-
P A R T
XSeries
30C H A P T E R
Series
30.1 APPROXIMATING A FUNCTION BY A POLYNOMIAL
Preview
Addition and multiplicationthese are our fundamental
computational tools. A high-
powered computer, for all its computational sophistication,
ultimately relies on these basic
operations. How then can a computer numerically approximate
values of transcendental
functions? How are values of exponential, logarithmic, and
trigonometric functions com-
puted?
Consider the sine function, for example. A calculator can
approximate sin 0.1 with a
high degree of accuracy, accuracy not readily accessible from
unit circle or right triangle
denitions of sin . How can such a good approximation be
obtained?
If we know the value of a differentiable function at the point ,
then we can usethe tangent line to at to approximate the functions
values near . The tangentline is the best linear approximation of
near ; higher degree polynomials offer thepossibility of staying
even closer to the values of near and following the shapeof over a
larger interval around . In this section we will improve upon the
tangent line
approximation, obtaining quadratic, cubic, and higher degree
polynomial approximations of
around . We will generally nd the t improving with the degree of
the polynomial.919
-
920 CHAPTER 30 Series
Such polynomial approximations are convenient because they
involve only the operations
of addition and multiplication; they are easily evaluated,
easily differentiated, and easily
integrated.
The process of approximating an elusive quantity, successively
rening the approxi-
mation, and using a limiting process to nail it down is at the
heart of theoretical calculus. In
this chapter we obtain successively better polynomial
approximations of a function about a
point by computing increasingly higher degree polynomial
approximations. By computing
the limit as the degree of the polynomial increases without
bound, we will discover that,
under certain conditions, we can represent a function as an
innite polynomial known
as a power series. The fact that sin , cos , and have
representations as power series is
remarkable in its own right. In addition, this alternative
representation turns out to be com-
putationally very useful. Power series representation of
functions was known to Newton
who used it as a computational aid, particularly for integrating
functions lacking elemen-
tary antiderivatives. It was the subject of work published by
the English mathematician
Brook Taylor in 1712 and was popularized by the Scottish
mathematician Colin Maclaurin
in a textbook published in 1742. Although mathematicians had
been using the ideas as early
as the 1660s, the names of Taylor and Maclaurin have been
associated with power series
representations of functions.
Polynomial Approximations of sin around 0In this section we will
use polynomials to numerically estimate values of some
transcen-
dental functions.
EXAMPLE 30.1 A calculator or computer gives sin 0.1 to ten
decimal places, displaying 0.0998334166.Obtain this result by using
a polynomial to approximate sin near 0 and evaluatingthis
polynomial at 0.1.
SOLUTION We will approach this problem via a sequence of
polynomial approximations to sin for near zero until we arrive at
the desired result. We denote by the th degree
polynomial approximation. is of the form 0 1 22 , where0, 1, ,
are constants. We must determine the values of these constants so
the
ts the graph of well around 0.
Constant Approximation
Because sin is continuous and 0.1 is near 0, we know sin 0.1 sin
0 0.
0 0; sin 0.1 00.1 0
Tangent Line Approximation
The tangent line passes through 0, 0 and has a slope of 0 cos 0
1.
1 ; sin 0.1 10.1 0.1
-
30.1 Approximating a Function by a Polynomial 921
y
y
x
x
y = x
y = x
f(x) = sin x
f(x) = sin x
2 211
1
.1
.1
1
(a)
(b)
sin .1
Figure 30.1
Rening the tangent line approximation: Any polynomial
approximation, , of sin
about 0 certainly ought to be as good a local approximation as
is the tangent lineapproximation, 1 . Therefore, like sin the graph
of must pass through 0, 0and must have a slope of 1 at 0. This
means that
0 0 and 0 1.
0 1 22 so 0 0 0. 1 22 332 1 so 0 1 1.
Therefore, is of the form 22 33 .Note that sin lies below the
tangent line for 0 and above the tangent line for 0.
Therefore the approximation sin must be decreased for 0 and
increased for 0in order to improve upon it.
Second Degree Approximation
2 must be of the form 22, where 2 is a constant. But 22 cannot
be negativefor 0 and positive for 0, as required; we cannot improve
upon the tangent line
approximation by using a second degree polynomial. We need at
least a third degree
polynomial to improve upon the tangent line approximation.
-
922 CHAPTER 30 Series
Before moving on, lets look at the second degree polynomial
approximation from a
geometric viewpoint. If 2 0 1 22 is to be the best parabolic
approximationto sin about 0, then it must satisfy the following
three conditions.
It has the same value as sin at 0. 20 0It has the same slope as
sin at 0. 20 0It has the same concavity as sin at 0. 2 0 0
Each one of these conditions determines the value of one
coefcient of 2. The rst two
result in 0 0 and 1 1, respectively. The second derivative of
sin at 0 is zero.2
2sin
0
cos
0 sin
0 0
The parabola must have a second derivative of zero;
consequently, it is not a parabola at
all.
Third Degree Approximation
To determine the coefcients of the third degree polynomial of
best t, we require that the
polynomial, 3 0 1 22 33, and sin agree at 0 and thateach nonzero
derivative of the polynomial is equal to the corresponding
derivative of sin
at 0. These four conditions determine the four coefcients.
30 0 30 0 3 0 0 3 0 0
We have already demonstrated that the rst two conditions result
in 0 0, 1 1. Asan exercise, show that the third and fourth
conditions require that 2 0 and 3 16 ,respectively.
3 0 02 1
63 1
63.
Notice that the 163 term is negative for 0 and positive for 0,
providing an
appropriate adjustment to the tangent line approximation. (See
Figure 30.2.)
y
x22
P1(x) = x
P3(x) = x
f(x) = sin x
x3
6
Figure 30.2
EXERCISE 30.1 Let sin and 3 0 1 22 33. Calculate , , , , , , and
evaluate each at 0. Show that the four conditions given above
-
30.1 Approximating a Function by a Polynomial 923
determine 0, 1, 2, and 3, respectively and that
0 0, 1 1, 2 0, and 3 1
6.
Using the third degree polynomial to approximate sin 0.1
gives
sin 0.1 30.1 0.10.13
6 1
10 1
6000 0.09983.
This matches the calculator estimate of sin 0.1 to six decimal
places. The actual value is a
bit larger than 30.1.
Higher Degree Approximations
To nd the th degree polynomial approximation we require that and
sin agree at
0 and each nonzero derivative of the polynomial matches that of
sin .The condition that the fourth derivatives agree ends up
meaning that 4 0, so we will
proceed directly with the fth degree polynomial.
Let 5 0 1 22 33 44 55. Requiring that all nonzero deriva-tives
of 5 match the derivatives of sin at 0 means the following
conditionsmust be satised.
50 0 50 0 5 0 0 5 0 0
45 0 40 where denotes the th derivative of
55 0 50
As an exercise, show that these conditions determine 0, 1, 2, 3,
4, and 5, respectively,
and that
5 1
63 1
1205.
EXERCISE 30.2 Let sin and 5 be the fth degree polynomial given
above. Show that the sixconditions stated mean that
0 0, 1 0, 2 0
2!, 3
03!
, 4 40
4!, 5
50
5!,
where ! 1 3 2 1. Conclude that
0 0, 1 1, 2 0, 3 13!1
6, 4 0 and 5
1
5! 1
120.
Using the fth degree polynomial approximation to sin to
approximate sin0.1 gives
sin 0.1 50.1 0.10.13
6 0.1
5
120 1
10 1
6000 1
12 106 0.0998334166.
This agrees with the 10 decimal places given for sin 0.1.
-
924 CHAPTER 30 Series
y
x22
P1(x) = x
P3(x) = x
f(x) = sin x
x3
6
P5(x) = x +x3
6
x5
120
Figure 30.3
The graphs of sin , 1, 3, and 5 are given in Figure 30.3.
EXERCISE 30.3 Using a computer or graphing calculator, graph sin
, 1 , 3 3
3!,
and 5 3
3! 5
5!. Zoom in around 0 and observe how well each polynomial
approximates the values of sin near 0. Try to guess formulas for
7, 9, and11. Graph these as well and decide how much condence you
have in your answers.
Below is a table of values given to 10 decimal places.
sin 1 3 5
0.2 0.1986693308 0.2 0.1986666667 0.19866933330.1 0.0998334166
0.1 0.0998333333 0.0998334167
0 0 0 0 0
0.1 0.0998334166 0.1 0.0998333333 0.0998334167
0.5 0.4794255386 0.5 0.4791666667 0.4794270833
1 0.8414709848 1 0.8333333333 0.8416666667
2 0.9092974268 2 0.6666666667 0.9333333333
2.5 0.5984721441 2.5 0.1041666667 0.7096354167
OBSERVATIONS From the graphical and numerical data gathered we
observe that
i. for a xed near zero, the higher the degree of the polynomial
approximation the better
its value approximates that of sin , and
ii. the higher the degree of the polynomial, the further away
from zero the approximation
is reasonable.
NOTE In all the work weve done with sin , must be in radians,
not in degrees.
sin cos only for in radians.
Taylor Polynomial Approximations
In the previous example we constructed polynomial approximations
to sin around 0 by choosing the coefcients of the polynomial such
that the polynomial and all itsnonzero derivatives matched and its
corresponding derivatives at 0. This method ofconstructing
polynomial approximations to a function about a number in its
domainis remarkably useful.
-
30.1 Approximating a Function by a Polynomial 925
Let be a function whose rst derivatives exist at . For the sake
of simplicity,we begin with the case 0.
