time is precious, but we do not know yet how precious it ... · stationary points for functions of...

Time is precious, but we do not know yet how precious it really is. We will only know when we are no longer able to take advantage of it…

Proverbs 21:5 The plans of the diligent lead to profit as surely as haste leads to poverty

STA 2204 CALCULUS FOR STATISTICS III

PURPOSE

To extend the concepts of differentiation and integration to functions of several variables, and also to

investigate inequalities and estimates and their use in understanding the convergence of sequences and

series.

OBJECTIVES

By the end of this course the student should be able to;

a) Apply an appropriate convergence test for a given sequence or series

b) Apply the appropriate form of Taylor’s or Maclaurin’s series for a given function

c) Apply the trigonometric and hyperbolic representation of complex numbers

d) Apply reduction formulae in integration

e) Identify improper integrals and determine their convergence

f) Appreciate the integral as a limit of a sum

g) Evaluate double integrals

h) Apply the techniques for determining stationary points for functions of more than one

variable

i) Apply the techniques for determining the terms of a Fourier series

DESCRIPTION

Sequences and series: convergence tests. Power series: Taylor's and Maclaurin's theorems including

applications to binomial, logarithmic, exponential, trigonometric and hyperbolic functions.

Trigonometric and hyperbolic representation of complex numbers. Integration: reduction formulae

and its applications. Improper integrals and their convergence. Integration as the limit of a sum

including pincer method for evaluation of simple integrals. Double integrals including change of order

of integration and change of variable, Jacobians. Stationary Points for functions of more than one

variable. Lagrange multipliers. First and second order partial derivatives for functions of two or more

variables. Techniques for Fourier series.

PRE-REQUISITES: STA 2105 Calculus for Statistics II

OURSE TEXT BOOKS

1. S J Salas and E Hille Calculus: One and Several Variables.7th ed. Wiley, 1995.

2. Thomas & Finney. Calculus and Analytical Geometry. 7th. Addison-Wesley, 1988

COURSE JOURNALS:

1. Journal of the American Statistical Association

2. Calculus of Variations and Partial Differential Equations ISSN: 0944-2669 (print version) ISSN: 1432-

0835 (electronic version) Springer

FURTHER REFERENCE TEXT BOOKS AND JOURNALS:

1. Edwards, C. H. Multivariable Calculus With Analytic Geometry, 5th. Prentice Hall, 1997 ISBN-10:

0137363311 ISBN-13: 978-0137363315

2. Larson, Ron; Hostetler, Robert P.; Edwards, Bruce H. Calculus With Analytic Geometry, 8th ed. Houghton

Mifflin College, 2005 ISBN: 0131246461

3. Stephenson, G. Mathematical Methods for Science Students,2nd ed. Longman, 1995 ISBN 0-582-44416-0

4. Advanced Calculus with Applications in Statistics (Second Edition)

5. Stewart. Calculus Concepts and Contexts: Multi-variable and Single Variable. Brooks/Cole Pub Co,

2004 ISBN-10: 0495011630 ISBN-13: 978-0495011637.

6. Journal of Applied Mathematical Sciences

7. Journal of Convex Analysis

8. ESAIM: Control, Optimisation and Calculus of Variations



SEQUENCES AND SERIES In this chapter we’ll be taking a look at sequences and (infinite) series. Actually, this chapter

will deal almost exclusively with series. However, we also need to understand some of the

basics of sequences in order to properly deal with series. We will therefore, spend a little

time on sequences as well.

Sequences A sequence is a list of terms written down in a definite or specific order using a simple rule.

The list may or may not have an infinite number of terms in them although we will be dealing

exclusively with infinite sequences in this unit. General sequence terms are denoted as;

. . . term,1)n( term,n- term,second - ,first term th

1

th

21 nn aaaa

Because we will be dealing with infinite sequences each term in the sequence will be

followed by another term as noted above. In the notation above we need to be very careful

with the subscripts. The subscript 1n denotes the next term in the sequence and NOT one

plus the thn term. In other words, 11 nn aa so be very careful when writing subscripts to

make sure that the “+1” doesn’t migrate out of the subscript! This is an easy mistake.

There is a variety of ways of denoting a sequence. The following are equivalent ways of

denoting a sequence. 1121 . . . ,, , . . . , ,

nnnnn aaaaaa

In the second and third notations above na is usually given by a formula.

Note the difference between the second and third notations above. If the starting point is not

important or is implied in some way by the problem it is often not written down as we did in

the third notation. Next, we used a starting point of in the third notation only so we

could write one down. There is absolutely no reason to believe that a sequence will start at

. A sequence will start where ever it needs to start.

Example 1 Write down the first few terms of each of the following sequences.

(a)

1

2n

1

n

n (b)

0

n

1

2

)1(

n

n

(c) 1n

nb , where

Solution

(a)

1

2n

1

n

nTo get the first few sequence terms here all we need to do is plug in values of

n into the formula given and we’ll get the sequence terms.

...,25

6,

16

5,

9

4,

4

3,2

n

1

1

2

n

n

Note the inclusion of the “…” at the end! This is an important piece of notation as it is the

only thing that tells us that the sequence continues on and doesn’t terminate at the last term.

(b)

0

n

1

2

)1(

n

n

This one is similar to the first one. The main difference is that this

sequence doesn’t start at

.

....,16

1,

8

1,

4

1,

2

1,1

2

)1(

0

n

1

n

n

Note that the terms in this sequence alternate

in signs. Sequences of this kind are sometimes called alternating sequences.



(c) 1n

nb where ofdigit n

thnb

This sequence is different from the first two in the sense that it doesn’t have a specific

formula for each term. However, it does tell us what each term should be. Each term should

be the nth digit of π. So we know that ...91415926535.3

The sequence is then, ....,9,5,3,5,6,2,9,5,1,4,1,3 ]

In parts a and b of the previous example note that we were really treating the formulas as

functions that can only have integers plugged into them. Or,

n

n

ngn

nnf

2

)1()(

1)(

1

2

This is an important idea in the study of sequences (and series). Treating the sequence terms

as function evaluations will allow us to do many things with sequences that we couldn’t do

otherwise. Before delving further into this idea however we need to get a couple more ideas

out of the way.

First we want to think about “graphing” a sequence. To graph the sequence n a we plot the

points n,an , as n ranges over all possible values on a graph. For instance, let’s graph the

sequence

1

2n

1

n

n. The first few points on the graph are,

..,,25

6,5,

16

5,4,

9

4,3,

4

3,2,2,1

The graph, for the first 30 terms of the sequence, is then,

This graph leads us to an important idea about sequences. Notice that as n increases the

sequence terms in our sequence, in this case, get closer and closer to zero. We then say that

zero is the limit (or sometimes the limiting value) of the sequence and write,

0n

12limlim

na

nn

n

If nn

alim

exists and is finite we say that the sequence is convergent. If nn

alim

doesn’t exist or is infinite we say the sequence diverges.

Sometimes we will say the sequence diverges to if

nn

alim and if

nn

alim we will sometimes say that the sequence diverges to .

Get used to the terms “convergent” and “divergent” as we’ll be seeing them quite a bit

throughout this chapter.



Example 2 Determine if the following sequences converge or diverge. If the sequence

converges determine its limit.

(a)

1

2

2

510n

13

nn

n (b)

1

2

nn

ne (c)

1n

)1(

n

n

Solution

(a)

1

2

2

510n

13

nn

n To evaluate the limiting value we need to divide both numerator and

denominator by the largest power of n, in the numerator and then take the limit.. ie

5

3

5

3

510n

13

n10

1

2

22

limlim

n

nn n

n

So the sequence converges and its limit is .

(b)

1

2

nn

ne We will need to be careful with this one. We will need to use L’Hospital’s

Rule on this sequence.

1

2

n

22

limlimn

n

n

n

ee

So, the sequence in this part diverges to .

More often than not we just do L’Hospital’s Rule on the sequence terms without first

converting to x’s since the work will be identical regardless of whether we use x or n.

However, we really should remember that technically we can’t do the derivatives while

dealing with sequence terms.

(c)

1n

)1(

n

n

We will also need to be careful with this sequence. We might be tempted to

just say that the limit of the sequence terms is zero (and we’d be correct). However,

technically we can’t take the limit of sequences whose terms alternate in sign, because we

don’t know how to do limits of functions that exhibit that same behaviour. Also, we want to

be very careful to not rely too much on intuition with these problems. As we will see in the

next section, and in later sections, our intuition can lead us astray in these problem if we

aren’t careful.

we’ll first need to compute, 0n

1

n

)1(limlim

n

n

n

Therefore, since the limit of the sequence terms with absolute value bars on them goes to zero

then, 0n

)1(lim

n

n

which also means that the sequence converges to a value of zero.

Terms and Definitions. Given any sequence n a we have the following.

(i) We call the sequence increasing if 1nn aa and we call the sequence decreasing if

1nn aa for every n.

(ii) If n a is an increasing sequence or a decreasing sequence we call it monotonic.

(iii)If there exist a number m such that nam for every n we say the sequence is bounded

below. The number m is sometimes called a lower bound for the sequence.



(iv) If there exist a number M such that Ma n for every n we say the sequence is bounded

above. The number M is sometimes called an upper bound for the sequence.

(v) If the sequence is both bounded below and above we call the sequence bounded.

Note that in order for a sequence to be increasing or decreasing it must be increasing or

decreasing for every n. In other words, a sequence that increases for three terms and then

decreases for the rest of the terms is NOT a decreasing sequence! Also note that a monotonic

sequence must always increase or it must always decrease.

Example 3 Determine if the following sequences are monotonic and/or bounded.

(a) 0

2

nn (b)

1

1)1(n

n (c)

5

2n

2

n

Solution

(a) 0

2

nn This sequence is a decreasing sequence (and hence monotonic) because,

22 )1( nn for every n. Also, since the sequence terms will be either zero or negative

this sequence is bounded above. We can use any positive number or zero as the upper bound,

M, however, it’s standard to choose the smallest possible bound if we can and it’s a nice

number. So, we’ll choose since, nevery for 02 n

This sequence is not bounded below however since we can always get below any potential

bound by taking n large enough. Therefore, while the sequence is bounded above it is not

bounded.

As a side note we can also note that this sequence diverges (to if we want to be specific).

(b)

1

1)1(n

n The sequence terms in this sequence alternate between 1 and -1 and so the

sequence is neither an increasing sequence nor a decreasing thus it is not monotonic

The sequence is bounded since it is bounded above by 1 and bounded below by -1.

Again, we can note that this sequence is also divergent.

(c)

5

2n

2

n

This is a decreasing sequence (and hence monotonic) since, 22 1)(n

2

n

2

The terms in this sequence are all positive and so it is bounded below by zero. Also, since

the sequence is a decreasing sequence the first sequence term will be the largest and so we

can see that the sequence will also be bounded above by . Therefore, this sequence is

bounded.

We can also take a quick limit and note that this sequence converges and its limit is zero.

Question Determine if the following sequences are monotonic and/or bounded.

(a)

11n n

n (b)

0

4

3

10,000nn

n

Theorem: If na is bounded and monotonic then na is convergent.

Be careful to not misuse this theorem. It does not say that if a sequence is not bounded

and/or not monotonic that it is divergent. .

Note as well that we can make several variants of this theorem. If na is bounded above

and increasing then it converges and likewise if na is bounded below and decreasing then

it converges.



Series the Basics In this section we will introduce the topic that we will be discussing for the rest of this

chapter. That topic is infinite series. So just what is an infinite series? Well, let’s start with

a sequence 1

nna (note is for convenience, it can be anything) and define the

following,

n

i

inn aaaaasaaasaasas1

321321321211 . . . ...., , ,

The ns are called partial sums and notice that they will form a sequence, 1

nns . Also

recall that the is used to represent this summation and called a variety of names. The most

common names are : series notation, summation notation, and sigma notation.

We want to take a look at the limit of the sequence of partial sums, 1

nns . Notation ally

we’ll define,

11

limlimn

n

n

i

in

nn

aas

We will call

1n

na an infinite series and note that the series “starts” at 1n because that is

where our original sequence, 1

nna , started. Had our original sequence started at 2 then

our infinite series would also have started at 2. The infinite series will start at the same value

that the sequence of terms (as opposed to the sequence of partial sums) starts.

If the sequence of partial sums, 1

nns , is convergent and its limit is finite then we also call

the infinite series,

1n

na convergent and if the sequence of partial sums is divergent then

the infinite series is also called divergent.

Note that sometimes it is convenient to write the infinite series as,

. . . 321

1

aaaan

n

We do have to be careful with this however. This implies that an infinite series is just an

infinite sum of terms and as we’ll see in the next section this is not really true.

First, we should note that in most of this chapter we will refer to infinite series as simply

series. If we ever need to work with both infinite and finite series we’ll be more careful with

terminology, but in most sections we’ll be dealing exclusively with infinite series and so

we’ll just call them series.

Now, in

1n

na the n is called the index of summation or just index in short and note that the

letter we use to represent the index does not matter. So for example the following series are

all the same. The only difference is the letter we’ve used for the index.

etckin nnn

12

12

12 1

3

1

3

1

3

It is important to again note that the index will start at any value. Here we’ve used 1n but

the index could have started anywhere. In fact, we will usually use na to represent an

infinite series in which the starting point for the index is not important. Do not forget

however, that there is a starting point and that this will be an infinite series.



Properties of Infinite Series

If na and nb are both convergent series then,

nca , where c is any number, is also convergent and nn acca

kn

n

kn

n ba is also convergent and,

kn

nn

kn

n

kn

n baba .

The first property is simply telling us that we can always factor a multiplicative constant out

of an infinite series The second property says that if we add/subtract series all we really need

to do is add/subtract the series terms. Note as well that in order to add/subtract series we

need to make sure that both have the same initial value of the index and the new series will

also start at this value.

Note that,

kn

nn

kn

n

kn

n baba rather ...... 321321

11

bbbaaaban

n

n

n

Index Shift The basic idea behind index shifts is to start a series at a different value for whatever the

reason (and yes, there are legitimate reasons for doing that). Consider the following series,

2 2

5

nn

n

Suppose that for some reason we wanted to start this series at 0n , but we didn’t want to

change the value of the series. This means that we can’t just change the 2n to 0n as

this would add in two new terms to the series and thus change its value.

Performing an index shift is a fairly simple process to do. We’ll start by defining a new

index, say i, as follows, 2 ni

Now, when 2n , we will get 0i . Notice as well that if n then 2i so

only the lower limit will change here. Next, we can solve this for n to get, 2 in

We can now completely rewrite the series in terms of the index i instead of the index n

simply by plugging in our equation for n in terms of i.

02

02

2 2

7

2

5)2(

2

5

ii

ii

nn

iin

To finish the problem out we’ll recall that the letter we used for the index doesn’t matter and

so we’ll change the final i back into an n to get,

02

2 2

7

2

5

nn

nn

nn

To convince yourselves that these really are the same summation let’s write out the first

couple of terms for each of them,

.2

10

2

9

2

8

2

7

2

7 and....

2

10

2

9

2

8

2

7

2

55432

025432

2

nn

nn

nn

So, sure enough the two series do have exactly the same terms.

There is actually an easier way to do an index shift. The method given above is the

technically correct way of doing an index shift. However, notice in the above example we

decreased the initial value of the index by 2 and all the n’s in the series terms increased by 2

as well.

This will always work in this manner. If we decrease the initial value of the index by a set

amount then all the other n’s in the series term will increase by the same amount. Likewise,



if we increase the initial value of the index by a set amount, then all the n’s in the series term

will decrease by the same amount.

Let’s do a couple of examples using this shorthand method for doing index shifts.

Example 4 Perform the following index shifts.

(a) Write

1

1

n

nar as a series that starts at .

(b) Write

1

1

2

31nn

n as a series that starts at .

Solution

(a) In this case we need to decrease the initial value by 1 and so the n’s (okay the single n) in

the term must increase by 1 as well.

00

1)1(

1

1

n

n

n

n

n

n ararar

(b) For this problem we want to increase the initial value by 2 and so all the n’s in the series

term must decrease by 2.ie

31

2

31)2(

2

11

2

31

)2(

31

)2(

31 nn

nn

nn

nnn

The next concept is about alternate ways to write series when the situation requires it.

Let’s start with the following series and note that the starting point is only for

convenience since we need to start the series somewhere.

...54321

1

aaaaaan

n

Notice that if we ignore the first term the remaining terms will also be a series that will start

at 2n instead of 1n So, we can rewrite the original series as follows,

.2

1

1

n

n

n

n aaa

In this example we say that we’ve stripped out the first term.

We could have stripped out more terms if we wanted to. In the following series we’ve

stripped out the first two terms and the first four terms respectively.

5

4321

13

21

1

and .n

n

n

n

n

n

n

n aaaaaaaaaa

Being able to strip out terms will, on occasion, simplify our work or allow us to reuse a prior

result so it’s an important idea to remember.

Notice that in the second example above we could have also denoted the four terms that we

stripped out as a finite series as follows,

5

4

15

4321

1 n

n

n

n

n

n

n

n aaaaaaaa

This is a convenient notation when we are stripping out a large number of terms or if we need

to strip out an undetermined number of terms. In general, we can write a series as follows,

111 kn

n

k

n

n

n

n aaa

Series Convergence/Divergence Let’s recap just what an infinite series is and what it means for a series to be convergent or

divergent. We’ll start with a sequence 1

nna and again note that we’re starting the

sequence at 1n only for the sake of convenience and it can, in fact, be anything.



Next we define the partial sums of the series as,

n

i

inn aaaaasaaasaasas1

321321321211 . . . ...., , , and these

form a new sequence, 1

nns .

An infinite series is then the limit of the partial sums. Or nnn

n sa lim1

If the sequence of partial sums is a convergent sequence (i.e. its limit exists and is finite) then

the series is also called convergent and in this case if sassn

nnn

1

then,lim ,

Likewise, if the sequence of partial sums is a divergent sequence (i.e. its limit doesn’t exist or

is plus or minus infinity) then the series is also called divergent.

Let’s take a look at some series and see if we can determine if they are convergent or

divergent and see if we can determine the value of any convergent series we find.

Example 5 Determine if the following series is convergent or divergent. If it converges

determine its value.

a)

1n

n b)

22 1

1

n n c)

0

)1(n

n d)

113

1

nn

Solution

a)

1n

n To determine if the series is convergent we first need to get our hands on a formula

for the general term in the sequence of partial sums.2

)1( . . . 32 1

nnnsn

To determine if the series is convergent we will first need to see if the sequence of partial

sums,

12

)1(

n

nn is convergent or divergent. That’s not terribly difficult in this case. The

limit of the sequence terms is,

2

)1(limlim

nns

nn

n

Therefore, the sequence of partial sums diverges to and so the series also diverges to .

b)

22 1

1

n n This is actually one of the few series in which we are able to determine a

formula for the general term in the sequence of partial sum using partial fractions. As we

shall see later .

The general formula for the partial sums is, )1(2

1

2

1

4

3

1

1

22

nni

sn

i

n and in this case

we have, 4

3

)1(2

1

2

1

4

3limlim

nns

nn

n

The sequence of partial sums converges and so the series converges also and its value is,

4

3

1

1

22

n n

c)

0

)1(n

n In this case we really don’t need a general formula for the partial sums to

determine the convergence of this series. Let’s just write down the first few partial sums.

etcssss 0111111110111 3210



So, it looks like the sequence of partial sums is, ...,0,1,0,1,0,1 1

nns and this

sequence diverges since nn

slim

doesn’t exist. Therefore, the series also diverges.

d)

113

1

nn

The general formula for the partial sums is.

n

n

iins

3

11

2

3

3

1

11

In this case the limit of the sequence of partial sums is,2

3

3

11

2

3limlim

n

nn

n

s

The sequence of partial sums is convergent and so the series will also be convergent. The

value of the series is,2

3

3

1

11

nn

So, we’ve determined the convergence of four series now. Two of the series converged and

two diverged. Let’s go back and examine the series terms for each of these. For each of the

series let’s take the limit as n goes to infinity of the series terms (not the partial sums!!).

converged. series this01

1 diverged. series this

2limlim

nn

nn

converged. series this03

1 diverged. series thisexistnot does )1(

1limlim

n

n

n

n

Theorem; If na converges then 0lim

nn

a

Proof First let’s suppose that the series starts at . If it doesn’t then we can modify

things as appropriate below. Then the partial sums are,

1321

1

1

11321

1

. . . and . . .

n

n

i

innn

n

i

in aaaaasaaaaaas

Next, we can use these two partial sums to write, 1 nnn ssa

Now because we know that na is convergent we also know that the sequence 1

nns is

also convergent and that for some finite value s. However, since 1n as

n we also have ssnn

1lim .

We now have, 011 limlimlimlim

ssssssa nn

nn

nnn

nn

Be careful not to misuse this theorem! This theorem gives a necessary but not sufficient

condition for a series to converge.. In other words, the converse of this theorem is NOT true.

