Chapter 12

Infinite series, improper integrals, andTaylor series

12.1 Introduction to series

In studying calculus, we have explored a variety of functions. Among the most basic are polynomials,i.e. functions such as

p(x) = x5 + 2x2 + 3x + 2.

These functions have features that make them particularly easy to handle: it is elementary tocompute the derivatives and integrals of polynomials. Further, the value of such a function canbe computed by simple multiplications and additions, i.e. by basic operations easily handled byan ordinary calculator. This is not the case for, say, trigonometric functions. For example, tofind the decimal value of sin(2.5) we require a scientific calculator. The same difficulty occurs forexponentials and logarithms. For this reason, being able to approximate a function by a polynomialis an attractive proposition. This idea forms the core topic of this chapter.

We have discussed linear approximations in differential calculus, Here we extend such elemen-tary approximation methods. Our goal is to increase the accuracy of the linear approximation byincluding higher order terms (quadratic, cubic, etc), i.e. to find a polynomial that approximatesthe given function.

We first review the idea of series introduced in Chapter 1.

12.2 The geometric series

Recall the geometric series seen earlier this term:

Sn = 1 + r + r2 + r3 + . . . + rn =

n∑

k=0

rk

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Math 103 Notes Chapter 12

We have seen a trick that helps to sum the series, namely multiplying both sides by r. This leadsto

Sn = 1+ r + r2 + . . . + rn

rSn = r + r2 + . . . + rn + rn+1

Subtracting the second line from the first and cancelling terms leads to

(1 − r)Sn = 1 − rn+1.

Solving for the sum Sn leads to the observation made previously, namely that

The sum of a finite geometric series,

Sn = 1 + r + r2 + · · ·+ rn =n∑

k=0

rk,

is

Sn =1 − rn+1

1 − r. (12.1)

We would like to investigate how this “sum” behaves when more and more terms of the seriesare included. Our goal is to find a way of attaching a meaning to the expression

Sn =∞∑

k=0

rk,

when the series becomes an infinite series. We will use the following definition:

12.2.1 The infinite geometric series

Definition

An infinite series that has a finite sum is said to be convergent. Otherwise it is divergent.

Definition

Suppose that S is an (infinite) series whose terms are ak. Then the partial sums, Sn, of this seriesare

Sn =

n∑

k=0

ak.

We say that the sum of the infinite series is S, and write

S =

∞∑

k=0

ak,

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Math 103 Notes Chapter 12

provided that

S = limn→∞

n∑

k=0

ak.

That is, we consider the infinite series as the limit of the partial sums as the number of terms n isincreased. In this case we also say that the infinite series converges to S.

We will see that only under certain circumstances will infinite series have a finite sum, and wewill be interested in exploring two questions:

1. Under what circumstances does an infinite series have a finite sum.

2. What value does the partial sum approach as more and more terms are included.

In the case of a geometric series, the sum of the series, 12.1 depends on the number of terms inthe series, n via rn+1. Whenever r > 1, or r < −1, this term will get bigger as n increases, whereas,for 0 < r < 1, this term decreases with n. We can say that

limn→∞

rn+1 = 0 provided |r| < 1.

These observations are illustrated by two specific examples below. This leads to the followingconclusion:

The sum of an infinite geometric series,

S = 1 + r + r2 + · · · + rk + · · · =

∞∑

k=0

rk,

exists provided |r| < 1 and is

S =1

1 − r. (12.2)

Examples of convergent and divergent geometric series are discussed below.

12.2.2 Example: A geometric series that converges.

Consider the geometric series with r = 12, i.e.

Sn = 1 +1

2+

(

1

2

)2

+

(

1

2

)3

+ . . . +

(

1

2

)n

=

n∑

k=0

(

1

2

)k

Then

Sn =1 − (1/2)n+1

1 − (1/2)

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Math 103 Notes Chapter 12

We observe that as n increases, i.e. as we retain more and more terms, we obtain

limn→∞

Sn = limn→∞

1 − (1/2)n+1

1 − (1/2)=

1

1 − (1/2)= 2.

