Example
Consider the conditionally convergent series formed from the divergent harmonic series by alternating every other sign:
1 −12 +1 3− 1 4 + 1 5 − 1 6+ . . . = ln 2, (2.18) which converges to the natural logarithm of 2. Multiply this equation term by term by 1/2: 1 2 − 1 4 + 1 6− 1 8 + 1 10− . . . = 1 2ln 2. (2.19)
10 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES
Add these two series:
1 + 1 3− 1 2 + 1 5 + 1 7− 1 4+ 1 9 + 1 11− 1 6 + . . . = 3 2ln 2. (2.20) Since the reciprocal of each integer occurs exactly once in the last series, we would be tempted to rearrange the series to obtain
1 −12 +1 3− 1 4+ 1 5 − 1 6 + . . . = ln 2, (2.21) which is identical to the original series. There is an obvious contradiction here! In order to obtain the rearrangement (2.21), we have to go further and further out in the series (2.20), which apparently is not permissible.
2.4.2
A Theorem About Absolutely Convergent Series
Not only can absolutely convergent series be rearranged without changing their value, but they can be multiplied together term by term: If two series
S = ∞ X i=1 ui, (2.22a) T = ∞ X i=1 vi (2.22b)
are both absolutely convergent, the series
P = ∞ X i=1 j=1 uivj (2.23)
formed from the product of their terms written in any order, is absolutely con-vergent, and has a value equal to the product of of the individual series,
P = ST. (2.24)
2.5
Convergence Tests
The following tests can determine whether a given series is absolutely convergent or not.
2.5.1
Comparison test
If bn > 0 for all n andP ∞
n=1bn is convergent, and if |an| ≤ bn for all n, then ∞
X
n=1
2.5. CONVERGENCE TESTS 11 Version of August 27, 2011
Also, if |an| ≥ bn> 0 for all n, andP ∞
n=1bn diverges, then ∞
X
n=1
an is not absolutely convergent. (2.25b)
2.5.2
Root test
The seriesP∞
n=1an converges absolutely if from a certain term onward
n
p|an| ≤ q < 1, (2.26)
where q ≥ 0 is independent of n.
Proof: If the inequality holds, |an| ≤ qn. ButP ∞
n=1qn converges for q < 1,
it being the geometric series, so by 2.5.1, P∞
n=1|an| converges.
2.5.3
Ratio test
The seriesP∞
n=1an converges absolutely if from a certain term onward
an+1 an ≤ q < 1, (2.27) where q ≥ 0 is independent of n.
Proof: Without loss of generality, we may assume the inequality holds for all n; otherwise, we renumber the {an} sequence so that 1 labels the first term
for which the inequality (2.27) holds. Then an a1 = an an−1 an−1 an−2 an−2 an−3 · · · a2 a1 ≤ q n−1. (2.28)
Convergence is again assured by comparison with the geometric series. (Whether these tests are satisfied by the first few terms of a series is immaterial, since a finite number of terms of an infinite seris has no effect on the convergence.)
Example When does P∞
n=1nqn converge? If we use the root test, we examine
lim n→∞ n p|an| = |q| lim n→∞ n √ n = |q|,1 (2.29a)
while if we use the ratio test, we look at
lim n→∞ an+1 an = |q| limn→∞ n + 1 n = |q|. (2.29b) In either case, we see that the series is absolutely convergent if |q| < 1, and divergent otherwise.
1Because ln√nn= 1
nln n, which tends to zero as n → ∞,
n
√ n → 1.
12 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES
The following are refinements of the ratio test, which fails (that is, fails to reveal whether the tested series converges or not) when
lim n→∞ an+1 an = 1. (2.30)
For example, this indeterminate limit results for the case an = 1/n, which
yields a divergent series, but also for an = 1/(n ln2n), which corresponds to a
convergent sum (see Sec. 2.5.8).
2.5.4
Kummer’s test
Choose a sequence of positive constants bn. If
bn an an+1 − bn+1≥ C > 0, (2.31) for all n ≥ N, where N and C are fixed numbers, then
∞
X
n=1
an converges absolutely. (2.32)
On the other hand, if
bn an an+1 − b n+1≤ 0, (2.33) and ∞ X n=1 b−1 n diverges, (2.34) then ∞ X n=1 |an| diverges. (2.35)
Proof: If the inequality (2.31) holds, take l ≥ N, so that
C|al+1| ≤ bl|al| − bl+1|al+1|. (2.36)
So we have the inequality
n X l=N +1 |al| ≤ bN|aN| C − bn|an| C ≤ bN|aN| C . (2.37)
Hence, the nth partial sum, for n > N , is
sn = n X i=1 |ai| ≤ N X i=1 |ai| + bN|aN| C . (2.38)
2.5. CONVERGENCE TESTS 13 Version of August 27, 2011
The right-hand side of this inequality is a constant, independent of n. There-fore, the positive sequence of increasing terms {sn} is bounded above, and
con-sequently possesses a limit. The series is absolutely convergent. If the inequality (2.33) holds,
|an| ≥ |a N|bN bn , n > N, (2.39) so sinceP∞ n=1b −1 n diverges, so does P∞ n=1|an|.
2.5.5
Raabe’s test
Raabe’s criterion for absolute convergence is
n an an+1 − 1 ≥ K > 1, (2.40) for all n ≥ N, where N and K are fixed. And if
n an an+1 − 1 ≤ 1, (2.41) then ∞ X n=1 |an| diverges. (2.42)
Proof: In Kummer’s test put bn= n.
2.5.6
Gauss’ test
If an an+1 = 1 + h n + B(n) n2 , (2.43)where h is a constant and the function B(n) is bounded as n → ∞, then P∞
n=1|an| converges for h > 1 and diverges for h ≤ 1.
Proof: For h 6= 1 we can use Raabe’s test:
lim n→∞n h n+ B(n) n2 = h. (2.44)
For h = 1, Raabe’s test is indeterminate. In that case use Kummer’s test with bn= n ln n: for large n, n ln n 1 + h n+ B(n) n2 − (n + 1) ln(n + 1) ≈ n ln n 1 + h n+ B(n) n2 − (n + 1) ln n + 1 n ≈ h +B(n) n ln n − ln n − 1 ≈ (h − 1) ln n − 1 < 0, if h ≤ 1. (2.45)
14 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES
f
1 2 3 4 5 6 7 8 9 10
Figure 2.1: Bounds on a monotone series provided by an integral.
Because ∞ X n=2 1 n ln n diverges (2.46) (see homework), the seriesP∞
n=1|an| diverges.
2.5.7
Integral test
If f (x) is a continuous, monotonically decreasing real function of x such that f (n) = |an|, (2.47) then ∞ X n=1 |an| converges if Z ∞ 1 dx f (x) < ∞, (2.48)
and diverges otherwise.
Proof: It is geometrically obvious that Z ∞ 1 dx f (x) < ∞ X n=1 f (n) < Z ∞ 1 dx f (x) + f (1), (2.49)
for this follows merely from the geometrical meaning of the integral as the area under the curve of the function. See Fig. 2.1.
2.5.8
Examples
• The Riemann zeta function is defined by the series
∞
X
n=1
1
nα = ζ(α). (2.50)
We can test for convergence using Gauss’ test, by examining n + 1
n α
≈ 1 +αn for large n. (2.51) Thus the series converges if α > 1, and diverges if α ≤ 1.
2.6. SERIES OF FUNCTIONS 15 Version of August 27, 2011
• Consider the series
∞
X
n=1
1
(ln n)α. (2.52)
Let’s use Raabe’s test: ln(n + 1) ln n α = ln n + ln(1 + 1/n) ln n α ≈ 1 + α n ln n. (2.53) Because n α n ln n = α ln n → 0 as n → ∞, (2.54) we conclude that the series is divergent.
