Chapter
SERIES
D
OF FUNCTIONS
already run into series developments of functions several times :
exponential, sine, cosine functions were expanded into power series;
Taylor's theorem provides a way to develop series expansions for suitable
functions; the exponential of a matrix gives us the only sure way to solve"
We shall see in this chapter
a system of constant coefficient linear equations.
that a general technique for solving a differential equation involves approxi
mation of the solution by series expansions.
We shall begin by formulating the definition of convergence of a series of
continuous functions and verifying the general criteria guaranteeing conver
One of the most important of series expansions is that of power series.
gence.
We shall say that a function is analytic if it can be locally developed into a
power series. We shall finally verify the fundamental theorem of algebra
and complete the discussion of constant coefficient equations. We have
delayed this until now because the kind of analytic techniques involved in the
fundamental theorem are those which are most appropriately developed for
the class of analytic functions. Further techniques for operating with
power series will be explored, as well as the question of estimation of the error
in replacing the power series by a partial sum.
We have
the
"
400
5.7
5.1
Convergence
401
Convergence
Definition 1.
Let
{/J be a sequence of continuous functions defined on a
The series formed of the {fk} is the sequence of partial sums
We say that the series converges if the sequence of these partial
subset X of R".
(Zfc=i /*}
sum
converges
(in the
sense
of Definition 19 of
Chapter 2),
and denote the
limitbyi?=iA.
Precisely then, /=
Xfc=iA
if. corresponding
to every
e >
0, there is
an
JV such that
/(x)
ZA(x)
-
< e
for all
n >
N and
xe
X
k=l
Since the limit of
a
(Theorem 2.14),
we can
uniformly convergent
tinuous functions is continuous.
sequences,
obtain
we
a
sequence of functions is continuous
convergent series of con
assert that the sum of a
Likewise, from the Cauchy criterion for
corresponding
criterion for the convergence of series.
Proposition 1. (Cauchy Criterion) Let {fk} be a sequence of continuous
functions. The series /k converges if and only if, to each e > 0 there cor
respond an N such that
X
Proof.
criterion.
m
>
n
h
<
for all
n,
m >
N
We must show that the sequence gn=^l=i fk satisfies the Cauchy
For a given e > 0, let N be as in the proposition. Then, for any x,
:> N
\gm(x)
Thus \\gm
-
g\\
g(x)\
< e, so the
I /*(x)
2 /.
<
proposition
<
is proven.
guaranteed if the series of real numbers
X*=i ||/J converges (for !!?=+ 1 /J !*=,,+ 1 ll/J). This gives us a
powerful technique for verifying convergence of series.
Notice that the Cauchy criterion is
Let {fk} be a sequence of continuous functions defined
The series is said to converge absolutely if ^f=1 ||/J < oo.
Definition 2.
set X.
on a
402
5
Series
of Functions
Of course, as remarked above, an absolutely convergent series is conver
gent. In the case of absolute convergence we can pose a comparison test,
just
as
for series of numbers.
Theorem 5.1.
(Comparison Test) Let {fk} be
a set X.
functions defined
Suppose there is a
a
on
numbers and
integer N
an
(0 \\fk\\<Pn
>
of continuous
{pk} of positive
sequence
sequence
0 such that
fork>N
00
Z Pk
k
(ii)
=
Then
/fc
Proof.
<
l
absolutely.
converges
The verification is the
same as
that for number series (Theorem 2.3).
Examples
zk
1.
r
<
1.
Zr"
\\zk\\ < rk in
For
=
Tzr
series is not
\fk
in
{|z|
<
1}.
<
r} for
any
that domain, and
< oo
zk does
2.
z) uniformly and absolutely
VU
=
a
not
converge
Cauchy
L,fk
=
Z
uniformly
in
{\z\
In
fact, the
sequence of functions, because for every n,
=
1
k=l
Thus, for
\, say, there is no JV such that || "=,,/& II
in
n + 1
N,
fact, not even for m
=
m > n >
3.
e7
<
i for all
=
.
=
R finite.
Rk
*T\
X^=1 zk/k\
converges
Again, by comparison
R"
and
zr!<0
uniformly
in any disk
{|z|
<
R}
with
5.1
Convergence
403
cos nx
,
4-n^
converges
uniformly
on
the whole real line.
For any x,
cos nx
^n2
1/rc2
Since
5. If
f(z)
=
{ak}
=1
< oo
the
comparison
test
easily applies.
is any sequence of numbers such that
\ak\ < oo, then
on the closed unit disk.
function
is
a
continuous
zk
ak
The series converges uniformly since \\akzk\\ < \ak\.
Finally, for the purpose of availability, we record the obvious extensions
to series of the
concerning integration and differentiation of
propositions
sequences of functions.
Proposition 2.
(i) Let {/} be a sequence of continuous functions on the
Let gn(x)
\xafn- V tne series offunctions / converges, so
=
interval
if/.=r(/.)
n=l
(ii)
Let
'a
Ja
{/}
interval [a, b~\.
for
some
c,
\n=l
<">
I
be
a sequence
Let gn =fc.
of continuously differ entiable functions on the
If the series of functions # converges, and
converges, then the series
/(c)
[_a,b~].
does the series
/
converges.
The limit is
continuously differentiable and
(5-2)
(Z/,y =z/;
Examples
6.
CO
y
ln(l-x)=Z-r
K
k
=
l
-1<*<1
404
5
Series of Functions
This follows
by integrating
the
geometric series (Example 1)
term
by
term
1
!*
CO
.*
j0T^~t=S'oLt
tk
jfc+1
co
in(i-x)= y
oo
-
=
"
cos nx
,.
=
-
n\
k=i
is
infinitely differentiable.
"
sin
For the differentiated series
nx
is also convergent.
the series
S
fk
-
k^k
k=ok + l
fix)
y
(5.3)
can
By Proposition 2(ii) the sum is/'(x). Similarly,
by term, and gives
be differentiated term
n cos nx
"
n
=
1
(fi
which is
1)!
-
again convergent.
8. We
can
following
1
develop
a
observations.
series expansion for arc tan x
From the geometric series
",*
r~x=^xk
i-
we
x
k
obtain
1 + x
o
=
by substituting -x2 for
t=o
Integrate:
v2[+l
arctanx=
(-1)'
n=o
zk + 1
x
according
to the
5.1
405
Convergence
EXERCISES
1. For what values of
(a)
do these series of functions converge absolutely:
x
22V
|
(d)
n=0
(jr+18)"
n=0
cos nx
N
2
(b)
n=
|
(c)
n
n=0
n
2. In which domains of the
| hz-
(a)
n
=
(b)
0
|
n
=
0
| x<"2>
(f )
e
""
2
(e>
X
1
0
=
3. Which of these series
=
0
complex plane do these series converge?
-il
| -V
(c)
(Z)!
n
=
0
be differentiated
can
or
integrated
domain of convergence ?
(a)
Exercise 1(a)
(c)
(b)
Exercise
(d)
Exercise
1(d)
cos nx
1(b)
2
n
4. Find the power series
=
0
expansion
1
(a)
(c)
(l+x2)
nm
(b)
<
fl
for these functions
f
e'1
:
dt
Jo
y-
PROBLEMS
expansion for sin
2x.)
Find power series expansions for sin2
2. Prove Proposition 2.
1. (a)
(Hint :
(b)
Find
2 sin
a
power series
x cos x
3. Show that
lim
-.-l
2
=
log 2
n=l
Can you conclude
2
L_ii=i0g2?
=
x cos x.
sin
x
and cos2
x.
on
their
406
5
5.2
Series
of Functions
The Fundamental Theorem of
For the remainder of this
complex-valued
functions
the
are
P(z)
functions of
(where the at
chapter
complex
restrict attention
we
variable.
The
exclusively to
simplest class of such
polynomial functions ; that is, functions of the form
az"
=
a
Algebra
a^iz"'1
+
+
+ axz + a0
,
complex numbers). We shall always assume a # 0; in
degree of P. It is a basic fact of mathematics that
every polynomial has a root; that is, there is a number c e C such that P(c)
0. The proof of this fact consists in a systematic investigation of the analytic
properties of polynomials. First, we recall de Moivre's theorem.
this
are
is called the
case n
=
Lemma.
Every
Let
Proof.
ce
C,
most convenient.
em*
ei8. tnat jSj
_
c
n^
<*2
.
(co)n
a>i
1
=
then
expO'aj),
=
These
.
(pe'*)n
=
distinct, and
Now,
we
reie
are
=
.
.
are
.
.
a*
,
.
w
=
,
.
.
.
,
a>
exp(z'a)
=
The numbers
c.
all nth roots of
such that
=
re'" is
=
r
and
Let
1)
=
a_!
a
,
n
,
pe'*
2tt(u
=
=
2n
n
are
called the nth roots of
need two
polynomials.
.
n
c
p"
polar representation
2nk
=
,
distinct nth roots.
integral multiple of 2tt.
an
47T
=
n
For this purpose, the
c is a number
q js
_
n
Then
^ 0.
number has
An nth root of
277
i
complex
nonzero
all
distinct
pe^ioi,
and
Now, if
unity.
...,
pe'^cu,,
,
p
have
the
property
8 In,
(r)1'" and <f>
pe'*co are then all
=
...,
=
c.
deeper facts depending on the continuity properties
intuitively clear: that \P(z)\ gets arbitrarily large
The first is
of
as
The second is the crucial fact for the fundamental theorem: the
where a polynomial has a minimum modulus must be a root.
z-*co.
place
Lemma.
lim
(i)
Let
P(z)
\P(z)\
=
=
anz"
+
oo, that
+ axz + a0 be a polynomial
is, given any M
>
0 there is
a
of degree
n>0.
K>0 such that
|z|->CO
|P(z)|
(ii)
are z
>
M whenever
\z\
>
K.
# 0, then z0 cannot be a minimum point for
close to z0 such that \P(z)\ < |P(z0)|.
If P(z0)
\P\; that is, there
5.2
The Fundamental Theorem
of Algebra
407
Proof.
(i) The point here is that the highest-degree term of P is
regards the behavior of P as z^ oo. For z ^ 0,
the
dominating
term
as
1^)1
|z|"-k>A: also for
If |z|>ii:>l, then
Let M > 0 be
Z
=
so
!a|-i("j>l)
>
tf
k <n,
given, and choose
max(l,2M|a|-1,2|fl|-1 (2
Mj\
Then, for \z\>K,
_l
ak
"v
>\a\[l
)>-2M
Thus
\P(z)\^\z\"-i\an\>.K-i\a\>M
(ii) Suppose
now
that
Let
P(z0) ^0.
