Representation of functions as power series

Dr. Philippe B. Laval

Kennesaw State University
November 19, 2008
Abstract
This document is a summary of the theory and techniques used to
represent functions as power series.
1 Representation of Functions as Power Series
(8.6)
1.1 Theory
In this section, we develop several techniques to help us represent a function as
a power series. More precisely, given a function f (x), we will try to nd a power
series
1

n=0
c
n
(x a)
n
such that f (x) =
1

n=0
c
n
(x a)
n
. Part of the work will
involve nding the values of x for which this is valid. Part of the reason for doing
this is that a power series looks like a polynomial (except that it has innitely
many terms). Polynomials are among the easiest functions to work with. They
are easy to dierentiate, integrate, ... So, if f is a complicated function, replacing
it with a power series amounts to replacing it with a polynomial. Therefore,
working with f becomes easier. There are some technical diculties to resolve,
we will address those as we develop the technique.
First, we will look at techniques which will allow us to obtain a series rep-
resentation by using known series representations. The techniques involved are
substitution, dierentiation and integration. Then, we will learn a technique
which will allow us to nd a series representation for a given function f directly,
without having to use known series.
We now look at each technique (substitution, dierentiation and integration)
in details, using examples. At this point, we only know the following series
representation:
1
1 x
= 1 + x + x
2
+ x
3
+ ::: in (1; 1)
=
1

n=0
x
n
1
When nding the power series of a function, you must nd both the series
representation and when this representation is valid (its domain).
1.2 Substitution
We derive the series for a given function using another function for which we
already have a power series representation. Then, we do the following:
1. Figure out which substitution can be applied to transform the function
for which we know the series representation to the function for which we
want a series representation.
2. Apply the same substitution to the known series. This will give us the
series representation we wanted.
3. The domain of the new function is obtained by applying the same substi-
tution to the domain of the known series.
Example 1 Find a power series representation for f (x) =
1
1 + x
and its do-
main.
We use the representation of
1
1 x
, replacing x by x. We obtain:
1
1 + x
=
1
1 (x)
= 1 + (x) + (x)
2
+ (x)
3
+ :::
= 1 x + x
2
x
3
+ :::
=
1

n=0
(1)
n
x
n
The series representation for
1
1 x
was valid if jxj < 1. If we apply the same
substitution, we see that this representation is valid if jxj < 1 or jxj < 1. In
other words, the interval of convergence is also (1; 1).
Example 2 Find a power series representation for f (x) =
1
1 x
2
.
Proceeding as above and remembering that
1
1 x
=
1

n=0
x
n
, we have:
1
1 x
2
=
1

n=0
_
x
2
_
n
=
1

n=0
x
2n
= 1 + x
2
+ x
4
+ :::
2
This is a geometric series which converges when

x
2

< 1 that is when x

2
< 1
or 1 < x < 1.
Example 3 Find a power series representation for
1
2 + x
and nd its domain.
First, we rewrite
1
2 + x
=
1
2
1
1 +
x
2
. Now, we nd a power series representation
for
1
1 +
x
2
, then we will multiply it by
1
2
.
1
1 +
x
2
can be obtained from
1
1 x
by
replacing x by
x
2
. Since
1
1 x
=
1

n=0
x
n
, we have:
1
1 +
x
2
=
1

n=0
_

x
2
_
n
=
1

n=0
(1)
n
_
x
2
_
n
Therefore,
1
2 + x
=
1
2
1

n=0
(1)
n
_
x
2
_
n
=
1

n=0
(1)
n
x
n
2
n+1
1
1 x
converges when jxj < 1, so this series converges when

x
2

< 1 =)jxj <

2. So, the interval of convergence is (2; 2), the radius of convergence is 2.
Remark 4 In the rst of these two examples, we applied the substitution to the
expanded form of
1
1 x
. In the second, we applied it to the compact form. You
can do it either way, it is simply a matter of choice.
Example 5 Suppose that you are given that a series repersentation for e
x
is
e
x
=
1

