Our main goal in this chapter is to study how some of these functions can
be constructed via the process of finding solutions to differential equations by
means of power series.
As a first example, we consider our old friend, the
equation of simple harmonic motion
d
2
y
dx
2
+
y
= 0
,
(2.11)
which we have already learned how to solve by other methods. Suppose for the
moment that we don’t know the general solution and want to find it by means
of power series. We could start by assuming that
y
=
a
0
+
a
1
x
+
a
2
x
2
+
a
3
x
3
+
· · ·
=
∞
X
n
=0
a
n
x
n
.
(2.12)
It can be shown that the standard technique for differentiating polynomials term
by term also works for power series, so we expect that
dy
dx
=
a
1
+ 2
a
2
x
+ 3
a
3
x
2
+
· · ·
=
∞
X
n
=1
na
n
x
n

1
.
(Note that the last summation only goes from 1 to
∞
, since the term with
n
= 0
drops out of the sum.) Differentiating again yields
d
2
y
dx
2
= 2
a
2
+ 3
·
2
a
3
x
+ 4
·
3
a
4
x
2
+
· · ·
=
∞
X
n
=2
n
(
n

1)
a
n
x
n

2
.
We can replace
n
by
m
+ 2 in the last summation so that
d
2
y
dx
2
=
∞
X
m
+2=2
(
m
+ 2)(
m
+ 2

1)
a
m
+2
x
m
+2

2
=
∞
X
m
=0
(
m
+ 2)(
m
+ 1)
a
m
+2
x
m
.
The index
m
is a “dummy variable” in the summation and can be replaced by
any other letter. Thus we are free to replace
m
by
n
and obtain the formula
d
2
y
dx
2
=
∞
X
n
=0
(
n
+ 2)(
n
+ 1)
a
n
+2
x
n
.
50
Substitution into equation(2.11) yields
∞
X
n
=0
(
n
+ 2)(
n
+ 1)
a
n
+2
x
n
+
∞
X
n
=0
a
n
x
n
= 0
,
or
∞
X
n
=0
[(
n
+ 2)(
n
+ 1)
a
n
+2
+
a
n
]
x
n
= 0
.
Recall that a polynomial is zero only if all its coefficients are zero. Similarly, a
power series can be zero only if all of its coefficients are zero. It follows that
(
n
+ 2)(
n
+ 1)
a
n
+2
+
a
n
= 0
,
or
a
n
+2
=

a
n
(
n
+ 2)(
n
+ 1)
.
(2.13)
This is called a
recursion formula
for the coefficients
a
n
.
The first two coefficients
a
0
and
a
1
in the power series can be determined
from the initial conditions,
y
(0) =
a
0
,
dy
dx
(0) =
a
1
.
Then the recursion formula can be used to determine the remaining coefficients
by the process of induction. Indeed it follows from (2.13) with
n
= 0 that
a
2
=

a
0
2
·
1
=

1
2
a
0
.
Similarly, it follows from (2.13) with
n
= 1 that
a
3
=

a
1
3
·
2
=

1
3!
a
1
,
and with
n
= 2 that
a
4
=

a
2
4
·
3
=
1
4
·
3
1
2
a
0
=
1
4!
a
0
.
Continuing in this manner, we find that
a
2
n
=
(

1)
n
(2
n
)!
a
0
,
a
2
n
+1
=
(

1)
n
(2
n
+ 1)!
a
1
.
Substitution into (2.12) yields
y
=
a
0
+
a
1
x

1
2!
a
0
x
2

1
3!
a
1
x
3
+
1
4!
a
0
x
4
+
· · ·
=
a
0
1

1
2!
x
2
+
1
4!
x
4
 · · ·
+
a
1
x

1
3!
x
3
+
1
5!
x
5
 · · ·
.
51
We recognize that the expressions within parentheses are power series expan
sions of the functions sin
x
and cos
x
, and hence we obtain the familiar expression
for the solution to the equation of simple harmonic motion,
y
=
a
0
cos
x
+
a
1
sin
x.
The method we have described—assuming a solution to the differential equa
tion of the form
y
(
x
) =
∞
X
n
=0
a
n
x
n
and solve for the coefficients
a
n
—is surprisingly effective, particularly for the
class of equations called secondorder linear differential equations.