:
\302\273n
I
*
lit:
\342\200\242
i
\\
11
*
i
\\
x
\\
.
\\
1
\342\200\242
i
*
/
\342\200\242
.
\342\200\242
\302\253
i
\302\253i
\342\200\242
i
i
\342\200\242
i
i
Mathematics
9
Topic
Level
Higher
-
Option:
Calculus
for the IB Diploma
Paul
Ben
Fannon,
Woolley
Vesna
and
Kadelburg,
Stephen
Ward
PRESS
UNIVERSITY
CAMBRIDGE
Cambridge,
New
Cambridge
Mexico
Delhi,
Cape Town,
Madrid,
Melbourne,
York,
Singapore, SaoPaulo,
City
Press
University
Building, Cambridge CB28RU,
The Edinburgh
UK
www.cambridge.org
Information
on
\302\251
is in
publication
www.cambridge.org/9781107632899
Press 2013
University
Cambridge
This
title:
this
Subject to
copyright.
provisions of relevant
no reproduction of any part may
and
to the
permission
First
in
Poland
the
without
written
Press.
by Opolgraf
this
ISBN 978-1-107-63289-9
Cambridge
exception
agreements,
2013
A catalogue recordfor
Coverimage:
statutory
licensing
take place
of Cambridge University
published
Printed
collective
is available
publication
from
the
British
Library
Paperback
Thinkstock
University
accuracy of URLs
for
Press has no
external
for
responsibility
or third-party
internet
the persistence
websites
referred
or
to in
this publication, and does not guarantee
that any content on such websites is,
or will remain, accurate or appropriate. Information regarding prices, travel
in this work is correct at
timetables
and other factual information
given
the time of first printing
but Cambridge
University Press does not guarantee
the accuracyof
such
information
thereafter.
NOTICE TO TEACHERS
and copies of them remain
such copies may not be distributed
Worksheets
and
purchasing
institution.
copyright of Cambridge University
or used in any way outside the
in the
Press
'a
Contents
How to use this book
IV
Introduction
1
of
Limits
1
I
The
1\302\243B
The
I
The
4
Theorem
8
limit of a function
11
Squeeze
19
Rule
functions
I
Continuous
I
Differentiate
I
Rolle's
24
functions
29
and the
Theorem
Mean Value
45
of improper
Approximation
46
of Calculus
Theorem
Fundamental
Convergent and divergent improper
50
integrals
57
integrals
65
66
69
88
series
Infinite
of series
Convergence
H Tests for
convergence and divergence
series
Power
and
Maclaurin
-rA^
34
Theorem
integrals
Improper
4
3
limit of a sequence
Hi L'Hopital's
The
functions
and
sequences
Taylor
99
99
series
series
Maclaurin
to the
Approximations
series of
Maclaurin
Maclaurin
104
series
110
composite functions
113
series
Taylor
117
Applications
Differential
Setting
I
Separation
I
Homogeneous
HI
123
equations
126
of variables
differential
differential
Linear
I
6
up differential
I
123
equations
Approximations
Summary
129
equations
equations
132
to solutions
136
and mixed examination
practice
149
Answers
162
Glossary
174
Index
178
Acknowledgements
180
Contents
c
XJ^\\
+ d.
in
book
the
of
Structure
book
this
use
to
How
This book coversallthe materialfor Topic 9 (Calculus Option) of the Higher Level
for the International Baccalaureate course.It assumesfamiliarity
with
the
syllabus
1to 6), particularly
Calculus
topic
(syllabus
tried to include in the maintextonly the
(syllabus topics
We have
1.1).
topic
There are many interesting
highlight some of thesein
main
five
The
of all
summary
another
'From
perspective'
material
material
Sequences and Series(syllabus
that will be examinable.
and
we have tried to
beyond the syllabus
and 'Research explorer' boxes.
best coveredin the orderpresented,
chapter
although
of
the
core
calculus
material.
Chapter6 contains
knowledge
are probably
chapters
last section)only
for the
ideas that go
and
applications
the
6) and
Mathematics
core
requires
the topics
several topics - a favourite
and further examinationpractice,
with
in IB
trick
starts with
of the
many
5 (except
a
mixing
questions
examinations.
objectives to give you an idea about what the chapter
at the start of the topicthat
illustrates
what
problem
introductory
you should
be able to do after you have completed
the topic. You should not expectto be able to solve the
at
the
but
want
to think about possible strategiesand what
sort
of new facts
start,
problem
you
may
and methods would helpyou. The solution
to the introductory problem is providedat the end of
6.
chapter
Each chapter
contains.Thereis
The
most
of learning
boxes
point
Key
a list
an
important
and formulae are emphasised in the 'KEY
the Formulabooklet,there will be an icon:
in the Formula booklet and you may need to
ideas
formulae are given in
the
are not
formulae
to derive
the
When
boxes.
POINT'
then
; if this icon is not present,
learn them or at least know how
them.
Worked
examples
Eachworkedexample
is split
into
the example
Sometimes
you
an
idea
of what
two
is required to scorefull
is about muchmorethan
On the
columns.
right is what
might include more detail then you
and
examinations
method
marks
remembering
strictly
you shouldwrite
need,
in examinations.
methods.
down.
is designed to give
However, mathematics
but it
So, on
the left
of
the
worked
and suggest which routeyou should
use
with
exercise
that
differ
from
hope
help you
questions
any
the worked examples. It is very deliberatethat
some
of the questions
require you to do morethan
in the worked examples.Mathematicsis about thinking!
the methods
repeat
examples are notes that
tackle the question. We
How
to
use this
book
the
describe
that
thought
these will
processes
to
Signposts
There are severalboxesthat
issues
of knowledge
Theory
the book.
throughout
appear
Theory of knowledge lesson, but sometimesthe links
may
Mathematics is frequently usedas an example
of certainty
and
not
the case. In these boxes we will
to highlight
some
truth, but this is often
try
of the weaknessesand ambiguitiesin mathematics
as well as showing how
is a
lesson
Every
be obvious.
not
X
mathematicslinks to other areasofknowledge.
another
From
perspective
The International Baccalaureate\302\256
encourages
As wellas highlightingsomeinternational
these boxes also look at other
perspectives
historical,
Research
on the
in different
at things
ways.
mathematicians
between
mathematics
we are covering:
cultural.
and
pragmatic
looking
differences
explorer
As part of your course,you
topicof
as
a jumping-off
for
point
to write
a report on a mathematical
know which topicsaresuitable
to show where a topic can act
further work. This can alsogive you ideas for an
difficult to
reports, and so we
for such
a basis
asked
be
sometimes
It is
choice.
your
as
will
have
tried
Extended essay. Thereis a lot of greatmathematics
out
there!
hint
Exam
we would
Although
learning
is all
Mathematics
will
try to
highlight
you
making links. You might be interestedto seehow
something
used elsewherein the course, or you may need to go backand
a previous topic. Theseboxesindicateconnections
with
other
sections
about
learned will be
have just
of
yourself
book to
to
How
we
just
it is
rewind
Fast forward /
of the
there are somecommonerrors
boxes.
it in these
remind
as more than
of mathematics
think
to be awareof. If thereisa commonpitfall
for you
useful
to
an examination,
to pass
order
in
encourageyou
help you
find
your
way around.
use the questions
The colour-coding
The
are
questions
colour-coded
Black
they
easy,
to distinguish
are
drill questions.
questions
They help
are usually not structured likethe questions
some of them are quite tough.
between the levels.
in the book,
you practise the methodsdescribed
in the examination.
This does not mean they
How
Ai
i
***\\
+ 0,.
to
use this
but
are
book
Each
are of
increasing
(ii),.-)
((i),
drill
numbered
differently
>
all
are of
different skill.Letteredsubparts
part there
lettered
each
Within
difficulty.
of which
tests a
question
a similar difficulty.
Unless
a question
of
may be multiple roman-numeralparts
you want to do lots of practice we would
that
part and then checkyour answer.
you
only do one roman-numeral
made a mistake then you may want to think about what went wrongbeforeyou try
Otherwise move on to the nextlettered
part.
recommend
Green
the
to getting
path
Blue
questions which
are examination-style
questions
are
questions
a grade 3 or 4.
harder
examination-style
should beableto makesignificant
progress
have
If you
more.
any
should be accessible
to students
questions. If you are aiming for
most of these.
through
5 or
a grade
Red questions are at the very top end of difficulty in the examinations.If you
are likelyto be on coursefor a grade 7.
on
these
do
can
6 you
then you
Gold
\342\226\241
are a
questions
thinking and
type that are not
set in the examination,
but
discussionin orderto helpyou
to
a better
are
to provoke
designed
understanding
of a particular
concept.
At
chapter you will see longerquestions
of each
end
the
International Baccalaureate\302\256
Of course,theseare
a green
if you
just
These
examinations.
If you
guidelines.
question
can do all the red questions
you have to deal with
examination
that
time
follow
are aiming
you cannot do. Peoplearenever
typical
equally
for
good
not be surprised if you find
areas of the syllabus.
Equally,
6, do
at all
not guarantee
you will
and
examination
pressure
does
section of
colour-coding scheme.
the same
a grade
second
of the
get a grade 7; after
all,
in the
stress!
These questions are gradedrelative
to our
of the final examination, so when
first
experience
you
start the course you will find all the questionsrelatively
but by the end of the coursethey
hard,
should
seem
more straightforward.
Do not get intimidated!
Option an interesting and enrichingcourse.You might also find
but
do
not
topics
only make sense after lots of
challenging,
getintimidated,frequently
revisionand practice.Persevere
and
will succeed.
you
We
you
hope
find
the Calculus
it quite
The author
How to
use
this
book
team.
Introduction
In this
\342\200\242
\342\200\242
about
a limit of
find
to
how
can
\342\200\242
to
decide
\342\200\242
to
use
\342\200\242
to
solve
how
how
series
Find
L'Hopital's
approach)
Rule
functions, and someuseful
allow one of the limits
a
to
tend
to infinity
limit
various functions (Maclaurinand Taylor
series)
and
types of differential equations, both
exactly
approximately.
certain
layer has
a radius
four
layers:
a thicknessof 1.The
11
of- and the
first
layer
the volume
covering
has a
radius of
1, the seconda radius
the
of\342\200\224,
2
fourth
a radius
of\342\200\224.
4
3
needs
including
sequence
problem
Introductory
third
functions,
of the
to approximate
Consider a wedding cakewith
Each
series have
infinite
whether
terms
and differentiability of
definite integration to
extend
we
of
limits
the
functions
differentiable
of
\342\200\242
how
for finding
of continuity
definitions
properties
a sequence (the value that
methods
various
\342\200\242
formal
how
you will learn:
Option
of the cakeand
the
surface
area
(excluding
the bottom of the first
layer)
that
with icing.
Now imagine there areinfinitely
to the cake. What can you say
many
layers
of the cake and the surfaceareathat needs
now?
icing
about
the
volume
Introduction
fit
C
XJ^\\
+ d.
option builds on many
The calculus
sequences
and
series,
and
differentiation
integration.
interested many of the greatestmindsin
and others;and
and
physics
The option
have
far
reaching
mathematics:
applications
course, using the theory of functions,
Some of the ideasyou will study here have
of the
areas
different
Euler,
Newton,
Leibniz,
both within mathematics
and other subjectssuch
as
engineering.
is split into five
chapters:
of sequences
of sequences and functions looksfirst at the behaviour
terms.
How can we determine which sequences
headtowards
(converge
1 Limits
more
Cauchy, Riemann
as we take moreand
to)
a finite
value
when we take infinitely
and which just become infinitely
We then examine
the
terms,
many
large?
values of functions as we get closerand closer to a particular value in their domain. Thefocusis
on quotients
where both the numerator and denominatoraretending
to 0 or where
particularly
on the idea of a limit of a function, we then lookat where
they are both tendingto \302\251o.
Building
functions
are continuous
and where they can be differentiated,
before
two important
considering
theorems for differentiable functions.
2 Improper
at what
looks
integrals
be oo. Will
this always give
an
happens
infinite
curve), or are there somecircumstances
we know which is the casefor a given
Theorem
of Calculus.
many
different
link
in
when we take the upperlimiton an
for the integral (and hencean infinite
which
integral?
that
area
And
area) is finite?
starts with a look at the
value (or
This chapter
to
integral
under
again
the
how
do
Fundamental
the previous chapters and examinesthe sumsof infinitely
many
seen
with
some
series
that
it
is
to
add
possible
already
geometric
together infinitely
terms
and yet get a finite
value
for the sum to infinity;
here
we look at a wide range of
series
and develop
a finite value) or not. We
also
(has
ways to test if the seriesconverges
of an
improper
integrals and infinite sums and lookat ways of placing bounds on the value
3 Infinite
terms.
value
infinite
builds
series
on both
We have
sum.
and Taylor
4 Maclaurin
series
establishes
a way
of representing many
familiar
functions
as infinite
series in increasing integer powers of x. We examine
for functions, and
approximations
polynomial
their accuracy,and usethe seriesrepresentation
of functions
for integration
and to calculatelimits,
those
where
both numerator
and denominator are tendingto 0 or <*>.
again particularly
5 Differential
looks at setting up and solving differential
which
are used
equations,
equations
in
to describe the behaviourof many
nature
and
We
look
at
three
processes
engineering.
methodsfor solving
different
of these equations and then considersomemethods
to find
types
in the
solutions
to
differential
that
are
difficult
not
to
solve
(if
approximate
equations
impossible)
standard way.
Introduction
*
-
**\\
In
of
Limits
this
\342\200\242
to use
some
of a
of
limits
sequences
\342\200\242
to find
and
rules
algebraic
to calculate
sequences
you
chapter
learn:
will
limit
the
sequence
by
it between
squeezing
that
two sequences
both converge to the
functions
same
limit
\342\200\242
to
similar
apply
to find
principles
of
be
sequences
in the
core course, and will
focus now
to be
form
some finite
if so
and
or perhaps
- and
\342\200\242
to determine
functions
by
the
of
limits
\342\200\224
oo
0
does
Rule
I'Hopital's
to find
them.
Our
general term i^to define
is to see what happens
as we take more and
more termsofa sequence:
doesthere seem
value (a limit) which the sequence
approaches,
the sequence do so by increasing,
decreasing,
oscillating either sideofthe value?
functions
\342\200\242
to use
using a
with
familiar
main
met
have
should
You
limits
where
are continuous
and where they
are
differentiable
\342\200\242
to
apply
Theorem
Rolle's
and the
Mean
Value Theoremto
differentiable
functions.
terms
increase
terms decreaseto a
to a limit
terms
approach
limit
limit
with
oscillation
nothave
Alternatively, does the sequence
or perhaps
just getting largerin magnitude
same magnitude eithersideofsome value?
a (finite)
limit,
oscillating
due to
with the
1 Limits
C
XJ^\\
+ d.
of sequences
and
functions
3
\342\200\242\342\200\242\342\200\242-
terms
get
progressively
greater
terms get progressively
it is
Sometimes
immediately obviouswhich of theseis
place.
taking
For
oscillate
terms
smaller
general term
the sequence with
example,
un
\342\200\224
2n~l
1,2,4,8,16,...
is clearly
just getting
larger term by
and
term
will never
approach a limit.
i
The sequencedefined
\342\200\224
3,
ux
by
inductively
un+1
=
4-w\342\200\236
,
w\302\273,
-L,
3
seemsto be decreasing,
but
limit or if it does,what this
,
it is
limit
chapter we start by looking
questions about sequencesand
at
might
at ways
to answer
widen
our focus
then
these
to look
We then use the ideaoflimits
where a function is continuousand
to consider
3 The limit
a
of
have a
be.
it can be differentiated and finally
apply
two wellknown and useful theorems.
these
of
where
to develop
ideas
sequence
If the terms of a sequencei^head towards
the sequence convergesto a limitL, as n
and
...
of functions.
limits
the
153
clear whether it will
not
In this
functions
,
11 41
a value
L, we
\302\260\302\260
\342\200\224\302\273
(cn tends
say that
to
\302\251o')
write:
we
-L
limw\342\200\236
This
does not
necessarily mean that
actually reaches the value
terms,
the
sequence
L,
becomes
any
that,
n
appears to
3V5V7'\"
be convergingto 0,that
lim-
although no
Topic 9
-
Option:Calculus
term actually
of the
sequence
by taking more
close to L.
arbitrarily
but
= \342\200\224
For example,the sequence[u\342\200\236}
2
term
is 0.
is:
= 0
and more
'a
we have
Graphically
(see alongside):
Ifthe sequence
doesnot
Often, the behaviour
a graph.
drawing
u, n
as \\un
sequence such
For a
diverge.
without evaluating lots of
a possible limit
to find
want
may
termsor
said to
so apparent and we
is not
a sequence
of
it is
then
converge
=
j
, we
might
n
consider
trying to take thelimitsofboththe numeratorand denominator
and then dividing one by the other.The problem
here
separately
\342\200\224
is that both numerator and denominatordiverge
and
is not
-**n
defined.
however, take this approachifthelimitof both
is finite, as then the limits do behave
much
sequences
might expect (or at least hope!).
We can,
as you
1.1
KEY POINT
Algebra of limits
If
the
sequence
a limit a
and the
= pliman+qlimbn= pa + qb
[p,q
to
converges
a limit fo, then:
{an}
sequence
to
{bn}
lim(pan +qbn)
- (liman
\\im{anbn)
\\iman
**^
lim
=
-a
The algebra of limits only
are finite)
to the
supplements
If the
constant
the limits
make use of the following
exist (they
two
final point.
1.2
POINT
KEY
= ab
when
applies
will often
we
but
R)
(6*0)
b
lim\302\243>
vV
I
](limfon
e
sequence {an} diverges,thenfor any
c e R (c ^ 0):
lim uu
n-*\302\260\\*n)
\\ =
lim
CXD
c J
\"^\302\260\302\260V
While
both
not seem helpful
this might
numerator
for
and denominator
example
a common
illustrates
allowingus to apply
the
algebra
\\uS\\
\342\200\224
where
diverge, the following
of dealing with this problem,
way
of limits
result.
1
C
XJ^\\
+ d.
Limits
of
sequences
and functions
orked
1.1
example
Find the following:
(a) lim
n2
lim
(b)
2n2-3n
+ 97
\302\273-+\302\2732n
Neither 6n + 72 nor
we must rearrange
a numerator
and
+ 97
of limits
the algebra
apply
6n
97
Zn +
,lim 6\"+72 =
2/i +
and
97
...
lim
2+
97
6 +
72
n
0n.#
\342\200\224
=
n^oo n
lim
by Key
97
-n
lim
97
lim
72
2+
6+ 72
results*
=
,.lim \342\200\224
= 0
n->\302\260o
n
72
6+
72
+
n
by
through
+8
(a)
converges so*
the expression
to give
denominator
that both
2n
converge, by dividing
Now
+ 5
I
point 1.2
2 +\342\200\224)
+ 0
_6
~
2 +0
=3
Neither n2+5 nor
so
- 3n
2n2
converges
denominator
and
converge, by dividing
the algebra
apply
(b)
by
through
of limits
1+ -
n2+5
to give a
that both
the expression
rearrange
numerator
Now
*
+8
2n2-3n
+
8>'
3
o
2
3
lim
\342\200\224
=
0 and
results*
1+ -
H2+5
=
/r
lim
2-^
8
lim
/i2
all by
Key point
1+
lim
\342\200\224
=
l^m
AWoofl
n2
n
/i
: 0,
&
\342\200\224
n2
lim
\342\200\236^oo
+ l
2n2 \342\200\224Sn
Mm
+
0,
1.2
2
lim
1+
3
0
n
n2 J
\342\200\224
+ \342\200\224
0
2-0 + 0
2
It is tempting to think that
To
of
limits
a
result*?
hadion,
highest
V
-
first
by
through
Topic 9
algebra
^e
apply
Option:
^de
the
power
otn.
Calculus
the
sequencesthat diverge to infinity
we must not apply the algebra of
- <*>'
sequences.(In fact c\302\260\302\260
might
other
finite
value
or it
difference of two
is 0, i.e. that coo - oo = 0', but
limits to non-convergent
limit
of the
be 0,
might even be
but it might
<*>!)
In
into
necessary to manipulate the expression
for applying the algebra of limits and/or Key
these
be some
cases
a convenient
point
1.2.
it is
form
orked
1.2
example
Find lim (V^2 +n-n).
Neither Vn2
we can't apply
nor n converges
so
of limits to
algebra
+ n
the
Neither do we have the
situation
of a quotient where
the difference.
familiar
we
divide
could
as
through
in
Worked
example
a
So create
and bottom by the
The
vo
of
presence
multiplying
by
quotient
1.1
top*
vn2
+h
-h
same expression.
-b here suggests
by
multiplying
=
(Vh2 + n
vn2
+ n
+
+n
+n
n
\342\200\224n)
\\ln2
\342\200\224n
(vh2+hJ
Va + b
(h2 +n)
Vn2 +n
\342\200\224
n2
+n
n
vn2
Divide
by
through
inside the
through
dividing
n (this
+ n
+
n
1
will
mean*
square root
- + \342\200\224+1
byn2)
1+
- +1
n
Now
(a)
the following
(i)
algebra
of limits0
:.
lim (Vn2
+n -n)=
,
\342\200\224
=
1A
Exercise
1. Find
the
apply
limits:
3n-7
lim
+ l
2n
lim
(ii) n^oo
ll
4n +
3-2n
n2 +
(b) (i) lim
3n2
3n -1
+5n-7
4n4+3n3-10
lim
(ii)
(c)
(i)
n4+5
5-3n
limn-^oo
2n2
(ii) lim
+ An
n2-6
\342\200\236^oo
n3 _|_
(d) (i)
lim
+ 9
_ 3
5^
^5-3n^
7-4n
1
r^
**.
Limits
of sequences
and
functions
9n2+2
(ii)
lim.
lim
3n-2
5n + 7
(
(b) Hence
lim
find
the
2n3 + 3n2
+ 4)
n(n
n +
For
2
{ un
lim
n2 +3
2
whose
},
general
v n
+1
\342\200\224
The
yjn
<c
<b
[4 marks]
Theorem
for
described
there is anotherresultwhich
we
limits
finding
to have at
need
K
sequences
for
all n
UK}
and
{c .}
such that
eeZ+
and
liman
then
\\imbn
= limcn = L<
that
between
converge
them,
samelimit.
OO
= L.