n=0
x
n
n!
in (1; 1). What is a series representation for e
x
2
and what is
its domain?
We go from e
x
to e
x
2
using the substitution x !x
2
. Thus
e
x
2
=
1

n=0
_
x
2
_
n
n!
=
1

n=0
x
2n
n!
also valid in (1; 1)
3
1.3 Dierentiation and Integration
The driving force behind the integration and dierentiation techniques is the
theorem below which we state without proof.
Theorem 6 If the power series
1

n=0
c
n
(x a)
n
has a radius of convergence
R > 0 then the function dened by f (x) =
1

n=0
c
n
(x a)
n
= c
0
+ c
1
(x a) +
c
2
(x a)
2
+ ::: is dierentiable (hence) continuous on (a R; a + R) and
1. f
0
(x) = c
1
+2c
2
(x a) +3c
3
(x a)
2
+::: In other words, the series can
be dierentiated term by term.
2.
_
f (x) dx = C +c
0
(x a) +c
1
(x a)
2
2
+c
2
(x a)
3
3
+::: In other words,
the series can be integrated term by term.
3. Note that whether we dierentiate or integrate, the radius of convergence
is preserved. However, convergence at the endpoints must be investigated
every time.
Remark 7 This theorem simply says that the sum rule for derivatives and in-
tegrals also applies to power series. Remember that a power series is a sum, but
it is an innite sums. So, in general, the results we know for nite sums do
not apply to innite sums. The theorem above says that it does in the case of
innite series.
Remark 8 The formula in part 1 of the theorem is obtained simply by dier-
entiating the series term by term. Since
f (x) = c
0
+ c
1
(x a) + c
2
(x a)
2
+ :::
then
f
0
(x) =
_
c
0
+ c
1
(x a) + c
2
(x a)
2
+ :::
_
0
= (c
0
)
0
+ (c
1
(x a))
0
+
_
c
2
(x a)
2
_
0
+ :::
= 0 + c
1
+ 2c
2
(x a) + 3c
3
(x a)
2
+ ::: (1)
Alternatively, one can also dierentiate the general formula. Since
f (x) =
1

n=0
c
n
(x a)
n
4
then
f
0
(x) =
_
1

n=0
c
n
(x a)
n
_
0
=
1

n=0
(c
n
(x a)
n
)
0
by the theorem
=
1

n=0
nc
n
(x a)
n1
The rst term of this series (when n = 0) is 0, thus we can start summation at
n = 1. Hence, we have
f
0
(x) =
1

n=1
nc
n
(x a)
n1
(2)
The reader should check that formulas 1 and 2 are identical.
Remark 9 The formula in part 2 of the theorem is obtained by integrating term
by term. It can also be obtained by integrating the general formula. Since
f (x) =
1

n=0
c
n
(x a)
n
then
_
f (x) dx =
_
_
1

n=0
c
n
(x a)
n
_
dx
=
1

n=0
_
(c
n
(x a)
n
) dx by the theorem
Since an antiderivative of c
n
(x a)
n
is C +
c
n
(x a)
n+1
n + 1
, we have
_
f (x) dx = C +
1

n=0
c
n
(x a)
n+1
n + 1
If we expand this, we get
_
f (x) dx = C + c
0
(x a) + c
1
(x a)
2
2
+ c
2
(x a)
3
3
+ :::
Which is the formula which appears in the theorem.
5
Example 10 Given that a power series representation for f (x) is
f (x) = 1 + x + x
2
+ x
3
+ :::
=
1

n=0
x
n
nd a power series representation for f
0
(x) and
_
f (x) dx.
First, we nd f
0
(x). From the theorem, we know it is enough to dieren-
tiate term by term. Thus,
f
0
(x) = 0 + 1 + 2x + 3x
2
+ :::
= 1 + 2x + 3x
2
+ :::
Note that we can also obtain the same result by using the general formula.
In this case,
f (x) =
1