In this case, we write∞∑

n=0

(

1

2

)n

= 1 +1

2+ (

1

2)2 + . . . = 2

and we say that “the (infinite) series converges to 2”.

12.2.3 Example: A geometric series that diverges

In contrast, we now investigate the case that r = 2: then the series consists of terms

Sn = 1 + 2 + 22 + 23 + . . . + 2n =

n∑

k=0

2k

Sn =1 − 2n+1

1 − 2= 2n+1 − 1

We observe that as n grows larger, the sum continues to grow indefinitely. In this case, we say thatthe sum does not converge, or, equivalently, that the sum diverges.

It is important to remember that an infinite series, i.e. a sum with infinitely many terms addedup, can exhibit either one of these two very different behaviours. It may converge in some cases, asthe first example shows, or diverge (fail to converge) in other cases. We will see examples of each ofthese trends again. It is essential to be able to distinguish the two. Divergent series (or series thatdiverge under certain conditions) must be handled with particular care, for otherwise, we may findcontradictions or seemingly reasonable calculations that have meaningless results.

12.3 Trading a function for an infinite series

The expression found above holds true for any number r whose magnitude is less than 1. Supposewe think of this number as a quantity we wish to vary - let us call it x to remind us that we considerit a variable. Then rewritten in terms of x, the expression we found is

1

1 − x= 1 + x + x2 + x3 + . . . when |x| < 1

We wish to understand this in a new way: on the left we have a function, namely

f(x) =1

1 − x.

On the right we have what looks like an “infinite polynomial”, i.e. a sum of terms that are all powersof x. This type of expansion is called a power series, since it consists of a sum of powers of x. We

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Math 103 Notes Chapter 12

will soon see that this power series is a particular example of a Taylor series, an expansion thatcan be helpful in approximating more complicated functions such as trigonometric or exponentialfunctions.

One way to think of this new connection between the function on the left and the series on theright is that it defines a whole set of polynomials, each one formed by keeping one more term inthe (infinite) expansion, and each one a better and better approximation of the given function. Forexample, we could list the successive approximations as

T0(x) = 1

T1(x) = 1 + x

T2(x) = 1 + x + x2

Tn(x) = 1 + x + x2 + . . . + xn

The first polynomial is just the value of the function at one point (at the point x = 0, it turns out).The second is the linear approximation, arrived at by using information about the derivative. Thesubsequent polynomials retain more and more details of the function, as we will shortly see. We soarrive at the Taylor polynomial of degree n, shown at the end of this list. As long as |x| < 1, theTaylor polynomials provide a better and better approximation for the function f(x) = 1/(1− x) asn gets larger.

But we must take care, since if |x| > 1, the expansion will fail to converge - as we have alreadyestablished previously. For example, the value x = 1 leads to

1 + 1 + 1 + . . . ,

which does not converge. As another example, if we let x = −1, the sum looks like this:

1 + (−1) + 1 + (−1) + . . .

If we add the first two terms, we get zero. If we add the third term, we get 1. Adding the fourthterm gives zero. As we continue term by term, the sum keeps flipping back and forth between 0and 1. In this case, we also say that the sum does not converge.

Thus this Taylor series is valid only so long as |x| < 1. The restriction |x| < 1 is required toensure that the infinite sum will converge to a finite result. We say that the radius of convergence

of this series is 1.

We have seen two ways in which a series can diverge: first, the sum may become unbounded,i.e. infinitely large. Second, the sum could jump from value to value without approaching any fixedvalue. In both cases, the series is divergent and will have little practical significance.

12.4 Improper integrals

We will see that there is a close connection between certain infinite series and improper integrals,i.e. integrals over an infinite domain. Recall the following definition:

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Math 103 Notes Chapter 12

Definition

An improper integral is an integral performed over an infinite domain, e.g.∫

∞

a

f(x) dx.

The value of such an integral is understood to be the limit

∫

∞

a

f(x) dx = limb→∞

∫ b

a

f(x) dx.

i.e. we evaluate an improper integral by first computing a definite integral over a finite domaina ≤ x ≤ b, and then taking a limit as the endpoint b moves off to larger and larger values.