• To test for convergence of
∞
X
n=2
1
n(ln n)α, (2.55)
let us use the integral test:
Z ∞ 2 dx x(ln x)α = Z ∞ ln 2 d(ln x) (ln x)α = 1 1−α 1 (ln x)α−1 ∞ x=2 , α 6= 1, ln(ln x) ∞ x=2, α = 1 = finite α > 1, ∞ α ≤ 1. (2.56) Thus the series converges if α > 1 and diverges for other real α.
2.6
Series of Functions
2.6.1
Continuity
A (complex-valued) function f (z) of a complex variable is continuous at z0if
f (z) → f(z0) as z → z0 (2.57)
from any direction. That is, given ǫ > 0 we may find a δ > 0 such that
|f(z) − f(z0)| < ǫ whenever |z − z0| < δ. (2.58)
16 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES f x = f = fn ↓ ↑ 2ǫ
Figure 2.2: Uniform convergence of the partial sum fn(x) to the limit f (x). For
all x, fn(x) is within a band of width 2ǫ about f (x).
2.6.2
Uniform Convergence
Consider the infinite series
f (z) =
∞
X
i=1
gi(z) (2.59)
constructed from the sequence of functions {gi}∞i=1. The condition that this
series converge is expressed in terms of the partial sums,
fn(z) = n
X
i=1
gi(z) (2.60)
thusly: given ǫ > 0 we can find an integer N so that for n > N
|fn+p(z) − fn(z)| < ǫ for all p > 0. (2.61)
This is Cauchy’s criterion. In general the N required for this to occur will depend on the point z. If, however, Eq. (2.61) holds for all z if n > N independent of z, we say that the series converges uniformly throughout the region of interest. Equivalently, there exists a function f (z) such that
|f(z) − fn(z)| < ǫ for all n > N, N independent of z. (2.62)
That is, the partial sum fn is everywhere uniformly close to f , the limiting
function. This situation is illustrated in Fig. 2.2 for a real function of a real variable.
Contrast absolute and uniform convergence through the following examples. The series ∞ X n=1 (−1)n n + z2 (2.63)
is only conditionally convergent, because asymptotically the terms become (−1)n/n.
On the other hand, for real z it is uniformly convergent because N+p X n=N +1 (−1)n n + z2 < 1 N + z2 ≤ 1 N, (2.64)
2.6. SERIES OF FUNCTIONS 17 Version of August 27, 2011
which is the Cauchy criterion with ǫ = 1/N . In contrast, consider, for real z, the series
S(z) = ∞ X n=0 z2 (1 + z2)n (2.65)
which converges absolutely. For z = 0, S(0) = 0; and for z 6= 0, S(z) = z2 ∞ X n=0 1 (1 + z2)n = z2 1 − 1 1+z2 = 1 + z2. (2.66)
Thus S(z) is discontinuous at z = 0. The following theorem shows that this series cannot be uniformly convergent there.
Theorem
If a series of continuous functions of z is uniformly convergent for all values of z in a given closed domain, the sum is continuous throughout the domain.
Proof: Let fn(z) = n X i=1 gi(z). (2.67) Since fn(z) → f(z) uniformly, (2.68)
we can find, for any ǫ > 0, a value of n such that
|fn(z) − f(z)| < ǫ for all z (2.69)
throughout the domain. Then
|f(z) − f(z′ )| = |f(z) − fn(z) + fn(z) − f(z ′ ) + fn(z ′ ) − fn(z ′ )| ≤ |f(z) − fn(z)| + |f(z ′ ) − fn(z ′ )| + |fn(z) − fn(z ′ )|.(2.70) Since the fn’s are continuous, we can find a δ for any given ǫ such that
|fn(z) − fn(z′)| < ǫ whenever |z − z′| < δ. (2.71) Therefore, |f(z) − f(z′ )| < 3ǫ whenever |z − z′ | < δ. (2.72) QED.
Even if the limit function is continuous, convergence to it need not be uni-form, as the following example shows:
18 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES @ @ @ @ @ f x 0 1/n 2/n
Figure 2.3: Sketch of the function fn(x) given by Eq. 2.73).
Example
Consider the sequence of continuous functions,
fn(x) = nx, 0 ≤ x ≤ 1/n, (2/n − x)n, 1/n ≤ x ≤ 2/n, 0, otherwise. (2.73)
This function in sketched in Fig. 2.3. Note that the maximum of the function fn(x) is 1. On the other hand, for all x,
lim
n→∞fn(x) = 0, (2.74)
which is certainly a continuous limit function. But the convergence to this limit is not uniform, for there is always a point, x = 1/n, for which
0 − f n 1 n = 1 (2.75)
no matter how large n is. So the convergence is nonuniform.
Properties of Uniformly Convergent Series Consider the series of functions of a real variable,
f (x) =
∞
X
n=1
gn(x). (2.76)
1. If the gn are continuous, we can integrate term by term ifPngn is
uni-formly convergent over the domain of integration: Z b a dx f (x) = ∞ X n=1 Z b a dx gn(x). (2.77) 2. If the gn and gn′ = d
dxgn are continuous, and
P
ng ′
n is uniformly
conver-gent, then we can differentiate term by term:
f′ (x) = ∞ X n=1 g′ n(x). (2.78)
2.7. POWER SERIES 19 Version of August 27, 2011
Condition for Uniform Convergence
The following condition is sufficient, but not necessary, to ensure that a series is uniformly convergent.
If |gn(z)| < an, where {an} is a sequence of constants such that P ∞ n=1an
converges, then P∞
n=1gn(z) converges uniformly and absolutely.
Proof: The hypothesis implies N+p X n=N gn(z) < N+p X n=N an, (2.79)
so that if N is chosen so thatPN+p
n=Nan< ǫ, then N+p X n=N gn(z) < ǫ ∀z. (2.80)
2.7
Power Series
By a power series, we mean a series of the form,
∞
X
n=0
cnzn= c0+ c1z + c2z2+ . . . , (2.81)
where the cns form a sequence of complex constants, and z is a complex variable.
If a power series converges for one point, z = z0, it converges uniformly and
absolutely for all z satisfying
|z| ≤ η, (2.82)
where η is any positive number less than |z0|.
Proof: Since P∞
n=0cnz0n converges, it must be true that the terms are
bounded,
|cnz0n| < M, (2.83)
where M is independent of n (but not of |z0|). Hence if Eq. (2.82) is satisfied, ∞ X n=0 |cnzn| ≤ ∞ X n=0 |cn|ηn< M ∞ X n=0 η |z0| n < ∞, (2.84) since η/|z0| < 1. This proves absolute convergence. Uniform convergence follows
from the theorem above.
2.7.1
Radius of Convergence
Use the root test to determine where the power series
∞
X
n=0
20 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES z-plane
&% '$
ρ
Figure 2.4: Circle of convergence of a power series. The series (2.85) converges inside the circle, and diverges outside. The radius of convergence ρ is given by Eq. (2.87).
converges. That test says if lim
n→∞
n
p|cn||z| < 1, the series converges, (2.86a)
while if
lim
n→∞
n
p|cn||z| > 1, the series diverges. (2.86b)
Therefore, the power series converges within a circle of convergence of radius ρ, the radius of convergence, where
ρ = 1 limn→∞ p|cn n|
, (2.87)
and diverges outside that circle, as shown in Fig. 2.4. More detailed examination is required to determine whether or not the series converges on the circle of convergence.
2.7.2
Properties of Power Series Within the Circle of
Con-vergence
1. The function defined by the power series is continuous. [This follows from the theorem in Sec. 2.6.2.]
2. It may be differentiated or integrated term by term. [This follows from the theorem above, togther with the fact that ifP∞
n=0cnzn converges, so
doesP∞
n=0ncnzn−1, by the ratio test,
(n + 1)cn+1zn ncnzn−1 = n + 1 n cn+1 cn |z|. (2.88)
Now if z lies within the circle of convergence,
lim n→∞ cn+1 cn |z| < 1. (2.89)
2.7. POWER SERIES 21 Version of August 27, 2011
Since limn→∞(n + 1)/n = 1, convergence of the differentiated series is
assured.]