Q(z) =(P(z0)Y1P(z + z0)
Then G is also a polynomial,
point for Q. Let
Q(z)
=
1 +
2 * z*
=
G(0)
! +
=
1, and
we
must show that 0 is not
a
minimum
zm(am + z#(z^
# 0, and g(z)
thus continuous (and that is all we
is
and
a
is
polynomial
2s=m+i a*z"-(",+1). ^
to use the fact that for small z, zm
need to know about g). Here again we want
+*
the
to
close
polynomial 1 + zmam which has no minimum
so Q is very
dominates zm
with r < 1).
zm
that
z
so
-rjam
0
modulus at
(choose
choose an mth root of -a1 ; call it z, and consider the function
where
m
is chosen
as
the least
positive integer
,
=
In
our
case,
we
k for which
ak
=
408
5
Series
of Functions
Q(rz0) of a real variable
Q(rz0)
where
h(r)
=
We have
r.
l+rm(-l+rh(r))
=
a^girzo)
is
a
continuous complex-valued function.
Thus
\Q(rzo)\<\l-n+r+1\h(r)\
Now
limrrt(r) =0,
so
choose
we can
1 small enough
r0 <
r-.0
so
that
\r0h(r0)\ <i.
Then
I Q(rz0) I <, 1
which proves part
Theorem 5.2.
r0m +
-
i-o-G)
< 1
-
ir0m
< 1
(ii).
(Fundamental Theorem of Algebra)
There is
positive degree.
a
e
z0
C such that
P(z0)
Let P be
=
a
polynomial of
0.
Proof. Let P(0)
c0
By part (i) of the lemma, there is a K > 0 such that for
\z\>K, \P(z)\>\c0\. Now A={zeC; |z| <K} is compact, so \P\ attains a
minimum value on A, say at z0
But then z0 is a minimum point for all of C.
For,
since OeA, \P(z0)\ < |P(0)|
and for z A, |P(z)| > c0 > |P(z0)|. Thus, even
c0
for z A, we have |P(z0) | :< IP(z) |
But then, by part (ii), there is no alternative : we
must have P(z0)
0.
=
.
.
=
,
.
=
Factorization
Theory
We should recall that if
(this is
degree
zero
c
is
a zero
of the
proven below in Theorem 5.3).
1 less than that of P.
If deg Q
of P.
Further, Q(z)
order to find
=
exactly deg P
(z
-
zeros
>
c')2'(z)
of P.
polynomial P, then z c factors P
P(z) (z c)Q(z) and Q has
0, Q has a zero c', which is also a
-
Thus
and
=
we can
-
repeat this argument in
This is the factorization theorem of
algebra.
Theorem 5.3.
n >
There
0.
P(z)
Proof.
a(z
=
The
P(z)
=
(Factorization Theorem)
complex numbers a #0,
are
-
zi)
(z
+
a0
=
ai
Iz
z1;
.
.
.
,
a
polynomial of degree
z such that
z)
-
proof is by induction
OiZ
Let P be
I
on n.
1)
If
n
=1 the situation is
simple:
The Fundamental Theorem
5.2
409
of Algebra
(since ai ^ 0). Now we consider the case of general degree n, assuming the corollary
for polynomials of degree n 1. By the theorem, there is a point c such that
P(c) 0. Then
=
P(z)
=
P(z)
P(c)
-
=
2 *(z"
c*)
-
n-l/
2 *(*
=
<o
-
( zW-1 )
"J
\
n
on the right is a polynomial of degree n
1, so the induction assumption
z-i.
z_i) for suitable a^O, zu
(z
applies: it can be written as a(z Zi)
we
obtain
z
Thus, writing c
The factor
...,
=
,
(5-4)
P(z)=a(z-zi)---(z-z)
clearly unique, except for the order of the z;'s: a is the
P and {zu
of
coefficient
z} are the roots of P. Of course,
leading
If
let
not
be
need
distinct;
rs be the set of distinct roots.
ru
z
Zjl,
the list {zu
zj,
we let m,. be the number of occurrences of the root r{ in
We can rewrite (5.4) as
is called the multiplicity of the root r{
This factorization is
.
.
.
,
.
.
.
.
.
.
,
,
...,
.
mt
P(z)
=
a(z
(z
rt)mi
-
(5.5)
rs)m
-
+ ms
n, the degree of P.
clearly mt +
Before concluding this section we should remark
and
real
=
=
Real
polynomials.
polynomials
0), but their complex roots
+ axz + a0 be
P(f)
so r
is also
(z
-
=
a
a
real
come
polynomial.
ani?)n +---+a1r
root of P.
r)(z -r)
=
on
the factorization of
(viz., z2 + 1
need not have real roots
conjugate pairs.
If P(r)
0, then
in
Let
P(z)
az"
=
-\
=
+
a0=(anr"+---
+ a1z +
a0)~
=
PirY
=
0
Since
z2-(r
+
r)z
+
rr
=
z2-2 Re(r)z
+
\r\2
has real coefficients. Thus, if we rearrange the roots of P
r, r, we
into the real roots ru ...,rk and the conjugate pairs rk + l, rk+u
and
linear
of
a
into
quadratic
product
can rewrite the factorization (5.5)
the
polynomial
.
real
.
.
,
,
polynomials.
P(z)
=
a(z
-
rj"1
(z
-
rk)m\z2
-
2
k+il2)
(z2-2Re(rt)z
Rfi(rt+1)z
+
'
'
+
|rr|2)
410
5
Series
of Functions
PROBLEMS
4. Let w i, ...,co be the n nth roots of unity.
Show that they are
arranged at n equidistant points around the unit circle. Show that the sets
{cui,
oi}, {a>i, oil2,
oj'c1} are the same, if u>, is the nearest such
point to 1.
2 is
5. Let a>u
Choose k so that kn
w be the nth roots of unity.
divisible by 4.
Show that ikoju
ikwn are the nth roots of 1.
6. Show that : (a) deg PQ
deg P + deg Q.
max(deg P, deg Q) if deg P # deg Q.
(b) deg(P + Q)
(c) When is the equation in (b) not true?
7. Given two polynomials P, Q show that there is a polynomial R which
factors both P, Q and is factored by any polynomial which factors both
P, Q. R is called the greatest common divisor of P and Q.
8. Show that a real polynomial of odd degree has a real root.
9. Prove that the polynomial 1 + zma (m > 0) has no minimum modulus
.
.
..,
.
.
.
.
.,
,
...,
=
=
at
z
=
0.
10. For
P'(z)=
P(z)
=
2S'=o
az"
a
polynomial,
let
Znanz"'1
1
n=
(a) Verify that the transformation P^P' is linear and satisfies
(PQY=PQ' + P'Q
(by induction
(P -*P' is
a
on
deg P).
complex analog of differentiation)
(b) Prove that
P'(r)
=
(c)
r
is
multiple
a
root of P if and
Define P"
=(P')',
m
P"
=
if and
(P")', and so on. Then
only if P(r) P'(r)
=
5.3 Constant Coefficient Linear Differential
Now that
we
complete
the
Let L be
tion.
L is
a
mapping
L(f)
=
=
0 and
0.
of at least multiplicity
to
only if P(r)
=
r
=
is
a
Pfm"
root of P
*>(r)
=
0.
Equations
know the factorization theorem for
study
a
of constant coefficient
polynomials we can return
equations in one unknown func
constant coefficient differential
operator of order k ; that is,
from functions to functions defined
fm
+
'. f{i)
i
=
0
^C
by
(5-6)
5.3
Corresponding
PLiX)
Constant
to L is the
Xk
=
Linear
Coefficient
polynomial
kZaiXi
+
=
>
o
called the characteristic polynomial of L.
know about such differential operators.
Let L be
Theorem 5.4.
the equation
functions.
Lf=0
If r is a
411
Differential Equations
is
root
ofPL(X)
=
0,
we
already
The collection
given by (5.6).
n-dimensional
an
We recall the facts that
space of
then erx e S(L).
vector
S(L) of solutions of
infinitely differ entiable
Now if all the roots of the characteristic
solutions of Lf
=
S(L). To examine the case of multiple roots,
closely the relationship between the given differential
operator and its characteristic polynomial. If P is a polynomial, we will let
LP represent the corresponding operator; that is, for P(X) j=0 a,-*', FP
is defined by
pendent.
Thus
polynomial are distinct, we have n
0, and it is easily verified (Problem 1 1) that they are inde
they
span
must examine more
we
=
i
=
Now, from what
0
we
already
know about these differential
equations
we can
guess that the factorization of P will tell us all we want to know about LP
In fact, we can factor the corresponding operator accordingly as the next
.
lemma shows.
Lemma 1.
Proof.
LP(LQ(f)).
Lp+q
=
LP
+
Lq\ LPq
=
LpLq
Of course, LPLQ is defined as the
The first equation is obvious.
.
composition
of operators : (LPLQ)(f)
We
a little work.
=
The second takes
a0 then
0, that is, P(x)
will prove it by induction on the degree of P. If deg P
differentiable
for
sufficiently
and
any
=LP(LQ(f)\
a0LQ(f)
LPQ(f)
PQ=a0Q
function / Now suppose the lemma is true for all polynomials of degree n. Let P
of degree n+ 1. If a is a root of P, we can write P(X)
be a
=
=
,
=
polynomial
(X-a)S(X), where
=
is
of
degree
Thus, by hypothesis,
polynomial
LsLq. We have left only to verify the lemma for polynomials of degree 1.
That is, we must show that if R is a polynomial of degree 1 and T is any polynomial,
X a, so that
For once this is verified, we take R(X)
then Lrt=LrLt.
Lsq
S
a
n.
=
=
P
=
RS.
Then
Lpq =LRSq =LrLSq =LrLsLq =LrsLq =LpLq
412
5
Series
So, let R(X)
=
RT(x)
X-a, T(X)
=
2?U b, X'.
=f(b,- abl+1)X,+1
(
Now
of Functions
=
-
Then
abo
0
compute LRLT :
we
\
(m
(
in
m
IW" =2(w,0)'-^i,r
=
o
(
/
=
o
i
=
o
m
tibi-abi+l)f'^-ab0f
=
1
=
0
The lemma is proven.
It follows from the lemma that if
of LPQ(f)
We can,
factors.
0 is
=
solution of LP(f)
the factorization
by
P(X)
a
=
in
Now let P be
write P
S(LP).
equation LP(f)
we
as a
given polynomial.
product of first-order
with mt +
(X- aj"' ---(X- as)m
Thus
factor of P, then any solution
a
0.
theorem,
Because of Lemma 1 the solutions
are
Q is
=
a
+ ms
=
to the factors
corresponding
deg P
(X
aj)1"1
need to discover the solutions of the differential
0, where P(X)
(X c)m.
Consider, for example, the differential operator corresponding to (X c)2.
We know one solution: ecx; we find another by the technique of variation of
parameters. iX
c)2 X2 2cX + c2. Test the operator on y zecx.
=
=
=
-
y'
=
-
=
z'ecx + zcecx
y"
=
z"ecx + 2z'cee* + zc2ecx
Then
/'
Thus
z
general
=
-
2cy'
+
c2y
=
z"e"
=
0
or
z"
=
0
x, and the second solution is xecx.
We
can
guess then that the
situation is this.
Lemma 2.
The solutions
of LiX_c)m(f)
=
0
are
spanned by e", xecx,
...,
xm~V*.
Proof. We
by induction.
have to show that the named functions
The
case m
=
1 is
are
solutions.
already known (by Lemma 1).