n=0
x
n
Thus
f
0
(x) =
1

n=0
(x
n
)
0
=
1

n=0
nx
n1
=
1

n=1
nx
n1
Which gives us the same answer.
Next, we nd
_
f (x) dx. From the theorem, it is enough to integrate term
by term. Thus since f (x) = 1 + x + x
2
+ x
3
+ :::, we have:
_
f (x) dx = C + x +
x
2
2
+
x
3
3
+
x
4
4
+ :::
Alternatively, we can work from the general formula. Since f (x) =
1

n=0
x
n
,
we have
_
f (x) dx =
1

n=0
_
(x
n
) dx
= C +
1

n=0
x
n+1
n + 1
Which is the same formula.
6
1.3.1 Dierentiation
This time, we nd the series representation of a given series by dierentiating
the power series of a known function. More precisely, if f
0
(x) = g (x), and if
we have a series representation for f and need one for g, we simply dierentiate
the series representation of f. The theorem above tells us that the radius of
convergence will be the same. However, we will have to check the endpoints.
Example 11 Find a series representation for
1
(1 x)
2
, nd the interval of
convergence.
We begin by noting that
_
1
1 x
_
0
=
1
(1 x)
2
. Since
1
1 x
= 1+x+x
2
+x
3
+::::,
we have:
1
(1 x)
2
=
_
1 + x + x
2
+ x
3
+ :::
_
0
= 1 + 2x + 3x
2
+ 4x
3
+ ::::
=
1

n=1
nx
n1
The radius of convergence is still 1. Since
1
1 x
converges for x in (1; 1), we
know that
1
(1 x)
2
will also converge in (1; 1). It might also converge at the
endpoints, so we need to check them. Do it as an exercise.
Remark 12 If you prefer to work from the compact form of the series, it can
be done.
1
1 x
=
1

n=0
x
n
1
(1 x)
2
=
_
1

n=0
x
n
_
0
=
1

n=0
(x
n
)
0
=
1

n=1
nx
n1
However, you have to be careful with the starting value of n.
Example 13 Suppose you know that a series representation for sinx is
sinx = x
x
3
3!
+
x
5
5!

x
7
7!
+ :::
=
1

n=0
(1)
n
x
2n+1
(2n + 1)!
7
Find a power series representation for cos x.
cos x = (sinx)
0
=
_
x
x
3
3!
+
x
5
5!

x
7
7!
+ :::
_
0
= 1
x
2
2!
+
x
4
4!

x
6
6!
+ :::
=
1

n=0
(1)
n
x
2n
(2n)!
Example 14 Find a power series representation for 2xe
x
2
given that a series
repersentation for e
x
is e
x
=
1

n=0
x
n
n!
in (1; 1).
We can do this problem two ways.
Method 1 Using substitution, we can nd a power series representation for
e
x
2
, then multiply what we nd by 2x. The rst part was done earlier,
and we found that e
x
2
=
1

n=0
x
2n
n!
on (1; 1). Thus,
2xe
x
2
=
1

n=0
2xx
2n
n!
=
1

n=0
2x
2n+1
n!
also in (1; 1).
Method 2 We note that 2xe
x
2
=
_
e
x
2
_
0
. So, using substitution, we can nd
a power series representation for e
x
2
, then dierentiate it to get a series
8
representation for 2xe
x
2
.
2xe
x
2
=
_
e
x
2
_
0
=
_
1

n=0
x
2n
n!
_
0
=
1

n=0
2nx
2n1
n!
=
1

n=1
2nx
2n1
n!
when n = 0,
2nx
2n1
n!
= 0
=
1

n=1
2x
2n1
(n 1)!
=
1

n=0
2x
2n+1
n!
So, either way, we nd the same answer.
1.3.2 Integration
This time we nd the power series representation of a function by integrating
the power series representation of a known function. If g (x) =
_
f (x) dx and we
know a power series representation for f (x), we can get a series representation
for g (x) by integrating the series representation of f.
Example 15 Find a power series representation for ln(1 x).
We know that ln(1 x) =
_
dx
1 x
. By our earlier work, we found that
1
1 x
= 1 + x + x
2
+ x
3
+ ::: in (1; 1), so
ln(1 x) = C + x +
x
2
2
+
x
3
3
+ :::
We also know that ln1 = 0, From the above equation, we get that C = 0 by
letting x = 0. Therefore
ln(1 x) = x +
x
2
2
+
x
3
3
+ :::
=
1