Ie If 0lim

nn

a the series may or may not converge! but it’s obvious that if 0lim

nn

a the

series diverges

Special Series a).Geometric Series

A geometric series is any series that can be written in the form,

1

1

n

nar (or,

0n

nar with an

index shift) It can be shown that the partial sums are,

r

ar

r

a

r

ras

nn

n

111

)1(

The series will converge provided 1r or if 0a in which case

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx#IndexShift



r

aarar

r

as

n

n

n

n

nn

11 01

1

lim

Remark: If 1r and 0a the series diverges

Note that in using this formula we’ll need to make sure that we are in the correct form.

In other words, if the series starts at 0n then the exponent on the r must be n. Likewise if

the series starts at 1n then the exponent on the r must be 1n .

Example 6 Determine whether the following series converge or diverge. If they converge

give the value of the series.

(a)

1

1249n

nn (b)

01

3

5

)4(

nn

n

Solution

(a)

1

1249n

nn This series doesn’t really look like a geometric series. However, notice that

both parts of the series term are numbers raised to a power. This means that it can be put into

the form of a geometric series. We will just need to decide which form is the correct form.

Since the series starts at 1n we will want the exponents on the numbers to be 1n .

It will be fairly easy to get this into the correct form. Let’s first rewrite things slightly. One

of the n’s in the exponent has a negative in front of it and that can’t be there in the geometric

form. So, let’s first get rid of that.

1

2

1

1

1)2(

1

12

9

44949

nn

n

n

nn

n

nn

Now let’s get the correct exponent on each of the numbers. This can be done using simple

exponent properties.

1

1

111

21

12

1

1

12

9

4144

99

44

9

449

n

n

nn

n

nn

n

n

nn

So, this is a geometric series with 144a and 194 r Therefore, the series converges and

its value is,

5

1296

1

14449

94

1

12

n

nn

(b)

01

3

5

)4(

nn

n

Again, it doesn’t look like a geometric series, but it can be put into the correct

form. In this case the series starts at so we’ll need the exponents to be n on the terms.

Note that this means we’re going to need to rewrite the exponent on the numerator a little

0001

3

01

3

5

645

5

645

55

)4(

5

)4(

n

n

nn

n

nn

n

nn

n

So, we’ve got it into the correct form and we can see that 5a and 1964 rr and

so this series diverges.

Back in the Series Basics section we talked about stripping out terms from a series, but didn’t

really provide any examples of how this idea could be used in practice. We can now do some

examples.

Example 7 Use the results from example 6 to determine the value of the following series.

(a)

0

1249n

nn (b)

3

1249n

nn

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Basics.aspx#StripingTerms



Solution

(a)

0

1249n

nn Let’s notice that if we strip out the first term from this series we arrive at,

1

12

1

1212

0

12 49324494949n

nn

n

nn

n

nn

From the previous example we know the value of the new series that arises here and so the

value of the series in this example is,

5

2916

5

129632449

0

12

n

nn

(b)

3

1249n

nn In this case we can’t strip out terms from the given series to arrive at the

series used in the previous example. However, we can start with the series used in the

previous example and strip terms out of it to get the series in this example. So, let’s do that.

We will strip out the first two terms from the series we looked at in the previous example.

3

12

3

123021

1

12 4920849494949n

nn

n

nn

n

nn

We can now use the value of the series from the previous example to get the value of this

series.5

256208

5

12962084949

1

12

3

12

n

nn

n

nn

Exercise

1) Determine the 6th and 7th terms of the series below. Also determine the sum of the first 10 terms

a) .....4

1

8

1

16

1

b) .....405.045.05.0

c) .....163264

d) .....28log14log7log

2) The first term of a GP is 54 and the fourth term is 16, find the limiting sum of this GP.

3) Express the sum using sigma notation. Hence or otherwise Find the limiting sum of the series

a) .....064.016.04.01

b) .....001.001.01.01

c) ....444444

5

13

5

12

5

1

5

1

d) .....1 4321 eeee

e) .....432 13

16

13

8

13

4132

4) Determine the limiting sum of the following series

a)

04

3

n

n b)

1

2

nen

n

c)

13 2

n

n

d)

04

3

n

n e)

02

51

n

n

f)

16

23

n

n

nn

g))

1

7

2

5

3

n

nn

n

5) Find the sum to infinity of the following series

a)

13

2)1( 1

n

n

nn

b)

3

1

325

n

n c)

16

1

n

n

n

b).Telescoping Series The name telescoping comes from what happens with the partial sums and is best shown in

an example.

Example 8 Determine whether the series

02 23

1

n nnconverges or diverges. If it

converges find its limiting value

Solution let’s notice that we can use partial fractions on the series term to get,

2

1

1

1

21

1

23

12

nnnnnn



So, what does this do for us? Well, let’s start writing out the terms of the general partial sum

for this series using the partial fraction form.

Notice that every term except the first and last term cancelled out. This is the origin of the

name telescoping series.

This also means that we can determine the convergence of this series by taking the limit of

the partial sums.

1

2

11limlim

ns

nn

n

The sequence of partial sums is convergent and so the series is convergent and has a value of

123

1

02

n nn

In telescoping series be careful to not assume that successive terms will be the ones that

cancel. Consider the following example.

Example 9

Is the series

12 34

1

n nn converges or not? If it converges what is its value?

Solution As with the last example we’ll leave the partial fractions details to you to verify. The partial

sums are,

In this case instead of successive terms cancelling out, a term will cancel with a term that is

farther down the list. The end result this time is two initial and two final terms are left.

Notice as well that in order to help with the work a little we factored the 2

1 out of the series.

The limit of the partial sums is,

12

5

3

1

2

1

6

5

2

1limlim

nns

nn

n

So, this series is convergent (because the partial sums form a convergent sequence) and its

value is, 12

5

34

1

12

n nn

Note that just because you can do partial fractions on a series term does not mean that the

series will be a telescoping series. The following series, for example, is not a telescoping

series despite the fact that we can get partial fraction the series terms.

112 2

1

1

1

23

23

nn nnnn

n



In order for a series to be a telescoping series we must get terms to cancel and all of these

terms are positive and so none will cancel.

Next, we need to go back and address an issue that was first raised in the previous section. In

that section we stated that the sum or difference of convergent series was also convergent and

that the presence of a multiplicative constant would not affect the convergence of a series.

Now that we have a few more series in hand let’s work a quick example showing that.

Example 10 Determine the limiting value of the series

1

12

249

34

4

n

nn

nn.

Solution To get the value of this series all we need to do is rewrite it and then use the previous results.

15

3863

5

1296

12

5449

34

1449

34

4

1

12

12

1

12

2

n

nn

nn

nn

nnnn

Exercise (continued)

6) Determine whether the following series is convergent or divergent. Use the knowledge of

telescoping series

a)

1)2(

5

nnn

b)

1

11lnn

n c)

1

2242lnk

knk d)

114

12ln

nn

e)

1

2

1

2

11

n

nn

c).The p-series

It’s a series of the form

knpn

1where 0p It converges if 1p and diverges if 10 p

Example 11 Determine if the following series are convergent or divergent.

(a)

47

1

n n (b)

1

1

n n

Solution

(a) This is a p-series with 17 p and so the series is convergent.

(b) For this series 121 p and so the series is divergen.

d). Harmonic Series

Here is the harmonic series.

1

1

n n

The harmonic series is divergent and we’ll need to wait until the next section to show that.

Example 12 Show that each of the following series are divergent.

(a)

1

5

nn

(b)

4

1

nn

Solution

(a)

1

5

nn

To see that this series is divergent all we need to do is use the fact that we can

factor a constant out of a series as follows,

1

1

1

5 5n

nn

n

http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx



Now,

1

1

n

n is divergent and so five times this will still not be a finite number and so the

series has to be divergent. In other words, if we multiply a divergent series by a constant it

will still be divergent.

(b)

4

1

nn

In this case we’ll start with the harmonic series and strip out the first three terms.

611

1

1

4

1

4

131

21

1

1 1

nn

nn

nn

nn

In this case we are subtracting a finite number from a divergent series. This subtraction will

not change the divergence of the series. We will either have infinity minus a finite number,

which is still infinity, or a series with no value minus a finite number, which will still have no

value. Therefore, this series is divergent.

Just like with convergent series, adding/subtracting a finite number from a divergent series is

not going to change the divergence of the series.

So, some general rules about the convergence/divergence of a series are now in order.

Multiplying a series by a constant will not change the convergence/divergence of the series

and adding or subtracting a constant from a series will not change the

convergence/divergence of the series. These are nice ideas to keep in mind.

Tests for Convergence of Series Generally, finding a formula for the general term in the sequence of partial sums is a very

difficult process. In fact in this section we’ll not be doing much with the partial sums of

series due to the extreme difficulty faced in finding the general formula. This also means that

we’ll not be doing much work with the value of series since in order to get the value we’ll

also need to know the general formula for the partial sums.

Integral Test

Suppose )(f x is a continuous, positive decreasing function in the interval ,k and let

nan )(f then if

kdxx)(f is; (i) convergent so is

kn

na (ii) divergent so is

kn

na

NB the lower limit on the improper integral must be the same value that starts the series.

Again this test does NOT give the value of a series. It will only give the

convergence/divergence of the series. That’s it. No value.

Example 13 Determine if the following series is convergent or divergent.

(a)

0

2

n

nne (b)

4

ln

n n

n

Solution

a)

0

2

n

nne The function that we’ll use in this example is, .)(2xxexf This function is

always positive on the interval that we’re looking at. Now we need to check that the function

is decreasing. It is not clear that this function will always be decreasing on the interval

given. We can use our Calculus I knowledge to help us however. The derivative of this

function is, .21)('22 xexxf



This function has two critical points (which will tell us where the derivative changes sign) at

2

1x . Since we are starting at we can ignore the negative critical point. Picking a

couple of test points we can see that the function is increasing on the interval 2

1,0 and it is

decreasing on ,2

1 . Therefore, eventually the function will be decreasing and that’s all

that’s required for us to use the Integral Test.

21

21

21

021

00

2222

limlimlim ..

n

n

nx

n

nx

n

x eedxxedxxe

The integral is convergent and so the series must also be convergent by the Integral Test.

b)

4

ln

n n

n In this case the function we’ll use is

x

xxf

ln)(

This function is clearly positive and if we make x larger the denominator will get larger and

so the function is also decreasing. Therefore, all we need to do is determine the convergence

of the following integral.

22

21

4

2

21

444lnlnln

lnlnlimlimlim nxdx

x

xdx

x

x

n

n

n

n

n

The integral is divergent and so the series is also divergent by the Integral Test.

Comparison Test / Limit Comparison Test In the previous section we saw how to relate a series to an improper integral to determine the

convergence of a series. While the integral test is a nice test, it does force us to do improper

integrals which aren’t always easy and in some cases may be impossible to determine the

convergence of. For instance consider the following series.

0 3

1

nn n

In order to use the Integral Test we would have to integrate

0.

3

1dx

xx and I’m not even

sure if it’s possible to do this integral. Nicely enough for us there is another test that we can

use on this series that will be much easier to use.

First, let’s note that the series terms are positive. As with the Integral Test that will be

important in this section. Next let’s note that we must have 0x since we are integrating on

the interval x0 . Likewise, regardless of the value of x we will always have 03 x .

So, if we drop the x from the denominator the denominator will get smaller and hence the

whole fraction will get larger. So, nn n 3

1

3

1

Now,

0 3

1

nn

is a geometric series and we know that since 13

1r the series will converge

and its value will be 2

3

1

1

3

1

31

0

nn

Now, going back to our original series and write down the partial sums we get

n

iin

is

0 3

1

Since all the terms are positive adding a new term will only make the number larger and so

the sequence of partial sums must be an increasing sequence.

1

1

00 3

1

3

1

n

n

ii

n

iin s

iis

Then since, nn n 3

1

3

1

and because the terms in these two sequences are positive we can

also say that,

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx#GeometricSeries



2

3

2

3

3

1

3

1

3

1

000

n

nn

n

ii

n

iin s

is

Therefore, the sequence of partial sums is also a bounded sequence. Then from the second

section on sequences we know that a monotonic and bounded sequence is also convergent.

So, the sequence of partial sums of our series is a convergent sequence. This means that the

series itself,

0 3

1

nn n

is also convergent.

So, what did we do here? We found a series whose terms were always larger than the

original series terms and this new series was also convergent. Then since the original series

terms were positive (very important) this meant that the original series was also convergent.

To show that a series (with only positive terms) was divergent we could go through a similar

argument and find a new divergent series whose terms are always smaller than the original

series. In this case the original series would have to take a value larger than the new series.

However, since the new series is divergent its value will be infinite. This means that the

original series must also be infinite and hence divergent.

We can summarize all this in the following test.

General Comparison Test

Suppose that we have two series na and nb with 0, nn ba for all n and nn ba for

all n. Then,

i) If nb is convergent then so is na . ii) If na is divergent then so is nb

In other words, we have two series of positive terms and the terms of one of the series is

always larger than the terms of the other series. Then if the larger series is convergent the

smaller series must also be convergent. Likewise, if the smaller series is divergent then the

larger series must also be divergent. Note as well that in order to apply this test we need both

series to start at the same place.

Do not misuse this test. Just because the smaller of the two series converges does not say

anything about the larger series. The larger series may still diverge. Likewise, just because

we know that the larger of two series diverges we can’t say that the smaller series will also

diverge! Be very careful in using this test

Note as well that the requirement that 0, nn ba and nn ba really only need to be true

eventually. In other words, if a couple of the first terms are negative or nn ba for a couple

of the first few terms we’re okay. As long as we eventually reach a point where 0, nn ba

and nn ba for all sufficiently large n the test will work.

To see why this is true let’s suppose that the series start at and that the conditions of

the test are only true for for 1 Nn and for Nnk at least one of the conditions is not

true. If we then look at na (the same thing could be done for nb ) we get,

1Nn

n

N

kn

n

kn

n aaa

The first series is nothing more than a finite sum (no matter how large N is) of finite terms

and so will be finite. So the original series will be convergent/divergent only if the second

infinite series on the right is convergent/divergent and the test can be done on the second

series as it satisfies the conditions of the test. Let’s take a look at some examples.

http://tutorial.math.lamar.edu/Classes/CalcII/MoreSequences.aspx#Series_MoreSeq_BndMon

http://tutorial.math.lamar.edu/Classes/CalcII/MoreSequences.aspx#Series_MoreSeq_BndMon




a)

122 cosn nn

n b)

14

2

5

2

n n

n

Solution

a)

122 cosn nn

nSince the cosine term in the denominator doesn’t get too large we can

assume that the series terms will behave like nn

n 12 which, as a series, will diverge. So,

from this we can guess that the series will probably diverge and so we’ll need to find a

smaller series that will also diverge.

Recall that from the comparison test with improper integrals that we determined that we can

make a fraction smaller by either making the numerator smaller or the denominator larger. In

this case the two terms in the denominator are both positive. So, if we drop the cosine term

we will in fact be making the denominator larger since we will no longer be subtracting off a

positive quantity. Therefore,nn

n

nn

n 1

cos 222

Then, since

1

1

nn

diverges (it’s harmonic or the p-series) by the Comparison Test our original

series must also diverge.

b)

14

2

5

2

n n

n In this case the “+2” and the “+5” don’t really add anything to the series and so

the series terms should behave pretty much like 24

2 1

nn

n which will converge as a series.

Therefore, we can guess that the original series will converge and we will need to find a

larger series which also converges.

This means that we’ll either have to make the numerator larger or the denominator smaller.

We can make the denominator smaller by dropping the “+5”. Doing this gives,

4

2

4

2 2

5

2

n

n

n

n

At this point, notice that we can’t drop the “+2” from the numerator since this would make

the term smaller and that’s not what we want. However, this is actually the furthest that we

need to go. Let’s take a look at the following series.

14

12

14

14

2

14

2 12

122

nnnnn nnnn

n

n

n

As shown, we can write the series as a sum of two series and both of these series are

convergent by the p-series test. Therefore, since each of these series are convergent we know

that the sum

14

2 2

n n

n is also a convergent series. Recall that the sum of two convergent

series will also be convergent.

Now, since the terms of this series are larger than the terms of the original series we know

that the original series must also be convergent by the Comparison Test.

The comparison test is a nice test that allows us to do problems that either we couldn’t have

done with the integral test or at the best would have been very difficult to do with the integral

test. That doesn’t mean that it doesn’t have problems of its own.



Consider the following series.

03

1

nnn

This is not much different from the first series that we looked at. The original series

converged because the 3n gets very large very fast and will be significantly larger than n.

Therefore, n doesn’t really affect the convergence of the series in that case. The fact that we

are now subtracting n off instead of adding n on really shouldn’t change the convergence.

We can say this because the 3n gets very large very fast and the fact that we’re subtracting n

off won’t really change the size of this term for all sufficiently large values of n.

So, we would expect this series to converge. However, the comparison test won’t work with

this series. To use the comparison test on this series we would need to find a larger series

that we could easily determine the convergence of. In this case we can’t do what we did with

the original series. If we drop the n we will make the denominator larger (since the n was

subtracted off) and so the fraction will get smaller and just like when we looked at the

comparison test for improper integrals knowing that the smaller of two series converges does

not mean that the larger of the two will also converge.

So, we will need something else to help us determine the convergence of this series. The

following variant of the comparison test will allow us to determine the convergence of this

series.

Qn Determine if the following series is convergent or divergent

03

32

210

4

n n

nn.

Limit Comparison Test

Suppose we have two series na and nb with 0, nn ba for all n. Define n

n

n b

ac lim

If c is positive (i.e. ) and is finite (i.e. ) then either both series converge or both

series diverge.

Note that it doesn’t really matter which series term is in the numerator for this test, we could

just have easily defined c as n

n

n a

bc lim

and we would get the same results. To see why this

is, consider the following two definitions.

n

n

nn

n

n a

bc

b

ac limlim

Start with the first definition and rewrite it as follows, then take the limit.

ca

b

a

b

b

ac

n

n

nn

n

nn

n

n

111

limlimlim

In other words, if c is positive and finite then so is c and if c is positive and finite then so is

c. Likewise if 0c then c and if c then c Both definitions will give the same

results from the test so don’t worry about which series terms should be in the numerator and

which should be in the denominator. Choose this to make the limit easy to compute.

Also, this really is a comparison test in some ways. If c is positive and finite this is saying

that both of the series terms will behave in generally the same fashion and so we can expect

the series themselves to also behave in a similar fashion. If 0c or c we can’t say this

and so the test fails to give any information.



The limit in this test will often be written as

n

nn b

ac1

lim since often both terms will be

fractions and this will make the limit easier to deal with.

Let’s see how this test works.

Example 15 Determine if the following series converges or diverges.

a)

0 3

1

nn n

b)

237

2

3

4

n nn

nn

Solution

a)

0 3

1

nn n

To use the limit comparison test we need to find a second series that we can

determine the convergence of easily and has what we assume is the same convergence as the

given series. On top of that we will need to choose the new series in such a way as to give us

an easy limit to compute for c.

We’ve already guessed that this series converges and since it’s vaguely geometric let’s use

0 3

1

nn

as the second series. We know that this series converges and there is a chance that

since both series have the n3 in it the limit won’t be too bad. Here’s the limit.

n

n

n

nn

nnc

313

3

1limlim

Now, we’ll need to use L’Hopital’s Rule on the second term in order to actually evaluate this

limit. 13ln3

11 lim

n

n

c So, c is positive and finite thus by the Comparison Test both

series must converge since

0 3

1

nn

converges.

b)

23 37

24

n nn

nn Fractions involving only polynomials or polynomials under radicals will

behave in the same way as the largest power of n will behave in the limit. So, the terms in

this series should behave as 3

13

7

12

3 7

2

nn

n

n

n and as a series this will diverge by the p-series

test. In fact, this would make a nice choice for our second series in the limit comparison test

so let’s use it.

4

1

4

11

4

11

4

1

43

34

1

34

73 37

2

37

37

34

37

31

limlimlim

nn

n

nn

nnn

nn

nnc n

nnn

So, c is positive and finite and so both series will diverge since

23

1

1

n ndiverges.

Finally, to see why we need c to be positive and finite (i.e. and ) consider the

following two series

2

1

n n and

22

1

n n The first diverges and the second converges.

Now compute each of the following limits.

01

1

1 and

1

1limlimlimlim 2

2

n

n

nn

n

n nnnn



In the first case the limit from the limit comparison test yields c and in the second case

the limit yields 0c . Clearly, both series do not have the same convergence.

Note however, that just because we get 0c or c doesn’t mean that the series will have

the opposite convergence. To see this consider the series,

13

1

n n and

12

1

n n

Both of these series converge and here are the two possible limits that the limit comparison

test uses.

nn

nn

n

n nnnnlimlimlimlim

3

2

2

3 1

1 and0

1

1

1

So, even though both series had the same convergence we got both and .

The point here is to remind us that if we get or from the limit comparison test we

will know that we have chosen the second series incorrectly and we’ll need to find a different

choice in order to get any information about the convergence of the series.

c) Ratio Test In this section we are going to take a look at a test that we can use to see if a series is

absolutely convergent or not. Recall that if a series is absolutely convergent then we will also

know that it’s convergent and so we will often use it to simply determine the convergence of

a series.

Before proceeding with the test let’s do a quick reminder of factorials. This test will be

particularly useful for series that contain factorials (and we will see some in the applications)

so let’s make sure we can deal with them before we run into them in an example.