The Squeeze Theorem says
Option: Calculus
= 2
Theorem
If we have
-
[7 marks]
\"->~Vw-3+2
1.3
POINT
Squeeze
9
given by
lim
In addition to the methodsalready
ic
term is
- v n -1
-j=
Squeeze
of sequences,
our disposal.
1
[6 marks]
y
algebra of limits to prove that
\302\243J
a
+6
[6 marks]
lim
KEY
2n2 +3n
un
Use the algebraoflimitsto find
Use the
+12n
\\
+ 4)
n(n
n +
sequence
n3 +
+3
n2
un
find
[3 marks]
lim
that
Show
5
7
>5n +
(b) Hence find
(a)
3
3n-2
(a) Showthat
to the
then
that
can find
two sequences
and squeeze anothersequence
sequence must also convergeto the
same limit
that
if we
u
A
\342\200\242
\342\200\242
#
\342\200\242
*\"2
we often
this result
To use
need
known
inequalities
for standard
functions.
orked example 1.3
the
Use
The
obvious
to find
Theorem
Squeeze
lim
sinn
\302\253->\302\260o
fl
-1<sinn<1
place to start is by bounding1
sinn
zl<\302\243i^<l
As both
can
to 0
\\ \342\200\224
\\
\\ \342\200\224
I and
converge
the Squeeze
apply
we\302\253
|
\342\200\224
=
nJ
by the
e'mn
lim
it can
you use the SqueezeTheorem
or
of
the
The
one,
both,
sequences
a^and
cn.
a common way of doing this.
When
be
difficult
following
fortunes
Since
lim
Theorem
neZ+
n
n
n
all
for
lim
\342\200\224
=
0
^\302\260\302\260
n
Squeeze Theorem
= 0
to choose
illustrates
orked example1.4
that
Show
lim
\342\200\224=
0.
\302\253->oo
yf
n\\
We
can
see
that
\342\200\224
>0
so we
nn
for
a sequence
largeas
Hl
but
that is always
which
tends
are looking
>0
fora\\\\ neZ+
at least as
toOasn^oo
.
nn
so that
we can
squeeze _
nn
1
XJ^\\
+ d.
Limits
of
sequences
and functions
i
h
continued...
To find
a sequence
such
the
out
start by
term of
general
*
Next,
writing
our sequence
n\\
321
_n(n-l)(n-2)(n-5)
n
n
nn
n
n n
long-hand
is less
numerator
Each
than
introduce
the
leave
We
final
term
the
Apply
the
inequality.
so that
unaltered
not everything
to1
use that
n so
cancels
n n 1
1
n n n n
n n n
n
n\\
1
\342\200\224
<-
0<
forallneZ+
n
n\"
out
Theorem1
Squeeze
n n n n
- =0
lim
Since
can
conclude
by the
Squeeze
\\we
n\\
that
lim
\342\226\240\342\226\2400
theorem
IB
Exercise
1. Usethe Squeeze
Theorem
(a) lim
cos n
-sin3n
4n2
(b) lim
\302\253->~
(c)
the following:
find
to
n2 +
8
nn
lim
\302\253->~(2w)!
(d)
lim^fc!)
(a) Show that
x
trick
a common
is
to use part
an
as
such
inequality
3.
at the core
for
lim\342\200\224=
0.
[6 marks]
(a) Show that
Look
\342\226\240in
question
course
6n
that
the
of
to introduce
of
use the SqueezeTheorem
to show
expansion
binomial
back
(b) Hence
> 6.
n
5!
n\\
It
- for n
\342\200\224
< \342\200\224
(b) By
>
+ x)n
(l
^ \"
^
taking x = I
show
x2
for
that
Ufn
allx>0,neZ+.
-1)
<
n^
1.
a
rejoinder
the binomial
(c) Hencefind
lim
[8 marks]
yfn.
expansion.
(a) Find
lim
(b) Hence
the
use
Theorem
Squeeze
to find
\\imS\\\342\200\224j==
n^tl^n2
+k
[7 marks]
(a) By considering (1 +
2n
n{n
x)n
for
a suitable
value of x, show that
\342\200\224 \342\200\224
\\){n
2)
>\342\226\240
for
all n
e Z+
3!
(b) Hence find
lim\342\200\224.
w_>oo2n
10
v
Topic
9
-
Option: Calculus
[8 marks]
(a) Find the value
Hence
(c)
Find
\342\200\224
lnx
passing
touches the curve.
inequality involving x and lnx for x> 0.
the origin
through
(b)
to y
the tangent
which
a at
an
form
3h3
lim
+lnn3
.
+2*
\342\200\236->oon3
\342\200\224
that
Show
(a)
n
(b) Showthat
<ln
that
show
\\+n
(i)
Find lim(l
(ii)
What
n>\\.
has
but
again
you made?
have
many
applications.
+-
assumptions
one
of the
of percentage
context
n)
V
is
most
important in maths.
often introduced informally
1Y
f 1 + \342\200\224
<lforalln>l.
n
(c) Hence
(d)
for all
+ ri)<n
ln(l
in
established
limit
Question 7, part (d)(i)
jfjkk
^^
*
n>\\.
for all
< Inn
The
in
chapter
It is
in
the
increase,
wide-ranging
We
will
3 of
meet
it
this option.
[15 marks]
a function
The limit of
a function f(x) at a point a, limf(x),
is like that of the limit of a sequence: it is eitherthe value that
the function
attains at a or approachesas x \342\200\224>
a. The
difference
is that the domain will not just bepositive
and that the
integers
limit can be taken as x tendsto any value,
not just as x \342\200\224>
<*>.
of the
The
idea
For
example:
limit of
limx2 =
32=9
lim tan2
x = \302\260\302\260
lim
e~x =
0
71
While we cannot input x = \342\200\224
into
secondexample
above
to get
e~x
functions
(as
it is
the limits of
and
to these limits
oo
tend
x to
get the limit in
the
domain) ori = oo
we can see that
respectively,
not in the
0
into
the
by referringto their graphs.
y
y
=
\342\231\246
I y
0
rZ
Using
tan2
a combination
of two
tan2 x
of these limits,we
-
could
also
say
limtan^x
= o
1 Limits
r^
+\302\253.
of sequences
and
functions
11
This process of taking thelimit'inside'a
the functions wewill meet,but
in
valid.
is not
general
for all
is valid
function
use the ideaofthe limit of a sequence to form a
definitionfor thelimitofa function.
If we take a convergent
the
function
at successive
sequence x1,x2,x3,...and evaluate
terms of the sequence,we get a new sequence:
We can
f(x1),f(x2),f(x3),...
This
definition;*
ohen
most useM
need to
for
and
xn
to
yn
and
to
tend
not
the\" same
limit.
xi
POINT
KEY
If for
Worked example
t^>
1.6
definition
can
be used
limit
show that
a function
.
1.4
any sequence
{xn}
L is said
\342\200\224>
L, then
f(xn)
x3
Xi
such that xn \342\200\224\302\273
x0 we have that
to be thelimitofthe function.
how
demonstrates
this
f(x)
but
x\302\253
(W
which
ty) do
=
y
two
find
sequences
tend
mat
the function
say that
so yoo
do
To
point.
we
at a
not exist
does
too,
converges.
limit
a
showingthat
If this sequenceconverges
fo
the
limit
of
to
of
does not
not necessarily
it need
that
Note
be the casethat
]^>
for example,considerthe function
x^\\
\\2x
-
f(x)
= L:
f(x0)
<
X
O
\342\200\224
J.
existat a point xQ.
lim
a;->l
any
Taking
/(I)
= 3
f(x)
= 2
sequence
that converges
to x = 1,we
the function converging to f(x) = 2,
though /(1) = 3.
a sequence
Finding
how
the
function
is
that
will
lim/(x)
*\"*
the
Topic 9
-
Option:
Calculus
following
and choose any
way, that is
sequence
from the left
that
or
even
terms
successive
sequence, raises an important questionabout
tends to the limit. For example,
consider
sequence
= tanx
get
= 2,
and then seeing
that converges to a limit
behaves
as we evaluate it at
of the
f(x)
clearly
tends
from
how
the
to x
below.
the
function
= \342\200\224
in
'a
y
would
Here we
what
However,
in
the
following
have
=
x
tan
\342\200\224\302\273
\302\251o.
tan(x)
had chosena sequencethat
way, that is from the right or from
if we
71
x
to
tended
\342\200\224
\342\200\224
above?
+- X
Now,
we would
Clearly we
same
cannot
value
this
If we
for a
limits
different
this casewe say
limit.
the
that
function
limit
at
the
does not
of x.
1.5
POINT
KEY
two
have
and so in
point
exist at
have tan(x) \342\200\224>
a different
\342\200\224<*>;
find the limit of a function
sequence
that
from
approaches
at
point
below,
xQ by having
we write:
a
lim f(x)
If we do soby having
above we write:
a sequence
that approaches
from
lim f(x)
X^Xq~
from aboveand thelimitfrom
belowto existand coincide
for the limit to exist at that
If
do
then:
point.
they
We
both
need
the limit
lim f(x)- lim f{x)-
lim
f(x)
1 Limits
C
XJ^\\
+ d.
of sequences
and
functions
13
The limits from aboveand belowmay
asymptotes (as in
also be
the caseof tanx at
not
be
at
equal
71
x = \342\200\224),
and
an obviousdifferencein thelimitsat
there
might
where
a point
the
definition of the function changes.
orked
1.5
example
f-3x+
For
f(x)-\\
0
Here
function.
the
to find
need
just
determine
(if it
limf(x)
x^iJ
we see
that
clearly
doesn't
the
limits
show they're
and
below
and
2
exists).
idea to start by sketching*
a good
is often
We
x >
-1
\\x1
It
x<2
7
limit
the
from
exist
above\342\200\242
lim f(x)
=1
lim
= 3
different to
this
justify
x^2+
f(x)
Since
lim
x^>2
often
a good
piecewise
behaviour
t^ir
different points
at
in
the
v
Topic
9
-
not exist.
example,
Option: Calculus
\302\2432*
+\342\226\240
functions,
a limit
example
different
may also not existat
1.5,
formulae,
a point
if the
function oscillates very rapidly as it approachesthat point.
This
is rather like the case mentionedat the beginning
of the chapter
for sequences that oscillate without ever tending to a limit.
For
domain.
14
other
With
functions
understand
doee
function, such as that in Worked
by
consisting of two or morepiecesdefined
is said to be 'piecewise'.
idea to sketch
to
lim f(x)
f(x)^
that a
Note
u is
f{x)
lim
f(x)
= cos
\342\200\224
behaves
in the
following
way:
orked
1.6
example
does not exist.
Show that limcos
If
we
find two
can
tend to 0
which
for
but
sequences,
two different limits,
f(xn)
the
then
yn, that*
and
xn
Let
=
f(x)
coe\342\200\224
x
and f(yn)tend to
limit of f(x) does not
exist
The
an =271,471,671,../
sequence
has cos(an) = l and
371
71
i
571
nas
n
=\"2'~2~'~2~'\"\"\"
So
take
cos(bn)
xn =
the
Let
*n =
and
yn
->0
2nn
sequence
= 0 for all
\342\200\224
and
y
=
n
=
-\302\2730
(2n-1)7i
\342\200\224
Then
\\
f
1
= coe
f(xn)
1
J
\\2nn
= coe(2nn)
=1
lim f{xn)
n
all
for
= \\
xn^0
f
\302\256ut
f{yn)
\\
1
= coe
(2h-1)tt
2n-1
:C06
71
2
W
= 0
for
all
n
= 0
\342\200\242\342\200\242\342\200\242
lim
f(yn)
yn^o
As
\\\\mf(x\342\200\236)*\\\\mf
the
(y\342\200\236),
limit
does
not exist.
Just as for
Theorem
sequences,the Squeeze
orked
The
lim
x cos
x->0
difficulty
7
here
is the
1.
\342\200\224
inside
xcoe
the
0
However, since | cosx
\342\200\224
and
functions.
(-1 = 0.
so we cannotevaluate
xcos
for
1.7
example
Showthat
holds
the
| < 1,we can bound
potentially
use the
cos function,
limit
directly
the
function
.e.-|x|<
=1*1coe
\342\200\224
xcoe\\
XJ^\\
+ d.
\\<\\x\\
Squeeze Theorem
1 Limits
C
\\\\x\\
of sequences
and
functions
continued...
We can now
since
x^O
let
both
clearly
\302\261|x|
by the
Therefore,
^0*
e'mce
lim
and
= 0
(-|x|)
=0
lim
x^O|x|
we must
Squeeze Theorem,
have
\342\200\224
= =0
\\
xooel
lim
x^O
The
orked
algebra
also holds for
of limits
Vx
functions.
1.8
example
Evaluate:
(a) lim{x3-5x-8|
L
x^3
can
We
algebra
apply
of
(x +
lim
(b)
J
2)2
x^~2}V*2+4x+ 13-3 I
limits
[a)
straight0
lim
- &}= lim
- 5x
{x5
away
=
- 5
x5
-
x
lim
35-(5x3)-<S
=4
Since
lim
x^-2
+ 4x
Vx2
+13
lim
(x
+ 2)
=0and#
(x + 2)2
+ 13-3
(b)
(x +
undefined.
the
immediately as
However,
quotient
we can
is an
This
notice
that
\342\200\224
is
the
quadratic under the
is also
(x + 2) which
Now apply the
(x
+
2)2
top
multiply
{(x
+
square
of the
will
us
root
in the
algebra
give
numerator
of limits0
= ^/(x
+ 2)2
+ 9
(x
lim
X^>-2
J(x
= V0+~9
=6
V
-
Option: Calculus
\302\2432*
+ .
+
+
2f+9
+3
+ 2)2
4x + 13-3
+ 2)2
+ 3
3)
9}-9
So,
=
9
+
+9
(x + 2)2
2|Vx2 +
Topic
+
2)2
(x + 2)2U(x
lim
16
+ 2)2
L/(x
2)2
2)2+9-3
^](x + 2f+9
2)2+9-3
simplify
bottom by Vx2 + 4x + 13+3.
even better option when we
completing
+
+
7(x + 2)2+9 +
2)2
\342\200\242v/(x
by using the trick from
Worked example 1.2 and
and
7(x +
- 3 = 0 we can'tapply
of limits
algebra
(x
Vx2 + 4x
+9
+3
3)
5
+5
lim
I
however, the
Sometimes,
'
'0
problem of
\342\200\224
can
with
be removed
0
much less algebraicmanipulation.
orked
lim
Find
1.9
example
*->2
x*
x-2
we
Since
a
have
form
the
of
limit
\342\200\242
'0'
\342\200\224
,
0
apply the algebra
we can't
is
useful
Even
of the
though
2
the
Therefore
x =
And
2 is not
= x+
factorises
the graph at this point*
clarify what is happening
in
domain0
the
we have
the graph:
y =
y
we
can
eee
f(x)
that:
y
<2-4
-, the
function
limit
x-2
EXAM
that
to draw
to
2)(x-2)
x-2
x-2
away
we recognise
numerator
It
+
limits
of
straight
However,
x2-4 _(x
lim:
x^2 x-2
x^2
as
\342\226\240
= 4
clearly exists
HINT
It is
a good idea to see
always
be simplified before attempting
1. For eachfunction,
f(x)
\342\200\242
lim
f(x)
an
any
fraction
algebraic
other
methods.
can
1C
Exercise
lim
\342\200\242
if
find
\342\200\242
lim
f(x)
for the
given values of
(if they
exist):
x0:
1 Limits
XJ^\\
+ d.
of
sequences
and functions
(a)
(i) f(x)
(b)
(i)/(x)
r,
(ii)
/(*)
= \342\200\224\342\200\224
,x0=0
=
(ii)
/(*)
=-
(ii)
f(x)
-,x0=l
/(x) = ln(x-2),x0=2
(c) (i)
,
x<-\\
[x2 +Ax-5
.
x>-l
x2+3x-18
(e) (i) /(*)
3
/(*)
x2
x^5
\342\200\2245x
x =5
2
Let
/(*)
=
6
x<-5
-2x-\\\\
-5<x<-3
-x2+4
-3<x<l
-3
x=l
x2 (a)
4x+
the
Sketch
(b) Showthat
=
=2
y/x-2yx0
for
xn =
xQ
-1
= 3
for x0 =
5
l<x<5orx>5
5
of f(x).
graph
lim
3-x ,x0=3
x-3
x2-25
=
- for
x^3
x2-3x
=
x
x<l
[x3-2x2+3
x
(ii) /(*) = <L2.x-5 _
(ii)
I-Ax
= 2x-3,x0=4
not exist.
does
f(x)
lim f(x) ifit exists.
(c) Find x-\302\273-3
does not exist. Does lim/(x) exist?
(d)
/(5)
(e)
Find
(a)
(b)
all points,
List
the
following
x0, where lim
limits, if
f(x) doesnot exist.[11marks]
they exist:
2x2-3x + 4
x^-2 x2 - 2x +1
lim
x2
- 4x
+3
x2
- 5x
+4
lim
*->i
[6 marks]
a
x<\342\200\2242
2
x =
-2
Let/(x) = x2 +bx + c -2<x<3
x>3
2x+ 16
Given
that
lim
find constants a, fo
18
Topic 9
-
Option:
Calculus
\302\273\\
+l
exists for
/(x)
and
c.
all x g D, and
that
lim
f(x)
= -3,
[6 marks]
,
v
x-4x
For/(x)=
:
r
(a) Showthat
lim
=
/(x)
-1.
x->0+
(b)
Does
lim f(x)
(a)
Show
that
exist? Explain
(b) Hence find
+
x2~6x
.
lim
your answer.
9\342\200\224
=
[5
+2
J(x-3f+4
=
.
[6 marks]
*-*W*2-6x + 13-2
'
x
marks]
\342\200\224
2
x^2
For
function
the
\342\200\242
\342\200\224
\\x-2\\
f(x)
x=2
0
find
(a)
lim/(x)
x->2+
(b) lim
f(x)
X^2~
(c)
(a) Showthat
El
^
3(Vx + 8+3)
=
2+ Vl+ 3x
.
lim
*-* Vx + 8-3
'
^
x3 sin
lim
Find
V*+8-3
/(*)
find
Hence
(b)
[9
lim-
Determine whether
Worked
- .
\342\200\224
?' It
much
very
denominator
similar
is no
there
Unfortunately
get
functions
is of
the
= 0, i.e.of
by manipulatingthe
is not always
possible.
general answerto the question,'What
depends on
is lim-
f(x)
x\342\200\224>a
. Again
in
of
how the numeratorand
to 0!
problem
\342\200\224
= lim^(x)
the problem
q(x)
that
your
[6 marks]
lim/(x)
form, but this
a different
into
quotient
where
we avoided
case
that
In
justifying
the potential difficulty
highlighted
f(x)
lim-
with
up
fully
'
marks]
Rule
1.8b
example
ending
exists,
'
L'Hopital's
g)
A
[8 marks]
71
cos
answer.
is
.
x->0
J
marks]
+ 3x-2
= Vl
'
that /(*)
Given
El
[6
lim/(x)
= limg(x) = \302\260\302\260,
is that it depends
and denominator.
the answer
numerator
where lim/(x)
x\342\200\224>a
x\342\200\224>a
on the specific
1
r^
**.
Limits
of
sequences
and functions
of
this
functions
at
looked
Y^.
^^
option,
we
will
4E
use
Taylor
Given functions
^^
lim f(x)
x\342\200\224>a
f(x) and g(x) such
= lim g (x) = 0
x\342\200\224>a
thenlim^
series.
provided
or
x^a g(X)
=
the
(a) lim
*->0
x^a
limits:
following
sinx
lim
(b)
x_>e
x
Since the
form
limit
'0'
of
this
\342\200\224
we
use
1-lnx
x
\\nx
(c)
\342\200\224
function
e
is of
I'Hopital's
the#
lim
x
(a)J
K
Rule
anc\\
= 0
\\'\\me'mx
x^O
\\\\mx =
0
3y I'Hopital'sRule
_d_
(emx)
= lim-^r-
lim
dx
=
(X)
coex
lim
x^O
\\
_1
\"1
= 1
Since the
form
limit
'0'
of
this
\342\200\224
we
use
function
is of
I'Hopital's
the#
=0
(b) lim(1-lnx)
Rule
and
= 0
Wm(x-e)
x^e
3y I'Hopital'sRule
,
lim
1 \342\200\224
In jc
x^e x
=
(1-lnx)
ax
,
lim-^j\342\200\224 x^e
e
_^_
dx
(x-e)
_\302\261
x^>e
e_
1
= _\\_
e
20
v
Topic
9
-
answer
Option: Calculus
\302\2432*
+ .
I
either
= lim
lim^
exists.
the latter limit
that
that
lim
f(x)
x\342\200\224>a
orked example 1.10
Find
The
Rule
circumstances)
using Maclaurin
and
form?
1.6
L'Hopital's
an ^.
calculating
(in certain
them
POINT
KEY
method
for
cannot
of
where
alternative
we
provided by:
again
in Section
this
manipulate
be
will
type
deal with limits likethesewhen
the quotient into a more helpful
do we
So how
Limits of
g\\X)
g (x)
x\342\200\224>a
= oo
is
continued...
the
Since
\342\200\224
we
form
is of the4
function
this
of
limit
use
lim
\\nx
x^<=\302\260
(c)
Rule
I'Hopital's
3y
x =
lim
and
=
OO
00
x^<=\302\260
Rule
I'Hopital's
d ,
v
\342\200\224
lim
\\nx
dx
\342\200\224
= lim
d ,
-W
x^
X
(Inx)
x^>\302\260\302\260
v
dx
=
X
lim
x^\302\260\302\260
1
lim 1
=
=0
X
HINT
EXAM
You can only
form
the
apply I'Hopital's
'0'
'00'
\342\200\224 \342\200\224
. It
or
valid
these
in
that is of
cases.
to use I'Hopital'sRule
be necessary
it might
Sometimes
is only
\302\260\302\260
0
a quotient
to
Rule
more
once.
than
orked
1.11
example
lim
Evaluate
the
Since
.
of
limit
this
'0'
\342\200\224
we
form
is of
function
use
the#
Rule
I'Hopital's
=0
lim
x^O(1-cosx)
and lim(x2) = 0
x^O
3y
I'Hopital's
Rule
,.lim 1-cosx
x^>0
,.'im ^0-ooex)
X2\342\200\224-x^>0
e'mx
\342\200\224
lim
Zx
In
to evaluate
trying
that
it is
also
this
limit,
of the form
we
'0'\342\200\224
so
realise4
we
can
0
apply I'Hopital's
and
Rule
=0
lim (sinx)
x^>0
again
Applying
lim(2x)
= 0
Rule again
I'Hopital's
d
\\-coex
lim
x^o
x^O
(e'mx)
d
dx
=
(Zx)
coex
lim
x^O
\"2
1
C
XJ^\\
+ d.