Definition

When the limit shown above exists, we say that the improper integral converges. Other wise wesay that the improper integral diverges.

With this in mind, we compute a number of classic integrals:

12.4.1 Example: A divergent integral

I =

∫

∞

1

1

xdx = lim

b→∞

∫ b

a

1

xdx

I = limb→∞

ln(x)

∣

∣

∣

∣

b

1

= limb→∞

(ln(b) − ln(1))

I = limb→∞

ln(b) = ∞

The fact that we get an infinite value for this integral follows from the observation that ln(b)increases without bound as b increases. Thus the area under the curve f(x) = 1/x over the infiniteinterval 1 ≤ x ≤ ∞ is not finite. We say that the improper integral of 1/x diverges (or does notconverge).

12.4.2 Example: A convergent integral

Now consider the related integral

I =

∫

∞

1

1

x2dx = lim

b→∞

∫ b

1

x−2 dx.

Then

I = limb→∞

(−x−1)

∣

∣

∣

∣

b

1

= − limb→∞

(

1

b− 1

)

= 1.

Thus, this integral converges and we see that its value is 1.

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Math 103 Notes Chapter 12

12.4.3 When does the integral of 1/xp converge?

More generally, consider

I =

∫

∞

1

1

xpdx.

By a similar calculation, we find that

I = limb→∞

x1−p

(1 − p)

∣

∣

∣

∣

b

1

I = limb→∞

(

1

1 − p

)

(

b1−p − 1)

Thus this integral converges provided that the term b1−p does not “blow up” as b increases. Thiscan only be true so long as p > 1. In this case, we have

I =1

p − 1.

To summarize our result,

∫

∞

1

1

xpdx

{

converges if p > 1

diverges if p ≤ 1

12.4.4 Comparisons

The integrals discussed above can be used to make comparisons that help us to identify whenother improper integrals converge or diverge. The following important result establishes how thesecomparisons work:

Suppose we are given two functions, f(x) and g(x), both continuous on some infinite interval[a,∞). Suppose, moreover, that, at all points on this interval, the first function is smaller than thesecond, i.e.

0 ≤ f(x) ≤ g(x).

Then the following conclusions can be made:

1.

∫

∞

a

f(x) dx ≤

∫

∞

a

g(x) dx. (This means that the area under f(x) is smaller than the area

under g(x).)

2. If

∫

∞

a

g(x) dx converges, then

∫

∞

a

f(x) dx converges. (If the larger area is finite, so is the

smaller one)

3. If

∫

∞

a

f(x) dx diverges, then

∫

∞

a

g(x) dx diverges. (If the smaller area is infinite, so is the

larger one.)

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Math 103 Notes Chapter 12

Example

We can determine that the integral∫

∞

1

x

1 + x3dx

converges by noting that for all x > 0

x

1 + x3≤

x

x3=

1

x2.

Thus∫

∞

1

x

1 + x3dx ≤

∫

∞

1

1

x2dx

Since the larger integral on the right is known to converge, so does the smaller integral on the left.

12.5 Infinite series

In this section we present a few classical series which are important landmarks and which carryspecial significance. These will be useful in our appreciation of the behaviour of infinite series.

Geometric series

We have already seen that

The geometric series converges when |r| < 1 and its sum is then

S =

∞∑

n=0

rn =1

1 − r.

If r ≥ 1 the geometric series diverges.

The harmonic series

The harmonic series is a sum of terms of the form 1/k where k = 1, 2, . . . . At first appearance,this series might seem to have the desired qualities of a convergent series, simply because thesuccessive terms being added are getting smaller and smaller, but this is a misleading and deceptiveobservation!

The harmonic series

∞∑

k=1

1

k= 1 +

1

2+

1

3+

1

4+ . . . diverges

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Math 103 Notes Chapter 12

The harmonic series

<==== the function y=1/x

0.0 11.0

0.0

1.0

Figure 12.1: The harmonic series is a sum that corresponds to the area under the staircase shownabove.

It can be shown that the harmonic series does not converge. by comparing it to an integralof the function

y = f(x) =1

x.