n=0
x
n+1
n + 1
Therefore,
ln(1 x) =
1

n=0
x
n+1
n + 1
9
Since the original series converges in (1; 1), this one will also converge there.
The end points should be checked, we leave it as an exercise.
Example 16 Find a power series rpresentation for tan
1
x.
We use the fact that tan
1
x =
_
dx
1 + x
2
. First, since
1
1 x
= 1+x+x
2
+x
3
+:::,
se see that
1
1 + x
2
= 1 x
2
+ x
4
x
6
+ ::: and therefore
tan
1
x = C + x
x
3
3
+
x
5
5

x
7
7
+ :::
Using the fact that tan
1
0 = 0 gives us C = 0. Thus
tan
1
x = x
x
3
3
+
x
5
5

x
7
7
+ :::
1.4 Things to Know
Be able to nd the series representation of a function by substitution,
integration or dierentiation.
Problems assigned: # 1, 3, 5, 7, 11, 13, 21, 35 on pages 604, 605
2 Taylor and Maclaurins Series (8.7)
2.1 Introduction
The previous section showed us how to nd the series representation of some
functions by using the series representation of known functions. The methods
we studied are limited since they require us to relate the function to which
we want a series representation with one for which we already know a series
representation. In this section, we develop a more direct approach. When
dealing with functions and their power series representation, there are three
fundamental questions one has to answer:
1. Does a given function have a power series representation?
2. If it does, how do we nd it?
3. What is the domain in other words, for which values of x is the represen-
tation valid?
Question 1 is more theoretical, we wont address it. We will concentrate on
questions 2 and 3.
Assuming f has a power series representation that is f (x) = c
0
+c
1
(x a)+
c
2
(x a)
2
+ :::, we want to nd what the power series representation is, that
is we need to nd the coecients c
0
; c
1
; c
2
; :::. It turns out that it is not very
dicult. The technique used is worth remembering. We rst nd c
0
. Having
found c
0
we next nd c
1
. Then, we nd c
2
and so on.
10
Finding c
0
. Since
f (x) = c
0
+ c
1
(x a) + c
2
(x a)
2
+ c
3
(x a)
3
+ c
4
(x a)
4
+ ::: (3)
it follows that
f (a) = c
0
+ c
1
(0) + c
2
(0)
2
+ :::
= c
0
Finding c
1
. Dierentiating equation 3 gives us
f
0
(x) = c
1
+ 2c
2
(x a) + 3c
3
(x a)
2
+ 4c
4
(x a)
3
+ :::
Therefore,
f
0
(a) = c
1
Finding c
2
. We proceed the same way. First, we compute f
00
(x), then
f
00
(a).
f
00
(x) = 2c
2
+ (2) (3) c
3
(x a) + (3) (4) c
4
(x a)
2
+ :::
Therefore
f
00
(a) = 2c
2
or
c
2
=
f
00
(a)
2
In general. Continuing this way, we can see that
c
n
=
f
(n)
(a)
n!
2.2 Denitions and Theorems
Theorem 17 If the function f has a power series representation, that is if
f (x) =
1

n=0
c
n
(x a)
n
= c
0
+ c
1
(x a) + c
2
(x a)
2
+ ::: for jx aj < R then
its coecients are given by:
c
n
=
f
(n)
(a)
n!
In other words
f (x) = f (a) + f
0
(a) (x a) + f
00
(a)
(x a)
2
2
+ f
000
(a)
(x a)
3
3!
+ :::
Denition 18 1. The series in the previous theorem
_
1

