Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps

We bound the tail of the series of |a
n
|.
∞

n=N
|a
n
| =
∞

n=N

|a
n
|
1/n

n
≤
∞

n=N
r
n
=
r
N
1 − r


∞
n=0
a
n
is absolutely convergent.
Example 12.1.8 Consider the series
∞

n=0
n
a
b
n
,
where a and b are real constants. We use the root test to check for absolute convergence.
lim
n→∞
|n
a
b
n
|
1/n
< 1
|b| lim
n→∞
n
a/n
< 1
|b|exp


lim
n→∞
1 ln n
n

< 1
|b|
e
0
< 1
|b| < 1
Thus we see that the series converges absolutely for |b| < 1. Note that the value of a does not affect the absolute
convergence.
Example 12.1.9 Consider the absolutely convergent series,
∞

n=1
1
n
2
.
534
We aply the root test.
lim
n→∞
|a
n
|
1/n

= lim
n→∞




1
n
2




1/n
= lim
n→∞
n
−2/n
= lim
n→∞
e
−
2
n
ln n
=
e
0
= 1
It fails to predict the convergence of the series.

Raabe’s Test
Result 12.1.5 The series

a
n
converges absolutely if
lim
n→∞
n

1 −




a
n+1
a
n





> 1.
If the limit is less than unity, then the series diverges or converges conditionally. If the limit
is unity, the test fails.
535
Gauss’ Test
Result 12.1.6 Consider the series


a
n
. If
a
n+1
a
n
= 1 −
L
n
+
b
n
n
2
where b
n
is bounded then the series converges absolutely if L > 1. Otherwise the series
diverges or converges conditionally.
12.2 Uniform Convergence
Continuous Functions. A function f(z) is continuous in a closed d omain if, given any  > 0, there exists a δ > 0
such that |f(z) − f(ζ)| <  for all |z −ζ| < δ in the domain.
An equivalent definition is that f(z) is continuous in a closed domain if
lim
ζ→z
f(ζ) = f(z)
for all z in the domain.
Convergence. Consider a series in which the terms are functions of z,


∞
n=0
a
n
(z). The serie s is convergent in a
domain if the series converges for each point z in the domain. We can then define the function f(z) =

∞
n=0
a
n
(z).
We can state the convergence criterion as: For any given  > 0 there exists a function N(z) such that
|f(z) − S
N(z)
(z)| =






f(z) −
N(z)−1

n=0
a
n
(z)







< 
for all z in the domain. Note that the rate of convergence, i.e. the number of terms, N(z) required for for the absolute
error to be less than , is a function of z.
536
Uniform Convergence. Consider a series

∞
n=0
a
n
(z) that is convergent in some domain. If the rate of convergence
is independent of z then the series is said to be uniformly convergent. Stating this a little more mathematically, the
series is uniformly convergent in the domain if for any given  > 0 there exists an N, inde pendent of z, such that
|f(z) − S
N
(z)| =





f(z) −
N

n=1

a
n
(z)





< 
for all z in the domain.
12.2.1 Tests for Uniform Convergence
Weierstrass M-test. The Weierstrass M-test is useful in determining if a series is uniformly convergent. The series

∞
n=0
a
n
(z) is uniformly and absolutely convergent in a domain if there exists a convergent series of positive terms

∞
n=0
M
n
such that |a
n
(z)| ≤ M
n
for all z in the domain. This condition first implies that the series is absolutely
convergent for all z in the domain. The condition |a
n

(z)| ≤ M
n
also ensures that the rate of convergence is independent
of z, which is the criterion for uniform convergence.
Note that absolute convergence and uniform convergence are independent. A series of functions may be absolutely
convergent without being uniformly convergent or vice versa. The Weierstrass M-test is a sufficient but not a necessary
condition for uniform convergence. The Weierstrass M-test can succeed only if the series is uniformly and absolutely
convergent.
Example 12.2.1 The series
f(x) =
∞

n=1
sin x
n(n + 1)
is uniformly and absolutely convergent for all real x because |
sin x
n(n+1)
| <
1
n
2
and

∞
n=1
1
n
2
converges.