In Figure 12.1 we show a graph of this function superimposed on a bar graph. We will compare theareas of the steps in the bar graph with the area under the graph of this function. For the area ofthe steps, we note that the base width of each step is 1, and the heights form the sequence

{1,1

2,1

3,1

4, . . .}

Thus the area of (infinitely many) of these steps is

A = 1 · 1 + 1 ·1

2+ 1 ·

1

3+ 1 ·

1

4+ · · · = 1 +

1

2+

1

3+

1

4+ . . . =

∞∑

k=1

1

k.

Thus the area of the bar graph is a geometric representation of precisely the series we are investi-gating, the harmonic series.

On the other hand, the area under the graph of the function y = f(x) = 1/x is given by theimproper integral

∫

∞

1

1

xdx

We have seen previously that this integral diverges!

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Math 103 Notes Chapter 12

¿From Figure 12.1 we see that the areas under the function, Af and under the bar graph, Ab,satisfy

Af < Ab.

Thus, since the smaller of the two is infinite, so is the larger. We have established, using thiscomparison, that the the sum of the harmonic series cannot be finite, so that this series diverges.

Other comparisons: The “p” series

More generally, we can compare series of the form

∞∑

k=1

1

kp

to the integral∫

∞

1

1

xpdx

in precisely the same way. This leads to the conclusion that

The “p” series,

∞∑

k=1

1

kp

{

converges if p > 1

diverges if p ≤ 1

For example, the series

∞∑

k=1

1

k2= 1 +

1

4+

1

9+

1

16+ . . .

converges. Notice, however, that the comparison does not give us a value to which the sum converges.It merely indicates that the series does converge.

Alternating series

We can modify the harmonic series to produce the alternating harmonic series in which the signalternates from term to term.

∞∑

k=1

(−1)k+1 1

k= 1 −

1

2+

1

3−

1

4+ . . .

The alternating harmonic series converges due to the alternation of signs. (Consecutive termsnearly cancel out). We will see later how to show that this series converges, remarkably to the valueln(2). However, the convergence is “slow”: it takes many terms added together to get close to the

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Math 103 Notes Chapter 12

“final value”. The speed of convergence is important for practical reasons: if we want to devise acalculation that will help us to find an approximate decimal expansion of some number (such asln(2)), we would certainly prefer to use a series that converges quickly, to reduce the number ofcalculations that have to be made.

More generally, for alternating series with terms that are decreasing to zero, we have convergence,as summarized below.

Suppose that successive terms ak and ak+1 have opposite signs. Then,provided the terms are getting smaller so that ak → 0, the alternatingseries

∞∑

k=0

ak

converges.

12.5.1 Example

In this example, we will consider the series

∞∑

k=0

1

k!= 1 + 1 +

1

2+

1

6+

1

24+ . . .

We can see that this series converges by noticing:

1

k!=

1

1·1

2·1

3· . . .

1

k≤ 1 ·

1

2·1

2· . . .

1

2=

1

2k−1=

(

1

2

)k−1

if k ≥ 2. In other words, the series eventually becomes smaller than a geometric series whichconverges.

∞∑

k=0

1

k!= 1 + 1 +

1

2+

1

6+

1

24+ . . .

< 1 + 1 +1

2+

1

4+

1

8+ . . .

1 + 1 +1

2+

1

4+

1

8+ . . . = 1 + (1 +

1

2+ (

1

2)2 + (

1

2)3 + . . .)

= 1 +∞∑

k=0

(

1

2

)k

= 1 + 2 = 3

Note that the series converges, and, moreover, converges to a number smaller than 3.

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Math 103 Notes Chapter 12

12.6 Finding Taylor series

Now that we understand a little bit about when series converge, we will turn our attention to findingthe series representations of some elementary functions.

Using one series to find another

Once we have identified the expansion of some function in terms of an infinite “Taylor polynomial”,we can build up many other expansions of related functions. In this section, we repeatedly useour geometric summation formula, together with substitution and/or some other operation to findinteresting new conclusions.