Limits
of
sequences
and functions
21
met
We
limits
the form
oo
of
\342\200\224oo'
for
sequences
First it is
- oo'
(oo
of the form
is the
had
oo \342\200\224
oo = \342\200\224
2
1.12
example
the following:
Evaluate
(a) lim(xlnx)
this
it
(a)
\\y\\x
x\\nx-
to write
need
we
here
a quotient and
conditions
for the
as
product
hope
\342\200\242
-1
so again think
Rule. However, to
I'Hopital's
apply
form
the
of
limit
ex
and
Ox(-oo)
of
x->0 V x
have a
We
lim
(b)
the
that
rule are then
of the
limit
The
now of
we can
is#
\342\200\224
and
form
the
satisfied
quotient
By
Rule:
I'Hopital's
so
oo
Rule
.
= lim
x^O
x\\y\\x
lim
apply I'Hopital's
x^O
x
d I 1 \\
dx\\x)
j_
=
lim^p
x^O
I
X2
X2
,.
= lim
=
x^O
X
lim
\342\200\224x
x^O
= 0
limit
The
this time is
'oo-oo'
and
obvious way of
combine
lead
may
form#
ex
-1
x(ex-l)
ex
we
proceeding
is of
of
the
Rule
the
quotient*
single
form
'0'
\342\200\224
so
and
0
I'Hopital's
Rule
is
-x
using
I'Hopital's
limit
-\\-x
xex
the two quotients. This
to some cancellation
or to the possibility of
The
X
no other
with
ex-\\-x
1
of the
needed
I3y
rule:
I'Hopital's
( d
,lim , 1
ex
\342\226\240o\\x
1
dx_
x^o
-1/
{e*-\\-x)'
d
(xex
V dx
=
Topic 9
-
Option:Calculus
0
Z&
+ 0,.
,.,
-
x)
ex -1
lim
xex
\342\226\2406
ex +
22
form
\342\200\224 \342\200\224
or
.
0
<^1
1
orked
of the
'
0 x oo'
necessary to rearrange theseexpressions
which
a quotient
limits
Worked
1.2
example
or
into
<^1 commented that this
was
not generally
0. Indeed
rule
we
when
be used to find
can also
L'Hopitals
\342\200\224
1
\302\260\302\260
continued...
The
limit
is still
end
so we
fa\342\204\242
= lim
of the#
need
~
'\302\247'
I'Hopital's
Rule
again
1
1+ 1+ 0
1
~2
Exercise ID
1.
the
Evaluate
using
following
-e
eA
(a) (i) lim
*^\302\260
lnx
(ii) lim
*-\302\273!
X
sinx
TT2 \342\200\224
V:
(ii) lim
*^3
(i) lim
(b)
Rule:
I'Hopital's
- 1
x2-9
sin(x-3)
cosg)
+X
X2
(c) (i) lim
*^\302\260
(ii)
tanx
2.
the
Evaluate
*^\302\260
(b)
^VJC-1
I'Hopital's Rule more than
once:
(x+ l)2-e2*
lim
(a)
by using
following
lim
1-cosx
(In*)2
lim
*-*
x->l
x3 + x2 -
5x
+ 3
1-cosx
lim
(c)
x->0
*^\302\260
3.x2
tanx-x
lim
(d)
*^\302\260
x-sinx
(e)
lim
e*
X\342\200\224>oo
X
6.x
1 +sin
(f)
lim
>-l + cos4x
4
3.
into a
By rearranging
quotient and applying
I'Hopital's
Rule,
find
the following:
(a) limx2lnx
x^O
lim
(b)
- tanx)
(secx
X^>\342\200\224
2
\342\200\224
(c)
limf\342\200\224
\342\200\224
l
x^>1\\x
the following
Find
,
x
(a)
lnx
..
lim-
limits:
x-cosx
*^\302\260
x
+ cosx
x+
(b) lim
*^\302\260
sinx
x-sinx
x-cosx
(c) lim
*^\302\260
x +
sinx
[5 marks]
1
S?i>* + d.
Limits
of
sequences
and functions
f
23
'a
is wrong with the following
'proof?
x3+x-2
3x2+l
6x
..
..
= lim
= lim\342\200\224= 3\342\200\236
*->i
*-*
2
x2
3x+ 2
2.x
3
(a) What
..
lim
*->i
(b) Find the correctlimit.
l-cos(x5)
lim
BFind
x->0
Find lim
Kg
wmm
x->0
[6 marks]
x
cos2
x^l
Q
[6 marks]
2x 10
C\302\260S2(3X)
Find lim
I
Once we have
[8 marks]
functions
Continuous
that
established
the limit
continuous
expect: if the function doesnot have
thenit is
(x-l)
there
so x
A function
is not
24
\302\253v
Topic
9
-
Option: Calculus
*-!\302\273*\"\342\200\242+,\302\273
to
there.
\342\200\224
gap'
is
as you
or cjump'
as the limit
(as long
is
might
at a point
exists in the first
of continuous as beingableto sketchthe
our pen off the paper.
taking
function
defined
a
exists
function
can think
without
function
y
there
continuous
We
Fora
of a function
x0, the next question is whetherthat
at x0. This idea of beingcontinuous
a point
place).
marks]
[8
sinx
\\x
Find limx2 arctan(x2)
at
marks]
[4
j
For
being
continuous
be
example,
such as /
continuous
at x
continuous
= 1is not in
certain point, it has to be
no senseto talk about
= 1,as it has an asymptote
domain.
its
[x)
at x
at a
it makes
=
= 1.
|2x
x^l
that
<
I
O
X
\342\200\224
J.
we met
in Section
1C
'a
x = 1is in
Here, although
= 2), there
(limf(x)
'
x^l
without
function
This
our pen from
taking
not being
of there
requirement
and we
a 'jump'
is
is stated moreformally
continuous
KEY
POINT
A
function
exists
the limit
and
domain
the
cannot draw the
the paper.
a 'jump' for
to be
a function
as:
1.7
at the
is continuous
f(x)
=
lim/(x)
point
Notice
that,
if
x0
to this
**
/(x0)
y
I
and the
limit
the
I
Both
|
be continuous
'
The
I
all
function
We now
causing
to
f(x)
it
to be
,i
continuous
there.
is said
of its
points
value f(x0) must existfor
- ,.
continuous if it is continuousat
is continuous
domain
(which
according
definition,
1
is
function,
domain.
of
at all points
is x ^ 1).
Sometimesan apparently
to results
definition
leads
contradict
our
here
and similarly
a
actually
as
its
sensible
that
intuition.
look at further examplesof the ideaof'jumps'or gaps'
with reference to the formal definition.
discontinuity,
orked example 1.13
where
Determine
the
following
are continuous.
functions
-2.x-6
(a) f(x) = \\x\\
(b)g(x)
=
x<\342\200\224\\
jc = \342\200\2241
2
x2
+2x-?>
x>-\\,x^\\
x-l
We only need to
where there may
consider
be
any points0
With
a problem.
| x | the only such point is x = 0
{a)
\\x\\
whether
lim
I
x I exists
*
lim
x^O~
lim |x|
|x|=
limf(x)
f(0)=
V 7
x^O
*
And
\\0\\
hence
= 0
= |0|
therefore
\\x\\\\e continuous
continuous
XJ^\\
+ d.
at x = 0 and
everywhere.
1
C
=0
=0
eo Hm|x|
x^O
and
here too.
x^O+
lim
whether
continuous
is clearly
x^O|x|
Check
x>0
continuous
\342\200\224
iAiA- ^> i_y
I .A-1
so
Check
= x
eo is clearly
Limits
of
sequences
and functions
25
continued...
As with the more
functions that we
The
the
idea to sketch
a good
but
just
start
as
So, we have:
to
V
=
f(x)
we
limits,
whether
by considering
+ Z)
\342\226\240\342\226\240x
x>-\\,x*\\
-1
looking for
when
3)(x-1)
x-\\
2x-3
+
sketch is the quotient
+
x-\\
first.
function
awkward piece
potentially
only
x2+2x-3_(x
(b)
complicated
piecewise1
met in Section
1C, it is
be
can
it
simplified.
need
We do
note
to
that
however,
here,
f(l) is not defined
J\\
Again, we only
to consider
need
points where there may be
Here these points are
a problem.
x = -1
is continuous
Clearly f[x)
any*
for:
x<-\\
1
and
x <1
-1 <
x>\\
the
Although
'jump' so the
is a
there
at x
exists
limit
= -1,
*
isn't
function
x =
For
lim
-\\
= -4
f(x)
x->-1-
continuous
butf(-1) = 2
limf(z)*f(-1)
Therefore
the
Although,
again,
from above and
this
time
as
it
is
'gap'
not
at
below
from
is a
there
in
the
defined
and
so f{x)
For
x =
both1
exists
limits
but
EXAM
HINT
As
the
a point
xn
sequence
The following
26
v
Topic 9
-
Option:
Calculus
\302\2432*
+\342\226\240
defined.
that
of limit, the
definition
is often
continuous. For
showing
= 4
f(x)
is not
continuous at x = \342\200\224\\
Therefore f(x) is not continuous
=1
with
lim
\\
f(1)
function
at
is not
x->-1
x = 1,
at x
x-\302\273-1
used
this
to
show
purpose
\342\200\224>
x0 such
that
it is
f(xn)
definition of
that
x = 1
at
>
continuity
a function
only necessary
is not
to
find
a
-\302\273
f(x0).
example demonstrates a standard methodfor
is not continuous at a point.
a function
orked
1.14
example
Show that:
cos|
/(*)=
\342\200\224
x>0
X)
x<0
0
is not
continuous
- 0.
like
to find
would
We
a sequence0
1
Take
Then
that f(xk)^f(0).
such
-^0,
xk
at x
2nk
f(x) cannotbe continuous
Then
xk ->0
but
f(xk) =
1
coe
f\342\200\224i
V2nk)
=
coe(2nk)
=1
I.e. xk^>0
eo f{x)
In
Worked
exist,
so we
that limcos
showed
1.6 we
example
*^\302\260
knew it could not be continuous
at
example is included to show that
that it is discontinuous at x = 0 than it is to
limit
does not exist.
the above
1. Determine
the
whether
help
(i)
/(*)
=
x2-l
x +
(ii) /(*) =
=
/(*)
(i)
not
easier
However,
to prove
the
[x2-\\
<
are continuous
at
some cases).
atx = 2
at*
=
at x
1
=
-2
1
x*-l
dXx-
x + \\
-1
x = -1
[-2
(c)
f(xk)-\302\273f(0)
continuous.
\342\200\224
0.
the graphs in
to sketch
= x3-3x2+5x-2
f(x)
(ii) g(x) = x4-16
(b)
x
does
establishthat
functions
following
the given points (it may
(i)
\\x)
B
is not
IE
Exercise
(a)
it is
\342\200\224
but
x2+2x-15
at
x =
4
x-4
(ii) g(x)
=
x-3
at
x2-x
^x
o
\342\200\224
3
+ 12
1 Limits
C
XJ^\\
+ d.
of
sequences
and functions
(d) (i)
(e)
x>2
[Ax-I
f(x) = lx + 3
\\
= 2
atx
\342\200\236
\342\200\236
x<2
x<-l
(ii) g(x)
= - \\x2-l
\\x-l
x>-l
(i)
f(x)
= t~2x+l
atx = -
(ii)
g(x)
= secx
atx =
x2-
at x -
-1
71
\342\200\24216
(f)
(i)
/(*)
x^4
'
x-
=
4
x=4
8,
X
(ii) g(x) =
x2
= 4
dXx
-1
atx
+x-2
x = l,-2
3,
=
l
For
X
x<0
1
x
1+ x
/\302\253=
x =
if they
(ii) x->0~
lim/(x)
lim
(iv)
x->0+*
f(x)
lim
x->l+
lim f(x)
(v)
(b)
At which
the
Find
limits existand find
the following
them
do:
lim/(x)
(i)
(iii)
l
x>l
whether
Determine
0
0<x<l
3
2x2
(a)
=
(vi)
points is /(x)
^0
of k
value(s)
for
/(x)
lim/(x)
[7 marks]
continuous?
which
the
function
x<\\
foc+2
/(*)=
x>l
kx
at x
is continuous
(a)
that
Given
= 1.
lim/(x)
x^xQ
W+fot
+ 13
the
constants
v
Topic 9
-
Option:Calculus
\302\2432*
+ .
for
the
function
-4<x<l
a and
b.
(b) Where is/(x) continuous?
28
xgD
x>l
[x + 9
find
for all
x = -4
7
;
exist
x<-4
f-2x-3
-/v
[4 marks]
[6
marks]
the function
that
Show
xcos
=
f(x)
\342\200\224
x^O
,
\\xz
x =
[0,
is continuous
x =
at
0
0.
[7 marks]
(jQFor
/(*)
X(
X
=
x<\302\243Q
[0
show
that
Once
then
that
0.
of a function exists at
the limit
that
established
have
we
x =
at
functions
Differentiable
a point
continuous
is only
/(x)
x0 and that the function is continuousthere,we
want to ask whether the function can be differentiated
might
at
point.
core course you
In the
first principles
should
and usedthe definition
of
from
at a
differentiation
studied
have
the
derivative
point x0:
We
also
can
an alternative
find
sometimes
work with, by a simplemanipulation
useful to
be more
definition, which will
of
the
variables.
If we
let h \342\200\224
x \342\200\224
x0, then
/(*\302\273+(*-*\342\200\242))-/(*\342\200\242>
Um
rW=
we get:
=1|m/(*)-/(*\342\200\242>
X-Xr,
*^*0
KEY POINT
1.8
r
The derivative
of
a function
/'(*0)
x0
is
= lim f(x0+h)-f(x0)
by
equivalently
/-(x0)=lim/W\"f(jCo)
*^*o
if the
If it
point
by:
given
or
at the
f(x)
X
-
Xn
limit exists.
does
not, then
the function is not differentiable
at
that
point.
1
-i.**^
Limits
of
sequences
and functions
that
function
any
continuous
at
that
point:
is differentiable,
If f(x)
Then
definitions of continuity and differentiability
that is differentiable at a point
x0 must be
from the
It follows
we have
lim
ff(x)-f(x0)^
x
lim
(f(x)
*->X0
-
lim
f (x0))= *->X0
=
\342\200\224
r{f(x)-f(x0))(x-x0)^
JC
JCq
lim (x -
7W-/U)
lim
x^x0
=
/'(*o)
xn
X^Xq
x0)
f'(x0)xO
= 0
.\\
lim
X^XQ
that
It follows
cannot
=
f[x)
and so f(x)
f(x0)
a function that
be differentiable
happen
\342\200\242
There
At
may
a point
such
in fact it
the
at
due to
\342\200\242
The
30
v
Topic
9
-
Option: Calculus
\302\273\\
+l
be a
tangent
it is
/'(*),
may
could
a point
for
a function
be differentiable
that is
there. This
can
factors.
'sharp point':
not clearhow the tangent
in
infinitely
said not
is simply
In this case Xq
function
two main
could be drawn
derivative
possible
not to
a point
is not continuousat
at that point.
However,it is alsoperfectly
continuous
is continuous.
many
to exist at
should
different
this
be
drawn;
ways and
point.
be vertical:
not
so again
be in the domain of the gradient
the derivative cannot existat this point.
so
1.9
KEY POINT
For a function f(x) to be differentiable
f(x)
must already
must
\342\200\242
f(x)
\342\200\242
the
tangent
as just
well
As
at x0 must
may
continuous
to
exist.
not
does
you will need
h
limits
the
when
above
1.10
If a
continuous
the
limits:
function f(x) is differentiable
/z->0_
fc->o+
\\x
and are
at
orked
xQ
then
h
equal.
not exist,orif they
is not differentiable.
limits does
of the
the function
x =
and Um f(x0+h)-f(x0)
lim f(x0+h)-f(x0)
If one
will mean
this
functions,
one way to do this is to find
and from below.
that
POINT
exist
not be vertical.
points where a function is not
also be required to use the definition
lim\342\200\224
fc-*>
\342\200\224>
0 from
KEY
to f(x)
f(x + h)-f(x)
that
show
Remember
h
a 'sharp point' at x0
not differentiable.
In the caseof
to
not have
you
it is
that
lim f(x)
hence
(and
x0
x0:
exist)
identifying
differentiable,
show
at
be continuous
must
\342\200\242
a point
at
not
are
then
equal,
1.15
example
Show that:
i
is not
(a) f(x) = {x\342\200\2241)3
(b) g (x)
need
We
=
I
x I is not
does
not
at x
differentiable
that
to show
at x
differentiable
so
first
= 0.
-\342\200\224\342\200\224
\\\\m\342\200\224
exist,
-1.
+
(a) f(1
+ h-1)3
h)-f(1)_(1
-(1-1)3
find
we
+ /i)-f(D
f(l
' h3_
h
h
1
It is
now
clear
and so
that
x = 1 is
limit
the
in the
not
the
We
to show
need
that
|im
is not finite*
9{*
+
f(1 +
lim\342\200\224
h)-f(1)^ = 00
of
domain
function
.'.
f'(x)
h)-g(x)t
And
(b)
hence
g(x
f(x)
is not
+ h)-g(x)
differentiable at
_\\0
\"
so first
g(x
we
find
+ ti)-g(x)
h
1 Limits
C
XJ^\\
+ 01.
+ h\\-\\0\\
h
h
does not exist,
x=1
of sequences
and
functions
continued...
This time the behaviour of
so obvious, but we suspect
we
=
-j\342\200\224
lim (-
h^0~
from
limits
=
So
limits
of these
each
find
lim
-r-=
lim
limit
being different.
below
and
above
that the
due to the
exist
not
might
is not#
limit
the
And
\"1
similarly,
lim ~r-=
=
g(x) = \\x\\
wise
a piece
the
that
1
\\h\\
-\342\200\224doee
not
exist
differentiate
is not
function,
and
at x
=0
we may want to join the
is differentiable. For
J\\
function
resulting
example,if the function
differentiable so that
lim 1
lim
Therefore
piecesso
-r
lim
=
If we have
-D
displacement,
represents
the
velocity
it needs
to be
is defined.
orked example 1.16
b so that
a and
constants
Find
the function
x<3
fin x
/(*)=\342\226\240
for all
is differentiable
a
For
the
that
differentiable, it*
continuous. This means
be
first
must
limits
x
when
and below
\342\200\224\302\273
3 from
the
be
must
above
and
same
equal to f(3)
Also, the
derivative
at
the
has
limits
definition of the#
to exist. We need to look
and below 3 again
above
limit
As x = 3 is
domain, the limit
3
x >
b
x > 0.
to be
function
ax+
f has
lim
in
the
derivative
first part
below
of
In
is
x at
of the
just
the
x =
3
So
a(3) +
b =
ln3
^>3a + b = ln3
Also,
f(3 +
lim
h)-f(3)
\342\200\224(\\nx)
'3
dxy
(ax +
b) = a^aaex^>'5
Hence,
1
a = \342\200\224
3
.b =
3fl-ln3
= 1-ln3
v
-
Option: Calculus
\302\2432*
+ .
= 3
above:
dx
9
lim
f(3
+ h)-f(3)
h^3H
From below:
1
d /, x \342\200\224
=
at x
_d_
Topic
=
h^3\"
From
32
= f(3)
lim f(x)
f(x)=
in the
from
to be continuous:
>
IF
Exercise
each of the followingfunctions and then find
1. Sketch
are any)
(if there
(a)
where the function
is not
all
points
differentiable.
= Ux-2 + 3
f(x)
(b) /(*)
=
x*l
{ x-1
x =l
[l
x2-9
(c)
=
/(*)
\342\200\224
3
x
x =
6
3
= +
(d) /(x)
|2x l|
(e)
(f)
,
s
/M
=
x<0
(x2
x>_0
{*
=
/(*)
1
7
x-lf
= y fx
(g) f{x)
(a) Sketch the function:
x<0
0
/(*)
=
0<x<4
y/x
X
-+
the function
where
Determine
(b)
(c) Determinewhere
the
is continuous.
is differentiable,
function
fully
answer.
your
justifying
x>4
1
[7 marks]
(a) Sketchthe function
x<0
2
=
/(*)
-x2 + 6x +1
0<x<2
x>2
x3+l
the function
where
Determine
(b)
(c) Determinewhere
function
is differentiable,
answer.
your
justifying
the
is continuous.
fully
[7 marks]
Determine where the function
(x3+ 3x2-x-3,
f
(x)
=
x2
<\\
+ 2x
-
31,
8x-10,
is
differentiable.
x<-3
-3 < x<3
x>3
[6
marks]
1 Limits
C
XJ^\\
+ d.
of sequences
and
functions
33
4
The
for
defined
function/is
%
e K.
by:
x<2
[x3
= \\
f(x)
v }
J
Find the values
x>2
that/is differentiable
c so
b and
of
+ c
+bx
[-X2
x e
all
for
D
.
[6 marks]
For:
xsin\342\200\224
x^O
x
/(*)=
x=0
0
(a) Showthat
(b) Find /'(*)
(c) Show that
[7J Assumingthat
Q
Showthat
at x
is continuous
f(x)
x *
when
0.
f(x) is not differentiable
x =
at
0.
/^10marks]
that:
is not differentiable at
= sin Lx
/(x)
show
exists,
/\"(a)
= 0.
x =
0. [5
marks]
(a) Given that
=
/(*)
{^faxTb
x<\\
\\x
x>\\
= 1, find a and b.
State the range of values of x for which/(x)
at x
is differentiable
(b)
is differentiable.
[7 marks]
x2sin\342\200\224
=
jj For/(*)
x=0
0
(a) Showthat f'{x)
(b) Find /'(0).
Find
(c)
Using
x ^
f'{x)
is not
for all
0.
continuous
in
are two
understanding
Topic9
*\\t>
^
-
Option:
Calculus
? l^v*-
[13
marks]
and the Mean
Theorem
important results that follow
of continuity and differentiability.
function on an interval
is continuous and differentiable
a non-constant
that function
34
- 0.
the previous question,constructa
/(*)
/'(()) >0 but there is no interval
which /(*) is an increasing function.