n=0
f
(n)
(a)
n!
(x a)
n
_
is called the Taylor series of the function f at a.
11
2. The nth partial sum is called the nth order Taylor polynomial It is denoted
T
n
: So,
T
n
(x) =
n

i=0
f
(i)
(a)
i!
(x a)
i
3. In the special case a = 0, the series is called a Maclaurins Series. So, a
Maclaurins series is of the form
1

n=0
f
(n)
(0)
n!
x
n
and the Maclaurins series
for a function f is given by
f (x) =
1

n=0
f
(n)
(0)
n!
x
n
= f (0) + f
0
(0) x + f
00
(0)
x
2
2
+ f
000
(0)
x
3
3!
+ :::
2.3 Examples
We now look how to nd the Taylor and Maclaurins series of some functions.
Example 19 Find the Maclaurins series for f (x) = e
x
, nd its domain.
The series will be of the form
1

n=0
f
(n)
(0)
n!
x
n
, we simply need to nd the coef-
cients f
(n)
(0). This is easy. Since all the derivatives of e
x
are e
x
, it follows
that f
(n
(x) = e
x
, thus f
(n)
(0) = e
0
= 1, hence
e
x
=
1

n=0
f
(n)
(0)
n!
x
n
=
1

n=0
x
n
n!
To nd where this series converges, we use the ration test and compute:
lim
n!1

a
n+1
a
n

= lim
n!1

x
n+1
(n + 1)!
x
n
n!

= lim
n!1
n! jxj
n+1
(n + 1)! jxj
n
= lim
n!1
jxj
n + 1
= 0
Thus the domain is all real numbers.
12
Example 20 Find a Taylor series for f (x) = e
x
centered at 2.
The series will be of the form
1

n=0
f
(n)
(2)
n!
(x 2)
n
, we simply need to nd the
coecients f
(n)
(2). This is easy. Since all the derivatives of e
x
are e
x
, it follows
that f
(n
(x) = e
x
, thus f
(n)
(2) = e
2
, hence
e
x
=
1

n=0
f
(n)
(2)
n!
(x 2)
n
=
1

n=0
e
2
n!
(x 2)
n
Example 21 Find the n
th
order Taylor polynomial for e
x
centered at 0 and
centered at 2. Plot these polynomials for n = 1; 2; 3; 4; 5. What do you notice?
We already computed the power series corresponding to these two situations.
We found that
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+
x
6
6!
+ :::
and
e
x
= e
2
+ e
2
(x 2) + e
2
(x 2)
2
2!
+ e
2
(x 2)
3
3!
+ :::
= e
2
_
1 + (x 2) +
(x 2)
2
2!
+
(x 2)
3
3!
+
(x 2)
4
4!
+
(x 2)
5
5!
+ :::
_
If we denote T
n
the n
th
order Taylor polynomial for e
x
centered at 0 and Q
n
the n
th
order Taylor polynomial for e
x
centered at 2, we have:
n
th
order Taylor polynomials for e
x
centered at 0.
T
1
(x) = 1 + x
T
2
(x) = 1 + x +
x
2
2!
T
3
(x) = 1 + x +
x
2
2!
+
x
3
3!
T
4
(x) = 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
T
5
(x) = 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
The graphs is shown on the next page.
13
Graphs of e
x
and the rst 5 Taylor polynomials centered at 0. The func-
tions have the following colors:
e
x
: black
T
1
: blue
T
2
: red
T
3
: green
T
4
: purple
T
5
: yellow
14
n
th
order Taylor polynomial for e
x
centered at 2:
Q
1
(x) = e
2
(1 + (x 2))
Q
2
(x) = e
2
_
1 + (x 2) +
(x 2)
2
2!
_
Q
3
(x) = e
2
_
1 + (x 2) +
(x 2)
2
2!
+
(x 2)
3
3!
_
Q
4
(x) = e
2
_
1 + (x 2) +
(x 2)
2
2!
+
(x 2)
3
3!
+
(x 2)
4
4!
_
Q
5
(x) = e
2
_
1 + (x 2) +
(x 2)
2
2!
+
(x 2)
3
3!
+
(x 2)
4
4!
+
(x 2)
5
5!
_
The graphs are shown on the next page..
15
Graphs of e
x
and the rst 5 Taylor polynomials centered at 2. The func-
tions have the following colors:
e
x
: black
Q
1
: blue
Q
2
: red
Q
3
: green
Q
4
: purple
Q
5
: yellow
We can see the following:
The Taylor polynomial approximates the functions well near the point at
which the series is centered. As we move away from this point, the ap-
proximation deteriorates very quickly.
The approximation is better the higher the degree of the Taylor polynomial.
More specically, the Taylor polynomial stay closer to the function over a
larger interval, the higher its degree.
16
Example 22 Find a Maclaurins series for f (x) = sinx and nd its domain
The series will be of the form
1