We do that
Thus
we
may
5.3
assume
Constant
the lemma for
need only verify that
a
Linear
Coefficient
given value of m, and prove it for
is
Llx_c)m+i(xme")
L(x-c->"<Lix-c,(xmeC')
413
Differential Equations
m
+ 1
By Lemma 2,
.
we
But this is
zero.
cxme"
=
La.c)m(mxm-'e" +
=
La_c)m(mxm-1e")=0
-
cxme")
by induction.
Theorem 5.5.
Let
p(X)
=
j=o fl<-^f
X" +
coefficients. Let au ...,asbe the roots
Then the space S(Lp)
ms, respectively.
polynomial with complex
multiplicities mt
of the differential equation
oe a
0 with
ofp(X)
of solutions
=
n-l
lp(/)=/
+
i
Ei/(0
<
=
=
o
0
of the functions xJea,x,
is the linear span
0 <j
<
m{
.
EXERCISES
5. Solve these differential
(a)
(b)
(c)
(d)
(e)
(f )
equations:
1
0, y"(0)
0, y'(0)
0, y(0)
4y
1
1 y'(0)
2, v"(0)
0, y(0)
5/
3y
y
y"
1.
1, /(0)
0, y"(0)
y- 6y" + 12/ ~Sy=0, y(0)
1
2, y"(0)
0, y(0)
3, /(0)
y
y~- 3y" + 3/
2.
2, y '(0)
2, v"(0)
2, y'(0)
yw + 2y" + ^ 0, v(0)
">
1,
I2y' + 9y 0, y(0) /(0) y"(0)
/4) + 4y('"> 2y<
y- 5y" + 8/
(g)
=
=
=
=
-
.
-
-
-
=
=
-
=
=
-
.
,
=
-
^"(0)=0.
3/ "' + 2y
y^
=
=
0, y(0)
=
-
.
=
0, y'(0)
=
=
=
-
-
=
=
=
=
=
=
=
=
=
y"(0)
=
=
=
y "(0)
=
1
.
PROBLEMS
11.
(a)
Show that if ru
.
.
.
,
r are n
distinct numbers, the matrix
1
r
r2
'It
is
nonsingular.
(Hint: If the
rows
are
dependent,
we
obtain
a
poly
1 with n distinct roots.)
nomial of degree n
functions
The
exp(/x) are independent. (Hint:
exp(rix),
(b)
If these functions were dependent, we would be able to prove that the
..
columns of the above matrix
are
.,
dependent.)
414
5.4
5
Series
of Functions
Solutions in Series
If now
given a linear differential equation which is not homogeneous,
we are
has variable coefficients, we have a problem of a much different magnitude.
In general, such problems cannot be solved explicitly.
Thus, we must seek
or
approximate solutions. This is one of the places where
representations of functions are usable. The procedure of series
approximation has two aspects. First, we must establish the theoretical
validity of such a technique and, secondly (and this is essential from the
practical point of view), we need a technique for effectively computing the
In this section we shall describe this procedure, deferring these two
error.
essential points (which turn out to be the same!) until Section 5.7.
ways to obtain
series
First,
an
example.
we want
Suppose
f"(x)+gl(x)f'(x)+g0(x)f(x)
=
0
the function / such that
/(0)
=
a0
f'(0)
=
a,
are defined in a neighborhood of 0.
We shall assume
sufficiently differentiable. Now our initial conditions
first two terms of the Taylor expansion of/ at 0 :
where g0 and gx
that
they
give
us
f(x)
=
are
the
a0 + atx +
higher-order
will be based
(5.7)
terms
"
the tacit
assumption that the higherorder terms" are computable, and knowing enough of them will give
a usable approximation to the solution.
Now, evaluating the differ
ential equation itself at 0 gives us the second-order term :
Our
technique
on
f"i0)+g1(0)f'(Q)+go(0)fi0)
=
0
or
f"(0)
=
-(g1(0)a1
+
go(0)ao)
so
f(x)
=
a0 + atx
i(aog0(0)
Differentiating the
+
fli#i(0))x2
differential
equation
+
higher-order terms
will
give
an
identity
(5.8)
express-
in terms of lower
ing/'"
Fix)
+
derivatives,
g[ix)f'(x)
+
<h(x)/"(x)
-(^(0)
g0(0))ai
+
415
Solutions in Series
5.4
so we
continue,
may
g'0ix)f(x)
+
g0ix)f'(x)
0
=
so
/'"(0)
=
(gi(0)2
=
and
a0 + ajx
=
9'M
~
g'0i0)ao
-
9oiO))at
~
+
have the third term of the
so we
/(x)
+
i(ao0o(O)
-
+
iCGhW2
+
higher-order
^1(O)(01(O)a1
(0o(O)fif!(O)
5o(0)ao)
+
g'oi0))a0
-
Taylor series of/:
0i0i(O))x2
+
3i(0)
-
+
fif0(0))ai+ 0o(O)<h(O)
-
-
3o(0))a]x3
terms
Example
Perhaps an explicit calculation
approximate solution of:
9.
/'
(x2
+
X0)
=
1)/
-
0
+ xy
y\0)
=
The solution thus
by substituting
f(x)
=
=
+
We shall find
an
(5.9)
x2
2
begins fix)
=
2x +
f (0) is easy
Equation (5.9):
.
the initial conditions into
to calculate
2x + x2 +
Differentiating (5.9),
y"
is in order.
2x/
+
(x2
1)/
-
we
obtain
+ y +
xy'
-
(5.10)
2x
Evaluating at 0 we find /""(O) =/"(0) -/(0) 2. Differentiating
(5.10) and evaluating at 0, we obtain /(4,(0) 2; once again gives
10. Thus, to five terms the Taylor expansion of the
/(5>(0)
=
=
=
-
desired solution is
fix)
=
Admittedly
2x + x2 +
\x3
+
this is not very
h x4
-
A x5
+
higher-order
glamorous, but it
is
terms
computable!
The
phrase
416
5
Series
of Functions
terms"
"higher-order
represents the
between the
error
nomial exhibited above and the actual solution.
fifth-degree poly
polynomial is com
That
pletely meaningless without some estimate on the error incurred. But our
method gives no hint as how to estimate. So, in the hope of being able to
give more form to the "higher-order terms," we will try a more brazen
approach : we begin by assuming that the desired solution is the sum of a
convergent power series (its full Taylor expansion") and we try to find the
coefficients. If /(x)
=0 ax", then differentiating term by term we
"
=
obtain
fn^x"-1
/'(x)=
n=l
n(-lK'x""2
/"(x)=
n
/w(x)
=
2
n(n
=
n
=
1)
-
(n
\)anx"-k
k +
-
k
We make these substitutions into the
solve for the
{a} by equating
Let
us
solution.
n(n
reconsider
lKx"-2
+
(x2
-
=
ax"-x
0
ay
=
ax"
+
+
Thus
2)(k
we
z2
=
0
(5.11)
2
l)ak+2
+
-
n=0
The coefficient of xk in the left-hand side of
ik
be the desired
"=0 ax"
becomes
n=l
=
a0
f(x)
problem
1)
=
n1=2
2
+
ik- 1K_!
have to solve these
-
(k
+
(5.11)
l)ak+1
is
+ ak_x
equations
a0 =0
ay
=
2
2a2~a1=2
3.2a3 + a0 2a2
4.3a4 + 2!
3a3
-
-
n(n
-
l)a
+
(n
-
(initial conditions)
ik 0)
(k 1)
(k 2)
=
=
=
0
=
0
2)a_3 -in- l)a_!
and
offm.
Let
(5.9).
The statement of the
-
given differential equations
the coefficients
=
=0
(k
-
n
-
2)
5.4
We
a
solve, because each equation
can
in- lK-i -(n-2K_3
1)
nin
=
Solutions in Series
can
417
be written in the form
n>2
(5.12)
-
However,
at
we
have
an
added
advantage
estimate for the general term a
an
.
in that
In fact,
we can
we
make
a
guess
assert
2"
[n/3]!
([x]
n
(5.13)
<
k,
=
largest integer less than or equal to x). This
0, 1, 2; we verify it in general by induction.
=
(fi-l)|aM-1|+(n-2)|fl.-3|
nin
1)
Kl<
-
<-(|an_!|
+
|a_3|)
n
W
2P_
<
n
\l(n
+
l)/3]!
-
[(n-3)/3]!
Now, since
in
-
3)
in
n
-
3)'
t
=
3
3
so
(?)
in
3)"
-
!
>
_
[513 I
Similarly,
my
~
3
Thus,
1
\a\<
3[n/3]!(
2"
;_W3]!
r?i ?
3
is in fact true for
418
5
Series
This
of Functions
now
tells
us a
lot.
For, the solution
to the
problem
in
(5.9)
differs from
ZX
+
+ ^"X
by
at most
X
dominated
T2~X
T^X
>6 ax",
where the an
Thus the
satisfy (5.13).
error
is
by
<
(1
LnL+
fe!
+
2|x|
Hence in the interval 0
+
~)lk + 2\ v|3Ji+2
<-)3k+l|v|3k+l
^3k\v\3k
^
Flrn W"= ^
a6[n/3]!
[,
k!
ka2
1
-
far
so
our
as
to discard the
-
(2|x|)3)
given by the above
(which is about 52) !
1 the solution is
polynomial except for our error of at most 7e2
The reader is forgiven if he is unimpressed with
should not go
the paucity of
*!
^
4|x|2) (exp(2|x|)3
< x <
I
+
our
estimate, but he
for this
technique
reason.
For
results is due to laziness rather than the uselessness
If we pushed this procedure up to 1000
a
task
for
(an easy
computer), then the error would be at most
which
is
less than (50)"90; a good estimate indeed.
2looo/l000!
of the
Taylor development.
terms
le2
Let
us
.
recapitulate
ym
+
i
the basic ideas.
iW*)/0
=
=
m, yiO)
We
=
c0
are
,
given
y'iO)
=
initial value
an
Cl,
.
.
.
,
y*-x>(0)
problem :
=
ck_,
0
"
the #'s and h by power series expansions and test the solution"
f(x)
jj= 0 a x". The first k terms are found from the initial conditions,
and the rest are found by equating the coefficient of x" on both sides of the
We
replace
=
equation.
This leaves
us
with these
problems
to
resolve:
(i) Can we represent the given #'s and /; by power series?
(ii) Can we differentiate a power series term by term ?
(iii) How do we multiply power series? (In the above illustration, the
g's were polynomials, so there was not much difficulty.)
(iv) Can the system of relations between the aks really be solved uniquely?
(v) Can we effectively estimate the error between the solution and a finite
part of its (supposed) Taylor expansion ?
Little
answer
by little, we will resolve these problems. Suffice it to say that the
(i) in general is No (see Section 5.8). However, in problems that
to
419
Solutions in Series
5.4
do arise
naturally, the given functions usually are sums of convergent power
If this is the case, all other questions can be satisfactorily answered;
that is, the solution also is the sum of a convergent power series whose co
efficients can be determined by the above technique and the estimate on the
series.
remainder
be
can
Let
effectively computed.
us
look at another illustration.
