Advanced Caclulus for Data Science

Chapter 3

Series

This chapter introduces the Taylor polynomial, which is a useful tool for approximating functions that

cannot be evaluated with arithmetic. Like with the derivative and integral before it, we would like to

send the error in these approximations to 0. This requires us to take a new kind of limit called a series.

We will develop the tools to work with series, with the ultimate goal of deﬁning and utilizing Taylor

series.

Contents

3.1 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

3.2 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

3.3 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

3.4 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

3.5 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

Section 3.1

Taylor Polynomials

Goals:

1 Approximate a function with a Taylor polynomial.

2 Compute error bounds for a Taylor polynomial.

When learning algebra and trigonometry, we learn to use exact values like

√

7 instead of decimal

approximations, like 2.646. This prevents us from introducing errors into our calculations. However,

there are also advantages to approximation. Decimal approximations give us a much better sense of the

size of a number than ln 873 or e

. (Which of these is larger?)

Unfortunately arithmetic does not give us methods for approximating many quantities. Ideally, we

would like a method of approximation whose accuracy is limited only by how much time we wish to

spend computing. An example of this is long division. We can compute as many decimal places of

as we want, getting closer and closer to the exact value. Of course, long division can only approximate

fractions.

The method we will develop in this section is called a Taylor polynomial. It gives us a way to

approximate otherwise incomputable functions. The beginning point is the tangent line. The tangent

line was the motivation for developing the derivative, but its greatest beneﬁt is not geometric. The

tangent line approximates the values of a function near the point of tangency. While the function may

be diﬃcult to evaluate, the equation of the tangent line is linear. We can evaluate it by hand.

Question 3.1.1

How Can We Improve on a Linearization?

Formula

The linearization or tangent line to a function f(x) at a has the equation.

L(x) = f(a) + f

′

(a)(x − a)

By design f and L have

1 Equal values at a.

2 Equal ﬁrst derivatives at a.

This means that for values of x near a, L(x) and f(x) will have similar values. L(x), which is easy

to compute, can be used as an approximation of f(x). As x travels away from a and y = f(x) curves

away from its tangent line, this method will lose accuracy. We could make a better approximation, if

we could match second, third, fourth derivatives of f (x). A line cannot do that, but a polynomial can.

170

Question 3.1.2

What Is a Taylor Polynomial?

A polynomial that mimics the ﬁrst n derivatives of a function is called a Taylor polynomial. Here is

the formal deﬁnition.

Deﬁnition

The n

Taylor polynomial of f (x) at x = a is a degree n polynomial that shares the value and ﬁrst

n derivatives of f at x = a. Its formula is

(x) =

k=0

(k)

(a)

(x − a)

Remarks

The variable is x. f

(k)

(a) is not a function but a number.

(0)

is the zeroth derivative, meaning f

(0)

(a) = f(a).

0! is deﬁned to be 1.

Example 3.1.3

Computing a Taylor Polynomial

Find the degree 3 Taylor polynomial of y =

√

x at x = 4.

Use it to estimate

√

171

Example 3.1.3

Computing a Taylor Polynomial

Solution

We will apply the equation of the Taylor polynomial where a = 4 and n = 3. Examining the

formula shows we need to know the value of ﬁrst three derivatives of f(x) at a = 4.

f(x) = x

1/2

f(4) = 2

′

(x) =

−1/2

′

(4) =

′′

(x) = −

−3/2

′′

(4) = −

′′′

(x) =

−5/2

′′′

(4) =

256

We can plug these into the summation formula:

(x) =

k=0

(k)

(4)

(x − 4)

f(4)

(1) +

′

(4)

(x − 4) +

′′

(4)

(x − 4)

′′′

(4)

(x − 4)

= 2 +

(x − 4) −

(x − 4)

256

(x − 4)

= 2 +

(x − 4) −

(x − 4)

512

(x − 4)

To approximate

√

5, notice

√

5 = f(5) and f(5) ≈ T

(5).

(5) = 2 +

(5 − 4) −

(5 − 4)

512

(5 − 4)

= 2 +

(1) −

(1) +

512

(1)

1024

512

128

512

−

512

1145

512

172

Example 3.1.4

Writing a Sum in Σ Notation

As our Taylor polynomials get longer, we would like to condense them into

notation. Part of

the challenge is choosing an expression that will produce all the terms of our sum. Write each of the

following sums in Σ notation.

4 + 7 + 10 + 13 + 16 + 19 + 22

2 + 6 + 18 + 54 + 162 + 486

−3 + 4 − 5 + 6 − 7 + 8 − 9 + 10

√

Solution

The terms increase by 3 each time. Repeated addition is multiplication, in this case 3k plus some

starting value. Starting with index k = 0 is convenient, because 3(0) = 0 at the starting value.

4 + 7 + 10 + 13 + 16 + 19 + 22 =

k=0

4 + 3k

The terms are multiplied by 3 each time. Repeated multiplication is exponentiation, in this case

times some starting value. Starting with index k = 0 is convenient, because 3

= 1 at the

starting value.

2 + 6 + 18 + 54 + 162 + 486 =

k=0

(2)(3

)

The absolute values of this sum could just be the values of the index variable. To create an

alternating + and − pattern, we can multiply by (−1)

−3 + 4 − 5 + 6 − 7 + 8 − 9 + 10 =

k=3

(−1)

In a fraction, we can model the numerator and denominator separately.

√

k=1

√

(k + 1)

173

Example 3.1.5

A Taylor Polynomial in

Notation

Write the 10th degree Taylor Polynomial for f(x) =

centered at x = 3.

Solution

Computing 10 derivatives seems excessive, so we will compute 4 and try to ﬁnd a pattern. We’ll write

f(x) = x

−2

and apply the power rule.

f(x) = x

−2

′

(x) = −2x

−3

′′

(x) = 6x

−4

′′′

(x) = −24x

−5

(4)

(x) = 120x

−6

We observe

The sign of these derivatives is alternating, which we can model with a (−1)

The coeﬃcients look like a factorial pattern, but oﬀset. For example when k = 2 we obtain 3!.

We model this with (k + 1)!.

The exponent of x decreases by the same amount each step. We model it with −2 − k.

This suggests a general formula for the kth derivative.

(k)

(x) = (−1)

(k + 1)!x

−2−k

We plug x = 3 into f

(k)

(x) and assemble the Taylor Polynomial:

(x) =

k=0

(−1)

(k + 1)!3

−2−k

(x − 3)

Question 3.1.6

How Accurate Is the Taylor Polynomial?

An approximation is much more useful, if we can put a bound on its error. We will present an error

bound theorem called “Taylor’s Inequality.” Taylor polynomials are eﬀective approximations because

they try to match the values and rates of change of the original function. In order to make a careful

argument, we begin with the basic principal that we can compare functions using the values of their

derivatives.

174

Theorem

Let f and g be diﬀerentiable functions. Consider an interval [a, b], and suppose f(a) = g(a).

1 If f

′

(x) = g

′

(x) on [a, b], then f(x) = g(x) on [a, b]

2 If f

′

(x) < g

′

(x) on [a, b], then f(x) < g(x) on (a, b]

Reasoning

Intuitive If two functions start at the same value at a, then the one that grows faster will have a higher

value at b.

Formal The Fundamental Theorem of Calculus says

f(x) − f(a) =

′

(t)dt g(x) − g(a) =

′

(t)dt.

Larger functions have larger integrals.

Figure: Two functions with a common value at a: f(x) with a smaller derivative and g(x) with a

larger derivative.

Notation

Given a function f(x) and its nth Taylor polynomial T

(x) centered at a, the remainder at x is

(x) = f(x) − T

(x)

175

Question 3.1.6

How Accurate Is the Taylor Polynomial?

If we are using T

(x) to approximate f(x),

(x) = −error of T

(x).

We should be very interested in knowing the value of R

(x). We will use our derivative comparison

theorem to make two arguments

1 If f

(n+1)

(x) is a constant M, then we can compute R

(x) exactly.

2 If |f

(n+1)

(x)| ≤ M then the error in 1 is the worst-case scenario.

Theorem

If f

(n+1)

(x) is a constant M on [a, b], then

f(x) = T

n+1

(x) = T

(x) +

(n + 1)!

(x − a)

n+1

Beginning with our assumption about the (n+1)th derivatives and the equality of the nth derivatives

at a, we can use our derivative comparison theorem to equate the nth derivatives on [a.b]. We can use

that equality to equate the (n −1)th derivatives on [a, b]. We continue this reasoning until we conclude

that the functions are equal.

n+1

f(x) =

n+1

(x) = M on [a, b]

f(a) =

n+1

(a)

f(x) =

n+1

(x) on [a, b]

n−1

f(a) =

n−1

n+1

(a)

n−1

f(x) =

n−1

n+1

(x) on [a, b]

f(a) =

n+1

(a)

f(x) =

n+1

(x) on [a, b]

f(a) = T

n+1

(a) f(x) = T

n+1

(x) on [a, b]

Because derivatives and values of

a Taylor polynomial match the function

Remark

This theorem tells us that when f

(n+1)

(x) is a constant M, R

(x) = f(x) −T

(x) =

(n+1)!

(x−a)

n+1

But what if f

(n+1)

(x) is not a constant? In this case we will settle for a bound on f

(n+1)

(x).