3. Two such power series may be multiplied together term by term, within the smaller of the two circles of convergence. [This follows from the theorem in Sec. 2.4.2.]
4. The power series is unique. [It suffices to show that if
f (z) = ∞ X n=0 cnzn= 0 ∀z, cn= 0 ∀n. (2.90) Indeed, f (0) = c0= 0, (2.91a) f′ (0) = c1= 0, (2.91b) . . . , f(n)(0) = n! cn = 0.] (2.91c)
2.7.3
Taylor Expansion
The Taylor expansion for a real function of a real variable is obtained from the above argument. If we write a function as a power series,
f (x) = ∞ X n=0 cnxn, (2.92) then cn= 1 n!f (n)(0). (2.93)
Hence, the power series is the Taylor series of the function it represents,
f (x) = ∞ X n=0 1 n!f (n)(0)xn. (2.94)
2.7.4
Hypergeometric Function
The hypergeometric function F is defined by the power series
F (a, b; c; z) = ∞ X n=0 Anzn= ∞ X n=0 (a)n(b)n (c)n zn n!. (2.95)
Here the coefficients are defined in terms of the Pochhammer symbol,
(a)n = a(a + 1)(a + 2) · · · (a + n − 1) =
Γ(a + n)
22 Version of August 27, 2011 CHAPTER 2. INFINITE SERIES
To determine convergence, we examine Anzn An+1zn+1 = Γ(a + n)Γ(b + n) Γ(c + n) Γ(c + n + 1) Γ(a + n + 1)Γ(b + n + 1) (n + 1)! n! zn zn+1 = 1 + c n 1 + n1 1 + a n 1 + b n 1 z = 1 z 1 + 1 n(c + 1 − a − b) + O 1 n2 , (2.97)
where O(1/n2) means that the next term goes to zero as n → ∞ at least as
fast as 1/n2. According to the ratio test, the radius of convergence of this series
is |z| = 1; that is, the series diverges for |z| > 1, and converges uniformly and absolutely for any z such that |z| ≤ η < 1. The remaining question is what happens on the circle of convergence, |z| = 1. According to Gauss’ test, the series is then absolutely convergent if c > a + b [if the constants are complex, if Re (c − a − b) > 0]. For the point z = 1 the series is certainly divergent if this condition is not satisfied; however, if −1 < Re (c − a − b) ≤ 0 the series is conditionally convergent on the unit circle except for the exceptional point z = 1. On the other hand, if Re (c − a − b) ≤ −1 the series is divergent on the unit circle because the terms in the series increase in magnitude.
Chapter 3
Elementary Transcendental
Functions
3.1
Exponential Function
Define, for all complex z, the exponential function by
exp(z) = ez= ∞ X n=0 1 n!z n. (3.1)
By the ratio test,
n!
(n + 1)!|z| = |z|
n + 1→ 0 ∀z, (3.2) the series converges everywhere. By the theorem of the Sec. 2.7, that means that the series converges uniformily in any finite closed region.
Note that the following property holds:
exp(z1+ z2) = ∞ X n=0 1 n!(z1+ z2) n = ∞ X n=0 n X m=0 1 n! n! m! (n − m)!z m 1 z2n−m = ∞ X k=0 1 k!z k 1 ! ∞ X l=0 1 l!z l 2 ! = exp(z1) exp(z2). (3.3) Then, by induction (ez)n = enz, (3.4)
where n is any positive integer.
24 Version of September 7, 2011CHAPTER 3. ELEMENTARY TRANSCENDENTAL FUNCTIONS
Hyperbolic and trigonometric functions are defined in terms of the exponen-tial function: sinh z = e z− e−z 2 cosh z = ez+ e−z 2 , (3.5a) sin z = e iz− e−iz 2i cos z = eiz+ e−iz 2 , (3.5b) so that
i sin z = sinh iz, (3.6a) cos z = cosh iz, (3.6b)
for all complex z. Note that
eiz = cos z + i sin z. (3.7) Therefore, the polar representation of a complex number,
z = r(cos θ + i sin θ)
= reiθ, (3.8)
becomes a most useful and compact representation. In particular,
zn = rneinθ (3.9)
implies De Moivre’s formula,
cos nθ + i sin nθ = (cos θ + i sin θ)n. (3.10)
3.1.1
Definition of π
There exists a positive number π such that
1.
eπi/2= i, and (3.11a)
2.
ez= 1 if and only if z = 2πin, (3.11b) where n is an integer.
Hence exp(z) is periodic with period 2πi,
3.2. THE NATURAL LOGARITHM 25 Version of September 7, 2011 θ z x iy @ @ @ @ @
“cut” or “branch line”
Figure 3.1: Cut plane for defining the logarithm.
3.2
The Natural Logarithm
If z = reiθ, we define
ln z ≡ log z ≡ ln r + iθ, (3.13) where ln r is defined as the inverse of the exponential function for real positive r,
r = eln r. (3.14) Thus we have
z = eζ where ζ = ln r + iθ = log z. (3.15) Recall that θ = arg z is a multivalued function, because θ is only defined up to an arbitrary multiple of 2π. [This is just the periodic property (3.12).] Re-call further that we defined the principal value of the argument as that which satisfied
−π < arg z ≤ π. (3.16) Correspondingly, we say that the single-valued logarithm function (also denoted log z) is defined in the cut plane shown in Fig. 3.1. In measuring θ from the +x axis, one is not allowed to cross the cut along the −x axis. (Where the cut is placed is an arbitrary convention.) The correspondingly defined single-valued functions arg z and
log z = log |z| + i arg z, (3.17) or
−π < Im log z ≤ π, (3.18) are also referred to as the principal values of the argument and logarithm, re-spectively.
Now we define complex powers of complex numbers as follows:
ζz≡ ez log ζ, (3.19) where log ζ is defined in the cut plane. Then
eξz
= ez log eξ
26 Version of September 7, 2011CHAPTER 3. ELEMENTARY TRANSCENDENTAL FUNCTIONS
when
arg eξ = Im ξ (3.21) lies between
−π < Im ξ ≤ π. (3.22) If this is not so,
log eξ = ξ + 2πin (3.23)
where n is so chosen that
−π < Im (ξ + 2πin) ≤ π, (3.24) and eξz = ez(ξ+2πin). (3.25) For example, √ z = z1/2= e12log z (3.26)
is defined as a single-valued function only in the cut plane
−π < arg z ≤ π. (3.27)
3.3
Inverse Hyperbolic and Trigonometric
Func-tions
The inverse hyperbolic and trigonometric functions are defined in terms of the logarithm:
arcsinh z = loghz + (z2+ 1)1/2i, (3.28a)
arccosh z = loghz + (z2− 1)1/2i, (3.28b) arctanh z = 1
2log 1 + z
1 − z, (3.28c) which are defined in the cut planes shown in Fig. 3.2.
arcsin z = −i arcsinh iz
= −i loghiz + (1 − z2)1/2i, (3.29a) arccos z = −i arccosh z
= −i loghz + (z2− 1)1/2i, (3.29b)
arctan z = −i arctanh iz = i 2log 1 − iz 1 + iz = i 2log i + z i − z, (3.29c) which are defined in the cut planes shown in Fig. 3.3. Note that the branch
3.3. INVERSE HYPERBOLIC AND TRIGONOMETRIC FUNCTIONS27 Version of September 7, 2011 x iy • • i −i arcsinh z: x iy • +1 arccosh z: x iy • • +1 −1 arctanh z:
Figure 3.2: Cut planes for defining the inverse hyperbolic functions. The thick lines represent the cuts.
x iy • • +1 −1 arcsin z: arccos z: x iy • • +i −i arctan z:
Figure 3.3: Cut plane for defining the inverse trigonometric functions.
lines (cuts) are chosen so as not to cross the region where both the range and the domain of the functions are real, because for real x,
sin x, cos x ∈ [−1, 1], (3.30a) tan x ∈ (−∞, ∞), (3.30b) sinh x ∈ (−∞, ∞), (3.30c) cosh x ∈ [1, ∞), (3.30d) tanh x ∈ [−1, 1]. (3.30e) An alternative notation for the inverse functions is provided by the superscript −1, as for example,
arcsinh z = sinh−1
z, (3.31)
Chapter 4
Bernoulli Polynomials
4.1
Bernoulli Numbers
The “generating function” for the Bernoulli numbers is
x ex− 1 = ∞ X n=0 Bn n!x n. (4.1)
That is, we are to expand the left-hand side of this equation in powers of x, i.e., a Taylor series about x = 0. The coefficient of xn in this expansion is B
n/n!.