D e f i n i t i o n
The th degree polynomial, , that is equal to 0 when evaluated at
0 andwhose rst derivatives are equal to those of when evaluated at
0 is calledthe th degree Taylor polynomial generated by at 0. The
polynomial is saidto be centered at 0, or expanded about 0.
More generally, we can expand a function about using a
polynomial in powers of .
0 1 2 2 3 3
D e f i n i t i o n
The th degree polynomial in powers of that is equal to when
evaluatedat and whose rst derivatives match those of at is called
the thdegree Taylor polynomial generated by at . We refer to as the
center ofthe polynomial.
When evaluated at its center, a Taylor polynomial is equal to
the value of its generating
function. Our hope is that for near the center the value of the
polynomial is close to the
value of the function.1
We now turn our attention to computing Taylor polynomials. In
the next section we
will look at the accuracy of Taylor polynomial approximations,
and subsequently will see
what we get by allowing the degree of the Taylor polynomial to
increase without bound.
Computing a Taylor Polynomial Centered at 0Suppose and its rst
derivatives exist at 0. We want to nd constants 0, 1, 2, , such
that
0 1 22
is the Taylor polynomial generated by about 0.We impose the
following 1 conditions; each enables us to solve for one coef-
cient.
0 0 0 0 0 0 0 0...
0 0
(30.1)
1 While this is not always the case, often we will nd it
true.
-
926 CHAPTER 30 Series
In short, 0 0 for 0, 1, , .2
We begin by nding the rst derivatives of .
0 1 22 33 44 1 22 332 443 1 2 2 3 2 3 4 3 42 1 2 3 2 3 4 3 2 4 1
2 3
4 4 3 2 4 1 2 3 4...
1 1 2 3 3 2 1 1 2 3 2
1 2 3 2
Next we evaluate each expression at 0.0 0 0 1 0 2 2 0 3 2 3 3!3
4 0 4!4
...
1 0 1!1 0 !
We summarize: 0 ! for 0, 1, , . Returning to (30.1), the
original 1
conditions, we obtain3
! 0 for 0, 1, , .
Solving for , the coefcient of , we obtain
0!
for 0, 1, , . We summa-rize our result.
The th degree Taylor polynomial generated by at 0 is given
by
0 0 0
2!2
03!
0
!.
That is,
0
0
!.
This work behind us, we compute the th degree Taylor polynomial
generated by
about 0 as follows.
2 Here we use the convention that 0 .3 Recall: 0! 1 and 0 .
-
30.1 Approximating a Function by a Polynomial 927
1. Compute the rst derivatives of .
Be alert to the possibility of patterns emerging. You improve
your chances of noticing
patterns by not multiplying out. For instance, 5 4 3 2 is easier
to recognize as 5!than is 120.
2. Evaluate and each of its derivatives at 0.3. The coefcient of
is the constant
0!
.
EXAMPLE 30.2 Find the th degree Taylor polynomial generated by
at 0.
SOLUTION
0 0 0
2!2
0
!
The derivative of is ; therefore 0 0 1 for 0, 1, 2, , . Thus
1 2
2!
3
3!
!.
Graphs of and several of its Taylor polynomials are shown in
Figure 30.4. Note that
1 1 is simply the tangent line approximation to at 0.
y
y
x
x
P4
P2
P3P4 P2
P1
P1(x)
P1(x)
P1(x) = 1 + x
P2(x) = 1 + x +
P3(x) = 1 + x +
P4(x) = 1 + x +
P2(x)
P3(x)
P4(x)
e x
P3
f(x) = ex
f(x) = ex
15
10
5
5
5
1
11 2
2 3 4
(a)
(b)
Magnification around x = 0
x2
2!
x2
2!
x3
3!
x2
2!
+
x3
3!+ x
4
4!+
Figure 30.4
On the following page is a table of values produced using Taylor
polynomials for . Values
are given to nine decimal places.
-
928 CHAPTER 30 Series
1 2 3 4 5
0.1 1.105170918 1.1 1.105 1.105166666 1.105170833
1.105170917
0.2 1.221402758 1.2 1.22 1.221333333 1.221400000 1.221402667
0.5 1.648721271 1.5 1.625 1.645833333 1.6484375 1.648697917
1 2.718281828 2 2.5 2.66666666 2.708333333 2.716666666
EXAMPLE 30.3 Find the 8th degree Taylor polynomial generated by
cos about 0.SOLUTION 8 0 0
02!
2 808!
8
cos 0 1 sin 0 0 cos 0 1 sin 0 0 4 cos 40 1 5 sin 50 0 6 cos 60 1
7 sin 70 0 8 cos 80 1
8 12
2!
4
4!
6
6!
8
8!
Notice that the coefcients of all the odd power terms are zero.
This makes sense; cosine is
an even function. Analogously, the coefcients of all even power
terms in the expansion of
sin about 0 are zero since sin is an odd function. The graph of
cos and some of its Taylor polynomials centered at 0 are given
in
Figure 30.5. (Graph them yourself and you can zoom in around
0.)
P4(x)f(x) = cos x
P2(x) = 1
P4(x) = 1
P6(x) = 1
P2(x) P6(x)
f(x) = cos x
P8(x)x2
2!
x2
2!
x4
4!
x2
2!
+
x4
4!+ x
6
6!
P8(x) = 1 x2
2!x4
4!+ x
6
6!+ x
8
8!+
22
y
x
Figure 30.5
Computing a Taylor Polynomial Centered at Suppose we want to
approximate ln 1.2 using a Taylor polynomial. We cant use a
Taylor
polynomial for ln expanded about 0 because neither nor any of
its deriva-tives exist at 0. We can, however, either center the
Taylor polynomial for ln1 at 0 or work with the Taylor polynomial
for ln expanded about 1. We will do thelatter. First we will look
at how to compute a Taylor polynomial centered at .
Recall that the th degree Taylor polynomial for at is an th
degreepolynomial in powers of ,
0 1 2 2 ,
-
30.1 Approximating a Function by a Polynomial 929
such that the values of and its nonzero derivatives are equal to
those of
when evaluated at . That is, the coefcients 0, 1, 2, , are
determined by theconditions
for 0, 1, 2, , . (30.2)
As an exercise, calculate the rst derivatives of . (Dont
multiply out ; usethe Chain Rule.) Evaluating these derivatives at
, conclude that
!.
This result, together with (30.2) enables us to solve for , 0,
1, 2, , .
!
The th degree Taylor polynomial generated by around is given
by
2! 2
! .
That is,4
0
! .
Given a particular function, , and center, , we compute the rst
derivatives of , evaluate
each at , and use !
as the coefcient of .
Note:
The equation of the line tangent to at is of the form 1 1,
where1, 1 , and . The equation is therefore ;this is 1, as
expected.
When 0 were back to the Taylor polynomial centered at 0.
EXAMPLE 30.4 (a) Find the th degree Taylor polynomial for ln
centered at 1.(b) Use 5 to estimate ln1.2.
SOLUTION (a) 1 1 1 12!
12 13!
13 1
! 1
4 In order to use this summation notation we must adopt the
convention that 0 1 even if .
-
930 CHAPTER 30 Series
Compute derivatives of ln , looking for a pattern.
ln 1 0 1
1 1 1
2 1 11! 2 3 1 2 2! 4 3 2 4 41 3 23! 5 4 3 2 5 51 4 3 2 4!
......
11 1! 1 11 1!
0 1 112!
12 2!3!
13 3!4!
14
11 1!
! 1
1 12
2 1
3
3 1
4
4 11 1
.
More compactly,
0
1
! 1
0
11 1!!
1
011 1)!
1)! 1
011 1
.
(b)ln 5 1
122
13
3 1
4
4 1
5
5
ln 1.2 51.2 0.20.22
2 0.2
3
3 0.2
4
4 0.2
5
5 0.1823306
Compare this with the actual value of ln 1.2; it matches for the
rst four decimal places.
Aside: Dealing with Factorials and Alternating Signs
Factorials: Parentheses are important.
2! 2 2 1 2 2 3 2 1 2 2 1!
On the other hand, 2! 2 ! 2[ 1 3 2 1]. Similarly, 2 1! 2 1! 2
1.
-
30.1 Approximating a Function by a Polynomial 931
Alternating signs:
1 and 11 can be used to indicate alternating signs. Which is
needed to do thejob is determined by the notational system you
happen to have chosen. The simplest
way of determining which you need is by trial and error. Try 1
and check it witha particular -value. If it doesnt work, switch to
11.
EXAMPLE 30.5 Approximate5
34 using the appropriate second degree Taylor polynomial.
SOLUTION Let 15 . We must center the Taylor polynomial at a
point near 34 at which the valuesof , , and can be readily
computed.
An off-the-cuff approximation of5
34 is5
34 5
32 2; we know that 5
34 is a bit
more than 2. Center the Taylor polynomial at 32.
2 32 32 32 32
2! 322
15 32 2
15
45 32 1
5
1
3245
15
1
24 1
80
425
95 32 4
25
1
29 1
25 27 1
3200
Therefore,
2 21
80 32 1
6400 322.
5
34 234 21
802 4
6400 2 1
40 1
1600 2.024375
This agrees with the actual value of5
34 to four decimal places.
If you study closely the numerical data in this section you can
start to get a sense of
the magnitude of the error involved in a Taylor polynomial
approximation. The size of the
error can be estimated by graphing using a calculator or
computer. In the nextsection we will state Taylors Theorem, which
will provide not only a method of estimating
errors independent of a calculator, but also an invaluable
theoretical tool.
P R O B L E M S F O R S E C T I O N 3 0 . 1
For Problems 1 through 7, do the following.