176

Theorem [Taylor’s Inequality]



(n+1)

(t)



≤ M for all x between a and b, then for all x between a and b,

(x)| ≤



(n + 1)!

(x − a)

n+1



To prove Taylor’s Inequality, we compare the derivatives of f(x) with the worst-case scenario w(x) =

(x) +

(n+1)!

(x −a)

n+1

. The derivatives w

(k)

(a) are the same as T

(k)

(a) and f

(k)

(a) for 0 ≤ k ≤ n,

and

n+1

w(x) = M.

n+1

f(x) ≤

n+1

w(a) = M on [a, b]

f(a) =

w(a)

f(x) ≤

w(x) on [a, b]

n−1

f(a) =

n−1

w(a)

n−1

f(x) ≤

n−1

w(x) on [a, b]

f(a) =

w(a)

f(x) ≤

w(x) on [a, b]

f(a) = w(a) f(x) ≤ w(x) on [a, b]

Because M is a bound on



n+1

f(x)



Because derivatives and values of

a Taylor polynomial match the function

To ﬁnish the argument we need to

1 Produce a lower bound for f using w(x) = T

(x) −

(n+1)!

(x − a)

n+1

2 Solve the inequality bounds for R

(x).

(x) −

(n + 1)!

(x − a)

n+1

≤ f(x) ≤ T

(x) +

(n + 1)!

(x − a)

n+1

−

(n + 1)!

(x − a)

n+1

≤ R

(x) ≤

(n + 1)!

(x − a)

n+1

3 Repeat for intervals of the form [b, a]. These work the same way with a sign reversed.

177

Example 3.1.7

A Taylor Approximation Error Bound

Let f(x) = sin x.

Give a general form for the n

Taylor polynomial for f at x = 0.

Find a bound on f

(n)

(x) for each n.

What happens to the error bound as x increases but n stays the same?

What happens to the error bound as n increases but x stays the same?

What does this tell us about the relationship between the T

(x) approximations and f(x)?

Solution

For the Taylor polynomial formula, we need to compute the derivatives of f(x).

f(x) = sin x f(0) = 0

′

(x) = cos x f

′

(0) = 1

′′

(x) = −sin x f

′′

(0) = 0

′′′

(x) = −cos x f

′′′

(0) = −1

(4)

(x) = sin x f

(4)

(0) = 0

(5)

(x) = cos x f

(5)

(0) = 1

In order to write a general Taylor polynomial, we would need a general expression for f

(k)

(0). The

pattern is obvious, but trying to express it as a formula is much more diﬃcult. The solution is a

trick worth remembering:

Since the even derivatives are zero, those terms do not appear in our Taylor polynomials. Since

we want to only have odd terms in our summation, we can let our index variable be k, but our

exponents in each term be 2k + 1. Thus as k goes from 0 to n, the summation will include only

the odd terms x

through x

2n+1

. We can produce the following chart to work out our coeﬃcients:

k f

(2k+1)

(0)

0 1

1 −1

2 1

3 −1

178

This is an easier pattern to express:

(2k+1)

(0) = (−1)

Now we are ready to write a formula. Since we intend to sum from k = 0 to k = n, we are

actually producing the (2n + 1)th Taylor polynomial.

2n+1

(x) =

k=0

(2k+1)

(0)

(2k + 1)!

2k+1

k=0

(−1)

(2k + 1)!

2k+1

These are the odd degree Taylor polynomials, but what about the even numbered ones? Since

(x) is just T

2n−1

(x) plus the 2nth term, and the 2nth term is zero, we can write

(x) =

n−1

k=0

(−1)

(2k + 1)!

2k+1

Given the chart above, we can see that the derivatives are sines and cosines. These are bounded

above by 1 and below by −1. Since Taylor’s inequality requires a bound of the form |f

(n+1)

(x)| ≤

M, we write

(n+1)

(x)| ≤ 1

And luckily, thus works for all x and all n.

Taylor’s Inequality says that |R

(x)| ≤



(n+1)!

n+1



. As x goes to ∞, this bound goes to ∞ as

well. This makes sense, since T

(x) is polynomial, while the function it is approximating stays

between −1 and 1.

When n increases x

n+1

increases by a factor of x. On the other hand, (n + 1)! increases by a

factor of n + 2. As n increases without bound, (n + 1)! grows faster than x

n+1

and their ratio

approaches 0.

Any T

(x) will eventually become inaccurate outside a certain distance from 0. On the other

hand, if we want to approximate sin(x) for a particular x, we can make T

(x) have as small an

error as we want by choosing suﬃciently large n.

179

Example 3.1.7

A Taylor Approximation Error Bound

Figure: f(x) = sin x approximated by its Taylor polynomials, T

(x)

Main Ideas

In order to understand how the error changes as n increases, we need to have an expression for

(n)

(x).

We can choose M to be the largest value of |f

(n+1)

| on the interval [a, x]. This may not be the

value of |f

(n+1)

(a)|.

In general, Taylor polynomials will become less accurate the farther you get from a.

We can often mitigate this inaccuracy by choosing larger values of n.

The (n + 1)! in Taylor’s Inequality might suggest that as n increases, the error in the nth Taylor

polynomial must shrink toward 0. However, this is not the case. Some functions are not well estimated

by their Taylor polynomial.

180

Example

f(x) =

(

0 if x ≤ 0

−

if x > 0

(k)

(0) = 0 for all k. So the Taylor polynomial

at x = 0 is

(x) =

k=0

No matter how large n gets, T

(x) will not get any closer to f(x) for any x > 0.

How can this happen, given Taylor’s Inequality? The derivatives of f get bigger and bigger. M

grows so fast that the error R

(x) gets no smaller even with an (n + 1)! in the denominator of Taylor’s

Inequality.

Figure: A function whose derivative bounds grow factorially

Despite examples like this, it turns out that Taylor polynomials often do a good job of approximating

functions. For numerical computations, an approximation is good enough. For more theoretical situ-

ations, we would like to let n go to ∞ so that the error goes to 0 and we can use the polynomial as

an exact replacement of the function. Unfortunately, with inﬁnitely many terms, we no longer have a

polynomial at all. Instead we have an object that we will call a Taylor series. We will develop the tools

to deﬁne and work with Taylor series over the course of this chapter.

181

Section 3.1

Exercises

Summary Questions

Why do we use Taylor polynomials?

Why is there a denominator of k! in the formula for a Taylor polynomial?

Explain why we’d always rather center a Taylor polynomial for y = ln x at x = 1.

What properties make a Taylor polynomial T

(x) a better approximation of f(x)?

3.1.1

Suppose we use the linearization of f (x) =

√

x at x = 8 to approximate

√

What is the relationship between f(x) =

√

x and

√

Suppose L(x) is the the linearization of f (x) at x = 8. Would you expect L(6) to overesti-

mate or underestimate

√

6? Explain in a sentence or two.

Suppose you were locked in a room with only a pencil and paper and asked to compute the ﬁrst

ten decimal places of the following numbers:

√

7 e

Which could you compute?

For the ones you can compute, how would you do it?

182

3.1.2

Is a tangent line a Taylor polynomial?

Suppose T

(x) is the Taylor polynomial for f(x) centered at x = 10. List what information T

(x)

and f(x) have in common, being as speciﬁc as possible.

If f (x) is a decreasing function, what can you say about the coeﬃcients of any Taylor polynomial

of f(x)?

Q10

Suppose f(x) has a Taylor polynomial

(x) = 5 + 3(x − 2) −

(x − 2)

+ 2(x − 2)

What is f (2)?

Is f increasing or decreasing at x = 2?

Is f concave up or concave down at x = 2?

3.1.3

Q11

Let f(x) = e

Find the degree 8 Taylor polynomial of y = f(x) centered at x = 0.

How could you use this to estimate the value of e?

Can you use sigma notation to write a general form for the degree n Taylor polynomial of

y = e

Q12

Let f(x) = ln x

Write the 5th Taylor polynomial of f(x) at x = 1.

Use your polynomial to approximate ln 2.

183

Section 3.1

Exercises

Q13

Write the 10th Taylor polynomial for f (x) = cos x centered at x = π.

Q14

Write the 4th Taylor polynomial for f (x) =

centered at x = 5.

3.1.4

Q15

Write each of the following sums in Σ notation.

15 − 45 + 105 − 315 + 945

24 + 19 + 14 + 9 + 4 − 1 − 6

Q16

Write each of the following sums in Σ notation.

11 − 13 + 15 − 17 + 19 − 21 + 23

384 + 192 + 96 + 48 + 24 + 12 + 6

100

1000

10000

3.1.5

Q17

Write an expression in Σ notation for the 53rd Taylor polynomial of f (x) = ln x centered at

x = 1

Q18

Write an expression in Σ notation for the 15th Taylor polynomial of f(x) = e

centered at x = 0

Q19

Write an expression in Σ notation for the 100th Taylor polynomial of f(x) = cos x centered at

x = 0

Q20

Write an expression in Σ notation for the 71st Taylor polynomial of f (x) =

centered at x = 10

184

3.1.6

Q21

Why don’t we have any theorems for a lower bound for error? Give your answer in a few sentences.

Q22

Suppose you are using Taylor polynomials of f(x) centered at x = 0 to approximate f(−3).

