6.7 Representation of Functions as Power Series

6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
6.7
6.7.1
343
Representation of Functions as Power Series
Theory
In this section and the next, 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
1
1
X
X
n
n
…nd a power series
cn (x a) such that f (x) =
cn (x a) . Part of the
n=0
n=0
work will involve …nding the values of x for which this representation is valid.
Part of the reason for doing this is that a power series looks like a polynomial
(except that it has in…nitely many terms). Polynomials are among the easiest
functions to work with. They are easy to di¤erentiate, 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 di¢ culties to resolve, we will address those as we develop the
techniques.
First, we will look at techniques which will allow us to obtain a series representation by using known series representations. The techniques involved are
substitution, di¤erentiation 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, di¤erentiation and integration)
in details, using examples. At this point, we only know the following series
representation:
1
1
x
=
=
1 + x + x2 + x3 + ::: in ( 1; 1)
1
X
xn
n=0
The reader will have recognized a geometric series.
When …nding the power series of a function, you must …nd both the series
representation and when this representation is valid (its domain).
6.7.2
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 substitution to the domain of the known series.
344
CHAPTER 6. INFINITE SEQUENCES AND SERIES
1
centered
1+x
Example 6.7.1 Find a power series representation for f (x) =
at 0 and its domain.
We use the representation of
1
1+x
1
=
, replacing x by
1
x
1
1
( x)
x. We obtain:
2
=
3
1 + ( x) + ( x) + ( x) + :::
1 x + x2 x3 + :::
1
X
n
=
( 1) xn
=
n=0
1
was valid if jxj < 1. If we apply the same
1 x
substitution, we see that this representation is valid if j xj < 1 or jxj < 1. In
other words, the interval of convergence is also ( 1; 1).
The series representation for
Example 6.7.2 Find a power series representation for f (x) =
at 0.
Proceeding as above, replacing x by x2 and remembering that
we have:
1
1
x2
=
=
1
X
n=0
1
X
x2
1
x2
1
1
1
x
centered
=
1
X
xn ,
n=0
n
x2n
n=0
=
1 + x2 + x4 + :::
This is a geometric series which converges when x2 < 1 that is when x2 < 1
or 1 < x < 1.
Example 6.7.3 Find a power series representation for
…nd its domain.
1
centered at 0 and
2+x
1
1 1
=
. Now, we …nd a power series representation
2+x
21+ x
2
1
1
1
1
for
x , then we will multiply it by 2 .
x can be obtained from 1 x by
1+
1+
2
2
First, we rewrite
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
replacing x by
345
1
X
x
1
. Since
=
xn , we have:
2
1 x n=0
1
1
X
=
x
1+
2
n=0
1
X
=
n
x
2
( 1)
x
2
n
n=0
n
Therefore,
1
2+x
=
=
1
1X
n x
( 1)
2 n=0
2
1
X
( 1)
n
n=0
n
xn
2n+1
x
1
converges when jxj < 1, so this series converges when
< 1 =) jxj <
1 x
2
2. So, the interval of convergence is ( 2; 2), the radius of convergence is 2.
Remark 6.7.4 In the …rst of these two examples, we applied the substitution to
1
the expanded form of
. In the second, we applied it to the compact form.
1 x
You can do it either way, it is simply a matter of choice.
x2
.
2+x
2
We note that this is our previous example multiplied by x hence
Example 6.7.5 Find a power series representation for
x2
2+x
=
x2
1
X
( 1)
n
n=0
=
1
X
( 1)
n
n=0
xn
2n+1
xn+2
2n+1
Example 6.7.6 Suppose that you are given that a series representation for ex
1
X
xn
2
is ex =
in ( 1; 1). What is a series representation for ex and what
n!
n=0
is its domain?
2
We go from ex to ex using the substitution x ! x2 . Thus
e
x2
n
1
X
x2
=
n!
n=0
=
also valid in ( 1; 1)
1
X
x2n
n!
n=0
346
CHAPTER 6. INFINITE SEQUENCES AND SERIES
6.7.3
Di¤erentiation and Integration of Power Series
The driving force behind the integration and di¤erentiation techniques is the
theorem below which we state without proof.
Theorem 6.7.7 If the power series
1
X
n
cn (x
n=0
a) has a radius of convergence
1
X
R > 0 then the function de…ned by f (x) =
n
cn (x
a) = c0 + c1 (x
a) +
n=0
2
c2 (x
a) + ::: is di¤ erentiable (hence) continuous on (a
0
2
1. f (x) = c1 + 2c2 (x a) + 3c3 (x
be di¤ erentiated term by term.
2.
R; a + R) and
a) + ::: In other words, the series can
2
R
3
(x a)
(x a)
+ c2
+ ::: In other words,
f (x) dx = C + c0 (x a) + c1
2
3
the series can be integrated term by term.