Value
For
at x
for which
Rolled Theorem
There
x.
the function in
function
]-a,a[
exists
when
f'{x)
(d) Showthat
(ll.)
x^O
x
from
[a,b],
and
our
if we
know
it starts
and
least one
same
the
at
finishes
y-value,
then it is clear that
in
the
turning point somewhere
must
there
be at
interval.
= f(b)
/(\302\253)
KEY
I
|
1.11
POINT
Theorem
Rolle's
For a
function, f(x\\ that is continuous
and differentiateon ]a,b[,
iff(a) that
then
f(b)
must
exist a
interval [a,b]
point c e ]a,b[such
= 0.
f'(c)
Obviously continuity
have a
there
on an
is importanthereas
could
we
otherwise
function such as:
y
/(I)
for which
However,
a continuous
1/
= /(3)
f (3) J
1)=
f(a) - f(b)
we
also
need
function
but
J\342\200\224
there
is no
f'(c) - 0.
otherwise we could have
point c where
differentiability,
that does not have
a turning
point,
such as:
that a
Remember
such
sharp point
as
in
modulus
this
is not the
function
same as a
/(l) = /(3)
turning
turn.ng
At a
- 0 but
point, fix)
point.
at
point
sharp
the
t =-2) the derive**
does
not
The mostcommonapplication
of Rolle's
Theorem
a maximum numberof possible
rootsofa polynomial.
is to
exist.
establish
1 Limits
-i.**^
even
of sequences
and
functions
orked
1.17
example
Prove that
the
polynomial
= x3 + 3x2 + 6x +1 has
f(x)
Since f[x) is continuous, we only
an interval on which the
changes
be a value,
to conclude
f(x)
of
sign
x0,
= 0, i.e. at leastoneroot
an
need
now
one root,
to show
so suppose there are two
look
This suggests that
to conclude
Theorem
point
in the
produce
might
we
and
for a
contradiction
could
apply
Rolle's
there is a
interval
the
Suppose that thereexistsa
is only*
there
that
X\\
\342\200\242
f(x0) =
we are
root,
f{x<[)
is continuous
f(x)
and differentiate,
exists ace ]x0,x,[
there
Theorem
Rolle's
jsuchthat f'(c)
which
Jx^x^,
Since
by
second
> X0
Then
turning
contradiction
exist
.'.
f (x0)
We
f{x) is continuous, there must
x0 g ]-1,1[ such that f(x0)= 0
has
at least one root.
f(x)
Since
that
such
interval
that
in
= 11>0
f(1)
must
there
that
f(-1) = -3<0
need#
to find
root.
one
exactly
=
0.
However,
f'(x)
looking for
= Z)X2+6x
+6
= 3(x2+2x+
=
3[(x
+
2)
1)2+3]*0
Thereforewe have
there is onl / the
a contradiction
one
root
and so
x0.
Rolle's Theorem can be useful
for these
kinds of question but
it is rather restrictiveto requirethe function
to have the same
value at both end points of the interval. If we removethis
restriction
then
we see that there must still be a point c e
at
but
the
which
now
of
This leads
is parallel
tangent
course that tangent
to the following
Value
generalisationof Rolle'sTheorem.
Theorem
For a function, f(x),
that
is continuous
[a,b] and differentiateon ]a,b[,there
ce]a,b[ such that
Topic
9
-
Option: Calculus
]a,b[
to f(b),
\342\200\242
12
KEY_PPJNTJ
Mean
to the linejoining f(a)
does not have 0 gradient.
f'(c)
= ;f(b)-f(a)
b
\342\200\224
a
on an
must
interval
exist
a point
from the diagramabove,
clearly follows
result
This
prove it formally using Rolles Theorem.We must
function based on f(x) that fulfils
the conditions
points of the interval
for
needed
Let
g(x) = f(x)-
Clearly
g(x)
Rolle s
( f(b)-f(a)
a
choose
at the
end
Theorem.
(x-a).
b-a
and differentiable
is continuous
can
we
but
since f(x) is, and
moreover,
=
g(a)
f(P)-f{a) (a-a)
f(a)-
b-a
= /(*)
and
f(b)-f(a)
g(b) = f(b)=
(b-a)
b-a
f(b)-{f(b)-f(a))
= f(a)
that
is,
=
g(a)
g(b).
Therefore, by Rolle'sTheorem,thereexists
such
that
ce
a point
]a,b[
= 0.
g'(c)
So
a
^r(c)=m-fM
Therearea
Theorem
b-a
of
number
of the
applications
possible
Mean Value
(MVT).
orked
Prove that
It looks
example
|
1.18
- sinfo <
|
sina
like we
can
just
|a
-
apply
b
|.
the
on f(x)
Since f
MVT*
= sinx
(x)
= e'mx
is continuous and
differentiablefor all x e R, by
Theorem there exists c e ]a,b?[
f{b)-f{a)
Value
that
such
= f\\c)
^-a
e'mb
Mean
the
\342\200\224
e'ma
\342\226\240
=
coec
b-a
We
now
to take
need
and
mod of
observe
that
both
\342\200\242
sides
|cosx| <
e'mb-e'ma
b-a
1.
e'ma
-
1 Limits
XJ^\\
+ d.
= |cosc|<1
e'mb\\<\\a
-
b\\
of sequences
and
functions
orked
If /(*) is suchthat
smallest
the
find
1.19
example
f'(x) > -2
so
We have information
of the interval
\342\200\242
point
to find out
want
that
know
the other
be
Mean
by the
so
ce
exists
there
]2,7[
and hence
Value
Theorem,
]2,7[ such that
f(7)-f(2)
useful
x e
all
for
is differentiate
f(x)
continuous and
end
at
something about the function
end point, so MVT might
derivative exists
Since the
we
continuous.
one
on
and
f (x) is
know
we
hence
and
/'(*) > -2 forallx e ]2,7[,
value for/(7).
possible
differentiable
-4 and
=
/(2)
=
7-2
f'(c)
f(7)-(-4)=nc)
Use the fact
that
and
> -2
P(c)
for
Since f'{x)>-2
\342\200\242\342\200\242
all
xe]2,7[,
rearrange
5
+ 4>-10
=> f(7)
=>
i.e. the
The Mean
Theorem
Value
-14
f(7)>
can
smallest possiblevalue
used in more
also be
for
f(7)
is -14
practical settings.
orked example1.20
and travelling belowthe speedlimitof70 mph passes
24 miles
the same car passes another police officer
further
police officer at 12:00. At 12:20
the
it was
at
less
than
70
The
driver
is
motorway; again
mph.
pulledover
along
travelling
second
and given a speeding ticket.
policeman
car
A
Use
how the policeknew
to show
Theorem
Value
Mean
the
the motorway
along
driving
had
the
driver
the
function
exceeded
a
by
the speed
limit during his journey.
that
a function
need
We
will
car's
the#
model
journey
Let f(t)
position
We need
and
time
to
be
measured
in
distance moved from
t = 0 (at 12:00) to
when
take
can
position
i.e.
hours\342\200\242
initial
be
0,
be
function
this
Assuming
t hours
time
at
differentiate,
there is a time
=0
f(0)
by
24-0
-0
=
Option:Calculus
c
ZL*>
+0,.
at
T
3
-
t0
fl-J-f(o)
n*o)=
Topic 9
the
72
-0
that gives the care
after 12:00
to be
Mean
which:
the
continuous and
Value Theorem
continued.
f'(t) = v(f),
car's
the
time
some
at
context the Mean Value
Theorem
have attained its average speed.
In this
Note:
must
car
is the cars velocity
t0 and so the car was
However, f\\t0)
velocity
72 mph
Its
during
travelling
journey.
that at some
is saying
at
point on the journey,
the
Exercise 1G
1.
the
Find
+ 8 xe[0,5]
(i) x2-5x
(a)
(b) (i)
the given intervals
xg[0,7t]
cos(2x)
+ l xe[-l,l]
(c) (i)x3-x
whose existenceis
Theorem:
Rolle's
by
guaranteed
2.
of c in
value(s)
(ii)
5-2x-x2
(ii)
sin
(ii)
y/lx-x2
xe[-5,3]
-
xg[0,87t]
xe[0,2]
value(s) of c in the given intervals, whose existence is
guaranteed by the Mean Value Theorem:
x2 - 5.x + 3
(a) (i) 3x2+x-7
(ii)
xe[l,3]
xe[-2,2]
Find the
(b)
(i)
(c)
(i)x3-3x
The
function
(ii) tanx xe
0,-
xe
sinx
2
+ 4
(ii) 7-Vx
xe[0,2]
for x
is defined
f(x)
0,3
xe[4,36]
e [1,10]and
has
= 12,
/(l)
and
f'(x)<2.
Usethe
for
possiblevalue
[4 marks]
= Ax2
\\
Let fix)
(a)
Show
(b)
Why
travelling
at
3
some
acceleration
xe [a,b]
show that:
45
[4 marks]
b-a
)
Jx .
this
a car
at
2
there
that
does
9 a.m.
+ C,
+Bx
fq + b}_f(b)-f(a)
f
that
the largest
/(10).
For f(x)
At
to find
Theorem
Value
Mean
does not
not
contradict
is travelling
mph.
exist a c e ]l,4[ such
at
Use the
point between
was exactly 24
that,
f'{c)
Rolle's Theorem?
- 0.
[4 marks]
39 mph.
15 minutes later it is
Mean Value Theoremto show
9 a.m.
mh~2.
and 9.15a.m.,the
car's
[5 marks]
1 Limits
C
XJ^\\
+ d.
of
sequences
and functions
In a
race, a hare and a tortoiseboth
unexpectedly,pass the
some
at
that
same
x3-
and,
as well.Show
they weremoving
solution of the equation
only real
0.
[6marks]
two positive
are exactly
there
that
^ (a)
- 7=
+ 9x
ex =
real numbersx
3x.
such
[9 marks]
that
Given
(b) Use
the
with
[6 marks]
1 is the
x \342\200\224
3x2
Prove
that
same time
same time
speed.
Q Prove that
2
the race,
during
point
at the
start
at the
line
finish
= Vl
f(x)
the Mean Value
+ x,
show that
-
c>
all
for
0.
that
to prove
Theorem
<
/'(c)
\342\226\240V*
y/l
< 1 + \342\200\224
for
+ x
0
x >
[6 marks]
Showthat:
| cos2a (a)
sin2 b
a +
sin2
ex >
1+
In
2,
x + \342\200\224
for
2
a
(l + x), use
ex > 1+ x.
the
that
14.] (a) If f(x) and g(x)
g'(x) ^ 0, show
(b) to
n. Prove that
are
that
a function
h{x) f{x)
kg(x)
constant,there
exists
k is
where
of the form
some appropriately
such
ce]fl,fc[
and
and differentiate
continuous
by using
\342\200\224
there are at
= 0.
p(x)
with
xe~R
\342\200\224
-
marks]
0.
of degree
polynomial
n
[7
place an upper boundon
answer to 3DP.
[11 marks]
your
giving
x >
to part
answer
(13.) Let p(x) be
most distinct
to prove
argument
x2
(c) Useyou
that
to prove
Theorem
a similar
Use
/ (x) = ex
the function
By considering
Mean Value
(b)
- cos2
b\\<2\\a-b\\
chosen
that:
f'(c)_f(b)-f(a)
g'(c)
(b) If/(a)
=
g(a)
= 0,
g(b)-g(a)
prove that
lim/M=lim\302\243M
providing thelatterlimitexists.
[15.]
Establish
the
fully justifying
number
of roots
your answer.
of the
equation cosx = x
(xg!)
Summary
\342\200\242
Limits
algebraically; that is if the sequence{an}converges
as expected
behave
the sequence {bn}to a limit b,
+ qbn)-
lim(pan
pliman +qlimbn
= liman
lim(anbn)
If the
lim ^
^
{an} diverges,
(<* ^
necessary
usually
limits
finite
have
both
dividing through by
\342\200\242
If a
c y
to manipulate an into a form where the numerator
so that the algebra of limits can be applied.
This
can
the
power of n.
highest
limit, then that
and denominator
be achieved
often
between two other sequences,both of which converge
and must converge to the limit too.
is sandwiched
sequence
to the
sequence
have
if we
Theorem:
an<bn<cn
same
sequences
forallneZ+
and
[an },{&\342\200\236}
{cn}
andliman = limcn
that
such
= L<\302\260o
lim\302\243> \342\200\224L.
If for any sequence
{xn} such
the limit of the function.
that
have
\342\200\224\302\273
x0 we
xn
that f(xn)
\342\200\224\302\273
L, then
L e
IR
is said
function does not existat a point x y find two sequences
x y but
for which
yn \342\200\224\302\273
f(xj and/(jn) do not tendto the samelimit.
If we find the limit of a function at point
x0 by having a sequence that
approaches
To
by
is squeezed
TheSqueeze
then
then for some constantc e IR,
lim 5
=0
V
\342\200\242
It is
(p,qeR)
^*\302\260)
sequence
'
= pa + qb
= ab
limfon
hfr^
limhf
a and
a limit
to
then:
that
show
the
of a
limit
xn
from
to be
\342\200\224>
and
xQ
below,
lim f(x).
we write
If we do soby
from above we write lim f(x).
that approaches
a sequence
having
x^x+
We
the limit
both
need
to exist at
that
If
point.
from aboveand thelimitfrombelowto existand
they do then:
lim f(x) = lim f(x)
The Squeeze
Theorem and
=
lim
the algebraoflimitshold for
use PHopitaPs Ruleto find
of functions
limits
functions
f(x)
g(x)^0, if either
and g(x)
lim
f(x)
x\342\200\224>a
then
limx^a
f(x)
g(X)
= lim- f\\x)
= lim
that
are
g (x)
x\342\200\224>a
that
x^a g'(x) provided
differentiable
of the
in the
= 0 or limf(x) =
x\342\200\224>a
the
latter
0
form -
'
coo'
and
neighbourhood
lim
g (x)
^A
\342\200\224:
oo
of a point a and
= ^
x\342\200\224>a
exists.
1 Limits
\"
\302\253
the limit
as well.
functions
0
Given
for
f(x)
<
We can
coincide
of
sequences
and functions
be necessaryto manipulatefunctions
It may
quotientsso
A
function
that
at the
is continuous
f(x)
Both the limit and the value
To
The function
is said to be continuous
of a
The
derivative
Or
equivalently
function
by
f(x
X
does
For a
not, then
-
to be continuousthere.
x ,
find a sequencex
for which
\342\200\224>
but
xQ
points of its domain.
at all
is given
by f'(x0)
= lim f(x0
+h)-f(x0)
\342\200\224
limit exists.
if the
the function is not differentiable
at
must be continuousat x0
f(x0).
XQ
function f(x) to be differentiable
f(x)
=
f(x)
is continuous
J\\x)~J{xo'
\\[m
lim
for f(x)
at the point x0
x^xo
If it
if
a point
at
if it
f(x)
\\-
x0
exist
continuous
is not
a function
that
show
point
must
f(x0)
form (0xoo'or(oo-oo' into
be applied.
can then
Rule
l'Hopital's
of the
limits
have
that
a point
hence
(and
at
that
point.
x0:
must already
lim f(x)
exist)
X^Xq
must
f(x)
The
A
tangent
that
function
Rolle's
not have
to f(x)
a 'sharp point' at x0
at x0 must
for differentiable
Fora
function,
f(a)
- f(b) then theremust
f(x),
The Mean Value
For a
Topic 9
-
that
that
point.
functions statesthat:
on an interval [a,b]
c e ]a,b[ such that
point
is continuous
a
exist
and differentiable
f'{c)
Theorem is a generalisationofRolle'sTheorem
that
function, f(x),
must exist
at a point is continuousat
is differentiable
Theorem
not bevertical.
that
is continuous
a point c e ]a,b[such
Option:Calculus
that
on an
f'(c)
-
interval [a,b]
J-^-*-\342\200\224
b-a
.
on
]a,b[,
if
on
]a,b[,
there
- 0.
states
and differentiable
that:
h
i
examination
Mixed
(a)
1
practice
that the
b so
and
a
Find
function
x<-\\
-(x + l)3-l
/m=
+ b
ax
+ 6x
-2x2
is
(b)
continuous
differentiate?
evaluate
By letting m = \342\200\224,
lim
[6 marks]
[6 marks]
\342\200\224
n tan
one real solutionto the equation
[6marks]
= 3x-2.
sinx
Q
is exactly
there
that
Prove
x >1
-1
everywhere.
is f(x)
Where
\342\200\224\\<x<\\
The
of a
term
general
sequence is given by the formula:
n2 + 3n
= \342\226\240
a\342\200\236
neZ+
2n2-\\
= L where LgR,
Given that liman
(a)
(b) Find
the smallestvalue
for all
10-3
-L\\<
\\an
of
N e
find
value
the
of L.
Z+ such that
n> N
[9 marks]
( IB Organization
(a)
Write
xx in the
(b)
Using
l'Hopitals
form ek.
evaluate
Rule,
(c) What assumption
3
2006)
have
you
limx*
made?
for
Consider the sequence\\ \342\200\224
\\
any
[10
marks]
[10
marks]
igR+
n\\
(a)
an integer m > x
By choosing
v-n
v-n
xtl
^
\342\200\224
< \342\200\224
xtl
and
letting
that:
show
n>m,
m\"
x-
mn
n\\
(b)
(m-l)!
the Squeeze
By using
Theorem show that
x\"
lim\342\200\224= 0.
\302\253^\302\260\302\260
n\\
Find:
s-V3s
(a) lim
4-s
s->4
(b)
+ 4
^cosx-l^
lim
X->0
V
xsinx
[9 marks]
J
1
C
S^
+ d.
Limits
of
sequences
and functions
Q
that
Suppose
(a)
that
Show
odd function
is an
f(x)
there is
some a e ]-x,x[
lim
Given that
1 +
I
that
show
(c) Therefore,
i\342\200\224
(b)
Prove
(c)
Hence
-
that
find
(4n
Find
(b)
Hence
(c)
What
Option: Calculus
r^
**.
=-.
nj
e
Theorem,
using the Squeeze
Show that (1 + xf
(a)
-
marks]
G^f
lim
(a)
an
9
[8
= lim
V
Topic
j\\a)
=e:
-]
| 1
lim
(b) Hence,
44
=
that
such
It
nj
(a) Showthat
jj
on
is differentiable
sin x<|x|
(b) Hence show that-|x|<
3
that
lim 2^
Yn
lim\342\200\224^
X->oo
=1
of the
sequencewhosegeneralterm
\342\200\242
is given
by
t14
marks]
'%
find lim(ex
assumption
[13 marks]
lim\342\200\224'-
> nx foralln e Z+.
the limit
+ 5n
find
+ x)x
have
you made
in part
(b)?
[9 marks]
In this
you
chapter
will learn:
mproper
\342\200\242
how
relate
to
and
differentiation
integrals
definite integration
using the Fundamental
Theoremof Calculus
\342\200\242
to establish
You
probability
and b by
from the
be familiar,
should
of a
Core syllabus,with
continuous random variable falling
finding
between
the
a
integrating the pdf betweentheselimits.
when
an
with
integrals
upper
of oo (improper
limit
integrals) converge
to a
when
\342\200\242
how
and
value
finite
they diverge
to
some
evaluate
improper
integrals
that
converge
\342\200\242
to establish
between
relationships
some improper
integrals and
infinite
series.
=
P(a<X<b)
But what if we want
real
P(X
> a),
ff(x)dx
where the
pdf is non-zerofor
all
numbers?
x
P(X>a)
=
J\302\260\302\260/(jc)cbc
2 Improper
r*a*
+*.
,,
integrals
45
'a
h
raises
this
but
this
take oo
be finite
and,
or not the value of
can find it.
the
how we
of this form.We
on integrals
our knowledge
of limits
a
finite
value
for the
get
use
will
from chapter 1to determine
when
integral
we
our integral,
limit on
upper
if so,
concentrate
we
chapter
the
as
about whether
questions
will now
integral
In
like to
we would
Here
and
link
value. First,however,we analyse
the
and definite integration morecloselyusing
find the
cases
some
in
differentiation
between
Fundamental
the
of Calculus.
Theorem
of
Theorem
Fundamental
The
Calculus
and x = b
x-a
the limits
between
A
area under a continuouscurve
that the
seen
have
will
You
y
is found
f(x)
by calculating
the
definite integral:
y =
f(x)y^^X
If
the
allow
we
Denoting
A(x) =
\302\243f(t)dt
\342\226\272
by x
we can
write:
A(x)=[Xf(t)dt
Noticeherethat, as x is a limit, we cannot usex as the variable
we are integrating with respect to, so we simply
use another
variable t (this could be anything
as once
the integration
is done
we are going to replace it with the limits x and a anyway).
We
refer
to this as a 'dummy variable'.
x.
the graph y
the upper limit, x, we will
area
the
consider
Now
By changing
under
shown in the
graph opposite.
Ifthe
in
change
x is
a
by
approximated
AA
a
upper limit.
limit
this
be
will
upper
of this
function
A= \\bf(x)dx
limit to vary then the area
-j{x) betweena and
change
area
the
as
very small, the new areacan be
with area:
rectangle
AA~ f(x)xAx
AA
\342\226\240M
Ax
In
the
limit,
/ (x)
= \342\200\224.
= lim
A*^o
Ax
dx
or alternatively:
-f
\\Xf(t)dt
dx
shows
This
with
area
But
r-v
46
Topic 9
-
Option:Calculus
s>,
+ 0,.
derivative of the areafunction
Hence, the processof integration
respect
under a curve) is the reverse
ofdifferentiation.
to x.
the
isf(x).
AA,
is the
that/(x)
= f(x)
function
A(x) is
not the only
function
A(x)
(finding
whose
the
derivative
For
function
any
A
(x) = g (x) + c for someconstant
c
our understandingof differentiation,
Value
Theorem.
proved formally using the Mean
can be
since
Then,
have
must
clear from
This is
A(a)
g (a)
have
also
we
- f(x\\ we
that g'(x)
such
g(x)
+ cc =
=>
.\\
x = fo, we
Letting
A
and
0
-#(a)
\302\243(x)-\302\243(a)
(b)
=
g (b)
-
g(a)
A(b)=\\ f(x)
Therefore
J
This
is all
KEY
POINT
g(b)
\342\200\224
g(a)
2.1
of Calculus
Theorem
For a continuousfunction
f(x)
on the
=
\342\226\240^-\\Xf(t)dt
dxJa
\\af(x)dx
can
we
Although
functions
integrated
with
find
already
at given
respect
Fundamental
example
following
this idea.
interval [a,
b]:
f(x)
any function g(x) suchthat
And for
of the
dx =
dx
in Key Point 2.1.
summarised
Fundamental
through
f(x)
=0
jaf(t)dt
=
A(x)
have
=
it
but
g'(x)
=
f(x):
= g(b)-g(a)
definite integrals and evaluate
limits, the idea of differentiating
to the upper limit as in the first statement
Theorem of Calculus (FTC)is new. The
illustrates a straightforward application of
orked example2.1
Find
\342\200\224[Xcos(t2)
dt.
dx-
At
this
sight
we
indeed
difficult;
the
first
function
cos(f2).