However, for each k, the best bound you can put on f

(k)

(x) on [−3, 0] is

. Will you be able

to guarantee a good approximation of f(−3) this way? Explain.

Q23

Suppose the fourth derivative of f(x) is f

(4)

(x) = e

. Suppose we have written T

(x), the

degree 4 Taylor polynomial of f(x) centered at x = 1. What can you say about the diﬀerence

between T

(5) and f(5)? Be speciﬁc and justify your answer with a computation. You do not

need to simplify any arithmetic in your calculations.

Q24

Sketch a graph of y = e

and several tangent lines. On which part of the graph do the tangent

lines appear to approximate the function better? Does Taylor’s Inequality conﬁrm this observa-

tion? Explain.

3.1.7

Q25

Here is the degree 3 Taylor polynomial of f(x) =

√

x centered at x = 4:

(x) = 2 +

(x − 4) −

(x − 4)

512

(x − 4)

Which derivative will let you bound the error of this approximation?

Can you put a bound on this derivative that holds for all x?

Can you put a bound on this derivative that holds for x in the interval [4, 5]?

What error bound does this suggest for using T

(5) to approximate

√

Q26

Let f(x) =

√

Write the degree 2 Taylor polynomial of f centered at x = 8.

If you wanted to use the Taylor polynomial to approximate

√

10, how would you do that?

185

Section 3.1

Exercises

What bound could you place on the error in the approximation in

Q27

Let f(x) = e

Write the degree 5 Taylor polynomial of f centered at 0.

How could we use this polynomial to approximate

√

Produce an error bound for your approximation in

Q28

Let f(x) = xe

Compute the Taylor polynomial T

(x) for f (x) centered at x = 0.

Compute the theoretical error bound for T

(2).

Explain the diﬃculties that would arise from this error bound, if your goal is to approximate

f(2) by hand. Can you resolve them?

Q29

Let f(x) = cos 3x

Write the degree 4 Taylor polynomial of f centered at x = 0.

How would you use that Taylor polynomial to approximate the value of cos

3π

What bound can you place on the error of such an approximation?

Q30

Consider the graph of y = f (x) below.

186

Suppose you wanted to produce the second degree Taylor polynomial of f centered at a =

−1. Indicate whether the constant term and each coeﬃcient would be positive or negative.

Provide evidence for your answer.

Would T

(4) underestimate or overestimate f(4)? Explain.

Synthesis and Extension

Q31

Let f(x) = x

− 3x + 5.

Write an expression for T

(x), the Taylor polynomial centered at x = 2.

What can you say about th error R

(x) for any x?

What relationship does this suggest between f (x) and T

(x)?

Can you verify this relationship algebraically?

Conjecture a general relationship between polynomial functions and certain Taylor polyno-

mials. Can you use Taylor’s inequality to justify your conjecture?

187

Section 3.2

Sequences

Goals:

1 Use notation to describe the terms of an inﬁnite sequence.

2 Calculate the limit of an inﬁnite sequence.

Sequences are the ﬁrst step in our development of Taylor series. While they appear to have little in

common with polynomials of inﬁnite degree, they are the scaﬀolding on which such objects are built.

Question 3.2.1

What Is a Sequence?

A sequence is an ordered set of numbers. If this set is inﬁnite, we can most rigorously deﬁne it by

giving a general formula for the n

term for some index variable n. Here are three diﬀerent notations

for the same sequence.



. . .

 

n + 1



∞

n=1

n + 1

Example

The ﬁrst three terms of





∞

n=0

are

= 0

= 1

Question 3.2.2

What Is the Limit of a Sequence?

Deﬁnition

If we can make the elements of a sequence a

arbitrarily close to some number L by considering only n

above a certain number, then we write

lim

n→∞

= L

and we say the sequence converges to L. If a

does not converge to any such L then we say it

diverges.

188

Remarks

The ﬁrst few or even the ﬁrst thousand terms of a sequence have no bearing on the limit. We

only care that we can eventually get close to L.

“Arbitrarily close” means any level of closeness than anyone could ask for. Eventually the sequence

must be within

100

of L, and

1000

and

1000000

Figure: A sequence converging to L = 3

Example 3.2.3

Computing a Limit

Calculate lim

n→∞

n + 1



. . .



189

Example 3.2.3

Computing a Limit

Solution

Writing the ﬁrst few terms suggests that this sequence approaches 1. To see that, we can measure the

distance to 1:

1 − a

= 1 −

n + 1

We can make this smaller than any positive number. For instance to make a

within

1000

of 1, we can

consider only n > 1000. We conclude lim

n→∞

n + 1

= 1

Figure: The sequence

n+1

converges to L = 1.

Question 3.2.4

How Are Limits of Sequences and Functions Related?

The deﬁnition of lim

n→∞

should look familiar. The deﬁnition of the limit of a function is similar.

In fact, the limit of a f(x) as x → ∞ has a nearly identical construction, except that n must be an

integer, while x can be any real number. The following theorem lets us use that connection to evaluate

limits.

190

Theorem

Suppose for a sequence a

, there is a function f (x) such that f(n) = a

for all n (or at least all n

suﬃciently large). If

lim

x→∞

f(x) = L

we can conclude that

lim

n→∞

= L.

Example 3.2.5

Sequence Limits Using Functions

Find limits of the following sequences:

lim

n→∞

n + 3

lim

n→∞

lim

n→∞

−n

lim

n→∞

lim

n→∞

(−1)

Solution

We will use x to denote a real number variable and n to denote natural numbers.

lim

x→∞

x + 3

= 2, so lim

n→∞

n + 3

= 2.

lim

x→∞

= 0, so lim

n→∞

= 0.

lim

x→∞

−x

= 0 so lim

n→∞

−n

= 0.

191

Example 3.2.5

Sequence Limits Using Functions

lim

x→∞

can be evaluated with L’hˆopital’s rule.

lim

x→∞

= lim

x→∞



∞

form, L’hˆopital’s again



= lim

x→∞

= 0so lim

n→∞

= 0

f(x) = (−1)

is not well deﬁned for real numbers so we can’t use its limit. Instead examine the

sequence directly. The sequence has the form

−1, 1, −1, 1, −1, 1, −1, 1, . . .

This does not approach arbitrarily close to any number. No matter how many early terms we

disregard, there will always be terms remaining that are not close to 1, or not close to −1 or not

close to any other number. Thus a

= (−1)

diverges.

The following limit laws for sequences should look familiar. They mirror the laws for limits of

functions.

Theorem [Limit Laws]

If lim

n→∞

= K and lim

n→∞

= L then the following sequences converge with the following limits:

lim

n→∞

+ b

) = K + L

lim

n→∞

− b

) = K − L

lim

n→∞

) = KL

If L = 0, then lim

n→∞

For any constant c, lim

n→∞

= cK

192

Synthesis 3.2.6

Indeterminate Forms with Factorials

We will encounter sequences of the form a

. If b

or c

both go to 0 or ±∞, then any attempt

to use

lim

n→∞

= lim

x→∞

f(x)

would require l’Hˆopital’s rule.

Dominance

We say f (x) dominates g(x) if lim

x→∞

f(x)

g(x)

= ±∞. We write

f(x) >> g(x)

Even if you include a constant multiple or add multiple functions together, the dominant function

will outgrow any combination of dominated ones. We have already established an order of dominance

using l’Hˆopital’s rule:

exponential

(larger base>>smaller base)

polynomial

(larger degree>>smaller)

root

(smaller power>>larger)

logarithm

(smaller base>>larger)

But n! is not a diﬀerentiable function. We cannot analyze it using l’Hˆopital’s rule. Where does it ﬁt

in the domincance pecking order?

Theorem

As n → ∞, n! will eventually dominate any exponential function (and thus any polynomial, root or

logarithm).

We will not provide a formal proof, but here is a useful thought experiment. Suppose we compare

n! to 63

. At ﬁrst 63

grows faster, multiplying by 63 every time we increase n. However, when n is

greater than 63, n! is multiplying by a higher number. When n reaches one billion, 63

increases by a

factor of 63 every step, while n! increases by a factor of 1, 000, 000, 000. By this point n! is much larger

and growing much faster.

193

Section 3.2

Exercises

Summary Questions

Why do we use n instead of x as an index for a sequence?

Describe three diﬀerent ways of denoting a sequence.

When is the limit of a sequence equal to the limit of a function?

If a

= b

+ 1000 for 1 ≤ n ≤ 2000000, what does that tell us about the limits lim

n→∞

and

lim

n→∞

3.2.1

Find a general expression for a

, the n

term of the following sequences. Use this to write the

sequences using both other types of notation.

{2, 5, 10, 17, 26, 37, 50, . . .}



, −

, . . .





, . . .



What is the fourth term in the sequence {n

− 5n}

∞

n=3

194

3.2.2

Show using the deﬁnition of the limit of a sequence that lim

n→∞

sin n

= 0.

Show using the deﬁnition of the limit of a sequence that lim

n→∞

− 1

= 1.

A sequence is increasing if every term is larger than the previous term. Must an increasing

sequence always diverge? Explain.

Q10

A sequence is alternating if its terms alternate between positive and negative values. Is it possible

that the limit of an alternating sequence exists? What would its value have to be?

3.2.3

Q11

Consider the sequence a

= 2

What function could we write such that f(n) = a

Does lim

x→∞

f(x) converge?