Note that we can write the left-hand side of this expression in an alternative form x ex− 1 = x ex/2 ex/2− e−x/2 = x e −x/2 2 sinhx2 = x 2 coshx 2 − sinh x 2 sinhx 2 = x 2coth x 2 − x 2. (4.2) Note that x 2coth x
2 is an even function of x, while x
2 is odd. Therefore we
conclude that all but one of the Bernoulli numbers of odd order are zero:
B1 = −1
2 (4.3a)
B2k+1 = 0, k = 1, 2, 3, . . . . (4.3b)
By writing x = iy and noting that
cothiy
2 = −i cot y
2, (4.4)
30 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS we conclude that iy 2 coth iy 2 = y 2cot y 2 = ∞ X n=0 B2n(iy)2n (2n)! , (4.5) or y 2cot y 2 = ∞ X n=0 (−1)n B2n (2n)!y 2n. (4.6)
By straightforward expansion in powers of x we can read off the first few Bernoulli numbers: x 2 coth x 2 = x 2 coshx2 sinhx2 ≈ x 2 1 +2!1 x22 +4!1 x24 +6!1 x26 +8!1 x28 + . . . x 2 + 1 3! x 2 3 +5!1 x25 +7!1 x27 +9!1 x29 + . . . ≈ 1 + 1 2! x 2 2 + 1 4! x 2 4 + 1 6! x 2 6 + 1 8! x 2 8 × 1 − 1 3! x 2 2 + 1 5! x 2 4 + 1 7! x 2 6 + 1 9! x 2 8 + 1 3! x 2 2 + 1 5! x 2 4 + 1 7! x 2 62 − 1 3! x 2 2 + 1 5! x 2 43 + 1 3! x 2 24 + O(x10) = . . . = 1 +x 2 2! 1 6 +x 4 4! −1 30 +x 6 6! 1 42 +x 8 8! −1 30 + . . . . (4.7)
So by comparison with Eq. (4.1) we find
B0= 1, B2= 1 6, B4= − 1 30, B6= 1 42, B8= − 1 30. (4.8) What is the radius of convergence of the series
z ez− 1 = − z 2 + ∞ X n=0 B2n (2n)!z 2n? (4.9)
Recall that a power series converges everywhere within its circle of convergence, and diverges outside that circle. Since a uniformly convergent series must con-verge to a continuous function, the power series must concon-verge to a well-behaved function within the circle of convergence. That is, the limit function must have a singularity somewhere on the circle of convergence, but must be singularity-free within the circle of convergence. The precise theorem, proved in Chapter
4.1. BERNOULLI NUMBERS 31 Version of September 16, 2011
n B2n Asymptotic value Relative error
0 1 −2 300% 1 16 π12 39% 2 −1 30 − 3 π4 7.6% 3 421 2π456 1.7% 4 −1 30 − 315 π8 0.41% 5 665 141752π10 0.099% 6 −691 2730 − 467775 2π12 0.025% 7 76 425675254π14 0.0061% 8 −3617510 −638512875π16 0.0015% 9 43867798 976924698752π18 0.00038% 10 −174611330 −92807846381252π20 0.0000095%
Table 4.1: The Bernoulli numbers B2n for n from 0 to 10, compared with
the asymptotic values (4.12). The last column shows the relative error of the asymptotic estimate. Note that the later rather rapidly approaches the true value.
5, is that the radius of convergence of a power series is the distance from the origin to the nearest singularity of the function the series represents.
In this case, it is clear that the generating function is singular wherever ez= 1, except for z = 0. Thus the closest singularities to the real axis occur at
±2πi, so that the radius of convergence is 2π. On the other hand
(2π)2= ρ2= lim n→∞ |B2n| (2n)! [2(n + 1)]! |B2(n+1)| = lim n→∞(2n + 2)(2n + 1) B2n B2n+2 , (4.10)
from which we can infer the fact that the Bernoulli numbers grow rapidly with n,
|B2n| ∼ (2n)!
(2π)2n, n → ∞. (4.11)
We cannot deduce the sign or overall constant from this analysis: The true asymptotic behavior of B2n is
B2n∼ 2(−1)n+1
(2n)!
(2π)2n. (4.12)
32 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS
4.2
Bernoulli Polynomials
The Bernoulli polynomials are defined by the generating function
F (x, s) = x e xs ex− 1 = ∞ X n=0 Bn(s) xn n!, (4.13)
that is, according to Eq. (2.94),
Bn(s) = ∂ ∂x n F (x, s) x=0 . (4.14)
From the properties of F (x, s) we can deduce all the properties of these poly-nomials: 1. Note that F (x, 0) = x ex− 1 = ∞ X n=0 Bn xn n!. (4.15) Therefore, we conclude that the Bernoulli polynomials at zero are equal to the Bernoulli numbers,
Bn(0) = Bn. (4.16)
2. Next we notice that
F (x, 1) = x e x ex− 1 = x 1 − e−x = −x e−x− 1 = F (−x, 0), (4.17)
so that by comparing corresponding terms in the generating function ex-pansion, we find
Bn(1) = (−1)nBn(0) = (−1)nBn. (4.18)
3. If we differentiate the generating function with respect to its second argu-ment, we obtain the relation
∂ ∂sF (x, s) = x2exs ex− 1 = ∞ X n=0 B′ n(s) xn n!. (4.19) But obviously x2exs ex− 1 = xF (x, s) = ∞ X n=0 Bn xn+1 n! , (4.20) so equating coefficients of xn/n! we conclude that
B′
4.3. EULER-MACLAURIN SUMMATION FORMULA33 Version of September 16, 2011
(Note that B′
0(s) = 0 is consistent with this if B−1(s) is finite.)
Again, by direct power series expansion of the generating function we can read off the first few Bernoulli polynomials:
F (x, s) ≈ x1 + xs + 1 2(xs)2+16(xs)3 x +12x2+ 1 3!x3 ≈ 1 + x s −1 2 + x2 s2 2 − s 2− 1 6 + 1 4 + . . . , (4.22)
from which we read off
B0(s) = 1, (4.23a)
B1(s) = s −1
2, (4.23b)
B2(s) = s2− s +1
6. (4.23c)
By keeping two more terms in the expansion we find
B3(s) = s3− 3 2s 2+1 2s, (4.23d) B4(s) = s4− 2s3+ s2− 1 30. (4.23e) Note that the properties (4.16) and (4.18) are satisfied. Note further we can use the property (4.21) to derive higher Bernoulli polynomials from lower ones. Thus from Eq. (4.23c) we know that
B′
3(s) = 3s2− 3s +
1
2. (4.24)
The expression for B3(s), (4.23d) is recovered, when it is recalled that B3= 0.