Examples
10.
x3y'"
y(0)
x2y"
+
=
1
xy'
+
y'iO)
+ y
=
(5.14)
ex
=
1
y"i0)
Let the solution be
f(x)
=
1/6
^=0ax". Substituting
=
in
(5.14),
we
obtain
n(n
l)(n
-
2)anxn
-
n(n
+
n=0
n
=
CO
CO
OO
QO
-
l)ax"
naxn
+
anx"
=0
n=0
0
+
CO
=
y"
-
t-0!
which
ain(n
gives
these
l)(n
-
equations
2)
-
+
n(n
-
for the coefficients :
1)
+
1)
for all
=
n
or
a.
"
(5.15)
=
n!(n3-2n2-n
l)
have not used the initial conditions and fortunately
conform to the requirements (5.15). That is, for this particular
there is a unique solution independent of any initial con
Notice, that
they
+
we
equation,
ditions.
This
does
not
Picard's theorems do not
contradict
any
apply (since
the
previous results because
leading coefficient is not
invertible).
11.
y
_
yiO)
xy'
=
1
+
2y
=
(5-16)
0
y'iO)
=
0
420
5
Series
of Functions
Here Picard's theorem does
with the
apply,
initial conditions.
given
(5.16) becomes
date.
nin
l)flx"-2
-
nanxn
-
n=0
so we
should get
Let/(x)
2ax"
+
n=0
=
=
"=0
unique solution
ax" be the candi
a
0
n=0
or
a0
(n
=
+
1
fli
=
0
2)(n+l)an+2-nan
2an
+
0
=
n>0
or
2)a
2)(n+l)
(n
fl"+2
(n
-
+
Thus a2
1, a3
zero.
The solution
=
In the next
is most
=
section,
we
0, aA
is/(x)
shall
advantageous (as
we
=
=
0 and thus all further coefficients
are
x2.
1
fully develop the theory of power series. It
already seen) to do so in the complex
have
domain.
EXERCISES
6. Find
y"
-xy
with
=
an
0
y(0)
with
=
=
0
y'(0)
1
=
of at most 10"* in the interval
an error
7. Do the
y-x2y
approximate solution for
same
\
y(0)=0
an error
[ 1, 1].
for
/(0)=0
/'(0)=0
of \0~* in [i, J].
recursive formula for the coefficients of the solution, and
reasonable estimate:
8. Find
(a)
(b)
(c)
(d)
(e)
a
2/ + v
1 /(0)
0.
0, v(0)
/'
1
1
+
/(0)
y"
2/ xy e\ y(0)
1, arbitrary initial conditions.
y(k) + v
/'-/c2v=0.
0.
/ x2 + xy, y(0)
=
-
=
=
,
=
=
=
-
,
=
=
=
.
a
5.5
Power Series
421
PROBLEMS
12.
Why doesn't Picard's theorem apply to Equation (5.14)?
equation xy" + y'=0 seems to have only one
solution by the series method, but two independent solutions by the method
of separation of variables. Explain that.
13. The second-order
5.5
Power Series
We have
already discussed at length the power series expansion of the
and trigonometric functions, the geometric series and some others.
We have also seen that the Taylor formula produces a power series expansion
exponential
for suitable functions.
We have observed that there is
a
certain disk
corre
to each power
series, called the disk of convergence. The series
We shall recollect all this
converges inside that disk and diverges outside.
information as the starting point of our discussion of complex power series.
sponding
Let cn be
Theorem 5.6.
a
numbers.
of complex
sequence
negative number R (called the radius of convergence
with these properties:
nM= 0 cn z" diverges if\z\>R.
"=0 cz" converges absolutely
(a)
(b)
\z\
<
a non-
uniformly
in any disk
c
z")
{zeC:
with r<R.
r)
R has these two
(i)
(ii)
and
There is
of the power series
R
=
P
=
descriptions:
l.u.b.
{t: \c\t"
is
bounded}.
(limsup(|c|)1/")"1.
Proof. For at least part of the proof we could refer
proposition we consider the set
[t
>0
:
there is
an
If this set is unbounded,
to
"
M such that M > | c 1 1 for all
we can
take R
=
<x>,
Proposition
9.
As in that
n)
otherwise, let R be the least
upper
bound of this set.
(a)
Suppose \z\
bounded.
>R.
Then there is
Since |c| |z"| > |c| t" for all
a
t,
\z\>t>R such that {\c\t"} is
n, we cannot
have lim cz"
un
=0sojrz"
diverges.
(b)
Let r<R, A
=
(ze C:\z\ <r).
Then there is
a
t,r<t<R
such
that
422
5
Series
of Functions
{|c|/"} is bounded,
If \z\ < r,
M.
by
say
'
\cz"\
\c\t"
=
r
'
t
Thus, letting || || be the uniform norm for C(A), we have ||cz"|| < M(r/t)". Since
rjt < 1, 2 (r/t)" < , so by comparison 2 c z" converges absolutely and uniformly
in
C(A).~
Further, by definition, R is given by (i); the
shall leave
as
more
esoteric formulation
(ii)
we
Problem 14.
Examples
If
12.
is
cnz"
a
power series with radius
given
R, the question may arise: what happens
is that
answer
(a) If the
comparison
2 (z"/n2)
practically anything
can
on
of convergence
P? The
\z\
=
happen.
is summable, that is,
converges uniformly in {\z\ <
{c}
sequence
2 cz"
has radius of convergence 1 and
{|z|<l}.
(z"/n)
(b)
the circle
\c\ < oo, then by
1}. Thus the series
converges uniformly in
also has radius of convergence 1, but
[( 1)"/"] does converge.
2 (l"/w)
does
not converge, whereas
(c)
z
=
general assertion
does not converge
21 z"
=
1
!).
the circle of convergence is possible,
needn't be concerned with the behavior of the series there (except
we
in
z" has radius of convergence 1 but
with \z\
1 at all (lim z" # 0 if \z\
,
for any
Since
no
on
particular cases).
The
13.
geometric series
=0 z" is a power series with radius of
This series converges to (1
z)_1 uniformly and
Thus
any disk {z e C: \z\ < r} with r < 1.
convergence 1.
absolutely
y~
1
f0
Now let
a e
C,
a
1
1
\z\
^ 0.
=
a
<
1
=
[1
-
1
Then
1
'
_
a~^~z
for
z"
=
z
-
on
(z/a)]
a
lz\"
h \a~)
z"
=
This convergence is assured in the disk
14.
The series
=0 (z"/0
io ^
{z
6
C:
|z|
<
|a|}.
has infinite radius of convergence.
5.5
Thus the
is
Power Series
423
continuous function on the whole plane, denoted
does converge to the exponential function for real
values of z. We have seen that, for real x, eix
cos x + i sin x.
We
can use this (or
to
obtain
series
for the sine and
Taylor's theorem)
cosine:
sum
ez since this
a
sum
=
,
+
cos x
i sin x
(ix)n
>
=
=
n!
o
x
^;
i
k%(4k)l
(4k
(-l^x2"
=
~
4^o
+
(2k)!
+
1)!
(4/c
lio
-2k
(2k
We also have the
elz
cosz
=
for all
numbers
z
We
iz)
cos(j'z)
+ i
+ i
can use
them to
z2k+1
(for the series will
ez
iz)
3)!
(5-17)
(5.18)
iz and
e~z
plane.
to-(-*c2FM)i
Replacing z by
interesting equations :
cos(
+
1)!
+ i sinz
complex
(4k
equation
15.
=
+
oo
""-R-tm
2)!
(-l)fcx2t+1
These series also converge on the entire
define the complex cosine and sine :
co
+
sin(
=
sum
iz, alternately,
cos(i'z)
i
we
again that way).
obtain these other
sin(z'z)
sin(z'z)
Thus
ez + e"
=
2~
cos(iz)
(5.19)
-isin(iz)
(5.20)
ez -e~
=
For real values of z, the left-hand sides of
Equations (5.19), (5.20)
are
424
Series
5
of Functions
We
hyperbolic cosine and hyperbolic sine, respectively.
these expressions to define the complex cosh and sinh :
the
ez + e~z
cosh
z
.
ez-e~z
,
sinh
=
z
=
(5.19) and (5.20), the complex trigonometric
imaginary axis, the hyperbolic functions :
Because of
on
the
cosh
sinh
cos(zz)
=
z
z
=
We should also note that the
bolic
Since
ones.
cosh2
sinh2
z
z
cos2(/z)
=
can use
+
-i
functions are,
(5.21)
sin(z'z)
trigonometric
sin2(/z)
=
identities
imply the hyper
(5.21) that
1, it follows from
(5-22)
1
(see Exercise 10).
So does the series
has radius of convergence 1.
that the sums of
see
later
k.
We
shall
"=1nk (z"/n!), for any integer
z"
closed
all these series can be given by
expressions (such as
"=1 (z"/!)
16.
=
(1-z)-1).
given by a power series. In
"=0 anz" is the same as giving
its power series expansion. What is more interesting is that any
point in C can be chosen as the center of a power series expansion for
A
17.
polynomial
function in C is
fact, writing the polynomial p(z)
p.
=
Let z0e C and write
N
KZ)
aniZ
=
=
n
Using
Z0 +
Z0)"
the binomial theorem this becomes
P(z)=
(nW-z0)'zS-''
=
n
All
~
0
i
0
=
0
being finite, we may arrange
(5.23) as a sum of powers of z
sums
rewrite
Kz)=
n
=
0
(\ /kr"(z-z0)"
m
=
(5.23)
V /
n
which is the desired
expansion.
terms at will.
z0
:
Thus
we can
5.5
generally, any series of the form =0
series
Can we
expansion centered at z0
power
centered at a point other than the origin? The
More
.
15),
and the
is like the
proof
one
convergence intervenes after the
above for
analog
Power Series
425
c(z z0)" will be called a
expand ez in a power series
-
is yes (cf., Problem
polynomials, but the question of
of
answer
Equation (5.23)
above.
It is
a
general fact that for any function given by a power series, we may move the
center of the expansion to any other point in the disk of convergence.
The
truly courageous student should try to prove this now; it can be done. We
will give a proof later which is simple and avoids convergence problems but
requires more sophisticated information about functions defined by power
series expansions.
Addition and
Multiplication of Power
Series
Suppose /, g are complex-valued functions defined by power series
Then we can find series expansions for
pansions centered at a point z0
functions /+ g and/# also. Addition is easy: if, say
.
f(z)
=
an(z
-
giz)
z0Y
bn(z
=
-
ex
the
z0)"
then
(f+g)iz)
=
yj(a
But to find the series
+
b)(z-z0y
expansion
for the
product requires
a
little
more care.
0 (this involves no loss of generality). To say that
Suppose that z0
that in a certain disk A, /is the limit of the polynomials
is
to
z"
say
/(z) 2 a
Thus,
g is the limit of the polynomials ?=0 bnz".
Similarly,
az".