537
Dirichlet Test. Consider a sequence of monotone decreasing, positive constants c
n
with limit zero. If all the partial
sums of a
n
(z) are bounded in some closed domain, that is





N

n=1
a
n
(z)





< constant
for all N, then

∞
n=1
c
n

a
n
(z) is uniformly convergent in that closed domain. Note that the Dirichlet test does not
imply that the series is absolutely convergent.
Example 12.2.2 Consider the series,
∞

n=1
sin(nx)
n
.
We cannot use the Weierstrass M-test to determine if the series is uniformly convergent on an interval. While it is easy
to bound the terms with |sin(nx)/n| ≤ 1/n, the sum
∞

n=1
1
n
does not converge. Thus we will try the Dirichlet test. Consider the sum

N−1
n=1
sin(nx). This sum can be evaluated
in closed form. (See Exercise 12.9.)
N−1

n=1
sin(nx) =

0 for x = 2πk

cos(x/2)−cos((N−1/2)x)
2 sin(x/2)
for x = 2πk
The partial sums have infinite discontinuities at x = 2πk, k ∈ Z. The partial sums are bound ed on any closed interval
that does not contain an integer multiple of 2π. By the D irich let test, the sum

∞
n=1
sin(nx)
n
is uniformly convergent
on any such closed interval. The series may not be uniformly convergent in neighborhoods of x = 2kπ.
538
12.2.2 Uniform Convergence and Continuous Functions.
Consider a series f(z) =

∞
n=1
a
n
(z) that is uniformly convergent in s ome domain and whose terms a
n
(z) are continuous
functions. Since the series is uniformly convergent, for any given  > 0 there exists an N such that |R
N
| <  for all z
in the domain.
Since the finite sum S
N
is continuous, for that  there exists a δ > 0 such that |S

N
(z) −S
N
(ζ)| <  for all ζ in the
domain satisfying |z −ζ| < δ.
We combine these two results to show that f(z) is continuous.
|f(z) − f(ζ)| = |S
N
(z) + R
N
(z) − S
N
(ζ) − R
N
(ζ)|
≤ |S
N
(z) − S
N
(ζ)| + |R
N
(z)| + |R
N
(ζ)|
< 3 for |z −ζ| < δ
Result 12.2.1 A uniformly convergent series of continuous terms represents a continuous
function.
Example 12.2.3 Again consider

∞

n=1
sin(nx)
n
. In Example 12.2.2 we showed that the convergence is uniform in any
closed interval that does not contain an integer multiple of 2π. In Figure 12.2 is a plot of the first 10 and then 50 terms
in the series and finally the function to which the series converges. We see that the function has jump discontinuities
at x = 2kπ and is continuous on any closed interval not containing one of those points.
12.3 Uniformly Convergent Power Series
Power Series. Power series are series of the form
∞

n=0
a
n
(z −z
0
)
n
.
539
Figure 12.2: Ten, Fifty and all the Terms of

∞
n=1
sin(nx)
n
.
Domain of Convergence of a Power Ser ies Consider the series

∞

n=0
a
n
z
n
. Let the series converge at some
point z
0
. Then |a
n
z
n
0
| is bounded by some constant A for all n, so
|a
n
z
n
| = |a
n
z
n
0
|




z
z

0




n
< A




z
z
0




n
This comparison test shows that the series converges absolutely for all z satisfying |z| < |z
0
|.
Suppose that the series diverges at some point z
1
. Then the series could not converge for any |z| > |z
1
| since
this would imply convergence at z
1
. Thus there exists some circle in the z plane such that the power series converges

absolutely inside the circle and diverges outside the circle.
Result 12.3.1 The domain of convergence of a power series is a circle in the compl ex plane.
Radius of Convergence of Power Series. Consider a power series
f(z) =
∞

n=0
a
n
z
n
540
Applying the ratio test, we see that the series converges if
lim
n→∞
|a
n+1
z
n+1
|
|a
n
z
n
|
< l
lim
n→∞
|a
n+1

|
|a
n
|
|z| < 1
|z| < lim
n→∞
|a
n
|
|a
n+1
|
Result 12.3.2 Ratio formula. The radius of convergence of the power series
∞

n=0
a
n
z
n
is
R = lim
n→∞
|a
n
|
|a
n+1
|

when the limit exists.
Result 12.3.3 Cauchy-Hadamard formula. The radius of convergence of the power series:
∞

n=0
a
n
z
n
is
R =
1
lim sup
n

|a
n
|
.
541
Absolute Convergence of Power Series. Consider a power series
f(z) =
∞

n=0
a
n
z
n
that converges for z = z

0
. Let M be the value of the greatest term, a
n
z
n
0
. Consider any point z such that |z| < |z
0
|.
We can bound the residual of