4.3
Euler-Maclaurin Summation Formula
Using the above recursion relation (4.21) we can deduce a very important for-mula which allows a precise relation between a discrete sum and a continuous integral. First note that since B0= B0(s) = 1 we can write
Z 1 0 f (x)B0(x) dx = Z 1 0 f (x) dx, (4.25)
valid for any function f . But now we can integrate by parts using B′ 1(x) = B0(x) : (4.26) Z 1 0 f (x) dx = Z 1 0 f (x)B′ 1(x) dx
34 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS = f (x)B1(x) 1 x=0 − Z 1 0 f′ (x)B1(x) dx = 1 2[f (1) + f (0)] − Z 1 0 f′ (x)B1(x) dx. (4.27)
Here, we have used the facts that
B1(0) = B1= −1
2, (4.28a)
B1(1) = −B1=
1
2. (4.28b)
Now we can continue integrating by parts by noting that
B1(x) = 1 2B ′ 2(x), (4.29) so that Z 1 0 f (x) dx = 1 2[f (1) + f (0)] − 1 2[f ′ (1)B2(1) − f′(0)B2(0)] +1 2 Z 1 0 f′′ (x)B2(x) dx = 1 2[f (1) + f (0)] − 1 2B2[f ′ (1) − f′ (0)] +1 2 Z 1 0 f′′ (x)B2(x) dx. (4.30)
A general pattern is emerging. Let us assume the following formula holds for some integer k (we have just proved it for k = 1):
Z 1 0 f (x) dx = 1 2[f (1) + f (0)] − k X m=1 B2m (2m)! h f(2m−1)(1) − f(2m−1)(0)i + 1 (2k)! Z 1 0 f(2k)(x)B 2k(x) dx. (4.31)
We shall then prove that the same formula holds for k → k + 1, thereby es-tablishing this formula, the Euler-Maclaurin summation formula, for all k. We proceed as follows. Note that
B2k(x) = B′ 2k+1(x) 2k + 1 = B′′ 2k+2(x) (2k + 1)(2k + 2), (4.32) so that by integrating by parts, we rewrite the last term in Eq. (4.31) as
1 (2k)! Z 1 0 f(2k)(x) 1 (2k + 1)(2k + 2)B ′′ 2k+2(x) dx
4.3. EULER-MACLAURIN SUMMATION FORMULA35 Version of September 16, 2011 = 1 (2k + 2)! f(2k)(1)B′ 2k+2(1) − f(2k)(0)B ′ 2k+2(0) − Z 1 0 f(2k+1)(x)B′ 2k+2(x) dx = 1 (2k + 2)! − f(2k+1)(1)B2k+2(1) + f(2k+1)(0)B2k+2(0) + Z 1 0 f(2k+2)(x)B2k+2(x) dx , (4.33)
where we have noted that for k > 0 B′ 2k+2(0) = (2k + 2)B2k+1(0) = 0, (4.34a) B′ 2k+2(1) = (2k + 2)B2k+1(1) = −(2k + 2)B2k+1(0) = 0. (4.34b) Hence Z 1 0 f (x) dx = 1 2[f (1) + f (0)] − k+1 X m=1 B2m (2m)! h f(2m−1)(1) − f(2m−1)(0)i + 1 (2k + 2)! Z 1 0 f(2k+2)(x)B2k+2(x) dx. (4.35)
This is exactly Eq. (4.31) with k replaced by k + 1; so since the formula is true for k = 1 it is true for all integers k ≥ 1. Notice that the last term in this formula, the remainder, can also be written in the form
− 1 (2k + 3)!
Z 1
0
f(2k+3)(x)B2k+3(x) dx. (4.36)
Now consider the integral (N a positive integer)
Z N 0 f (s) ds = N −1 X k=0 Z k+1 k f (s) ds = N −1 X k=0 Z 1 0 f (k + t) dt, (4.37)
where we have introduced a local variable t. For the latter integral, we can use the Euler-Maclaurin sum formula, which here reads
Z 1 0 f (k + t) dt = 1 2[f (k + 1) + f (k)] − n X m=1 B2m (2m)! h f(2m−1)(k + 1) − f(2m−1)(k)i + 1 (2n)! Z 1 0 f(2n)(k + t)B2n(t) dt. (4.38)
36 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS
Now when we sum the first term here on the right-hand side over k we obtain
N −1 X k=0 1 2[f (k + 1) + f (k)] = N X k=0 f (k) −1 2[f (0) + f (N )], (4.39)
while the second term when summed on k involves
N −1 X k=0 h f(2m−1)(k + 1) − f(2m−1)(k)i= f(2m−1)(N ) − f(2m−1)(0). (4.40) Thus we find Z N 0 f (s) ds = N X k=0 f (k) −1 2[f (0) + f (N )] − n X m=1 1 (2m)!B2m h f(2m−1)(N ) − f(2m−1)(0)i + 1 (2n)! Z 1 0 N −1 X k=0 f(2n)(t + k)B2n(t) dt. (4.41)
Equivalently, we can write this as a relation between a finite sum and an integral, with a remainder Rn: N X k=0 f (k) = Z N 0 f (s) ds +1 2[f (0) + f (N )] + n X m=1 1 (2m)!B2m h f(2m−1)(N ) − f(2m−1)(0)i+ R n, (4.42)
where the remainder
Rn= − 1 (2n)! Z 1 0 N −1 X k=0 f(2n)(t + k)B2n(t) dt. (4.43)
is often assumed to vanish as n → ∞. Note that the remainder can also be written as Rn = − 1 (2n)! Z N 0 f(2n)(t)B2n(t − ⌊t⌋) dt, (4.44)
where ⌊t⌋ signifies the greatest integer less than or equal to t.
4.3.1
Examples
1. Use the Euler-Maclaurin formula to evaluate the sumPN
n=0cos(2πn/N ). N X n=0 cos2πn N = Z N 0 dn cos2πn N + 1 2(1 + 1) + 0 = 1, (4.45)
4.3. EULER-MACLAURIN SUMMATION FORMULA37 Version of September 16, 2011 because f(2m−1)(0) = f(2m−1)(N ) = 0 (4.46) and Z N 0 dn cos2nπ N = N 2π Z 2π 0 dx cos x = 0. (4.47)
Of course, the sum may be carried out directly,
N X n=0 cos2πn N = 1 2 N X 0 ei2πn/N+ e−i2πn/N = 1 2 1 − e2πi(N +1)/N 1 − e2πi/N + 1 − e−2πi(N +1)/N 1 − e−2πi/N = 1 2(1 + 1) = 1. (4.48)
2. The following sum occurs, for example, in computing the vacuum energy in a cosmological model:
∞
X
l=0
(2l + 1)e−l(l+1)t. (4.49)
How does this behave as t → 0? We will answer this question by using the Euler-Maclaurin formula assuming that the remainder Rn tends to zero
as n → ∞. Thus we will write the limiting form of that sum formula as
∞ X l=0 f (l) = Z ∞ 0 dl f (l) + 1 2[f (∞) + f (0)] + ∞ X k=1 B2k (2k)! h f(2k−1)(∞) − f(2k−1)(0)i. (4.50) Here f (l) = (2l + 1)e−l(l+1)t, (4.51) so that f (∞) = f(2k−1)(∞) = 0, (4.52) while a very simple calculation shows
f (0) = 1, (4.53a) f′ (0) = 2 − t, (4.53b) f′′′ (0) = −12t + 12t2− t3, (4.53c) f(5)(0) = 120t2− 180t3+ 30t4− t5, (4.53d) f(7)(0) = −1680t3+ 3360t4− 840t5+ 56t6− t7, (4.53e) f(2k−1)(0) = O(t4), k ≥ 5. (4.53f)
38 Version of September 16, 2011CHAPTER 4. BERNOULLI POLYNOMIALS
Thus Eq. (4.50) yields
∞ X l=0 (2l + 1)e−l(l+1)t = Z ∞ 0 dl (2l + 1)e−l(l+1)t+1 2 −B2 2 f ′ (0) −B4 4!f ′′′ (0) − . . . = 1 t Z ∞ 0 du e−u+1 2+ 1 2 1 6 (t − 2) + 1 4! −1 30 [12t + O(t2)] + O(t2) = 1 t + 1 3 + t 15+ 4 315t 2+ 1 315t 3+ . . . . (4.54)
Here the integral was evaluated by making the substitution u = l(l + 1)t, du = (2l + 1)t dl, and in the last line we have displayed the next two terms in this asymptotic expansion for small t.