All examples are based on the fact (discussed earlier) that

1

1 − u= 1 + u + u2 + u3 + . . . =

∞∑

k=0

uk

when |u| < 1.

12.6.1 Example: Simple substitution

We make the substitution u = −t, then |u| < 1 means that | − t| < 1, so the above formula for thesum of a geometric series implies that:

1

1 − (−t)= 1 + (−t) + (−t)2 + (−t)3 + . . .

1

1 + t= 1 − t + t2 − t3 + t4 + . . .

This means we have produced the series for the function 1/(1 + t).

12.6.2 Example: Term by term integration

We will use the results of Example 12.6.1, but we follow our substitution by integration. On theleft, we integrate the function f(t) = 1/(1 + t) (to arrive at a ln type integral) and on the right weintegrate the power terms of the expansion. We are permitted to integrate the power series termby term provided that the series converges.

∫ x

0

1

1 + tdt =

∫ x

0

(1 − t + t2 − t3 + t4 − . . .) dt

ln(1 + x) = x −x2

2+

x3

3−

x4

4+ . . . =

∞∑

k=1

(−1)k+1xk

k

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Math 103 Notes Chapter 12

This procedure has allowed us to find a series representation for a new function, ln(1 + x).

ln(1 + x) = x −x2

2+

x3

3−

x4

4+ . . .

When is this series expected to converge? We obtained this by starting with the geometric serieswhich only converged when |u| < 1. We made the substitution u = −t, and then integrated for0 < t < x. To assure convergence, the value of t is not permitted to leave the interval |t| < 1 so wecertainly cannot expect the series for ln(1 + x) to converge when |x| > 1 . Indeed, for x = −1, wehave ln(1 + x) = ln(0) which is undefined. We also get a right hand side of

−(1 +1

2+

1

3+

1

4+ . . .)

which is the divergent harmonic series preceded by a negative sign. Thus, we must avoid x = −1,since the expansion will not converge there, and neither is the function defined.

An expansion for ln(2)

Strictly speaking, our analysis does not predict what happens if we substitute x = 1 into theexpansion of the function found in Example 2, because this value of x is outside of the permittedrange −1 < x < 1 in which the Taylor series can be guaranteed to converge. It takes some deepermathematics (Abel’s theorem) to prove that the result of this substitution actually makes sense,and converges, i.e. that

ln(2) = 1 −1

2+

1

3−

1

4+ . . .

We state without proof here that the alternating harmonic series converges to ln(2).

12.6.3 Example

Suppose we make the substitution u = −t2 into the geometric series formula, and recall that weneed |u| < 1 for convergence. Then

1

1 − (−t2)= 1 + (−t2) + (−t2)2 + (−t2)3 + . . .

1

1 + t2= 1 − t2 + t4 − t6 + t8 + . . . =

∞∑

k=0

(−1)nt2n

This series will converge provided |t| < 1. Now integrate both sides, and recall that the antiderivativeof the function 1/(1 + t2) is arctan(t). Then

∫ x

0

1

1 + t2dt =

∫ x

0

(1 − t2 + t4 − t6 + t8 + . . .) dt

arctan(x) = x −x3

3+

x5

5−

x7

7+ . . . =

∞∑

k=1

(−1)k+1 x(2k−1)

(2k − 1)

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Math 103 Notes Chapter 12

An expansion for π

For a particular application of this expansion, consider plugging in x = 1. Then

arctan(1) = 1 −1

3+

1

5−

1

7+ . . .

But arctan(1) = π/4. Thus we have found a way of computing the irrational number π, namely

π = 4

(

1 −1

3+

1

5−

1

7+ . . .

)

= 4

(

∞∑

k=1

(−1)k+1 1

(2k − 1)

)

.

While it is true that this series converges, the convergence is slow. (This can be seen by adding upthe first 100 or 1000 terms of this series with a spreadsheet.) This means that it is not practicalto use such a series as an approximation for π. (There are other series that converge to π veryrapidly.)