∞
n=0
|a
n
z
n
|,
R
N
(z) =
∞

n=N
|a
n
z
n
|
=

∞

n=N




a
n
z
n
a
n
z
n
0




|a
n
z
n
0
|
≤ M
∞

n=N





z
z
0




n
Since |z/z
0
| < 1, this is a convergent geometric series.
= M




z
z
0




N
1
1 − |z/z

0
|
→ 0 as N → ∞
Thus the power series is absolutely convergent for |z| < |z
0
|.
Result 12.3.4 If the power series

∞
n=0
a
n
z
n
converges for z = z
0
, then the series c onverges
absolutely for |z| < |z
0
|.
Example 12.3.1 Find the radii of convergence of the following series.
542
1.
∞

n=1
nz
n
2.
∞


n=1
n!z
n
3.
∞

n=1
n!z
n!
1. We apply the ratio test to determine the radius of convergence.
R = lim
n→∞




a
n
a
n+1




= lim
n→∞
n
n + 1
= 1

The series converges absolutely for |z| < 1.
2. We apply the ratio test to the series.
R = lim
n→∞




n!
(n + 1)!




= lim
n→∞
1
n + 1
= 0
The series has a vanishing radius of convergence. It converges only for z = 0.
543
3. Again we apply the ration test to determine the radius of convergence.
lim
n→∞




(n + 1)!z
(n+1)!

n!z
n!




< 1
lim
n→∞
(n + 1)|z|
(n+1)!−n!
< 1
lim
n→∞
(n + 1)|z|
(n)n!
< 1
lim
n→∞
(ln(n + 1) + (n)n! ln |z|) < 0
ln |z| < lim
n→∞
−ln(n + 1)
(n)n!
ln |z| < 0
|z| < 1
The series converges absolutely for |z| < 1.
Alternatively we could determine the radius of convergence of the series with the comparison test.
∞


n=1


n!z
n!


≤
∞

n=1
|nz
n
|

∞
n=1
nz
n
has a radius of convergence of 1. Thus the series must have a radius of convergence of at least 1.
Note that if |z| > 1 then the terms in the series do not vanish as n → ∞. Thus the series must diverge for all
|z| ≥ 1. Again we see that the radius of convergence is 1.
Uniform Convergence of Power Series. Consider a power series

∞
n=0
a
n
z
n

that converges in the disk |z| < r
0
.
The sum converges absolutely for z in the closed disk, |z| ≤ r < r
0
. Since |a
n
z
n
| ≤ |a
n
r
n
| and

∞
n=0
|a
n
r
n
| converges,
the power series is uniformly convergent in |z| ≤ r < r
0
.
Result 12.3.5 If the power series

∞
n=0
a

n
z
n
converges for |z| < r
0
then the series converges
uniformly for |z| ≤ r < r
0
.
544
Example 12.3.2 Convergence and Uniform Convergence. Consider the series
log(1 − z) = −
∞

n=1
z
n
n
.
This series converges for |z| ≤ 1, z = 1. Is the series uniformly convergent in this domain? The residual after N terms
R
N
is
R
N
(z) =
∞

n=N+1
z

n
n
.
We can get a lower bound on the absolute value of the residual for real, positive z.
|R
N
(x)| =
∞

n=N+1
x
n
n
≤

∞
N+1
x
α
α
dα
= −Ei((N + 1) ln x)
The exponential integral function, Ei(z), is define d
Ei(z) = −

∞
−z
e
−t
t

dt.
The exponential integral function is plotted in Figure 12.3. Since Ei(z) di verges as z → 0, by choosing x su fficien tly
close to 1 the residual can be made arbitrarily large. Thus this series is not uniformly convergent in the domain
|z| ≤ 1, z = 1. The series is uniformly convergent for |z| ≤ r < 1.
545
-4
-3
-2
-1
-3
-2
-1
Figure 12.3: The Exponential Integral Function.
Analyticity. Recall that a sufficient condition for the analyticity of a function f(z) in a domain is that