(a) Compute the fourth degree Taylor polynomial for at 0.(b) On
the same set of axes, graph , 1, 2, 3, and 4.
(c) Use 1, 2, 3, and 4 to approximate 0.1 and 0.3. Compare
these approximations to those given by a calculator.
1.
2. ln1
3. tan1
-
932 CHAPTER 30 Series
4. 1 4
5. 1
6. 24 32 1
7. 1 2
8. Below is a graph of . For each quadratic given, explain why
the quadratic could
not be the second degree Taylor polynomial for at 0.(a) 2 3
1
22
(b) 1 5 22(c) 2 2 1
32
y
x
y = f(x)
9. Let 2 0 1 22 be the second degree Taylor polynomial generated
by at 0, where the graph of is the one given in Problem 8 above.
Use the graphto determine the signs of 0, 1, and 2.
10. (a) Find the second degree Taylor polynomial generated by
sec at 0.(b) Graph 2 and sec on the same set of axes.
11. (a) Compute the third degree Taylor polynomial for tan about
0.(b) Why is it reasonable to expect the coefcient of the 2 term to
be zero?
12. The graph of a differentiable function is given. Use the
graph to determine the
signs of the coefcients of the second degree Taylor polynomials
indicated. has a
minimum at 0 and a point of inection at 2.(a) 2 0 1 22(b) 2 0 1
1 2 12(c) 2 0 1 2 2 22(d) 2 0 1 3 2 32
-
30.1 Approximating a Function by a Polynomial 933
y
x
y = f(x)
1 2 3
13. Let ln1 . Find the th degree Taylor polynomial generated by
about 0.
14. Compute the th degree Taylor polynomial expansion of 1
about 1. Graph and 1, 2, 3, and 4 on a common set of axes.
In Problems 15 through 18, use a second degree Taylor polynomial
centered appropri-
ately to approximate the expression given.
15.3
8.3
16.
103
17. tan10.75
18.3
29
19. Compute the third degree Taylor polynomial generated by sin
at 4
.
20. Find the fth degree Taylor polynomial for
centered at 9.
21. Write the third degree Taylor polynomial centered about 0
for 11 ,
where is constant.
Introduction to Error Analysis: Problems 22 and 23.
22. Let . Use the data given in the table on page 928 to compute
the following.(a) 0.1 0.1 for 1, 2, , 5(b) 0.2 0.2 for 1, 2, , 5(c)
0.5 0.5 for 1, 2, , 5(d) 1 1 for 1, 2, , 5
Compare the size of the difference between the actual value of
the function and the poly-
nomial approximation with that of the rst unused term of the
Taylor polynomial
that is, the last term of the next higher degree polynomialand
observe that they have
the same order of magnitude.
-
934 CHAPTER 30 Series
23. Let and let be its th degree Taylor polynomial about 0.
Graph for 1, 2, , 5.
24. Use a third degree Taylor polynomial to approximate ln
0.9.
25. 12 3 1 5 12 7 13. Find the following.(a) 1 (b) 1 (c) 1 (d)
1
26.
3 12 53 17 56. Find the following.(a) 5 (b) 5 (c) 5 (d) 65
27. Compute the sixth degree Taylor polynomial generated by sin
about .
28. Compute the sixth degree Taylor polynomial generated by cos
about 2
.
29. Let 1 , where is a constant, 0, 1, 2, 3, 4, 5.(a) Compute
the third degree Taylor polynomial for around 0.(b) Compute the fth
degree Taylor polynomial for around 0.
30. Using the results of Problem 29(a), approximate the
following. Compare your results
with the numerical approximations given by a calculator.
(a)
1.002 (b) 11.03
(c)3
1.001
30.2 ERROR ANALYSIS AND TAYLORS THEOREM
An approximation is of limited use unless we have a notion of
the magnitude of the error
involved. Every Taylor polynomial has an associated error
function, , dened
by
function
polynomial
approximation
associated
error
is referred to as the Taylor remainder; .For a Taylor polynomial
centered at we expect the magnitude of the remainder
to decrease as increases and as approaches . Because each
successive renement
of a Taylor polynomial involves a higher derivative, we might
expect to involve
the 1st derivative of . While Taylors Theorem does not pin down
the remainderprecisely, it provides a means of putting an upper
bound on the magnitude of the error.
-
30.2 Error Analysis and Taylors Theorem 935
T a y l o r s T h e o r e m
Suppose and all its derivatives exist in an open interval
centered at . Thenfor each in
2!
2 3!
3
! ,
where
1 1!
1 for some number in , between and .
Note that has the same form as the next term of a Taylor
polynomial except that
the 1st derivative is evaluated at some between and instead of
at itself. Itsform agrees with the expectations laid out before the
statement of Taylors Theorem. When
applying the theorem we do not expect to be able to nd ; if we
could, an approximation
wouldnt have been needed.
In practice, we look for a bound, ! , such that 1 ! for all
between and
and use the inequality
!
1! 1.
This is referred to as Taylors Inequality.
A sketch of the proof of Taylors Theorem is given in Appendix
H.
Lets revisit some of the problems from the previous section and
see what information
Taylors remainder provides about the accuracy of
approximations.
EXAMPLE 30.6 Give a good5 upper bound for the error involved in
estimating sin 0.1 using the approxima-tion sin 3
3!.
SOLUTION We can call 33!
either 3 or 4, the two being equal. Well call it 4 as this
will give a better bound on the error.
0.1 40.1 40.1
sin0.1
0.1 0.13
3!
40.1
Taylors Theorem says 11! 1 for some between and . In this
example 4, sin , 0, and 0.1.
40.1 5
5!0.15
5 We say good because 1 million, for instance, is an upper
bound, but not what we are aiming for.
-
936 CHAPTER 30 Series
The derivatives of sin are sin and cos , so 5 1.
0 40.1 1
5!
1
105 1
120 105 1
1.2 107 8.3 108.
EXAMPLE 30.7 We want to use an th degree Taylor polynomial for
centered at 0 to approximate .How large must be to assure that the
answer differs from by no more than 107? Assumewe know 3.
SOLUTION 11! 1 for some between and . In this example ,
0, and 1.
1 for between 0 and 1. increases with , so 1 3.
1 3
1! 13
1!
We must nd an integer such that 31! 1107 , or, equivalently,
1! 3 107.
We nd by trial and error. 11! 39,916,800 3 107, whereas 10! is
not large enough.
1 11, so 10.
We must use the 10th degree Taylor polynomial: 1010
0
!.
Checking, we see that10
01! 1 1 1
2! 1
3! 1
10! 2.718281801, which
differs from by less than 107. In fact, we solved the problem
efciently; had we usedone less term of the expansion, the error
would have been more than 107.
EXAMPLE 30.8 In Example 30.5 we approximated5
34 using a second degree Taylor polynomial centered
at 32. Find a reasonable upper bound for the magnitude of the
error.
SOLUTION 11! 1 for some between and . In this example 2,
15 , 32, and 34.
234
3!34 323
234
6 8 for some between 32 and 34.
We must nd ! such that ! .
15
45 ; 4
25
95 ; 36
12514
5
36125
1
5
14for some between 32 and 34.
-
30.2 Error Analysis and Taylors Theorem 937
The smaller , the larger , so
0 36125
1
5
3214 36
125
1
214 9
125 212
0 234 9
125 2121
68 3
125 210 3
128000 2.344 105.
The error is less than 2.4 105.Taylors Theorem gave a good
estimate of the error; the actual error involved in
Example 30.5 is approximately 2.24 105.
If a computer or graphing calculator is at our disposal, error
estimates can be readily
available. Suppose, for example, that we plan to use the third
degree Taylor polynomial for
ln centered at 1 in order to approximate ln for [0.3, 1.7]. We
want an upper boundfor the error involved in doing so. In other
words, for [0.3, 1.7] we use the approximation
ln 1 12
2 1
3
3
and want an estimate of 3 ln 1 12
2 13
3
. We can simply graph
3 on [0.3, 1.7], obtaining the graph shown in Figure 30.6. Using
the tracer we estimatethat the magnitude of the error is less than
0.145.
As an exercise, use Taylors Remainder to estimate the error.
x
y
0.3 1.710.1
0.10.20.3
(.3, .14464)
Graph of |R3(x)| = ln x (x1) (x1)2
2
(x1)3
3+ ] || ]
Figure 30.6
EXAMPLE 30.9 Use graphical methods to nd an upper bound for the
error involved in using the tangentline approximation 1 1
2 to approximate 1
1 for 0.001.
SOLUTION Graph 2 1 12
1 1
2
on the domain [0.001, 001]. (Play around with therange to obtain
a useful graph.) The graph is given in Figure 30.7 on the following
page.
-
938 CHAPTER 30 Series
3.8 107
.001.001
y
x
(.001, 3.8 107) (.001, 3.8 107)
R2(x) = (1 + x) [1 x]1
2
1
2
Figure 30.7
For 0.001, the approximation 11 1
12 produces an error of less than
4 107. Any physicist will attest to the fact that physicists
often use Taylor polynomials to
simplify mathematical expressions. In fact, they often use only
rst or second degree
polynomials. While this may at rst strike you as a dubious
strategy, the following example
will demonstrate that in certain situations the error introduced
is minimal.
EXAMPLE 30.10 According to Newtonian physics, an objects kinetic
energy, " , is given by
" 120#
2,
where 0 is the mass of the object at rest and # is its
velocity.
Einsteins theory of special relativity produces a more involved
expression for " .
According to Einstein, the mass of an object is a function of
its velocity, 01#22
.