12.6.4 Finding the coefficients directly from the function

As indicated in the introduction, approximating functions by the sum of power functions (i.e. bya polynomial or a series) is helpful in practical calculations. For one thing, we can evaluate apolynomial (or finite series) more readily than functions such as arctan or ln. In the precedingdiscussions, we found some series representations of functions fortuitously. Here we set out toidentify the connection by starting with the function of interest.

Suppose we have a function which we want to write as

f(x) = a0 + a1x + a2x2 + a3x

3 + . . . =

∞∑

k=0

akxk.

We can use information about the function to directly determine the coefficients. To determine a0,let x = 0 and note that

f(0) = a0 + a10 + a202 + a30

3 + . . . = a0

We conclude that

a0 = f(0).

By differentiating both sides we find the following:

f ′(x) = a1 + 2a2x + 3a3x2 + . . . + kakx

k−1 + . . .

f ′′(x) = 2a2 + 2 · 3a3x + . . . + (k − 1)kakxk−2 + . . .

f ′′′(x) = 2 · 3a3 + . . . + (k − 2)(k − 1)kakxk−3 + . . .

f (k)(x) = 1 · 2 · 3 · 4 . . . kak + . . .

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Math 103 Notes Chapter 12

Here we have used the notation f (k)(x) to denote the k’th derivative of the function. Now evaluateeach of the above derivatives at x = 0. Then

f ′(0) = a1, ⇒ a1 = f ′(0)

f ′′(0) = 2a2, ⇒ a2 = f ′′(0)2

f ′′′(0) = 2 · 3a3, ⇒ a3 = f ′′′(0)2·3

f (k)(0) = k!ak, ⇒ ak = f(k)(0)k!

This gives us a recipe for calculating all coefficients ak. This means that if we can compute all thederivatives of the function f(x), then we know the coefficients of the Taylor series as well. Becausewe have evaluated all the coefficients by the substitution x = 0, we say that the resulting powerseries is the Taylor series of the function about x = 0.

12.6.5 Example: A Taylor series for ex

Consider the function f(x) = ex. All the derivatives of this function are equal to ex. In particular,

f (k)(x) = ex ⇒ f (k)(0) = 1.

So that the coefficients of the Taylor series are

ak =f (k)(0)

k!=

1

k!

Therefore the Taylor series for ex about x = 0 is

a0 + a1x + a2x2 + a3x

3 + . . . + akxk + . . . = 1 + x +

x2

2+

x3

6+

x4

24+ . . . +

xk

k!+ . . . =

∞∑

k=0

xk

k!

This is a very interesting series. We state here without proof that this series converges for all valuesof x. Further, the function defined by the series is in fact equal to ex that is,

ex = 1 + x +x2

2+

x3

6+ . . . =

∞∑

k=0

xk

k!

The implication is that the function ex is completely determined (for all x values) by its behaviour(i.e. derivatives of all orders) at x = 0. In other words, the value of the function at x = 1, 000, 000is determined by the behaviour of the function around x = 0. This means that ex is a very specialfunction with superior “predictable features”. If a function f(x) agrees with its Taylor polynomialon a region (−a, a), as was the case here, we say that f is analytic on this region. It is known thatex is analytic for all x.

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Math 103 Notes Chapter 12

Remark: How the exponential function grows

We can use the results of this example to establish the fact that the exponential function grows“faster” than any power function xn. That is the same as saying that the ratio of ex to xn (for anypower n) increases with x.

We can see this fact directly from the results of our Taylor series for ex and the following simple“division”:

ex

xn=

1 + x + x2

2!+ x3

3!+ . . . xn

n!+ xn+1

(n+1)!+ . . .

xn

=1

xn+

1

xn−1+ . . . +

1

n!+

x

(n + 1)!+ . . .

>x

(n + 1)!

We see that the last term becomes arbitrarily large as x → ∞, implying that

limx→∞

ex

xn= ∞

This shows that the exponential function ex grows faster than any power function, and hence alsofaster than any (finite) polynomial.

12.6.6 Example: Other (related) exponential functions

We can also easily obtain a Taylor series expansion for functions related to ex, without assemblingthe derivatives. We start with the result that

eu = 1 + u +u2

2+

u3

6+ . . . =

∞∑

k=0

uk

k!