If n is an integer such that 0n then n factorial is defined as

definitionby 1!0

on so and )!3)(2)(1()!2)(1()!1(!

1n if123...)3)(2)(1(!

nnnnnnnnnn

nnnnn

When dealing with factorials we need to be very careful with parenthesis. For instance,

!2)!2( nn as we can see if we write each of the following factorials out.

123...)3)(2)(1(2!2

123...)32)(22)(12(2)!2(

nnnnn

nnnnn

Okay, we are now ready for the test.

Suppose we have the series na . Define n

n

n a

aL 1

lim

Then,

i) if 1L the series is absolutely convergent (and hence convergent).

ii) if 1L the series is divergent.

iii) if 1L the series may or may not converge.

Notice that the absolute value bars in the definition of L are absolutely required. If they are

not there it will be possible for us to get the incorrect answer.

Example 16 Determine if the series below is convergent or divergent.

a)

11241

10

nn

n

n b)

0 5

!

nn

n c)

2

2

!12n n

n d)

11

2

9

nn

n

n

Solution

a)

11241

10

nn

n

n With this first example let’s be a little careful and make sure that we have

everything down correctly.

Here are the series terms

1241

10

n

n

nn

a



Recall that to compute an+1 all that we need to do is substitute n+1for all the n’s in an.

32

1

!)1(2

1

142

10

411

10

n

n

n

n

nnn

a

Now

18

5

2

1

16

10

)2(4

)1(10

10

41

411

10limlimlimlim 2

12

!)1(2

1

1

n

n

n

nn

na

aL

nnn

n

n

n

nn

n

n

So, by the Ratio Test the series converges absolutely and hence will converge.

As seen in this example there is usually a lot of cancelling that will happen in these. Make

sure that you do this cancelling. If you don’t do this kind of cancelling it can make the limit

fairly difficult.

b)

0 5

!

nn

n Now that we’ve worked one in detail we won’t go into quite the detail with the rest

of these. Here is the limit.

15

1

!

5

5

)!1(limlim 1

n

n

nL

n

n

nn

So, by the Ratio Test this series diverges.

c)

2

2

!12n n

n In this case be careful in dealing with the factorials

10

1

212

1!12

!12

)1(!12

!112

)1(2

2

2

2

2

limlimlim

n

n

nnn

n

n

n

n

n

n

nL

nnn

So, by the Ratio Test this series converges absolutely and so converges

d)

11

2

9

nn

n

n Do not mistake this for a geometric series. The n in the denominator means

that this isn’t a geometric series. So, let’s compute the limit.

1

2

9

12

9

12

9

9

2

12

9limlimlim

1

2

1

n

n

n

nn

nL

nnn

n

n

n

n

Therefore, by the Ratio Test this series is divergent.

d) Root Test This is the last test for series convergence that we’re going to be looking at. As with the

Ratio Test this test will also tell whether a series is absolutely convergent or not rather than

simple convergence.

Suppose that we have the series na Define n

nn

nn

n

aaL1

limlim

then,

i) if 1L the series is absolutely convergent (and hence convergent).

ii) if 1L the series is divergent.

iii) if 1L the series may be divergent, or onvergent.

We will need the following fact in some of these problems. Fact: 11

lim

nnn


a)

1123n

n

nn b)

13

3

27

35

n

n

n

nn c)

3

12

n

n

n

http://tutorial.math.lamar.edu/Classes/CalcII/Series_Special.aspx#GeometricSeries



Solution

a)

1123n

n

nn There really isn’t much to these problems other than computing the limit and

then using the root test. Here is the limit for this problem.

1333 2212 limlimlim 1

1

nn

n

n

n n

n

nnL

So, by the Root Test this series is divergent.

b)

13

3

27

35

n

n

n

nn Again, there isn’t too much to this series.

17

3

7

3

27

35

27

353

3

3

3

limlim

1

n

nn

n

nnL

n

n

n

n

Therefore, by the Root Test this series converges absolutely and hence it converges.

Note that we had to keep the absolute value bars on the fraction until we’d taken the limit to

get the sign correct.

c)

3

12

n

n

n Here’s the limit for this series.

1

1

1212121

1

limlim

n

n

nnL

n

n

n

After using the fact from above we can see that the Root Test tells us that this series is

divergent.

Strategy for Series Now that we’ve got all of our tests out of the way it’s time to think about organizing all of

them into a general set of guidelines to help us determine the convergence of a series. here is

the set of guidelines for determining the convergence of a series.

a) With a quick glance does it look like the series terms don’t converge to zero in the limit,

i.e. is 0lim

nn

a If so, the test for divergence

b) If the series looks like a modified p-series or a modified geometric series, try the

Comparison Test.

c) Is the series a rational expression involving only polynomials or polynomials under

radicals (i.e. a fraction involving only polynomials or polynomials under radicals)? If so,

try the Comparison Test and/or the Limit Comparison Test. Remember however, that in

order to use the Comparison Test and the Limit Comparison Test the series terms all need

to be positive.

d) Does the series contain factorials or constants raised to powers involving n? If so, then

the Ratio Test may work. Note that if the series term contains a factorial then the only

test that we’ve got that will work is the Ratio Test.

e) Can the series terms be written in the form n

nn ba )( ? If so, then the Root Test may work.

f) If )(nfan for some positive, decreasing function )(xf and

adxxf )( is easy to

evaluate then the Integral Test may work.

Remember that these are only a set of guidelines and not a set of hard and fast rules to use

when trying to determine the best test to use on a series. If more than one test can be used try

to use the test that will be the easiest for you to use.



Exercise

1) Determine the convergence or divergence of the following series using the integral test, if possible.

a)

125

1

nn

b)

1

1

nn

c)

1)(ln

12

nnn

d)

1)1( 22

nn

n e)

1

ln2

nn

n f)

13

1

n

n g)

11

12

nn

h)

112

nn

n i)

154

12

nnn

j)

14)(ln

12

nnn

k)

1

543

n

nnn

2) Determine the convergence or divergence of the following series using the direct comparison test, if possible.

a)

13

12

nn

b)

1

2

nnn

n c)

04

12

nnn

d)

15

sin2

n

nn e)

11

14 3

nn

f)

21ln

1

nnn

g)

11

cos2

3

nnn

n

h)

010

sin1

n

nn i)

1

23

n

ne n

j)

1

ln2

nn

n k)

1

ln

nnn l)

1

tan3

1

nn

n

3) Determine the convergence or divergence of the following series using the limit comparison test, if possible.

a)

134

3

n

n b)

34

22

nn

c)

112

nn

n d)

116

1

nn

e)

01

32

nn

f)

01

252

nn

n g)

34

12

nn

h)

0)12(

134

nn

n

i)

21

nn

n j)

05

32

nn

n k)

119

2 2

nn

n l)

2

ln

23

nn

n m)

2

)ln( 43

nn

nn

4) Determine the convergence or divergence of the following series using the ratio test, if possible.

a)

02

!

n

nn b)

092

2

nn

n c)

13

4

nn n

n

d)

13

52

n

n

n

e)

145

n

nn f)

1

32

kk

k

g)

34

)3(1

n

nn

n

h)

0)3(ln

3

n

nn

i)

2)!2(

2)1( 2

nn

nn

j)

2!

)()1( 221

nn

n nn

k)

0)!12(

3 2

nnn l)

1)3(

91

nn n

n

m)

13 31

n

nn

n

n)

1)!2(

)!(4 2

nn

nn

5) Determine the convergence or divergence of the following series using the root test, if possible.

a)

112

12

2

n

n

n

n b)

1

3n

nn e c)

1

3

nn

en

n

d)

185

142

2

n

n

n

n e)

1)1(

3

kk k

k

f)

13

)1(

12

n

n

n

nn

g)

11

2

n

n

nn h)

322

1

n

nn

n

i)

1)(tan

)1(1

nn n

n

j)

13 31

n

nn

n

Power Series

It, is any series that can be written in the form

0n

n

n axc where a and cn are real numbers.

The cn’s are often called the coefficients of the series. The first thing to notice about a power

series is that it is a function of x. That is different from any other kind of series that we’ve

looked at to this point. In all the prior sections we’ve only allowed numbers in the series and

now we are allowing variables to be in the series as well. This will not change how things

work however. Everything that we know about series still holds.

In the discussion of power series convergence is still a major question that we’ll be dealing

with. The difference is that the convergence of the series will now depend upon the values of

x that we put into the series. A power series may converge for some values of x and not for

other values of x.



Radius and Interval of Convergence Before we get too far into power series there is some terminology that we need to get out of

the way. First, as we will see in our examples, we will be able to show that there is a number

R so that the power series will converge for, Rax and will diverge for Rax . This

number is called the radius of convergence for the series. Note that the series may or may

not converge if Rax . What happens at these points will not change the radius of

convergence. Secondly, the interval of all x’s, including the endpoints if need be, for which

the power series converges is called the interval of convergence of the series.

These two concepts are fairly closely tied together. If we know that the radius of

convergence of a power series is R then we have the following.

diverges seriespower and

converges seriespower

RaxxRa

RaxRa

The interval of convergence must then contain the interval RaxRa since we know

that the power series will converge for these values. We also know that the interval of

convergence can’t contain x’s in the ranges RaxxRa and since we know the

power series diverges for these value of x. Therefore, to completely identify the interval of

convergence all that we have to do is determine if the power series will converge for

or . If the power series converges for one or both of these values then

we’ll need to include those in the interval of convergence.

It is important to note that no matter what else is happening in the power series we are

guaranteed to get convergence for .ax The series may not converge for any other value of

x, but it will always converge for .ax

Let’s work some examples. We’ll put quite a bit of detail into the first example and then not

put quite as much detail in the remaining examples.

Example 18 Determine the radius and interval of convergence for the following power series.

a) n

nn

n

xn

)3(4

)1(

1

b) n

n

n

xn

)84(2

1

c)

1

)6(

nn

n

n

x

Solution

a)n

nn

n

xn

)3(4

)1(

1

Okay, we know that this power series will converge for 3x , but that’s

it at this point. To determine the remainder of the x’s for which we’ll get convergence we

can use any of the tests that we’ve discussed to this point. After application of the test that

we choose to work with we will arrive at condition(s) on x that we can use to determine

which values of x for which the power series will converge and which values of x for which

the power series will diverge. From this we can get the radius of convergence and most of

the interval of convergence (with the possible exception of the endpoints).

With all that said, the best tests to use here are almost always the ratio or root test. Most of

the power series that we’ll be looking at are set up for one or the other. In this case we’ll use

the ratio test.

n

xn

xn

xnL

nnn

n

n

nn

n 4

)3)(1(

)3()1(

4

4

)3)(1()1(limlim 1

11

Before going any farther with the limit let’s notice that since x is not dependent on the limit

it can be factored out of the limit. Notice as well that in doing this we'll need to keep the

absolute value bars on it since we need to make sure everything stays positive and x could

well be a value that will make things negative. The limit is then,

33411

lim

xxLn

n

n



So, the ratio test tells us that if 1L the series will converge, if 1L the series will diverge,

and if 1L we don’t know what will happen. So, we have,

diverges series4313

converges series4313

41

41

xx

xx

We’ll deal with the 1L case in a bit. Notice that we now have the radius of convergence

for this power series. These are exactly the conditions required for the radius of

convergence. The radius of convergence for this power series is 4R .

Now, let’s get the interval of convergence. We’ll get most (if not all) of the interval by

solving the first inequality from above.

17434 xx

So, most of the interval of validity is given by . All we need to do is determine if

the power series will converge or diverge at the endpoints of this interval. Note that these

values of x will correspond to the value of x that will give 1L .

The way to determine convergence at these points is to simply plug them into the original

power series and see if the series converges or diverges using any test necessary.

7x : In this case the series is

11

2

11

)1(4)(4

)1()4(

4

)1(

nn

nnn

nn

nn

nn

n

nnnn

This series is divergent by the Divergence Test since 0lim

nn

n= .

1x : In this case the series is,

11

)()4(4

)1(

n

nn

nn

n

nn

This series is also divergent by the Divergence Test since nn

n

)(lim

doesn’t exist.

So, in this case the power series will not converge for either endpoint. The interval of

convergence is then, 17 x

In example a the power series didn’t converge for either endpoint of the interval. Sometimes

that will happen, but don’t always expect that to happen. The power series could converge at

either both of the endpoints or only one of the endpoints.

b) n

n

n

xn

)84(2

1

Let’s jump right into the ratio test

8421

)284

)84(2)1(

)84(2limlim

11

xn

nx

x

n

n

xL

nnn

nn

n

So we will get the following convergence/divergence information from this.

diverges series21842

converges series21842

81

81

xx

xx

We need to be careful here in determining the interval of convergence. The interval of

convergence requires Rax and or for diverge Rax . Now, let’s find the interval of

convergence. Again, we’ll first solve the inequality that gives convergence above.

817

815

81

81 2 xx

Now check the endpoints.

8

15x : The series here is,

121

1

12

n

nn

n

n

nn

This is the alternating harmonic series and we know that it converges.



817x : The series here is,

1

121

1

2

nn

n

n

n

n

This is the harmonic series and we know that it diverges.

So, the power series converges for one of the endpoints, but not the other. This will often

happen so don’t get excited about it when it does. The interval of convergence for this power

series is then,8

178

15 x

c)

1

)6(

nn

n

n

x In this case the root test seems more appropriate. So,

01

66)6(

limlimlim

1

n

xn

x

n

xL

nnn

n

n

n

So, since 10 L regardless of the value of x this power series will converge for every x.

In these cases we say that the radius of convergence is R and interval of convergence is

x

Remark. If the power series only converges for then the radius of convergence is

and the interval of convergence is . Likewise if the power series converges for every x

the radius of convergence is and interval of convergence is .

Qn Determine the radius and interval of convergence for the power series

1

2

)3(nn

nx

Taylor and Maclaurin Series Suppose the function )(f x does in fact have a power series representation about ,

....)(f3

3

2

210

0

axcaxcaxccaxcxn

n

n

Also assume that the function )(f x has derivatives of every order and that we can in fact find

them all.

Then plugging ax gives, 0)(f ca

Next take the derivative of the function (and its power series) then plug in we get,

1

3

4

2

321 )('f....432)('f caaxcaxcaxccx and we now know c1.

Let’s continue with this idea and find the second derivative.

2

)("f2)("f....34232)("f 22

2

432

accaaxcaxccx

Using the third derivative gives,

23

)("'f23)("'f....23423)("'f 3343

accaaxccx

Using the fourth derivative gives,

234

)("''f234)("''f....2345234)("''f 4454

accaaxccx

Hopefully by this time you’ve seen the pattern here. It looks like, in general, we’ve got the

following formula for the coefficients. !

)(f )(

n

ac

n

n . This even works for n=0 if you recall

that and define )(f)(f )0( xx

So, provided a power series representation for the function )(xf about ax exists the

Taylor Series for )(f x about ax is,



....!3

)("'f

!2

)("f)('f)(f

!

)(f)(f

32

0

)(

axa

axa

axaaaxn

ax

n

nn

If we use 0a , so we talking about the Taylor Series about , we call the series a

Maclaurin Series. Therefore the Maclaurin Series for )(f x is,

....!4

)0("''f

!3

)0("'f

!2

)0("f)0('f)0(f

!

)0(f)(f 432

0

)(

xxxxxn

xn

nn

Example 19 Find the Maclaurin Series for; a) xex )(f b) xx cos)(f

Solution

a) xex )(f This is actually one of the easier Taylor Series that we’ll be asked to compute.

To find the Taylor Series for a function we will need to determine a general formula for

)(f )( an . This is one of the few functions where this is easy to do right from the start.

To get a formula for )0(f )(n all we need to do is recognize that ....,3,2,1)(f )( nex xn

and so, ....,3,2,11)0(f 0)( nen

Therefore, the Taylor series for xex )(f about x=0 is,

00 !!

1

n

n

n

nx

n

xx

ne

b) xx cos)(f First we’ll take some derivatives of the function and evaluate them at x=0. ie

xx cos)(f xx sin)('f xx cos)("f xx sin)("'f xx cos)(""f xx sin)('""f

Thus 1)0(f 0)0('f 1)0("f 0)0("'f 1)0(""f 0)0(""'f and so on

In this example, unlike the previous one, there is not an easy formula for either the general

derivative. However, there is a clear pattern to the evaluations. So, let’s plug what we’ve got

into the Maclaurin series and see what we get,

...!6

1

!5

0

!4

1

!3

0

!2

101

...!5

)0(""'f

!4

)0(""f

!3

)0("'f

!2

)0("f)0('f)0(fcos

65432

5432

xxxxxx

xxxxxx

Let’s drop the zeroes, we can get.

0

2642

)!2(

)1(...

!6!4!21cos

n

nn

n

xxxxx

Qn Find the Maclaurin Series for; a) xex )(f b) 234)(f xexx c) xx sin)(f

Ans a)

0 !

)1(

n

nnx

n

xe b)

0

4234

!

)3(2

n

nnx

n

xex c)

0

12

)!12(

)1(sin

n

nn

n

xx

Example 20 Find the Taylor Series expansion of ;

a) xex )(f about 4x . b) xx ln)(f in powers of 2x

Solution

a) xex )(f about 4x

Here are the first few derivatives and the evaluations at 4x xex )(f

xex )('f xex )("f

xex )("'f xex )(""f . . .

xnn ex )1()(f )( Thus

4)0(f e 4)0('f e

4)0("f e 4)0("'f e

4)0(""f e

. . . . 4)( )1()0(f enn

The Taylor Series is then,

0

4

4!

)1(

n

nn

x xn

ee

b) xx ln)(f in powers of 2x

Here are the first few derivatives and the evaluations



xx ln)(f , 1)('f xx ,

2)("f xx , 32)("'f xx ,

46)(""f xx , 524)(""'f xx

. . . nnn xnx )!1()1()(f 1)(

Thus 2ln)2(f , 2

1)2('f , 4

1)2("f , 8

2)2("'f ,

16

6)2("''f , 32

24)2("'"f . . . . nnn n 2)!1()1()2(f 1)(

Note that while we got a general formula here it doesn’t work for . This will happen on

occasion so don’t worry about it when it does.

In order to plug this into the Taylor Series formula we’ll need to strip out the term first.

1

1

1

1

1

)(

0

)(

22

)1(2ln

2!2

)!1()1(2ln2

!

)2(f)2(f2

!

)2(fln

n

n

n

n

n

n

n

n

n

nn

n

nn

xn

xn

nx

nx

nx

Notice that we simplified the factorials in this case. You should always simplify them if

there are more than one and it’s possible to simplify them.

Also, do not get excited about the term sitting in front of the series. Sometimes we need to

do that when we can’t get a general formula that will hold for all values of n. Qn Find the

Taylor Series expansion of 2)( xxf about 1x . Ans

0

1)1(n

nxn

Applications of Series There are in fact many applications of series, unfortunately most of them are beyond the

scope of this course. One application of power series (with the occasional use of Taylor

Series) is in the field of Ordinary Differential Equations when finding Series Solutions to

Differential Equations..

Many of the applications of series, especially those in the differential equations fields, rely on

the fact that functions can be represented as a series. In these applications it is very difficult,

if not impossible, to find the function itself. However, there are methods of determining the

series representation for the unknown function.

While the differential equations applications are beyond the scope of this course there are

some applications from a Calculus setting that we can look at.

Example 21 Determine a Taylor Series about for the integral dxx

sin x

Solution

To do this we will first need to find a Taylor Series about for the integrand. This

however isn’t terribly difficult. We already have a Taylor Series for sine about so we’ll

just use that as follows,

0

2

0

12

)!12(

)1(

)!12(

)1(1

x

sin x

n

nn

n

nn

n

x

n

x

x

We can now do the problem .

0

12

0

2

)!12)(12(

)1(

)!12(

)1(

x

sin x

n

nn

n

nn

nn

xCdx

n

xdx

So, while we can’t integrate this function in terms of known functions we can come up with

a series representation for the integral.

This idea of deriving a series representation for a function instead of trying to find the

function itself is used quite often in several fields. In fact, there are some fields where this is

http://tutorial.math.lamar.edu/Classes/DE/SeriesIntro.aspx

http://tutorial.math.lamar.edu/Classes/DE/SeriesIntro.aspx



one of the main ideas used and without this idea it would be very difficult to accomplish

anything in those fields.

Another application of series isn’t really an application of infinite series. It’s more an

application of partial sums. In fact, we’ve already seen this application in use once in this

chapter. In the Estimating the Value of a Series we used a partial sum to estimate the value

of a series. We can do the same thing with power series and series representations of

functions. The main difference is that we will now be using the partial sum to approximate a

function instead of a single value.

We will look at Taylor series for our examples, but we could just as easily use any series

representation here. Recall that the nth degree Taylor Polynomial of f(x) is given by,

n

r

rr

axr

ax

0

)(

n!

)(f)(T

Let’s take a look at example of this.

Example 22 For the function xx cos)(f plot the function as well as )(T,)(T 42 xx and

)(T8 x on the same graph for the interval [-4 , 4].