3. Note that whether we di¤ erentiate or integrate, the radius of convergence
is preserved. However, convergence at the endpoints must be investigated
every time.
Remark 6.7.8 This theorem simply says that the sum rule for derivatives and
integrals also applies to power series. Remember that a power series is a sum,
but it is an in…nite sums. So, in general, the results we know for …nite sums do
not apply to in…nite sums. The theorem above says that it does in the case of
in…nite series.
Remark 6.7.9 The formula in part 1 of the theorem is obtained simply by
di¤ erentiating the series term by term. Since
f (x) = c0 + c1 (x
2
a) + c2 (x
a) + :::
then
f 0 (x)
=
c0 + c1 (x
=
(c0 ) + (c1 (x
=
0 + c1 + 2c2 (x
2
a) + c2 (x
0
a) + :::
0
a)) + c2 (x
a) + 3c3 (x
a)
2
0
0
+ :::
2
a) + :::
Alternatively, one can also di¤ erentiate the general formula. Since
f (x) =
1
X
n=0
cn (x
n
a)
(6.4)
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
347
then
0
f (x)
1
X
=
=
=
cn (x
n=0
1
X
n=0
1
X
n
a)
!0
n 0
(cn (x
a) ) by the theorem
ncn (x
a)
n 1
n=0
The …rst term of this series (when n = 0) is 0, thus we can start summation at
n = 1. Hence, we have
1
X
f 0 (x) =
n 1
ncn (x
a)
(6.5)
n=1
The reader should check that formulas 6.4 and 6.5 are identical.
Remark 6.7.10 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
X
cn (x
a)
n
n=0
then
Z
f (x) dx
=
=
Z
1
X
n
cn (x
a)
n=0
1 Z
X
(cn (x
!
dx
n
a) ) dx by the theorem
n=0
n+1
Since an antiderivative of cn (x
Z
n
a) is C +
f (x) dx = C +
cn (x a)
n+1
, we have
1
n+1
X
cn (x a)
n+1
n=0
If we expand this, we get
Z
f (x) dx = C + c0 (x
a) + c1
2
(x
a)
2
+ c2
Which is the formula which appears in the theorem.
3
(x
a)
3
+ :::
348
CHAPTER 6. INFINITE SEQUENCES AND SERIES
Example 6.7.11 Given that a power series representation for f (x) is
f (x)
=
=
1 + x + x2 + x3 + :::
1
X
xn
n=0
R
…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 di¤ erentiate term by term. Thus,
f 0 (x)
=
0 + 1 + 2x + 3x2 + :::
=
1 + 2x + 3x2 + :::
Note that we can also obtain the same result by using the general formula.
In this case,
1
X
f (x) =
xn
n=0
Thus
f 0 (x)
=
=
=
1
X
n=0
1
X
n=0
1
X
0
(xn )
nxn
1
nxn
1
n=1
Which gives us the same answer.
R
Next, we …nd f (x) dx. From the theorem, it is enough to integrate term
by term. Thus since f (x) = 1 + x + x2 + x3 + :::, we have:
Z
x3
x4
x2
+
+
+ :::
f (x) dx = C + x +
2
3
4
Alternatively, we can work from the general formula. Since f (x) =
1
X
n=0
we have
Z
f (x) dx
=
(xn ) dx
n=0
=
Which is the same formula.
1 Z
X
C+
1
X
xn+1
n+1
n=0
xn ,
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
349
Finding a Series Representation Using Di¤erentiation
This time, we …nd the series representation of a given series by di¤erentiating
the power series of a known function. Suppose that we want to …nd the series
representation of a function g (x). If we can …nd a function f (x) such that
f 0 (x) = g (x), and if we have a series representation for f then the series representation for g is obtained by di¤erentiating 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.
Remark 6.7.12 It might not always be easy to know whichRfunction f satis…es
f 0 (x) = g (x). However, we know a formula for it. f (x) = g (x) dx.
Example 6.7.13 Find a series representation for
convergence.
0
1
We begin by noting that
1
=
x
1
(1
2.
x)
1
(1
Since
2,
x)
1
1
x
…nd the interval of
= 1+x+x2 +x3 +::::,
we have:
1
(1
x)
1 + x + x2 + x3 + :::
=
2
0
1 + 2x + 3x2 + 4x3 + ::::
1
X
=
nxn 1
=
n=1
The radius of convergence is still 1. Since
1
1
1
converges for x in ( 1; 1), we
x
2 will also converge in ( 1; 1). It might also converge at the
(1 x)
endpoints, so we need to check them. Do it as an exercise.
know that
Remark 6.7.14 If you prefer to work from the compact form of the series, it
can be done.
