10.6 - Representations of Functions as Power Series
• In this section, we aim to find an alternative way to represent functions
as power series by using what we have learned thus far about series. Let’s
begin with a formula we introduced in Section 10.2: the sum of a geometric
series.
∞
∑
1
= 1 + x + x2 + · · · + xn =
xn for |x| < 1.
1−x
n=0
Given what we’ve learned since first seeing this equation, we see that this
1
is a series representation of the function f (x) = 1−x
. If we replace the x
with different values, we can find series representations for a good number
of rational functions.
• Suppose we want to find a power series representation of the function
f (x) =
1
.
1 + 4x2
One of the first things you may notice is that f (x) looks somewhat similar
to the left-hand side of the formula for the geometric series. Can we modify
f (x) in such a way that we can apply the geometric series formula?
First notice that
1
1
=
,
1 + 4x2
1 − (−4x2 )
and, if we replace the x by a −4x2 in the geometric series formula, we
obtain
∞
∞
∑
∑
1
2 n
=
(−1)n 4n x2n .
(−4x
)
=
1 − (−4x2 ) n=0
n=0
Since the interval of convergence for the geometric series is |x| < 1, the
interval of convergence for this series is
)
(
1
1
1
1 1
2
2
2
| − 4x | < 1 ⇒ 4x < 1 ⇒ x < ⇒ − < x < , or − ,
.
4
2
2
2 2
• Another important concept we learned in this section was∑
that of term-byterm differentiation and integration. If the
power
series
cn (x − a)n has
∑∞
n
radius of convergence R > 0, then f = n=0 cn (x − a) is differentiable
(and thus continuous) on the interval (a − R, a + R), and
f ′ (x) =
∞
∑
ncn (x − a)n−1
n=1
∫
f (x) dx = C +
∞
∑
n=0
cn
(x − a)n+1
.
n+1
In both of the previous equations, the radius of convergence is still R. Note
that these formulas are consistent with what we know about differentiating
or integrating polynomials.
1
Using this knowledge, let’s try and find a power series representation and
the radius of convergence of
f (x) =
Note first that
x2
.
(1 − 2x)2
x2
= x2
(1 − 2x)2
(
1
(1 − 2x)2
)
,
which looks a lot like the formula for our geometric series, except the
denominator is squared. However, if we differentiate both sides of the
formula for a geometric series, we get
[∞ ]
[
]
∞
∑
d
1
d ∑ n
1
=
x ⇒
=
nxn−1 .
dx 1 − x
dx n=0
(1 − x)2
n=1
Now we have an expression with the denominator squared, and if we replace x with 2x and don’t forget to multiply by the x2 we pulled out of
the numerator, we get
∞
∞
∑
∑
x2
n−1 n−1 2
=
2n−1 nxn+1 .
n2
x
x
=
(1 − 2x)2
n=1
n=1
Now the interval of convergence for this series is
1
1
|2x| < 1 ⇒ − < x < , or
2
2
(
1 1
− ,
2 2
)
.
• We can also use this notion of term-by-term differentiation and integration
to integrate functions for which we do not know an antiderivative. For
example, let’s say we want to compute the following integral:
∫
arctan x
dx.
x
First, note that
∫
arctan x =
1
dx =
1 + x2
∫ ∑
∞
∞
∑
x2n+1
n 2n
.
(−1) x dx = C1 +
(−1)n
2n + 1
n=0
n=0
To determine C1 , notice that arctan 0 = 0, so C1 = 0.
Now we can say
arctan x =
∞
∞
∑
arctan x ∑
x2n
x2n+1
⇒
=
(−1)n
.
(−1)n
2n + 1
x
2n + 1
n=0
n=0
Integrating this expression, we get
∫
arctan x
dx =
x
∫ ∑
∞
∞
∑
x2n+1
x2n
dx = C2 +
(−1)n
(−1)n
2n + 1
(2n + 1)2
n=0
n=0
2
10.7 - Taylor and Maclaurin Series
• In this section, we learned that the Taylor expansion of the function f at a
is given by
f (x) =
∞
∑
f (n) (a)
n=0
n!
(x−a)n = f (a)+
f ′ (a)
f ′′ (a)
f ′′′ (a)
(x−a)+
(x−a)2 +
(x−a)3 +· · · .