Solution

Here is the general formula for the Taylor polynomials for cosine.

n

r

rr

axr

x0

2

n)!2(

)1()(T

The three Taylor polynomials that we’ve got are then,

403207202421)(T and

2421)(T,

21)(T

8642

8

42

4

2

2

xxxxx

xxx

xx

Here is the graph of these three Taylor polynomials as well as the graph of cosine.

As we can see from this graph as we increase the degree of the Taylor polynomial it starts to

look more and more like the function itself. In fact by the time we get to )(T8 x the only

difference is right at the ends. The higher the degree of the Taylor polynomial the better it

approximates the function.

Also the larger the interval the higher degree Taylor polynomial we need to get a good

approximation for the whole interval.

Before moving on let’s write down a couple more Taylor polynomials from the previous

example. Notice that because the Taylor series for cosine doesn’t contain any terms with odd

powers on x we get the following Taylor polynomials.

http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx



403207202421)(T and

2421)(T,

21)(T

8642

9

42

5

2

3

xxxxx

xxx

xx

These are identical to those used in the example. Sometimes this will happen although that

was not really the point of this. The point is to notice that the nth degree Taylor polynomial

may actually have a degree that is less than n. It will never be more than n, but it can be less

than n.

The final example in this section really isn’t an application of series and probably belonged in

the previous section. However, the previous section was getting too long so the example is in

this section. This is an example of how to multiply series together and while this isn’t an

application of series it is something that does have to be done on occasion in the

applications. So, in that sense it does belong in this section.

Example 3 Find the first 3 terms in the Taylor Series for xex x cos)(f about 0x .

Solution

Before we start let’s acknowledge that the easiest way to do this problem is to simply

compute the first 3-4 derivatives, evaluate them at 0x , plug into the formula and we’d be

done. However, as we noted prior to this example we want to use this example to illustrate

how we multiply series.

We will make use of the fact that we’ve got Taylor Series for each of these so we can use

them in this problem.

0

2

0 )!2(

)1(

!cos

r

rr

r

rx

r

x

r

xxe

We’re not going to completely multiply out these series. We’re going to do enough of the

multiplication to get an answer. The problem statement says that we want the first three non-

zero terms. That non-zero bit is important as it is possible that some of the terms will be

zero. If none of the terms are zero this would mean that the first three non-zero terms would

be the constant term, x term, and x2 term. However, because some might be zero let’s assume

that if we get all the terms up through x4 we’ll have enough to get the answer. If we’ve

assumed wrong it will be very easy to fix so don’t worry about that.

Now, let’s write down the first few terms of each series and we’ll stop at the x4 term in each.

...

2421...

24621cos

42432 xxxxxxxex

Note that we do need to acknowledge that these series don’t stop. That’s the purpose of the “

... ” at the end of each. Just for a second however, let’s suppose that each of these did stop

and ask ourselves how we would multiply each out. If this were the case we would take

every term in the second and multiply by every term in the first. In other words, we would

first multiply every term in the second series by 1, then every term in the second series by x,

then by x2 etc.

By stopping each series at x4 we have now guaranteed that we’ll get all terms that have an

exponent of 4 or less. Do you see why?

Each of the terms that we neglected to write down have an exponent of at least 5 and so

multiplying by 1 or any power of x will result in a term with an exponent that is at a

minimum 5. Therefore, none of the neglected terms will contribute terms with an exponent

of 4 or less and so weren’t needed.

So, let’s start the multiplication process.

xseriesseries

x

ndnd

xxx

xxxxxxxxxe

2

53

12

4242432

...242

...242

1...242

1...2462

1cos



......5764824

...144126

...4842

4

2413

612

21 2

864

2

753

2

642

xseriesxseriesxseries ndndnd

xxxxxxxxx

Now, collect like terms ignoring everything with an exponent of 5 or more since we won’t

have all those terms and don’t want them either. Doing this gives,

...63

1...1cos43

4

24

1

4

1

24

13

6

1

2

12

2

1

2

1 xx

xxxxxxex

There we go. It looks like we over guessed and ended up with four non-zero terms, but that’s

okay. If we had under guessed and it turned out that we needed terms with x5 in them all we

would need to do at this point is go back and add in those terms to the original series and do a

couple quick multiplications. In other words, there is no reason to completely redo all the

work.

TRIGONOMETRIC AND HYPERBOLIC FORMS OF

COMPLEX NUMBERS Introduction

The equation 012 x has no solutions, because for any real number x the square 2x is

nonnegative, and so 12 x can never be less than 1. In spite of this it turns out to be very

useful to assume that there is a number i for which one has 12 i

Any complex number is then an expression of the form biaz where a, b ∈ ℝ and

1i is an imaginary number. The number a is called the real part of z, and b is called its

imaginary part. We often write

azzz )()Re(2

1 for the real part of z and

bzzz )()Im(21 for the imaginary part of z

A real number like a can often be interpreted as a special complex number by writing it as

ia 0 . Similarly, a way of writing i is i10 .

Definition (Conjugate) the complex conjugate to biaz is biaz

The Complex Number Plane Since any complex number is specified by

two real numbers, one can visualize them by

plotting a point with coordinates (a, b), ie the

real part (a) along the horizontal axis, and the

imaginary part (b) along the vertical axis, in

the plane for a complex number bia . The

plane in which one plot these complex

numbers is called the Complex plane, or

Argand plane.

Note that i is the unit of measurement on the

vertical axis.

For any given complex number biaz ,

we defines the absolute value or modulus of

z to be the distance from the origin O to the

point z in the complex plane (see figure)

alongside). The modulus of z denoted as z

is given by zzbaz 22

The angle Oz makes with the positive real axis is called the argument of z and is denoted as

)arg( z . )arg( z is given by . . . . 3, 2,,1, k tan)arg( 1 kzab



Note that the argument is not uniquely defined but we’ll conventionally choose )arg( z to

be in region 2)arg(0 z

Note: In quadrants I and III, tan is positive. In the quadrants II and IV, 0tan . In

quadrant I, a

bz 1tan)arg( . But, for quadrant III, the positive ab1tan must be upgraded

with the addition of π. Similarly, for z in quadrant II, ab1tan is negative. To bring it to the

second quadrant we have to add π. For z in the fourth quadrant, ab1tan is again negative,

but its location is correct. To keep it there we now have to add 2π, so that the value falls into

the interval [0,2π).

To sum up, denote a

b1

0 tan . Then

00

0

00

0

00

0

00

0 2)arg()arg()arg()arg(

babababa

zzzz

From trigonometry we can see that for any complex number biaz one has

sin and cos rbra where zr and )arg( z

Therefore, any complex number can be represented in the trigonometric form (or in polar

coordinates) as

)sin(cossincos irirrz

Example 1: Find argument and absolute value of iz 3 hence express z in polar form.

Solution:

Since 3a and 1b then 21322 bazr

z lies in the 4th quadrant so 2)arg(23 z but

611

63

1 2arctan)arg( z

Therefore )sin()cos(2611

611 iz

Exercise:

1) Find argument and absolute value of the following complex numbers

a) iz 31 b) iz 1 c) iz 31 d) i•z2

2

2

2 e) iz2

3

2

1

2) The imaginary part of a complex number is known to be twice its real part. The absolute

value of this number is 4. Which number is this?

3) The real part of a complex number is known to be half the absolute value of that number.

The imaginary part of the number is 1. Which number is it?

The trigonometric form is intimately related to the operation of multiplication.

Theorem 1: Let )sin(cos11 irz and )sin(cos22 irz . Then;

i) )sin()cos(2121 irrzz …………………… *

ii) )sin()cos(2

1

2

1 ir

r

z

z

Proof

)sin()cos()sincoscos(sin)sinsincos(cos

sinsincossinsincoscoscos)sin(cos)sin(cos i)

2121

2

212121

irrirr

iiirririrzz

)sin()cos(sincos

)sincoscos(sin)sinsincos(cos

sincos

sinsincossinsincoscoscos

)sin)(cossin(cos

)sin)(cossin(cos ii)

2

1

22

2

1

222

2

2

1

2

1

2

1

ir

ri

r

r

i

iii

r

r

ii

ii

r

r

z

z

Theorem 1 tells us that the modulus of the product of two numbers is the product of their

moduli, something we already knew. But it also tells us something we did not: the argument

http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbers.shtml#extendedArg

http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbers.shtml#trigForm

http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbers.shtml#m123



of the product is the sum of the arguments of the multiplicands. (The latter of course needs to

be taken modulo 2π.)

).2 (mod)arg()arg()arg( and 21212121 zzzzzzzz

In other words, when you multiply complex numbers, their lengths get multiplied and their

arguments get added.

Remark: If rrrzzz 2121 and then 2sin2cos22

21 irzzz

By this remark For any complex number z the argument of its square

)arg(2)arg()arg()arg()arg( 2 zzzzzz

Similarly, the argument of it’s cube is )arg(3)arg( 3 zz and in general )arg()arg( znzn

for any integer n.

Theorem 2: (De Moivre’s Theorem)

Let )sin(cos irz and n ∈ ℤ+ then nnrrz nnn sincos)sin(cos

Proof is by induction and is left as exercise to the learners

Basically the De Moivre's formula for any integer n≠0 is nini n sincos)sin(cos

For 0n and 0z , we define 10 z and then De Moivre's formula holds for all integer n,

with no exception

Example 2: Given )sin()cos(32

32

21

1 iz and )sin()cos(2611

611

2 iz find;

a) 21zz b) 2

1

z

z c) 321zz . d) 2

1

2

z

z

Solution:

a) 226

156

15611

32

611

32

21

21 sincos)sin()cos()sin()cos(2 iiizz

b) )sin()cos()sin()cos()sin()cos(6

5

6

5

4

1

6

7

6

7

4

1

6

11

3

2

6

11

3

2

2

21

2

1 iiiz

z

c) )sin()cos()3sin()3cos(sincos12

3

2

3

22

3

22

33

21 iiizz

d) )sin()cos(4)sin()cos(

67

67

32

611

32

611

5.02

1

2 iiz

z

)sin()cos(16)2sin()2cos(16)sin()cos(437

37

67

672

67

6722

1

2 iiiz

z

Example 3: Use De Moivre’s theorem to show that

2

3

tan31

tantan33tan

Solution

3sin3cos)sin(cos 3 ii Expanding the left hand side binomially we get

3223

3223

3322233

sinsincos33sin and sincos3cos3cos

sinsincos3sincos3cos

sinsincos3sincos3cos)sin(cos

i

iiii

23

32

sincos3cos

sinsincos3

3cos

3sin3tanBut

Dividing every term in the numerator and

denominator by 3cos we get

2

3

23

32

tan31

tantan3

sincos3cos

sinsincos33tan

Exercise:

1) Given )sin()cos(234

34

1 iz and )sin()cos(247

47

2 iz find;

http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumbers.shtml#argModulo



a) 2

1 2zz b)

2

1

z

z c) 32

1 2zz . d) 6

1

2

z

z

2) Use De Moivre’s theorem to show that

42

3

tantan61

tan4tan44tan

Euler's Form of Complex Numbers Complex numbers are very conveniently denoted as exponentials; this makes multiplication

easy. Recall the Taylor series for .....!3!2

132

hh

heh replacing h with i we have

sincos.....!5!3

....!6!4!2

1

.....!6

)(

!5

)(

!4

)(

!3

)(

!2

)()(1

53642

65432

ii

iiiiiiei

Thus any complex number can be represented as jrez which is called the Euler form

The conjugate of z in this form is jrez

Generally, theorem 1 leads to Euler's formula sincos iei such that, quite naturally, )( iii eee . In particular, nnrerr ninnn

sincos)sin(cos

Exercise:

1) Show that iteetiiteet itititit sinhsin andcoshcos21

21

2) Compute the absolute value and argument of )1)(2(ln iez

IMPROPER INTEGRALS Introduction

So far we have only considered definite integral b

adtt)(f under the conditions that )(f t is

defined and bounded on the bounded interval [a, b]. In this section we will extend the theory

of integration to bounded functions defined on unbounded intervals and also to unbounded

functions defined on bounded or unbounded intervals.

A definite integral b

adtt)(f is said to be improper if;

i) at least one of the integration limits is a and/or b is infinite

ii) there exist at least one point of discontinuity in the interval [a, b]

Let’s start with the first kind of improper integrals in which at least one of the integration

limits is infinite

.

Improper Integrals with Infinite Interval of Integration In this kind of integral one or both of the limits of integration are infinity. There are

essentially three cases that we’ll need to look at. That is:

adtt)(f ,

b

dtt)(f and

dtt)(f

In these 3 cases the integration is said to be over an infinite interval.

Let’s take a look at an example that will also show us how we are going to deal with these

integrals.

Example 1 Evaluate the following integral.

1

12 dt

t

Solution



This integral looks very simple. However, because infinity is not a real number we can’t just

integrate as normal and then “plug in” the infinity to get an answer. To see how we’re going

to do this integral let’s think of this as an area problem. So instead of asking what the integral

is, let’s instead ask what the area under 2

1)(ft

t on the interval ) , [1 is.

1

12 dt

t

We still aren’t able to do this, however, let’s step back a little and instead ask what the area

under )(f t is on the interval ] , [1 n where 1n and n is finite. This is a problem that we can

do.

22

1

11

1 11n

nn

tn tdtA

Now, we can get the area under )(f t on ) , [1 simply by taking the limit of nA as n goes to

infinity.

11limlim 2

1 nn

nn

AA

This is then how we will do the integral itself.

11lim1limlimlim 222

1

11

1

1

1

nn

n

n

n

tnn

nt tdtAdtA

So, this is how we will deal with these kinds of integrals in general. We will replace the

infinity with a variable (usually n), do the integral and then take the limit of the result as n

goes to infinity.

On a side note, notice that the area under a curve on an infinite interval was not infinity as we

might have suspected it to be. In fact, it was a surprisingly small number. Of course this

won’t always be the case, but it is important enough to point out that not all areas on an

infinite interval will yield infinite areas.

Let’s now get some definitions out of the way. We will call these integrals convergent if the

associated limit exists and is a finite number (i.e. it’s not plus or minus infinity) and

divergent if the associated limit either doesn’t exist or is (plus or minus) infinity.

Let’s now formalize up the method for dealing with infinite intervals.

i) If n

adx f(x) exists for every an then,

n

anadx f(x)limdx f(x) provided the limit

exists and is finite.

ii) If b

ndx f(x) exists for every bn then,

b

nn

b

dx f(x)limdx f(x) provided the limits

exists and is finite

iii) If

c

c

dx f(x) and dx f(x) are both convergent then,

c

c

dx f(x)dx f(x)dx f(x) where c is any number. Note as well that this

requires BOTH of the integrals to be convergent in order for this integral to also be

convergent. If either of the two integrals is divergent then so is this integral.

Example Determine if the following integral is convergent or divergent and if it’s

convergent find its value.

a)

1

1 dxx

b)

0

x-3

1 dx c)

dx xe

2x- d)

2sin xdx

Solution

a) So, the first thing we do is convert the integral to a limit, do the integral and then evaluate

the limit.

1lnlnlimlimlim ln11

1

1

1 ndxdxn

n

n

n

xn

x x



Here, the limit is infinite and so the integral is divergent

b) There really isn’t much to do with these problems once you know how to do them. We’ll

convert the integral to a limit/integral pair, evaluate the integral and then the limit.

3232n-32limlimdx limdx x-320

0

x-3

10

x-3

1

nn

nnn

Again, the limit is infinite and so this integral is divergent

c) In this case we’ve got infinities in both limits and so we’ll need to split the integral up

into two separate integrals. We can split the integral up at any point, so let’s choose 0a

since this will be a convenient point for the evaluation process. The integral is then,

0

x-0

x-x- dx xedx xedx xe222

We’ve now got to look at each of the individual limits

21

21n-

21

0

0x-

0x- )01(e1lim

2x-

21limdx xelimdx xe

222

e

n

n

nnn

So, the first integral is convergent. Now let’s take a look at 2nd integral.

21

21n-

21

0

0

x-

0

x- )01(e1lim2x-

21limdx xelimdx xe

222

e

n

n

n

n

n

This integral is convergent and so since they are both convergent the integral we were

actually asked to deal with is also convergent and its value is

0dx xedx xedx xe21

21

0

x-0

x-x- 222

d) nosxxdxn

n

n

n

nx c2coslimlimdx sinlimsin cos

222

This limit doesn’t exist

and so the integral is divergent

Remark If 0a then

a xdxp

1 is convergent if 1p and divergent if 1p

Exercise:

Determine whether each integral is convergent or divergent. Evaluate those that are

convergent.

a) )0(0

kdxe kx b)

0

1 dxe x

x c)

0

4

12 dx

x d)

0

24 x

dx e)

0 2dx

x1

x

f)

0

2 dxex x g)

xx ee

dx h)

0)1( dxex x

Discontinuous Integrand We now need to look at the second type of improper integrals that we’ll be looking at in this

section. These are integrals that have discontinuous integrands. The process here is basically

the same with one subtle difference. Here are the general cases that we’ll look at for these

integrals

a) If )(f t is continuous on the interval bt a and not continuous at bt then,

n

abn

b

adttdtt )(flim)(f provided the limit exists and is finite. Note as well that we do

need to use a left hand limit here since the interval of integration is entirely on the left

side of the upper limit.



b) If )(f t is continuous on the interval bt a and not continuous at at then,

b

nan

b

adttdtt )(flim)(f provided the limit exists and is finite. In this case we need to use a

right hand limit here since the interval of integration is entirely on the right side of the

lower limit.

c) If f(t)is not continuous at ct where bc a and b

c

c

adttdtt )(f and )(f are both

convergent then, b

c

c

a

b

adttdttdtt )(f )(f)(f

As with the infinite interval case this requires BOTH of the integrals to be convergent in

order for this integral to also be convergent. If either of the two integrals is divergent then

so is this integral.

d) If )(f t is not continuous at at and bt and if b

c

c

adttdtt )(f and )(f are both

convergent then, b

c

c

a

b

adttdttdtt )(f )(f)(f Where c is any number. Again, this

requires BOTH of the integrals to be convergent in order for this integral to also be

convergent

Note that the limits in these cases really do need to be right or left handed limits. Since we

will be working inside the interval of integration we will need to make sure that we stay

inside that interval. This means that we’ll use one-sided limits to make sure we stay inside the

interval. Let’s do a couple of examples of these kinds of integrals.

Example Determine if the following integral is convergent or divergent. If it is convergent

find its value. a) 3

0 x-3

1 dx b) 3

2

3- dx x c)

0 x

1 dx 2

Solution

a) The problem point is the upper limit so we are in the first case above.

32n-3232limlimdx limdx 3

030 x-3

1

3

3

0 x-3

1 x-32

n

n

n

n

n

The limit exists and is finite and so the integral converges & the integral’s value is 32 .

b) This integrand is not continuous at x=0 and so we’ll need to split the integral up at that

point

3

0

3-0

2

3-3

2

3- dx xdx xdx x

Now we need to look at each of these integrals and see if they are convergent.

22

181

02

02

3-

0

0

2

3- lim2-

21limdx xlimdx x x

nn

n

n

n

n

At this point we’re done. One of the integrals is divergent that means the integral that we

were asked to look at is divergent. We don’t even need to bother with the second

integral.

c) This is an integral over an infinite interval that also contains a discontinuous integrand.

To do this integral we’ll need to split it up into two integrals. We can split it up anywhere,

but pick a value that will be convenient for evaluation purposes.

1 x

11

0 x

1

0 x

1 dx dx dx 222

In order for the integral in the example to be convergent we will need BOTH of these to

be convergent. If one or both are divergent then the whole integral will also be divergent.

We know that the second integral is convergent by the fact given in the above remark.

So, all we need to do is check the first integral.



1lim1-

limdx xlimdx 1

0

1

0

12-

0

1

0 x

1 x2 nn

nnnn

Since the first integral is divergent then the whole integral is divergent

Exercise:

Evaluate each of the following integral if possible.

a) 2

1x-1

1 dx b)

3

0 2dx

x-9

1 c)

2

2 2dx

1-x

1 d)

2

0dx sec

x e)

0dx tan x

e)

2

0 2dx

4-x

2x f)

2

2dx 1-x 3

2

g) 1

0

2 dx lnl3x x h) 3

1 2dx

3x-x

3

Comparison Test for Improper Integrals Oftenly we aren’t concerned with the actual value of the improper integrals but we are rather

interested in whether the integral is convergent or divergent. Also, there will be some

integrals that are difficult to integrate and yet we would still like to know if they converge or

diverge.

To deal with this we’ve got a test for convergence or divergence that we can use to help us

answer the question of convergence for an improper integral. We will give this test only for a

sub-case of the infinite interval integral, however versions of the test exist for the other sub-

cases of the infinite interval integrals as well as integrals with discontinuous integrands.

Comparison Test

Suppose 0 g(x)f(x) on the interval xa

1. If

adx f(x) converges then so does

adx g(x)

2. If

adx g(x) diverges then so does

adx f(x)

Note that if you think in terms of area the Comparison Test makes a lot of sense. If f(x) is

larger than g(x) then the area under f(x) must also be larger than the area under g(x). So, if the

area under the larger function is finite (i.e.

adx f(x) converges) then the area under the

smaller function must also be finite (i.e.

adx g(x) converges). Likewise, if the area under

the smaller function is infinite (i.e.

adx g(x) diverges) then the area under the larger

function must also be infinite (i.e.

adx f(x) diverges).