Does the theorem equating limits of functions and sequences apply to this function?

Can we argue that lim

n→∞

diverges anyway?

Q12

Consider the sequence a

= n sin(πn)

What is lim

x→∞

x sin(πx)?

Compute the ﬁrst few values a

, a

, and a

What is lim

n→∞

n sin(πn)?

Does this contradict one of our theorems? Explain.

195

Section 3.2

Exercises

3.2.4

Q13

Compute lim

n→∞

log n

Q14

Compute lim

n→∞

Q15

Compute lim

n→∞

+ 3

− 9

Q16

Compute lim

n→∞

sin n

log n

Q17

Compute lim

n→∞

√

Q18

Compute lim

n→∞

tan

−1

3.2.5

Q19

Compute lim

n→∞

Q20

Compute lim

n→∞

+ 3n + 1

Q21

Does n

grow faster or slower than n!? Explain.

Q22

Yuran knows that lim

n→∞

= ∞ because n! growns faster than a

. However, he thinks he can

make the denominator grown faster than the numerator if he uses a product like

Will he eventually obtain a non-inﬁnite limit by this method? Explain how you know.

196

Synthesis & Extension

Q23

Suppose we have a sequence a

(

f(n) if n ≤ 342

g(n) if n > 342

. Which of the following could help us

evaluate lim

n→∞

lim

x→∞

f(x)

lim

x→∞

g(x)

Q24

Let T

(x) be the nth Taylor polynomial of f(x) = ln x centered at x = 1.

Write an expression for T

(x) using Σ notation.

Write an expression for the error bound of T

(x) for some x between 0 and 1.

For what values of x will the error bound shrink to 0 as n goes to ∞?

197

Section 3.3

Series

Goals:

1 Identify partial sums of a series.

2 Recognize harmonic and alternating harmonic series.

3 Apply the divergence test.

4 Evaluate geometric series.

5 Apply the ratio test.

The ﬁrst step in understanding a Taylor polynomial of inﬁnite degree is understanding how to add

up inﬁnitely many of anything. This proposition is mechanically absurd. Addition is an operation for

two numbers at a time. Adding three or four numbers requires us to add two or three times. Adding

inﬁnitely many requires us to add inﬁnitely many times, something no one has time to do.

Yet there are some intuitive exercises we could perform. Suppose we lay a length of

m next to

next to

m. If we continued indeﬁnitely, we could imagine these lengths extending an entire meter.

Figure: One meter expressed as a sum of inﬁnitely many smaller lengths

What reasoning could we use to make this exercise rigorous? How could we add up lengths or

numbers where the pattern is not so intuitive? The formal object that does this is called a series. A

series is the ﬁrst step on our way to push the Taylor polynomial to inﬁnite degree. It is also the most

general. While we are concerned with one speciﬁc (and very useful) type of series, there are other

applications worth exploring as well.

Question 3.3.1

What Is a Series?

You have been encountering series since you ﬁrst learned about decimals. You likely have not seen

a rigorous description of what they mean.

0.33333333 . . . 3.1415926...

We can write

0.3333 . . . =

100

1000

10000

+ ···

3.1415 . . . = 3 +

100

1000

10000

+ ···

You may have an intuitive sense of what these quantities are, but what does it mean to add up

inﬁnitely many numbers?

198

Deﬁnition

A series is a sum of the form

∞

k=1

where a

is an inﬁnite sequence. If it is more convenient, we can

give k a diﬀerent initial value. If the context is clear, we can write

as a shorthand.

Example

0.33333 . . . =

∞

k=1

The harmonic series is

∞

k=1

This tells us what a series is but not how to evaluate it. How do we know that, for example

0.333 . . . =

We evaluate a series by associating it with a sequence of partial sums.

Deﬁnition

The n

partial sum of the series

∞

k=1

= a

+ a

+ ··· + a

A series

∞

k=1

converges to L if

lim

n→∞

= L.

A series that does not converge to any L diverges.

Vocabulary Note

Do not confuse a sequence with a series. One is a list of numbers. The other is the sum of a list of

numbers.

199

Example 3.3.2

Computing Partial Sums

Consider

∞

k=1

Compute the ﬁrst few partial sums s

, s

of this series.

Compute lim

n→∞

Solution

100

1000

333

1000

100

1000

10000

3333

10000

In order to use our usual methods of limits, we would need an algebraic expression for s

. It isn’t

immediately clear how to produce one. Given our knowledge of decimals, we expect the answer

to be

. We will use this as a hint. We expect

− s

to approach 0.

− s

300

− s

3000

− s

30000

extrapolating suggests

− s

3(10)

Assuming this pattern holds, we have

lim

n→∞

− s

= lim

n→∞

3(10)

= 0

and we conclude that

∞

k=1

200

Main Idea

Often, we can show that

∞

k=1

= L by computing L − s

and seeing that it converges to 0.

Figure: The partial sums s

converging to L =

Example 3.3.3

The Harmonic Series

We have seen examples of series in which the terms approach 0 as k → ∞. These have allowed us

to add inﬁnitely many terms and obtain a ﬁnite sum. Does this always work? No. A series can have its

terms approach 0, and yet the partial sums go to ∞. The most famous example of this is the harmonic

series:

∞

k=1

. Rather than computing the partial sums directly (which would be a lot of computation)

we will compare the partial sums to an expression that is easier to calculate. We will replace each term

by a fraction with a power of 2 in the denominator. Here’s what we’ll do with s

| {z }

=1 +

Since we replaced each term with something smaller and obtained a sum of

, we can conclude that

. Continuing this pattern, the terms

sum to more than

so s

. In general we can

201

Example 3.3.3

The Harmonic Series

make s

bigger than any integer c by setting n = 2

where

1 +

m > c.

This tells us that the harmonic series diverges.

Question 3.3.4

What Is a Geometric Series?

The two series so far that we have been able to evaluate belonged to a larger family. These are the

geometric series.

Deﬁnition

A geometric series is a series of the form

∞

k=1

k−1

a is the initial term. r is the common ratio between terms.

Example

∞

k=1





k−1

= 1 +

+ ···

∞

k=1





k−1

100

1000

+ ··· =

Unlike many other series, geometric series are simple enough that we can write a formula for their

sum. We can get a convenient expression for s

by performing a cute algebra trick. We’ll multiply s

by r and subtract rs

from s

. Most of the terms cancel and we obtain an equation that we can solve

for s

= a + ar + ar

+ ··· + ar

n−1

−rs

= −ar − ar

− ar

− ··· − ar

(1 − r)s

= a − ar

a(1 − r

)

1 − r

The last step requires that 1 − r = 0, since we cannot divide by 0. As long as r = 1, we can evaluate

the series by taking a limit.

∞

k=1

k−1

= lim

n→∞

a(1 − r

)

1 − r

202

To evaluate this limit, we need to understand the behavior of r

as n → ∞

If −1 < r < 1 then higher powers of r get smaller and smaller and r

→ 0.

If r > 1 then higher powers of r get larger and larger and r

→ ∞.

If r < −1 then higher powers of r get larger but alternate signs. lim

n→∞

does not exist.

If r = 1 then the series is a + a + a+ a + a + ···, then s

= an which diverges to ±∞, depending

on the sign of a.

If r = −1 then the series is a − a + a − a + a − ···, then s

alternates between a and 0. This

sequence does not converge.

We can apply the above to completely solve the problem of evaluating a geometric series. Our result is

the following theorem:

Figure: The partial sums of

k−1

for various r

Theorem

Geometric series have the following partial sums

k=1

k−1







a(1 − r

)

1 − r

if r = 1

an if r = 1

These converge to

1 − r

when |r| < 1 and diverge when |r| ≥ 1.

203

Example 3.3.5

Evaluating Geometric Series

Identify a and r in the following geometric series. Then evaluate the series.

+ ···

∞

n=2

0.999999 . . .

Solution

a is the initial term, which is

. The common ratio is the ratio between any two terms.

4/15

2/3

Since |r| < 1, the sum of the series is

∞

k=1







k−1

1 −

The initial term of this series is 9. The common ratio is 3. Since |3| ≥ 1,

∞

n=2

diverges.

0.999999 . . . =

100

1000

+ ···. This has an initial term of

and a common ratio of

|r| < 1 so

0.999999 . . . =

1 −

= 1

204

Question 3.3.6

What Does the Size of a

Tell Us About

The discussion of the geometric series suggests that certain properties of a series make convergence

impossible. Speciﬁcally, in the cases in which the terms were not shrinking to 0, the partial sums were

growing without bound or oscillating. This intuition can be formalized in the following theorem, which

applies to more than just geometric series.

Theorem [The Divergence Test]

Let a

be a sequence. If lim

k→∞

= 0, then the series

∞

k=1

diverges.

Remark

The divergence test does not tell us anything, if lim

k→∞

= 0. The series might converge, and it might

not. In this case we say the test is inconclusive.

Example 3.3.7

Applying the Divergence Test

What does the divergence test tell us about each of the following series?

∞

k=2

∞

k=2

∞

k=2

− 1

+ 7

∞

k=2

205

Example 3.3.7

Applying the Divergence Test

Solution

The sequence is a

= 3

. lim

k→∞

= ∞. This limit is not 0, so by the divergence test, the series

diverges.