Einsteins theory says energy, $, equals 2, where is the speed of
light. He concludes
that an objects kinetic energy is given by the difference 2 02.
Using the expressionfor , Einsteins theory says
" 02
11 #22
102
1 #2
2
12
1
. (30.3)
Our goal in this example is to show that if an object is
traveling much slower than the speed
of light, then according to Einsteins theory, the error involved
in using the Newtonian
expression for " is small.
SOLUTION We begin by noting that if # is substantially less than
, then #
is small, and#
2is even
smaller. From Example 30.9 we know that 1 12 can be well
approximated by its rstdegree Taylor polynomial, 1 1
2, for small. Let #2
2. Using the approximation
1 #
2
2
12
1#
2
2
12
1 12
#
2
2
in Equation (30.3) we obtain
-
30.2 Error Analysis and Taylors Theorem 939
" 02
1 #2
2
12
1
02
1 12
#2
2 1
02
1
2
#2
2 1
20#
2.
Lets estimate the size of the error introduced by using the
Newtonian expression for " for
an object traveling at speeds of 300 m/s or less. 3 108 m/s.Well
nd an upper bound for the error in replacing 1 12 by 1 1
2 for 3002
2
and multiply the answer by 02.
1
2!2 for some between 0 and 300.
1 12 ; 121 32 ; 3
41 52
1 3
2 41 52 3
8
1 3002
2
5 30044 3.75 1025
Multiplying by 02 gives 03.375 108.
Therefore, for speeds of up to 300 m/s, the error incurred in
computing " using
Newtonian physics is less than 3.4 1080, where 0 is the mass of
the body at rest.
P R O B L E M S F O R S E C T I O N 3 0 . 2
1. Find a good upper bound for the magnitude of the error
involved in approximating
cos by 1 22! 4
4!for 0.2. Do this using Taylors Inequality; then check your
answer by graphing the remainder function.
2. Use the third degree Taylor polynomial for at 0 to estimate.
Then use TaylorsTheorem to get a reasonable upper bound for the
remainder.
3. We will use the th degree Taylor polynomial for ,
0
!, to approximate 1
.
What should be in order to guarantee that the approximation is
off by less than 105?
4. Use the third degree Taylor polynomial for ln centered at 1,
1 122
13
3, to approximate ln1.5. Then give an upper bound for the
remainder using
Taylors Theorem.
5. The second degree Taylor polynomial for 1 is 1 12!
2. If
the second degree Taylor polynomial is used to approximate
1 for 0.2, ndan upper bound for the magnitude of the error. Use
the Taylor Inequality; then check
your answer by graphing 2.
-
940 CHAPTER 30 Series
6. For near zero, cos 1 22! 4
4! 1 2
2!. What degree Taylor polyno-
mial must be used to approximate cos0.2 with error less than
1108
?
7. Approximate3
27.5 using an appropriate second degree Taylor polynomial. Find
a
good upper bound for the error by using Taylors Inequality.
8. The second degree Taylor polynomial generated by ln1 about 0
is 22
.
Use Taylors Theorem to nd a good upper bound on the error
involved in using this
polynomial to approximate the following.
(a) ln1.2 (b) ln0.8
9. By graphing 2, estimate the values of for which the
approximation
ln 1 122!
can be used without producing an error of magnitude greater
than 103.
10. For near zero, 1 22! 3
3!. Find a reasonable upper bound for the magni-
tude of the error involved in using this approximation for 0.5.
Use TaylorsInequality and check your answer by graphing 3.
11. A hyena is loping down a straight path away from a stream.
The hyena is 6 m from the
stream, moving at a rate of 2 m/s and decelerating at a rate of
0.1 m/s2. Use a second
degree Taylor polynomial to estimate its distance from the
stream 1 second later.
12. What degree Taylor polynomial for about 0 must be used to
approximate 0.3with error less than 105?
13. (a) Find the th degree Taylor polynomial for 11 centered at
0.
(b) How many nonzero terms of the polynomial in part (a) must be
used to approximate
12
with error less than 105?
14. According to Einsteins theory of special relativity, the
mass of an object moving with
velocity # m/s is given by
01 #2
2
,
where 0 is the mass of the object at rest and is the speed of
light, 3 108 m/s.(a) Use the rst degree Taylor polynomial for 1
1 to arrive at the estimate
0 0
2
#2
2.
(b) If an object is moving at 100 m/s, nd an upper bound for the
error involved in
using the approximation given in part (a).
-
30.3 Taylor Series 941
30.3 TAYLOR SERIES
Dening Taylor Series
In many examples in this chapter weve observed that the higher
the degree of the Taylor
polynomial generated by at , the better it approximates for near
. Forfunctions such as sin and cos , the higher the degree of the
Taylor polynomial the longer
the interval over which the polynomial follows the undulations
of the functions graph.
Letting the degree of the polynomial increase without bound
gives us the Taylor series
for .
D e f i n i t i o n
If a function has derivatives of all orders at , then the Taylor
series of at(or about) is dened to be
2!
2
! ,
that is,
0
! .
We refer to this series as the Taylor expansion of about or
centered at .In the special case where 0, the series 0 0! can be
called the Maclaurin
series for .
From the work weve done with Taylor polynomials, we can easily
nd the Maclaurin
series for , sin , and cos .
EXAMPLE 30.11 Find the Maclaurin series for .SOLUTION All
derivatives of are . When evaluated at 0, is 1. Maclaurin series
for :
1 2
2!
!
0
!
EXAMPLE 30.12 Find the Maclaurin series for sin .SOLUTION Even
order derivatives Odd order derivatives
sin 0 0 cos 0 1 sin 0 0 cos 01 4 sin 40 0 5 cos 50 1
......
......
2 1 sin 20 0 21 1 cos 210 1
Maclaurin series for sin :
3
3!
5
5! 1
21
2 1! 0
1 21
2 1!
-
942 CHAPTER 30 Series
EXAMPLE 30.13 Find the Maclaurin series for 11 .
SOLUTION 1 1 0 1 1 2 0 1 21 3 0 2 3 21 4 0 3!
......
!1 1 0 !
Maclaurin series for 11 :
1 22
2! 3!
3
3! 4!
4
4! !
!
1 2 3 0
The Maclaurin series for 11 should look familiar.
0
is a geometric series with 1and % . Therefore we know that it
converges to 1
1 for 1 and diverges for 1.
This observation at the end of Example 30.13 highlights the
important question What is
the signicance of the Taylor series for ? For instance, for what
does the Maclaurin
series for sin converge? When it converges, to what does it
converge? In particular, does
sin 0.1 0.1 0.133!
0.155!
1 0.12121! ?
Or, more generally, for which values of is it true that
sin 3
3!
5
5!
7
7! 1
21
2 1! ?
These latter questions can be answered using Taylors
Theorem.
Taking the limit as increases without bound gives
lim lim .
Therefore, is the sum of its Taylor series if and only if lim 0.
We statethis more precisely below.
T h e o r e m o n C o n v e r g e n c e o f T a y l o r S e r i
e s
If is innitely differentiable on an interval centered around ,
then the Taylorseries for at converges to for all if and only if
lim 0for all , where is the Taylor remainder.
-
30.3 Taylor Series 943
In applying this theorem we frequently use the fact that lim! 0
for every .
Think about this; it should make sense that eventually ! will be
much larger than for
xed . We prove this below.
Fact: lim! 0 for every real number .
Proof: 0 !
1
2
3
Let be a positive constant integer such that 0
1. Then
0 1
2
3
! " positive terms, each
less than or equal to
1
! "
positive terms,each less than
So 0
!
(30.4)
If 0 % 1, then lim % 0. Therefore lim % 0 for 0 % 1 and
constant.
But 0
1, so lim
0.
Return to (30.4) and let increase without bound.
lim 0 lim
!
lim
0 lim
!
0 0
Therefore lim
!
0, by the Sandwich Theorem.
We are now ready to show that sin and are equal to their
respective Taylor series.
EXAMPLE 30.14 Show that sin
01 21
21! for all .
SOLUTION For each there exists a between 0 and such that
0 1 1!
1.
Therefore 11
1! .The latter inequality holds because 1 sin or cos and both
are
bounded by 1.
lim 0 lim lim
1 1!
0 lim 0
From the Sandwich Theorem we conclude that lim 0 and
thereforelim 0 for all . Thus, sin is equal to its Taylor expansion
about zero forall .
-
944 CHAPTER 30 Series
EXAMPLE 30.15 Show that
0
!for all .
SOLUTION For each there exists a between 0 and such that
0 1 1!
1 1
1! is an increasing function, so .
0 1 1!
lim 0 lim lim
1 1!
But lim
1 1!
lim
1 1!
0 0.
0 lim 0
So lim 0 by the Sandwich Theorem. Therefore, lim 0.
We conclude that 1 22! 3
3! for all .
EXERCISE 30.4 Show that cos is equal to its Maclaurin series for
all .
Take a moment to reect upon the rather remarkable results we
have accumulated. Not
only can we express , sin , and cos as innite polynomials
(called power series), but
we determined the coefcients using information about derivatives
evaluated only at 0.We think of a derivative as giving local
information, yet somehow information generating
the entire function is encoded in the set of innitely many
derivatives. This is philosophically
intriguing.
Lets take inventory on convergence issues.
A Taylor series might converge to its generating function for
all .
For example, consider the Maclaurin series for , sin , and cos
.
A Taylor series might converge to its generating function only
over a certain interval.
