7. R
EPRESENTATIONS OF
F
UNCTIONS BY
P
OWER
S
ERIES
7.1.
Introduction.
We start by recalling Theorem 4.4. Replacing
r
by
x
, we can restate (4.4) in
the form
1
1

x
=
∞
summationdisplay
n
=
0
x
n
(

x
<
1),
(7.1)
that is, the geometric series
represents the function f
(
x
)
=
(1

x
)

1
when

x
<
1. In this lecture,
we will learn how to derive new power series representations from old ones.
7.2.
Substitution.
We start with a very simple idea: if
f
(
x
)
=
∞
summationdisplay
n
=
0
c
n
(
x

a
)
n
(

x

a
<
R
),
then
f
[
g
(
x
)]
=
∞
summationdisplay
n
=
0
c
n
(
g
(
x
)

a
)
n
(

g
(
x
)

a
<
R
).
If
g
(
x
) is chosen wisely, this simple idea can be used to a great advantage.
Examples.
(A) Substituting
(

x
2
)
mapstochar→
x
in (7.1), we find that
1
1
+
x
2
=
1
1

(

x
2
)
=
∞
summationdisplay
n
=
0
(

x
2
)
n
=
∞
summationdisplay
n
=
0
(

1)
n
x
2
n
,
provided that
vextendsingle
vextendsingle

x
2
vextendsingle
vextendsingle
<
1
⇔

x

2
<
1
⇔

x
<
1.
That is, we obtain the power series representation
1
1
+
x
2
=
∞
summationdisplay
n
=
0
(

1)
n
x
2
n
(

x
<
1).
(B) By (7.1) with (

x
/3)
mapstochar→
x
,
1
3
+
x
=
1/3
1

(

x
/3)
=
1
3
∞
summationdisplay
n
=
0
parenleftBig

x
3
parenrightBig
n
=
∞
summationdisplay
n
=
0
(

1)
n
x
n
3
n
+
1
,
provided that

x
/3
<
1
⇔

x
<
3.
(C) We now want to represent the function
f
(
x
)
=
x
+
1
x
2
+
5
x
+
6
37
as a power series in (
x

1). We know that
f
has a partial fraction decomposition of the
form
x
+
1
x
2
+
5
x
+
6
=
A
x
+
2
+
B
x
+
3
(7.2)
for some constants
A
and
B
. Replacing
x
by (
x

1) in the previous example, we obtain
1
x
+
2
=
1
3
+
(
x

1)
=
∞
summationdisplay
n
=
0
(

1)
n
(
x

1)
n
3
n
+
1
(

x

1
<
3),
and similarly
1
x
+
3
=
1
4
+
(
x

1)
=
∞
summationdisplay
n
=
0
(

1)
n
(
x

1)
n
4
n
+
1
(

x

1
<
4).
Combining the last three equations, we deduce that
f
(
x
)
=
A
∞
summationdisplay
n
=
0
(

1)
n
(
x

1)
n
3
n
+
1
+
B
∞
summationdisplay
n
=
0
(

1)
n
(
x

1)
n
4
n
+
1
=
∞
summationdisplay
n
=
0
bracketleftbigg
A
(

1)
n
3
n
+
1
+
B
(

1)
n
4
n
+
1
bracketrightbigg
(
x

1)
n
(

x

1
<
3).
(Note that
x
must satisfy the more restrictive of the conditions

x

1
<
3 and

x

1
<
4:
namely, the former.) Thus, it remains to find the coefficients
A
and
B
in (7.2), or equiva
lently, in the identity
x
+
1
=
A
(
x
+
3)
+
B
(
x
+
2).
Plugging in
x
=
2 and
x
=
3, we obtain the equations

1
=
A
and

2
=
B
,
respectively. Therefore,
A
=
1,
B
=
2, and
f
(
x
)
=
∞
summationdisplay
n
=
0
bracketleftbigg
(

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

1)
n
4
n
+
1
bracketrightbigg
(
x

1)
n
(

x

1
<
3).
7.3.
Termbyterm differentiation and integration.
In this section, our main tool will be the
following theorem.
Theorem.
Suppose that
f
(
x
)
=
∞
summationdisplay
n
=
0
c
n
(
x

a
)
n
(

x

a
<
R
).
(7.3)
Then f is differentiable (and therefore continuous) in the interval
(
a

R
,
a
+
R
)
and satisfies
(i)
d
dx
[
f
(
x
)]
=
∞
summationdisplay
n
=
0
c
n
d
dx
[(
x

a
)
n
]
=
∞
summationdisplay
n
=
1
nc
n
(
x

a
)
n

1
(

x

a
<
R
)
;
(ii)
integraldisplay
x
a
f
(
t
)
dt
=
∞
summationdisplay
n
=
0
c
n
integraldisplay
x
a
(
t

a
)
n
dt
=
∞
summationdisplay
n
=
0
c
n
n
+
1
(
x

a
)
n
+
1
(

x

a
<
R
)
.
38
In particular, the series appearing on the right sides of (i) and (ii) have radii of convergence R.
Remarks.
1.
Conceptually, the theorem says that we can differentiate and integrate a power series
termbyterm: this is the meaning of the first equalities in (i) and (ii). The second equali
ties in (i) and (ii) simply record the almost trivial observations that
d
dx
[(
x

a
)
n
]
=
n
(
x

a
)
n

1
and
integraldisplay
x
a
(
t

a
)
n
dt
=
(
x

a
)
n
+
1
n
+
1
.