Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
8.6
Representing Functions as Power Series
2
Functions such as ex do not have an elementary antderivative. One way to
overcome that is to express the function as a power series and then integrate
the terms of this power series to find a power series representation of the
2
antiderivative of ex . Also, some type of differential equations have solutions
that are repsresented as power series.
Our first power series representation of a function is the representation of
∞
X
1
= 1 + x + x2 + · · · =
xn , |x| < 1.
1−x
(1)
n=0
Example 8.6.1
Find a power series representation of the function f (x) =
mine its interval of convergence.
1
1+4x2
and deter-
Solution.
Replacing x by −4x2 in (1), we find
∞
X
1
=
(−4x2 )n = 1 − x2 + x4 − x6 + x8 − · · · .
1 + 4x2
n=0
This geometric series converges for all x such that | − 4x2 | = 4|x2 | < 1
or x2 < 14 . Solving this quadratic inequality, we find |x| < 21 which is the
interval of convergence
Example 8.6.2
Find a power series representation of the function f (x) =
mine its interval of convergence.
Solution.
We first note that f (x) =
1 1
3 1+ 2x
3
1
3+2x
and deter-
. Replacing x by − 2x
3 in (1), we find
∞ X
2x 9 2
8
1
2x n
−
=1−
+ x − x3 + · · · .
2x =
3
3
4
27
1+ 3
n=0
Thus,
1
1 2x
9
8
= −
+ x2 − x3 + · · · .
3 + 2x
3
9
12
81
This geometric series converges for all x such that | 2x
3 | < 1. Thus, the interval
3
of convergence is |x| < 2
1
Example 8.6.3
Find a power series representation of the function f (x) =
mine its interval of convergence.
x2
3+2x
and deter-
Solution.
From the previous example, we find
∞ x2 X
2x n
1 x2
=
−
3 1 + 2x
3
3
3
n=0
∞
2 n n+2
1X
−
x
=
3
3
n=0
=
∞
X
(−1)n
n=2
2n−2 n
x .
3n−1
This geometric series converges for all x such that | 2x
3 | < 1. Thus, the inter3
val of convergence is |x| < 2
As discussed above, one can represent some known functions as power series
by relating them to geometric series. We can obtain power series representation for a wider variety of functions by exploiting the fact that a convergent
power series can be differentiated, or integrated, term-by-term to obtain a
new power series that has the same radius of convergence as the original
power series. The new power series is a representation of the derivative,
or antiderivative, of the function that is represented by the original power
series. More formally, we have
Theorem 8.6.1
If R > 0 is the radius of convergence of a power series
∞
X
cn (x − a)n then
n=0
the function
f (x) = c0 + c1 (x − a) + c2 (x − a)2 + · · · =
X
cn (x − a)n
n=0
is differentiable on the interval (a − R, a + R) with derivative
0
f (x) =
∞
X
ncn (x − a)n−1 .
n=1
2
Since f (x) is differentiable, it is therefore integrable with antiderivative
Z
∞
X
(x − a)n+1
f (x)dx = C +
cn
n+1
n=0
where C is a constant of integration. Both new series have radii of convergence R (but not necessarily the same interval of convergence).
Example 8.6.4
Find a power series representation of the function f (x) =
tiating (1).
1
(1−x)2
by diferen-
Solution.
We have
1
d
=
2
(1 − x)
dx
1
1−x
=
∞
X
nx
n−1
∞
X
=
n=1
(n + 1)xn .
n=0
The radius of convergence of this geometric series is 1
Example 8.6.5
Find a power series representation of the function f (x) = ln (1 + x).
Solution.
We have
Z
ln (1 + x) =
dx
=
1+x
Z X
∞
(−1)n xn dx = C+
n=0
∞
X
∞
X
xn+1
xn
= C+ (−1)n−1 .
n+1
n
(−1)n
n=0
n=1
Replacing x = 0 we find C = 0. Hence,
ln (1 + x) =
∞
X
(−1)n−1
n=1
xn
.
n
The radius of convergence is R = 1
Example 8.6.6
Find a power series representation of the function f (x) = tan−1 x.
Solution.
We have
−1
tan
Z
x=
dx
=
1 + x2
Z X
∞
n 2n
(−1) x dx = C +
n=0
∞
X
(−1)n
n=0
3
x2n+1
.
2n + 1
Replacing x = 0 we find C = 0. Hence,
−1
tan
x=
∞
X
(−1)n
n=0
The radius of convergence is R = 1
4
x2n+1
.
2n + 1