FTC says
We
apply
might appear0
cannot
integrate
the
However,
we don't need to.
the
theorem
I3y
Theorem
Fundamental
the
\342\200\224
I
coe(t2)c\\t
=
of Calculus:
coe(x2)
(XX
with
f(f) = cos(f2)
2 Improper
C
XJ^\\
+ d.
integrals
The variable
place
orked
with respect
to which the differentiation
dummy variable do not have
and the
to
is taking
be x
and t.
2.2
example
Find:
d
(a)
2.1,
point
except
the
the
that
ra
d
=
--Jf(y)dy
limit.
lower
the
in
we use
Therefore
\342\200\242
variable
is w.r.t. the
differentiation
in
statement
the
dz
m
J
dx
This is similar to
Key
rb
(b)
^-\\\"f(y)dy
= -f(a)
relationship
ro
rb
Theorem
Fundamental
the
by
of Calculus
(b)
Here the differentiation
x but
x is not
one of the
Therefore
Differentiate
and
g(b)
just
w.r.t. x,
\342\200\242
Then
J
f{z)
dz =
g(b)-g(o)
[ f(z)dz
Jo
\342\200\242
that
noting
=
are not functions
g(o)
Let g'(z) = f(z)
limits.
with an
proceed
expression for
is w.r.t.
J\"f(z)dz ^-(0(b)-g(a))
dxJa
dx
of x
\342\226\240
0
could also apply the Fundamental TheoremofCalculus
In such caseswe
to a
with
two variables,
f(x,y).
needto make that we integrate (or differentiate) with
and
treat
the other as a
respect to just oneof these
constant.The
looks
at this and also the idea
example
of
the
function
before we integrate it; sometimes
We
function
sure
variables,
following
differentiating
known as differentiating
orked
rb
\342\200\224
Av
Ja
x2cosy
integral
2.3
example
Find:
d
(a)
the
under
(b)
dy
(*
f fdx
Ja
COS y)dy
(a)
The integration here is w.r.t.
y, so we treat the x2 as
constant,
unaffected
by
the
integration
Topic
9
-
Option: Calculus
C
XJ^\\
+ d.
a
rb
Ja
x2coeydy
=
=
rb
x2\\
Ja
coey
dy
x2[e\\nyfa
= x2(e'\\nb-e\\na)
sign.
continued...
Then differentiate
sin a
that
remembering
are
sin b
w.r.t. x,
.*.\342\200\224\\x2coeydy
(XX
\342\200\242
and
=\342\200\224(x2(sinb-sina))
(XX
=
2x(e'\\Y\\b-e'\\Y\\a)
constants
just
(b)
need to differentiate
time we
This
\342\200\242
\342\200\224
integrate.
As the
differentiation is w.r.t.
x, cosy is
we
before
(x2
dxK
coe y)
y)
= 2x
coe yy
unaffected
Now
w.r.t. y,
integrate
is now
x
.'.
\342\200\242
that
noting
\342\200\224(x2
coey)dy
\\
Ja
dx
=
=
=
that
see
integration
Zxcoey
dy
unaffected
=
We
\\
Ja
we get the same answer,whetherwe do the
or the differentiation first. This is true using
2x\\
J a coeydy
2x[e'mxfa
2x(e\\Y\\b-e\\Y\\a)
any
function.
2A
Exercise
1. Find
the following:
(a)
(i)
(b)
(i)
(C)
(i)
Ar^H^a,
(d)
(i)
^-\\\"e-y3
d
(i)
c3
\342\200\224
f
. /-..\\
(ii) j-\\ j2
dx
2cos5\302\243
dt
.... d pi.
.
\342\200\224
(ii)
2sm(2t3)dt
dx-
f
5yjt2+3dt
m At'^-te
dtJbx2+7
dtJax2+2x+5
2. Findthe
(a)
dt
dx JoJV
-f-
(ii)
dy
-^f
3cos2(2f2) dt
following:
(ii)
j-\\Xf(t)dt
\342\200\224\\yg(t)dt
dyJc
d
(b)
(c)
(i) ^-\\bf{x)dx
dbJa
(ii)
daJa
d
(i)
\302\261ff(x)dx
dxJa
rb
\"H
gMdx
rfc
dtJa
2 Improper
XJ^\\
+ d.
integrals
d<
(i)
<e)
(ii) \\b-^-g(x) dx
dx
(i) f^-f(x)
(d)
w
ifM't
f f
W
d<
Find:
d
rx d
d
\342\200\224\342\200\224
]nt
(a)
dt
dt
dxJa
d
rx
\342\200\224
\342\200\224f(t)dt
(b)
[4
dxJa dt
marks]
Q (a) Find:
d
dx^a
(b) Is
rb
rbx3
t
dt =
\342\200\224
f
f(x,t)
dtJa
d
fx3^
*a dx\\ t
\342\200\224f(x,t)
\\
Ja
for
dt
J
any function
/(*,*)?
r^
dt
[6
(a)
iC^t2)dt
d
rym
\342\200\224
(b)
7
rule, find:
the chain
Using
i
marks]
4
e~x
dx
[5 marks]
dyJy
Let f(t) be continuous
Use the Mean Value
on
\\a,b\\
andF(x)=
f[t)
that there
to show
Theorem
I
dt.
exists c e ]a,b[
that:
such
J
f(t)
dt =
[7 marks]
f(c)(b-a)
and divergent
Convergent
improper integrals
return
now
We
Jf a
In
See
1A and
Sections
<^[ IB for convergence
<^[
of sequences.
the
core
What
if we
Can
we still
an
Option: Calculus
v
i-A
= f(x)
the limits of integration wereoften
along this line of reasoning and let b
evaluate the integral?
continue
integral
can either
much the sameway
chapter 1.
-
course,
whereg'(x)
g(b)-g(a)
convenient, relatively small numberssuch as 0,1, tt. However,
can
also be very large and this makesno difference
to the
method used to evaluating the integral.
to oo, in
9
=
f(x)dx
familiar part of the FTC:
limits
Well,
Topic
to the
converge to a finite
that
we
saw with
limit
oo ?
\342\200\224\302\273
or diverge
sequences in
'a
of the
Integrals
f(x) dx are knownas improperintegrals.
2.2
POINT
KEY
poo
form
poo
The
dx is convergentif thelimit
f(x)
integral
improper
limf/(*)
ck
b^ooJa '
=
we
When
with
work
limits of
following
fc->\302\253
Otherwise the
is finite.
and
exists
-
lim{/(&))
1(a)
integral diverges.
use
the
improper integrals, we often
functions (familiar to us from the previous
chapter):
orked
\342\200\242
lim
e~x =
\342\200\242
lim
at*
Jo
b and
with
limit
but
normal
the#
replace
take the
Jpb
0
lim
dx-
e~5x
<s~5x
dx
b^ooJO
limit
after the integration is
t^oo
as
(p>0)
e~3x dx.
Integrate as
upper
=0
2.4
example
poo
Evaluate
0
=
e
lim
3
completed
=
1
lim
e~5b+
I
b^oo [
=
2
2
1
lim
e
|
1A
\342\200\224
,,1 + 1
3
As
e~3b
b^>\302\260\302\260,
-\302\2730.
the*
Therefore
is convergent
integral
3
orked example2.5
Determine
Replace
take
the
the
upper
limit as
different cases:
that
if
p
after
b^oo
\342\200\242
b and
with
limit
h[
xp dx
=
lim
b^ooJi
[
xp dx
the
is completed
integration
We realise
|.oo
p e R, J xp dx is convergent.
values of
which
for
be two#
there
will
= -l
we need to
integrate
with
Inx
lim
p + 1
ifp^-1
ib
lim [Inx],,
\\fp
\342\200\224
\342\200\224\\
2 Improper
XJ^\\
+ d.
integrals
f
-
51
'a
continued.
/^F+i
A
1F+i
lim
lim
\\fp\302\261-\\
p+1
p + 1
(Inb-ln1)
=
ifp
-\\
V+1-1A
lim
\\fp\302\261-\\
p + 1
Inb
lim
this
At
case
two
see that
needs to be split
we can
stage
p^-1
bp+i^oo
p +
when
but
the#
\\fp>-\\
into
p + l>0,
When
cases.
separate
= -\\
ifp
\\fp<-\\
p + \\
l<0,
ifp
=
-\\
i.e.
xp dx
j
converges
p <
for
only
-1
Or e^uivalently
r \302\260\302\260
1
\342\200\224
dx cower-ges
J
There are
the
or
many
in
integrals
many
Section
cannot
be expressed in
and yet lookat first sight as
methods. Examples that we will
that
meet
last
(the
function e~*2) is
Gaussian
it is
Although
perhaps
we said above that
can
be expressed in
they
the sine integral
function
these
terms
integrals
of special
S\\(x) defined
impossible
error
function
We
of specially
kind
this
see a
will
that
integrals
]^> not find
and
be expressed
functions defined just
in
in Section
--v
Topic9
-
Option:
as
important
indefinite
it
the
forms
it
integral,
in
for
of standard
terms
this
purpose;
can
for the
that
basis
shown
be
functions,
for example,
function
aid
JXe-f2df
our
here at
understanding
integration is difficult
need anotherway
]^>
the
4E.
Calculus
+ 0,.
convergent
without
of
The followingis oneway
2B,
attempting
this
can
not
impossible
question
whether
determining
actually
all?
(if
of standard functions,seeExercise
estimation,
s>,
4fJ\302\245^
chapter
i
Sometimes the
we can-
quantifying
error
52
method
otherwise,
this
in
\342\231\246w
of
one
to
as
defined
approximating
for
either
Option,
an
find
cannot
erf(x) =
Does using
(an indefinite
as
defined
erf(x)
functions
susceptible
= yfe
w Jo
Si(*)=f*^df
or the
this
most
the
to
j: e-x (jX
When
in
>1
dx,jex2dx,je_x2dx
jsin(x2)dx,jcos(x2)dx,j
pdf for the normal distribution.
should be
if they
integration
4E, include
Of these, the
of standard
terms
be found)
cannot
integral
only for p
2(a))
or not
to evaluate
be achieved:
in terms
and so
we
an integral is
it explicitly.
2.3
POINT
KEY
Test for
Comparison
If 0 < f(x) < g(x) for
poo
dx is
J a f(x)
poo
convergent if
(has finite
area boundedby
below g(x), then
(and
poo
g(x) dxis convergent.
f(x) dx is divergent.
know
the
that
dx
g(x)
is convergent
is always
that f(x)
and
x-axis)
course
of
a smaller
have
a then:
Ja
poo
if
if we
is intuitive:
result
This
x >
all
is divergent
dx
g(x)
improper integrals
also be convergent
the
x-axis).
by
dx must
f(x)
area bounded
orked example2.6
that
Show
The
integration
consider
using
find
dx is
[
Jo
1+ X2
a function
convergent.
S
here is complicated
the
result and
above
g(x)>-
1+
x2
such that
so we#
\\
foraWxeU
+ x2>x2
to
try
[ ^v
q(x)
Jo
1+
J
X5'2
X2
X2
converges
We
know
example
(Worked
poo
I
that
1
j>
Then, e'mce
\342\200\224rydxconver9es#
2.5) and
so we apply
above
JO
dx converqee,
J
j3/2
the
result
JO 1 +
X2
by the
converges
Comparison Test for
Improper Integrate.
The following result is alsouseful,
often
with the
in conjunction
Comparison Test:
KEY
2.4
POINT
As with
the Comparison
is just what
Test for
we mightexpect:I f
(x)
I
is
this
integrals,
improper
always
whereas f(x) mightgo belowthe x-axis.Therefore,
above
result
the x-axis
I
|
f(x)
|
dx
2 Improper
r^
*a.
integrals
will always
have a value
there might
latter
as
as large
least
at
|.oo
Ja
f(x) dx, as for
be cancellationdueto 'negative
the
areas'.
y
A
=
y
y=
/O)
1/0*01
is the
I\302\260\302\260
\\f(x)\\dx
sum of
red and blue areas
dx is the sum
f(x)
J
areaslessthe
of blue
areas
of red
sum
Of courseif J | f(x) | dx
of
value
orked
01
ax converges.
5
Test*
negative and so
be
we can't apply
test
the
directly
we can remove this*
taking the modulus and
However,
by
problem
to show
attempting
converges
abov<
dx.
f(x)
the Comparison
can
cosx
but
be the smaller
1
Ji 1+ x3
to use
want
We
COSX
f\302\260\302\260
i
so must
then
2.7
example
that
Show
|.oo
is finite,
cosx i
that
coex
1 x3
,
+
dx
j
1+
1+ x3 > x3
|cosx|<1and
x3
(and hence, using the
cosx
e resu
J] 1+
converges)
oo
From
Worked
converges
example
and so
2.5
[
Ji
we can apply
Comparison
*
1
\342\200\224
dx
x3
Since
\342\200\224
dx
J
converges,
coex
the
1+ x3
Test
by
the
Test for
Comparison
cosx
C\302\260\302\260
COi
Hence,
54
Topic 9
-
Option:
r*a*
Calculus
+\302\253.
so does
Ji
dx
1+
x\302\260
converges.
dx
improper integrals,
the limit of a convergentimproperintegral
encountered
may need to use any of the integrationmethods
the Core syllabus, such as integration by parts
and substitution,
and also the methods from the previous
chapter.
to find
need
If we
orked
in
2.8
example
|.oo
the
Evaluate
we
convergent
Integrateby
parts,
use of
to
lim
integral J xe~x dx.
improper
the#
remembering
with the
deal
xe~x
h[
dx =
upper
=
limit
lim
b^ooJi
lim
{{-be~b
-
= 2e-1
faced
now
are
with
a
of the#
limit
3y
'
type
'\302\243
and
so apply
lim
-e~xdx\\
-J
+ e~')
1
+
-
-
(e~b e~')}
b'
lim
Rule:
VHdpltaVe
1 + fr
I'Hopital's
dx
[-xe-*]^
= lim
We
xe~x
[
=
lim
\342\200\224
=o
Rule
xe~x dx
In the above
be quoted in
i.
the examin
cases
such
\342\200\224
=
0,
using
can just
results
standard
as Worked
example 2.8.
^y.
See
VHopitals
Section
ID for
^u,
Rule.
XP
\342\200\236
\342\200\224
=
lim
.
lim
2.5
POINT
r
established
Rule. This and the following
l'Hopitals
KEY
we
example
= 2e~
0
xp
\342\200\242
lim
*^\302\260\302\260ln.x
Exercise
2B
1. Test the followingfor convergence.
Forthosethat
the
value
of
the
improperintegral.
give
(a)
(c)
T\342\200\224l\342\200\224-dx
fe
dx
(b)
Jo
f
e
(d)
fJO
xe
do
converge,
4 dx
x
dx
2 Improper
r^
**.
integrals
'a
2.
for
following converge using the ComparisonTest
of the
which
Find
integrals:
improper
(b) JV*2
dx
,
cosf
\342\200\224
ax
poo
dx
<d)
3+5
;
I.
2+4x
x+2
(e) J3r-^i-d*
3. Findfor
sin2x
(f)
+ 1
X2~3x
of p
values
what
dx
f_?\302\2432
Jl X4+x + 2
the following
dx
improper integrals
converge:
(a)
lnx
dx
f\302\260V*
I Evaluate
Evaluate
(a)
ji
xp
[3 marks]
dx.
I
e~x
f
\342\200\224\342\200\2247
[3marks]
dx.
JO
x2+1
Show that
= (x2+3)2+5
x4+6x2+14
(b)
n^d*
(b)
Jo
all
for
xg]
show that
Hence
x2+3
dx
r x4+6x2+14
(a)
[5 marks]
converges
that
Show
for 0
= 4
\342\200\224=
dx
\\
Ja
- 2\\]a
yjx
< a < 4.
4.
1
f4
(b) Hence
explain why Jo
\342\200\224==
dx
converges
find
and
its value.
Vx
[5 marks]
ri
(a) Showthat
(b) Hencefind
the
3
rJ3
Q
^
56
Topic
9
-
Option: Calculus
r^
4 a.
of:
value
dx
[6
improper integral:I
roo sin
Show that
Does
exact
x2+x-2
Evaluate the
Show
x +2
x-1
x2+x-2
Jo
by induction
1
the
.
improper
(Ax)
dx
that
.1
I
integral
xne~x
dx
=
f\302\260\302\260sinx
,
Jl
x
[6 marks]
[10 marks]
.
1+ A2
e
dx
X
=
-\342\200\224-
\342\200\224sin
marks]
n\\
dx converge?
[11marks]
r^
T
7
[9 marks]
Approximation of improper integrals
last section that
in the
saw
We
we
that
integrals
exactly. It makes
evaluate
find an
approximate
there are someimproper
are convergent
can prove
but that
sense to see whetherwe
of
value
the
integral
in such
we
can
cannot
at least
cases.
of integration
as the area
interpretation
Using the geometrical
between a curve and the x-axis,we shall try to place bounds
on a convergentimproperintegral;that is, to find values L (the
lower bound)and U (the upper bound) such that:
L< J7(*) dx<u
To
achieve
this,
we consider
of
rectangles
of the
at the
As
the
we
a function f(x) and construct
integer value of x > a. The height
rectangle is determined by the height
left hand endofthe interval.
of
the
function
f(a)
some rectangles will overestimatethe areaunder
In general it is not
and some will underestimate.
the
total
area of the rectangles is an over- or
can see,
function
clear
underestimate
of
whether
area
the
value
each
1 at
width
of
the
f(x)
integral
the function
under
(and hence of the
dx).
see from the diagramthat the area of the rectangles
is an overestimatewherethe function
is decreasing
and an
ifwe restrict
underestimate where it is increasing.Therefore,
ourselves
to the specific case of either an increasingor a
we can use the area of theserectangles
to
function,
decreasing
form
the
bounds
we are looking for on the integral.
We can
So,consider
a decreasingfunction
firstly
f (x)
y
.
Upper
y =
r^
Sum U
>
fix)
**.
,,
dx
f(x)
Ja\302\260\302\260
for allx
> a.
'a
The
area of
total
the rectanglesis
(/(a)xl) + (/(a + l)xl) + (/(a
=
+ /(a + l) + /(a + + ...
/(a)
+
+
2)xl)
\342\200\224
2)
And
the diagram
shows that
is an
this
\\j(x)dx<jj(k)
which
to as
upper
The
soms
as
a^lower
are also known
Riemann
sums.
is the
upper
To form
I
f(x)
consider the rectangles
formed
to as
referred
the lower
sum,
the right-hand
end
the rectangle of height
using
by
with
is starting
referred
dx.
a lower boundL,usually
of each interval; that
giving
k=a
bound U mentioned above,usually
sum of
the upper
overestimate,
/(a + l).
y
Sum L <
Lower
\\y
= f(x)
a+l
total
The
area
/ (x) dx
Ja\302\260\302\260
a+2
a+3
a + 4
rectangles is:
of these
(/(a + l)xl) + (/(a +2)xl) (/(a
=/
(a +1)+ / + 2) + / (a + 3)+
+
+
3)xl)
+ \342\200\242
\342\200\242
\342\200\242
\342\200\242
(a
=
\302\261f(k)
k=a+l
And
this is
an understimate, giving:
f^f(k)<[f{x)dx
k=a+l
consider an increasingfunction wewould have
the same bounds but the other way round. However,
note
exactly
that in order for an increasingfunction to have
a convergent
to
improper
integral (if it is divergentthereisno pointus trying
bounds
on
the
function
must
bebelow
the
x-axis.
it),
put
If we
were
to now
\\y
58
Topic 9
-
Option:
r^
Calculus
+\302\253.
= g(x)
L
dx < U
g(x)
<Ja\302\260\302\260
'a
2.5
KEY POINT
For a decreasingfunction /(x)for allx> a, we
upper and lower sum such that:
these
use
will
We
relationshipsbetween
an
have
^w^
^^
improper
integrals
and
sums
infinite
Section 3B in
X/(/c)<j7(x)dx<X/(/c)k=a
the
in
^^^
^^
next
chapter.
k=a+l
function g(x)for
an increasing
For
an
have
sum such that:
and lower
upper
allx > a, we
fj(k)<rg(x)6x<fig{k)
k=a+l
k=a
the upper and lowersumsaregeometric
series
them
and give numerical
actually be able to evaluate
When
the value
the
of
orked
2.9
example
Usuch
and
to establish
need
first
We
f 1V
is
L<\\
that
- -
or decreasing
increasing
and upper sum the
round. We can do this
y =
to the graph
reference
by
-(Z)~x) is a
in the
reflection
x-ax\\e of y
= 3~
y
i
the lower
way
dx<U.
for
our expressions
to get
order
V
whether*
v\302\260y
in
bounds on
integral.
improper
Find constantsL
will
we
right
either
or by
differentiating
Therefore,
Now
lower
sums
of
have
now
We
for upper and \342\200\242
an increasing function
the result
apply
the
forming
geometric series
upper and lower
-(Z)~x)
is increasing for
x
all
Hence:
n
i-GT<r-Gr*<i-G
k=\\
k=2
V
\342\200\242
But Z\"
3
1
*
bounds so we can evaluate
these
Both
have
/\" =
using
1
the
\342\200\224;
lower
S^ =-.
a
\\-r
sum has
and2^-
1
the
upper
sum a
= \342\200\224
6
Therefore:
1
a = \342\200\224
and
3
\342\200\224
<
r - fiY
-
h dx< 6
2 Improper
s^
+
01.
integrals
f -
59
Often we will not be able to evaluate the infinite
sums
and we
will only be ableto give expressions
for the upper and lower
bounds. In such cases, however, we may be able to evaluate
the
itself and thereby form bounds on the value
of the
integral
in the next chapter
infinite
sum. We shall look at this approach
which dealswith infinite
series;
an introductionto the idea.
but the
following example
gives
orkedexample2.10
Find
(a)
and lower
upper
sums for
the
integral
improper
1
fee \342\200\224
dx.
