Power Series
Lecture Notes
A power series is a polynomial with infinitely many terms. Here is an example:
0 aB b œ "  B  B #  B $  â
Like a polynomial, a power series is a function of B. That is, we can substitute in different
values of B to get different results. For example,
0 a!b œ "  !  !  !  â œ ".
and
"
"
"
"
0Œ  œ " 


 â œ #.
#
#
%
)
Though power series may seem complicated, they are actually not much more difficult to deal
with than polynomials. For example, it is easy to take the derivative of a power series:
.
’"  B  B#  B$  B%  â“ œ "  #B  $B#  %B$  â
.B
and it is just as easy to take the integral:
#
$
( ˆ"  B  B  B  â‰.B œ ŒB 
" #
"
"
B  B $  B %  â  G Þ
#
$
%
As you can see, a power series is not much more complicated than a polynomial.
Taylor Series
Consider again the power series
0 a B b œ "  B  B #  B $  â.
As we have seen, it is easy to compute the sum of this series for different values of B. For
example,
"
"
"
"
"
$
0Œ  œ " 


â œ
œ .
$
$
*
#(
"  "Î$
#
In fact, no matter what value of B we plug in, the result is always a geometric series. The first
term of this series is ", and the common ratio is always B.
Indeed, the entire power series
"  B  B#  B$  â
can be thought of as a geometric series with a common ratio of B. This gives us a simple formula
for the sum:
"  B  B#  B$  â œ
"
"B
This is our first example of a Taylor series—a power series that adds up to a known function.
So which functions can be expressed as power series? The answer may surprise you:
/B œ "  B 
" #
"
"
B  B$  B%  â
#
$x
%x
cos B œ " 
" #
"
"
B  B%  B'  â
#x
%x
'x
sin B œ B 
" $
"
"
B  B&  B(  â
$x
&x
(x
tan" B œ B 
" $
"
"
B  B&  B(  â
$
&
(
lna"  Bb œ B 
" #
"
"
B  B$  B%  â
#
$
%
All of these functions—the exponential, the sine, the cosine, the logarithm, the inverse tangent—
all of them are really just polynomials!
This result seems to good to be true. How is it possible that something like /B is a
polynomial? According to the table above,
/B œ "  B 
" #
"
" %
" &
B  B$ 
B 
B  â.
#
'
#%
"#!
Why would this be true? Well, the defining property of /B is that it is equal to its own derivative:
. B
a / b œ /B .
.B
However, the Taylor series for /B is also equal to its own derivative:
.
" #
" $
" %
" &
B 
B  â
Œ"  B  B  B 
.B
#
'
#%
"#!
œ "B
" #
"
" %
B  B$ 
B  â.
#
'
#%
Based on this observation alone, it seems reasonable that the two functions might be the same.
Unfortunately, we are not yet in a position to fully explain the formulas above. The goal of
the next two sections is to learn how to express various functions as power series. By the time
we are done, you will understand all five of these formulas.
Geometric Power Series
Recall the formula for the sum of a geometric series:
+  +<  +<#  +<$  â œ
+
"<
As we have seen, this formula works perfectly well when + and < are functions of B. For
example, plugging in + œ " and < œ B gives the formula
"
œ "  B  B #  B $  â.
"B
There are many more series we can get this way. For example, using + œ " and < œ B gives
"
œ "  B  B #  B $  â,
"B
and using + œ B and < œ #B gives,
B
œ B  #B#  %B$  )B%  â
"  #B
EXAMPLE 1 Find a formula for the sum of the following series:
B#  #B&  %B)  )B""  "'B"%  â
SOLUTION
This is a geometric series with a common ratio of #B$ . The first term is B# , so
B#  #B&  %B)  )B""  "'B"%  â œ
B#
"  #B$
è
EXAMPLE 2 Find a power series representation for each of the following functions:
(a)
B
"  $B
(b)
"
"  B#
SOLUTION
(a) This should be the sum of a geometric series with + œ B and < œ $B:
B
œ B  $B#  *B$  #(B%  â
"  $B
(b) This is the sum of a geometric series with + œ " and < œ B# :
"
"  B#
œ "  B#  B%  B'  â
è
Differentiation and Integration
You can differentiate an integrate power series term-by-term, just as you would a polynomial:
.