C
f(z) dz = 0
for all simple, closed contours in the domain.
Consider a power series f(z) =

∞
n=0
a
n
z
n
that is uniformly convergent in |z| ≤ r. If C is any simple, closed
contour in the domain then

C

f(z) dz exists. Expanding f(z) into a finite series and a residual,

C
f(z) dz =

C
(S
N
(z) + R
N
(z)) dz.
Since the series is uniformly convergent, for any given  > 0 there exists an N

such that |R
N

| <  for all z in |z| ≤ r.
Let L be the length of the contour C.





C
R
N

(z) dz





≤ L → 0 as N

→ ∞
546

C
f(z) dz = lim
N→∞

C

N−1

n=0
a
n
z
n
+ R
N
(z)

dz
=

C
∞


n=0
a
n
z
n
=
∞

n=0
a
n

C
z
n
dz
= 0
Thus f(z) is analytic for |z| < r.
Result 12.3.6 A power series is analytic in its domain of uniform convergence.
12.4 Integration and Differentiation of Power Series
Consider a power series f(z) =

∞
n=0
a
n
z
n
that is convergent in the disk |z| < r
0

. Let C be any contour of finite
length L lying entirely within the closed domain |z| ≤ r < r
0
. The integral of f(z) along C is

C
f(z) dz =

C
(S
N
(z) + R
N
(z)) dz.
Since the series is uniformly convergent in the closed disk, for any given  > 0, there exists an N

such that
|R
N

(z)| <  for all |z| ≤ r.
We bound the absolute value of the integral of R
N

(z).






C
R
N

(z) dz




≤

C
|R
N

(z)|dz
< L
→ 0 as N

→ ∞
547
Thus

C
f(z) dz = lim
N→∞

C
N


n=0
a
n
z
n
dz
= lim
N→∞
N

n=0
a
n

C
z
n
dz
=
∞

n=0
a
n

C
z
n
dz
Result 12.4.1 If C is a contour lying in the domain of uniform convergence of the power

series

∞
n=0
a
n
z
n
then

C
∞

n=0
a
n
z
n
dz =
∞

n=0
a
n

C
z
n
dz.
In the domain of uniform convergence of a series we can interchange the order of summation and a limit process.

That is,
lim
z→z
0
∞

n=0
a
n
(z) =
∞

n=0
lim
z→z
0
a
n
(z).
We can do this because the rate of convergence does not depend on z. Since differentiation is a limit process,
d
dz
f(z) = lim
h→0
f(z + h) − f(z)
h
,
we would expect that we could differentiate a uniformly convergent series.
Since we showed that a uniformly convergent power series is equal to an analytic function, we can differentiate a
power series in it’s domain of uniform convergence.

548
Result 12.4.2 Power series can be differentiated in their domain of uniform convergence.
d
dz
∞

n=0
a
n
z
n
=
∞

n=0
(n + 1)a
n+1
z
n
.
Example 12.4.1 Differentiating a Series. Consider the series from Example 12.3.2.
log(1 − z) = −
∞

n=1
z
n
n
We differentiate this to obtain the geometric series.
−

1
1 − z
= −
∞

n=1
z
n−1
1
1 − z
=
∞

n=0
z
n
The geometric series is convergent for |z| < 1 and uniformly convergent for |z| ≤ r < 1. Note that the domain of
convergence is different than the series for log(1 − z). The geometric s eries does not converge for |z| = 1, z = 1.
However, the domain of uniform convergence has remained the same.
549
12.5 Taylor Series
Result 12.5.1 Taylor’s Theorem. Let f(z) be a function that is single-valued and analytic
in |z − z
0
| < R. For all z in this open disk, f(z) has the convergent Taylor series
f(z) =
∞

n=0
f

(n)
(z
0
)
n!
(z −z
0
)
n
. (12.1)
We c an also write this as
f(z) =
∞

n=0
a
n
(z −z
0
)
n
, a
n
=
f
(n)
(z
0
)
n!