J3
x2
oo
Hence
(b)
and lower boundson V
^
place upper
k=3
To
the
sum
way
right
\342\200\224
whether
is
do this
We can
need
round we
function,
decreasing
limit of x = 3 on
lower
the
evaluate
can't
can evaluate
time we
infinite
the
sums,
1
\342\200\224
fora\\\\x>3,
is
X5
X2
for
x >
all
1
Hence Y
3
1
\342\200\224
\342\200\224
dx
<[
x
k=A K
the
< Y
1
\342\200\224
k=3 R
integral
(b)
this*
but
1
integral
improper
<0
lower\342\200\242\342\200\242
noting
the
\342\200\224
decreasing
and
for upper
of a
sums
We
the result
apply
2
As
graph or by differentiating
Now
dx\\X2
to the
reference
by
(a) AM
establish
to
or decreasing
increasing
either
k2
l
and upper*
for the lower
the expressions
get
j
\342\200\224.
J3 x2
itself
dx =
lim
x~2 dx
b->\302\260\302\260J3
=
lim
=
[-x-1l
T
J3
lim(-b-1+3-1)
_\\_
~3
This
us a
yqives
lower bound
^
\302\253
1
of
\342\200\224
immediately
1
1^1
*-^ \\,2
y
^
\342\200\242*
h2
3
which
gives
a
bound
\\ower
for
V
1
Zu'Zz
k=z>^
To
an upper
form
oo
bound we
need to
But
change
oo
-|
-J
into
Xli\"
by adding the previous
^t^-
^
1
\302\243k2
1
1^1
+
32
3
11
^>
-
\342\200\224
\342\200\224<
+
h2
-X2
\342\200\242*
k=3'
k=Al
term
of the
series,
\342\200\224
=^>
which
Put
these
bounds
together*
1
f,
an upper
gives
1^14
< >
Therefore \342\200\224
3
4
\342\200\224
< \342\200\224
>
bound
\342\200\224
< \342\200\224
f-ik2
9
y
Topic 9
-
Option:
C
Calculus
XJ^\\
+ d.
2C
Exercise
1. Findthe upperand lower sums
functions are decreasing):
(a)
(O
2.
dx
f\302\260V2
f^
J2
for the
(b)
(the
following integrals
dx
1\342\204\242^=^=
d*
xz
the upper and lower sumsforthe following
functions
are increasing):
Find
(a)
r~\342\200\224
(c)
f
dx
(b)
-^\342\200\224
dx
I
Plnf
(the
integrals
1
r\302\260\302\260
Find
(a)
dx
j2 e
x \342\200\224
ex
upper
1
Jo
(x + 2)3
dx
the
(b) Showthat
f(x)
Find the
(a) Find
1
r-
f-
(a) Show that
(c)
sums of:
and lower
(b)
Jo
improper
integral
=
\342\200\224
ex+ e~x
is
x2+8x
J3 ex+e_x
sums
lower
1
ex +
[6 marks]
dx exists.
for all
decreasing
fupper and lowersumsof Js
the upperand
dx
+ 17
e~x
x > 0.
dx. [7
marks]
of:
dx
\\\302\260\302\2602-x
(b)
By evaluating
the two
series, find
such that:
roo
a <\\
2~x dx
J10
two
constants
a and b
<b
Give exactvaluesof a and b.
(c)
Using
the
result
that
f
i
a*
\\ax ax-
J
and your
^
\\-C
Ina
answer from
\342\200\224
<ln2<l
part (b),show
that:
[10 marks]
2
2
r^
+\302\253.
Improper
integrals
61
Summary
of Calculus
Theorem
Fundamental
\342\200\242
The
area under a curve).Fora
(to find the
links the conceptsof differentiation
function
continuous
on the
f(x)
and
integration
interval [a,b],
=
-fj-[Xf{t)dt f{x)
dxJa
g(x) such
any function
for
and
that
\342\200\224
g'{x)
f(x)
dx =
\\j(x)
\342\200\242
The
integral
improper
\\
is convergent
\342\200\242
Convergence
known
dx
dx
g(x)
if
is convergent
if
is divergent
poo
Ja
1
poo
\342\200\242 \342\200\224
dx
poo
f(x)
a '
'
\342\200\242
dx converges,
of infinitely
For
for p
poo
Ja
dx
f(x)
is divergent
(does
not exist).
g(x) dx is convergent.
> 1.
does <Jpoo f(x) dx.
a
functions or
many rectanglesof width
a decreasing
such
function
other
f(x) dx is divergent.
then so
of increasing
integrals
Improper
sum
only
converges
J
J
\\
(exists) if the limit is finite. Otherwisethe integral
poo
\342\200\242
If
lim
of improper integrals can be established
with
by
comparison
<
<
>
if
or
that
0
for
all*
a
then:
is,
convergent
divergentintegrals;
f(x)
g(x)
poo
a
dx =
f(x)
or divergence
J a f(x)
J
g(b)-g(a).
/(x)for
decreasingfunctions canbe approximated
by
the
1.
all x > a, we
an
have
upper
and lower
(Reimann) sum
that:
fJf{k)<[f(x)dx<f^f(k)
k=a+l
For an
such
increasing function
g(x)for all*> a , we
k=a
have
that:
fJg{k)<[g{x)dx<fJg(k)
k=a
AJL
r-v
*
62
Topic 9
-
Option:Calculus
1*
*\302\273>\\
+ 0,.
k=a+l
an
upper
and lower
(Reimann) sum
examination
Mixed
o
the
Evaluate
j;
Find the
(a)
[6 marks]
,d\302\273
+
(x2+8x
5)2
following:
^rin(c\302\260s23x+i)dx
(b)
B
integrals:
improper
following
x+i
w
Q
2
practice
[4 marks]
^rin(c\302\260s23x+i)df
Write In n in the
(a)
where
a,b
x >
and
1 and
(b)
For
(c)
By choosing
rb
form Inn =
are to
f(x)
f(x)
dx
be found.
c> 0, show that
a suitablevalue
\342\200\224
<
xc~l.
c, find
for
the limit
of the sequence
whose
generaltermis:
O
(a) Using
poo
poo
dx =
cosx
e~x
Jo
Find
the value
that:
dx
e~x sinx
of these two
[9marks]
n
integration by parts,show
Jo
(b)
Inn
=
un
[11
marks]
integrals.
IB
(\302\251
Q
Show
Q
(a)
Show
(b)
Find
upper
(a)
Show
that
x >
(b)
Find
lim
\342\200\224
a
(c)
that
poo
the improper
that
Determine
Jl
jc
dx existsfor
Xa
all
3 >1.
c
2007)
Organization
[8 marks]
2
f\302\260\302\260
e~x dx
Jo
*->~
integral
sin
converges.
and lower
vx
sums for the integral.
[7marks]
-1 for x > 2
ex
x
whether
\302\243*
poo
or not J
2
dx
converges.
\342\200\224j=\342\200\224
v*
-l
[8
2 Improper
*V
-
^
v>A
+ 01.
marks]
integrals
El
(a) Showthat
*\342\200\242
2x
- -JAx2
+21 =
, where
,
2x
k is
a constant to
be
+ V4x2+21
determined.
(b)
Evaluate
4x
r\302\260\302\260 qx
\342\200\224,
Jl
2
(a) Usea sketchto illustrate
(b)
Hence,
determine
roo(lnx)
I
Topic9
-
Option:
Calculus
that
[9 marks]
In x
<x
for x > 1.
the following
whether
converges or diverges.
cosx
\342\200\224
x*
64
dx.
V4x2+21
dx.
[6 marks]
In this
Infinite
\342\200\242
how
convergence
from
of
this
1
a
sequence
geometric
conditions (|r | < 1)the
that
sum
positive
converge to a limit;
For example,for
the
series
27
9
2+ - +3
+
32
8
2
+ ---
sum
converges
2^2
\342\200\242
to
If | r | > 1
for
a
2
1-r
1--
_
series
a geometric
types of series that
or
convergent
divergentvery easily.
For
example
the series
for
T,
what
or a
no such finite
can
we
categorise
^k
series
integer
as
clearly does
not
k=1
series such
as 1H
111
1
^1,\342\200\224?
=
1\342\200\224
>
1
3
4
tlk
\\
\342\200\224
>
to 0,
converges
so perhaps
that
our sum will
too?
Unsurprisingly,
the idea
of the convergence
of seriesisvery
convergenceof sequences.We will
start
reference
by defining convergence of series generallywith
to convergence
of sequences and then lookat ways in which we
can determine whether series(including
the
above!)
converge
or diverge and in some caseswhat
their
limits
are too.
closely
infinite
which
of x
powers of x converge.
limit.
This looks as if it couldbe convergent;
each
term
is getting
smaller and indeed we know from chapter1ofthis option
converge
values
in increasing
1+ 2+ 3+ 4H\342\200\224=
2
the sequence
some
series
the
\342\200\242
to find
converge.
But
of
limits
convergent
is)
there is
are other
There
or
on
bounds
put
U
k=\\
to (its sum to infinity
c
negative
diverge
^,(3
the
the
and
terms converge
=
\342\200\224
both
with
positive
series
geometric
whether
\342\200\242
to determine
number.
never reach that
of tests
number
a
using
more and more termsofthe sequence,
the
nearer
and nearer to a finite
number
but could
approached
series
converge or diverge
added
as we
is,
said to
was
series
option
whether
\342\200\242
to determine
infinite series where the
were summed.
Under certain
you studied
to
of sequences
chapter
course
of
series
infinite
the
the
the
relate
to
convergence
series
In main
termsof
you
chapter
will learn:
related
to that
of the
See
1A
Section
<^[
IB for
convergence <^[
of sequences.
3
c
xj^\\
+ d.
and
Infinite
series
65
'a
9 Convergence of series
infinite
From
an
series
given
by
S=
(S =
divergent
+ u2
ux
+ u3
+
<*>) infinite
+...
u4
sense of the ideaofa
to make
In order
we have the infinite
u1,u2,u3,u4,...
sequence
convergent
series,
up the sequence
or divergent
S^S^S^S^...
where
Sx
of a
idea
familiar
Sn, n
each
Here
+
S3-ux
+ u2
Sn-ul
+ u2+...
u2
+
u3
+
e Z+, is known as the nth
un
partial
sum.
definition:
3.1
POINT
KEY
setting
ux
S2-ux
have the following
We then
or
\302\251o)
the now
sequenceby
convergent
=
(S <
relate this to
we will
r
If the
a limit
sequence Sx,S2,S3,S4,...of partialsumsconverges
series
is convergent
S, then the infinite
(to S).
to
S = 1imSn= ^uk
k=\\
Otherwiseit is
orked
3.1
example
By finding
the
divergent.
nth
of the
sum
partial
determine whetherthe seriesconverge
following series,
diverge.
3-Jfc
(a)
(b)
X
Y?A%
k=\\
We
k=\\
series
a =
with
2 and
r=
2
\342\200\224
and
the
first
formula
n terms
to get
partial
(a) This is a
geometric eerleewith
a
can
3
use the
therefore
sum of
a geometric*
this as
recognise
4
the
the
nth
nth partial
_fl(l-r\
1-r
sum
_f(iJ)
! 2
1
3
=
6
hd:
66
Topic
9
-
Option: Calculus
S>,
+ 0,.
r=
2
\342\200\224
3
So the
for
\342\200\224
2 and
sum
is given
by
or
continued...
If
this
sums
of partial
sequence
converges,
converges
so find
recognisable,
the
now see
so doee
S\342\200\236
converges,
- + \342\200\224
+ 0
4
2
(b)
terms
few
first
series*
arithmetic
is an
it
= 6
\342\200\224
\342\200\224
a \342\200\224
ana a aa
2
to get
n terms
So, the
the*
for
formula
known
the
series
arithmetic
is an
This
2
4
1
We can use a
sum of the first
^2 x
isn't perhaps so#
This series
We
therefore
3,
5n^6(]-0)
Since
immediately
\302\260o.
I \342\200\224
0 and
I \342\200\224\302\273
n
series
the
then
Letting
nth
with
1
\342\200\224
4
sum
partial
is given by:
nth
sum
partial
2xi+(\342\200\236-,)(-l
n-1
5n-n2
If
this
sums##
of partial
sequence
converges,
the
then
Letting
series
n^>\302\260\302\260
5n-n2
=
\302\260o
> \342\200\224
3\342\200\236
converges
&
3-k
Since
the
With
use
our
certain
work
for
are after
just
Za
when
establish
of partial
sequences
about what
think
is to
step
a convergent
all
so does
diverges,
series,we can now
of sequencesto helpus
on convergence
series (which
converge.The first
uk
of infinite
of convergence
definition
S>n
sums)
we know about
series ^uk.
k=\\
If
^uk
then we
converges
know
the
that
sequence
of partial
sums
k=\\
[Sn
} converges
We
also
have
to a limit
that:
S; that
un=(u1+u2-\\
=
Sn
is,
lim
Sn =
S.
n\342\200\224>oo
\342\200\224
Vun)
{ul+u2-\\
1-
un_x)
\342\200\224
bn_!
3 Infinite
C
XJ^\\
+ d.
series
= lim(Sw
So, limi^
=
=
-
S^)
]imSn\342\200\2241imSn_1
s-s
= 0.
This givesus the importantresultthat:
3.2
POINT
KEY
series 2juk is convergentthen
If the
= 0.
lim%
k=\\
The
not necessarily
it is
is not
however,
converse,
true: just becauselim%=
k^~
the casethat
0
is convergent.
^juk
k=\\
In
section
next
the
we will use Key
]^> 3.2 to
point
divergence
An
the
determine
of
latter is ]T-.
k
Here
=
0
\\imuk
k^\302\260\302\260
but,
as we will
k=i
]^>
~
some
i
see in the next section^\342\200\224diverges.
serfes
Exercise
of the
example
k=i k
3A
1. Evaluate
(correct to 3SF)the first
four
sums
partial
for the
series
^uk where:
k=\\
(a)
=
uk
\342\200\224
\342\200\224-
Kt=sin
(b)
y 3
k2
(c)
2.
= -\342\200\224-
uk
an
Write
expression
ofthe
series,
or
convergent
given
uk
(d)
k+5
(in terms
and
hence
J
= e~k
of n) for the nth partial sum
decide whether the seriesis
divergent.
(a)
+ l
f>fc
(b)
k=\\
w
ii
k=l
5
(d) ii.2*
\302\260
hV
|>-^
k=\\
k=\\
2y
p
i
3.
Given
S\342\200\236=3\302\2532-l
(b)
S\342\200\236=3B-1
(c)
Sn=^--5
(d)
S\342\200\236=lnn
2\"-l
=
S\342\200\236
3x2w
Topic 9
-
Option:
Calculus
1
r^
**,
series
2J4k,
k=1
(a)
(e)
68
partial sum ofthe
(and simplify)
4\302\253
r*
for the nth
the expression
find
uk:
for
Tests
[sj
and
convergence
divergence
Divergence Test
From
be
3.2 we
point
Key
a series
of
checking
diverges.
Test
Divergence
If
cannot
3.3
POINT
KEY
the series
have a possibleway
We therefore
convergent.
whether
note that if lim uk^0
or if
^ 0
limuk
the limit does not exist,the
series
is
V%
divergent.
this will
Of course
be
might
but
not
series (limuk
k^oo
could divergeas mentionedabove),
thing to check.
the series
0 and
often the first
it is
a divergent
identify
always
orked example 3.2
series
the
that
Show
To show that
diverges.
>
we need
^0
limu^
\302\2434=-
4k2
to manipulate uk
that enables us to
into
find
a form
its
1+
limit
^ +^
k
4 +
as/c^oo
limu, =
Hence
+ 3
k2
4
\342\200\224
^0
^
2
r~
diverges.
now present a number of tests for determining
or divergence
of series, which can be established
convergence
the principles
discussed in Section 3A.
using
We will
The
The
first
of these
improper
used
integrals
We
like
would
to know
or not.
Divergence
before
tests is
cons.der.ng
other tests
any of the
similar to the ComparisonTestfor
in chapter
2 of this option.
whether the following
from
series converges
chapter.
this
See Section 2B (Key
point
<l
2.3)
for
Comparison
Test
improper
XJ^\\
+ d.
the
for
^~
^^
integrals.
3
c
a
to check
good idea
Test
Test
Comparison
it is
exam
foe
In
Infinite
series
'a
1+-11111
\342\200\224
\342\200\224
\342\200\224
+
+
While we cannotsay
recognise
it is
that
+
82
much
about
this
1 -11111
+ - +
244
series immediately,
\342\200\224
\342\200\224
+
+
\342\200\242
+
+
243
81
27
9
3
we do
geometric series:
to the
similar
\342\200\242
+
+
28
10
4
1
which we know convergesbecauser =
-
(S^ =
3
term of our seriesis lessthan or
corresponding term of the geometricseries:
note that each
3
-).
also
We
2
equal
to the
1<1
4 3
10
1
9
1
-<\342\226\240
27
28
sum
each partial
Hence,
the
(after
correspondingpartial
must therefore
the
of
sum
a limit
(to
converge
This idea is generalisedby
the
first
than the
series and our series
is less
one)
geometric
less than
3
-).
result:
following
3.4
KEY POINT
Comparison Test
series
two
Given
that
< bk for
ak
\342\200\242
all k e
is convergent
^bk
terms 2Jik
of positive
Z+,thenif:
to a limit
k=\\
to
T where
is divergent,
^Jik
2
comparing
70
Topic
9
-
Option: Calculus
C
XJ^\\
+ d.
such
S, ^Jik is alsoconvergent
so is
^bk
k=\\
the Comparison
of this option, this
as with
of partial
\302\243pk
k=\\
T <S
k=\\
chapter
and
k=\\
a limit
\342\200\242
Just
k=\\
areas under
sums.
Test for improper integralsin
result is intuitive.
instead
of
Now,
a
functions, we are comparing
sequence
convergesto S and
If ^bk
<
ak
all k
for
bk
e Z+ we have
the
k=\\
sums of each series:
the partial
for
following
If
=
Sn
Tn = YX=i
converges
bk
Y^k=i
converges
ak
to
S then
toT
< S
**UJ
m
S1
\342\200\242rp
T2
Ti
\342\200\242n
In looking
and
if
diverges, we have:
^jik
forji
convergent ^bk
k=\\
k=l
Tn =
If
then Sk =
first
k,
a\302\253
bk
Z)fe=i
diverges
\342\200\242
\342\200\242
\342\200\242\342\200\242\342\200\242
\342\200\242
ak<h^
it is often
that
such
Y2=i ak diverges
worth considering
geometric
series.
\342\200\242
\342\200\242
*
S3
S2
\342\200\242*
-\342\200\242 \342\200\242
\342\200\242
We
will
see
shortly
the p-series
that
]^> also
\342\200\242*t2
commonly
in the
is
used
]^>
Comparison
Test
Ti
\342\200\242n
orked
3.3
example
Establish whether
or not the seriesV \342\200\224
converges,
Jx2k+?>
We
will
first
try
series
the
this
we
satisfies the conditions
not have
do
we
approach
trying
finding
and
converges
Comparison
If
known
of the
Test
any success with
will
to establish
a
that
by finding*
converges
a series we know
which
show
of all to
then
consider
divergence by
divergent
series
3
XJ^\\
+ d.
Infinite
series
'a
continued...
for
2k>0
The
we
is similar
series
know
converges
(it
to \\
\342\200\224-
is a
geometric
which
k
all
and
fora\\\\ k
2^+3>2^
\342\200\224-\342\200\224<\342\200\224
fora\\\\keZ+
series
with
r = \342\200\224
), so
let's start by
considering
3
2^ +
2k
this
and elnce z^~^k
the
Limit
converqee,eo doeeZ^ 2k + 3
test.
Comparleon
Test
Comparison
we meet a serieswhich is very
Sometimes
convergent (or divergent) series,
conditions of the Comparison Test.
but
For
series
the
example
by
^>
1
= 1H
1
\302\243?2*-l
1
7
the
not fulfil
does
11111
3
known
to a
similar
which
1
1
63
31
15
+ \342\200\242
oo
to our
similar
terms
has
known convergentseriesV
k=\\
each
of its
series
and
terms are larger than thoseofthe known
not smaller.
is thereforeno use,
the
I
Limit
Comparison
Given
two
convergent
Although the standard ComparisonTest
extension
of the test is.
following
series
Test
of positive
terms ^jik and^fc^, where
k=\\
lim
but
z
3.5
POINT
KEY
j
\342\200\224
/ >
other and
k=\\
0, then if one seriesconverges
\342\200\224
=
if one seriesdiverges
so
does
so
does
the
the other.
EXAM HINT
Choose as b^
had
72
Topic 9
-
Option:Calculus
xj^\\
+
d.
hoped
the
general
to apply the
series
Test.
Comparison
term
of the
to
which
you
orked
3.4
example
Show that the
series
We know
for the
isn't suitable*
series
this
that
is convergent.
l
z
k=\\
Let
1
ComparisonTest
with
akk =
^
fc=i
but
we
of the
Limit
u
anaa bk=
1
\342\200\224
2k
Then
conditions
the
satisfies
it
hope
1
2k -1
1
-x
2k
\342\200\224
Test
Comparison
2*-1
V
and so
lim
Hence
\342\200\224
1
V
^^
7
by the
converges
Limit Comparison
Test.
Test
Integral
In the
chapter
previous
(sums)forthe improper
caseswhere was increasing
use this resultforthe case
and
f(x)
where
function and a \342\200\224
1, but
not the
now
dx in
f(x)
integral
f(x)
decreasing.
is a
bounds
put bounds on
method
graphical
usedin Section2Cto
the separate
sum
and
and
upper
lower sums,^^
example
have:
on
the
^3*
Worked
at
and
2.10 where
we placed
improper integral.