’-!  -" B  -# B#  -$ B$  â“ œ -"  #-# B  $-$ B#  â
.B
#
( ˆ-!  -" B  -# B  â‰.B œ Š-! B 
EXAMPLE 3 Find a power series representation for
SOLUTION
-" #
-#
B  B $  â‹  G
#
$
"
.
a"  B b#
Observe that:
"
.
"
# œ .B ” "  B •
a"  B b
But:
"
œ "  B  B#  B$  â
"B
Therefore:
"
a"  B b#
œ "  #B  $B#  %B$  â
We now come to our first major task: finding a power series for the natural logarithm.
è
EXAMPLE 4 Find a Taylor series for lna"  Bb.
SOLUTION
Observe that
(
But
"
.B œ lna"  Bb  G .
"B
"
is the sum of a geometric power series:
"B
"
œ "  B  B#  B$  â
"B
Integrating both sides gives:
"
"
"
lna"  Bb œ ŒB  B#  B$  B%  â  G
#
$
%
Plugging in B œ ! reveals that G œ !. Therefore:
_
" #
" $
" %
a"b8" 8
"
lna"  Bb œ B  B  B  B  â œ
B
#
$
%
8
8œ"
è
The formula we just derived is our first really important result:
lna"  Bb œ B 
" #
"
"
B  B$  B%  â
#
$
%
We can find a power series for the inverse tangent using the same method:
EXAMPLE 5 Find a Taylor series for tan" B.
SOLUTION
Recall that
(
But
"
.B œ tan" B  G .
#
"B
"
is the sum of a geometric series with + œ " and < œ B# :
"  B#
"
"  B#
œ "  B #  B %  B '  â.
Integrating both sides gives:
"
"
"
tan" B œ ŒB  B$  B&  B(  â  G
$
&
(
Plugging in B œ ! reveals that G œ !, so:
tan" B œ B 
" $
"
"
B  B&  B(  â
$
&
(
è
The result was:
" $
"
"
B  B&  B(  â
$
&
(
tan" B œ B 
EXAMPLE 6 Express the integral:
"
(
"
.B
"  B(
!
as the sum of an infinite series.
Note that this integral would be very difficult to evaluate on its own. However, the
integrand is the sum of a geometric series:
SOLUTION
"
"  B(
œ "  B(  B"%  B#"  â
Therefore:
(
!
"
"
"
ˆ"  B(  B"%  B#"  ≠.B
.B
œ
(
"  B(
!
"
"
"
"
œ ”B  B)  B"&  B## •
)
"&
##
!
œ "
"
"
"
"



â
)
"&
##
#*
è
Substitution and Multiplication
There are two more important tricks for working with power series. The first is substitution:
EXAMPLE 7 Find a power series representation for tan" aB# b.
SOLUTION
The power series for tan" B is:
tan" B œ B 
" $
"
"
B  B&  B(  â
$
&
(
All we need to do is substitute B# in for B:
tan" ˆB# ‰ œ ˆB# ‰ 
œ B# 
" # $
"
"
ˆB ‰  ˆB# ‰&  ˆB# ‰(  â
$
&
(
" '
"
"
B  B"!  B"%  â
$
&
(
è
Though this method is very simple, it often comes off as confusing because of the two
different B's. The idea here is that:
tan" ? œ ? 
" $
"
"
?  ?&  ?(  â
$
&
(
for any ?. All we are doing is substituting in ? œ B# .
You can think of many geometric series this way. For example, the series:
"
œ "  B(  B"%  B#"  â
"  B(
can be obtained by substituting B( into the power series for
"
.
"B
EXAMPLE 8 Find a power series representation for lna"  B& b.
SOLUTION
We know that:
lna"  Bb œ B 
" #
"
"
B  B$  B%  â
#
$
%
Substituting in B& for B yields:
lnˆ"  B& ‰ œ B& 
" "!
"
"
B  B"&  B#!  â
#
$
%
è
Power series can also be added, subtracted, and multiplied like polynomials.
EXAMPLE 9 Find a power series representation for B$ tan" B.
SOLUTION
We have:
B$ tan" B œ B$ ŒB 
œ B% 
" $
"
"
B  B &  B (  â
$
&
(
" '
"
"
B  B)  B"!  â
$
&
(
è
Summations for Power Series
It is sometimes difficult to express a power series in summation notation. We give a few
examples.