The sequence is a

. lim

k→∞

= 0. The divergence test is inconclusive. It cannot tell us

whether this series diverges or converges. By our earlier work, we happen to know this series

diverges.

The sequence is a

−1

. lim

k→∞

− 1

+ 7

. This limit is not 0, so by the divergence test,

the series diverges.

The sequence is a

. We need L’Hˆoptial’s rule to evaluate the limit.

lim

k→∞

= lim

k→∞



still

∞

form



= lim

k→∞

= 0

The divergence test is inconclusive. It cannot tell us whether this series diverges or converges. It

turns out that this series converges, but we do not have a method to verify that yet.

Question 3.3.8

What Is the Ratio Test?

So far we have two tests to determine the convergence of a series. One test is very speciﬁc, applying

only to geometric series. The other is very imprecise. The divergence test is often inconclusive. It

does not help us to evaluate a series at all, only recognizing some series that diverge. Unfortunately,

these shortcoming are typical of series tests. A rigorous study of inﬁnite series requires learning almost a

dozen tests. On a randomly chosen series, most of these tests will be inconclusive, and none of them will

give a numerical value, even if the series happens to converge. Because we are interested in extending

Taylor polynomials to have inﬁnitely many terms, some of these tests are much more useful than others.

The most useful is the ratio test, though it is still no help in evaluating a series and is still sometimes

inconclusive.

In the case of a geometric series,

k−1

, the common ratio between terms determines whether this

series grows out of control, or whether the terms shrink quickly enough that the partial sums converge.

Even when a series is not geometric, we can attempt to apply similar reasoning to determine whether it

converges. A non-geometric series does not have a constant ratio. The ratio between successive terms

will change as we progress through them. We will instead compute the limit of these ratios.

206

Theorem [The Ratio Test]

If lim

k→∞



k+1



= L < 1, then

converges absolutely.

If lim

k→∞



k+1



= L > 1 or is inﬁnite, then

is divergent.

If lim

k→∞



k+1



= 1, then the ratio test is inconclusive.

Remark

Converges absolutely is a term for series with both positive and negative terms. It means the series

would converge, even if the signs of all the terms were all positive. The alternative is conditional

convergence, meaning the series’s convergence may require the positive and negative terms partially

canceling each other out.

Example

The series

1 −

−

− ···

converges (we won’t prove this). If we made all the terms positive, it would be the harmonic series,

which diverges. This series converges conditionally, not absolutely.

Absolute versus conditional convergence can be interesting to play with. You may see references to

it in other math books, but we won’t have any further use for it.

Example 3.3.9

Applying the Ratio Test

Does

∞

k=1

(−1)

k−1

converge or diverge?

Does

∞

k=1

converge or diverge?

Does

∞

k=1

k converge or diverge?

207

Example 3.3.9

Applying the Ratio Test

Solution

First we will compute and simplify the ratio. Then we will take its limit and draw a conclusion.



k+1



(−1)

(k+1)!

(−1)

k−1



(−1)

k−1

(k + 1)!



(−1)

(1)(2)(3) ···(k)

(−1)

k−1

(1)(2)(3) ···(k)(k + 1)



(expand the factorials)



(−1)

k−1

(k + 1)



(cancel the matching factors)



−1

k + 1



(cancel k − 1 powers of − 1)

k + 1

(absolute value of a negative number is its negatve)

Now we take the limit

lim

k→∞

k + 1

= 0

0 < 1 so by the ratio test,

∞

k=1

(−1)

k−1

converges.

We will apply the ratio test. First we compute the ratio, and then we take a limit.



k+1



k+1

(k+1)



k+1

+ 2k + 1)



+ 2k + 1



(cancel the 2s)

+ 2k + 1

lim

k→∞

+ 2k + 1

= 2

2 > 1 so by the ratio test, this series diverges.

208

We will apply the ratio test. First we compute the ratio, and then we take a limit.



k+1



k + 1



k + 1

lim

k→∞

k + 1

= 1

Here the ratio test is inconclusive. It cannot tell whether this series converges or diverges. However,

we can probably ﬁgure this out another way. The terms of this series are increasing, which means

the partial sums will grow faster and faster. This was the reasoning behind the divergence test.

lim

k→∞

k = ∞

Since lim

k→∞

k = 0, the divergence test concludes that the series diverges.

Main Ideas

When applying the ratio test, be sure to replace every k with k + 1 for the a

k+1

term.

Familiarize yourself with the algebra rules that allow you to simplify ratios of exponentials and

factorials.

Example 3.3.10

A Strategy for Series Tests

209

Example 3.3.10

A Strategy for Series Tests

Strategy

Given the three ways we have to test for divergence and convergence and the relative ease of applying

each, here is a reasonable approach to testing a series.

Check lim

n→∞

by dominance

Compute

n+1

Compute lim

n→∞

n+1

converges

diverges

Inconclusive, look up another test

not zero

zero

hard to tell

constant |r| ≥ 1 constant |r| < 1

not constant

< 1> 1

= 1

hard to tell

Let’s apply our strategy to see what we can tell about

∞

n=1

Solution

First we’ll check that the terms go to zero. If they don’t we quickly classify this as a divergent series.

lim

n→∞

= 0

They do, so we need another check. Now we’ll compute the ratio between terms.

n+1

(n+1)

+ 2n + 1

This is not a constant; it depends on n. Thus a

is not a geometric series. We’ll try the ratio test.

lim

n→∞



n+1



= lim

n→∞



+ 2n + 1



= lim

n→∞

+ 2n + 1

= 1

This means that the ratio test is inconclusive. We do not know whether this series converges or diverges.

We have exhausted all our tests. If we want the answer, we need to look up another test.

210

Section 3.3

Exercises

Summary Questions

What is the diﬀerence between a sequence and a series?

How do we evaluate a series?

What is a geometric series. How do we evaluate one?

What does it mean to say that a series test is inconclusive?

How do each of the following factors behave in the ratio



k+1



(p a constant)

(c a constant)

How would the ratio test apply to a geometric series

k−1

3.3.1

Give a more common name for each of the following series.

2 +

100

1000

10000

100000

+ ···

100

1000

10000

+ ···

Use a calculator to get a decimal approximation of

and write it as a series of fractions with

powers of 10 as denominators.

211

Section 3.3

Exercises

3.3.2

Consider the series

∞

k=1

k(k + 1)

Compute the ﬁrst four elements in the series.

Compute the partial sums: s

, s

What do the partial sums appear to be converging to?

Can you use algebra to generalize your answer to

to s

Q10

Compute the ﬁrst 3 partial sums of

∞

k=1

k + 1

. Don’t simplify the arithmetic.

Q11

Compute the ﬁrst four partial sums of

∞

k=1

(−1)

. What do you think this suggests about the sum

of the series?

Q12

Compute the ﬁrst ﬁve partial sums of

∞

k=0

(−2)

. Use them to make a prediction about the value

of the series.

3.3.3

Q13

Give an example of an n such that you know the nth partial sum of the harmonic series is greater

than 20.

Q14

Modify our argument for the harmonic series to show that

∞

k=0

√

diverges?

212

3.3.4

Q15

+ ··· a geometric series? How can you tell?

Q16

Is 1 + 4 + 9 + 16 + 25 + ··· a geometric series? How can you tell?

Q17

The ﬁrst two terms of a geometric series are 5 and 7.5. What is the third term?

Q18

The ﬁfth term of a geometric series is 17. The eigth term is 51. What is the sixth term?

3.3.5

Q19

Evaluate

∞

k=0

5(0.3)

Q20

Evaluate

∞

k=0





Q21

Evaluate

∞

j=3

Q22

Evaluate

∞

k=1

0.8

Q23

Evaluate

∞

k=4

(18)

Q24

Evaluate

∞

k=1

100

. What decimal does this represent?

Q25

For what values of z does

∞

k=0

converge?

Q26

For what values of p does

∞

k=3

12p

converge?

213

Section 3.3

Exercises

3.3.6

Q27

If a

100

for all k, then what can you say about the value of s

k=1

Q28

If lim

k→∞

100

, use the deﬁnition of a limit and the reasoning in the previous exercise to

show that

∞

k=1

diverges.

3.3.7

Q29

What does the divergence test say about

∞

k=1

Q30

What does the divergence test say about

∞

k=1

+ 1

+ 3k

Q31

What does the divergence test say about

∞

k=2

ln k?

Q32

What does the divergence test say about

∞

k=2

ln k

3.3.8

Q33

Will the divergence test detect every series that “fails” the ratio test (L > 1)? Explain.

Q34

If lim

n→∞



an + 1



does not exist, the ratio test is inconclusive. Give examples of two series where

this limit does not exist, one series that diverges and one that converges.

214

3.3.9

Q35

Apply the ratio test to

∞

k=1

. What can you conclude?

Q36

Apply the ratio test to

∞

k=1

(k + 1)

. What can you conclude?

Q37

Apply the ratio test to

∞

k=1

(−1)

k−1

. What can you conclude?

Q38

Apply the ratio test to

∞

k=1

(−8)

. What can you conclude?

Q39

Apply the ratio test to

∞

k=1

. What can you conclude?

Q40

Apply the ratio test to

∞

k=3

+ 4k − 2

. What can you conclude?

Q41

Apply the ratio test to

∞

k=1

√

k + 1

What can you conclude?