1!
2!
3!
If a = 0, we call this special case of the Taylor series the Maclaurin series.
• Let’s use this definition to find the Maclaurin series for f (x) = sin 2x. The
table below gives us the first three derivatives and their values at 0:
n
0
1
2
3
4
5
f (n) (x)
sin 2x
2 cos 2x
−4 sin 2x
−8 cos 2x
16 sin 2x
32 cos 2x
f (n) (0)
0
2
0
−8
0
32
Then the Maclaurin series for f (x) is
sin 2x =
∞
∑
2
8
32
(−1)n (2x)2n+1
x − x3 + x5 + · · · =
.
1!
3!
5!
(2n + 1)!
n=0
n
2n+1
(2x)
Now, if we let an = (−1)(2n+1)!
, then
)
(
an+1 1
(2x)2n+3 (2n + 1)!
lim
·
= 4x2 lim
= 0 < 1,
= lim
n→∞ an n→∞ (2n + 3)(2n + 2)
n→∞ (2n + 3)! (2x)2n+1
so R = ∞.
Let’s use the same approach to find the Taylor series for f (x) =
a = 1. A table similar to the one in the last problem is given below:
n
0
1
2
3
f (n) (x)
x−1
−x−2
2x−3
−6x−4
f (n) (0)
1
−1
2
−6
Then the Taylor series about x = a is
1
1
2
6
= 1 − (x − 1) + (x − 1)2 − (x − 1)3 .
x
1!
2!
3!
Notice that the coefficients cancel and we get
1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · =
∞
∑
(−1)n (x − 1)n .
n=0
3
1
x
at
Now, for an = (−1)n (x − 1)n ,
n+1
an+1 = lim |x − 1| n = |x − 1| < 1 ⇒ 0 < x < 2 ⇒ R = 1.
lim n→∞ an n→∞ |x − 1|
• It may seem like a pain to get the Taylor series for these “easy” functions
and that may make you fear having to find series for more complicated
functions. However, knowing the Taylor series for an elementary function
can aid in finding the Taylor series for a more complicated, yet related
one.
Suppose we want to find the Maclaurin series for f (x) = cos x3 . Page 630
in the textbook gives the Maclaurin series for cos x as
cos x =
∞
∑
x2n
(−1)n
.
(2n)!
n=0
Now, if we replace the x with an x3 , we obtain
cos x3 =
∞
∑
x6n
(−1)n
.
(2n)!
n=0
It can be shown that R = ∞.
• Another advantage of Taylor series is being able to compute otherwise
difficult limits where L’Hopital’s rule can’t come to the rescue. Consider
the limit
x − arctan x
lim
.
x→0
x3
Referring to the table on page 630, we can replace arctan x by its Maclaurin
series and we get
lim
x→0
x − (x −
x3
3
5
+ x5 −
x3
x7
7
+ ···)
(
1 1 2 1 4
= lim
− x + x − ···
x→0 3
5
7
)
=
1
.
3
We are able to make this substitution because power series are continuous
functions.
• Just as we were able to find antiderivatives for “difficult” functions using
Taylor series, we are also able to approximate definite integrals of “difficult” functions in the same manner by finding the antiderivative and using
the Fundamental Theorem of Calculus to evaluate the integral with the
first few terms of the Taylor expansion.
∫1
2
Suppose we want to estimate 0 e−x dx correct to within an error of 0.001.
2
Let’s first compute the antiderivative of e−x . The Maclaurin series for
f (x) = ex is given on page 630 as
ex =
∞ n
∑
x
n=0
4
n!
.
Then, replacing x with −x2 ,
e−x =
2
∞
∑
x2n
(−1)n
.
n!
n=0
Now, using term-by-term integration, we get
∫
e−x dx = C +
2
∞
∑
(−1)n x2n+1
n=0
(2n + 1)n!
.
Applying the Fundamental Theorem of Calculus,
∫
1
e
0
−x2
[
x3
x5
x7
x9
dx = x −
+
−
+
− ···
3
10 42 216
]1
≈ 0.7475.
0
The Alternating Series Estimation Theorem shows us that
1
1
=
< 0.001.
11 · 5!
1320
5