3. The Riemann zeta function (2.50) is defined by
ζ(α) = ∞ X n=1 1 nα, Re α > 1. (4.55)
Suppose we approximate this by the first M terms in the sum occurring in the Euler-Maclaurin formula (4.42):
ζ(α, M ) = 1 α − 1+ 1 2− M X m=1 B2m (2m)!f (2m−1)(1), (4.56)
where f (n) = n−α, and the first two terms here come from the integral
and the 1
2f (1) terms in the EM formula. It is easy to see that
f(2m−1)(1) = −Γ(α + 2m − 1)
Γ(α) . (4.57) Given the asymptotic behavior of the Bernoulli numbers in (4.12), it is apparent that the limit M → ∞ of ζ(α, M ) does not exist. This limit is an example of an asymptotic series. However, in Table 4.2 we compare the sum of the first N terms of the series in (4.55) with the first N terms in the series defined by (4.56), that is ζ(α, N ), for α = 3, where ζ(3) = 1.2020569. The original series converges monotonically to the correct limiting value, but not spectacularly fast. For N = 9 terms, the relative error is about −0.5%. The asymptotic series is divergent; however, the N = 1 term is in error by only 4%, and the average of the N = 1 and N = 2 is larger than the true value by only +0.5%. This illustrates a characteristic feature of asymptotic series: A few terms in the series approximates the function rather well, but as more and more terms are included the series deviates from the true value by an ever increasing amount.
4.3. EULER-MACLAURIN SUMMATION FORMULA39 Version of September 16, 2011 N PN n=1n −3 ζ(3, N ) 1 2[ζ(3, N ) + ζ(3, N + 1)] 1 1 1.25 1.208 2 1.125 1.1667 1.208 3 1.1620 1.25 1.175 4 1.1777 1.1 1.308 5 1.1857 1.5167 0.694 6 1.1903 −0.1286 7 1.1932 8.6214 8 1.1952 −51.6619 9 1.1965 470.564
Table 4.2: Two approximations compared for ζ(3) = 1.20206 . . .: N terms in the defining series (4.55) and N terms (without the remainder) in the Euler-Maclaurin sum (4.56). The former converges monotonically to the limit from below, while the later diverges, yet approximates the true value to better than 1% for low values of N .
Chapter 5
Analytic Functions
5.1
The Derivative
Let f (z) be a complex-valued function of the complex variable z. The derivative of f is defined as f′(z) = df dz = limδz→0 f (z + δz) − f (z) δz = limδz→0 δf δz, (5.1) if the limit exists and is independent of the way in which δz approaches zero. This is illustrated in Fig. 5.1
5.1.1
Examples
What is the derivative of zn?
d dzz n = lim δz→0 (z + δz)n− zn δz = limδz→0 nzn−1δz δz = nzn−1. (5.2) -? 6 @@R @ @ I
Figure 5.1: In the complex plane, δz, as indicated by the arrows in the figure, can approach zero from any direction.
42 Version of October 1, 2011 CHAPTER 5. ANALYTIC FUNCTIONS
Then, since ez 1s represented by a power series which converges everywhere,
and therefore converges uniformly in any finite bounded (compact) region, it is also differentiable everywhere,
d dze z = d dz ∞ X n=0 1 n!z n = ∞ X n=1 1 (n − 1)!z n−1 = ez. (5.3)
The derivative of the exponential function is the function itself.
5.2
Analyticity
Whenever f′(z
0) exists, f is said to be analytic (or regular, or holomorphic) at
the point z0. The function is analytic throughout a region in the complex plane
if f′ exists for every point in that region. Any point at which f′ does not exist
is called a singularity or singular point of the function f .
If f (z) is analytic everywhere in the complex plane, it is called entire.
Examples
• 1/z is analytic except at z = 0, so the function is singular at that point. • The functions zn, n a nonnegative integer, and ez are entire functions.
5.3
The Cauchy-Riemann Conditions
The Cauchy-Riemann conditions are necessary and sufficient conditions for a function to be analytic at a point.
Suppose f (z) is analytic at z0. Then f′(z0) may be obtained by taking δz
to zero through purely real, or through purely imaginary values, for example. If δz = δx, δx real, we have, upon writing f in terms of its real and imaginary parts, f = u + iv, f′(z0) = ∂u ∂x+ i ∂v ∂x z=z0 . (5.4)
On the other hand, if δz = iδy, δy real, we have similarly,
f′(z0) = ∂u i∂y + i ∂v i∂y z=z0 = −i∂u ∂y + ∂v ∂y z=z0 . (5.5)
Since the derivative is independent of how the limit is taken, we can equate these two expression, meaning that they must have equal real and imaginary parts, ∂u ∂x = ∂v ∂y, ∂v ∂x = − ∂u ∂y. (5.6)
5.4. CONTOUR INTEGRALS 43 Version of October 1, 2011
These are the Cauchy-Riemann conditions.
These conditions are not only necessary, but if the partial derivatives are continuous, they are sufficient to assure analyticity. Write
f (z + δz) − f (z) = u(x + δx, y + δy) − u(x, y) + i[v(x + δx, y + δy) − v(x, y)] = u(x + δx, y + δy) − u(x, y + δy) + u(x, y + δy) − u(x, y)
+ i[v(x + δx, y + δy) − v(x, y + δy) + v(x, y + δy) − v(x, y)] = δx∂u ∂x + δy ∂u ∂y + i δx∂v ∂x+ δy ∂v ∂y (5.7)
which becomes, if the Cauchy-Riemann conditions hold
f (z + δz) − f (z) = δx∂u ∂x − δy ∂v ∂x+ i δx∂v ∂x+ δy ∂u ∂x = (δx + iδy) ∂u ∂x+ i ∂v ∂x , (5.8)
so since δz = δx + iδy, we see δf δz → ∂u ∂x+ i ∂v ∂x (5.9) independently of how δz → 0, so f′(z) = ∂u ∂x + i ∂v ∂x (5.10) exists. Example
Consider the function z∗ of z; that is, if z = x + iy, z∗= x − iy. The
Cauchy-Riemann conditions never hold, ∂x ∂x = 1 6=
∂(−y)
∂y = −1, (5.11) so z∗ is nowhere an analytic function of z.
5.4
Contour Integrals
Suppose we have a smooth path in the complex plane, extending from the point a to the point b. Suppose we choose points z1, z2,. . . , zn− 1 lying on the curve,
and connect them by straight-line segments. Likewise connect a = z0 with z1
amd b = zn with zn−1. See Fig. 5.2. Then the contour integral of a function f
is defined by the following limit, Z b a C f (z) dz = lim ∆zi→0 n→∞ n X i=1 f (zi)∆zi, ∆zi= zi− zi−1, (5.12)
44 Version of October 1, 2011 CHAPTER 5. ANALYTIC FUNCTIONS • • • • • • • • • • • • • • • a = z0 z1 z2 z3 b = zn H BB
Figure 5.2: Path C in the complex plane approximated by a series of straight-line segments.
and the limit taken is one in which the number n of straight-line segments goes to infinity, while the length of the largest one goes to zero. Whenever this limit exists, independently of how it is taken, the integral exists. Note that in general the integral depends on the path C, as well as on the endpoints.