Q42

Apply the ratio test to

∞

k=1

√

−k

What can you conclude?

3.3.10

Q43

Use one of the tests from this section to deterine whether

∞

k=1

k+1

converges.

Q44

Use one of the tests from this section to deterine whether

∞

k=1

3(4

)

converges.

Q45

Use one of the tests from this section to deterine whether

∞

k=1

k+1

converges.

Q46

Use one of the tests from this section to deterine whether

∞

k=1

7k9

2k+1

converges.

215

Section 3.3

Exercises

Synthesis & Extension

Q47

In a paragraph or two, explain: How is evaluating an improper integral similar to evaluating an

inﬁnite series. How are they diﬀerent?

Q48

Suppose we have a sequence a

such that lim

n→∞

= 0 and

∞

n=1

= 437. Suppose we then

increase the values of the ﬁrst ﬁve terms of a

by 10, 000 each.

Explain how this will aﬀect the value of lim

n→∞

Explain how this will aﬀect the value of

∞

n=1

Q49

Suppose we wanted to approximate

∞

dx by rectangles of length ∆x = 1, with heights

measured at the left endpoints.

What are the areas of the ﬁrst 5 rectangles, starting from x = 0?

How many rectangles will you need in total?

Express the sum of the areas of these rectangles as a series.

Does this series converge? To what value?

Does your series over- or underestimate the true value of the integral?

Q50

Suppose we wanted to approximate

∞

dx by rectangles of length ∆x = 1, with heights

measured at the right endpoints.

What are the areas of the ﬁrst 5 rectangles, starting from x = 1?

Express the sum of the areas of all the the rectangles you’ll need as a series.

Does your series over- or underestimate the true value of the integral?

What is the true value of the integral? What does this suggest about whether your series

converges or diverges?

Q51

Suppose that a discrete random variable X has distribution function

(x) =

(

if x is a positive integer

0 otherwise

216

Verify that f

(x) is a valid probability distribution function.

Compute P (X > 4).

Compute E[X] (this is diﬃcult).

Q52

Suppose that a discrete random variable X has distribution function

(x) =

(

−

x+1

if x is a positive integer

0 otherwise

Verify that f

(x) is a valid probability distribution function.

Compute P (3 ≤ X ≤ 5).

Explain why you can’t compute E[X].

217

Section 3.4

Power Series

Goals:

1 Use series tests to determine for what values of x a power series converges.

2 Identify the radius of convergence of a power series.

3 Recognize functions that can be rewritten as a power series.

The inﬁnite degree polynomials we seek to deﬁne are series. The tools we’ve developed so far

provide the foundation for understanding the objects we want to construct, but there is more to do. A

polynomial also contains a variable. In this section we deal with the ramiﬁcations of including a variable

in an inﬁnite series.

Question 3.4.1

What Is a Power Series?

So far we have studied inﬁnite series of numbers. If instead of just numbers, our terms include

variables, then we’ve created a function. Plugging in diﬀerent values for the variable gives us a diﬀerent

series of numbers.

Example

The expression

1 + x + x

+ x

+ ···

becomes

1 + 2 + 4 + 8 + ···

when we evaluate it at x = 2. It becomes

1 −

−

+ ···

when we evaluate it at x = −

Deﬁnition

An inﬁnite series of the form

∞

k=0

(x − a)

is called a power series centered at a.

It is a function of x whose domain is all values of x that make the series converge.

For the purposes of this deﬁnition, we deﬁne x

= 1 even when x = 0.

218

Example 3.4.2

A Geometric Series as a Power Series

Use the geometric series formula to write f (x) =

1 − x

as a power series and ﬁnd its domain.

Solution

1 − x

is the sum of a geometric series. In this case, the initial term a = 1 and the common ratio r is

x. If we write out the ﬁrst few terms we obtain 1 + x + x

+ x

+ ···. We see this is a power series

centered at 0. The coeﬃcients c

are all equal to 1. We could write it as

∞

k=0

The domain of a power series is the values of x that make it converge. We know that this geometric

series converges if and only if the common ratio x has absolute value less than 1. Those values of x,

the open interval (−1, 1), are the domain of f.

Example 3.4.3

The Domain of a Power Series

What is the domain of

∞

k=1

(x − 5)

Solution

The domain is the set of x values that make the series converge. The ratio test will be helpful here.

The ratio between terms is



k+1



(k+1)

k+1

(x − 5)

k+1

(x − 5)



(k + 1)

(x − 5)

k+1

(x − 5)



+ 2k + 1)(x − 5)



Notice this entire computation is invalid if x = 5, because we cannot divide by 0. We can examine

this case directly. If x = 5 then every term of the series is 0, and the series converges. For the rest of

the real numbers, we compute the limit as k → ∞, but x will remain in the result.

lim

k→∞



+ 2k + 1)(x − 5)



(x − 5)



lim

k→∞



+ 2k + 1



(x − 5)



219

Example 3.4.3

The Domain of a Power Series

The ratio test can tell us whether the series converges for some values of x. If



(x−5)



< 1 the series

converges. We can solve for x



(x − 5)



< 1



x − 5



< 4 (since 4 > 0)

−4 < x − 5 < 4

1 < x < 9 (add 5 to all three expressions)

On the other hand, if



(x−5)



> 1 the series diverges. Solving for x follows a similar procedure.



(x − 5)



> 1



x − 5



> 4 (since 4 > 0)

x − 5 < −4 or x − 5 > 4

x < 1 or x > 9

What about when x = 1 or x = 9?



(x−5)



= 1 so the ratio test is indeterminate. We would

need another test to resolve these points. In this case, we are lucky. If x = 9 the series becomes

∞

k=1





(4). The divergence test is useful here: lim

k→∞

= ∞. Since the terms do not approach 0,

the series diverges. A similar argument works for k = 1.

Main Idea

The ratio test is usually successful in ﬁnding where a power series converges. Generally it is inconclusive

at only two points. We will not always have a test that can tell us whether the series converges at these

points.

You may notice a pattern in the types of domains we have computed for power series. That pattern

is formalized in the theorem below, which tells us that the domain of a power series must take a very

particular form.

220

Theorem

Given a power series

∞

k=0

(x − a)

centered at a, one of the following is true.

1 The series converges only when x = a.

2 The series converges when x is any real number.

3 There is a radius of convergence R such that

a The series converges when |x − a| < R, and

b The series diverges when |x − a| > R.

In case 3 , the inequality |x − a| < R solves to a − R < x < a + R, which means the domain is an

interval centered at a and extending a distance R to either side. The theorem does not state whether

this is a closed, open or half open interval. This reasoning extends intuitively, if not formally, to the

other cases. 1 can the thought of as a (closed) interval extending distance 0 on either side. 2 would

then be an interval extending inﬁnitely on either side.

Figure: The domain |x − a| < R of a power series.

Remark

The main consequence of this theorem is that when solving for the domain of a power series, we

can simplify our use of the ratio test. The interval of convergence will always be the solution to

lim

k→∞



ak + 1



< 1. The endpoints may or may not lie in the domain. The points beyond the endpoints

will never be part of the domain.

Question 3.4.4

Can We Integrate or Diﬀerentiate a Power Series?

When f(x) is a polynomial, we can ﬁnd the derivative and anti-derivative of f(x) by computing the

(anti-)derivative of each term. The following theorem says that we can do this for a power series too.

221

Question 3.4.4

Can We Integrate or Diﬀerentiate a Power Series?

Theorem

If f(x) is the power series

∞

k=0

(x − a)

and f(x) has radius of convergence R > 0 then f(x) is

diﬀerentiable and continuous on the interval (a − R, a + R), and

1 f

′

(x) =

∞

k=1

(x − a)

k−1

f(x) dx = C +

∞

k=0

(x − a)

k+1

k + 1

Both of these functions also have radius of convergence R.

Remark

Notice that we remove the k = 0 term from the derivative. The derivative of that term is 0, but

(x − a)

−1

is undeﬁned at x = a.

Example

We have seen that

1 − x

∞

k=0

on the interval (−1, 1). From that we can compute:

∞

k=0

∞

k=1

k−1

∞

k=0

dx =

∞

k=0

k+1

k + 1

+ c

Both have domain (−1, 1).

222

Section 3.4

Exercises

Summary Questions

What is the diﬀerence between a polynomial and a power series?

What test is useful for establishing the domain of a power series? What form can this domain

have?

How can we integrate or diﬀerentiate a power series?

How does diﬀerentiation aﬀect the radius of convergence of a power series?

3.4.1

Use Σ notation to express the following series

10 + 15x + 20x

+ 25x

+ 30x

+ ···

−

− ···

Use Σ notation to express the following series

1 −

−

720

40640

− ···

+ 4x

+ 9x

+ 16x

+ 25x

+ ·

223

Section 3.4

Exercises

3.4.2

Consider f(x) =

1−4x

If f is the sum of a geometric series, what is r?

Write f(x) as a geometric series centered at 0.

What is the domain of your answer in

Write

1 − 3(x − 2)

as a power series centered at x = 2.

Can the power series p(x) =

∞

k=1

(x + 7)

be evaluated using the sum of a geometric series

formula? Explain.

Q10

Evaluate f(x) =

∞

k=3

(x − 2)

at x = 6 using the formula for the sum of a geometric series.