For example, 11
0
only for 1, 1.At minimum a Taylor series will be equal to the
value of its generating function at its
center.6
Power Series
Well put Taylor series in a broader context by discussing power
series.
6 It is possible for a Taylor series to converge, but not to its
generating function, except at its center. This pathology is
illustrated
in Problem 35 at the end of this section.
-
30.3 Taylor Series 945
D e f i n i t i o n
A power series in is an innite series of the form
0 . A power series in
( ), or a power series centered at , is a series of the form 0
.
U n i q u e n e s s T h e o r e m f o r P o w e r S e r i e s E
x p a n s i o n s
If has a power series expansion (or representation) at , that
is, if
0 # for , then that power series is the Taylor seriesfor at
.
The Uniqueness Theorem can be veried by repeatedly
differentiating the power series
expansion term by term and evaluating each successive derivative
at .The Uniqueness Theorem carries with it computational power. For
example, we could
have avoided computing derivatives in Example 30.13 by using the
fact that
1
1 1 2 for 1, 1.
This is a power series expansion of 11 , and therefore it must
be the Taylor series for
11
at 0.
Convergence of a Power Series7
T h e o r e m o n t h e C o n v e r g e n c e o f a P o w e r S
e r i e s 8
For a given power series
0 , one of the following is true:i. The series converges for all
.
ii. The series converges only when .iii. There is a number , 0
such that the series converges for all such that
( is within of the center) and diverges for all such that .
is called the radius of convergence. If the series converges for
all , we say ;if the series converges only at its center, we say
0.
The set of all for which a power series converges is called the
interval of converge
of the series. From the theorem stated above we see that a power
series in will havean interval of convergence centered around . At
the endpoints of the interval the seriescould either converge or
diverge; further investigation is necessary. In other words, if
the
radius of convergence is , the interval of convergence will be
one of the following:
7 The student or instructor who prefers a thorough discussion of
convergence before a discussion of the convergence of a power
series can turn to page 964 (Section 30.5), and, after
completing that section, return to this point.8 Justication is
given in Appendix H.
-
946 CHAPTER 30 Series
(b R, b + R]
b R
[b R, b + R)
b R
[b R, b + R]
b R
(b R, b + R)
b R
The behavior of a power series at the points and can be tricky,
but for we will nd the behavior reassuringly like that of
polynomials in many respects.We will use substitution, integration,
and differentiation of power series on toobtain new Taylor series
from familiar ones. Before moving in this direction we must add
one more very important Taylor series to our list of familiar
ones.
The Binomial Series
EXAMPLE 30.16 THE BINOMIAL SERIES Find the Maclaurin series
generated by 1 , where is constant. This series is called the
binomial series.
SOLUTION The Maclaurin series is given by
0 0
!.
1 0 1 1 1 0 11 2 0 1 1 21 3 0 1 2
......
1 2 11 0 1 1...
...
Therefore the Maclaurin series is
1 12!
2 1 23!
3
1 2 1!
9
Fact: The Maclaurin series for 1 converges to 1 for 1, 1 and
divergesfor 1.
1 1 12!
2 1 23!
3
1 1!
for 1, 1
Proving this fact by showing that lim 0 is difcult, but
possible. We omit theproof.10
REMARKS CONCERNING THE BINOMIAL SERIES
1. In the case that is a positive integer the series terminates
with the term; subsequent
coefcients all contain a factor . We are left with an expansion
of the polynomial
9 The coefcients match those given by Pascals Triangle.10 By the
end of Section 30.5 you will be able to show that the radius of
convergence of the binomial series is 1.
-
30.3 Taylor Series 947
1 . As an exercise, show that if 4 the binomial series becomes 1
4 1 4 62 43 4.
2. The notation
is often used as an abbreviation for the binomial coefcients
where
121!
for 1 and 0
1. Using this notation we can write1
0
for 1, 1.
3. The binomial expansion is valuable to know, as applications
of it abound. Examples
30. 9, 30.10, and 30.13 all involve binomial expansions. Often
one uses the rst and
second order approximations,
1 1 or 1 1 12!
2
for small, in computations in applied science.
EXERCISE 30.5 By letting 1, use the binomial series to nd the
Maclaurin series for 11 . Then let
to arrive at the Maclaurin series for 11 .
Below we list some commonly used Taylor expansions together with
their intervals of
convergence.
1 2
2!
3
3!
! for all
sin 3
3!
5
5! 1
21
2 1! for all
cos 1 2
2!
4
4! 1
2
2! for all
1 1 12!
2 1 1!
for 1
1
1 1 2 3 for 1
You will nd it useful to know these series off the top of your
head because other series can
be derived directly from these.
Obtaining New Taylor Series From Familiar Ones: Substitution
EXAMPLE 30.17 Find the Taylor expansion for 2
about 0.
SOLUTION Calculating this series by computing derivatives very
quickly becomes unwieldy. Instead,
well use substitution.
-
948 CHAPTER 30 Series
1 2
2!
3
3!
! for all . Let 2.
2 1 2
22
2!
23
3!
2
!
2 1 2
4
2!
6
3 1
2
! for all .
By the Uniqueness Theorem, this is the Taylor series for 2
about 0.
EXAMPLE 30.18 Find the Maclaurin series for 2 sin cos .SOLUTION
2 sin cos sin2
sin 0
1 21
2 1! for all . Let 2.
sin20
1 221
2 1! 0
1 22121
2 1! for all
sin2 0
1221 21
2 1! 0
1 22122
2 1!
We can write this out as
22 234
3! 2
56
5! 1221
22
2 1! .
By the Uniqueness Theorem, this is the Maclaurin series for 2
sin cos . Note the
difference between substituting 2 for in the rst step and
multiplying the whole series
by in the second step.
EXAMPLE 30.19 Find the fourth degree Taylor polynomial for
9 2 about 0. For what -values does the Taylor series converge to
?
SOLUTION Lets transform this function so that we can use the
binomial series.
9 2
#9
1
2
9
3
#1
2
9 3
1
2
9
12
From the binomial series we know
1 1 12!
2 for 1.
so
1 12 1 12
12
1
2
2!
2
1 12 1 12 1
82 for 1.
Let 29
.
-
30.3 Taylor Series 949
1
2
9
12
1 12
2
9
1
8
2
9
2 for
2
9
1#1
2
9 1 1
182 1
6484 for 2 9
3
#1
2
9 3 1
62 1
2164 for 3, 3
Thus, the fourth degree Taylor polynomial is 3 162 1
2164. The Taylor series for
converges to on 3, 3. Note that in Examples 30.17 and 30.18 the
old series being used converge for all
real numbers. In Example 30.19 this was not the case; the new
interval of convergence was
obtained by substitution.
P R O B L E M S F O R S E C T I O N 3 0 . 3
1. Find the Maclaurin series for cos and show that it is equal
to cos for all .
2. (a) Find the Maclaurin series for ln1 .(b) On the same set of
axes, graph ln1 and 6. Observe that the polynomial
approximation to ln1 is good for 1.(c) Graph 6 ln1 6. Observe
that 6 is close to zero on 1.
In the next section we will show that the radius of convergence
of the Maclaurin series
for ln1 is 1.
3. The interval of convergence of the Maclaurin series for ln1
is 1, 1]. Onthis interval the series converges to ln1 .(a) Find the
Maclaurin series for ln1 .(b) By setting 1 in part (a), nd the
Taylor series for ln centered at 1.(c) Find the Taylor series for
ln at 1 by taking derivatives. Make sure your
answers to parts (b) and (c) agree.
(d) What is the interval of convergence for the Taylor series
for ln centered at 1?(e) Graph ln and several of its Taylor
polynomials at 1 to be sure your answer
to part (d) is reasonable.
In Problems 4 through 9, nd the Taylor series for centered at
the indicated value
of .
4. sin ,
5. 2 cos , 2
6. 10, 0
7. 1
, 1
-
950 CHAPTER 30 Series
8. 3 23, 0
9. 1 5, 0
10. A power series centered at 0 has a radius of convergence of
5. For each value of given below, determine whether the series
converges, diverges, or there is not enough
information available to determine.
(a) 0 (b) 3 (c) 5 (d) 7(e) 1.8 (f)
5 (g) 5 (h) 6
11. A power series of the form
0 2 has a radius of convergence of 3.(a) For what values of can
you say with condence that the series converges?
(b) For what values of can you say with condence that the series
diverges?
(c) For what values of are you given inadequate information to
determine conver-
gence?
12. The interval of convergence of a power series is 2, 5].(a)
What is the radius of convergence?
(b) What is the center of the series?
13. A power series is of the form
0 3. Which of the intervals given belowcould conceivably be the
interval of convergence of the series? For each option ruled
out, explain the rationale.
(a) 0, (b) 2, 4 (c) [10, 4 (d) [3, 3](e) 4, 2 (f) 5,1] (g) ,
In Problems 14 through 21, use your knowledge of the binomial
series to nd the th
degree Taylor polynomial for about 0. Give the radius of
convergence of thecorresponding Maclaurin series. One of these
series converges for all .
14. 1 3, 3
15. 11 , 2
16. 1 23 , 3
17. 3
1 2, 5
18. 1 35, 6
19. 112 , 5
20. 29 12 , 3
21. 4 , 3
-
30.3 Taylor Series 951
22. (a) Expand 4 by multiplying out or by using Pascals
triangle.(b) Rewrite as [1
]4 4 1
4. Use the binomial series to expand
1
4, multiply by 4, and demonstrate that the result is the same as
in part (a).
23. Find the Maclaurin series for 112 . What is the radius of
convergence?
24. Use the binomial series to nd the Maclaurin series for
112
. What is the radius of
convergence?