So, from chapter2 we
^^
^3*
positive
the infinite
the
establish
Here we will
decreasing
at the
back
Look
upper and lower
found
we
sum
bounds
V
\342\200\224
k=3 k2
fjf{k)<jj{x)dx<fjf(k)
k=a+l
For
the
lower
k=a
sum, letting
a = 1 and
noting
that
f{\\)
is positive:
f^f(k)<[f(x)dx
k=l+l
=>X(/(*))-/(1)<f/Md*
k=\\
^i/{k)<f{i)+jj{X)dx
k=\\
with
this
Combining
the upper
sum J /(*) dx< 2jf(k)
k=\\
we
now
have:
f f(x)dx<
^J(k)<
/(l)+f
f(x)dx
3 Infinite
*V
-
^
v>A
+ d.
series
'a
h
/\342\200\242oo
if the
Then,
may be
this
use
place
an
I
L, the sumis boundedbetween
you
exam,
the
In
integral
improper
dx
f{x)
L and/(l)
constants
two
the
to a limit
converges
+ L:
to
\302\253q\302\253'\302\253fjf
method
bounds on
infinite
L
k=\\
the sum increasesas moreand more
terms
are added. It seems clear that
an increasing
series that is
+ L) must converge to
bounded above (hereby the constant/(l)
be evaluated
cannot
+
L<fj(k)<f(l)
that
som
Also,
directly.
is positive,
as/(x)
a limit
Area
or equal
than
is less
that
curve:
blue
under
Area
+
/(l)
red curve:
under
Area
to this bound:
in grey
dx
f(x)
J^\302\260
boxes:
dx
f{x)
J^\302\260
/W
Y^kLi
y = J\\x)
y
= f(x
+ i)
\"^
1
1
1
r-^
1
10
Do we
need
formal
proof of
have
to
a
The general
a
'
decreasing
result that any increasing seriesboundedabove
or
series bounded
below has a limit is known as the
MonotoneConvergenceTheorem.
result
Convergence Theorem
when it seems so clear that it
true?
Does a proof add to our
part of the syllabus
is
I
is not
the following test for
us understand
convergenceof series:
3.6
POINT
KEY
understanding?
it helps
result
this
Although
like the Monotone
Test
Integral
a positive
Given
decreasing function f(x), x>l,
poo
if
I
f(x)dxis:
\342\200\242
then
convergent
^f(k)
is convergent
k=\\
divergent
then
^f(k)
is divergent.
k=\\
The
closely
following
the
of the
example
of Worked
method
use of the IntegralTestresembles
example 2.5 when we examined
the convergenceofthe improper
integral
Ajl
r-v
74
Topic 9
-
Option:
Calculus
s>,
+ 0,.
I
xp
dx.
orked example3.5
of p
values
which
for
Determine
the series ^kp converges.
k=\\
the
Test
Divergence
it
For p
looks*
divergent so
must be
the series
use
p >0
that when
note
We
like
>0
14 =
lim
For
this
p
In
=
see
p<0.
and decreasing for
We
to#
turn our attention
that
k?
k>
1
[
xp
the eerlee
different
there
that
realise
cases:
if
= -l
p
be
with
+ 1
p
\\fp\302\261-\\
1
two
lim [inxl.
|V+i-0
lim
lim Inb
lb->\302\260\302\260
At this
two
bP+]
->
stage we
p^-1
separate
oo but
can see that
the#
CX)
needs to be split into
cases.
When p + l>0,
when p
+1 < 0,
b^+1
= -\\
ifp
In x
,
case
the
b
lim
we need to
integrate
by
[ x^dx
lim
dx=
is positive
so we can
will
diverges
Test.
use the Integral Test.
We
=1#0
Iim1
cases
both
Divergence
we
=\302\260\302\260^0
= 0
\\\\muk
Now
\\lmkp
to confirm
->
If
p #
-1
/f
p #
-1
\\f-\\<p<0
1
\\fp<-\\
p + 1
0
ifp = -\\
ex:
/.
Integral Test
by the
\342\200\242
for
divergent
\342\200\242
convergent
So,
^^p
for
^/p is:
-1<p<0
p < \342\200\2241
is convergent
only
when
p <
\342\200\2241
or
equivalently,
is convergent
only
when
p >
1.
3
S?i>*
+ d.
Infinite
series
'a
The series ^
k=\\
br
The result
is
p-senes
and
this
/er?
often
v/i\302\253
ueV.n
pother
such
to
but
<*
lor otherwise^
of other
a
seen that the
have just
about the
what
\342\200\224
\\
oo
j
harmonic series^\342\200\224diverges,
k=\\
establish
the convergence
We
p
series
Alternating
conation
tests
the p-series. When
to as
referred
series.
harmonic
is the
be
will
kp
be
as the Cordon
Test
\342\200\224
same series with
*
and
positive
alternating
terms?
negative
number
;
^V
series.
2 3 4
k
k=\\
greater chance of being convergentdueto
the negative
terms
but we cannot use any of the tests we have
seen
thus
far as these all require positive terms.Any
series
with
positive and negative signs is known as an
alternately
series.
The
result
allows us to analyse
alternating
following
has a
It certainly
convergenceof
KEY POINT
this
kind
the
of series.
3.7
Alternating Series Test
I
If for
series ^juk:
an alternating
k=\\
\342\200\242
| uk+\\
\342\200\242
lim
I
k^J
|
< | uk |
uk*' I
then the
f\302\260r
sufficiently
k
large
= 0
seriesis convergent.
The
fact
that the magnitude
of each
the signs are alternately
and
positive
sequence of partial sumsSk alternates
w
to 0
is decreasing
negative)
either
means
side of
(while
that the
its eventual
limit:
s6
\342\226\240k
76
Topic 9
-
Option:Calculus
S>,
+ 0,.
case ofour alternating
harmonic
series
(in Worked
example 3.6 belowwewill showthis is convergentby the
Series Test) we have the first
few
sums:
partial
Alternating
In the
S1 =
S2
S3
=
=
=
l--
l--
1> S
2
-<S
2
+
- = ->S
2 3 6
2 3 4 12
In
S =
fact
this
\\^>find
while we will not be required to
approximate the limit
using
ln2 and
we can
exactly,
L^>
point 3.8.
Key
posWve
Alternating
It is stated in the conditionsofthe Alternating
I
uk+\\
|
< | uk |
f\302\260r
to betrue for
k e
finite sum
some
have
all
Z+ (although it
often will be) as we will
from the finitely
terms
many
situation occurs. This clearlydoesnot affect
the series.
the
terms are
need
does not actually
k. This
large
sufficiently
just
by
H)
\"sed br
coslM
the
same purpose.
of
convergence
usudhr
generated
this
before
negative
and
that
Test
Series
orked example3.6
whether
Determine
Establish
in
expressions
to apply
readiness
1
\\k+l
\342\200\224
^(\342\200\224l)
k=\\
for
uk
K
and
is convergent.
\342\200\242
\\uk\\
uk+]
1
= -and
=
k+i|
1
\342\200\224
the Alternating
Series Test
now
We
conditions
need
of the
to check
test are
that
0<
the*
\342\200\224<-
k+1
satisfied
i.e.
K+i|<KI
and
lim
Therefore,
by the
~
Z*i\\ty
for
alike
Z+
k
for alike Z+
i
i
\\uk \\=
lim
-1 =
Alternating
0
Series Test
\\e convergent.
3
C
XJ^\\
+ d.
Infinite
series
'a
Once we know we have a convergent
series, we can
alternating
for the limit.
go on to think about an approximation
If we wereto stop,or truncate,
the
series
at the nth partial
sum,
how accurate would this approximationbeto thetruelimit S?
The
distance
between
the nth partial
sum
and S, | S - Sn |, known
as the truncation
between
the
error, is lessthan the distance
sum
(n + l)th partial sum and the nth partial
| Sn+1 -Sn\\.
Sk
A
t
T\"
-
S\\
|S\342\200\236
\342\200\224
i_
Sn+i\\
\\Sn
TSVi+i
+
nn-h 1
But
Sn+i-Sn=(ul+u2+'\" + un+un+l)-(ul+u2+'\" + un)
=
un
therefore:
and
=
|s-s\302\253|<|s\302\253+i-s\302\253|
That
is, we
KEY
POINT
have the following:
3.8
is the sum
= \\uk
If S
K+il
an
of
series that satisfies:
alternating
k=\\
uk+11
< | uk |
for all k
e Z+
lim|% | = 0
k^
then
error
the
the
(n +
the first
in taking
to S (thetruncation
error)
as an
n terms
is less
than the
approximation
absolute value
of
l)thterm:
|s-s\342\200\236l<h+i|
For
example,
an
9 terms of our alternating
error of lessthan the 10thterm
of the limit:
taking the first
will give an
series
harmonic
approximation
|S-S9|<|u10| =
If we
increase
the error
the number
is lessthan
= 0.01.
100
78
Topic 9
-
Option:
C
Calculus
XJ^\\
+ d.
of termsin
1
10
our
=
for
0.1
approximation
to 99,
orked
How many
is it
^ (-1) k2
k=\\
to take to find an approximationthat
J--1-
Vf-iV\"-1
*
actually is:
this series
what
see
\342\200\224
series
of the
terms
necessary
First
3.7
example
fc=i
~r is an
IB\"\"
L
J-
J
22
32
42
2
(fc+i):
lim \\uA=
^
and
So, we have an alternating
can consider using the
if the
result
point
expression
for
previous
3.8
to find
the
truncation
~1
k-i
inr
an*
with
\342\200\224
\342\200\224
0
Alternating
Series Test
convergesto 5
of using
error
truncation
The
series
alternating
fc->oofc2
by the
Therefore,
are satisfied
conditions
Use Key
series*
0.001?
it2
lim
fc_\302\273J
and
to within
is accurate
the first
n
terms
to
5 satisfies
approximate
error
|s-s\342\200\236|<k+1|:
(\302\253+T
the error to
We need
We therefore
be#
0.001
than
less
require
1
=>
(n
=^>
+
-<
0.001
l)2>1000
n>Vl000-1
= 30.62
Therefore,
seen
have
need
to
take
31 terms.
and conditional convergence
Absolute
We
we
that
series
the harmonic
^11111
2 3 4 5
f^k
while
diverges,
the
alternating
series
5
4
3
2
k
\302\2431
of the
version
converges.
Serieslikethis
terms
are
that
are
negative
whereas thosesuch
become
only
as
known
>
as conditionally
\342\200\224
=
\302\243ik2
when some
convergent
lH
1
22
convergent,
1
y
of their
1
42
1\342\200\224
52
3
S>,
+ 0,.
Infinite
series
'a
which are convergent
negative,areknown
as
any of
making
the terms
convergent.
absolutely
3.9
KEY POINT
series
A
before
already
convergent if the seriesZ^\\uk
is absolutely
2J*k
k=\\
is
k=l
convergent.
If a
*k
series 2j*k
is converi
but
gent
k=\\
If
> 0
uk
the same
are
k then
for all
the
option
series is absolutely
if ^J
result
to that for
point
(Key
uk
convergent,
|
is convergent
then
then
it is
so is
follows immediatelyfrom the Comparison
Test,
e
for
all
k
Z+.
uk
\\
|
this
not seem that useful, it can
might
establish the convergenceof someserieswith
all positive but which are not alternating
series
-
nl
^i
Snow that
k=i
note
We
^.
^
.
is convergent.
k2
that
terms
positive
is
this
or an
not alternately
are
terms
3.8
sin/c
^
0.841+
a series
with
only1
alternating series as the
positive and negative:
si
:
not
0.227+
0.0156-0.0473+
to prove
convergence,
\342\200\242\342\200\242\342\200\242
i
k2
In order
therefore hoping that
it is
absolutely
we are*
convergent
(and therefore convergent)
80
Topic 9
-
Option:Calculus
r*a*
**.
^juk
k=\\
While
example
convergent, i.e.
as
uk <
clearly
orked
improper integrals
2.4).
k=\\
This result
convergence
3.10
POINT
If a
then
convergent.
absolute convergenceand
analogous
in chapter2 of this
r
is divergent,
thing.
We also have
KEY
I
k=\\
is conditionally
series
the
2j\\ uk
sin/c
k2
allow
us to
terms
that
either.
are not
continued...
In
our disposal now#
and therefore
\342\200\224
taken the modulus
have
we
tests at
other
the
reviewing
that
|<
.\\uk
<1)
{e'mce\\e\\nk\\
K
all
have
positive
series
convergent
terms,
we recognise
^tj
would be suitable for
ae
and
is convergent,
Zu~^
Test
the Comparison
with
use
the
that
Comparison
Test so
is
by
the
elnk
2*i
elnk
since
Finally,
^
2~i~j~^T
be absolutely
to
nae been
convergent,
shown
it is also
convergent.
The answer
is that
for
the same way
series we
convergent
absolutely
way we like without
in any
terms
the
or
convergent
convergent?
conditionally
reorder
whether a seriesis absolutely
it matter
does
Why
can
the
affecting
sum
for any finite series).But,
for
surprisingly,
conditionally convergent
of the same terms of a
series;that is, different arrangements
in different
series
can
result
values for
conditionallyconvergent
(in just
could
we
that
do this
cannot
we
the sum of the series!
For
+
^(-l)
\342\200\224
S
let S
and
again
=
convergent series
the conditionally
consider
example,
^ 0
be its sum,i.e.
1111111
+
+
l--
2 3 4
+ ...
+
6
5
8
7
consider the followingrearrangementofthe terms
that ensures each occurs still just the onceand
pattern
And then
a
into
with
the
correct
sign:
1
-+
12
7
(--\342\200\224)
K
2)
:-1
4
U
1
1
+ -
8
6)
1
-1^, 1--
+
+
+
1
+
12
10
8
2345678
2^
1
\342\200\224
+
4 6
1111111
2
10J
+U
-
14
_1
14
+ ...
16
1_
16
+ .
1-S
2
Thus, since S ^ 0,we
dependingon
how
the sum
choose
to order the terms, which
we cannot re-order the terms of a
have
we
a different
value for
is clearly
hence
absurd;
conditionallyconvergentseries.
3
C
XJ^\\
+ d.
Infinite
series
'a
where
case
The
inconclusive
^^
\"*4
lim
divergence.
Uk+\\
lim
The Ratio Testjs
the
when
kl or
< 1, then
hence
(and
evolves
next
section.
if:
k=\\
convergence/
useful
the
determine
to
tests
Given a series ^juk>
of the other
to one
study of power seriesin
Ratio Test
resort
to
have
will
useful and will
is particularly
convergence
KEYPOINT3.il
^.
you ^^
occurs
this
for
in our
central
next section. When
^w
test
in
the
in
detail
more
The final
will
examined
be
Ratio Test
Test is
Ratio
the
the seriesis absolutely
convergent
convergent)
> 1,
then the
seriesis divergent
uk
often
uk
oK
Uk+1
\342\200\242
lim
= 1,
then
Test is
Ratio
the
inconclusive.
Uk
orked example 3.9
the
Using
Ratio
Test,
not the following
whether or
establish
series
k\\
(a)
I
(b)
10*
k=\\
<c>
K-if1
(b) you
may assume that
1+
lim
v
Establish
(2fc)l
I
k=\\
f
In part
(absolutely):
converge
for
expressions
readiness
to apply
-1 | =e
n
u^and uk+] in#
the Ratio Test
(a)
w
(-3)V
ak\" = -\342\200\224
and
\\0k
^+1
(-3)t+i(fc+iy\302\260
\342\200\224
\\0k
Jk+]
Find
ensuring
rearranged
into
a
that the
convenient
the
expression is##
form
limit
to
to enable
be
So,
}0k
+ lf
(-3)k+\\k
\\0k
found
(-3)V
+
(-3)k+\\k
\\0k
tf\302\260 \342\226\240
X
\\0k
(-3)kk'\302\260
k + \\\\\302\260
-3
1
10
\342\226\2405
k + \\
101
k
1+ 5_
10
v
82
Topic
9
-
Option: Calculus
r^
**.
j
be
continued...
the
Take
k
as
limit
oo and
\342\200\224>
the criteria
with
compare
of the
:
Hence,
Test
Ratio
lim
fc-\302\273oo
1
101
uk
Therefore
Test the
Ratio
the
by
10
J
series
convergesabsolutely.
Establish
for
expressions
ujk+i
Find
that the
ensuring
uk
and
k\\
uk+]j>
kk
form
convenient
a
the
to
limit
to enable
(k+])M
{k +
=
(k +
\"k
(*t +
found
be
+ l)l
So,
expression is*
\"fc+1
rearranged into
(fc
J
rivi
the Ratio Test
to apply
\\)\\
kk
1)fc+1
k!
1)I
k\\
*r*
+ lf1
X(fc
k
1
=
X
(fc + 1)x
k+1
-&J
1
1+Hence,
We are
given
in
the
that*
question
lim
-<1
1 + 7-
Mm
Uv
absolutely.
converges
Establish
for u^and uk+]
to apply the Ratio
readiness
-^
Find
in*
expressions
^ =
Test
that the expression
ensuring
(c)
(2fc)l
a
into
the
Note
that
the terms
all
as
form
convenient
is no need for
limit
to
+
1)!)2
I
k\\
found.
modulus
{2k)
\\2
{(k +
(2k)!
are positive there
the
1))'
(2(k+^x(^
So,
to enable
be
+
uk+^
^^u\"i\"((k+i)!)2
((k
rearranged
(2(^
and
-j\342\200\224^
(^r
is##
Ratio Test the series
by the
Therefore
+ i)
+ 2)(2k
_(2k
sign
i)\\
_4k2+6k + 2
k2+2k
+
A
4 +
,1+
Take the
limit
k
as
the
oo and
\342\200\224\302\273
criteria
compare
of the Ratio
with
Test
\342\200\242*
Hence,
Therefore
lim
6
2
\\
\342\200\224
\342\200\224+
k
k2
2
1
k
k2
\342\200\224
\342\200\224+
-^1
*->-
= 4-> 1
Uk
by the
Ratio Test
the seriesdiverges.
>^>
3 Infinite
C
XJ^\\
+ d.
series
'0>
The
lim
result
1Y
(
1+ -
<\302\243
was
in Exercise
established
in Worked
, used
example 3.9,
n)
V
<*&
7.
IB, question
Exercise 3B
1. Showthat
the
(a)
(i)
(b)
a)
y^-
(c)
(i)
X
following
~^\342\200\224
Y
converge
(ii)
using the
Comparison Test:
iw^)
(ii)
I
^W+lk
k=\\
(ii)
x
+1
Jfc-2
Jt=3
fc3+3
k2+3k+ 5
k=2
\302\260\302\260
1
(ii)
X1
?**
2t+3fc
(e)
(i)
X
fc=l
(f) (i)
2.
(ii)
a
quesfon
requ.res
exam
CO
fc=4
(b)
wffl
yoU
be
Jed
However,
\342\200\242lt
for
J
fc=2
this
(i)
y^z\302\261
62^+1
+ 1
fc
X
(ii)
(u)
>
(ii)
X
Vfc4-2
X
;n.*\302\253
so look out
{K + kf
3k
fc=l
it!
(e) (i)
Vfc2+i
Vfc4+i
X
(ii)
fc=l
(i)
k\\
(ii)
X
X
+ k
M\\x
k=\\
3.
X
fc=l
x +
(f)
whether the followingconvergeor diverge
Test.
Comparison
Determine
Limit
(a) (i) X
Jt=2
4
(ii) XJ+3
Sfc
]fc3-2
9*
<\302\273\302\273
w
(ii)
i^1
^5^-4
(c)
a)
Topic 9
-
Option:
r^
Calculus
*a.
+
X^\342\200\224
\302\2433*-l
y\342\200\224^\342\200\224
ti2k
84
or
-j
^2fc2+7fc + 4
(c) (i)
(d)
occasionally
ik2+2
Test
following divergeusingthe Comparison
X
not
to knov/
expecteat\"
exercise
(i)
given
you are
Ls_
|
use
lim\\H-l=e'
of
S
Test:
Divergence
tf
*2
?
Show that the
the
in
| sin /c
cos2fc
I
yfk
+ 3
(ii)
y\342\200\224-\342\200\224
\302\24324k2-k-l
using
the
4k2-k + 5
k2+5
(i)
(d)
\302\243
k=2
k4-2
(\")
I
(\")
I
3fc+ 2
tls/3 +
4.
the
Using
k5
whether the following
determine
Test,
Integral
tlj4k7-3
convergeor diverge.
\302\260\302\260
i
w
(i)
(i)
(b)
(ii)
l^
Ivfe
\302\243-
(ii)
I
(n)
i
(ii)
V\342\200\224sin
*=4*2+9
k=4
k=i(k2+9)
\302\260\302\260
3
i
\342\200\224\342\200\224
(c)
a)
y
(d)
(i)
V\342\200\224cos
(e)
(i)
I
\342\200\224
Infc
' i
>^
U2y
(ii) !>\"
\342\226\240/c2
fc=l
5.
Series Test to determine whetherthe
or not. Where they do, find
an upper
converge
the first 10 terms of the
(to 3SF) on the error in taking
the Alternating
Use
series
following
bound
seriesas
an
approximation.
k+1
(a)
(-if
(-1)'
(i)
(ii)
\302\243
a=i
(b) (o
i(-if^
(\")
3A: +
4
(-2f
inr^1
*=i
(\")
\302\243
fc=2
(-3)'
I
k=2
COs(/c7l)
(d)
(i)
Jk
t
fc3
fc=l
(c) (i)
I
COs(/c7r)/c
(ii)
\302\243
fc=2
In*
\302\243
k=\\
k\\
fc+1
(e)
{-kf
(\"*)'
(i)
(\")
\302\243
fc=l
6. Determine
whether the following
k3k
i
absolutely,
converge
or diverge.
conditionally
(a) (i)
I
cos/c
\302\243
k=\\
/#.x
>A
(n)
>
sin/c
K
4k2
(b) (i)
g^'\"^
S(-)'^
COS
(d) (i)
\302\243
i-Jk
(fat)
+ lk-l
(ii)
(-1)'
I iilk2+2k
k+1
(-2)'
(e)
(i)
(-) l(-l)%
\302\243
i(2k+i)\\
r*a*
+3
<\"\302\273
k=2
+*.
...
+l
7.
Use
establish whether
Test to
Ratio
the
the following
series
or diverge.
converge
k\\
(a)
(i)
+ l)
S(2*
^,
(b)
(i)
(ii)
Svs
k\\
Jt=l K
e
(2*01
(c) (i)
k=\\
(klf
2kk\\
(d) (i)
(\")
I
(ii)
I lk
k3k
(i)
(e)
\342\226\240J
rtk
-'
=1
Determine whether
+ 4
each of the following
series
or
converges
diverges.
fc
(a)
+ 1
arctan/c
(b)
I
I
a
(c)
i(-ir
[8 marks]
/c3+l
*=i
(a) Find the values
of
which
x for
whether the series
(b) Determine
CO
^
-j
[5
kin
marks]
or diverges.
converges
Determine
VA:5+3
whether the
series
^ cos(kn)4k
k\\
k=\\
diverges,
or converges
conditionally
converges
absolutely.
