Review Power Series
Example
3
Example:
Since
∞
X
n
=0
[(
n
+ 2)(
n
+ 1)
a
n
+2

a
n
]
t
n
= 0
,
it follows that
(
n
+ 2)(
n
+ 1)
a
n
+2

a
n
= 0
The first two coefficients,
a
0
and
a
1
are
arbitrary
, then all other
coefficients are specified by the
recursive relation
:
a
n
+2
=
a
n
(
n
+ 2)(
n
+ 1)
Thus, with
a
0
arbitrary
a
2
=
a
0
2!
,
a
4
=
a
2
4
·
3
=
a
0
4!
,
...,
a
2
n
=
a
0
(2
n
)!
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (7/24)
Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Example
4
Example:
Similarly, with
a
1
arbitrary
a
3
=
a
1
3
·
2
,
a
5
=
a
3
5
·
4
=
a
1
5!
,
...,
a
2
n
+1
=
a
1
(2
n
+ 1)!
It follows that we have two
linearly independent
solutions
y
1
(
t
) =
∞
X
n
=0
t
2
n
(2
n
)!
and
y
2
(
t
) =
∞
X
n
=0
t
2
n
+1
(2
n
+ 1)!
,
with the
general solution
y
(
t
) =
a
0
y
1
(
t
) +
a
1
y
2
(
t
)
Note:
y
1
(
t
) = cosh(
t
)
and
y
2
(
t
) = sinh(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (8/24)
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Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Review Power Series
Review Power Series:
Consider the power series:
∞
X
n
=0
a
n
(
x

x
0
)
n
The series converges at
x
if
lim
k
→∞
k
X
n
=0
a
n
(
x

x
0
)
n
exists for
x
. It clearly converges at
x
0
, but may or may not for
other values of
x
The series
converges absolutely
if the following converges:
∞
X
n
=0

a
n
(
x

x
0
)
n

Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (9/24)
Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Ratio Test
Ratio Test:
For the power series:
∞
X
n
=0
a
n
(
x

x
0
)
n
The
ratio test
provides a means of showing
absolute
convergence
. If
a
n
6
= 0,
x
fixed, and
lim
n
→∞
a
n
+1
(
x

x
0
)
n
+1
a
n
(
x

x
0
)
n
=

x

x
0

lim
n
→∞
a
n
+1
a
n
=

x

x
0

L,
then the power series
converges absolutely
at
x
, if

x

x
0

L <
1.
If

x

x
0

L >
1, then the series
diverges
.
The test is
inconclusive
with

x

x
0

L
= 1.
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (10/24)
Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Example
Example:
For the power series:
∞
X
n
=1
(

1)
n
+1
n
(
x

2)
n
The
ratio test
gives:
lim
n
→∞
(

1)
n
+2
(
n
+ 1)(
x

2)
n
+1
(

1)
n
+1
n
(
x

2)
n
=

x

2

lim
n
→∞
n
+ 1
n
=

x

2

.
This
converges absolutely
for

x

2

<
1.
It
diverges
for

x

2

L
≥
1.
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (11/24)
Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Radius of Convergence
Radius of Convergence:
For the power series:
∞
X
n
=0
a
n
(
x

x
0
)
n
,
typically, there is a positive number
ρ
, called the
radius of
convergence
, such that the series
converges absolutely
for

x

x
0

< ρ
and
diverges
for

x

x
0

> ρ
Generally, we are not concerned about convergence at the endpoints
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Power Series
Ordinary Point
— (12/24)
Introduction
Series Solutions of Differential Equations
Example
Review Power Series
Properties of Series
1
Consider the series
∞
X
n
=0
a
n
(
x

x
0
)
n
=
f
(
x
)
and
∞
X
n
=0
b
n
(
x

x
0
)
n
=
g
(
x
)
converging for

x

x
0

< ρ
with
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