In Problems 25 through 34, use any method to nd the Maclaurin
series for .
(Strive for efciency.) Determine the radius of convergence.
25.
26. sin 3
27. cos 2
28. 32
29. cos2
30. 3
31. 2 cos
32. cos2 (Hint: use a trigonometric identity)
33. , where and are constants and is not a positive integer.
34. 123
35. Pathological Example: Let
1
2 for 0,0 for 0.
(a) Graph on the following domains: [20, 20], [2, 2], and [0.5,
0.5]. (Agraphing instrument can be used.)
(b) It can be shown that is innitely differentiable at 0 and
that 0 0for all . Conclude that the Maclaurin series for converges
for all but only
converges to at 0.
36. Find the Maclaurin series for 1
. What is its radius of convergence?
37. For 1, 1], ln1 22 3
3 4
4 1 1
1 .(a) Find the Maclaurin series for ln1 2. What is its interval
of convergence?(b) Find the Maclaurin series for ln . What is its
interval of convergence?(c) Find the Maclaurin series for log101
.
-
952 CHAPTER 30 Series
38. Discover something wonderful. We know 1 22! 3
3!
! for
all real . Now dene raised to a complex number, where 1, to be
where 1 2
2! 3
3!
! .
(a) Use the fact that 2 1, 3 , and 4 1 to simplify the
expression for .Gather together the real terms (the ones without s)
and the terms with a factor of
. Express as a sum of two familiar functions (one of them
multiplied by ).
(b) Use your answer to part (a) to evaluate .
39. The hyperbolic functions, hyperbolic cosine, abbreviated
cosh, and hyperbolic sine,
abbreviated sinh, are dened as follows.
cosh
2sinh
2
(a) Graph cosh and sinh , each on its own set of axes. Do this
without using a
computer or graphing calculator, except possibly to check your
work.
(b) Find the Maclaurin series for cosh .
(c) Find the MacLaurin series for sinh .
Remark: From the graphs of cosh and sinh one might be surprised
by the choice
of names for these functions. After nding their Maclaurin series
the choice should
seem more natural.
(d) Do some research and nd out how these functions, known as
hyperbolic functions,
are used. The arch in St. Louis, the shape of many pottery
kilns, and the shape of
a hanging cable are all connected to hyperbolic trigonometric
functions.
30.4 WORKING WITH SERIES AND POWER SERIES
Absolute and Conditional Convergence
There are many ways in which power series can be treated very
much as we treat polyno-
mials, but there are ways in which they can behave differently
and must be treated with
caution. This makes sense; there are ways in which series and
nite sums behave very dif-
ferently. In order to sort this out a bit, not only do we need
to steer clear of divergent series
and power series outside their interval of convergence, but we
need to rene our notion of
convergence to distinguish between absolute and conditional
convergence.
D e f i n i t i o n
A series
1 is absolutely convergent if
1 converges.
Note that if the terms of a series are either all positive or
all negative, then convergence
implies absolute convergence. There is only an issue when some
terms are positive and
some terms are negative.11
11 Actually, there is not an issue provided there exists a
constant such that is either positive for all or negative for
all .
-
30.4 Working with Series and Power Series 953
Fact: If a series converges absolutely, it converges. This is
proven in Appendix H.
D e f i n i t i o n
A series
1 is conditionally convergent if it is convergent but not
absolutelyconvergent.
Why is this distinction handy? Well, one might hope that the
order of the terms in a sum could
be rearranged without altering the sum, yet for innite series
this is true only if the series
converges absolutely. In fact, it can be proven that if
0 is conditionally convergent,then the order of the terms can be
rearranged to produce any nite number. This unsettling
fact is enough to make one wary of conditionally convergent
series.
Its hard to be wary of something without a concrete example, so
we will take this
opportunity to look at alternating series. You will nd that
alternating series are fascinating
in their own right, and that this excursion into the topic of
alternating series will produce
as a by-product an error estimate that will prove useful when
dealing with many Taylor
polynomials.
Alternating Series
D e f i n i t i o n
A series whose successive terms alternate in sign is called an
alternating series.
For any xed the Maclaurin series for sin and cos are alternating
series. The Maclaurin
series for will alternate when is negative and the one for ln1
will alternate when is positive.
There is a simple convergence test, proved by Leibniz, that can
be applied to alternating
series. We know that for a general series,
1 , the characteristic lim 0is necessary but not sufcient for
convergence. The divergence of the harmonic series
1 12 1
3 1
illustrates this fact. However, if a series is alternating, then
if
the magnitude of the terms decreases monotonically towards zero,
this is enough to assure
convergence.
Alternating Series Test
An alternating series,
11 or
111 for 0, converges ifi. 1 , the terms are decreasing in
magnitude and
ii. lim 0, the terms are approaching zero.
The Basic Idea Behind the Alternating Series Test12
Consider the series 1 2 3 4 11 for 0. Suppose thatconditions (i)
and (ii) are satised. In Figure 30.8 we plot partial sums.
12 This is not a rigorous argument, but it can be made rigorous
using the theorem that every bounded monotonic sequence is
convergent.
-
954 CHAPTER 30 Series
0
S2 S6 S5
S
S3 S1
S1 = a1
S4
a6
a5
a4
a2
a1
a3
Figure 30.8
1 1 is to the right of zero.2 1 2 lies between 0 and 1 because 2
1.3 1 2 3 lies between 2 and 1 because 3 2.
...
Picture starting at the zero. Take a big step forward to 1, then
a smaller step backward to
2, then an even smaller step forward to 3, and so on. The
partial sums oscillate; is
between 1 and 2 because 1. The distance between 1 and is andlim
0. Therefore the sequence of partial sums is approaching a nite
limit , withsuccessive partial sums alternately overshooting then
undershooting .
This argument can be made rigorous by considering the increasing
but bounded se-
quence of partial sums, 2, 4, 6, , and the decreasing but
bounded sequence of partial
sums 1, 3, 5, , and showing that both sequences converge to the
same limit.
Our analysis provides us with an easy-to-use error estimate. If
an alternating series
satises the two conditions of the Alternating Series Test and if
we approximate the sum, ,
using a partial sum , then the magnitude of the error will be
less than 1, the magnitudeof the rst unused term of the series.
Furthermore, if the last term of the partial sum is
positive, then the partial sum is larger than ; if its last term
is negative, then the partial
sum is smaller than . We refer to this as the Alternating Series
Error Estimate.
EXAMPLE 30.20 Consider the alternating harmonic series 1 12 13
14 15 11 1 .(a) Show that this series converges conditionally.
(b) It can be shown that
111 1 converges to ln 2. How many terms of the seriesmust be
used in order to approximate ln 2 with error less than 0.001?
SOLUTION (a) The series
111 1 is alternating. It satises the conditions of the
AlternatingSeries Test:
i. The terms are decreasing in magnitude: 11
1.
ii. The terms approach zero: lim lim 1 0.Therefore the series
converges. But
1
11 1 1 1 is the harmonic se-ries, which diverges. Therefore the
alternating harmonic series converges conditionally.
(b) By the Alternating Series Error Estimate we know that the
magnitude of the error is
less than the magnitude of the rst omitted term. Therefore we
use the estimate
-
30.4 Working with Series and Power Series 955
ln 2 9991
11 1
;
we need 999 terms. This series for ln 2 converges very
slowly!
EXAMPLE 30.21 Estimate 1 with error less than 103.
SOLUTION 1 22! 3
3!
! Thus
1 12 1 1
2 1
22 2! 1
23 3! 1 1
2 ! .
This series is alternating, its terms are decreasing in
magnitude, and its terms tend toward
zero. Therefore, we can apply the Alternating Series Error
Estimate. We must nd such
that
1
2 ! 1
1000, or equivalently, 2 ! 1000.
We do this by trial and error. 24 4! 384 but 25 5! 3840
1000.1
255! 1
1000, so we dont need to use this term.
12 1 1
2 1
22 2! 1
23 3! 1
24 4! 11
2 1
8 1
48 1
384
1 .6068.
Notice that the Alternating Series Error Estimate is simpler to
apply than Taylors
Remainder.
Lets return to the disturbing remark made before introducing
alternating series. The
assertion was that if a series converges conditionally, then
rearranging the order of the terms
of the series can change the sum. Were now ready to demonstrate
this.
1 12 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 ln 2
Multiplying both sides by 2 gives
2 22 2
3 2
4 2
5 2
6 2
7 2
8 2
9 2
10 2
11 2 ln 2 ln 4 (30.5)
Rearrange the order of the terms in Equation (30.5) so that
after each positive term there
are two negative terms as follows.
2 1 24 2
3 2
6 2
8 2
5 2
10 2
12 2
7 2
14 2
16 2
19
2 1 24
2
3 2
6
2
8
2
5 2
10
2
12
2
7 2
14
2
16
1 12 1
3 1
4 1
5 1
6 1
7 1
8
ln 2.By rearranging the order of the terms we changed the sum
from ln 4 to ln 2. Riemann proved
that by rearranging the order of the terms we can actually get
the sum to be any real number.
-
956 CHAPTER 30 Series
On the other hand, it can be proven that if a series converges
absolutely to a sum of , then
any rearrangement of the terms has a sum of as well. This is one
of the reasons we prefer
to work with absolutely convergent series whenever possible.
Manipulating Power Series
Having dened absolute convergence, we can return to the theorem
on the convergence of
a power series and state a stronger form. (See Appendix H for
justication.)