Be careful not to misuse this test. If the smaller function converges there is no reason to

believe that the larger will also converge (after all infinity is larger than a finite number…)

and if the larger function diverges there is no reason to believe that the smaller function will

also diverge.

Let’s work a couple of examples using the comparison test. Note that all we’ll be able to do is

determine the convergence of the integral. We won’t be able to determine the value of the

integrals and so won’t even bother with that.

Remark: When using the comparison test, the following inequalities are useful:

xeexxxxx xxp asln and122

Example Determine if the following integral is convergent or divergent.

a)

2 x

cos dx 2

2x b)

3 ex

1 dx x c)

3 e-3

1 dx x- d)

1 x

)2(sin31dx

4 x e)

1 xe dx

-x



Solution

a) Let’s take a second and think about how the Comparison Test works. If this integral is

convergent then we’ll need to find a larger function that also converges on the same

interval. Likewise, if this integral is divergent then we’ll need to find a smaller function

that also diverges. So, it seems like it would be nice to have some idea as to whether the

integral converges or diverges ahead of time so we will know whether we will need to

look for a larger (and convergent) function or a smaller (and divergent) function.

To get the guess for this function let’s notice that the numerator is nice and bounded and

simply won’t get too large. Therefore, it seems likely that the denominator will determine

the convergence or divergence of this integral and we know that

2 x

1 dx 2converges since

12 p by the remark in the previous section. So let’s guess that this integral will

converge.

So we now know that we need to find a function that is larger than 2

2

x

cos x and also

converges. Making a fraction larger is actually a fairly simple process. We can either

make the numerator larger or we can make the denominator smaller. In this case we can’t

do a lot about the denominator. However we can use the fact that 1cos02

x to make the

numerator larger (i.e. we’ll replace the cosine with something we know to be larger,

namely 1). So, .22

2

x

1

x

cos x Now, as we’ve already noted

2 x

1 dx 2 converges and so by the

Comparison Test we know that

2 x

cos dx 2

2 x must also converge.

b) Let’s first take a guess about the convergence of this integral. As noted after the remark in

the last section about

a xdxp

1 if the integrand goes to zero faster than x1 then the integral

will probably converge. Now, we’ve got an exponential in the denominator which is

approaching infinity much faster than the x and so it looks like this integral should

probably converge.

So, we need a larger function that will also converge. In this case we can’t really make

the numerator larger and so we’ll need to make the denominator smaller in order to make

the function larger as a whole. We will need to be careful however. There are two ways to

do this and only one, in this case only one, of them will work for us.

First, notice that since the lower limit of integration is 3 we can say that x

we know that exponentials are always positive. So, the denominator is the sum of two

positive terms and if we were to drop one of them the denominator would get smaller.

This would in turn make the function larger.

The question then is which one to drop? Let’s first drop the exponential. Doing this gives,

x1

ex

1x

. This is a problem however, since

3 x1 dx diverge. We’ve got a larger function

that is divergent. This doesn’t say anything about the smaller function. Therefore, we

chose the wrong one to drop.

Let’s try it again and this time let’s drop the x. x1

ex

1x

. Also

3-n-3-

33

x-

3

x- eeelimx-limdx elimdx e e

n

n

n

n

n

Since

3

x- dx e is convergent then by the Comparison test

3 ex

1 dx x is also convergent

c) This is very similar to the previous example with a couple of very important differences.

First, notice that the exponential now goes to zero as x increases instead of growing larger



as it did in the previous example (because of the negative in the exponent). Also note that

the exponential is now subtracted off the x instead of added onto it.

The fact that the exponential goes to zero means that this time the x in the denominator

will probably dominate the term and that means that the integral probably diverges. We

will therefore need to find a smaller function that also diverges.

Making fractions smaller is pretty much the same as making fractions larger. In this case

we’ll need to either make the numerator smaller or the denominator larger. This is where

the second change will come into play. As before we know that both x and the

exponential are positive. However, this time since we are subtracting the exponential

from the x if we were to drop the exponential the denominator will become larger and so

the fraction will become smaller. In other words, x1

ex

1x-

and we know that

3 x1 dx

diverges and so by the Comparison Test we know that

3 e-3

1 dx x- must also diverge

d) First notice that as with the first example, the numerator in this function is going to be

bounded since the sine is never larger than 1. Therefore, since the exponent on the

denominator is less than 1 we can guess that the integral will probably diverge. We will

need a smaller function that also diverges.

We know that 1)2(sin04

x . In particular, this term is positive and so if we drop it from the

numerator the numerator will get smaller. This gives, x

1

x

)2(sin31 4

x

and

1 x

1 dx diverges so

by the Comparison Test

1 x

)2(sin31dx

4 x also diverges.

e) Normally, the presence of just an x in the denominator would lead us to guess divergent

for this integral. However, the exponential in the numerator will approach zero so fast that

instead we’ll need to guess that this integral converges.

To get a larger function we’ll use the fact that we know from the limits of integration that

1x . This means that if we just replace the x in the denominator with 1 (which is always

smaller than x) we will make the denominator smaller and so the function will get larger. x e

xe-x

and we can show that

1dx e x converges. In fact, we’ve already done this for a

lower limit of 3 and changing that to a 1 won’t change the convergence of the integral.

Therefore, by the Comparison Test

1 xe dx

-x

also converges.

Exercise:

Determine whether each of the following integral is convergent or divergent.

a)

0 10x

1 dx 2

b)

1 xdx

xe

1 c)

1 5 2dx

3x

1

x d)

2 1

sin2 dx x

x e)

1

x- dx 2

e

Limit Comparison Test Theorem: If positive functions f(x) and g(x) are continuous on xa and

llx

0wherelimg(x)

f(x) then either both integrals

aadx g(x) anddx f(x) converge or

both integrals diverge

Example: Does the integral

1

1x

dxconverge or diverge?

Solution:



We note that f(x) looks a lot like x

1 = g(x) and

1 x

1 dx diverges by the p-test. Furthermore,

11x

xlim

g(x)

f(x)lim

xx so the Limit Comparison Test says

1dx

1x

1 diverges.

Revision Problems

1. Determine whether the integral

1

1 dt sint

converge or diverge?

2. Evaluate each improper integral; determine whether it converges or not. Reuse some part

of your work in (a) for (f) .a)

2 2dx

1x

2 b) c)

0

2 dxx12

3

d)

1

0

ln dx x

x e)

5

4dx

4x

x f)

1

0 2dx

1x

2 g)

0x dx sine x h)

22x

dx2 x

2. Evaluate each improper integral; determine whether it converges or not

a)

0xdx ex b)

0 x-

-x

dx e1

e c)

2

-2dx1-x d)

2

0 2

1 dx x

e )

2

0 4

1 dx 2x

f)

4

2 2

4x

dx

3. Using either comparison or limit comparison test, determine whether the improper integral

below is converge or diverge . a)

1 6

1x

dx b)

3

lncos2

dx

xx c)

3 35

xdx

xx

d)

1 3

2

dx 1

sin

x

x e)

1

1

4

2

dx tan1

sinx

x

x

REDUCTION FORMULAR

Here we are going to use integration by parts ie dx

duvuv

dx

dvu where u and v are

functions of x

Example Derive the reduction formular for;

a) dxxn

n exI hence compute 3I b) dxxsinI n

n hence compute 4I

Solution

a) Lei xnxn evnx

dx

due

dx

dvx and and u 1

1-n

xnx1xnx1xnxn

n nIexeexe-exexI

dxxndxnxdx nn

)Ixe6(ex3ex)I2ex3(ex3IexI 0

xx2x3

1

x2x3

2

x3

3

cdx

xx

0 eeIbut

cc x23xxx2x3

3 e)6x6x3x(e6xe6ex3exI

b) dxdx sin xx sinxsinI 1-nn

n

xvxinxndx

dux

dx

dvxu nn cos and scos)1(sin and sinput 21



2-n

1

nn2-n

1

nn

n2-n

121

221

221n

n

I1sincos

III)1(sincosI)1(I

II)1(sincosdx sindx sin)1(sincos

dx ssin1)1(sincos

dx scos)1(sincosxsinI

n

n

n

xxnnxxn

nxxxxnxx

xinxnxx

xinxnxxdx

nn

nnnn

nn

nn

02

1

4

3

3

24

3

3

4 I2

sincos

4

sincosI

4

sincosI Now

xxxxxx

cxdxdx xsinIbut 0

0 cxxxxx 8

3

8

33

4

14 sincossincosI

Exercise

1) Derive the reduction formular for; a) xdxsinxI n

n hence compute 4I b)

xdxcosxI n

n hence compute 3I c) dxxcosI n

n hence compute 5I

2) Given ,xtanI n

n dx show that 2-n

11n ItanI xn

n hence compute 4I

3) Given ,xsecI n

n dx show that 2-n

2

n I1

2

1

sectanI

n

n

n

xx n

hence compute 6I (Hint

xxx nn 22 secsecsec )

4) Suppose ,(lnx)xI nm

n dx show that 1-n

1

n I11

)(lnI

m

n

m

xx nm

TECHNIQUE OF FOURIER SERIES Definition (a periodic function)

A periodic function is a function that repeats itself in regular intervals or periods. The most

important examples are the trigonometric functions, which repeat over intervals of length 2π.

Periodic functions are used throughout science to describe oscillations, waves, and other

phenomena that exhibit periodicity. Any function which is not periodic is called aperiodic

The graph of a periodic function f(x) that has period L exhibits the same pattern every L units

along the x-axis, so that L)f(xf(x) for every value of x. If we know what the function

looks like over one complete period, we can thus sketch a graph of the function over a wider

interval of x (that may contain many periods)

These notes develop Fourier series on the

level of calculus and we will only consider

working out Fourier series for functions f(x)

with period .2L

The idea behind the Fourier series is to

decompose a function, f(x), that is periodic

on the interval 2 (ie: )2f(xf(x) n for

...,3,2,1n ) can be decomposed into

contributions from nxnx cos andsin

http://www.wikipedia.org/wiki/Trigonometric_functions

http://www.wikipedia.org/wiki/Oscillation

http://www.wikipedia.org/wiki/Wave

http://www.wikipedia.org/wiki/Frequency



Introduction The general Fourier series of a function f(x),with period ,2 may be written as an infinite

series of the form

*sincosasincosa

.....3sin2sinsin.....3cos2coscosaf(x)

11

021

1

021

321321021

n

n

n

n

n

nn nxbnxanxbnxa

xbxbxbxaxaxa

The constants nn baa and ,0 are called the Fourier coefficients

Virtually any periodic function that arises in applications can be represented as the sum of a

Fourier series. For example, consider the three functions whose graph are shown below:

These are known, respectively, as the triangle wave (x), the sawtooth wave N(x), and the

square wave (x), Each of these functions can be expressed as the sum of a Fourier series:

022222 )12(

x)12( cos.....

9

9x cos

7

xcos7

5

5x cos

3

3x cos x cos(x)

n n

n

1

nxin .....

5

5xsin

4

4xsin

3

3xsin

2

2xsin x s(x)

n n

sinN

0 12

)x12(in .....

9

9xsin

7

7xsin

5

5xsin

3

3xsin x s(x)

n n

nsin

Fourier series are critically important to the study of differential equations, and they have

many applications throughout the sciences. In addition, Fourier series played an important

role in the development of analysis, and the desire to prove theorems about their convergence

was a large part of the motivation for the development of Lebesgue integration.

Useful Integrals and Trigonometric Results When calculating the Fourier coefficients nn baa and ,0 , for which ...,3,2,1n the

following trigonometric results are useful. Each of these results, which are also true for

...,3,2,1,0 n , can be deduced from the graph of sin x or that of cos x

0sin n

nn )1(cos

...,11,7,31

...,9,5,11

even0

sin2

n

n

n

n

...,10,6,21

...,8,4,01

odd0

cos2

n

n

n

n



When calculating the Fourier coefficients nn baa and ,0 , the following five integrals are useful.

a) 0)cos()cos()cos()cos(cos

sin 11

-

nnnn

n

nxnxdx

nn

b) 000)sin()sin(sin

cos 11

-

nnnn

n

nxnxdx

c)

n

nxxdxnxnxdx

2

sin

2

12cos1

2

1sin

--

2

d)

n

nxxdxnxnxdx

2

sin

2

12cos1

2

1cos

--

2

e) 0)(sin)(sin2

sinsincos 22

21

2

-

nn

n

nxdxnxnx

n

Finding the Fourier Coefficients To determine ,0a we integrate both sides of equation * and then make 0a the subject of the

formular . Ie

11

021

1

021

dxsindxcosdxa

sincosaf(x)dx

n

n

n

n

n

nn

nxbnxa

dxnxbnxa

f(x)dx

1aa)0()0()2(a 00

11

021

n

n

n

ba

To determine ,na we multiply the whole Fourier series by nxcos them integrate both sides

over the interval x and finally make na the subject of the formular . Ie

n of any valuefor a)0()()0(a

dxsincosdxcosdxcosa

cossincosadxcosf(x)

n

11

02

1

11

2

02

1

1

02

1

n

n

n

n

n

n

n

n

nn

ba

nxnxbnxanx

nxdxnxbnxanx

dxcosf(x)

1an nx



To determine , we multiply the whole Fourier series by nxsin them integrate over a complete

period leads to terms which are zero apart from one which corresponds to the coefficient ,nb ,

that is:

dxsinf(x)1

n of any valuefor )()0()0(a

dxsindxsincosdxsina

sinsincosadxsinf(x)

n

n

11

02

1

1

2

1

02

1

1

02

1

nxb

bba

nxbnxnxanx

nxdxnxbnxanx

n

n

n

n

n

n

n

n

nn

Example 1 Let f(x) be a function of period 2 such that .23f(x) 2 xxx Obtain

the Fourier series representation of f(x). Solution

We need to determine the Fourier coefficients and then plug them in the complete Fourier series. Now

2232

0 21

dx)23(1

f(x)dx1

a

xxxx

2222

32

22

n

)1(12

cos120

)cos()26(00

cos)26(0

1

sin6

cos)26(

sin)23(

1dxcos)23(

1a

nn

n

n

n

n

n

n

nx

n

nxx

n

nxxxnxxx

n

23

2

3

2

32

22

n

)1(4

cos60

cos)23(

cos60

cos)23(

1

cos6

sin)26(

cos)23(

1dxsin)23(

1b

nn

n

n

n

n

n

n

n

n

nx

n

nxx

n

nxxxnxxx

n

Therefore the Fourier series representation of xx 23f(x) 2 is f0llows

Recall equation *

11

021 sincosaf(x)

n

n

n

n nxbnxa

112

1222 sin)1(

4cos)1(

23n

n

n

n

nxn

nxn

xx

Example 2

Express the square-wave function illustrated in the figure below as a Fourier series.

Solution



We need to determine the Fourier coefficients but what is f(x)? From the graph we can see that

x

x

00

01f(x) Now

1011

dx1

0dxdx1

f(x)dx1

a00

0

0

0

x

0sin)sin(0sinsin

nxdx cos1

nxdx cos0nxdx cos11

dxcosf(x)1

a

0

0

0

0

n

n

n

n

n

n

nx

nx

odd isn if

even isn if01)1(0cos)cos(cos

nxdxsin 1

nxdx s0nxdx s11

dxsinf(x)1

2

0

0

0

0

n

n

n

nn

n

n

nx

ininnxb

Therefore the Fourier series representation of f(x) is

.....7xsin 5xsin 3xsin x s-sin1)1(

)(f71

51

312

21

121

innxn

xn

n

Exercise

1) Let f(x) be a function of period 2 such that

x

x

01

01f(x)

a) Sketch a graph of f(x) in the interval 22 x

b) Show that the Fourier series for f(x) in the interval x is

.....7xsin 5xsin 3xsin x sf(x)71

51

314 in

c) By giving an appropriate value to x, show that .....171

51

31

4


xx

x

0

00f(x)


b) Show that the Fourier series for f(x) in the interval x is

.....

3

3xsin

2

2xsin x s.....

5

5x cos

3

3x cos x cos

2

4f(x)

22in


1

5

1

3

1 1

8 222

2


2

0f(x)

x

xx


b) Show that the Fourier series for f(x) in the interval 20 x is

.....

3

3xsin

2

2xsin x s.....

5

5x cos

3

3x cos x cos

2

4

3f(x)

22in

c) By giving an appropriate value to x, show that

.....7

1

5

1

3

11

8 (ii) and.....

7

1

5

1

3

11

4 (i)

222

2

4) Let f(x) be a function of period 2 such that 20f(x) 21 xx





.....4xsin 3xsin 2xsin x sf(x)41

31

21

2 in


51

31

4


20

0f(x)

x

xx



.....

3

3xsin

2

2xsin x s.....

5

5x cos

3

3x cos x cos

2

4f(x)

22in


1

5

1

3

11

8 222

2

Odd and Even Functions A function is said to be even if f(x)f(-x) for all 𝑥 ∈ ℝ eg cos x nx2 for ...,3,2,1n and it

is said to be odd if f(x)f(-x) for all 𝑥 ∈ ℝ eg sin x 12 nx for ...,3,2,1n . All functions

symmetric about the y-axis (or line 0x ) are even functions.

Note: It does not necessarily mean that if a function is not odd then it’s even. A function with

odd and even terms is neither odd nor even. Eg xx 23 2 is neither odd nor even.

If f(x) is an even periodic function of period 2 defined over the interval , x then

the Fourier series representation of f(x) has 0a and cosine terms only. Ie 0bn for all n and

there is no need for calculating it.. Since f(x) is even it is possible to rewrite the the formulas

for n0 a and a as

lyrespectivedxcosf(x)2

a and f(x)dx2

a0

n0

0

nx

If f(x) is an odd periodic function of period 2 defined over the interval , x then

the Fourier series representation of f(x) has sine terms only. Ie 0a and 0a n0 for all n

and there is no need for obtaining them.. The formulas for nb reduces to

0n dxsinf(x)

2b nx

Example 3 Determine the Fourier series to represent the 2 periodic function graphed

below. Use the knowledge of odd and even functions

Solution

a) The graph clearly shows that the functions is periodic with period 2 . Again over the nterval

x , the function is a straight line with gradient 1 and y-intercept 0. xx )(f . We

now check whether the function is even or odd. )(f)()(f xxxx these

functions is odd and 0a and 0a n0



nn

n

n

nx

n

nxxnxnx

n 1

0

200n

)1(2

cos2sincos2dxsinx

2dxsinf(x)

2b

.....7xsin 5xsin 3xsin x s2sin)1(

2)(f71

51

31

1

1

innxn

xn

n

b) The graph clearly shows that the functions is periodic with period 2 and it’s symmetric

about the y axis )(f x is an even function and thus 0bn for all n. Now over the interval

x0 , the function is given by

x

x

2

2

1

01f(x)

022

21dx-1dx

2f(x)dx

2a

2

2

000

n

n

n

n

n

n

n

nx

n

nxnxnxnx

222

000

n

sin4sin0

sin2

sinsin2dxcos-dxcos

2dxcosf(x)

2a

2

2

2

2

.....5x cos3x cos x coscossin

f(x)51

314

1

24

n

nxn

n

Half Range series Sometimes a function of period 2 may be defined over the interval x0 instead of

x Thus it’s only possible to obtain the half range series.

The half range cosine series is geven by

1

021 cosaf(x)

n

n nxa where

0n

00 dxcosf(x)

2a and f(x)dx

2a nx

And the half range sine series is given by

1

21 sinf(x)

n

n nxb where

0n dxsinf(x)

2b nx

Example 4 Let f(x) be a 2 periodic function, determine the Fourier series to represent f

xx 0f(x) 5

Solution

nn

n

n

nx

n

nxxnxnx

n 1

2

0

220

5

0n

)1(10

cos10sincos10dxsinx

2dxsinf(x)

2b

.....7xsin 5xsin 3xsin x ssin)1(

)(f71

51

3110

1

1

10

innxn

xn

n

Exercise

1. Let f(x) be a function of period 2 such that xx,3f(x) Determine whether

f(x) is even or odd hence Show that its Fourier series is given by

1

sin)1(

6n

n

nxn

2. Determine the Fourier series to represent the 2 periodic function xx ,f(x)

. Use the knowledge of odd and even functions

3. Let f(x) be a function of period 2 such that xx22f(x)

a) Determine whether f is even or odd hence



b) Show that the Fourier series representation of f(x) in the interval x is

12

2

32 cos

)1(4f(x)

n

n

nxn

4. Express the square-wave function

x

x

00

01f(x) as a Fourier series

5. Determine the Fourier series to represent the 2 periodic function graphed below. Use

the knowledge of odd and even functions

6. Let f(x) be a function of period 2 determine the half range sine series of

xx 0,f(x) 3

FUNCTIONS OF SEVERAL VARIABLES Introduction

In many applications in science and finance, a function of interest depends on multiple

variables. We will now show how to extend the analysis of functions of a single variable to

functions of multiple variables. We will restrict ourselves to the case of two variables, i.e.

functions of the form ),(f yxz .

The extension to larger numbers of variables being relatively straightforward, apart from the

fact that functions of three and more variables are somewhat harder to visualise.