1
1
x
1
(1
x)
2
=
1
X
xn
n=0
=
=
=
=
0
1
1
x
1
X
x
n
n=0
1
X
(xn )
n=0
1
X
n=1
nxn
!0
0
1
350
CHAPTER 6. INFINITE SEQUENCES AND SERIES
However, you have to be careful with the starting value of n.
Example 6.7.15 Suppose you know that a series representation for sin x is
x3
x5
x7
+
+ :::
3!
5!
7!
1
2n+1
X
n x
( 1)
(2n + 1)!
n=0
sin x =
x
=
Find a power series representation for cos x.
cos x =
=
=
=
(sin x)
0
x5
x7
x3
+
+ :::
x
3!
5!
7!
x2
x4
x6
1
+
+ :::
2!
4!
6!
1
2n
X
n x
( 1)
(2n)!
n=0
0
2
Example 6.7.16 Find a power series representation for 2xex given that a
1
X
xn
series representation for ex is ex =
in ( 1; 1).
n!
n=0
We can do this problem two ways.
Method 1 Using substitution, we can …nd a power series representation for
2
ex , then multiply what we …nd by 2x. The …rst part was done earlier,
1
X
x2n
2
and we found that ex =
on ( 1; 1). Thus,
n!
n=0
2xex
2
1
X
2xx2n
n!
n=0
=
1
X
2x2n+1
=
n!
n=0
also in ( 1; 1).
2
Method 2 We note that 2xex = ex
2
0
. So, using substitution, we can …nd
2
a power series representation for ex , then di¤ erentiate it to get a series
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
351
2
representation for 2xex .
2xex
2
=
=
ex
2
0
1
X
x2n
n!
n=0
!0
1
X
2nx2n
=
n!
n=0
=
=
=
1
X
2nx2n
n!
n=1
1
1
when n = 0,
2nx2n
n!
1
=0
1
X
2x2n 1
(n 1)!
n=1
1
X
2x2n+1
n!
n=0
So, either way, we …nd the same answer.
Finding a Series Representation Using Integration
This time we …nd the power series representation of a function by integrating
the power series representation of a known function. Suppose that we are given
g (x) and we want a series
R representation for g (x). If we can …nd a function
f (x) such that 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 .
R
Remark 6.7.17 The function f (x) such that g (x) = f (x) dx is f (x) =
g 0 (x).
Example 6.7.18 Find a power series representation for ln (1 x).
R dx
We know that ln (1 x) =
. By our earlier work, we found that
1 x
1
= 1 + x + x2 + x3 + ::: in ( 1; 1), so
1 x
ln (1
x) = C + x +
x2
x3
+
+ :::
2
3
We also know that ln 1 = 0, From the above equation, we get that C = 0 by
letting x = 0. Therefore
ln (1
x)
=
=
x2
x3
+
+ :::
2
3
1
X xn+1
n+1
n=0
x+
352
CHAPTER 6. INFINITE SEQUENCES AND SERIES
Therefore,
ln (1
1
X
xn+1
n+1
n=0
x) =
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 6.7.19 Find a power series representation for tan 1 x.
R dx
1
We use the fact that tan 1 x =
. First, since
= 1+x+x2 +x3 +:::,
1 + x2
1 x
1
= 1 x2 + x4 x6 + ::: and therefore
we see that
1 + x2
1
tan
Using the fact that tan
1
6.7.4
x7
+ :::
7
0 = 0 gives us C = 0. Thus
1
tan
x3
x5
+
3
5
x=C +x
x
x3
x5
x7
+
+ :::
3
5
7
1
2n+1
X
n x
=
( 1)
2n + 1
n=0
= x
Things to Know
Be able to …nd the series representation of a function by substitution,
integration or di¤erentiation.
Remember the power series representations we found in this section and
when the representation is valid. So far, we know:
1
1
x
=
1
X
n=0
xn when jxj < 1
1
X
1
n
=
( 1) xn when jxj < 1
1 + x n=0
1
X
1
n
=
( 1) x2n when jxj < 1
1 + x2
n=0
1
(1
ln (1
tan
1
2
x)
=
nxn
n=1
x) =
x=
1
X
1
X
n=0
1
when jxj < 1
1
X
xn+1
when jxj < 1
n+1
n=0
( 1)
n
x2n+1
when jxj < 1
2n + 1
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
6.7.5
353
Problems
1. If the radius of convergence of the power series
1
X
radius of convergence of
1
X
cn xn is 10, what is the
n=0
cn nxn
1
? Why?
n=1
2. Find a power series representation for the functions below using substitution and determine the interval of convergence.
(a) f (x) =
1
5
x
1
(b) f (x) =
4 + x2
x
(c) f (x) =
9 + x2
1
(d) f (x) =
1 + 9x2
3. Consider f (x) =
1
2.