n=0
f
(n)
(0)
n!
x
n
, we simply need to nd the coe-
cients f
(n)
(0). That is we need to nd all the derivatives of sinx and evaluate
them at x = 0. We summarize our ndings in the table below:
n f
(n)
(x) f
(n)
(0)
0 sinx 0
1 cos x 1
2 sinx 0
3 cos x 1
4 sinx 0
So, we see that we will have coecients only for odd values of n. In addition,
the sign of the coecients will alternate. Thus, since
f (x) = f (0) + f
0
(0) x + f
00
(0)
x
2
2
+ f
000
(0)
x
3
3!
+ f
(4)
(0)
x
4
4!
+ :::
we have
sinx = 0 + x + 0
x
3
3!
+ 0 +
x
5
5!
:::
= x
x
3
3!
+
x
5
5!

x
7
7!
+ :::
=
1

n=0
(1)
n
x
2n+1
(2n + 1)!
To nd where this series converges, we use the ration test and compute lim
n!1

a
n+1
a
n

.
Since a
n
=
x
2n+1
(2n + 1)!
, a
n+1
=
x
2n+3
(2n + 3)!
. Therefore,
lim
n!1

a
n+1
a
n

= lim
n!1

x
2n+3
(2n + 3)!
x
2n+1
(2n + 1)!

= lim
n!1
jxj
2n+3
(2n + 1)!
jxj
2n+1
(2n + 3)!
= jxj
2
lim
n!1
(2n + 1)!
(2n + 3)!
= jxj
2
lim
n!1
1
(2n + 2) (2n + 3)
= 0
Thus the series representation is valid for all x.
17
Here are some important functions and their corresponding Maclaurins se-
ries
1
1 x
= 1 + x + x
2
+ x
3
+ x
4
+ ::: (1; 1)
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+ ::: (1; 1)
sinx = x
x
3
3!
+
x
5
5!

x
7
7!
+ ::: (1; 1)
cos x = 1
x
2
2!
+
x
4
4!

x
6
6!
+ ::: (1; 1)
tan
1
x = x
x
3
3
+
x
5
5

x
7
7
+ ::: [1; 1]
Remark 23 The n
th
order Taylor polynomial (T
n
) associated with the series
expansion of a function f is the n
th
degree polynomial obtained by truncating the
series expansion and keeping only the terms of degree less than or equal to n.
In the case of sinx, since
sinx = x
x
3
3!
+
x
5
5!