Then the substitution u = −x leads to the result

e−x = 1 − x +x2

2−

x3

6+ . . . =

∞∑

k=0

(−x)k

k!.

The substitution u = x2 leads to

ex2

= 1 + x2 +(x2)2

2+

(x2)3

6+ . . . =

∞∑

k=0

(x2)k

k!

In other words,

ex2

= 1 + x2 +x4

2+

x6

6+ . . . =

∞∑

k=0

x2k

k!

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Math 103 Notes Chapter 12

sin(x)

T1

T2

T3

T4

0.0 7.0

-2.0

2.0

Figure 12.2: An approximation of the function y = sin(x) by successive Taylor polynomials,T1, T2, T3, T4. The higher Taylor polynomials do a better job of approximating the function ona larger interval about x = 0.

12.6.7 Example: Taylor series of simple trigonometric functions

In this example we determine the Taylor series for the function f(x) = sin x.

The derivatives of y = sin(x) are

f(x) = sin xf ′(x) = cos x

f ′′(x) = − sin xf ′′′(x) = − cos xf (4)(x) = sin x

and then the cycle repeats. This means that f(0) = 0, f ′(0) = 1, f ′′(0) = 0, f ′′′(0) = −1 and thenthe pattern repeats. In other words,

a0 = 0, a1 = 1, a2 = 0, a3 = −1

3!, a4 = 0, a5 =

1

5!, . . .

Now it is a fact that sin x is analytic for all x so that

sin x = x −x3

3!+

x5

5!−

x7

7!+ . . . =

∞∑

n=0

(−1)n x2n+1

(2n + 1)!

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Math 103 Notes Chapter 12

Consider the sequence of polynomials

T1(x) = x

T2(x) = x −x3

3!

T3(x) = x −x3

3!+

x5

5!

T4(x) = x −x3

3!+

x5

5!−

x7

7!

Then these polynomials provide a better and better approximation to the function sin(x) closeto x = 0. The first of these is just a linear (or tangent line) approximation that we had studiedlong ago. The second improves this with a quadratic approximation, etc. Figure 12.2 illustrateshow the first few Taylor polynomials approximate the function sin(x) near x = 0.

To determine the Taylor series for cosx, we can simply differentiate the series for sin x.

cos x =d

dxsin x

=d

dx(x −

x3

3!+

x5

5!−

x7

7!+ . . .)

= 1 −3x2

3!+

5x4

5!−

7x6

7!+ . . .

We find that

cos(x) = 1 −x2

2+

x4

4!−

x6

6!+ . . . =

∞∑

n=0

(−1)n x2n

(2n)!

12.7 Application of Taylor series

In this section we illustrate some of the applications of Taylor series to problems that may bedifficult to solve using other conventional methods

12.7.1 Evaluating definite integrals

Some functions do not have an antiderivative that can be expressed in terms of other simple func-tions. Integrating these functions can be a problem, as we cannot use the convenience of evaluatingthat antiderivative at endpoints, as the Fundamental Theorem of Calculus specifies. In some cases,we can approximate the value of the definite integral using a Taylor series, as the following exampleillustrates.

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Math 103 Notes Chapter 12

Example 1

Evaluate the definite integral

∫ 1

0

sin(x2) dx

A simple substitution (e.g. u = x2) will not work here, and we cannot find an antiderivative. Hereis how we might approach the problem using Taylor series: We know that the series expansion forsin(t) is

sin t = t −t3

3!+

t5

5!−

t7

7!+ . . .

Substituting t = x2, we have

sin(x2) = x2 −x6

3!+

x10

5!−

x14

7!+ . . .

In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylorseries, just as we would a polynomial:

∫ 1

0

sin(x2) dx =

∫ 1

0

(x2 −x6

3!+

x10

5!−

x14

7!+ . . .) dx

=

(

x3

3−

x7

7 · 3!+

x11

11 · 5!−

x15

15 · 7!+ . . .

)∣

∣

∣

∣

1

0

=1

3−

1

7 · 3!+

1

11 · 5!−

1

15 · 7!+ . . .