=
1
ı2π

C
f(z)
(z −z
0
)
n+1
dz, (12.2)
where C is a simple, positive, closed contour in 0 < |z − z
0
| < R that goes once around the
point z
0
.
Proof of Taylor’s Theorem. Let’s see why Result 12.5.1 is true. Consider a function f (z) that is analytic in
|z| < R. (Considering z
0
= 0 is only trivially more general as we can introduce the change of variables ζ = z − z
0
.)
According to Cauchy’s Integral Formula, (Result ??),
f(z) =
1
ı2π

C
f(ζ)

ζ − z
dζ, (12.3)
where C is a positive, simple, closed contour in 0 < |ζ −z| < R that goes once around z. We take this contour to be
the circle about the origin of radius r where |z| < r < R. (See Figure 12.4.)
550
Im(z)
Re(z)
r
C
R
z
Figure 12.4: Graph of Domain of Convergence and Contour of Integration.
We expand
1
ζ−z
in a geometric series,
1
ζ − z
=
1/ζ
1 − z/ζ
=
1
ζ
∞

n=0

z
ζ


n
, for |z| < |ζ|
=
∞

n=0
z
n
ζ
n+1
, for |z| < |ζ|
We substitute this series into Equation 12.3.
f(z) =
1
ı2π

C

∞

n=0
f(ζ)z
n
ζ
n+1

dζ
551
The series converges uniformly so we can interchange integration and summation.

=
∞

n=0
z
n
ı2π

C
f(ζ)
ζ
n+1
dζ
Now we have derived Equation 12.2. To obtain Equation 12.1, we apply Cauchy’s Integral Formula.
=
∞

n=0
f
(n)
(0)
n!
z
n
There is a table of some commonly encountered Taylor series in Appendix H.
Example 12.5.1 Consider the Taylor series expansion of 1/(1 − z) about z = 0. Previously, we showed that this
function is the sum of the geometric series

∞
n=0

z
n
and we used the ratio test to show that the series converged
absolutely for |z| < 1. Now we find the series usin g Taylor’s theorem. Since the nearest singularity of the function is
at z = 1, the radius of convergence of the series is 1. The coefficients in the series are
a
n
=
1
n!

d
n
dz
n
1
1 − z

z=0
=
1
n!

n!
(1 − z)
n

z=0
= 1
Thus we have

1
1 − z
=
∞

n=0
z
n
, for |z| < 1.
552
12.5.1 Newton’s Binomial Formula.
Result 12.5.2 For all |z| < 1, a complex:
(1 + z)
a
= 1 +

a
1

z +

a
2

z
2
+

a
3


z
3
+ ···
where

a
r

=
a(a −1)(a −2) ···(a −r + 1)
r!
.
If a is complex, then the expansion is of the principle branch of (1 + z)
a
. We define

r
0

= 1,

0
r

= 0, for r = 0,

0
0


= 1.
Example 12.5.2 Evaluate lim
n→∞
(1 + 1/n)
n
.
First we expand (1 + 1/n)
n
using Newton’s binomial formula.
lim
n→∞

1 +
1
n

n
= lim
n→∞

1 +

n
1

1
n
+

n

2

1
n
2
+

n
3

1
n
3
+ ···

= lim
n→∞

1 + 1 +
n(n − 1)
2!n
2
+
n(n − 1)(n − 2)
3!n
3
+ ···

=


1 + 1 +
1
2!
+
1
3!
+ ···

We recognize this as the Taylor series expansion of
e
1
.
=
e
553
We can also evaluate the limit using L’Hospital’s rule.
ln

lim
x→∞

1 +
1
x

x

= lim
x→∞
ln


1 +
1
x

x

= lim
x→∞
x ln

1 +
1
x

= lim
x→∞
ln(1 + 1/x)
1/x
= lim
x→∞
−1/x
2
1+1/x
−1/x
2
= 1
lim
x→∞


1 +
1
x

x
=
e
Example 12.5.3 Find the Taylor series expansion of 1/(1 + z) about z = 0.
For |z| < 1,
1
1 + z
= 1 +