=
=
2^=o
fg is
the limit of the sequence of
we can
polynomials
easily,
polynomials
multiply
/
\n
N
\
I
N
=
=
0
If we collect terms in this
=
n
=
Now,
N
N
\
nz"
/
0
/)(lKz"
\n
("=0 z")(^=0 bz").
0
afcmz"+'"
m
expression
=
0
to form a series of powers of
z we
do not
the next few
get a very aesthetic expression, but if we take some terms from
we obtain a reasonable expression.
the
in
sequence
polynomials
g(
k=0
\n+m=k
anbm)zk
(5.24)
I
hope that fg is the limit of this sequence of polynomials. This is a
reasonable hope; for even though we have modified the original sequence
We could
426
of
5
Series
polynomials
of Functions
have neither added
we
by that sequence. In fact, by
that fg is the limit of (5.21).
3.
Proposition
nor
making
deleted from the series
careful
Let f(z)
=0 az", g(z)
of convergence of both series.
=
less than the radii
use
=
of this
fact,
=0 bz"
represented
verify
we can
and suppose
r
is
Then
(0 (/+ 9)iz) "=o ian + b)z* uniformly and absolutely in
{zeC: \z\ <r),
(ii) ifd)iz) f=o(.+m=jt anbm)zk uniformly and absolutely
=
A
=
=
in A.
Proof. Let pn(z)
'2}=0akzk, q(z) =^l=0bkzk. By hypothesis p-+f, q -^g
uniformly in A. Thus p + q ->/+ g, pq -+fg uniformly in A (Problem 2.55).
=
Since
Pn(z)
q(z)
+
2 ia- + b*)z"
=
k=l
(i) is proven.
(ii) Let
rJLz)
=
k
we
=
0\t + J
=
k
want to show that r
seem
worth
Pnq
J
-+fg uniformly
while to compute pq
our
I
rn
k
=
(
=
t>n
Now, each
computing
term on the
norms on
\\pnq,~
A
r.
We know that pqn
->fg
so
it would
But that is easy,
"'bX
2
0\l + J
in A.
k
/
right is of the form aibjz'+J with i>n
{z e C: \z\ <r},
or
j>n.
=
r\\<y+\\aiz'\\\{ Ijl^^llj
+(,i0iia'z'ii)r_ifin^^ii)
<tl+Mr'\(f\bj\A (J>k')( _|+ \bj\A
+
Thus,
5.5
Now,
we
know that
than both.
Given
2?=o \a,\r', Jj=0
>
e
0, there
\bj\ r>
are
finite.
427
Power Series
Let M be
a
number larger
are
(1) iVi>0 such that 2<=-.+ i \a,\r' < e if n > Nu
(2) N2 > 0 such that 2j=+i \bj\ ri<eifn>N2,
(3) N3 > 0 such that \\pq-fg\\ <e if n>N2.
These assertions follow from the known convergence of each
n > max(7V~i, N2
Ns),
case.
Thus, if
,
ll'-n-/^ll< llpn?n-/^ll+ II.P4,.-rJ|
the
proposition
2
|a,|r'
<
+
<
+ e-M+M-
(21^1^
=
2 \bj
+21^1'-'
(2M+ l)e
is concluded.
EXERCISES
9. Verify, in the way suggested in the text that cos2 z + sin2 z =1 is true
1 is always true.
sinh2 z
for all complex numbers z, and thus cosh2 z
10. Find a power series expansion for these functions:
=
exp(z2)
(e)
e~'
(b)
ez sin
(f)
cosh
(g)
sinhz
z
cos z
(c)
1
-dt
t
z
.
z
f
(d)
11.
J" ^Z
(a)
exp(Z2)<fr
Jo
Verify by multiplying
the power series that ez+w
12. From the addition formula for the exponential
the addition formulas for cos, sin, sinh, cosh.
=
ezew-
(Exercise 11), deduce
PROBLEMS
14. If
{c} is
neighborhood
Show
that
maximum
any bounded sequence, then the
of which has
the
radius
(limsupflcl)1'")-1.
of
infinitely
many
convergence
number, every
members, is denoted lim sup c
.
of
the
series
2CZ"
is
R
=
5
428
of Functions
Series
15.
Expand ez
Assuming
16.
in
a
point z0
represented by
power series about any
(1 +X2)"1'2
that
be
can
.
power series
a
centered at 0, find it. Find the power series for
17. Assuming that tan x can be represented by a power series centered
sin x, find the power series ex
at 0 and using the equation tan x cos x
arc cos x.
=
pansion of
5.6
tan
x.
Complex Differentiation
An easy property of the
exponential function
is
(5-25)
=1
lim
z
i-+0
For
n=o
n\
so
e'-l
_n-l
oo
7n-2\
oo
/
-V-1+'(^r)
parenthesis is a convergent
Thus, writing the parenthesis as g(z) :
Now the term in
at 0.
ez
-
1
lim
2->0
From
(5.25)
=
zg(z)
=
so
is continuous
1
z->0
and the
exp(z0
lim -^-2
z-0
1 + lim
z
series,
power
+
properties
z)
-
-
of the
exp(z0)
^-^
z
exponential
.
=
N
,
.
exp(z0) lim
z->0
ez
-
it follows that
I
=
exp(z0)
z
recognize the limit on the left as a difference
real ex
quotient and the entire equation as a replica of the behavior of the
more
to
consider
idea
a
be
It
generally such
good
might
ponential function.
turns
out to be a
This
the
on
a process of differentiation
complex plane.
The student of calculus will
5.6
very
significant idea,
because there
represent functions which
Definition 3.
ofz0inC.
,.
lim
Let/be
complex-valued function defined in a neighborhood
/(z)-/(zo)
Z
In this
Zq
write the limit
case we
set U and differentiable at every
on
many beautiful and useful ways to
are
differentiable.
are so
differentiable at z0 if
/is
z-zo
exists.
a
429
Complex Differentiation
as
point
/'(z0).
of U,
we
If/ is
defined in
an
open
say that / is differentiable
U.
The usual
facts
algebraic
on
differentiation hold true in the
complex
domain.
Proposition 4.
(i) Suppose fi
g
are
the derivatives given
differentiable
at z0
Then
.
so
are
and fg with
f+g
by
if+g)'izo)=f'izo)+9'izo)
ifg)'izo)
=
f'izoMzo) +fiz0)9'iz0)
differentiable at z0 and /(z0) # 0. Then 1// is differ
(l/f)'(z0)
-f'(z0)/fiz0)2.
(hi) Suppose f is differentiable at z0 and g is differentiable at f(z0). Then
of is differentiable at z0 andig f)'(z0) g'(f(zQ))f'(z0).
(ii) Suppose f
is
entiable at z0 and
g
=
=
These
Proof.
propositions are so much like the corresponding propositions
proofs will be left to the reader.
in
calculus that their
Examples
18. The function
z0
any
z
polynomial
operations
as
z
=
1 for all
zero.
and constant functions by
described in
Proposition 4(i),
all
Since
a suc
polynomials
differentiable.
19. The function
as z
takes
on
z
is nowhere differentiable.
For the difference
z^)/(z z0), for zj= z0, is a point on the unit circle
ranges through a neighborhood of z0 this difference quotient
all values on the unit circle, so it could hardly converge.
quotient (z
and
clearly differentiable, and z'(z0)
is obtained from
cession of
are
is
A constant function is differentiable with derivative
.
-
-
430
5
Sum
of a
Series
of Functions
Power Series is
Infinitely Differentiable
Our introduction to this section
where differentiable.
was
essentially
It is in fact true that the
differentiable in its disk of convergence.
Theorem 5.7.
f
is
Let
differentiable
f'(z)=
/(z)
at every
=
j=0 anz"
We
proof
of
a
that ez is every
power series is
verify this basic fact.
now
have radius
in the disk
point
a
sum
of
convergence R.
Then
of radius R and
Y,nanz"-1
(5.26)
n=l
has the
same
radius
lim
Proof,
of convergence
sup(|a|)1/n
"=0 az".
as
lim()1/n lim supGa,,!)1'"
=
=
lim
sup(|a|)1/n,
so
the series
radius of convergence as the given series. We must show
2
that it represents the derivative off. Fix a z0 \z0\<R and choose r > |z0|. The
series 2"la"l'"""1 converges absolutely, so given e>0 there is an N such that
nanz"'1 has the
same
,
2">n "
lank""1
Now consider the difference
<e.
/z"-z"\
f(z)-f(z0)=
z
z0
11
=
quotient defining f'(z0) :
=
0
\
z
z0
/
fai^z"-kzoA
\k=l
J
n=l
If \z\
<
r as
well
as
|z0|
< r,
2aitz"-kzk0A
\fc=
J
n>N
Similarly,
1
then
<2 Mj4r'-''rk-1<2 nMr'-'Ke
|2> nanz0~ 1\<e.
/(z)-/(z)
lt>N
k
=
l
Thus,
-
b-i
<
2e +
Zo
2
\Cn
z
z0
Now, by continuity, as z -> z0 the last term tends to zero. Thus, there is a 8 > 0
such that if | z
z0 1 < 8, the last term is less than e.
Thus, for \z\ <rand |z z0|
<
8
we
have
m-fco)
Z0
-2anzo"1
<3e
which proves that the limit of the difference
quotient exists
and is
given by (5.26).
5.6
431
Complex Differentiation
In particular, since /' is given by a convergent power series, it also is
differentiable, with derivative f"(z)
(
\)az"~2, and so forth. We
thus obtain these results, which form the complex version of Taylor's theorem
=
for
-
of convergent power series.
sums
Let
Corollary 1.
f(z) Y^azn have radius of convergence R.
infinitely differentiable in {z: \z\ <R). Furthermore, for every
derivative, fm is given by a convergent power series,
/<(z)
n(n
=
Then
=
1)
-
(n
-
k +
f
is
k the kth
l)az"-"
n>k
coefficients {an}
a
f(z) j=0 az" be convergent
uniquely determined by f:
Let
Corollary 2.
are
Notice that the definition of complex derivative is
laries hold for functions of
Corollary 3.
If f(x)
=
infinitely differentiable
an
a
disk about 0.
The
/<>(0)
=
of the differentiation of functions of
is
in
=
=
a
a
real variable
"=0
at x0
a(x
-
a
real variable.
genuine generalization
Thus the
represented by
in
x0)"
a
same
corol
power series :
neighborhood of
x0
,
then
f
and
/(n)(*o)
:
/(x)
<n
=
1)
-
in
-
k)an(x
-
x0f-k
n>k
from the theorem and their
proofs are
implication of Corollary 2 is that the
coefficients of a power series representation of a function are uniquely and
directly determined by the function. In particular, a function cannot be
These corollaries
are
written
as
the
tion allows
cos
2
us
z
sum
to
+
easily derived
Notice that the
left to the student.
of
a
power series in more than
the identity
one
way.
This observa
easily verify
sin2
z
=
1
(5-27)
sums
For the function cos2 z + sin2 z is a polynomial in functions which are
can
of power series and thus is the sum of a power series. Its coefficients
432
be
5
Series
of Functions
computed according
right-hand side
But the
for real z, thus it must
always
The
Corollary 3 just by letting z take on real values.