EXAMPLE 10 Express the series:
B$ 
" (
"
"
"
B  B""  B"&  B"*  â
#
$
%
&
using summation notation.
For this series, it seems easiest to have the first term be 8 œ ", the second term be
8 œ #, and so on:
SOLUTION
8
"
#
$
%
&
â
8th term
B$
" (
B
#
" ""
B
$
" "&
B
%
" "*
B
&
â
"
. The power of B is increasing by % each time, so it
8
should be similar to %8. Indeed, it looks like the power of B is %8  ", so:
As you can see, the coefficient is always
B$ 
_
" (
"
"
"
"
B  B""  B"&  B"*  â œ " B%8"
#
$
%
&
8
8œ"
è
EXAMPLE 11 Express the series:
tan" B œ B 
" $
"
"
B  B&  B(  â
$
&
(
using summation notation.
SOLUTION
This time we start with 8 œ !:
8
!
"
#
$
â
8th term
B
"
 B$
$
" &
B
&
"
 B(
(
â
The alternating  and  signs can be taken care of with a a"b8 . It is important here that the
even-numbered terms are positive (which is why we decided to start at 8 œ !). If the oddnumbered terms were positive, we would need a a"b8" .
The power of B is an arithmetic sequence that increases by # each time. In particular, the
formula for the power is, so it should be similar to #8  ", so
tan" B œ "
_
8œ!
a"b8 #8"
B
#8  "
è
The following table shows the summation notation for each of our five primary series:
œ "
_
/
B
8œ!
cos B œ "
_
8œ!
sin B œ "
_
8œ!
tan" B œ "
_
8œ!
lna"  Bb œ "
_
8œ"
" 8
"
"
B œ "  B  B#  B$  â
8x
#
$x
a"b8 #8
"
"
"
B œ "  B#  B%  B'  â
a#8bx
#
%x
'x
a"b8 #8"
"
"
B
œ B  B$  B&  â
a#8  "bx
$x
&x
a"b8 #8"
"
"
"
B
œ B  B$  B&  B(  â
#8  "
$
&
(
a"b8" 8
"
"
"
B œ B  B#  B$  B%  â
8
#
$
%
EXERCISES
1–4 ç Find a power series representation for the function using
the formula for the sum of a geometric series.
1. 0 aBb œ
3. 0 aBb œ
B
"B
2. 0 aBb œ
"
"  %B#
4. 0 aBb œ
21. (
!
B#
"  B%
$
%B
"  B&
5–16 ç Find a power series representation for the given function.
1
lna"  Bb
.B
B
22. (
"
"
"
.B
"  B"!
23. Use power series to estimate the integral (
!Þ&
!
"
.B to
"  B'
within !Þ!!!!".
24. Use power series to estimate the integral (
!Þ#
#
/B .B to
!
5. 0 aBb œ lna"  Bb
6. 0 aBb œ tan" ˆB& ‰
within !Þ!!!!".
7. 0 aBb œ lna&  Bb
8. 0 aBb œ lna"  #Bb
25–30 ç Express the given series using summation notation.
9. 0 aBb œ /$B
11. 0 aBb œ cosˆB# ‰
10. 0 aBb œ sin #B
12. 0 aBb œ /B
25. B 
#
"
"
" #
" $

B
B 
B â
"†#
#†$
$†%
%†&
13. 0 aBb œ B# /B
14. 0 aBb œ B tan" a$Bb
26.
15. 0 aBb œ
16. 0 aBb œ
27. %B 
tan" B
B
" $
" &
" (
B 
B 
B â
È$
È&
È(
sin B  B
B$
) $
"'
$#
'%
B  # B&  # B(  # B*  â
##
$
%
&
17–22 ç Express the integral as an infinite series.
28.
%$ #
%% %
%& '
%' )
B 
B 
B 
B â
&
(
*
""
17. (
"
.B
"  B%
18. (
29.
" $
" (
"
"
B 
B 
B"" 
B"&  â
$
$x † (
&x † ""
(x † "&
19. (
B
20. (
!
sin >
.>
>
!
B
.B
"  B&
B
lnˆ"  ># ‰ .>
30. B# 
" )
"
"
B 
B"% 
B#!  â
#x † (
%x † "$
'x † "*