3.4.3

Q11

What is the domain of

∞

k=1

(x − 3)

Q12

Compute the domain of

∞

k=0

(x + 2)

Q13

Compute the domain of

∞

k=0

(x − 6)

Q14

Compute the domain of

∞

k=0

Q15

Compute the radius of convergence of

∞

k=0

k(x+ 3)

. What interval does this guarantee the series

converges on?

Q16

Compute the radius of convergence of

∞

k=0

k!x

. What interval does this guarantee the series

converges on?

224

Q17

Compute the radius of convergence of

∞

k=1

(x − 5)

. What interval does this guarantee the

series converges on?

Q18

Suppose you are told that a given power series p(x) centered at x = a converges at x = −4 and

diverges at x = −7.

If a = 1, what can you say about the domain of p(x)?

What are all of the the possible values of a? Explain your reasoning (brieﬂy).

3.4.4

Q19

Compute the antiderivative of

∞

k=0

(x − 3)

Q20

Compute the derivative of

∞

k=0

(x + 2)

. What is its domain?

Q21

Compute the derivative of

∞

k=0

(x − 6)

. What is its domain?

Q22

Compute the antiderivative of

∞

k=0

Q23

What is the domain of the ﬁfth deriative of

∞

k=0

k(x + 3)

Q24

Compute the radius of convergece of the antiderivative of

∞

k=4

(x − 5)

225

Section 3.4

Exercises

Synthesis & Extension

Q25

Consider the power series

p(x) =

∞

k=0

+ k

(x + 3)

Compute the domain of P . You do not need to check any endpoints of your answer.

Write an expression for

p(x) dx.

Q26

Consider the series S =

∞

k=1

How is S related to the power series p(x) =

∞

k=1

k−1

Compute the an avtiderivative P (x) of p(x).

Write P (x) as ratio F (x), using the sum of a geometric series formula.

Compute F

′

(1). What is the signiﬁcance of this value?

Q27

Write a power series for f (x) = tan

−1

x by

Diﬀrentiating f(x)

Writing f

′

(x) as a geometric series

Taking an antiderivative of the geometric series

226

Section 3.5

Taylor Series

Goals:

1 Use a combination of power series and algebra to work with functions.

2 Integrate and diﬀerentiate power series.

Our goal has been to understand how to extend a Taylor polynomial to have inﬁnite degree. We are

now ready to deﬁne the object rigorously. In general we will not know how to evaluate Taylor series.

If all we want to do is approximate values, they oﬀer no advantages over Taylor polynomials. The

applications of Taylor series are more abstract. After deﬁning these objects, we collect some tricks and

applications for working with them.

Question 3.5.1

What Is a Taylor Series?

Deﬁnition

The Taylor series of f (x) at x = a is

T (x) =

∞

k=0

(k)

(a)

(x − a)

The Taylor series’s notation simply swaps an n for an ∞ in the expression of a Taylor polynomial. If

we wanted to describe the mathematical relationship precisely, we would say its partial sums s

are the

Taylor polynomials T

(x) of f at x = a.

Remark

Several mathematicians contributed to the discovery of Taylor series. Taylor series centered at x = 0

were popularized by Colin Maclaurin, and so are often called Maclaurin series.

This deﬁnition is built upon a stack of more general deﬁnitions, and the methods we have for working

with those apply here.

A Taylor series is a type of power series.

A power series is a type of series

A series is equivalent to a sequence of partial sums.

This list should make us feel better about our hard work over the last few sections. It also gives us

information that helps us understand Taylor series better. For example, since Taylor series are power

series, their domains are also intervals of radius R centered at a.

227

Question 3.5.1

What Is a Taylor Series?

Limitations of Taylor Series

Taylor polynomials were designed to approximate f(x). We might hope that T (x) would be the perfect

approximation, that T (x) and f(x) are equal. Unfortunately, there are obstacles to this.

The Taylor series might not converge for all x.

The Taylor polynomials might not approximate f (x) very well at all. Recall our example

f(x) =

(

0 if x ≤ 0

−

if x > 0

For this function T (x) = 0.

Example 3.5.2

Writing a Taylor series

Let f(x) = e

Find the Taylor series for f (x) centered at x = 0.

On what interval does it converge?

Solution

We have seen previously that f

(k)

(x) = e

for all k and thus f

(k)

(0) = 1. We plug this into the

Taylor series formula.

T (x) =

∞

k=0

A Taylor series is a power series. We will use the ratio test to identify the interval of convergence.

228

The ratio of successive terms is



k+1



(k+1)!

k+1



k!x

k+1

(k + 1)!x



k + 1



lim

k→∞



k + 1



= 0

This limit is zero no matter what value of x we choose. Since 0 < 1, the ratio test concludes that

this series converges for any value of x. In other words, the domain is all real numbers.

Synthesis 3.5.3

Is a Taylor Series Equal to the Function it Approximates?

Let f(x) = ln x

Find a pattern in the derivatives and write a general expression for the kth derivative: f

(k)

(x).

Use your answer to

to write expressions for the Taylor polynomials T

(x) and the Taylor series

T (x) of ln x centered at 1. Simplify the coeﬃcients if possible.

What does the ratio test tell you about where T (x) converges?

If we wanted to apply Taylor’s inequality to T

(x), we would need to know where the derivative is

largest (in absolute value). Where is the (n + 1)th derivative largest on the interval [x, 1]? (Here

0 < x < 1).

Where is the (n + 1)th derivative largest on the interval [1, x]? (Here x > 1).

What does Taylor’s inequality say about where R

(x) → 0 as n → ∞?

What does our answer to the previous question tell us about T (x)?

229

Synthesis 3.5.3

Is a Taylor Series Equal to the Function it Approximates?

Solution

Let’s compute some derivatives and see if we can ﬁnd an expression for f

(k)

(x)

f(x) = ln(x) f(1) = 0

′

(x) = x

−1

′

(1) = 1

′′

(x) = −x

−2

′′

(1) = −1

′′′

(x) = 2x

−3

′′′

(1) = 2

(4)

(x) = −6x

−4

(4)

(1) = −6

(5)

(x) = 24x

−5

(5)

(1) = 24

These answers look like factorials, but they’re shifted by 1. They’re also alternating signs, which

we can model with (−1)

, except that the even powers are negative. The power of x is −k. One

way to model this is f

(k)

(x) = (−1)

k+1

(k − 1)!x

−k

Plugging in x = 1 gives f

(k)

(1) = (−1)

k+1

(k − 1)! except at k = 0. For that case we compute

ln 1 = 0. This means we can leave it out of the summation. The form for the remaining terms

allows for some nice simpliﬁcation.

T (x) =

∞

k=1

(−1)

k+1

(k − 1)!

(x − 1)

∞

k=1

(−1)

k+1

(x − 1)

We’ll apply the ratio test



k+1



(−1)

k+2

(k+1)

(x − 1)

k+1

(−1)

k+1

(x − 1)



(−1)

k+2

k(x − 1)

k+1

(−1)

k+1

(k + 1)(x − 1)



−k(x − 1)

k + 1



k|x − 1|

k + 1

230

Now we’ll solve for when the limit of this ratio is less than 1.

lim

k→∞

k|x − 1|

k + 1

< 1

|x − 1| lim

k→∞

k + 1

< 1

|x − 1| < 1

−1 < x − 1 < 1

0 < x < 2

The Taylor series converges on the interval (0, 2).

To apply Taylor’s inequality to bound |R

(x)|. We need a bound on |f

(n+1)

(x)| on the interval

from 1 to x. Looking back at our earlier computation, we obtain f

(n+1)

(x) = (−1)

n+2

n!x

−n−1

In this case that x > 1, the derivative f

(n+1)

decreases in magnitude from x to 1 so it is largest

at x. We can use M = n!x

−n−1

In this case, f

(n+1)

decreases in magnitude from 1 to x so it is largest at 1. We can use M = n!.

The easier case is x ≥ 1. In this case Taylor’s inequality states

(x)| ≤



(n + 1)!

(x − 1)

n+1



≤

n + 1

(x − 1)

n+1

As n approaches inﬁnity, this bound goes to inﬁnity if x − 1 > 1 and to 0 if x − 1 ≤ 1. In the

case that 0 < x < 1, we need some clever algebra to write this as a multiple of an exponential.

(x)| ≤



n!x

−n−1

(n + 1)!

(x − 1)

n+1



≤

n + 1





x − 1



n+1



This goes to 0 if

x−1

≤ 1 and inﬁnity otherwise. Solving this (and assuming x > 0) gives x ≥

Putting these together, we can state that the error bound from Taylor’s inequality approaches 0

as we takes higher degree Taylor polynomials, as long as

≤ x ≤ 2.

The answer to the previous question tells us that T (x) converges to ln x on



, 2



, since the error

bound and hence the error goes to 0. On the other hand, outside this interval, the error might

still go to 0 on





, even though the error bound does not. The series diverges outside (0, 2)

so it cannot converge to ln x there.

231

Synthesis 3.5.3

Is a Taylor Series Equal to the Function it Approximates?

Remark

It turns out that T(x) = ln x on (0, 2], which is a larger interval than we were able to establish using

Taylor’s inequality. This should not bother us. Taylor’s inequality produces a bound on the error. The

fact that the bound on the error is going to inﬁnity, doesn’t mean the actual error does. In this case,

for x between 0 and

, the actual error approaches 0.