This is an alternating series so we know that it converges. If we add up the first four terms, thepattern becomes clear: the series converges to 0.31026.

12.7.2 Solving differential equations

Some differential equations cannot be solved in terms of familiar functions (just as some functionsdo not have antiderivatives which can be expressed in terms of familiar functions). However, Taylorseries can come to the rescue again. Here we will present two examples of this idea.

Example

We are already familiar with the differential equation and initial condition that describes unlimitedexponential growth.

dy

dx= y,

y(0) = 1.

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Math 103 Notes Chapter 12

Indeed, from previous work, we know that the solution of this differential equation and initialcondition is y(x) = ex, but we will pretend that we do not know this fact in illustrating theusefulness of Taylor series. In some cases, where separation of variables does not work, this optionwould have great practical value.

Let us express the “unknown” solution to the differential equation as

y = a0 + a1x + a2x2 + a3x

3 + a4x4 + . . .

Our task is to determine values for the coefficients ai

Since this function satisfies the condition y(0) = 1, we must have y(0) = a0 = 1.

Differentiating this power series leads to

dy

dx= a1 + 2a2x + 3a3x

2 + 4a4x3 + . . .

But according to the differential equation, dy

dx= y, Thus, it must be true that the two Taylor series

match, i.e.

a0 + a1x + a2x2 + a3x

3 + a4x4 + . . . = a1 + 2a2x + 3a3x

2 + 4a4x3 + . . .

This can only happen if the coefficients of like terms are the same, i.e. if the constant terms oneither side of the equation are equal, if the terms of the form Cx2 on either side are equal, and soon for all powers of x. Equating coefficients, we obtain:

a0 = a1 = 1, ⇒ a1 = 1a1 = 2a2, ⇒ a2 = a1

2= 1

2

a2 = 3a3, ⇒ a3 = a2

3= 1

2·3

a3 = 4a4, ⇒ a4 = a3

4= 1

2·3·4

an-1 = nan, ⇒ an = an-1

n= 1

1·2·3...n= 1

n!

This means that

y = 1 + x +x2

2!+

x3

3!+ . . . +

xn

n!+ . . . = ex

which, as we have seen, is the expansion for the exponential function that is the real solution wehave been expecting. In the example here shown, we would hardly need to use series to arrive atthe right conclusion, but in the next example, we would not find it as easy to discover the solutionby other techniques discussed previously.

Example 2: Airy’s equation

Airy’s equation arises in the study of optics, and (with initial conditions) is as follows:

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Math 103 Notes Chapter 12

y′′ = xy

y(0) = 1

y′(0) = 0

As before, we will write the solution as a series:

y = a0 + a1x + a2x2 + a3x

3 + a4x4 + a5x

5 + . . .

Using the information from the initial conditions, we get y(0) = a0 = 1 and y′(0) = a1 = 0. Nowwe can write down the derivatives:

y′ = a1 + 2a2x + 3a3x2 + 4a4x

3 + 5a5x4 + . . .

y′′ = 2a2 + 2 · 3x + 3 · 4x2 + 4 · 5x3 + . . .

The equation then gives

y′′ = xy

2a2 + 2 · 3a3x + 3 · 4a4x2 + 4 · 5a5x

3 + . . . = x(a0 + a1x + a2x2 + a3x

3 + . . .)

2a2 + 2 · 3a3x + 3 · 4a4x2 + 4 · 5a5x

3 + . . . = a0x + a1x2 + a2x

3 + a3x4 + . . .

Again, we can equate the coefficients of x, and use a0 = 1 and a1 = 0, to obtain

2a2 = 0 ⇒ a2 = 02 · 3a3 = a0 ⇒ a3 = 1

2·3

3 · 4a4 = a1 ⇒ a4 = 04 · 5a5 = a2 ⇒ a5 = 05 · 6a6 = a3 ⇒ a6 = 1

2·3·5·6

This gives us the first few terms of the solution:

y = 1 +x3

2 · 3+

x6

2 · 3 · 5 · 6+ . . .

If we continue in this way, we can write down many terms of the series.

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