(5.27) is the Taylor expansion of cos2 z + sin2 z,
be the Taylor expansion for all z. Hence (5.27) is
to
of
true.
Cauchy-Riemann Equation
It is of value to compare the notion of
complex differentiation with that
R2, since R2 C. Suppose that
is
a
function
defined
in
a
/
complex-valued
neighborhood of z0 x0 + iy0
is
differentiable
as
a
function
of
two
real
variables, then the differential
If/
is
defined
and
is
a
valued
linear function on R2. Iff
dfix0,y0)
complexis also complex differentiable, then
of differentiation of functions defined
on
=
=
ru
\
f'(z0)
fiz)fizo)
r-
lim
=
Z
z->zo
exists.
Let
z
fft(z0)\
->-z0
/e /,0-.
(5.28)
Zq
along the horizontal
i;fix>yo)-fixo>yo)
hm
X
Xq
=
line.
we
let
z
along
-+z0
fft(z0)\
v
hm
=
a
vertical,
Sf
(5.28) specializes
(x0 v0)
,
OX
to
(5.29)
also have
fixo,y)-fixo,y0)
Cv J'o)
~
y-j-o
we
Then
=
x-*xo
If
.
ldf
=
T
(x0 y0)
,
Sy
(5.30)
Thus the
right-hand sides of (5.29) and (5.30) are the same. In conclusion,
complex differentiable function must satisfy (when considered as a function
of two real variables) this relation
a
(DM!)
(5.31)
This is called the
Cauchy-Riemann equation. More precisely, the Cauchyequations are found by writing /= u + iv and splitting into real
and imaginary parts. Let us record this important fact.
Riemann
Theorem 5.8.
Split f
Let
into real and
f be
a
complex differentiable function in a domain D.
a function of two real
imaginary parts and consider f as
5.6
variables
(z
=
x
Then these
iy).
+
433
Complex Differentiation
partial differential equations
hold in D:
dyji
dx
i
du
dv
(5.32)
dy
du
dv
(5.33)
=
dx
d~y
dy
dx
Proof. Equation (5.32)
(5.32) and the identities
du
.dv
d~x~~8~x
~dx
df
,
+
is))
,8v
'
^ (z0)r
=
=
Thus the differential of
f'(z0)r
f'izo)ir
a
follows from
Equation (5.33)
Hy
is
complex differentiable, its differential
=
complex-valued
8u
Bf
~d~y~~d~y
Notice that when /is
dfizo ir
observed above.
was
+
+
+
given by
-jt (z0>
if'iz0)s
is)
complex differentiable function
is
a
complex
linear
function.
We shall show, via the techniques of the next few chapters, that a complex
differentiable function can be written as the sum of a convergent power
series. Thus, just by virtue of the differential being complex linear, the
function has derivatives
orders and is the
of all
sum
of its power
series.
PROBLEMS
18. Prove
Proposition
4.
19. Prove Corollaries 1 and 2 of Theorem 5.7.
20. Show that if / is an infinitely differentiable function
( , e), and there is an M > 0 such that
|/w(x)|
then
<M
/is the
sum
all
of
21. Suppose /is
plane. Show that:
(a)
(b)
n
a
a
allx
interval
-s<x<e
the unit disk.
power series which converges in
complex differentiable function in a domain in the
if /is real-valued it is constant.
|/| is constant, then /is constant.
if
on an
434
J
Series
of Functions
22. Write the
Cauchy-Riemann equations in polar coordinates.
Differentiate along the ray and circle through a point.)
23. Suppose / ,...,/, are given by convergent power series in
If F is
a
a
(5.34)
F(Mz),...,fk(z))=0
for real z, then (5.34) is true for all z in A.
24. Compute the limits of these quotients
arc
tan
(d)
->
0:
r
sin hz
sin
as z
1
cos z
z
z
(b)
disk A.
in k variables such that
polynomial
(a)
(Hint:
cos z
(e)
TT72
z
sin
1
cos z
z
tanz
n=0, 1,2, 3, 4
(f)
(c)
complex- valued function of two real
that/is a complex differentiable if and only
if the differential df(z0) is complex linear for all z0 e D.
26. Suppose that / is twice differentiable in D, and is complex differ
entiable.
Show also that /' is complex differentiable. Supposing that
that /is a quadratic polynomial in z.
show
0,
(/')'
27. If f=u+iv is complex differentiable in D and twice differentiable,
Suppose / is
25.
variables in
a
differentiable
a
Show
domain D.
=
then
5.7
82u
d2u
d2v
82v
dx2
dy2
dx2
dy2
Differential
Equations
A function which
be
Analytic Coefficients
represented
as
the
sum
of
C will be said to be analytic at
of linear differential equations in order to answer
series at
study
posed
can
with
and to
a
point
a.
a e
in Section 5.5.
provide
following fact.
the
We
can use
sought-for
convergent power
a
We
now return to
some
of the
the
questions
the information in Section 5.6 to do this
estimates.
In
particular,
we
shall
verify
the
5.7
5.
Proposition
Suppose h, g0,
differential equation
solution of the
ym
g,yw
+
i
is also
Differential Equations
0
=
#fc_l5 gk
y(0)
=
435
Analytic Coefficients
are
analytic
a0,...,
Then the
at 0.
^""(O)
=
ak^
0
=
analytic
h(x)
+
...,
with
at
0; that is,
in
some
disk centered at 0, it is the sum of a con
can be recursively calculated from the
vergent power series whose coefficients
differential equation.
We
already know,
of the solution;
our
from Section 5.5, how to compute the Taylor coefficients
business here is to show that the resulting series does in
fact converge. This, of course, involves
required by Theorem 5.7.
Suppose then that h,g0,...,gk-i
2
=
the kind of estimate
analytic in the interval |x|
Then
< R.
oo
oo
h(x)
are
producing
g,(x)
" x"
2
=
a* x"
n=0
n=0
We
some positive number M, \a\ <MR~", |a'| <MR'n for all ;' and n.
shall obtain the desired estimate in terms of M, R and the initial conditions. For
0), leaving the general case
simplicity, we shall do the homogeneous case only (h
and for
=
for the reader
If
(Problem 27).
CO
fix)
2
=
n
=
^ X"
0
is the desired solution,
we
have
ak-i
Co
=
Oo
,
,
ck
-
1
=
.
.
.
.
(
and the rest of the coefficients
are
found from these
equations :
2(n-l)---(n-k)cx"-"
n
=
k
+
(
l(f "n'x")/ (fn(n-l)---(ni)cx-)/
\n=l
=
0\n
=
=
0
(5.35)
0
now has a pain in his stomach similar to that of the author as
this
he wrote
equation. Patience, dear readerthe fun has just begun ! Equating
of equations for the
coefficients of xm to zero, we obtain this recursive system
Surely,
the reader
436
5
Series
of Functions
coefficients :
m(m
(m
\)
k)cm
k-l m-k
2 2 aa'(m +
+
1
=
0
a
=
i
(k+a)) (m (/c + a))cm +,_,*+) =0
0
or
1
k-l m-k
(m
m
2 2 ( + '-(* + ))
k)
i=o a=o
(m
By the restraints on i and
the highest subscript of c
solve
assume
C
max
=
M>
We
on
C is written
(CM\J
for all
an
c0
,
.
.
.
,
=
m
1,
so
ck-i we can
estimate.
For this
assure
that
fory =0,..., k-l
m,
(5.37)
by induction.
Thus
we use
(5.36), assuming
k-l m-k
<tn
2 2
\'\ km+l-(k + )l
i=o <z=o
-/Ml
=
0fb^
1 ATC"-1
J}=o
try to find
1 k
(5.36)
n < m:
1
I Cm I
(k + a)<m + k
1.
Thus, given
now
to
so as
inequality for all
this
now prove
a))a'cm+i-(t+1,)
1, and let
\cj\<iJ
Now
m
+
^,P(i?-l)-1(Pt-l),^,^y/J,0<y</c-l)
The last condition
(5.37)
have
+ i
the right is m
we
on
Equations (5.36) successively.
purpose
We
a
(k
R-'
=
(R-"
-
\~r)
"^
1XP-1
m-k+\ Mm
m
(/n-k+l)
Rm
-
I)"'
=
R(Rh
R
R(R
1)
-
1
-
l)-'(Rk
-
\)R-".
Thus
(~r
by definition of C. By this estimate we see that f(x)
2k*Lo c *" converges in the
interval {x: |x| < R(CM)''}, and in that interval is the solution to our problem.
=
5.7
We
might perhaps
our
above estimates
we
could
usually
Differential Equations
with
Analytic Coefficients
437
have made
a better estimate by more clever substitutions
; but
sufficient for the results desired. In any particular case
be clever and obtain even better estimates.
were
Examples
20.
y'
exy'
+
We will find
on
tion.
n(n
=
0, /(0)
1
=
.
a
l)cx"-2
-
lt-Ml
"^x""1)/
\n
n\]
+
2
=
give
cx"+1
+
\=i
0
c0
=
0,
q
=
1.
=
-
m(m
1)
|
=
0
(5. 38)
o
Equation (5.38)
lm-2 1
-1
=
=
the interval
The initial conditions
cm
0, y(0)
=
polynomial which approximates the solution to within
[0.01, 0.01]. Let 2 cx" be the supposed solu
By substituting the series we obtain
10~5
n
+ xy
becomes
\
-(m-l-i>m-i-i
+ cm_3
I
Thus
c2
=
c3
=
c4
=
Cs
=
-\cx
=
-\
c0) 0
\Qc2
-J1ji3c3 + 2c2 + Iq + cx) 5L
-2Xo(4c4 + 3c3 + c2 + iq + C2)
+ ct +
=
=
=
TJo
so forth.
The question is not really what the coefficients are
(that is to be left to a machine) but how many coefficients need to
be computed. The coefficients appear to be bounded (we could in
fact show that they must be, cf. Problem 29). Let's try to prove that
|cj < K for all n by induction. We have
and
m_2
1
|cJ
^
^73T) fS 7T('"
m
so
1
long
-
X
~
0|Cm-1-il
+
|Cm-31
my
as
m >
4.
four terms, that is, k
Thus
=
we
1/2.
may take for K
a
bound for the first
Then the difference between the solution
438
of Functions
Series
5
and the kth
partial
|c||x|"<^|xr
for
|x|
bound
on
the
10"
Taylor expansion
is dominated
by
i|x|kl
1
L
The interval
1.
<
i.i_.l
2
=
Z >k
n>k
of its
sum
|X|
we are
concerned with is |x|
<
10_1
so our
is
error
iio
=
18
9
This is less than 10"
5
if k
6, thus (computing also c6) the solution
=
differs from
lv2
2X
v
X
by
-L
"T
1 V4
34X
i_ V5
JT
1
y6
60A
_
JJ0A
3
10-5 for all values of x in [-0.01, 0.01].
at most
good an estimate in the interval
just knowing that the coefficients are
easy to see
bounded is not good enough. We have to know that \c\ < Kr" for
Let's try r
some r < 1 and some K.