Functions of Two Variables A function f of two variables is a rule that assigns to each ordered pair (x, y) in the domain of

f a unique number f (x, y). As with functions of a single variable, if the function is specified

by a formula, the domain is taken to be the largest set of ordered pairs for which the formula

is defined;

Example 1: For the function 22416),(f yxyx the domain is given by

1164

041622

22 yx

yx on or inside an ellipse in xy plane

The range is given by: 40or 160 zz

Example 2: Suppose2

),(fy

xyx , find the domain and the value )1,9(f

Solution

We must have 0 and0 yx ; therefore, the domain is the set

0 and0:),((f) Dom yxyx

Finally, 3)1,9(f 2)1(

9

Example 3: Suppose xeyx xy ln),(f , find the domain and the value )2,1(f .



Solution

We must have x > 0; Therefore, the domain is the set 0:),((f) Dom xyx

Finally, 2)2(1 1ln)2,1(f ee

Example 4: (An Applied Example)

It costs $100 to a bike company to make a three-speed bike and $150 to make a ten-speed

bike; The company’s fixed costs are $2,500. Find the company’s cost function and use it to

compute the cost of producing 15 three-speed and 20 ten-speed bikes;

Solution

Suppose that x is the number of 3-speed and y the number of 10-speed bikes that the

company produces; Then

costs fixedbikes speed-10bikes speed-3

2500150100),(C yxyx

Thus, the cost for producing 15 ‘3-speed’ and 20 ‘10-speed’ bikes is

000,7$2500)20(150)15(100)20,15(C

The graph of z = f(x, y) is a surface in space; the domain is the projection onto x-y plane

It’s not easy to draw a graph of a functions of 2 variables but here are some plots of functions

of two variables:

Figure 1: Functions of two variables: xyz and ysin x cos z .

Partial derivatives

Recall the definition of derivative ie if f(x)y then

h

f(x)-h)f(xlim (x)'f

0hwhich is a

measures rate of change of f(x) with respect to a unit change in x

In the case of TWO or MORE independent variables, we are looking at the rate of change of

),(f yxz with respect to a change in ),( yx . There are an infinite amount of directions to

change the point ),( yx . Which direction do we choose? We will choose the paths parallel to

the coordinate axes. This is called Partial Differentiation. Partial derivatives are distinguished

from ordinary derivatives by using a instead of a d.

First Order Partial derivatives:

For a function of two variables there are two 1st order partial derivatives

The partial derivative of f with respect to x denoted xx fsimply or ),(for yxx

z

is defined by

x

xyx

x

y) ,f(x -y) , f(xlim),(f

0x



The partial derivative ),(fx yx represents the instantaneous rate of change of f with respect

to x when y is held constant. To compute ),(fx yx we take the derivative of ),(f yx with

respect to x assuming that y is constant;

The partial derivative of f with respect to y denoted yy fsimply or ),(for yxy

z

is defined by

y

yyx

y

y) ,f(x -)y ,f(x lim),(f

0y

The partial derivative ),(fy yx represents the instantaneous rate of change of f with respect

to y when x is held constant. To compute ),(fy yx we take the derivative of ),(f yx with

respect to y treating x as a constant.

Example 5 Obtain the 1st order partial derivatives of the function yxyxyx cos53),(f 2

Solution

yxxyxyxyyx sin53),(f and cos56),(f 2

yx

Example 6

Show that 4224 2 CyyBxAxz satisfies the partial differential equation zy

zy

x

zx 4

Solution

3223 44 and 44 CyyBxy

zBxyAx

x

z

zCyyBxAxCyyBxAx

CyyBxyBxyAxxy

zy

x

zx

424484

4444

42244224

3223

Example 7 A company’s profit from producing x radios and y televisions per day Is

xyyxyx 23

23

64),(P .

a) Find the marginal profit functions; b) Find and interpret 36) (25,Py

Solution

xyyyyxyxyxyx 960),(P and 604),(P 21

21

2

3y2

3x

792536936) (25,Py This is the approximate increase in profit per additional

television produced when 25 radios and 36 televisions are produced;

Second Order Partial derivatives:

Remark, the first order partial derivatives are functions of x and y too.

For a function of two variables there are three second order partial derivatives, defined as

yyyy2

2

xxxx2

2

fsimply or ),(forfsimply or ),(for yxy

z

yy

zyx

x

z

xx

z



have usually1 wewherefsimply or ),(forderivative mixed theand xyxy

2

yxx

z

yyx

z

y

z

xx

z

yie

yx

z

xy

z

22

Example 8 Obtain the 1st and 2nd order partial derivatives of the following functions

a) 1262),(f 2 yxxyxyx

b) xyeyx ),(f

c) 4),(f 222 yxyxyx

Solution

a) 2f0f2f22f622f xyyyxxyx andxyx

b) xyxyxyxyxy exyandexeyxeye )1(fffff x

2

yy

2

xxyx

c) xandyxyxyx 2f2f22f2f22f xyyyxx

2

yx

Exercise

1) Find y

z

x

z

and for the functions; a) 1432 2334 xyyxxz b) yez x ln

c) 2324 425 yyxxz d) 22 yxez e) )ln( 22 yxz f)

22 yx

xyz

g)

yx

xyz

cos

2) Obtain the 1st and 2nd order partial derivatives of the following function a)

432 xxyez xy b) xyyxz tantan c) 22cos yxyxyez x d)

xyyxz sin3 32 e) 432 yxxyz f) )sin(2 yxez yx g) )(sin 1 xyz h)

)(tan 1 xyxz i) 22 2ln yxz

3) Determine fand fr for the function; a) 23),(f rrr

b) rerr 22),(f c) rerr cos),(f 2

4) Compute )3,1(f if 22

),(f yxeyx

5) Show that;

a) yx

yxz

22

satisfies the partial differential equation zy

zy

x

zx 3

b) xzzyyxu 222 satisfies the partial differential equation

2)( zyxz

u

y

u

x

u

c) 22

1

yxz

satisfies the partial differential equation 0

2

2

2

2

y

z

x

z

d) yx

xyz

satisfies the partial differential equation 02

2

22

2

2

22

y

zy

yx

zxy

x

zx



6) If

y

x

yz

4exp

1 2

, show that 2

2

x

z

y

z

7) For function yx

yxyx

3

2),(f

verify that; a) 0fxf yx y b) 0ff2f yy

2

xyxx

2 yxyx

Total Derivatives

Suppose for the function )( and)(,),(f thytgxyxz then z is a composite function of t

and we can obtain it’s total derivative with respect to t either by 1st expressing explicitly z as

a univariate function of t or by using chain rule (ie )(')(' thy

ztg

x

z

dt

dz

).

Example 9 Obtain the total derivative of yxez wrt t given that 23 and12 tytx .

Use both methods

Solution

a) If we 1st express z explicitly as a univariate function of t we get

1313)23()1( 222

)32( tttttt etdt

dzeez

b) Using Chain rule .y

ze

x

z yx

Also 3 and 2

dt

dyt

dt

dx

yxyxyx etetedt

dy

y

z

dt

dx

x

z

dt

dz )32(32

Exercise

1) Find the total derivative of z with respect to t given that ;

a) yxz 3 , 2 and12 tytx Use both methods

b) )ln( 23 yxz , tt eyex 32 and1 .

2) Let 22 yxz . Find the maximum value of z on the ellipse given parametrically by

tytx sin2,cos

3) Let 22 yyxz where teytx ,sin . Find dtdz when t = 0.

4) Let .),,(f 22 zyyxzyxw Consider the curve given parametrically by

tztytx ln and, 2 . Find dtdw at 2t

5) The radius of the base of a cone is 6 cm and it is increasing at a rate of 2 cm/sec. The

height of the cone is 10 cm and it is increasing at a rate of 3 cm/sec. At what rate is the

volume increasing? (Ans /seccm116 3 )

Implicit Partial Differentiation

Recall if f(x, y) = 0 defines y implicitly, then ),(f

),(f

y

x

yx

yx

dx

dy

Similarly Let z be a function of x and y defined implicitly be the function: 0 = z) y, f(x,

(meaning not solved for z), then

),,(f

),,(f and

),,(f

),,(f

z

y

z

x

zyx

zyx

y

z

zyx

zyx

x

z



Example 10 Find y

z

x

z

and for the function; .586222 yzxzzyx

Solution

0586),,(f 222 yzxzzyxzyx

yxzzyxzyzyxzxzyx 862),,(f82),,(f,62),,(f zyx

z-4yx

4z-y

8y-6x-2z

8z-2y and

z-4y3x

3z-x

8y-6x-2z

6z-2x

y

z

x

z

Exercise:

1) Find y

z

x

z

and where z is implicitly defined by: a) 032 zexzyx

b) 0)2cos( 2 zxy c) 0)sin(32 xyzzxy d) xyzxyzzyx )sin(423

2) Compute all first order partial derivatives of each of the following functions.

)32sin(),,(f 2zyxzyx

3) Find all first and second order partial derivatives of z with respect to x and y, where z is

defined by the equations 3zxyzyx

4) z is defined as a function of x and y by the equation 012 32 zyzx Use implicit

differentiation to compute 2

2

and x

z

x

z

5) For the following implicit function ),(fz yx defined by the equation

)(2222 zyxzyx , find 2

22

and y

z

yx

z

Stationary points: (Maxima, minima and saddle points)

Definitions A Function ),(f yx has a local maximum (local minima) at point (a, b) if

),(f),(f yxba ),(f),(f yxba for any point (x , y) sufficiently close to (a , b).

Needless to say a function may have both relative maxima and relative minima at various

points of its domain:

Definitions A Function ),(f yx has a global/ absolute maximum (minima) at point (a, b) if

),(f),(f yxba ),(f),(f yxba for every point (x , y) in the entire domain of ),(f yx .

Saddle point: You know that in one-variable calculus, critical points need not be local

maxima or minima; they can be inflection points. In several variables there is not only a

critical point that behaves like an inflection point, but an entirely new type of critical point

called a saddle point. Near this point, the function takes larger or smaller values depending

on the direction in which one travels away from the point.

Types of stationary points:

Before giving a similar procedure for finding maxima and minima for functions of two

variables, it is necessary to consider the types of critical points we may encounter

There are three types of stationary points that can be exhibited by functions of two variables

as illustrated below.



Condition for a stationary point:

The function ),(f yxz has a stationary point" at a point (a; b) in the domain of f if

0 and0),(),(

baba y

z

x

z

This condition provides two equations with two unknowns x and y. These equations can have

No solution, in which case the function ),(f yxz has no stationary points.

A unique solution, in which case the function ),(f yxz has a single stationary point.

Multiple solutions, in which case the function ),(f yxz has multiple stationary points.

Classification of stationary points:

The nature of a stationary point is determined by the function's second derivatives.

Here is a recipe for the classification of stationary points.

Suppose that f (x; y) has continuous second-order partial derivatives in some open disk containing the

point (a; b) and that 0),(f),(f yx baba . 1 For each stationary point (a , b), determine the

three second partial derivatives and evaluate them at the stationary point:

),(

2

),(

2

2

),(

2

2

and

bababayx

zC

y

zB

x

zA

Classify the stationary point according to the following cases:

a) If 02 CAB at point (a , b) then the function has a saddle point

b) If 0Bor 0Aeither and02 CAB at (a , b), then the function has a local

(relative) minima at this point

c) If 0Bor 0Aeither and02 CAB at (a , b), then the function has a local

(relative) maxima at this point

Example 11 Locate and classify the relative extreemas of the following functions

a) 1262),(f 2 yxxyxyx

b) xyeyx ),(f

c) 4),(f 222 yxyxyx

d) xyyxyx 33),(f 33

Solution

a) We need to first find the 1st order partial derivatives, equate them to zero and then solve

4y and1022

0622 points criticalat

022f

0622f

y

x

x

x

yx

x

yx

0)2()0(22fand0f2f 22

xyyyxx CAB

the function has a saddle point at (1 , 4).


Prov 16:3 Commit to the LORD whatever you do, and your plans will succeed

b) We again find the 1st order partial derivatives, equate them to zero and then solve

0y and0 points at turning0f

0f

y

x

x

xe

ye

xy

xy

point saddle01)0(0)1(fandff 22

x

2

yy

2

xx CABexyexey xyxyxy

c) We again find the 1st order partial derivatives, equate them to zero and then solve

12000)(2

2

02f

022f

2

2

2

y

x

yxoryx

xxx

iieqniexy

xy

xyx

There are 3 critical points where both yx f and f are equal to zero. They are (0 , 0), ( 2 , -1)

and (- 2 , -1)

The 2nd order partial derivatives are xandy 2f2f22f xyyyxx and we will use a

tabular way to classify the 3 critical points as shown below.

points A=2+2y B=2 C=2x 2CAB classification

(0 , 0) 2 2 0 4 Local minimum

( 2 , -1) 0 2 22 -8 Saddle pt

(- 2 , -1) 0 2 22 -8 Saddle pt

d) We need to first find the 1st order partial derivatives, equate them to zero and then solve

1 and 10)1)(1(3

0)1)(1(3

033f

033f

2

y

2

x

yx

yy

xx

y

x There are 4 critical points

where both yx f and f are equal to zero. They are ,)1- (1,,)1 (1, )1- (-1, and )1 (-1,

Now the 2nd order partial derivatives are 0f and 6f6f xyyyxx yx and we will use

a tabular way to classify the 4 critical points as shown below.

Critical point x6A y6B C=0 2CAB classification

(1 , 1) 6 -6 0 -36 Saddle pt

(1, -1) 6 6 0 36 Local minimum

(-1 , 1) -6 -6 0 36 Local maximum

(-1 , -1) -6 6 0 -36 Saddle pt

Example 2

A motor company makes compact and midsized cars. The price function for compacts is

x217p (for 80 x ) and for midsized y 20q (for 200 y ), both in thousands

of dollars, where x, y are the number of compact and midsized cars produced per hour;

Assume that the company’s cost function is 521615),(C xyyxyx thousand dollars;

How many of each type of car should be produced and how should each be priced to

maximize the company’s profit? What will be the maximum profit?

Solution First find the profit function

54222521615()20217(

)521615()(),(C),(R),(P

2222

cos

yxxyyxxyyxyyxx

xyyxyqxpyxyxyx

trevenue

Compute first derivatives 224),(Px yxyx and 422),(Py xyyx then find critical

points:

5

3

2

12

0422

0224

0),(P

0),(P

y

x

y

x

xy

yx

xy

yx

yx

yx

Thus )5,3(),( yx is the only critical point;

Finally, we verify that at (3, 5) we indeed have a local max; We have



Now 4),(Pxx yx , 2),(Pxy yx and 2),(Pyy yx

Therefore, 042)2(4PPP 22

xyyyxx and 04Pxx which show that at (3, 5) P has a

maxima. The prices and the maximum profit are given by

thousands113217217p x thousands1552020q y

thousands8$554325325)3(2

54222),(P

22

22

yxxyyxyx

Exercise

1) Find the partial differentials of; a) 1262),(f 2 yxxyxyx b) )1(),(f xyxyx

c) 4_),(f 222 yxyxyx d) xyeyx ),(f e) xyyxyx 36),(f 44 . Also

determine their stationary points

2) Find the location and nature of all stationary points of; a) 542),(f 22 yyxxyx

b) 215108),(f 22 yyxxyx c) 233),(f 22

4

13

3

1 yxxyxyxyx

d) 22),(f yxyx e) yxyxyx 33),(f 33 f) 22

)(),(f 22 yxeyxyx

g) 4103),(f 53 yxyxyx h) xyxyyx 24128),(f 23

3) Locate the stationary points of the following functions, and find their nature.

a) 102015),(f 223 yxxyx b) 5628),(f 232 yyxyx

c) 83532),(f 22 yxxyyxyx d) yxxyyxyxyx 1863362),(f 23

e) 2121092),(f 223 xyxyxyx f) xyyxyx 224),(f

g) 1222),(f 222244 yxyxyxyx h) 1)(2)(),(f 22222 yxyxyx

4) Let z be defined implicitly as a function of x and y by the equation

0784),,(f 223 yyxxzzzyx . Find, and test, all critical points of z.

5) A revenue function is: 22

2

3

2

3800500),(R yxxyyxyx , where x and y are prices

of two competing products. Find x and y to maximize R.

6) The owner of a business advertises in the newspaper and on the radio. He has found that

the number of units that he sells is modelled by 22

2

1128240),(f yxyxyx ,

where x (in thousands of dollars) is the amount spent on newspaper adverts and y (in

thousands of dollars) is the amount spent on radio adverts.How much should he spend on

each to maximize the number of units sold? Find the maximum number of units sold?

Constrained Optimization and Lagrange Multipliers One of the most common problems in calculus is that of finding maxima or minima (in

general, "extrema") of a function, but it is often difficult to find a closed form for the function

being extremized. Such difficulties often arise when one wishes to maximize or minimize a

function subject to fixed outside conditions or constraints. The method of Lagrange

multipliers is a powerful tool for solving this class of problems without the need to explicitly

solve the conditions and use them to eliminate extra variables.

A constrained optimization problem is a problem of the form maximize (or minimize) the

function ),(f yx subject to the condition 0. = y) ,g(x The function to be optimized ie

),(fz yx is called the objective function. There are 2 approaches to this constrained

optimization problem:

Use the constraint g to solve for one variable in terms of the other(s), then plug into f and

find its extrema.



New method: Lagrange Multipliers

From two to one

In some cases one can use the additional constraint to solve for y as a function of x and then

find the extrema of a univariate function. ie, if the equation 0 = y) ,g(x is equivalent to

h(x), =y then we may set h(x)) f(x, = (x) and then find the values a =x for which

achieves an extremum. The extrema of f are at (a, h(a)).

Example 12 Find the relative extrema of the function 222),(f yxyx subject to the

condition 0. 1-yx= y) ,g(x

Solution Using the additional constraint to solve for y as a function of x ie

h(x)x-1y0 1-yx

Now set 123)1(2 h(x)) f(x, = (x) 222 xxxx and then find the extrema of (x)

02x6 = (x)' at the critical point 3

2

3

1

3

1 -1y and x at that point. So the only

extrema of ),(f yx is at ),(3

2

3

1

Notice that 06 = (x)" the critical point is a minima.

the required constrained relative minima is 3

22

3

22

3

1

3

2

3

1 2),(f

Exercise:

1) Find the relative extrema of the function;

a) xyyx 2),(f subject to the constraint 0.42y3x= y) ,g(x

b) 2),(f xyyx subject to the constraint 0.42yx= y) ,g(x

c) 22),(f yxyx subject to the constraint 0. x-y= y) ,g(x 2

2) Minimize 22),(f yxyx subject to the constraint 0.3y2x= y) ,g(x

3) Farmer Brown wishes to enclose a rectangular coop of 1000 square feet. He will build

three sides of brick, costing $25 per linear foot, and the fourth of chain link fence, at $ 15

per linear foot. What should the dimensions be to minimize the cost?

4) An open rectangular box (5 sides but no top) has volume 3cm 500 .What dimensions give

the minimum surface area, and what is that area?

Lagrange’s Multipliers Method More often it is cumbersome to express y explicitly as a function of x or the resulting

univariate function may be complicated to optimize. Consequently an easier method exists

called the Lagrange’s Multipliers Method

The method of Lagrange multipliers (for a function of two variables) can be stated as follows:

To find the maximum or minimum of ),(fz yx subject to the constraint 0. = y) ,g(x ;

i) Form an arbitrary function called the Lagrange’s function ),(g),(f),,(F yxyxyx .

Where the Greek letter lambda, ( ) is called the Lagrange Multiplier and 0. = y) ,g(x is

the constraint. Economists call the Lagrange multiplier obtained in a production

function the marginal productivity of money. !



ii) Solve the system comprising of the partial differential equations

0),(f0),(f,0),(f yx yxyxyx for all values of x and y y (and if desired) and

determine all the critical points;.

iii) Find ),(f yx at each of the points ),( yx found in step ii above. The largest (smallest) of

the values is the required constrained relative maxima (minima) of ).,(f yx

Example 13 Maximize and minimize xyyx 4),(f subject to the constraint 5022 yx

Solution

The objective function is xyz 4 and the constraint function is 050),(g 22 yxyx

thus, the Lagrange’s function is; )50(4),,(F 22 yxxyyx .

Compute the three partial derivatives and set the partials equal to zero to find the critical

points:

22

1 fromeqn 2Subtract

22

eqns 2 thefrom Eliminate

y

xx0x4-4

)024(

)024(

024F

024F

stnd

yy

xyx

yxy

yx

xy

eqn

55502xbut 50050F 2222222 yxxyyxyx

Thus, there are four critical points: )5,5( and )5,5(,)5,5(,)5,5(

Since 100)5,5( f)5,5(f and 100)5,5( f)5,5(f we conclude that 100fmax

occurring at )5,5( and )5,5( and 100fmin occurring at )5,5( and )5,5( ;

Example 14

Maximize and minimize yxyx 3012),(f subject to the constraint 815 22 yx

Solution

The objective function is yxyx 3012),(f and the constraint function is

0815),(g 22 yxyx . Thus, the Lagrange’s function is;

)815(3012),,(F 22 yxyxyx .