T h e o r e m o n t h e C o n v e r g e n c e o f a P o w e r S
e r i e s
For a given power series
0 , one of the following is true:i. The series converges
absolutely for all .
ii. The series converges only when .iii. There is a number , 0,
such that the series converges absolutely for all
such that and diverges for all such that .
The points and must be studied separately. At these endpointsthe
series could converge conditionally, converge absolutely, or
diverge. For the sake of
simplicity we will generally restrict our attention to the
interval , in whichthe power series converges absolutely.
Differentiation and Integration of Power Series
D i f f e r e n t i a t i o n a n d I n t e g r a t i o n o f P
o w e r S e r i e s
Let
0 be a power series with radius of convergence , where 0,
possibly. Then the function 0 can be differentiated termby term or
integrated term by term on , . That is,
0
1
1 with radius of convergence
and
0
0
1
1with radius of convergence .
This result, whose proof is omitted, says that the radius of
convergence remains the same
after integration or differentiation; it gives no information
about convergence or divergence
at .13
13 The original series may diverge at an endpoint and yet
converge once integrated, or vice versa.
-
30.4 Working with Series and Power Series 957
This Theorem gives us convenient ways of generating new Taylor
series from familiar
ones and provides a tool for integrating functions that dont
have elementary antiderivatives.
EXAMPLE 30.22 Find the Maclaurin series for arctan . What is the
radius of convergence?
SOLUTION This is unwieldy to compute by taking derivatives.
Instead, well use the fact that1
1 2 arctan .
We know 11 1 2 3 for 1.
Let 2.1
1 2 1 2 22 23 2 for 2 1
1
1 2 1 2 4 6 12 for 1
1
1 2
1 2 4 6 12
arctan 3
3
5
5 1
21
2 1
To determine , evaluate both sides at 0. arctan 0 , so 0.
arctan 3
3
5
5
7
7 1
21
2 1
The radius of convergence is 1, so the series converges
absolutely for 1, 1 anddiverges for 1.
In fact, although we have only shown convergence for 1, 1, the
series convergesto arctan for 1 as well. When evaluated at 1, the
series is
4 1 1
3 1
5 1
7 .14
EXAMPLE 30.23 Find the Maclaurin series for ln1 by integrating
the series for 11 . What advantagedoes this approach have over
computing the series by taking derivatives?
SOLUTION We know 11 1 2 for 1.
Let .1
1 1
1 1 2 3 1 for 1, i.e., 1
1
1 2
2
3
3
4
4 1
1
1 for 1
So ln1 2
3
3
4
4 1
1
1 .
14 You will nd this series is carved in stone at the entrance to
Coimbra Universitys department of mathematics building in
Coimbra, Portugal.
-
958 CHAPTER 30 Series
To determine , evaluate at 0. ln1 0 , so 0.
ln1 2
2
3
3
4
4 1
1
1
An advantage of this method of arriving at the series is that we
know the radius of conver-
gence is 1, and that the series converges to ln1 for 1.
Once we know ln1 01 11 for 1 we can set 1, 1 and nd ln 01 111 1
122 133 1 11
1 . When 1 1, we know that 0 1 2, so the series forln about 1
must converge on 0, 2. In fact, it can be shown that both of
theseseries converge at the right-hand endpoint of the respective
interval of convergence.
ln1 2
2
3
3 1
1
1 for 1, 1]
ln 1 12
2 1
3
3 1 1
1
1 for 0, 2]
REMARK We saw in Example 30.20 that the series 1 12 1
3 1
4 converges very
slowly. Similarly, observe that 1 13 1
5 1
7 converges to
4very slowly. This series
is aesthetically pleasing but computationally inefcient. For
practical purposes the rate at
which a series converges is important. For instance, it is more
efcient to approximate ln 2
by looking at the following:
ln 2 ln
1
2
ln
1 1
2
1
2 1
22 2 1
23 3 1
24 4
12 1
8 1
24 1
64 1
160 .
12
is closer to the center of the series than is 1, so the series
converges more rapidlyat 1
2than at 1. For even more efciency in approximating ln 2 we can
nd the Maclaurin
series for ln
11
and evaluate it at 1
3. This is the topic of one of the problems at the
end of this section.
One reason that it is so useful to be able to represent a
function as a power series is that
a power series is simple to integrate. The use of power series
expansions as an integration
tool gured prominently in Newtons work and continues to be
important in the integration
of otherwise intractable functions. Consider, for example, 2, a
function that hasno elementary antiderivative. The graph of is a
bell-shaped curve which, with minor
modications, gives the standard normal distribution that plays
such an important role in
probability and statistics. It is crucial to know the area under
the normal distribution, and
-
30.4 Working with Series and Power Series 959
for this we must compute a denite integral. The following
example indicates how Taylor
series can be used in such a computation.
EXAMPLE 30.24 Approximate 0.2
0 2 with error less than 108.
SOLUTION From Example 30.17 we know that
2 1 2
4
2!
6
3! 1
2
! for all .
0.20
2
0.20
1 2
4
2!
6
3! 1
2
!
3
3
5
5 2! 7
7 3! 1
21
2 1! 0.2
0
0.2 0.23
3 0.2
5
5 2! 0.27
7 3! 1 0.2
21
2 1!
We can apply the Alternating Series Error Estimate because the
series above is alternating,
its terms are decreasing in magnitude, and its terms tend toward
zero. We look for a term
whose magnitude is less than 108
0.27
7 3! 27
7 3! 107 3 107 : not small enough
0.29
9 4! 29
9 4! 109 2.4 109 108
Therefore 0.20
2 0.2 0.2
3
3 0.2
5
5 2 0.27
7 6 with error less than 108.
0.20
2 0.197365029
There are three main reasons for our interest in representing
functions as power series.
Such representations are useful in
approximating functions by polynomials and approximating
function values numeri-
cally,
integrating functions that dont have elementary antiderivatives,
and
solving differential equations.
Although we have illustrated the rst two applications of power
series, we have yet to give
an example of the third. The Theorem on Differentiation of a
Power Series plays the major
role in this application.
Power Series and Differential Equations
The next example illustrates how power series can be used in
solving differential equations.
-
960 CHAPTER 30 Series
EXAMPLE 30.25 Use power series to solve the differential
equation .
SOLUTION Let be a solution to the differential equation. Assume
that has a power series
expansion.
0 1 22 33 1 22 332 443 1 22 3 2 3 4 3 42 12
If is a solution to , then .
22 3 2 3 4 3 42 12 0 1 22
The key notion is that for these two polynomials to be equal the
coefcients of corresponding
powers of must be equal. In other words, the constant terms must
be equal, the coefcients
of must be equal, and so on.
22 03 2 3 14 3 4 25 4 5 3...
1 2...
We can solve for all the coefcients in terms of 0 and 1.
Let 0 0, 1 1. Well solve for , 2, 3, in terms of 0 and 1.
2 0
2 0
2!3
13 2
13!
4 24 3
0
4 3 2! 0
4!5
35 4
1
5 4 3! 1
5!
6 46 5
06 5 4!
06!
7 57 6
17 6 5!
17!
8 68 7
0
8 7 6! 0
8!9
79 8
1
9!...
...
2 22
22 1 1
0
2!21
212 12 1
1
2 1!
-
30.4 Working with Series and Power Series 961
0 1 0
2!2 1
3!3 0
4!4 1
5!5 0
6!6 1
7!7
0
1 2
2!
4
4! 1
2
2!
cos
1
3
3!
5
5!
7
7! 1
21
2 1!
sin
0 cos 1 sin
EXERCISE 30.6 Verify that 0 cos 1 sin is a solution to the
differential equation .We have shown that if a solution to has a
power series representation, then thatsolution must be of the form
0 cos 1 sin , where 0 and 1 are constants.
In the example just completed, we recognized the Maclaurin
series for sin and cos .
It is entirely possible that we can solve for all the coefcients
of a power series and
simply have the solution expressed as and dened by the power
series expansion. There are
well-known functions dened by power series that arise in
physics, astronomy, and other
applied sciences. An example of such functions are the Bessel
functions, named after the
astronomer Bessel who came up with them in the early 1800s while
working with Keplers
laws of planetary motion. The Bessel function 0 is dened by
0
1 2
!222.
As is often the case in mathematics, while Bessel functions
arose in a particular astronomical
problem they are now used in a wide array of situations. One
such example is in studying the
vibrations of a drumhead. A graph of the partial sum 013
01 2
!222is given
in Figure 30.9.
.5
.5
1
1
22x
y
Graph of (1)kk = 0
13 x2k
(k!)2 22k
Figure 30.9
-
962 CHAPTER 30 Series
Transition to Convergence Tests
Because this chapter began with Taylor polynomials, it was
natural to move on to Taylor
series directly, without the traditional lead-in of convergence
tests for innite series. Taylors
Theorem enables us to deal with some convergence issues quite
efciently. Not only are
we able to show that the series for , sin , and cos converge,
but we can determine
that each converges to its generating function. Our previous
work with geometric series
allows us to conclude that the series for 11 converges to its
generating function on
1, 1. When we nd a Taylor series by manipulating a known Taylor
series, whetherby substitution, differentiation, or integration, we
can calculate the radius of convergence.
But, faced with a generic power series, we have few tools at our
disposal with which to
determine convergence and divergence. More fundamentally, we
have no systematic way
of determining the convergence or divergence of an innite series
of the form . The next
section will remedy this situation.
P R O B L E M S F O R S E C T I O N 3 0 . 4
For each series in Problems 1 through 9, determine whether the
series converges
absolutely, converges conditionally, or diverges.
1.
11 !1!
2.
111 !1!
3.