x
7
7!
+ :::
It follows that
T
1
= x
T
2
= x
T
3
= x
x
3
3!
T
4
= x
x
3
3!
T
5
= x
x
3
3!
+
x
5
5!
T
6
= x
x
3
3!
+
x
5
5!
and so on.
2.4 Summary
We now have four dierent techniques to nd a series representation for a func-
tion. These techniques are:
1. Substitution
2. Dierentiation
3. Integration
18
4. Taylor/Maclaurins series.
It may appear that the last technique is much more powerful, as it gives
us a direct way to derive the series representation. In contrast, the rst three
techniques require we start from a known series representation. However, the
rst three techniques should not be ignored. In many cases, they make the work
easier. We illustrate this with a few examples.
Example 24 Find a Maclaurins series for f (x) = e
x
2
.
Of course, this can be done directly. But consider the problem of nding all the
derivative of e
x
2
. One can also start from e
x
and substitute x
2
for x. Since
e
x
=
1

n=0
x
n
n!
= 1 + x +
x
2
2
+
x
3
3!
+
x
4
4!
it follows that
e
x
2
=
1

n=0
_
x
2
_
n
n!
=
1

n=0
(1)
n
x
2n
n!
This was pretty painless. To convince yourself that this is the way to do it, try
the direct approach instead!
Example 25 Find a Maclaurins series for cos x.
Again, one can try the direct approach as in example 22. It is not too dicult.
One can also realize that (sinx)
0
= cos x. Thus, one can start with the series for
sinx (which we already derived) and nd the series for cos x by dierentiating
it. We get
cos x = (sinx)
0
=
_
1

n=0
(1)
n
x
2n+1
(2n + 1)!
_
0
=
1

n=0
_
(1)
n
x
2n+1
(2n + 1)!
_
0
=
1

n=0
(1)
n
x
2n
(2n)!
= 1
x
2
2
+
x
4
4!

x
6
6!
19
2.5 Things to Remember
Given a function, be able to nd its Taylor or Maclaurins series.
Be able to nd the radius and interval of convergence of Taylor or Maclau-
rins series.
Section 8.7: # 3, 5, 7, 9, 11, 13, 19, 21, 23, 25, 31 on pages 615, 616.
3 Applications
The purpose of this section is to show the reader how Taylor series can be used
to approximate functions. The approximation can then be used to either evalu-
ate a function at specic values of x, to integrate or to dierentiate the function.
Of course, we would only use this technique with functions for which the tra-
ditional calculus methods do not work. For example, we may need to compute
1
_
0
e
x
2
dx. This cannot be done using the integration techniques learned in a
traditional calculus class since e
x
2
does not have an antiderivative which can
be expressed in terms of elementary functions. Another application might be to
approximate sin(0:01), without a calculator. The diculty does not lie in the
series representation of a given function, we now know how to represent func-
tions as power series. However, series have innitely many terms, for practical
purposes, we can only use a nite number of them. Thus, we replace the in-
nite series by the corresponding Taylor polynomial (see denition 18) of order
n (T
n
), for some n. We then use the Taylor polynomial instead of the function.
This can be used to evaluate a function, integrate a function or dierentiate a
function. However, when we perform the following approximation
f (x) f (a) + f
0
(a) (x a) +
f
00
(a)
2!
(x a)
2
+ ::: +
f
(n)
(a)
n!
(x a)
n
there are several questions to answer before we can carry it out:
1. How do we pick a, the number around which the series is centered?
2. How do we pick n so the Taylor polynomial T
n
approximates the given
function f within the desired accuracy?
Answer to question 1 There are several factors to take under consideration. First, since we have
to evaluate f
(n)
(a), a must be picked so we can do this evaluation easily.
Second, you will recall that the accuracy of the approximation decreases
as x gets further away from a. Therefore, we need to pick a so that the
values at which f (x) will be approximated are in the domain of the series,
and not too far from a. For example, if we had to approximate sin(:001),
then a = 0 would be a good choice because it satises both criteria.
20
Answer to question 2 Once we have selected a and have a series representation for f, we use the
techniques studied to approximate a series and nd the error.
The examples below illustrate these applications.
Example 26 Find the nth order Taylor polynomial for f (x) = cos x when
n = 2; 3; 4; 5; 6. Sketch the graph of cos x as well as the Taylor polynomials
found.
We already know the power series for cos x.
cos x = 1
x
2
2!
+
x
4
4!