Compute the three partial derivatives and set the partials equal to zero to find the critical

points:

yxxyyxy

x

y

x236

36

3

6

01030F

0212F

y

x

819815)2(2but 8150815F 2222222 yyyyxyxyx

Thus 392 yy when 6,3 xy and when 6,3 xy

Therefore, there are only two critical points: )3,6( and )3,6(

Since 162)3,6(f and 162)3,6(f we conclude that 162fmax occurring at )3,6( and

162fmin occurring at )3,6( ;

Example 15 A company wants to design an aluminium can that requires the least amount of

aluminium but that can hold exactly 12 fluid ounces (3m3.21 ). Find the radius r and the

height h of the can.

Solution

The objective function is the surface area of the can rhS 2r2A. 2

The constraint has to do with the volume 03.21r3.21r 232 hmhv

Therefore the Lagrange’s function is )3.21r(2r2),,( 22 hrhhrF



Take the partial derivatives and set them equal to zero to find critical points of F:

rhrhhr

rhh

r

hh

eqn

202-r42)0r2(

)0r22r4(

0r2F

0r22r4F

stnd 1 from 2Subtract

2

2

first two from Eliminate

2

h

r

2

3.2133222 r3.21r22r2Subst3.21r03.21rF rrhhh

in 3 h hence, and, in;5.12

3.21r 3

;

Exercise

1) Find the relative extrema of the function 222),(f yxyx subject to the condition

0. 1-yx= y) ,g(x

2) Find the potential extrema of the function:

a) xyxyx ln),(f 2 subject to the constraint that 038),(g yxyx

b) yxyx 2),(f subject to the constraint that 01),(g 2 xyyyx

c) yxyxyxyx 33),(f 22 subject to 01),(g 22 yxyx

d) 222),,(f zyxzyx subject to 0191532),(g zyxyx

3) Use Lagrange’s multipliers to find the maximum and minimum value(s) of the function

zyxzyx 53),,(f on the sphere 1222 zyx

4) Find the largest product the positive numbers x, y and z can have if 162 zyx

5) Find the maximum and minimum values of the function on zyxw 52 the sphere

36222 zyx

6) A storage tank is designed to hold 3000,1 m of hot liquid and the heat loss is to be

minimized by making the surface area of the tank as small as possible. If the tank is in the

shape of a right circular cylinder, use Lagrange Multiplier to determine its optimal

dimensions.

7) A standard oil drum has a capacity of 10 m3. Use the method of Lagrange Multiplier to

find the dimensions of the drum that requires the minimal amount of material to

construct. (Ans rh 2 and r 3 5

)

8) Find the rectangle with largest area that can be inscribed in the ellipse

3649),(f 22 yxyx

9) Maximize xyazyx ),,(f subject to the constraint that 0191532),(g zyxyx

10) A rectangular box is to be made to enclose a volume of 36 cubic metres. The bottom of

the box is made from scrap material costing $1 per square metre, while the material for

the top and the sides of the box costs $8 per square metre. Find the dimensions of the

most economical box.

11) The Acrosonic Company manufactures a bookshelf loudspeaker system that may be

bought full assembled or as a kit. The total weekly profit (in dollars) that the company

realized in producing and selling the systems is given by the function

5000100120),(P4

12

8

32

4

1 xyyxyxyx where x denotes the number of full

assembled units and y denotes the number of kits produced and sold per week. The

companyʼs management has decided that the production of these systems should be

restricted to a total of 230 per week. Under this constraint, how many assembled units and

how many kits should be produced per week to maximize their weekly profit?

12) The total monthly profit of Robertson Controls Company in manufacturing and selling x

hundred of its standard mechanical setback thermostats and y hundred of its deluxe

electronic setback thermostats per month is given by the total profit function

2804013),(P4

12

2

12

8

1 xyyxyxyx , where P is in hundreds of dollars.



If the production of setback thermostats is to restricted to a total of exactly 40 per month

a) how many of each model should the company manufacture in order to maximize

its monthly profits?

b) what is the maximum monthly profit?

c) find the value of the marginal productivity of money and explain.

13) The USPS will accept a package if the length plus its girth is not more than 84 inches.

What are the dimensions and the volume of the largest package with a square end that can

be mailed? (Ans the Lagrange’s eqn is )844(),,(F 2 yxxyyx ,

in 14 andin 28x y max volume is 3

max in 5,488 =414128,V )

14) A plot of land in the shape of a pentagon is composed of a rectangle surmounted by an

isoceles triangle. If the area of the land is required to be 1000 square metres, what will be

the dimensions for which the perimeter is a minimum?

15) It is known that the temperature T at any point ),,( zyx is given by

yzxzxyzyxzyx 164201152),,(T 222 Find the hottest point(s) on the unit

sphere 1222 zyx

DOUBLE INTEGRALS

Double integrals over Rectangular Regions Recall the area under the curve y=f(x) between the ordinates x=a and x=b and above the x-

axis can be approximated using the Rieman sum as illustrated below.

Let f(x) be a function of one variable. To

calculate the Area we partition the interval

[a, b] into a large number of subintervals of

width Δ x and form the Riemann Sum

n

k

k xx1

n )(fs which is really nothing

more than a sum of rectangles.

We then define the Area as the limit of this sum as the number of rectangles goes to i.e.

n

k

kn

nn

xxs1

)(flimlim

The exact area is also the definition of the definite integral ie

b

a

n

k

kn

dxxxxx )(f)(f)(f area1

lim

The above concept can be extended to functions of several variables but for now we will

restrict ourselves to functions of 2 variables.

Let y) ,f(x be a function of 2 variables

defined over a closed rectangular domain

dycbxa ,:y) ,(x d] , [c b] , [aR

as shown and let R be covered by a network

of lines parallel to both x and y axes as

shown below. These lines divide R into

small rectangles of area y.A x

Define S to be the part of the surface y) ,f(x z over the region R. What is the volume of the

region under S and above the x-y plane?



To approximate this volume we partition

the domain into a large number of

rectangles of area yA x . Choose a

point )y , (x ji on each rectangle and use

the value of )y , f(x jito approximate the

value of y) ,f(x on this rectangle. The

volume of the solid over each small

rectangle is

yx )y , f(xA)y , f(xV jiji Now we

simply sum these volumes and then let the

number of these rectangles go to

The question is how do we sum over these rectangles?

We start by fixing i and summing over j: ie

m

1

ji )y , f(xj

y

We then sum over i: ie

n

i j

n

i j

xyxy1

m

1

ji

1

m

1

ji )y , f(x)y , f(x

We then define the Volume as the limit by first taking the limit as m goes to and then the

limit as n goes to

n

i jmn

n

i jmn

xyxy1

m

1

ji

1

m

1

ji )y , f(x)y , f(xV limlimlimlim

b

a

d

c

n

i

d

cn

dxdyxdy y) ,f(x y) , f(x1

ilim

Definitions

If b),f(a,y)f(x,limb)(a,y)(x,

then f is continuous at point b)(a,

If f is continuous at all point R,b)(a, then f is continuous on R

Eg If ][0,2][0,Ry)(x, andsinxy y)f(x, then f is continuous on R

If )[0,)[0,Ry)(x,andx y xy)f(x, 2 f is continuous on R

Properties of double Integrals

i) R R

y)dAf(x,cy)dAcf(x,



ii) R R R

y)dAg(x,y)dAf(x,y))dAg(x,y)(f(x,

iii) R R

y)dAg(x,y)dAf(x, then R,y)(x, y)g(x,y)f(x, If

iv) If R = [a, b]×[c, d] is a rectangle and f (x,y) = g(x)h(y) is a product of two functions of one

variable, then b

a

d

cR

dyyhdxxgdAyxf ,

Example 1

Evaluate the integral R

4dA where R is the rectangular region defined by 3] , [-3 0] , [-1R

by 1st identifying it as a volume of a solid

Solution

Required is the volume of a box with base R and height 4

241)4(6R of area4V cubic units

Generally, we define the definite double integral as:

n

i jnm

xydA1

m

1area baseheight

ji,R

)y , f(xy) ,f(x lim

where ydydxA dxdd //order matters//. Definitely, just like in the case of univariate

functions, we don’t attempt to evaluate the limit but rather we apply the partial integration

together with the iterated integrals to evaluate double integrals.

Theorem If y) ,f(x is continuous on a plane region R then f is integrable over R.

Partial Integration

For the double integral b

a

d

cdxdyy) ,f(x , we first integrate the inside integral with respect to

y treating x as a constant. This is referred to as Partial Integration. Once that is done we will

be left with univariate definite integral (ie integrating a function of one variable).

Definition

Suppose y) ,f(x is integrable over the rectangular region d] , [c b] , [aR , then we define

the partial integration of f with respect to x as b

adxy) ,f(x .

Similarly the partial integration of f with respect to y is defined as d

cdyy) ,f(x

Example 2 Evaluate the integral: a) 4

0

23x dxy b) 4

1e dyxy

Solution

a) 24

0

22

21

4

0

2 1283x3x yxydxyx

b) xx

y

xyxy eedy

4

x1

4

1x1

4

1ee

Iterated Integral Suppose y) ,f(x is integrable over the rectangular region d] , [c b] , [aR , then

d

c

b

a

d

c

b

a

b

a

d

c

b

a

d

cdydxdydxdxdydxdy y) ,f(x y) ,f(x and y) ,f(x y) ,f(x

Theorem 2: If 0y) ,f(x and f is continuous on a rectangular region d] , [c b] , [aR then

the volume of the solid that lies above the region R and below the surface y) ,f(x z is given

by



Integral IteratedIntegral Double

R

y) ,f(x y) ,f(x y) ,f(x d

c

b

a

b

a

d

cdydxdxdydA

Example 3

Evaluate the integral 4

1

1

01 yxI dydx and

1

0

4

12 yxI dxdy . what do you notice?

Solution

37

4

131

4

1 21

4

1 21

4

1

1

0

4

1

1

01

23

21

yyyxyyxI

dydydydxdydx

37

1

0 314

1

0

4

132

1

0

4

1

1

0

4

12

23

21

yyyxI

xdxdxxdxdyxdxdy 21 II

Example 4

Evaluate R

xydx Acos where R is the rectangular region defined by 5] , [1 ] , [-R22

Solution

0cos5cos

sin5sinsincosAcos

2

2

2

2

2

2

2

2

51

5

1

5

1

xx

dxxxdxxyxydydxxxydxyy

R

Example 5

Find the volume of the solid s lying under the circular paraboloid 22z yx and above the

rectangle 3] , [-3 ]2 , 2[-R

Solution

The volume of the solid is given by;

units cubic104182186

Av

2

2

32

2

2

2

2

3

3

3

312

2

2

3

3

3

312

2

2

3

3

2222

xxdxx

dxyyxdxyyxdydxyxdyxy

R

Fact: If g and h are continuous functions of one variable and d] , [c b] , [aR then

d

c

b

)h()g()dA)h(g( dyydxxyxa

R

Note: This fact only holds if R is a rectangular region.

Example 6 If 5]ln , [0 2]ln , [0R find R

e dAy-2x

Solution

56

51

21

5ln

0

y-2ln

0

2

21

5ln

0

y-2ln

0

2x2ln

0

5ln

0

y-2xy-2x 114e-dyedAedA

x

R

edxeee

Exercise

1) Evaluate 4

0

8

0

23

0

4

1

23

0

4

1

2 )dydxy8x-(644

1 (iii)ydydxx (ii) ydxdyx (i)

3

0

2

1

8

0

4

0

2 2y)dxdy(3x (v) )dxdyy8x-(644

1 (iv)

2) Evaluate R

dAy) ,f(x where R is the rectangular region defined by;

a) 3] , [0 2] , 0[R and y31y) ,f(x

b) 2] , [1 ]4 , 0[R and y2xy) ,f(x

c) 1] , [-1 ]1 , 1[-R and 22 y-x-8y) ,f(x



d) 2] , [0 ] , [0R4 and xyxcosy) ,f(x e) 4] , [1 2] , 0[R and

23 4yx6y) ,f(x

3) Evaluate each of the iterated integrals 1

0

1

0

xy2

0

3

1

2 dxdyxe b) ydydx,x a)

2

2-

1

1-

2

2-

31

1-

22

2-

1

1-

322

0

2

0 20

1

0dydx|y|[x] g) dydxy][x f) dydx|yx| e)dydx

x1

y d) xsinydxdy c)

4) Find; a) ][0,[0,2]R where,ysinxydA b) [1,3][0,2]R wheredA,)3y-(xRR

2

1}y0 2,x-1|y){(x,R where,dAx1

y1 b). ][0,][0,R where,xcosxydA)

RR

c

5) Calculate the double integrals R

32 3y0 2,x1y)(x,R )dA,3xy-(2y a)

2}y,13x0 |y){(x,R ,dAx1xy e) [0,1][0,1]R ,dAxye d)

1}y0 2,x-1|y){(x,R dA,y2

x1 c) ]

3[0,]

6[0,R y)dA,xcos(x b)

R

2

R

yx

RR

22

6) Find the volume of the solid lying under the plane 15y2xZ and above the

rectangle

4)y1 0,x(-1 |y){(x,R

7) Let [0,2][0,2]R ie 2}y1 2,x0|y){(x,R 1},y0 2,x0|y){(x,R 21

& suppose that R R R2

2y)dAg(x, 6,y)dAg(x, 4,y)dAf(x, , evaluate, 1R

y)dA3g(x, a).

RRR

y))dA4g(x,y)(3f(x, d). 4)dAy)(2f(x, c). y))dAg(x,-y)(4f(x, b).

8) 1y)dAsin(x0 thatshow [0,1],[0,1]R IfR

9) Let ,2y0 3,x1|y)(x,R evaluate R

y)dA,f(x, where f is the given function

2y1 4,x3 4,

2y1 3,x1 1,

1y0 4,x1 2,

y)f(x, b). 2y0 4,x3 3,

2y0 3,x1 2,y)f(x, a)

Double integrals over General Regions

In most cases the domain (ie region R) is non rectangular and therefore we need to now look

at the following double integral D

y) ,f(x dA where D is any region. There are two types of

regions that we need to look at. Here is a sketch of both of them.

Type I Regions (Vertical) A plane region D is said to be a type I or

y-simple if it lies between the graphs of

two continuous functions of x.ie

, )()(,:y) ,(x D 21 xgyxgbxa

where )( and)( 21 xgxg are continuous on

[a, b]. Some examples of type I regions are

shown.



Theorem: If D is a region of type I such that )()(,:),( 21 xgyxgbxayxD then

b

a

xg

xgD

dydxd)(

)(

2

1

y) ,f(x Ay) ,f(x But how do we get the limits of y ie )( and)( 21 xgxg ?

When evaluating double integrals it is very common not to be told the limits of integration

but simply told that the integral is to be taken over a certain specified region D in the x- y

plane. In this case you need to work out the limits of integration for yourself. Great care has

to be taken in carrying out this task. The integration can in principle be done in two ways:

i) integrating first with respect to x and then with respect to y, or

ii) first with respect to y and then with respect to x.

The limits of integration in the two approaches will in general be quite different, but both

approaches must yield the same answer.To determine the integration limits for a region

of type I, we follow the 3 steps below

i) Sketch the region D and draw a

vertical line in the increasing direction

of y. Call it l

ii) Integrate for values of y from where l

enters the region D to where it leaves

D. ie from )(y t)( 21 xgoxgy

iii) Choose x values that includes all vertical

lines that can pass through D. ie from

boax xt

Remark: When doing the inner integral, we regard x as a constant not only in f(x, y) but also

in the limits of integration, )( and)( 21 xgxg .

Type II Regions (Horizontal) A plane region D is said to be a type II or x-simple if it lies between the graphs of two

continuous functions of y.ie



dycyxy ,)(h)(h:y) ,(x D 21 Where h1 and h2 are continuous on [c, d]. Some

examples of type II regions are shown.

Theorem: If D is a region of type II such that dycyxy ,)(h)(h:y) ,(x D 21 then

d

c

y

yD

dxdyd)(h

)(h

2

1

y) ,f(x Ay) ,f(x

For a type II region, the arrow is drawn horizontally from the left boundary to the right

boundary. We then iIntegrate for values of x from where the line enters the region D to where

it leaves the region D. ie from )( xt)( 21 xhoxhx . Finally we choose y limits that includes all

horizontal lines that can pass through D. ie from docy y t

Remarks: 1) When we set up a double integral, it is essential to draw a sketch of the domain D.

2) Some domains can be considered as type I or type II ie they qualify both types.

Example 7

Evaluate the integral

R

dAe yx where D is bounded by the lines 1and,0 xxyy

Solution

A sketch of the region is shown below.Note that the region can either be a type I region or a

type II region but we choose to treat it as a type I region

212

21

1

0

x2x

21

1

0

x2x

1

00

yx1

0 0

yxyx

-eeeeee

eeAe

dx

dxdydxdx

y

x

yR

Example 8

Evaluate R

dA4xy where R is the plane region on the x-y plane enclosed by the curve

22 xy and the line 42 xy and in the 1st quadrant.

Solution

We note that the region D is a type I region but not a type II region. Sketch the region D and

draw a vertical arrow as shown below.

At intersection, 242 2 xx , that is, 0181632)42( 222 xxxx 3

10,2x

The lower boundary is 42 xy and the upper boundary is 22 xy . So, we can write:

242,2:y) ,(x D 2

310 xyxx



units cubic18

36632

)42()2(2

2

4A4xy

Therefore,

310

310

310

310 2

310 2

2

24

2

33

3

32

2

32

2

22

2

2

42

2

2

2

42

xxx

xdxxx

dxxxx

dxxy

xydydxd

x

xy

x

xyD

Example 9 Evaluate D

dA2y)(x where D is the region bounded by the parabolas

22 1 and2 xyxy

Solution

We note that the region D is a type I region but not a type II region.

Sketch the region D and draw a vertical arrow as shown.

The parabolas intersect when22 12 xx , that is, 112 xx .

The lower boundary is y = 2x2 and the upper boundary is y = 1 + x2. So, we can write:

22 12,11:y) ,(x D xyxx

.

15

321

1

5

5

34

4

13

3

22

2

1

1

1

432

1

1

43222

1

1

1

2

2

1

1

1

2

321

)42)1()1(

)(

)2(A2y)(x

2

2

2

2

xxxxx

dxxxxx

dxxxxxx

dxyxy

dydxyxd

x

xy

x

xyD


3y)dA(x where }x1y2x 1,x-1|y){(x,D 22



Solution

The sketch is the same as the one in the above example. Thus

22

1-1

2

3

1-

1)x

2

1-xx

2

3x

4

1-x

2

1( dx4x-x

2

33x

2

32x-xx

)dx)(2x-)x((12

3)2x-xx(1 3y)dydx(x3y)dA(x

53421

1-

44233

1

1-

222222

D

1

1-

x1

2x

2

2

Example 11 Find the volume of the solid that lies under the paraboloid 22x yz and

above the region D in the xy–plane bounded by the line x2y and the parabola 2xy Solution

From the figure, we see that D is a type I region and D = {(x, y) | 0 ≤ x ≤ 2, x2 ≤ y ≤ 2x}

35216

2

0

7

2115

514

67

2

0

6

3143

314

2

0

6

3143

383

2

0

23

312

2

0

22222

2

yx

xx

2

2

xxx

dxxxx

dxxxxx

dxy

dydxydAy

x

xy

x

xyD

Example 12

Evaluate D

xydA where D is the region bounded by the line 1-xy and the parabola 6x22 y

Solution

Here the description of D as a type I region is more complicated because the lower boundary

consists of two parts.

Therefore we will treat D is a type II region.Now 11-x yxy and

)6(6x2 2

212 yxy

At intersection, 082)6(1 22

21 yyyy 4,2y

4}y2- 1,yx)6(:y){(x,D 2

21 y

36yyy4-

dyy4y2y8y-

dy)3y(-1)y(y

dyyx

xydxdyxydA

4

2

6

24143

322

21

4

2-

5

4132

21

4

2-

22

21

212

21

4

2-

1

)6(x

2

21

D

4

2-

1y

)6(x

2

21

2

21

y

y

y

y

y

Reversing the order of Integration



Although Fubini's Theorem guarantees that integrating in either order will yield the same

result, it can happen that one order of integration is easier to evaluate than the other. In such

situations, reversing the order of integration can be useful.

To change, reverse or switch the order of integration in a double integral, follow these steps:

i) Step 1: Sketch the region of integration.

ii) Step 2: Set up the region according to the new order, and determine the new limits of

integration one at a time, starting with the outer variable. Note that the region may need to

be split into several pieces, if the boundary of the region is complicated. See next section

iii) Step 3: Evaluate the new integral.

Example 13

Evaluate the iterated integral

1

0

12sin

xydydxy by first reversing the order of integration

Solution

If we try to evaluate the integral as it stands,

we are faced with the task of first evaluating

dyy2sin . However, it’s impossible to do so

in finite terms since dyy2sin is not an

elementary function. Hence, we must change

the order of integration.

we need to re-sketch the region D here.

Then, from this figure, we see that an

alternative description of D is:

1}y0 y,x0:y){(x,D

Therefore

)1cos1(cossinsinsinsin21

1

0

2

21

1

0

21

00

21

0 0

21

0

12

ydyyydyyxdxdyydydxy

y

x

y

xxy

Example 14 Reverse the order of integration in 1

0 1

xe

dydx .