−1
1

z +

−1
2

z
2
+

−1
3

z
3

+ ···
= 1 + (−1)
1
z + (−1)
2
z
2
+ (−1)
3
z
3
+ ···
= 1 − z + z
2
− z
3
+ ···
Example 12.5.4 Find the first few terms in the Taylor series expansion of
1
√
z
2
+ 5z + 6
about the origin.
554
We factor the denominator and then apply Newton’s binomial formula.
1
√
z
2

+ 5z + 6
=
1
√
z + 3
1
√
z + 2
=
1
√
3

1 + z/3
1
√
2

1 + z/2
=
1
√
6

1 +

−1/2
1

z

3
+

−1/2
2


z
3

2
+ ···

1 +

−1/2
1

z
2
+

−1/2
2


z
2

2

+ ···

=
1
√
6

1 −
z
6
+
z
2
24
+ ···

1 −
z
4
+
3z
2
32
+ ···

=
1
√
6


1 −
5
12
z +
17
96
z
2
+ ···

12.6 Laurent Series
Result 12.6.1 Let f(z) be single-valued and analytic in the annulus R
1
< |z − z
0
| < R
2
.
For points in the annulus, the function has the convergent Laurent series
f(z) =
∞

n=−∞
a
n
z
n
,
where
a

n
=
1
ı2π

C
f(z)
(z −z
0
)
n+1
dz
and C is a positively oriented, closed contour around z
0
lying in the annulus.
To derive this result, consider a function f(ζ) that is analytic in the annulus R
1
< |ζ| < R
2
. Consider any point z
555
in the annulus. Let C
1
be a circle of radius r
1
with R
1
< r
1
< |z|. Let C

2
be a circle of radius r
2
with |z| < r
2
< R
2
.
Let C
z
be a circle around z, lying entirely between C
1
and C
2
. (See Figure 12.5 for an illustration.)
Consider the integral of
f(ζ)
ζ−z
around the C
2
contour. Since the the only singularities of
f(ζ)
ζ−z
occur at ζ = z and at
points outside the annulus,

C
2
f(ζ)
ζ − z

dζ =

C
z
f(ζ)
ζ − z
dζ +

C
1
f(ζ)
ζ − z
dζ.
By Cauchy’s Integral Formula, the integral around C
z
is

C
z
f(ζ)
ζ − z
dζ = ı2πf(z).
This gives us an expression for f(z).
f(z) =
1
ı2π

C
2
f(ζ)

ζ − z
dζ −
1
ı2π

C
1
f(ζ)
ζ − z
dζ (12.4)
On the C
2
contour, |z| < |ζ|. Thus
1
ζ − z
=
1/ζ
1 − z/ζ
=
1
ζ
∞

n=0

z
ζ

n
, for |z| < |ζ|

=
∞

n=0
z
n
ζ
n+1
, for |z| < |ζ|
556
On the C
1
contour, |ζ| < |z|. Thus
−
1
ζ − z
=
1/z
1 − ζ/z
=
1
z
∞

n=0

ζ
z

n

, for |ζ| < |z|
=
∞

n=0
ζ
n
z
n+1
, for |ζ| < |z|
=
−1

n=−∞
z
n
ζ
n+1
, for |ζ| < |z|
We substitute these geometric series into Equation 12.4.
f(z) =
1
ı2π

C
2

∞

n=0

f(ζ)z
n
ζ
n+1

dζ +
1
ı2π

C
1

−1

n=−∞
f(ζ)z
n
ζ
n+1

dζ
Since the sums converge uniformly, we can interchange the order of integration and summation.
f(z) =
1
ı2π
∞

n=0

C

2
f(ζ)z
n
ζ
n+1
dζ +
1
ı2π
−1

n=−∞

C
1
f(ζ)z
n
ζ
n+1
dζ
Since the only sin gularities of the integrands lie outside of the ann ulu s, the C
1
and C
2
contours can be deformed to
any positive, closed contour C that lies in the annulus and encloses the origin. (See Figure 12.5.) Finally, we combine
the two integrals to obtain the desired result.
f(z) =
∞

n=−∞

1
ı2π


C
f(ζ)
ζ
n+1
dζ

z
n
For the case of arbitrary z
0
, simply make the transformation z → z −z
0
.
557