\. That is, we attempt to
<
n.
2~"
for
all
that
Now, using the equation
\c\
verify by induction
defining {c},
21.
Suppose
[1, 1]. It is
needed that
we
that
=
1
1
\<
S
|Cml
m(m
1)
-
K
1
e2
+ 4
(m
(my
I ik
1
~
K
K
\
2m-i-i^2n,-3J
~
1
i\
/m-22;
\
K
<
7
z
m
This is less than 2~mK
as soon asm>
the induction step proceeds
K so that the inequality holds for all
2(e2
as soon as m >
'
\cn\ < 1/2"" for
[ 1, 1] from its
all
^
n.
kth-order
lc*l
^
26 ;
26
4),
we
(K
=
26.
Thus
only
do).
choose
orm>
need
2 will
Thus
The desired solution differs in the interval
Taylor polynomials by
1
c(D"
n>k
m <
+
~fc-i
1
y
-
on
ok-2
2"-1t'o2"^2''
5.7
5
This is less than 10
of the
with
Differential Equations
if k is 19.
Thus
we
need to compute 19 terms
to find the desired
Taylor expansion
439
Analytic Coefficients
approximation.
22. Compute the solution of
(l^y'
/
+
to
an
exy
+
Let
f(x)
y(0)
1
=
y'(0)
=
0
3
in the interval [ i, ].
cnx" be the solution. We have these
accuracy of 10"
=
0
=
=0
equations
for the solution :
1,
c0=
Ci
=
0
n-2
-1
c
(n-2
(n- l-i)c-i-i+
i=o
=
n(n
1) \i=
-
We will show
Ic.l
by inductions
K for all
<
X(-i-0K+
.
1) \i=o
m(m
/""- x
K
<
m(m
-1)1
1
j
=
m(m
/
1
>=o
X
\
.
Suppose that
bounded.
-
\
t,k)
'!
/
|"m(m
-
1)
+
e
1)
e
<K
K
2
1).
m(m
as soon as w >
terms.
are
m-2 1
lm-2
i
<
{c}
(5.39)
Then
n <m.
1
|cm
that the
\
J
-.cn-2-i)/
=oi!
3.
Thus
We have from
we can
(5.39)
take A"
c2=
as a
bound for the first four
-\,c3= 0,
so we
may take K=\.
Then the estimate of the remainder after k terms in the interval
[-i, H
is
klW<^ ^
=
n>k^
>k
This is less than 10"
Z.
3
when k
=
10,
so we
compute
c4
=
h
c$ =20
c6
T20
etc.
need 11 coefficients.
We
440
Series
5
Up
of Functions
to six terms
(giving
at most an error of
1/64),
our
solution is
x^ + x^ + x^ 13x6
L2
20+720+'"
~T
EXERCISES
13. Find
power series
a
expansion
equation,
hood of 0 for this
(1
x2) /'
-
2x/
-
14. Find
yh
-
a
+
k(k
+
power series
Ixy' + 2k v
=
1) v
=
for the
general solution in
neighbor
0
expansion
for the
general solution of
0
15. Find the power series for the function y
l,
v(0)
y + e-y x2,
v,
y(0)=0
(/)2
y(0)=0,
(y')2=yy",
/ + 2x^2=0
(a)
(b)
(c)
(d)
a
=
=
=/(x) such
that
v'(0)=0
=
y'(0)
=
1
16. How many terms of the power series for the solution y do we need:
3
(a) for an accuracy of 10" in the interval ( i, ) in Exercise 3(a)?
5
in the interval ( 10, 10) in Exercise 3(a) ?
of
10"
for
an
accuracy
(b)
(c)
3(b)?
for
accuracy of 10"a in the interval
an
17. For what k
are
the solution of
(0.1, 0.1) in Exercise
Equations (5.1), (5.2) polynomials?
PROBLEMS
Generalizing the argument
28.
Theorem.
origin, then
/k>
+
be
=
=
are analytic in an interval (R,R) about the
of the differential equation
...,gk-i
any solution
k2Zglyi"
r
can
Ifh,g0,
in the text prove this theorem:
h
0
expressed
as
the
sum
of a convergent power
series in
neighborhood
a
of the
origin.
29. Suppose that
with R>\.
y" + gy' + hy
g, h are
convergent power series in
some
disk (|z| < R)
Show that the solution of the linear differential
=
0
equation
(5.40)
5.8
441
Flat Functions
Infinitely
sum of a convergent power series with bounded coefficients.
30. If the power series g(x)
~2,ax", h(x)=^bx" both have infinite
radius of convergence, then so does the series expansion of the solution
is the
=
of (5.40).
5.8
Infinitely
Flat Functions
Not all functions
which
susceptible to the kind
doing. A first requirement
are
have been
we
Taylor series analysis
of
is that the function have
derivatives of all order; even that however is insufficient. Another glance at
Theorem 5.6 will remind the reader that there is a behavior requirement on
these successive derivatives in order that the given function be the sum of its
We shall show
Taylor expansion.
by example
that there
differentiable functions which are not sums of power series.
make the notion of analyticity precise.
Let /be
Definition 4.
Let
a e
U.
/is analytic
such that /is the
in U if /is
complex- valued
at a if there is
a
function defined in
ball
{z: \z
-
a\
<
an
r}
convergent power series in this ball.
at every point of U.
sum
analytic
a
are
First,
of
a
infinitely
we
shall
open set U.
centered at
a
/ is analytic
We have deliberately stated this definition without reference to the domain
of definition of the function; it applies equally well to functions of a real or
are
complex variable. The only functions which we know to be analytic
For example, if/ is the sum of a convergent power
the polynomials and ez.
know that we can expand / in a series of
we do not
series at the
yet
origin,
a) with a any other point in the disk of convergence. We
powers of (z
shall see in the next chapter that this is the case. We have already seen that
and now we will
an analytic function has derivatives of all orders (is C)
The clue to this function is
a C function which is not analytic.
-
produce
given by
the
following fact,
Proposition
6.
lim
which follows from
P(t)e~
'
=
l'Hospital's
rule.
0, for any polynomial P
t.-0O
Proof.
Problem 31.
The function
we
have in mind
-
!exp(
V
0
-
1
(Figure 5.1)
x >
is defined
by
0
*/
(5.41)
x^O
442
5
Series
of Functions
Figure 5.1
a
is
differentiable at any
certainly infinitely
consider its behavior at 0.
(7(n)(x)
=
0
derivatives from the
a
0
all
0,
so we
need
only
n
from the left exist at 0.
exist and
right
x0 #
Now,
x <
Thus all derivatives of
point
are zero.
We have to show that all
More
precisely
we
must prove
that for all n,
<7<">(x)
hm
ffM(Q)
-
=
0
(5.42)
x->0
x>0
by induction.
We do this
lim
=
To do the
x >
0.
lim
-
x->0 X
x>0
X
X-+0
x>0
general
The
expl
\
case we
case n
)
X/
=
=
0 is easy :
lim te~'
=
0
(->oo
must have some idea what
<7(n)(x)
looks like for
Now
? (x)
A pattern
-
-
seems
i
exp( i)
to be
-
(J, i) exp( i)
^-yy)yi)
developing.
'W
-
+
-
5.8
For each
n
ff(n)(x)
This
can
there is
=
a
polynomial P
P (- exp (
j
be verified
-
^j
Functions
443
such that
for
by induction.
Infinitely Flat
x >
0
(5.43)
Assuming (5.43),
we
compute
'""<yyyy(yM-i)
'p-yi)
where
Pn+1(X)
-Z2(P'(Z) + Pn(X)).
follows immediately from Proposition 6 :
=
,.
lim
<T(n)(x)
(T(n)(0)
-
X
x->0
x>0
1
,.
=
lim
-
x-o X
x>0
P
Now that
/1\
-
1\
/
exp
\X/
-
\
-
X]
we
,.
=
lim
have this,
(5.42)
,
/N
rP(r)*T'
=
0
f-oo
Thus
a is also infinitely differentiable at 0.
But it is certainly not analytic.
Taylor expansion is "=0 0 x" which converges to o(x) only for x <, 0
and provides a poor means for approximating the value of <r(x) for x > 0.
However, the fact that infinitely differentiable functions exist with this
property has its bright side. The following construction will prove to be
Its
'
useful.
Lemma.
0
(i)
(ii)
(iii)
Given
xab(x)
xah(x) > 0
<
t^x)
Proof.
=
a
<
<b, there
1
0
(See Figure 5.2.)
is
a
C00
for all x,
if a<x<b,
if x > b or
ct(x(1
x))
function
xab such that
x < a.
has the
required properties
then define
Toli(x)
=
T0
(H) "((K)('-r3)
of
t0i.
We
444
5
Series
of Functions
Figure 5.2
Theorem 5.9.
tion
(i)
(ii)
(iii)
0
[a, b~] be a given interval and Uu
Un a finite collec
There
exist
C
[a, b~\.
functions pt such that
.
..,
for all xeR, all i,
ifx$Ut,
for all x e [a, U].
pix) < l
Pi(x) 0
l,
pt(x)
<
=
=
Proof.
(a, b).
x e
Let
of open intervals covering
Let
[/,
=
(at bi), and take
,
rt
=
Tib{
.
Then
t(x)
=
2 Tt(x) > 0
if
Let
(ti(x)
Pix)
x 6
<*)
{
=
0
x
(a, bi)
,
^ (a, bi)
The pi then have the desired
,
properties.
PROBLEMS
31. Prove that for any
polynomial P, lim P(t)e~'
=
0.
1-.QO
32. Let
a
be denned
by (5.31).
Define
\\t(\-t))dt
<d(x)
\\t(\-t))dt
Jo
Show that
(a) w is C, (b) 0 ^ o(x) < 1, for all x, (c) to(x) 0, if x < 0,
l,ifx>l.
33. Using Theorem 5.9 it can be shown that any continuous function is
the limit of C" functions.
Let/e C([0, 1]) and e > 0. Find a C function
e.
such
that
<
#
||/ g\\
Here's how to do it. First pick an integer N>0 such that |/(x) f(y)\
< e/2 if \x
y\<ljN. Now cover the interval [0,1] by the intervals
(d)a>(x)
=
=
5.9
U0,
...,
445
Summary
UN, where
/i-l /+1\
u'-(n- n)
and let px,...,pN be the
corresponding functions of Theorem 5.9.
Let
^-l/^H
For any x,
is in
x
only
l/(*)-*(*)l=M*)
5.9
two of the intervals
**>-'()
+ Pr+1
{U,},
say
Then
UT, U,+1.
e
/-,(^)
<2
e
+
2
Summary
Let
{/J
be
a
sequence of continuous functions.
The series formed from
the/k is the sequence of sums Q=1/J. If this sequence converges, we
The series
say that the series converges and denote the limit by = t fk
.
converges
absolutely
if
"=1||/t||
criterion for series
The
Cauchy
only if the sums ||=ll+1 /til
n sufficiently large.