Example
Consider I
K
dz
z (5.13)
where K is a circle about the origin, of radius r. (The circle on the integral sign signifies that the path of integration is closed.) From the polar representation of complex numbers, we may write
z = reiθ, (5.14a)
so since r is fixed on K, we have
dz = reiθi dθ. (5.14b)
Let us assume that the integration is carried out in a positive (counterclockwise) sense, so then I K dz z = i Z 2π 0 dθ = 2πi, (5.15)
which is independent of the value of r.
5.5
Cauchy’s Theorem
Chauchy’s theorem states that if f (z) is analytic at all points on and inside a closed contour C, then the integral of the function around that contour vanishes,
I
C
5.5. CAUCHY’S THEOREM 45 Version of October 1, 2011
C BBM Ci
Figure 5.3: The integral around the contour C may be replaced by the sum of integrals around the subcontours Ci.
Proof: Subdivide the region inside the contour in the manner shown in Fig. 5.3. Obviously I C f (z) dz =X i I Ci f (z) dz, (5.17)
where Ci is the closed path around one of the mesh elements, since the
con-tribution from the side common to two adjacent subcontours evidently cancels, leaving only the contribution from the exterior boundary. Now because f is analytic throughout the region, we may write for small δz
f (z + δz) = f (z) + δz f′(z) + O(δz2), (5.18)
where O(δz2) means only that the remainder goes to zero faster that δz. We
apply this result by assuming that we have a fine mesh subdividing C—we are interested in the limit in which the largest mesh element goes to zero. Let zi be
a representative point within the ith mesh element (for example, the center). Then I Ci f (z) dz = f (zi) I Ci dz + f′(zi) I Ci (z − zi) dz + I Ci O((z − zi)2) dz. (5.19)
Now it is easily seen that for an arbitrary contour Ci
I Ci dz = I Ci (z − zi) dz = 0, (5.20)
so if the length of the cell is ε, I
Ci
f (z) dz = O(ε3) = AiO(ε), (5.21)
which is to say that the integral around the ith cell goes to zero faster than the area Ai of the ith cell. Thus the integral required is
I
C
f (z) dz =X
i
46 Version of October 1, 2011 CHAPTER 5. ANALYTIC FUNCTIONS y C @ @ @ @ @ @ @ @ @ @ R
Figure 5.4: A multiply connected region R consisting of the area within a tri-angle but outside of an circular region. The closed contour C cannot be contin-uously deformed to a point without crossing into the disk, which is outside the region R.
where A is the finite area contained within the contour C. As the subdivision becomes finer and finer, ε → 0 and so
I
C
f (z) dz = 0. (5.23)
To state a more general form of Cauchy’s theorem, we need the concept of a simply connected region. A simply connected region R is one in which any closed contour C lying in R may be continuously shrunk to a point without ever leaving R. Fig. 5.4 is an illustration of a multiply connected region. C lies entirely within R, yet it cannot be shrunk to a point because of the excluded region inside it.
We can now restate Cauchy’s theorem as follows: If f is analytic in a simply
connected region R then
I
C
f (z) dz = 0 (5.24)
for any closed contourC in R.
That simple connectivity is required here is seen by the example of the function 1/z, which is analytic in any region excluding the origin.
Here is another proof of Cauchy’s theorem, as given in the book by Morse and Feshbach. If the closed contour C lies in a simply-connected region where f′(z) exists then
I
C
f (z) dz = 0. (5.25)
Proof: Let us choose the origin to lie in the region of analyticity (if it does not, change variables so that z = 0 lies within C). Define
F (λ) = λ I
C
f (λz) dz. (5.26)
Then the derivative of this function of λ is F′(λ) = I C f (λz) dz + λ I C zf′(λz) dz
5.6. CAUCHY’S INTEGRAL FORMULA 47 Version of October 1, 2011 ' & $ % ↑ C •i ↓ ↑ C′ C′′ γ z0
Figure 5.5: Distortion of a contour C to a small one γ encircling the singularity at z0. = I C f (λz) dz + zf (λz) z=end of C z=beginning of C − I C f (λz) dz = 0, (5.27)
where we have integrated by parts, because the function f is single valued. Thus F (λ) is constant. But F (0) = lim λ→0λ I C f (λz) dz = lim λ→0 I λC f (z) dz = 0 (5.28)
because f (0) is bounded because f is analytic at the origin. (We have deformed the contour to an infinitesimal one about the origin.) Thus we conclude that F (1) = 0. This proves the theorem.
5.6
Cauchy’s Integral Formula
If f (z) is analytic on and within the closed contour C, and z0 lies within C,
then the value of f at z0 is given in terms of its boundary values by
f (z0) = 1 2πi I C f (z) z − z0 dz, (5.29)
where the contour is traversed in the positive (counterclockwise) sense.
Proof: f (z)/(z − z0) is not analytic within C, so choose a contour inside of
which this function is analytic, as shown in Fig. 5.5. Here we have connected the contour C to the small contour γ by two overlapping lines C′, C′′which are
traversed in opposite senses. Now f (z)/(z − z0) is analytic on the inside of the
contour C + C′+ C′′+ γ. (By inside, we mean that if you follow the path in the
direction indicated by the arrows, the inside is only your left, and the outside is on your right.) Thus, by Cauchy’s theorem
I
C+C′+C′′+γ
f (z) z − z0
48 Version of October 1, 2011 CHAPTER 5. ANALYTIC FUNCTIONS
Now because we choose the lines C′, C′′ as overlapping, since f is continuous
in the neighborhood of those lines those two integrals cancel, Z
C′
+C′′
f (z)
z − z0dz = 0. (5.31)
And since the circle γ may be chosen arbitrarily small I γ f (z) z − z0 dz = f (z0) I γ dz z − z0 = −2πif (z0), (5.32)
since γ is traversed in a negative or clockwise sense. Thus the theorem (5.29) is proved.
(Implicit in the above is the assumption that the contour does not cross itself to wind around z0 more than once. If this happens, Cauchy’s formula is
modified. See homework.)
It is now easily shown from the definition of the derivative that if f is analytic on and within C, we may express the derivative by
f′(z 0) = 1 2πi I C f (z) (z − z0)2 dz, (5.33)
and in fact the nth derivative is given by
f(n)(z0) = n! 2πi I C f (z) (z − z0)n+1 dz. (5.34)
That is, if f is analytic, so is its derivative. An analytic function is infinitely differentiable, a property which is not true for a differentiable function of a real variable.
5.7
Morera’s Theorem
The converse to Cauchy’s theorem is the following:
If f (z) is continuous in a region R, and for all contours C lying in R I
C
f (z) dz = 0, (5.35)
then f (z) is analytic throughout R.
Proof: If f satisfies the above hypotheses, then the integral Z z2
z1
f (z) dz = F (z2) − F (z1) (5.36)
is a function of the endpoints only, and not of the path, as is evident from Fig. 5.6. But now the function F has a unique derivative,
F′(z) = f (z), (5.37) so that F (z) is analytic. Hence, so is its derivative f (z). QED.
5.8. THE LOGARITHM 49 Version of October 1, 2011 • • C1 ↑ ↑ C2 z2 z1
Figure 5.6: Two paths C1 and C2 connecting the point z1 with the point z2.
BecauseH C1−C2f (z) dz = 0, we conclude that Rz2 z1C1f (z) dz = Rz2 z1C2f (z) dz. • • • 1 z t-plane -θ |z|
Figure 5.7: Path of integration in the cut t plane used in defining the logarithm in Eq. (5.38).
5.8
The Logarithm
An alternative definition to that given in Sec. 3.2 is given by the path integral
log z = Z z
1
dt
t , (5.38)
over any contour connecting 1 with z which does not cross the cut line shown in Fig. 5.7. The cut is present so the contour cannot encircle the singularity of the integrand at t = 0. Because the arg function must be single-valued, the cut supplies the restriction
−π < arg(z) ≤ π. (5.39) The last equality means for negative z we approach the cut from above.