Solution

This integral is written as Type I, since we

first integrate on vertical intervals [1, ex ],

with boundaries y = ex , y = 1, while

1]. [0, x Invert the first equation and

from the figure we get the left and right

boundaries:

lny x , 1x and e]. [1, y

Therefore, we conclude that

e

y

e

dxdydydxx

1

1

ln

1

0 1

Non-simple Regions

Recall If 21 DDD , where D1 and D2

don’t overlap except perhaps on their

boundaries, then

21 DDD

y)dAf(x,y)dAf(x,y)dAf(x,



There are many regions that do not belong to any of the two previous types, such as the one

indicated below.

In this case, we have to partition our

region D into several disjoint sub-regions

so that each one is either x-simple or y-

simple. The integral will then be the sum

of several integrals, as stated in the next

theorem.

Ttheorem: If 221 D,...,.D,D partition our region D into n disjoint sub-regions,

221 D....DDD then

n21 DDDD

y)dAf(x,.....y)dAf(x,y)dAf(x,y)dAf(x,


6xydA where D is the region bounded by the graphs of the

equations. 0y and 9xx,y 9,xy

Solution

First let’s sketch the bounded area.The region is neither y-simple nor x-simple.

It looks like we might need to divide this

into two problems.ie the region to the left

and to the right of x=3 as shown

3ln486dx486xdx3x

6xydydx6xydydx6xydA

4

2439

3

1-3

0

3

9

3 0

3

0 0D

9

xx

Example 16 Reverse the order of the integral

1 2

240

2

0 21

21

y)f(x,y)f(x,x

xy

x

ydydxdydx

Solution

We 1st sketch the region of integration.

Reverse the equations xy 2 and

24 xy to read 4

2yx and

4

2

yx

as shown. At intersection

024

2

4

22

yyyy

. 2 y

and 1x ie (1 , 2) is the only point in our

region that makes sense

Thus

2

0

1 2

240

2

0

4

2

4

2

21

21

y)f(x,y)f(x,y)f(x,y

yx

x

xy

x

ydydxdydxdydx

Example 17 By first reversing the order evaluate

1

1

2

1

1

2

2

6xydydx6xydydx2

xx

x

Solution

We 1st sketch the region D as shown

.



We then write the integral as an iterated integral

with the order reversed

4

1354

1

4

4

332

4

1

324

1

2

4

1

21

1

2

1

1

2

2

y5y-6y

dy3y15y-12ydyy-y)-(23y

6xydxdy6xydydx6xydydx2

y

y

xx

x

.Exercise

1) Evaluate the iterated integrals

1

0

4

0

3

2x

yy

0

x1

0

1

y

3

1

0

3

3y

x1

0

1

0

x2

x-1

2x

0

2

dydxe f) dxdy 2ye e) dxdyx1 d)

dxdye c) )dydxy-(3x b) dydxcosx a)

22

2

2) Evaluate D

dA,e yx

where D is bounded by the lines y xand-y x2,y 1,y

3) Evaluate D

dA,exy where D is bounded by the lines 1 xandx y 0,y

4) Evaluate the double integral;

a) D

dA,)( yx where D is region bounded by xy 0,x and the hyperbola 1xy

b) D

dA,)cos( yx where D is bounded by 1 xand xy 0,y 2 .

c) D

2y)dA,(3x where D is quarter of the unit circle centered at the origin in the

positive x and y-directions. Ans 35

d)

D

- dA,e yx where D is the region in the first quadrant in which 1yx

e)

D

2- dA,e yx where D is the region in the first quadrant in which yx

f) D

22 dA,yx where D is the region 1yx0

g) D

1)dA,y(x where D is the region inside the unit square in which 0.5yx

5) Find the limits of integration of D

y)dA,f(x, in the domain 1}yx:y){(x,D 2

412

91

when D is considered first as a Type I and then as a Type II.

6) Evaluate the integral a) D

22 dAx-yx where 1}y0y,x0:y){(x,D

1}yx|y){(x,D wheredA,y-x-1 b) 22

D

22

D

dA22 yx c) , where x},10:y){(x,D 2 yxx

D

2xydA d) where D is the triangular region with vertices (0,0), (1,2), and (0,3).



D

22 dAy-x-16 e) where 16}yx|y){(x,D 22 by identifying it as a known solid

7) Find the volume of the solid that lies under the paraboloid 22 yxz and above the

region enclosed by 22 yx and xy .

8) Find the volume of the solid bounded by;

a) The cylinder 4yx 22 and the planes 0z and zy 0,x in the 1st octant.

b) The cylinder 4zy and 4yx 2222

c) the plane 1zy xand 0z 0,y 0,x

d) the planes 2zy2 xand 0z 2y, x0,x Ans 31 cubic units

9) Find the volume of the solid Enclosed by;

a) the paraboloid 22 3xz y and the planes x=0, y=1, z=0, and y=x.

b) Bounded by the planes z=x, y=x, z=0, and x + y = 2.

c) Bounded by the cylinders 222x ry and 222 rzy

10) By first reversing the order of integration, evaluate;

a) 4

1

3y

ydxdyxy b)

2

0

2 2

y

x dxdye c) 1

0

12cos

xdydxy d)

1

0 ln

e

e ydxdy

x

x e)

1

0

1

xdydxe y

x

f) 8

0

2

3

4

y

x dxdye g)

1

0

1sin

yx xx dxdy h)

1

0

153 cos

xydydxyx

11) Sketch the region of integration hence reverse the order of the integration a) a)

1

0

1

y)f(x,xy

dydx b)

2

0

4

2y)f(x,

xydydx c)

1

0y)f(x,

x

xydydx d)

2

1

2

2y)f(x,

y

dxdy e)

2

1

ln

0y)f(x,

x

ydydx f)

2

1

)2(11

0

1

0 0

23 2

y)f(x,y)f(x,x

y

x

ydydxdydx

12) By first reversing the order of integration, evaluate;

a 1

0

3

3

2

ey

x dxdy b)

4

0

2

2y1

x

xdydx c)

0 y

ysin

xdydx d)

2 2

0

2 sin

yxydydxx

e)

4

2 2

2

0 0)(6)(6

y

yx

y

ydydxxydydxxy

13) Evaluate the double integral D

dydx,exy in the given order where D is the plane region

bounded by the lines x2y x,y and the parabola 2y x

Double Integrals in Polar Coordinates Sometimes it’s necessary to change the variables in a double integral. A very commonly used

substitution is conversion into polar. This substitution is particularly suitable when the region

of integration D is a circle or an annulus (i.e. region between two concentric circles). Polar

coordinates r and are defined by cosrx and sinry

If f is continuous on a polar rectangle R given by b, ra 0 where

2 0 then

b

aD

rrfdAyxf rdrd)sin,cos(),(

Generally, If f is continuous on a polar region D of the form where 2 0 ,

)(h r)(h 0 21 then

)(h

)(h

2

1

rdrd)sin,cos(),(r

D

rrfdAyxf



Suppose our domain D is an annulus defined by the polar rectangle bra 0 and

where 2 0 . Partition D into small polar rectangles given by

kkkikkkkkij grgrD ,)()(:),( 21 where )()( 12 kk ggr and

kk as shown below

The area of rectangle ijD is given by kA

area of larger sector - area of smaller sector

which leads to the formular rrA kk

where *

ir is the average radius of the rectangle.

Then the typical Riemann sum is

m

ii

n

j

kkkkk rrrrf1

)sin,cos( .

Taking limits as we make both and r

approach zero we get

b

a

m

ii

n

j

kkkkkr

ddrrrrfrrrrf )sin,cos()sin,cos(lim10

0

To determine the limits of integration;

Draw an incidence line from the origin outwards to cut through the region

Integrate for values of r from where the line enters the region to where it leaves the region

Choose values that will include all incidence lines that can pass through the region


dAyx )4(3 2 where R is the region in the upper half-plane bounded by

the circles 122 yx and x2 + y2 = 4.

Solution

The region R can be described as 4}yx10,y :y){(x,R 22



In polar coordinates, it is given by: 1 ≤ r ≤ 2, 0≤

θ ≤ π

215

021

215

0

2

0

2

1

243

0

2

1

232

0

2

1

222

)2sin(sin7

sin15cos7

)sincos(

)sin4cos3(

)sin4cos3()4(3

d

drr

drdrr

rdrdrrdAyx

r

D

Example 19

Use a double integral to find the area enclosed by one loop of the 4-leaved rose cos2r .

Solution

From this sketch of the curve, we see that a loop is given by the region

}cos20, :){(r,D44

r

So, the area is:

units square)1(

2sincos2

=A(D)

2

2

12

cos2

0

2

2

1cos2

0

4

4

4

4

4

4

4

4

d

drrdrddAr

D

Example 20 Express explicitly the Cartesian integral as an equivalent polar integral ie find the

integration limits in polar form R

dAy) f(x, where R is the region bounded by the lines

3,1 xyy and the circle 4yx 22

Solution

A sketch of the region R is as follows

Substitute

rdrddAryrx sinandcos

Notice that

ecrry cos1sin1 ,

33tan

cos3sin3

rrxy and

24yx 22 r

At intersection

65.0sincos2 ecr

3

6

2

cos)rsin ,f(rcosy) f(x,

ecR

rdrddA

Example 21

By first expressing the Cartesian integral as an equivalent polar integral, evaluate

1

1

1

0 22

22

yx

1x

x

dydx

Solution

A sketch of the region R is as shown



8

9

016

1

8

1

0

2

4

1

0

1

0

23

0

1

0

22

22

2

sin2

1cos1cosr

1cosr

yx

1x

ddrd

rdrdr

dAR

Exercise

1) Evaluate R

dAyx )43( 2 where R is the region

20,21:),( rrR

2) Evaluate the iterated integral

0

Sin

0 0

Siny

0

22

0

Cos-1

0

2

0

Cos

0

2 dxdy x d) drdr c) drdrSin b) drdSinr a)

3) Use polar coordinates to evaluate

3

0

9

0

3222

a)x

dydxyx

)dxdyySin(x b)1

0

y-1

0

222

2

1

x-2x

0 22

2

dydxyx

1 c)

1

0

1

y-1- 2222dxdy

)yx(1

2 d)

4) Use a double integral to find the area enclosed by one leaf of the three-leaved rose 3cos3r .

5) Compute D

dAy , where D is the region in the first quadrant that is outside the circle 2r and

inside the cardiod cos22r .

6) Compute the volume of the solid that lies under the hemisphere2216 yxz , above the x-y

plane, and inside the cylinder 0422 xyx .

7) Use polar coordinates to evaluate

a) D

dAxy where D is the portion of the circle 122 yx that lies in the first quadrant.

b)

D

yx dA)-( 22

e where D is the region between the circles 122 yx and 422 yx .

c)

D

yx dAe22

where D is the region inside the circle 4y x 22 .

d) D

2 )dAsin(xy where 4}yx1|y){(x,D 22

e) dA,yxSinD

22

where 222 4yx|y)(x,D .

f) D

2dA,x where D is the region between the ellipse 42y x 22 and the circle 4yx 22

8) Switch to rectangular coordinate and then evaluate 3

4

43

-5Sec

0

23 drdSinr

Applications of Double Integration 1) Area, Volumes, and Average Values One straightforward but still very useful application of double integration is to computing

areas of regions in the plane, and volumes of regions in space: we have



D

dA1 = Area(D) , and D

dAy) ,f(x = Volume(D)

A closely related problem is to calculate the average value of a function on a region. To

calculate this, we simply integrate the function over the region, and then divide by the

region's area. Thus

D

dAy) ,f(x Area(D)

1= valueAverage where

D

dA1 = Area(D)

Example 22 Find the average value of xyy)f(x, over the region D enclosed by ,1x

1y and xy 1

Solution

A sketch of the region D is as follows

6

11

0

3

3

12

2

11

0

2

1

0

1

x-1

2

2

11

0

1

x-121

x)-(1dxx)-(1-x

dxxy2xydydx 11

1=

y) ,f(x Area(D)

1= valueAverage

x

dAD

Example 23 Find the average value of yxy)f(x, 2 over the region D lying between the

curves 2xy and 2 xy .

Solution

A sketch of the region is as below:

At intersection 2,12 2 xxx

3

102

1

3

312

21

2

1

22

1

2

2

)2(= Area(D)2

xxx

dxxxdydxx

x

700

16232

1

7

713

3445

51

203

2

1

6234

203

2

1

422

203

2

1

22

103

44

)2(=

y) ,f(x Area(D)

1= valueAverage

2

xxxx

dxxxxx

dxxxxydydxx

dA

x

x

D

Example 24 Find the average value of22 yx

1y)f(x,

x over the region D lying outside the

circle 1r and inside the cardioid cos1r .

Solution

A sketch of the region is as below:



Dregion over the f of integral double thedo Next we

22sinsin

coscos=1)cos1(=

=1 = Area(D)

cos11on intersectiAt

42

1

4

1

2

2

12

2

1

cos1

1

2

1

2

2

2

2

2

2

2

2

dd

rdrddAD

32

43

41

121

613

31

21

21

2

6122

2

6122

4

31323

31

cos1

1

3

31

cos1

1

1

2

2

2

2

2

2

2

2

2

2

2

2

2

2

)4sin()2sin(sin)2sin(

)2cos2cos21(sincoscos=

cos)2cos1()sin1(coscos=

coscoscoscos=cos1)cos1(cos=

coscosy) ,f(x

d

d

dd

drrrdrdrdAr

D

324

89

2y) ,f(x

Area(D)

1= valueAverage

4

32

43

D

dA

Exercise

1) Use a double integral to find the area bounded by 2xy and xy .

2) Use a double integral to find the volume of the solid in the first octant bounded by the

paraboloid 22 yxz and the planes 1,0 yxz .

3) Find the average value of xy81y)f(x, over the rectangular region defined by

21,30:),( yxyxD

4) Suppose that the temperature (in degrees Celcius) at a point (x, y) on a flat metal plate is 22 2810 y) T(x, yx , where x and y are in meters. Find the average temperature of the

rectangular portion of the plate for which 20,10:),( yxyx Ans 14/3

5) Find the average value of xyy)f(x, over the region D defined by

xeyxyxD 0,10:),(

6) Find the average value of 22

12y)f(x,yx

y

over the region D that lies above the x-axis

and between the circles 122 yx and 922 yx

7) Find the average value of the function yx24y)f(x, 2 over the region enclosed by the

curves 234 xy and 2xy

8) Find the average value of the function yy)f(x, over the region enclosed by the line

1 yx and the parabola 12 yx

9) Find the average value of the function xyey)f(x, over the region enclosed by the

parabolas xy 92 and yx 92



10) Calculate the average value of xey)f(x, over the region

eyyxyxD 1,ln0:),(

11) By1st changing to polar coordinates find the average value of 22

1y)f(x,yx

over the

region D that lies above the x axis but inside the circle

.

2) Masses, Centroids, and Moments of a Plate and Solid The center of mass of a physical object is its ‘balancing point’, where, if the object is

supported only at that point, gravity will not cause it to tip over. The center of mass is also

called the centroid of an object. Physically, an object behaves as if all its mass is

concentrated at its center of mass (hence the name): the forces exerted by an object are the

same as an equivalent point mass concentrated at the object's center of mass.

Suppose R represents the shape of a thin (flat) plate, with a surface density ),( yx

(‘surface density’ means mass per unit area and if the plate is the same thickness and the

same material throughout that will be a constant ie areamass . But if it is not constant, we find

the density at point (x, y) by taking a very tiny sample of the plate at (x, y) and taking the

mass per unit area of that. So if the sample is a rectangular

piece of (infinitesimally short) sides dx in the x-direction times dy in the y-direction, then it

has area dxdy . Say we call its mass dm (for a tiny bit of mass), then the surface density is

dxdy

dmyx ),( .To get the total mass we should add up the little small masses dm , or

actually integrate them. We get RR

dAyxdm ),(= Mass total

If we are given the shape and density of the lamine in 2 dimensions, then we can find its mass

and the location of its centre of mass. Let the density at any point (x, y) in D be given by ρ(x,

y), where ρ is a continuous function on D, then the total mass m is given by D

dAyx ),( = m

The centre of mass of such a plate is at a point ),( yx which is also computed with double

integrals. The centre of mass ),( yx has is given by;

D

mdAyxxx ),(1 and

D

mdAyxyy ),(1

Note: If the lamina is symmetric about the line 0x then 0x and if is symmetric about the

line 0y then 0y

Remark: If the density is a constant then the lamina is said to be homogeneous and the center

of mass in this case is called the center of gravity or the centroid.

Example 25

Find the total mass of a lamina occupying the region that lies in the first quadrant and inside

one loop of the rose 2cos42 r . It’s density at any point ),( yx is given by xyyx ),(

Solution

We can easily use polar transform rdrddAryrx sinandcos

A sketch of the lamina is as follows



3

1

0

3

3

1

0

2

2

1

0

2cos4

0

4

4

1

0

2cos4

0

2

44

4

4

2cos)2cos4(2sin

sincos

)sincos(

),( =

d

dr

rdrdr

dAyxm

r

D

Example 26 Find the centre of gravity of a homogeneous lamina which occupies the region

lying between the line xy and the curve 2xy .

Solution

A sketch of the lamina is as follows The lamina is homogeneous kyx ),(

6

1

0

3

312

21

1

0

2

1

0

)(

),( = m2

k

x

xD

xxkdxxxk

kdydxdAyx

21

1

0

4

233

1

0

2

1

0

61

2)(6

),(2

xxdxxxx

kxdydxdAyxxxx

xk

D

m

52

1

0

5

533

1

0

421

0

21

0

61 333),( 22

xxdxxxdxykydydxdAyxyy

x

xy

x

xk

D

m

Therefore the coordinates of the centre of gravity are ),(),(52

21yx

Example 27 Find the center of mass of a triangular lamina whose vertices are (0 ,0), (1 , 0)

and (0 , 2) if it’s density at any point (x , y)is given by xyyx 3),( .

Solution

A sketch of the lamina is as follows

2

11

0

4

4

632

1

0

32

1

0

2

2

31

0

22

0

2

2

3

1

0

22

0

43

6126

)22(

3),( = m

xxx

dxxxx

dxxxdxxy

xydydxdAyx

x

y

x

D

5

21

0

5

5

12431

0

432

1

0

221

0

22

0

221

0

22

0

21

64122412

)22(3332),(

xxxdxxxx

dxxxdxyxydydxxdAyxxxx

y

x

D

m

1

0

31

0

22

0

31

0

22

0

21 )22(2232),( dxxxdxxydydxxydAyxyyx

y

x

D

m

54

1

0

5

58432

1

0

432 68428242482 xxxxdxxxxx



Therefore the coordinates of the centre of mass are ),(),(5

4

5

2yx

Example 28

The density at any point on a semicircular lamina is proportional to the distance from the

center of the circle. Find the center of mass of the lamina.

Solution

Let’s place the lamina as the upper half of the circle 222 ayx

Then, the distance from a point (x, y) to the

center of the circle (the origin) is: 22 yx

Therefore, the density function is:

22),( yxkyx where k is a constant.

Both the density function and the shape of the lamina suggest that we convert to polar

coordinates. Then, ryx 22 and the region D is given by:

0,0:),( aryxD

Thus, the mass of the lamina is:

3

30

3

31

0 0

22 )(),( = adkardrdkrdAyxkdAyxm ka

DD

Both the lamina and the density function are symmetric with respect to the y-axis. So, the

center of mass must lie on the y-axis, that is, 0x

The y-coordinate is given by:

2

3cos

4

3sin

3sin

3

)sin(33

),(

00

4

4

1

30 0

3

3

0 03

22

3

1

aada

adrdr

a

rdrdrra

dAyxkyak

dAyxyy

a

a

DD

m

Thus, the center of mass ),( yx is located at the point ),0(2

3

a

Exercise

1) Find the centre of mass for the shape bounded by: 0 and6, yyxxy , with

density yyx 2),(

2) Find the centre of mass of a triangular lamina with vertices (0, 0), (0, 1), and (1, 0) if it’s

density function xyyx ),( .

3) Find the centre of gravity of a lamina that occupies the region bounded by 22 and xyxy

4) Find the centroid of the semicircular region in the xy-plane bounded by the x-axis and the

curve 22 xay ..

5) Redo example 2 using the density yxyx 31),( instead of xyyx 3),(

6) A plate with density xyyx 1),( . Occupies the region

4,20:),( 2 yxxyxD find the centre of mass of this plate.

7) A thin plate covers a triangular region D of the xy-plane bounded by the x-axis and the

lines xyx 2 and1 in the first quadrant. The plate’s density at any point Dyx ),( is

)1(6),( yxyx . Find the plate’s center-of-mass about the coordinate axes.



8) Find the centre of mass of the lamina bounded by;

a) The lines 0 ,4, yx and the curve xy

b) 0 and3,3 yxxy , 22),( yxyx . Use polar transformation.

9) If the joint density function for X and Y is given by

elsewhere0

100,100for )2(),(f

yxyxCyx

Find the value of the constant C. and then find )2,7( YXP .

10)



x

y

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