< oo.
asserts that the series converges if and
made
arbitrarily
small
comparison test.
integer
(0
()
then
by choosing
If there is
a
m,
sequence
{pk}
of
positive
can
be
numbers and
an
N > 0 such that
fork>iV
IIAII<P*
YjPk
<
/k converges absolutely.
integration.
If
/
converges,
so
does
Jj/
and
j;(z/.)-i({
fundamental theorem of algebra.
P(z)
=
anz',
+
---
+ a1z + a0
If P is
a
polynomial :
(5.44)
446
5
Series
of Functions
with a # 0. Then P(z) has a complex root. If r1;
rk are all the roots
of P, then corresponding to each root there is a positive integer m( (called
.
the
mt +
Piz)
+ mk
(z
=
iz
rk)m*
-
(5.45)
polynomial P given by (5.44)
differential operator LP:
=
afM
+
---
P is called the characteristic
Lp+Q
If
=
LP
+
Lq
a1f'
+
+
=
associate the constant
of LP
These formulas
.
xmi~1erix
r2x
xm2-ler2X
-ritx
mji-lgi-kjc
{cn}
be
a
(called
these properties :
sequence of
cz" diverges for \z\
(c)
P=[limsup(|C|)1/"]-1.
If/(z)
f(z)
converges
anz", g(z)
=
+
g(z)=
n
fiz)giz)=
k=0
in that
same
disk.
valid :
LpLq
complex numbers.
of the
the radius of convergence
(a)
(b)
cnz"
are
of
LP the collection of
,
=
erix
number R
coefficient
a0f
polynomial
LPq
we
(5.45) is the factorization of P, then the kernel
0, is spanned by the functions
solutions of LP(f)
Let
,
n
=
r^"'
-
To the
LP(f)
.
such that
multiplicity)
(i)
(ii)
.
>
YJbnzn
in
(
0
\n+m=k
{|z|
<
in the disk
t(an + bn)z"
=
a
nonnegative
power series
c z") with
R.
absolutely
=
There is
abm)zk
1
/}
for
{\z\
<
r <
r},
R.
then
5.9
If/ is a complex- valued
complex differentiable at z0
fiz)-fizo)
,.
hm
z
exists.
The
-
function defined
C,
we
say that
/
is
V
,,,
=/(z0)
z0
of
sum
a
convergent power series is complex differentiable
GO
Furthermore, the derivative is the
at
sum
00
/(z)=az"
n=0
of
sum
z0 in
if
every z0 in its disk of convergence.
of the derived series :
Thus the
near
447
Summary
/'(z)=
az"-1
n=l
convergent power series is infinitely differentiable.
a
A
differentiable
complex-valued function of two real variables is complex
differentiable if and only if it satisfies the Cauchy-Riemann equations :
f--4f)
\dy!
dx
If h, g0
gk can be represented as sums of convergent power series
a disk centered at zero, then the same is true for all solutions of the differ
,
in
ential
.
.
.
,
equation
ym
+
kZgiyw
+ h
=
o
i=0
Furthermore,
y(0)
=
once
given
the initial conditions
ao,...,/k-i\0)
=
ak-1
the coefficients of the power series can be recursively calculated using the
differential equation.
Given any finite covering of the interval [a, b~] by open intervals Uu
...,
U we can find C00 functions pt,
(a)
(b)
(c)
p such that
0<p,<l
0, outside Ut
pt
=
Piix)
Uu
...,
1 for all
=
These functions
cover
...,
are
U.
x e
called
a
[a, U]
partition of unity
on
[a, 6] subordinate
to
the
448
5
Series
of Functions
FURTHER READING
I. I.
Hirshman, Infinite Series, Holt, Rinehart and Winston, New York,
1962.
H.
Cartan, Elementary Theory of Analytic Functions of One or
Complex Variables, Addison-Wesley, Reading, Mass., 1963.
This text develops the subject of complex analysis from the point
Several
of view
of power series. It also contains a complete discussion of the theorem
existence of solutions of analytic differential equations.
Further material
can
T. A. Bak and J.
Lichtenberg,
Inc.,
New
on
be found in
Mathematics for Scientists, W. A.
Benjamin,
York, 1966.
Kreider, Kuller, Ostberg, Perkins, An Introduction to Linear Analysis,
Addison-Wesley, Reading, Mass., 1966.
W. Fulks, Advanced Calculus, John Wiley and Sons, New York, 1961.
M. R. Spiegel, Applied Differential Equations, Prentice-Hall, Englewood
Cliffs, N. J., 1958.
MISCELLANEOUS PROBLEMS
34. Find
defined
x e
sequence {/} of continuous nonnegative real-valued functions
the interval (0,1) such that f(x) =2=i /(*) exists for all
a
on
[0, 1], but /is not continuous.
|z| < 1, define
35. For
7k
oo
lnz=2r
k
k=l
Show that for all such z, z
36. Show that the series
=
1
exp(ln(l
z)).
i(z-n)2
converges to
C-{1,2,
a
complex differentiable function in the domain
...,,
...}.
37. Show that
exp(z)
=
lim
(1 + z/m)m.
(Hint:
Compute the
power
m-^oo
series expansion of (1 + zjm)m.)
38. Show that a real polynomial of odd degree always has
39. Let oj
exp(277f'//j). Show that
=
1
-
z"
=
(1
-
ojz)(\
-
a>2z) (!- wnz)
a
real root.
5.9
Summary
449
40. Let P be
a polynomial.
Show that P is the square of another poly
only if every root of P occurs with even multiplicity.
41 If P, Q are two polynomials, P divides Q if and only if
every root of
P is a root of Q with no larger multiplicity.
42. Suppose P is a polynomial of degree at least two. Show that there
is a c such that P(z)
c
0 has at least one multiple root. (Hint: Con
sider P' as defined in Problem 10. If P'(d)
0, take c
P(a).)
43. Let /i,
/ be functions in C(X). Show that these functions are
independent if and only if there are points x,,...,x such that the matrix
(f(xj)) is nonsingular.
44. Show that the functions e'x, xe",
x"e'x are independent.
45. Show that if P, Q are polynomials, S(LP) c S(LQ) if and only if P
divides Q.
46. If P is a polynomial of degree at least two, there is a c e C such that
the equation LPf= c/has a solution of the form xe'x.
47. If a linear differential equation has polynomial coefficients, it has
global solutions on all of R.
48. Let {c} be a sequence of complex numbers such that 2 knl < o.
Let f(z) =2=o cz". Prove that |c0| is not a relative maximum of |/|,
nomial if and
.
-
=
=
.
.
.
=
,
.
..,
unless all other coefficients vanish.
49. Suppose the function
f(x, y)
=
x2
-
y2 + iv(x, y)
is complex differentiable. Find v.
50. If / is a polynomial in x, y which is complex differentiable, then
/(x, v) has the form Q(x + iy), where Q is a polynomial. (Hint: Substi
tute
(z + z)/2, y
(z
z)/2, and use the Cauchy-Riemann equation.)
Suppose /is a C2 complex-valued function defined on a domain D
Show that if / and fz are both harmonic, then / is complex differ
x
=
51.
in C.
=
entiable.
52.
/,
g
If/
are
g
are
complex differentiable and |/|2 + \g\2 is constant, then both
constant.
Suppose that / is
53.
a
one-to-one
mapping of a domain D <= C onto
/. Show that if / is complex
Let g : A -*- Z> be the inverse of
differentiable, so is g.
A
<=
C.
54. We may consider the function ez
x + iy, u
Re ez, v
plane. Let z
the
u
=
e"
cos
v
y
(a) Show
origin,
mapping
mapping from the plane
maps the lines
the lines y
Show that in any interval
every value
a
to
\mez; that is
e* sin y
that this
centered at the
origin.
(b)
=
as
=
=
=
precisely
once.
=
x
=
const,
on
const, go onto the rays
{a
<y < a
+ 2-n-} this
the circles
through the
mapping
takes
450
5
Series
of Functions
(c)
particular,
In
ez maps the horizontal
strip { tt <y < it) one-toplane
negative real axis. Call this
domain D.
Define the complex logarithm log z: Z) - { tt <y <tt} as
the inverse of this mapping.
Show that log z is complex differentiable
onto the entire
one
except for the
and
(log)'
z
=
-
z
Show also that log
be
z can
represented by
a
power series centered at 1
in the disk {|z 1| < 1}. (Recall Miscellaneous Problem 35.)
that this provides a way for extending real functions to the
Notice
complex
domain besides that of power series. For example, the power series
about 1 of log x extends it only to the unit disk centered at 1
expansion
.
The above extension of
negative real axis.
55. Consider z2
maps the open
Let V
a
z
log is defined
in the entire
plane except the
This process is called analytic continuation.
mapping of the plane into the plane. Show that it
as a
right half plane
one-to-one onto the domain D of Problem 54.
be the inverse, and show that Vz is complex differentiable.
similar discussion for the mapping z".
56. Discuss the mapping properties of
cos
z,
sin
Provide
z.
57. Show that the power series expansion of the solution of Exercise 8(d)
with initial values y(0)
0 does not converge outside the unit disk.
1, y'(0)
=
58.
I.
Suppose that/
g
=
are
complex-valued functions defined
on
the interval
Show that
/(r)
f
j
dt
hz-g(t)
is
complex differentiable and can be represented by a power series at any
point of the image of g.
59. If /is C'inCx Xand for each fixed x,f(z, x) is differentiable in z,
then
F(t)=jf(z,x)dx
is also complex differentiable.
60. The equation of Exercise 6 is called Legendre's equation and the
solutions {/*} for integral k are called the Legendre polynomials. They have
this
interesting property:
fm(x)f(x)
dx
=
0
if
m
#
n
5.9
To prove this
((l-x2)/)'
we
must
observe that Legendre's equation may be written
k(k+l)v
+
451
Summary
=
as
0
Thus
\\
f-f-
/.
=
^TT)
[(1
~
by integration by parts. Now do the
61. Let P be a polynomial of degree
d.
differentiable.
is
that/(n)(z)e"',<z)
Show
62. Show that the
dx
*'>/<*>]'/<*>
i
same,
interchanging
Show that /(z)
a
polynomial
of
m
and
n.
ep(2' is complex
=
degree n(d 1).
polynomial
d"
exp(x2)
(exp(-x2))
solves the differential
(a) Find
properties :
63.
a
equation y"
2/ + 2ny
C real-valued function
0.
/ defined
R" with these
on
(i) 0</(x)<l
(ii) f(x)>0if \\x-x0\\<R
(iii) /(x)=0if ||x-Xo!l>P
(By C" we
tinuous.)
(b) Let
mean
A' be
R" such that X
Bi,
.
.
.
,
B is
a
all
a
closed set in R", and suppose Bt,
(iii)
=
0 if
2;=i/<
Find such
a
.
.
.
,
B
are
are con
balls in
A partition of unity on X subordinate
u B
Bi u
collection {/,...,/} of C" functions such that
<=
.
(i)0</<l
(ii) /,(x)
exist and
higher-order partial derivatives
x
=
$ Bi
lif*e*
partition of unity.
to