Since the integral is path independent, we may chose the path to consist of a segment along the positive z axis and an arc of a circle, as also shown in Fig. 5.7. Then the logarithm may be written as
log z = Z |z| 1 dt t + Z θ 0 |z| i eiθ′ dθ′ |z| eiθ′
50 Version of October 1, 2011 CHAPTER 5. ANALYTIC FUNCTIONS
= log |z| + iθ
= log |z| + i arg z, (5.40)
which coincides with the previous definition.
The logarithm is analytic in the cut plane, and its derivative is d
dzlog z = 1
z. (5.41)
If ξ = log z, define the inverse function by z = exp ξ. Since when z = 1, ξ = 0, we have exp(0) = 1. (5.42) Also we have d dξexp ξ = dz dξ = dz d log z = z = exp ξ. (5.43) These two properties uniquely define the exponential function.
5.9
A Theorem for Functions Represented by
Series
Let us suppose that the function Φ defined by the series
Φ(z) =
∞
X
n=0
fn(z) (5.44)
converges uniformly on a closed contour C, and that each fn is analytic on and
within C. Then, on and within C
Φ(z) =
∞
X
n=0
fn(z) (5.45)
converges and Φ is analytic.
Proof: Since a uniformly convergent series may be integrated term by term, we have for z0 within C
1 2πi I C Φ(z) z − z0dz = ∞ X n=0 1 2πi I C fn(z) z − z0dz = ∞ X n=0 fn(z0), (5.46)
by Cauchy’s integral formula. So this last sum exists; call it
Φ(z0) = ∞
X
n=0
5.9. A THEOREM FOR FUNCTIONS REPRESENTED BY SERIES51 Version of October 1, 2011 Now Φ′(z 0) exists as well: Φ′(z 0) = 1 2πi I C Φ(z) (z − z0)2 dz = ∞ X n=0 f′ n(z0), (5.48) so Φ is analytic within C.
Chapter 6
Taylor and Laurent
Expansions—
Analytic Continuation
6.1
Taylor expansion
Let f (z) be analytic within and on a circle C with center at z0. Let z be a point
within the circle. Then Cauchy’s integral formula can be written as
f (z) = 1 2πi I C f (z′ ) z′ − zdz ′ = 1 2πi I C f (z′ ) (z′ − z0) − (z − z0)dz ′ . (6.1)
Because z lies inside the circle,
|z′
− z0| > |z − z0|, (6.2)
we can expand the denominator,
1 (z′ − z0) − (z − z0) = 1 z′ − z0 1 1 − z−z0 z′−z0 = 1 z′ − z0 ∞ X n=0 z − z0 z′ − z0 n . (6.3)
This series converges absolutely and uniformly for z′
on the circle and z fixed inside, so it may be integrated term by term:
f (z) = ∞ X n=0 (z − z0)n 1 2πi I C f (z′ ) (z′ − z0)n+1dz ′ = ∞ X n=0 (z − z0)n 1 n!f (n)(z 0), (6.4)
using the result of Eq. (5.34).
54 Version of October 12, 2011CHAPTER 6. TAYLOR AND LAURENT SERIES
• • ρ
z0
ξ0
Figure 6.1: Circle of convergence for the Taylor series (6.4). Here z0is the point
about which the Taylor expansion is performed, ξ0is the closest singularity of f
to z0, and ρ = |ξ0− z0| is the radius of convergence. The Taylor series converges
within the circle of convergence, and diverges outside the circle of convergence. It may either diverge or converge on the circle of convergence.
This Taylor series will converge inside a circle having radius equal to the distance from z0to the nearest singularity, and diverge outside such a circle, as
illustrated in Fig. 6.1.
Proof: For |z − z0| < ρ, we can choose C in the above derivation to have
radius r, where |z−z0| < r < ρ, so the above expansion converges. For |z−z0| >
ρ, suppose it were true that the Taylor series converged. Then, according to the theorem in Sec. 2.7, it would converge at z = ξ0, to an analytic function
(Sec. 5.9). This is contrary to the assertion that ξ0is a singular point. QED.
Example
Consider the function
f (z) = 1
1 − z, (6.5)
which is analytic except at z = 1. The Taylor series about the origin, 1
1 − z = 1 + z + z
2+ . . . , (6.6)
converges only for |z| < 1. We may obtain a larger circle of convergence by expanding about some other point, say z = −1:
1 1 − z = 1 2 − (z + 1) = 1 2 1 1 − z+1 2 = 1 2 " 1 + z + 1 2 + z + 1 2 2 + . . . # , (6.7)
which converges inside a circle of radius 2, centered about z = −1. In both cases the singularity at z = 1 lies on the circle of convergence.
6.2. ANALYTIC CONTINUATION 55 Version of October 12, 2011
6.2
Analytic Continuation
The process of extending a power series representation of an analytic function is called analytic continuation. It can be done whenever there are only isolated singular points. The general idea is as follows.
Suppose we have a power series about z0
f (z) =
∞
X
n=0
an(z − z0)n, (6.8)
which has radius of convergence ρ. (That is, it converges if |z − z0| < ρ and
diverges if |z − z0| > ρ.) The function f has a singular point somewhere on
the circle of convergence. Since this power series represents an analytic function inside its circle of convergence, it can, by the above, be Taylor expanded about any other point lying within the circle of convergence, say z1,
f (z) =
∞
X
n=0
bn(z − z1)n. (6.9)
In general,1 the circle of convergence of this series will lie partly outside the
original circle. Thus f is now defined in a larger domain. In the new region, f may be expanded once again, and usually the new circle of convergence will lie partly outside both the first two circles, so again the meaning is extended. And so on. The idea is sketched in Fig. 6.2.
Entire functions may be represented by power series (Taylor expansions) valid everywhere, since they have no singular points.
6.3
Laurent Expansion
Let f (z) be analytic in the annulus defined by two concentric circles C1and C2,
both centered on z0, including the bounding circles. See Fig. 6.3. If z lies in
the annulus, Cauchy’s integral formula says (the interior boundary C1 must be
traversed in a clockwise sense—hence, the minus sign)
f (z) = 1 2πi I C2 f (z′ ) z′ − zdz ′ − 1 2πi I C1 f (z′ ) z′ − zdz ′ = 1 2πi I C2 f (z′ ) dz′ (z′ − z0) − (z − z0)− 1 2πi I C1 f (z′ ) dz′ (z′ − z0) − (z − z0).(6.10)
For the C2 integral, |z − z0| < |z′− z0| so we expand in (z − z0)/(z′− z0); for
the C1 integral |z − z0| > |z′− z0|, so we expand in (z′− z0)/(z − z0). Thus we
have f (z) = 1 2πi I C2 f (z′ ) ∞ X n=0 (z − z0)n (z′ − z0)n+1 dz′
56 Version of October 12, 2011CHAPTER 6. TAYLOR AND LAURENT SERIES • • z0 ξ0 • •• • z1 ξ1 z2 ξ 2
Figure 6.2: The process of analytic continuation of a function defined by a power series. The original series is a Taylor expansion about the point z0,
which converges inside a circle having radius equal to the distance to the near-est singularity ξ0. If the function is instead expanded about the point z1, it
converges in a different circle, having radius equal to the distance from z1to the
singular point closest to z1, namely ξ1. Instead the function can be expanded
about z2, lying outside the first circle of convergence, but inside the second,
which will define the function in a different circle of convergence, with radius of convergence equal to the distance to the singularity closest to z2, namely, ξ2.
This process may be repeated indefinitely. f is defined in the union of all such circles of convergence. • &% '$ z0 • z C1 C2