(1 + x)
(a) Use di¤erentiation to …nd a power series representation for f . What
is the radius of convergence?
1
(b) Use part a) to …nd a power series representation of g (x) =
3.
(1 + x)
What is the radius of convergence?
(c) Use part b) to …nd a power series representation of g (x) =
x2
(1 + x)
3.
What is the radius of convergence?
4. Consider f (x) = ln (1 + x).
(a) Find a power series representation for f (x). What is the radius of
convergence.
(b) Use part a) to …nd a power series representation of x ln (1 + x).
(c) Use part a) to …nd a power series representation of ln 1 + x2 . What
is the radius of convergence.
5. Find a power series representation for the functions below and determine
the interval of convergence.
R
(a) tan 1 x2 dx
R ln (1 x)
(b)
dx
x
354
CHAPTER 6. INFINITE SEQUENCES AND SERIES
1
X
xn
6. Let f (x) =
. Find the intervals of convergence for f , f 0 and f 00 .
n2
n=0
7. Consider the function f (x) =
1
X
xn
.
n!
n=0
(a) Find its interval of convergence.
(b) Find a power series representation for f 0 (x) and its interval of convergence.
(c) What is the relationship between theseries representation of f (x)
and that of f 0 (x)? What famous function satis…es this relationship?
6.7.6
Answers
1. If the radius of convergence of the power series
radius of convergence of
1
X
1
X
cn xn is 10, what is the
n=0
cn nx
n 1
? Why?
n=1
It is also 10.
2. Find a power series representation for the functions below using substitution and determine the interval of convergence.
(a) f (x) =
1
5
x
1
5
(b) f (x) =
x
=
1
X
xn
when jxj < 5
5n+1
n=0
1
4 + x2
1
2n
X
1
n x
=
(
1)
when jxj < 2
4 + x2
4n+1
n=0
(c) f (x) =
x
9 + x2
1
2n+1
X
x
n x
=
(
1)
when jxj < 3
9 + x2
9n+1
n=0
(d) f (x) =
1
1 + 9x2
1
X
1
1
n
=
( 1) 9n x2n when jxj <
1 + 9x2
3
n=0
6.7. REPRESENTATION OF FUNCTIONS AS POWER SERIES
3. Consider f (x) =
1
355
2.
(1 + x)
(a) Use di¤erentiation to …nd a power series representation for f . What
is the radius of convergence?
1
(1 + x)
2
=
1
X
n 1
nxn
( 1)
1
n=1
when jxj < 1
(b) Use part a) to …nd a power series representation of g (x) =
1
(1 + x)
3.
What is the radius of convergence?
1
(1 + x)
3
=
1
X
n
( 1)
n=2
n (n 1) n
x
2
2
when jxj < 1
(c) Use part b) to …nd a power series representation of g (x) =
x2
(1 + x)
3.
What is the radius of convergence?
x2
(1 + x)
3
=
1
X
( 1)
n=2
n
n (n 1) n
x when jxj < 1
2
4. Consider f (x) = ln (1 + x).
(a) Find a power series representation for f (x). What is the radius of
convergence.
ln (x + 1) =
1
X
n
( 1)
n=0
xn+1
when jxj < 1
n+1
(b) Use part a) to …nd a power series representation of x ln (1 + x).
x ln (x + 1) =
1
X
n
( 1)
n=0
xn+2
when jxj < 1
n+1
(c) Use part a) to …nd a power series representation of ln 1 + x2 . What
is the radius of convergence.
ln 1 + x2 =
1
X
n=0
( 1)
n
x2n+2
when jxj < 1
n+1
5. Find a power series representation for the functions below and determine
the interval of convergence.
356
CHAPTER 6. INFINITE SEQUENCES AND SERIES
(a)
(b)
R
tan
Z
1
tan
x2 dx
1
x2 dx = C +
1
X
n
( 1)
n=0
R ln (1 x)
dx
x
Z
ln (1 x)
dx = C
x
1
X
x4n+3
when jxj < 1
(2n + 1) (4n + 3)
n=0
xn+1
(n + 1)
2
when jxj < 1
1
X
xn
. Find the intervals of convergence for f , f 0 and f 00 .
2
n
n=0
The interval of convergence of f is [ 1; 1], the one for f 0 is [ 1; 1) and the
one for f 00 is ( 1; 1).
6. Let f (x) =
7. Consider the function f (x) =
1
X
xn
.
n!
n=0
(a) Find its interval of convergence.
( 1; 1)
(b) Find a power series representation for f 0 (x) and its interval of convergence.
1
X
xn
f 0 (x) =
in ( 1; 1)
n!
n=0
(c) What is the relationship between the series representation of f (x)
and that of f 0 (x)? What famous function satis…es this relationship?
Tell your teacher what your answer is.

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