Exercise
6.4.102
:
Solve (find the impulse response) x
0
+
ax
=
δ
(
t
)
, x
(0)
=
0
, x
0
(0)
=
0
.
Exercise
6.4.103
:
Suppose that
Lx
=
δ
(
t
)
,
x
(0)
=
0
,
x
0
(0)
=
0
, has the solution
x
(
t
)
=
cos
(
t
)
for
t
>
0
. Find (in closed form) the solution to Lx
=
sin(
t
)
, x
(0)
=
0
, x
0
(0)
=
0
for t
>
0
.
Exercise
6.4.104
:
Compute
L

1
n
s
2
s
2
+
1
o
.
Exercise
6.4.105
:
Compute
L

1
n
3
s
2
e

s
+
2
s
2
o
.
294
CHAPTER 6. THE LAPLACE TRANSFORM
Chapter 7
Power series methods
7.1
Power series
Note: 1 or 1.5 lecture, §8.1 in [EP], §5.1 in [BD]
Many functions can be written in terms of a power series
∞
X
k
=
0
a
k
(
x

x
0
)
k
.
If we assume that a solution of a di
ff
erential equation is written as a power series, then perhaps we
can use a method reminiscent of undetermined coe
ffi
cients. That is, we will try to solve for the
numbers
a
k
. Before we can carry out this process, let us review some results and concepts about
power series.
7.1.1
Definition
As we said, a
power series
is an expression such as
∞
X
k
=
0
a
k
(
x

x
0
)
k
=
a
0
+
a
1
(
x

x
0
)
+
a
2
(
x

x
0
)
2
+
a
3
(
x

x
0
)
3
+
· · ·
,
(7.1)
where
a
0
,
a
1
,
a
2
, . . . ,
a
k
, . . .
and
x
0
are constants. Let
S
n
(
x
)
=
n
X
k
=
0
a
k
(
x

x
0
)
k
=
a
0
+
a
1
(
x

x
0
)
+
a
2
(
x

x
0
)
2
+
a
3
(
x

x
0
)
3
+
· · ·
+
a
n
(
x

x
0
)
n
,
denote the socalled
partial sum
. If for some
x
, the limit
lim
n
→∞
S
n
(
x
)
=
lim
n
→∞
n
X
k
=
0
a
k
(
x

x
0
)
k
295
296
CHAPTER 7. POWER SERIES METHODS
exists, then we say that the series
(7.1)
converges
at
x
. Note that for
x
=
x
0
, the series always
converges to
a
0
. When
(7.1)
converges at any other point
x
,
x
0
, we say that
(7.1)
is a
convergent
power series
. In this case we write
∞
X
k
=
0
a
k
(
x

x
0
)
k
=
lim
n
→∞
n
X
k
=
0
a
k
(
x

x
0
)
k
.
If the series does not converge for any point
x
,
x
0
, we say that the series is
divergent
.
Example 7.1.1:
The series
∞
X
k
=
0
1
k
!
x
k
=
1
+
x
+
x
2
2
+
x
3
6
+
· · ·
is convergent for any
x
. Recall that
k
!
=
1
·
2
·
3
· · ·
k
is the factorial. By convention we define 0!
=
1.
In fact, you may recall that this series converges to
e
x
.
We say that (7.1)
converges absolutely
at
x
whenever the limit
lim
n
→∞
n
X
k
=
0

a
k
 
x

x
0

k
exists. That is, the series
∑
∞
k
=
0

a
k
 
x

x
0

k
is convergent. If
(7.1)
converges absolutely at
x
, then it
converges at
x
. However, the opposite implication is not true.
Example 7.1.2:
The series
∞
X
k
=
1
1
k
x
k
converges absolutely for all
x
in the interval (

1
,
1). It converges at
x
=

1, as
∑
∞
k
=
1
(

1)
k
k
converges
(conditionally) by the alternating series test. But the power series does not converge absolutely at
x
=

1, because
∑
∞
k
=
1
1
k
does not converge. The series diverges at
x
=
1.
7.1.2
Radius of convergence
If a power series converges absolutely at some
x
1
, then for all
x
such that

x

x
0
 ≤ 
x
1

x
0

(that is,
x
is closer than
x
1
to
x
0
) we have
a
k
(
x

x
0
)
k
≤
a
k
(
x
1

x
0
)
k
for all
k
. As the numbers
a
k
(
x
1

x
0
)
k
sum to some finite limit, summing smaller positive numbers
a
k
(
x

x
0
)
k
must also have a finite
limit. Therefore, the series must converge absolutely at
x
. We have the following result.
Theorem 7.1.1.
For a power series
(7.1)
, there exists a number
ρ
(we allow
ρ
=
∞
) called the
radius of convergence
such that the series converges absolutely on the interval
(
x
0

ρ,
x
0
+
ρ
)
and
diverges for x
<
x
0

ρ
and x
>
x
0
+
ρ
. We write
ρ
=
∞
if the series converges for all x.
7.1. POWER SERIES
297
x
0
x
0
+
ρ
x
0

ρ
diverges