Sin x Taylor

P A R T

XSeries

30C H A P T E R

Series

30.1 APPROXIMATING A FUNCTION BY A POLYNOMIAL

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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.


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


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.


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.


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.


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 .


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.


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 ,


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 .


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.


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


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


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.


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.


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.


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.


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


" 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.


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.


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!


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.


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 .


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.


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.


(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.


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.


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

.


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.


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


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


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 .


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.


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


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.


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.


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.


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


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.


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!


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


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.


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.


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?


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.


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


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

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