11 13
4.
21 ln
5.
10cos
10
6.
0
11
12
7.
11
100sin
2
8.
11 2
9.
01
210225
10. Is it possible for a geometric series to converge
conditionally? If it is possible, produce
an example.
11. How many nonzero terms of the Maclaurin series for ln1 are
needed to approx-imate ln
32
with an error of less than 104?
12. Approximate 1
with error less than 105.
-
30.4 Working with Series and Power Series 963
13. Arrive at the series for cos by differentiating the
Maclaurin series for sin .
14. Find the Maclaurin series for arcsin using the fact that
112
sin1 .What is the radius of convergence of the series?
In Problems 15 through 17, write the given integral as a power
series.
15.
cos2
16.
3
17.
115
18. Approximate 0.5
0 sin2 with error less than 108. Is your approximation an
overestimate, or an underestimate?
19. Approximate 0.1
0
13 with error less than 1010.
20. Find the Maclaurin series for ln2 along with its radius of
convergence.
21. (a) Find the Maclaurin series for ln
11
by subtracting the Maclaurin series for
ln1 from that for ln1 .(b) Show that when 1
3,
11
2.
(c) Use the rst four nonzero terms of the series in part (a) to
approximate ln 2.
Compare your answer with the approximation given by the rst four
terms of the
series for ln1 evaluated at 1, and the value of ln 2 given by a
calculatoror computer.
22. Show that
02
!is a solution to the differential equation 2 . What
familiar function does this series represent?
23. Show that if 0 is a power series solution to , then 01 ! .
What function does this series represent?
24. Use power series to solve the differential equation 9 . What
familiarfunction(s) does this series represent?
25. The Bessel function 0 is given by 0
01 2
!222. It converges for
all .
(a) If the rst three nonzero terms of the series are used to
approximate 00.1, will the
approximation be too large, or too small? Give an upper bound
for the magnitude
of the error.
(b) How many nonzero terms of the series for 01 must be used to
approximate 01
with error less than 104?
-
964 CHAPTER 30 Series
30.5 CONVERGENCE TESTS
In this section we focus on ways of determining whether or not a
given series converges.
We begin by looking at series of constants; in the last
subsection we apply our results to the
convergence of power series.
The Basic Principles
A series
1 converges to a sum if the sequence of its partial sums
converges to ,where is a nite number. In other words, if lim ,
where
1 , then the
innite series converges to . Otherwise, the series diverges.
Note that if lim ,then lim . The converse is not true.
Our rst case study was geometric series. (Refer to Chapter 18.)
For a geometric series
we are able to express in closed form and directly compute lim .
We nd that
0
converges to
1 if 1 and
diverges if 1.
Once we leave the realm of geometric series it can be difcult or
impossible to express
in closed form, so we generally cant compute lim directly.
Instead, we mightdetermine convergence or divergence by comparing
the series in question to a geometric
series or an improper integral.
We have already established one test for divergence; if the
terms of the series dont
tend toward zero then the series diverges.
nth Term Test for Divergence. If lim 0, then
1 diverges.If lim 0, we have no information and must turn our
attention back to the
sequence of partial sums.
D e f i n i t i o n
A sequence is increasing if 1 for all 1. It is decreasing if
1for all 1. If a sequence is either increasing or it is decreasing
it is said to bemonotonic.
A sequence is bounded above if there is a constant such that
forall 1. It is bounded below if there is a constant such that for
all 1.A sequence is said to be bounded if it is bounded both above
and below.
A bounded sequence may or may not converge. It could oscillate
between the bounds like
1. However, if the sequence is bounded and increasing, then its
terms must clusterabout some number . A similar statement can be
made for a decreasing sequence.The following theorem will prove
very useful.
-
30.5 Convergence Tests 965
B o u n d e d M o n o t o n i c C o n v e r g e n c e T h e o r
e m 15
A monotonic sequence converges if it is bounded and diverges
otherwise.
Suppose the terms of the series
1 are all positive. Then the sequence of partial sumsis
increasing: 1 1. Because the terms are positive, the sequence of
partialsums is bounded below by zero. Therefore, if is bounded
above, then converges andconsequently
1 converges; otherwise they diverge. We will use this line of
reasoning
repeatedly. Well refer to it as the Bounded Increasing Partial
Sums Theorem.
T h e B o u n d e d I n c r e a s i n g P a r t i a l S u m s T
h e o r e m
A series
1 , where 0, converges if and only if its sequence of partial
sumsis bounded above.
Our focus in this section is on the question of convergence
versus divergence and not on
the sum of a convergent series. Therefore, the starting point of
the series is not important; the
rst hundred or thousand terms of the innite series can be
chopped off without impacting
convergence issues. Keep this in mind when applying the results
of this section. For example,
if the sequence of partial sums is eventually monotonic, then
the Bounded Increasing Partial
Sums Theorem can be applied.
In the next few subsections we will discuss convergence tests
with the specication that
the terms of the series are positive. From the observation made
above, you can see that what
is really required is that the terms are positive for all
greater than some xed number,
or, more generally, have any of the required specications in the
long run.
The Integral Test
We revisit the idea of comparing an innite series and an
improper integral in the next
example.16
EXAMPLE 30.26 Determine whether the following series converge or
diverge.
(a)
113 1
13 1
23 1
33 1
43
(b)
11 1
1 1
2 1
3
SOLUTION In both of these series the terms are positive,
decreasing, and going toward zero, but the terms
of the series in part (b) are heading toward zero much more
slowly than those in part (a).
The values of some partial sums are given in the table on page
966. The information in the
table is inconclusive, but it leads us to guess that
113
might converge and
11
might diverge.
15 A formal proof of this theorem rests on the Completeness
Axiom for real numbers, which says that if a nonempty set of
real
numbers has an upper bound it must have a least upper bound.16
This was rst introduced in Section 29.4.
-
966 CHAPTER 30 Series
123456789
1011121314151617181920
1.01.1251.1620371.1776621.1856621.1902911.1932071.1951601.1965311.1975311.1982831.1988621.1993171.1996811.1999771.2046071.2048111.2049821.2051281.205253
11.7071062.2844572.7844573.2316703.6399184.0178834.3714364.7047705.0209975.3225095.6111845.8885346.1557956.4139946.6639946.9065307.1422327.3716487.595255
n Sn = n
k =1
1k3
Sn = n
k =1
1k
(a) (b)
Partial sums are recorded up to six decimal planes.
(a) To prove that
113
converges it is enough to show that the increasing sequence
of
partial sums is bounded. We do this by comparing the partial
sums to
113
, as
shown in Figure 30.10.
y
x10 2 3 4 5
y =
(1, 1)
(4, )
The areas of the shaded rectangles
correspond to the terms of the series1
13
1
x3
1
43(3, )
1
33
(2, )1
23
1
23+
1
33+
. . . This figure is not drawn to scale.
Figure 30.10
Each of the shaded rectangles has a base of length 1. The area
(base) (height),so the areas of the rectangles, from left to right,
are 1
13, 1
23, 1
33, . The sum of the
areas of the rectangles is
113
. Chop off the rst rectangle.
213
113
;
the rectangles lie under the graph of 13
. Consequently, lim
213 1 13 .
But
113
converges.
1
1
3 lim
1
3 lim
2
2
1
lim
122
12 1
2
Therefore, the partial sums of
113
are bounded by 1 12, and the series converges
by the Bounded Increasing Partial Sums Theorem. The sum of the
series is greater than
1 and less than 1.5.
(b) To prove that
11
diverges it is enough to show that the sequence of partial
sums
is unbounded. We do this by comparing the partial sums to
11 as shown in
Figure 30.11.
y
x1 2 3 4 5
y = 1x
(2, 12
(1, 1)
)
(3, 13
)(4,
14
)
The areas of the shaded rectangles
correspond to the terms of the series11
12
+13
+ +. . .
Figure 30.11
-
30.5 Convergence Tests 967
1
1
lim
1
12 lim
2
12
1
lim
2
2
Because we want to show that the partial sums are unbounded, we
draw rectangles that
lie above the graph of 1
. The areas of the shaded rectangles are, from left to
right,
11, 1
2, 1
3, , so the sum of the areas of the shaded rectangles is1 1 . We
see
that 1
11 , so
lim
11
1
lim
.
lim
.
The series diverges.
REMARKS
To show that a series with positive terms converges we show that
the increasing
sequence of partial sums is bounded, that is, is less than some
constant . To show
that it diverges, we show that where lim .Suppose we compare the
series
1 with the improper integral
1 . If
is positive, continuous, and decreasing on [1,, then by
including or omittingthe rst term of the series, we can depict the
area corresponding to the sum as lying
above or below the area corresponding to the improper integral.
(See Figures 30.12 and
30.13.)
y
x1 2 3 4 5 6
The sum of the area of the shaded
rectangles is a1 + a2 + a3 + a4 + a5
a1a2
a3 a4 a5
y = f (x)
(2, f (2))
Figure 30.12
y
x1 2 3 4 5 6
The sum of the area of the shaded
rectangles is a2 + a3 + a4 + a5 + a6
a2a3 a4 a5 a6
y = f (x)
(2, f (2))
Figure 30.13
-
968 CHAPTER 30 Series
Using the reasoning given we can obtain the Integral Test.
T h e I n t e g r a l T e s t
Let
1 be a series such that for 1, 2, 3 , where the function is
positive, continuous, and decreasing on [1,. Then
1
and
1
either both converge or both diverge.
The proof of the Integral Test is constructed along the lines of
Example 30.26 and makes a
nice exercise for the reader. Con