Figure: The Taylor polynomials approach ln x only on (0, 2].

Example 3.5.4

Mixing Taylor Series and Algebra

Let f(x) = x

sin x. Compute a Taylor series for f (x) centered at x = 0.

Solution

We could try to work out a pattern in the derivatives of f, but even evaluating at x = 0 the computations

become intractable.

′

(x) = 2x sin x + x

cos x

′′

(x) = 2 sin x + 4x cos x − x

sin x

′′′

(x) = 6 cos x − 6x sin x − x

cos x

(4)

(x) = −12 sin x − 8x cos x + x

sin x

232

Instead we can write the Taylor series for sin x. Our earlier work gave us an expression for the Taylor

polynomials and showed that their error goes to 0 as the degree goes to inﬁnity.

sin x =

∞

k=0

(−1)

)

(2k + 1)!

2k+1

We can obtain an expression for x

sin x by multiplying both sides by x

. Since we’re only multiplying

by a power of x, the resulting series will still be a power series centered at 0.

sin x = x

∞

k=0

(−1)

)

(2k + 1)!

2k+1

∞

k=0

(−1)

)

(2k + 1)!

2k+3

Main Idea

When constructing a Taylor series for f (x) = x

g(x) centered at 0, construct the Taylor series of g(x),

and then distribute the x

Example 3.5.5

Integrating a Taylor Series

Let f(x) = e

Write a Taylor polynomial T

(x) for f (x) at x = 0.

Find a better way to produce the Taylor series for f(x).

Compute a Taylor series for

dx.

233

Example 3.5.5

Integrating a Taylor Series

Solution

We will compute the ﬁrst four derivatives of f (x). We will need the chain rule and later the

product rule.

f(x) = e

f(0) = 1

′

(x) = 2xe

′

(0) = 0

′′

(x) = 2e

+ 4x

′′

(0) = 2

′′′

(x) = 12xe

+ 8x

′′′

(0) = 0

(4)

(x) = 12e

+ 48x

+ 16x

(4)

(0) = 12

We can plug these values into our T

(x) formula.

(x) =

= 1 + x

We can see that our derivative calculations would quickly get out of hand as we take higher order

derivatives. Even if there is a discernible pattern, it might take more computation to determine it.

A better approach is to start with a simpler Taylor series that we know.

∞

k=0

is a composition of e

and x

, so we will plug in x

for x in our e

Taylor series.

∞

k=0

)

∞

k=0

Taylor series are also power saeries. By our theorem on power series, we can integrate term by

term.

dx =

∞

k=0

k!(2k + 1)

2k+1

+ c

Note that

dx is not a function we can express algebraically or compute. A Taylor series gives

us some way to represent this function, but we shouldn’t be too satisﬁed. If we actually wanted

to evaluate it, the best we could do is approximate it with a partial sum.

234

Figure: The graph of e

dx, and the partial sums of its Taylor series.

Main Ideas

Compositions of functions can be composed through Taylor series.

Taylor series allow us to integrate functions that are otherwise impossible to integrate.

Application 3.5.6

Euler’s Formula

Recall i is an imaginary number that satisﬁes i

= −1.

Find an expression for f(x) = e

Write your answer in terms of the Taylor series for sin x and cos x.

Write two diﬀerent expressions for e

i2x

. How is this equation useful?

235

Application 3.5.6

Euler’s Formula

Solution

We can express e

by replacing x by ix in the Taylor series for e

T (x) =

∞

k=0

(ix)

To make much sense of this, we should try to simplify i

= 1 i

= 1

= i i

= i

= −1

= −i

We can write out the terms of T (x) as follows:

T (x) = 1 + ix −

−

The terms with a factor of i are the Taylor series for sin x multiplied by i. The terms without a

factor of i are the Taylor series for cos x. We can write

= cos x + i sin x

One way to write this would be to substitute 2x for x:

i2x

= cos 2x + i sin 2x

Another way would be to square our original formula.

i2x





= (cos x + i sin x)

= cos

x + 2i cos x sin x − sin

x (i

= −1)

Setting these equal to each other, we note that for two complex numbers to be equal, their real

parts must be equal and their imaginary parts must be equal.

cos 2x + i sin 2x = cos

x + 2i cos x sin x − sin

cos 2x = cos

x − sin

and sin 2x = 2 cos x sin x

These are the double angle formulas for sine and cosine.

236

We can take higher powers of e

to produce triple or quadruple angle formulas. This converts a

diﬃcult geometry problem into something a high school algebra student could compute.

Remark

You would expect a relationship like this to be very famous, and it is. e

= cos x + i sin x is called

Euler’s Formula. In addition to trigonometric formulas, it gives us insight into the complex numbers.

This connection between an exponential and a periodic function is so powerful that it is used in such

concrete applications as electrical engineering and signal processing.

Section 3.5

Exercises

Summary Questions

How can we be sure that a Taylor series converges to the function it is approximating?

What is the domain of a Taylor series?

How can we produce the Taylor series for x

f(x) or f(x

)? Where does the center need to be

for the result to be a Taylor series?

What is a Maclaurin series?

3.5.1

If we wanted to compute a decimal approximation of ln(1.25) by hand, would the Taylor polyno-

mial or the Taylor series be more useful?

If T (x) is a Taylor series centered at x = a, what are the possible forms that the domain of T (x)

could take?

237

Section 3.5

Exercises

3.5.2

How would the Taylor series of f(x) = e

change if we centered it at x = 1 instead of x = 0?

Let T (x) be the Taylor series of f(x) = e

centered at 0. Verify that T

′

(x) = T (x).

Write a Taylor series of f(x) =

centered at 4.

Q10

Write a Taylor series of f(x) =

centered at −5.

Q11

Write a Taylor series of f(x) = cos x centered at 0.

Q12

Write a Taylor series of f(x) = sin x centered at 0.

3.5.3

Q13

Show that the Taylor series of f(x) = e

centered at x = 0 is equal to f (x) for all real numbers

Q14

Show that the Taylor series of f(x) = sin x centered at x = 0 is equal to f(x) for all real numbers

Q15

Show that the Taylor series of f (x) =

centered at 4 is equal to f(x) for all x in the interval

(2, 6).

Q16

Suppose for a function f we are able to place a bound of

on the kth derivative of f over

any interval. For what values of x can we conclude that T (x), the Taylor series centered at 2, is

equal to f (x)?

Q17

We didn’t have a series test to determine whether

∞

k=1

(−1)

k+1

converges. How does our analysis

of the Taylor series of ln x allows us to conclude that this series converges? Hint: what is T (2)?

Q18

For a general function f and its Taylor polynomials and series, how are the following sets of points

related? Does every number belonging to one of these sets belong to one of the others?

The set of numbers x where T (x) converges.

The set of numbers x where |R

(x)| → 0 as n → ∞.

The set of numbers where f(x) = T (x).

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3.5.4

Q19

Write a Taylor series for f(x) = x

cos x centered at x = 0.

Q20

Write a Taylor series for f(x) = x

centered at x = 0.

Q21

Can we use our Taylor series for f(x) = ln x centered at 1 to write a Taylor series for g(x) =

ln x? Explain.

Q22

Write a Taylor series for f(x) =

(x+5)

centered at −5.

3.5.5

Q23

Let g(x) be an antiderivative of e

. Write the Taylor series for g(x) centered at x = 0.

Q24

Let g(x) be an antiderivative of cos(x

). Write the Taylor series for g(x) centered at x = 0.

Q25

Let f (x) = cos x. Let T (x) be the Taylor series of f centered at x = 0. Compute T

′′

(x). Why

does your answer make sense?

Q26

Write the Taylor series for f(x) =

centered at 1. Verify that one of its antiderivatives is a

Taylor series for ln x.

3.5.6

Q27

Rewrite our formula for cos(2x) to be entirely in terms of cos x.

Q28

Use Euler’s formula to compute a formula for cos 3x in terms of cos x and sin x.

Q29

According to Euler’s formula, what is the value of e

2πi

Q30

Use the Taylor series of ln x centered at x = 1 to compute ln(1 + i). Do you think this series

converges?

239

Section 3.5

Exercises

Synthesis & Extension

Q31

Let h(x) =

Compute the Taylor polynomial T

centered at x = 4.

If you wanted to use your Taylor polynomial from

to approximate

2.5

, what bound would

Taylor’s inequality put on the error? Don’t simplify the arithmetic.

What does the ratio test tell you about the domain of the Taylor series of h(x) centered at

x = 4?

Q32

Let X be a normal random variable with mean 0 and standard deviation 1. Write a series whose

value is P (0 ≤ X ≤ 1).

Q33

Suppose we produce the Taylor series T (x) for some f(x) centered at x = 10.

If the Taylor series converges at x = 5, must it also converge at x = 7? Explain.

If the errors of the Taylor polynomials T

(2) converge 0 as n goes to ∞ for some x, must

T (2) converge? If T (2) converges, must the errors converge to 0?

If you wanted to approximate f (7) as accurately as possible, which would be more useful, a

Taylor polynomial or a Taylor series?

Q34

Suppose we have a function f(x) and two diﬀerent numbers a and b. Suppose further that the

Taylor series for f(x) centered at a is equal to the Taylor series for f(x) centered at b. What

can you say about the domain of this Taylor series?

240