x
6
6!
+
x
8
8!
::::
So,
P
2
(x) = 1
x
2
2!
P
3
(x) = 1
x
2
2!
P
4
(x) = 1
x
2
2!
+
x
4
4!
P
5
(x) = 1
x
2
2!
+
x
4
4!
P
6
(x) = 1
x
2
2!
+
x
4
4!

x
6
6!
You will note that because every other coecient in the series expansion of cos x
is 0; P
3
= P
2
, P
5
= P
4
. The graph of these polynomials is shown on gure 1.
Remark 27 You will notice that as n increases, P
n
gets closer to the graph
of cos x. In other words, the accuracy of our approximation increases with n.
However, P
n
is a good approximation for cos x in a neighborhood of 0, as we
move away from 0, P
n
gets further and further away from cos x. This is impor-
tant. When one approximates a function with a Taylor polynomial about a, the
approximation is good only for values of x close to a.
Example 28 Approximate cos 0:01 with an error less than 10
20
.
First, we note that since :01 is close to 0, we can use a Taylor polynomial
centered at 0 to approximate cos (:01). Therefore, using the Taylor polynomial
for cos x centered at 0, and replacing x by 0:01, we get:
cos 0:01 =
n

i=0
(1)
i
0:01
2i
(2i)!
We need to nd n so that if we approximate
1

i=0
(1)
i
0:01
2i
(2i)!
by
n

i=0
(1)
i
0:01
2i
(2i)!
,
the error is less than 10
20
. We notice that
1

i=0
(1)
i
0:01
2i
(2i)!
is an alternating
21
Figure 1: Graph of cos x and some Taylor polynomials
series with b
n
=
0:01
2n
(2n)!
, so we know how to estimate its error. The error is
always less than b
n+1
. So, if we want the error to be less than 10
20
, it is
enough to solve:
b
n+1
< 10
20
0:01
2(n+1)
(2n + 2)!
< 10
20
We solve this by trying various values of n . The table below shows this proce-
dure:
22
n
0:01
2(n+1)
(2n + 2)!
1 4 10
10
2 1 10
15
3 2 10
21
We see that n = 3 is enough. Therefore:
cos 0:01
3

n=0
(1)
n
0:01
2n
(2n)!
0:999950000416665
Example 29 Approximate
_
1
0
e
x
2
dx with an error less than 0:001.
First, we nd a series representation for e
x
2
. Since
e
x
=
1

n=0
x
n
n!
it follows that
e
x
2
=
1

n=0
(1)
n
x
2n
n!
Therefore
_
e
x
2
dx =
1

n=0
(1)
n
x
2n+1
(2n + 1) n!
and therefore
_
1
0
e
x
2
dx =
1

n=0
(1)
n
x
2n+1
(2n + 1) n!

1
0
=
1

n=0
(1)
n
1
(2n + 1) n!
This is an alternating series
1

n=0
(1)
n
b
n
with b
n
=
1
(2n + 1) n!
. From our
knowledge of alternating series, we know that if we approximate
1

i=0
(1)
i
1
(2i + 1) i!
by
n

i=0
(1)
i
1
(2i + 1) i!
, the error will be less than b
n+1
=
1
(2n + 3) (n + 1)!
. So,
we nd n such that
1
(2n + 3) (n + 1)!
< :001
23
We try several values of n, we nd that when n = 3,
1
(2n + 3) (n + 1)!
= :00463
and when n = 4,
1
(2n + 3) (n + 1)!
= :0007576. So,
4

n=0
(1)
n
1
(2n + 1) n!
= 0:747 49
gives us the desired approximation.
3.1 Problems
The problems assigned for power series, Taylor series and representation of
functions as power series were:
Section 8.6: # 1, 3, 5, 7, 11, 13, 21, 35 on pages 610, 611.
Section 8.7: # 3, 5, 7, 9, 11, 13, 19, 21, 23, 25, 31 on pages 615, 616.
In addition, for the applications discussed in this section, do # 3, 5, 7 on
page 628.
24