[6marks]
(a)
Show
the series
that
H)fc
tt{k\\y
converges,
(b) Determine
how
approximate
the
many
sum
terms
series need
1
to within
Determine whetherthe following
clearly stating your reasoning.
(a)
of the
[6 marks]
1000
series
be taken to
converge
or diverge,
|;{^/^TT-^/^T}
k=\\
(2*)!
(b)
I(-l)'
k=0
3kkl
[9 marks]
(a) Show that
series
the
k-1
y
K-
k=\\
is convergent,
(b) Find
(a)
Show
the sum of the series.
[7
general term
the sequence with
that
uk
marks]
4k2
=
k3+9
for k> n stating the value
is decreasing
(b) Show that
the
of
n.
integer
series
the
a+i 4k2
i(-ir
converges.
(c) Determine
this
whether
or
is conditional
convergence
absolute.
[8 marks]
Forthe series
^> (3/c)!
/
v
c>0
S(Jfc!)V
determine for which valuesof the constant
c
[7 marks]
gives convergence.
Showthat
the
Test
Ratio
the
series
sin
A;
i-
[6 marks]
converges.
(a)
Given
write
(b)
a positive,
down
Hence
give an
partial
sum
Jn
upper bound
as an
poo
sums for
and lower
the upper
f(x) for all x > 1,
function
decreasing
the
for
error
dx.
/ (x)
the
in using
nth
to ^f(k).
approximation
k=\\
(c)
bound on
Find
an upper
sum
to estimate^>
the error in
using
the
5th partial
1
\342\200\224
k=\\ K
(d)
How many
approximate
3
terms of the seriesisit necessary
to take
^\342\200\224
k-
10~6?
within
to
[9 marks]
^5
(a) Show that
00
Ak+l
Y\342\200\224\342\200\224
\302\243(3*
+
l)!
converges.
(b) Test
for convergence:
> V^
r^
**.
^\342\200\224
(3/c + l)!
ti
,,
[10
marks
series
the
For
(a)
k=2k(lnk)
determine for which valuesa the seriesconverges
and
for
it diverges.
which
If
or diverge.
(b) Determine whetherthe following
converge
isconditionalor
they converge state whetherconvergence
absolute.
00
w
I
W
k=2
^jik is absolutely
Show that if
,x2
Xw,
kink
k=2
J
(-if
(-i\302\243
[12 marks]
Jfc(lnJfc)
then
convergent,
k=\\
(ak +1 ak
^
[6 marks]
is convergent.
Ij
k=\\
the Ratio
that
Show
Test is inconclusivefor
Ix3x5x...x(2fc-l)
^
power
series as a
function of x.
]^> chapter
that
are
series
of
variety
powert^>
used
to
wide
a
represent
In this chaptersofar,
we
series
been
has
we
will
convergent,
infinite
series generated
by
a general term; when sucha
the sum has been a number. Now
into
terms also includepowersofa
series whose
consider
if convergent,
igI;
variable
integers
positive
substituting
seen
have
We
next
in the
see
will
series
Power
of a
think
can
We
[5marks]
+ 2)
S4x6x8x...x(2A:
the value of the sum will
therefore
on x.
depend
common
functions.
KEY
r
3.12
POINT
A power
series is an infinite
^jik{x-b)k
series
of the
form:
\342\200\242
\342\200\242
+\342\200\242
-aQ + ax{x-b)+ a2(x-b)2+ a3(x-b)3
k=0
Often
b =- 0 and
this reduces
to
\342\200\224
t*Q ~i U-t J\\> \\~ t*9
J*
+ a3x3
+
...
k=0
As for the otherinfinite
series
in this
chapter,
we would
know when (for which valuesof x) a powerseriesconverges.
We
are certainly
familiar with this in some cases already,
example
Topic9
-
Option:
C
Calculus
XJ^\\
+ 01.
the geometric
series:
like to
for
= 1+ ex+ c2x2
(cx)
^
7
h
^
T<-
+
+ c3x3
constant
\342\200\242
\342\200\242
\342\200\242
for
some
ce
R
fc=o
to be
know
which
we
when
\342\200\224<x<-.
convergent only when | ex| < 1,i.e.
c
c
More generally, it
may be the casethat
series,
a power
converges:
^ak(x-b)k,
k=0
x g
\342\200\242
for all
\342\200\242
only
R
value b e R
a single
for
\342\200\242
forallxeIR
DIVERGES
- b <
\\
|x
that
such
Re
R, where
R+.
DIVERGES
CONVERGES
** x
1
is the
The latter
and
R
KEY
POINT
=
case for
our
above, with
series
geometric
b
- 0
- .
c
3.13
Re
number
The largest
IR+
such
that
series
a power
< R and diverges
for | x - > R is
converges
\\x
called the radius of convergenceofthe powerseries.It
-
for
may
be
b
b\\
by the
determined
\\
Ratio Test. If:
\342\200\242
R =
oo then
the series
converges for
\342\200\242
R =
0 then
the series
convergesonly
all
xgR.
when
x
\342\200\224
b.
a complete description of the rangeof values
for
series will converge but sincethe Ratio
Test
does
not help at the points x--R
and x = R we need to consider
these separately each time.
is nearly
which
a power
This
3.14
POINT
KEY
The interval
all
for
points
all pointssuch
point(s)of
r^
this
of convergence of a power seriesisthe setof
which the series converges. It always includes
- b < R but
that
|x
\\
may also include end
interval.
*c
...
orked
values of
what
For
3.10
example
x is ^X
(x
Test to
start
^~
radius
convergent?
the Ratio*
applying
by
the
find
2)
v/c+i
n
We
\342\200\224
i
Let
uk
{x-2f
=
of convergence
7
(x-2f1
v
Then
X-
4k + 2
VkTT
(x-2)k
2
1
+
= (x-2)x
1
k
1+ i
So,
Jk+\\
lim
fc-\302\273oo
I3y the
= \\x-2\\
Ratio Test,
^jjk is convergent
for
k=0
|x-2|<1
^> 1<x<3
We know
the
that
when |x-2| = l,
so we
Ratio
i.e.
when
x =
inconclusive*
l and x = 3,
When
these two
to examine
need
now
Test is
k=0
x = 1 :
k=0 VK -r I
possibilities separately
have
We
will
this
an
alternating
be
convergent
series, so we
by the Alternating
Series
hope*
Test
This
is an
alternating
series
with:
(_r
UWi =
Jk + 2
1
\\[k + 2
1
4k +
Apply
result
that'\342\200\224=
0'##
and
\\
=
lim luJ
lim
Therefore, it is
Series Test.
consider
Now
what happens
series
to
the
power*
When
when x = 3
Topic
9
-
Option: Calculus
s>,
+ 0,.
3:
1
k=0
90
x =
k=0 Vk+T
.
:=
0
convergent by
the
Alternating
continued...
is very
This
to the p-series
similar
*
with
\302\260\302\260
1
1
=
p
\342\200\224
know
. We
2
's divergent
X-T
and
Then
*=!
P2
like to show
so would
series is larger
term
of
this
each
than
corresponding
therefore
p-series,
of our
term
each
that
our
making
series divergent by the Comparison
Test.
each
term
of
our
series
is
However,
actually
smaller
than that of the p-series. We therefore
to use the
attempt
Limit
Test
Comparison
of
algebra
Apply
limits1
1+
Therefore,by
= \\>0
0
the
Limit
\302\260\302\260
1
since
d'werqee, so
X-j=
doee X
information
this
is it
then
Why
of a
So, the
together with that from4
the radius of convergence
to know the interval
so important
power series? In part, the answer
to
this
of
1
r\342\200\224-
+ 1
tW/c
tW/c
Put
Test,
Comparison
\302\260\302\260
series is convergent
for
1<
x <
3
convergence
is:
3.15
KEY POINT
differentiated or integrated term
contained within its interval of
by
at the end points
of the interval).
convergence(andpossibly
A
can be
series
power
any interval
over
term
Outsideof
this
interval
differentiate
cannot
we
or integrate in
this way
orked
3.11
example
1
Find an
expression for
Since
example
3.10
we are
lies entirely
here
term
integral
(x-2)
interval
within
the
of
iz h
iz
to
^/k+T
Jk + i
series
power
to
>.(x-2f dx
Worked*
in
that the interval
asked
ax
,
over
=1
(x-zr
{k
integrate
+ X)Jk
+
~\\
k+1
interval
we are able to
by term, i.e. pass the
inside the sum
sign
of convergence,
integrate
>
we established
convergencefor this
is [1, 3[ and the
which
f5/2~
=1
=
1
(*+T
s(fc+ir
3
m&x*'**
\\
-0
Infinite
series
'a
3C
Exercise
the radius of convergence
and
of
the
convergence
following:
1. Find
(a)
(i)
(c) (i)
2. Find
(ii)
fjclx\"
the
give
interval
of
+ 3)k
fjcl{x
(u)
|^-
lirr
the valuesof x
the
which
for
following
converge:
\\k+\\
K
k=\\
*J x
K) yl
^
(b)
fc=o
W
I
(2Jk
+
l)!
(-l)*s2*
S (2*)!
Findthe
power series
which the
x for
of
values
fcV+1
y
[6 marks]
converges.
Find the
interval of convergencefor the powerseries:
^5/c
+
(x
[6 marks]
4)
k=\\
(a)
of convergence
the interval
Find
*yZ,
f(x)
}
v
(b) Find
= l
*yO
the
for
power
series
*yQ.
+ x +\342\200\224
+ \342\200\224
+ \342\200\224
+ ...
3!
2!
f'{x) stating for
which
4!
values
of x
this holds.
[6 marks]
iS For
the
series
>
-y
S(fc
(a)
Find
the radius
(b) Find the values
^\342\200\224
+
1)4\302\273
of convergence.
of
x for
which the
series converges.
[9marks]
. 92
\\t>
I
Topic
9
-
Option: Calculus
/v
-
^
v*\\
+ d.
of convergenceof the following,
the interval
Find
of any
nature
end
points
the
discussing
of the
interval.
k5
K
k=o
00
X5k+1
(-l)k
(c)
convergence
the
at
J
X
[24 marks]
1
Summary
\342\200\242
The
nth
sum, Sn, of a
partial
seriesis the sumofthe first
Sn =
series
infinite
An
n terms:
+ un
u1+u2+-~
to S if the sequenceofits partialsums S1,S2,S3,S4,...
converges
converges
toS:
S =
limSw =
y\\u
k
k=\\
\342\200\242
Test
Divergence
If lim%
^ 0
\342\200\242
or if thelimitdoesnotexist,
the
two series
Given
of positive terms ^jik
and
k=\\
to a limit
is convergent
^bk
k=\\
so is
is divergent,
k=\\
^bk
ak <
that
such
all k,
for
bk
then if:
k=\\
S, ^Jik is alsoconvergent
to
a limit
T where
T <S
^bk.
k=\\
Test
Comparison
Given two
series of positive terms V%
so
converges
\342\200\242
the other
does
and
\\bk,
where
and if one seriesdiverges
so
=
lim\342\200\224
/ >
0, then
if one series
the other.
does
Test
Integral
-
is divergent.
k=\\
^jik
Given
V%
Test
Comparison
\342\200\242
Limit
series
a positive
convergent
decreasing function f(x), x > 1,if J
then
^f(k)
f(x)
dx
is:
is convergent
k=\\
-
divergent
then
^f(k)
is divergent.
k=\\
3 Infinite
r^
**.
series
'a
\342\200\242
An
\342\200\242
Test
Series
Alternating
alternately positive and negativeterms.
is one with
series
alternating
an alternating series ^juk:
If for
k=\\
|<|uk
Iuk+i
and
keZ+
f\302\260r
aU
|
lim|uk
|
=0
then the seriesis convergent.
If
S =
sum of an alternatingseriesthat
is the
^uk
satisfies:
k=\\
I
then
the error in
than
the absolute
uk+\\
I
< I uk
taking the first n
value of the (n +
aU
I f\302\260r
I
k^oo
as an
terms
lim
and
keZ+
uk
I
= 0
to S (thetruncationerror)
approximation
term:
l)th
|S-Sn|<|nn+1|
\342\200\242
A series
convergent if the series ^J%
is absolutely
^uk
k=\\
If a
series
is
convergent.
k=\\
is convergent
^Jtk
but ^J
uk
|
then
is divergent,
the series
is conditionally
k=\\
k=\\
convergent.
convergent series
An absolutely
\342\200\242
Ratio
is convergent.
Test
^juk, if:
a series
Given
k=\\
lim
uk+\\
then
< 1,
the series
is absolutelyconvergent(and
hence
convergent).
uk
Uk+\\
lim
> 1, then
uk
uk+\\
lim
A
then
-1,
k^oo
is an
series
for
-
94
Topic
all
for all
9
-
(x
-
b)k
\342\200\224aQ+ax{x-b)
be
IR
+ a2
(x
converge:
only for
-
test is inconclusive.
infinite series of the form:
^jik
k=0
-
the Ratio
Uv
power
It may
the seriesis divergent.
x e
R
a single
xgR
value
such that
Option: Calculus
r^
+\302\253.
\\x
b\\
< R,
where
Re
R+.
-
b)2
+a3(x-
b)3 +
\342\200\242
is less
'a
h
The largest numberR e IR+ such that a power series convergesfor \\x - b\\
| x b | > R is called the radius of convergenceofthe powerseries.It may
Ratio Test. If:
-
R
= \302\260\302\260
then
the
-
R
= 0
The
interval
then
series
the series
converges
be
and
for
diverges
by the
determined
for allxgR.
b.
converges only when x \342\200\224
of a power seriesisthe setofall
of convergence
converges. It always includesall pointssuch
of
< R
that
|x
- b <
\\
points
R but
for
which
may also
the series
include end point(s)
interval.
this
A power
series can
within its interval
be differentiatedor integratedtermby
of
term
over
any interval
contained
convergence.
t-
r-v
3 Infinite
.
i*
i~)A
+ oi.
,. *
I
series
95
'a
practice 3
Mixed examination
the
whether
Establish
following
or diverge:
converge
3k
(a) T-^-
1
*=iK
...
^
(b)
2j
5^ + 81
[7 marks]
t3k2Jk+7y[k
Find the
radius of convergenceofthe following:
k3xk
(a)
I-
k\\
k=0
X2k
(klf
(b)
X
u-o
[6 marks]
(2k)l
(a)
Show
(b)
Determine
that
\342\200\224
x2t
f{x)
x
for all x > 2.
is decreasing
whether
2>
2e-k
k=2
[8 marks]
or diverges.
converges
y
A
\\y
X
3
4
5
6
The diagram shows part of the graph
parallel to the coordinateaxes.
(a)
Using
the diagram
y
\342\200\224
\342\200\224
with
together
line
segments
x
show that,
r l
111
of
111
\342\200\224
+ \342\200\224
+ \342\200\224
+ \342\200\242\342\200\242\342\200\242<
+ \342\200\224
+ \342\200\224+ \342\200\242\342\200\242\342\200\242
\342\200\224dx<\342\200\224
43
Hence
(b)
53
63
find upper
J3 x3
33
43
53
and lower boundsforthe sum ^
\342\200\224
IB
(\302\251
96
Topic 9
-
Option:
Calculus
m&x*'**
Organization
[15
marks]
2008)
Consider
the sequence of partial sums {Sn}given
K
k=\\
Show
(a)
that for
by
all positive integersn
s2n>sn+\302\261
Hence
(b)
sequence {Sn} is not convergent.
prove that the
[7
IB
(\302\251
^
(a) Use the IntegralTestto show
S
2002)
Organization
series
the
that
marks]
1111
= lH
1
22
1
h...
1
42
32
converges. &
52
Let
(b)
,,1111
Show
that
= 2|
S-A
\342\200\224
\342\200\224
\342\200\224
+
22
(c) Hence find
Determine
the sum A,
(a) Show that,
/1
x
2 y
where
down successive
+2
f^yk
2
(d)
[9
marks]
[6
marks]
6
diverges.
terms of the series,
A
^~
\342\200\224.
B
A
2
k2+5k + 6 k + 2 A; + 3
A and B are constantsto be found.
that,
to
converges.
1
show
+ .
62
convergesor
ni
By writing
+
42
S converges
that
v
\\
(b) Show that
(c)
given
arctanx
whether ^\\
k=\\
El
52
42
32
22
2
k^k2+5k + 6 3
B
k+3
2
n
+ 3
Hence find,
txk2+5k + 6
[11marks]
3
m&x*'**
Infinite
series
97
3
Test the convergence
or divergence
reasoning
(a)
the
of
your
explaining
following,
fully
K-D\"V
n=0
cos (mi)
(b)
(c)
\302\243
n=\\
Un
n=2
J
[13marfcsJ
^(-l)-1\342\200\224^_
n(lnn)
The coefficientsof the powerseries^Jikxk
the
satisfy
relation,
k=0
ak
(a)
+ A%_!
+ Bak_2 =
0, where/c
Show that for | x \\ <
(l + Ax
+ Bx2
= 2,3,4,...
of convergence,
the radius
R,
=
)^akxk
a0-\\-(a1 + Aa0
)x
k=0
(b)
establish
Hence
that,
n
j
=
Xxk
for
Use the
2^
(a)
Ratio Testto determinewhether
[8 marks]
or
not
[9 marks]
Find an
expression
for
the
(b) Hence determinewhether
Find the interval
of
nth
or
not
the series
of the
convergence
sum of
partial
Show that the
[8 marks]
converges.
series
( \342\200\224
n}
(x
^sin
(a)
< 1
is convergent.
3+2
k=i
Ix I
x
-1
k=o
k
[12
\342\200\224l)
function
\342\200\224
is
decreasing
for all
marks]
x >3
xlnx(ln(lnx))
(b)
By
making
the substitution
u-\\nx,
l
evaluate Jr
j
dx
^2/3\"
xlnx(ln(lnx)j
(c)
Hence
determine whether Zu
,, ,/, /, , \\\\2/3
k=3kmk[m{mk))
98
Topic9
-
Option:
Calculus
converges.
[15
marks]
\"\"
(i
^
\342\200\224
~~h
-+
In this chapter
Maclaurin
\342\200\242
and
find
to
how
series in
functions
Taylor
\342\200\242
how
series
power
certain
x for
the error
find
to
the
only
taking
terms of
as
an
of the
end
the
previous chapter
were
that these
noted
they were actually
of a
functions
^Jikxk
=
power
series as
igI:
\342\200\242
to find
x-o
\342\200\242
\342\200\242
+ a3x3 + \342\200\242
its interval of convergenceand get a finite
that is a value for the function at that point.
we are already
function with which
sinx? Canwe
can
We
relate
power
a start
make
already
course that,
the main
for
/(x)
series
a power
has
series of first
geometric
(l +
expansion
powerseries,
powerseries.
of
series
to
functions
form lim-f(x)
it be
a
know
from
function
(l + 3x)-1
since it is the sum to infinity
function
be possible
should
limits
the sum;
of
term 1 and commonratio -3x;that
If it is possibleto expressthe
find
power
as ex, lnx or
such as these?
+ 9x2-27x3+81x4-243;c5+...
3x)\"1=l-3x
it
=
of x within
such
at answeringthis;we
the
certain
'gM'
Could
familiar,
example,
for
in
is
functions
of the
to functions
series
for
value
would we be evaluating?
function
what
But
a value
aeIR
(where
\342\200\242
to use
series is convergent,we can input
power
results
series
power
a constant)
k=0
If a
series
functions
using standard
series
power
infinite
other
variable
+ a2x2
+ axx
a0
we lookedat
from
different
first n
a power series
of composite
and
in
approximation
\342\200\242
to form
At
you
will learn:
=
/ (x)
to express
a
is:
|x|<-
3
(l + 3x) asa
other functions as
In chapter we look at methods for finding power series
and see how we can use themto
representations functions
make
for these
functions
and to evaluate limits.
this
for
approximations
series
Maclaurin
As
are
we
able to
manipulate,
series
their
idea to
express other types
(within
analyse
them
in the
radius
differentiate and integratepower
of convergence),
it seems
of function
as series so
a good
that we can
same way.
4 Maclaurin
C
XJ^\\
+ d.
and
Taylor
series
99
'a
To this end, consider
a
series
with radius of convergence
power
R > 0 and
suppose we define a function:
f(x) = a0+
x =
Letting
+ a3x3
+ a2x2
axx
we
\342\200\242
\342\200\242
\342\200\242
\\x\\
< R
0 we immediatelyhave:
/(0) =
Since
+
Oo
the power
series for
|jc|
f'{x) = a1+2a2x+ 3a3x2
+ 4a4x3
+
can differentiate
< R,
we have:
\342\200\242
\342\200\242
\342\200\242
= a1
/./'(0)
again:
Differentiating
/\"(*) = 2a2
+
2)a3x +
(3 x
=
.-./\"(0)
\342\200\242
\342\200\242
(4 x 3)a4x2+ \342\200\242
2!a2
/\"(0)
= \342\226\240
\342\200\242a,
2!
And
again:
= (3 x 2)a3+ (4 x 3x
f'\"(x)
=
.-./'\"(0)
+
2)a4x
\342\200\242
3!a3
/'\"(0)
= \342\226\240
\342\226\272a^
3!
Continuing
way we will get:
in this
nx)=m+imx+rmx,+rmx,+...
0!
1!
2!
KEY POINT
series
/(*) = /(o)+/'(o)*+
known
as
the
see that
]^>
will
it is useful to
be able to
find series ]^>
expansions about
points other than 0.
As we
the
that
are evaluating
series
Maclaurin
Topic
9
-
C
Option: Calculus
XJ^\\
+ d.
2,
-xz+;
series
f(k\\0)
3,
x3+-
of the function
exists
for all k
f(x).
e N.
all of the derivatives
at x = 0, we say that
of a function is centredat 0, or talk of the
series expansion around0.
we have
=
met
so far (for example f(x)
lnx doesnotsatisfy the above
condition
of f(k\\0) existing for all k e N; indeedit doesnot
exist
for any k e N) but we can find
Maclaurin
series
for many
We
cannot
find
a Maclaurin
familiar functions.
100
/\"(0).2^/'\"(0)
Maclaurin
It existsprovided
In Section4D we
jrMx\302\253
Jfc!
it=o
4.1
The power
is
=
3!
series for every function
orked
4.1
example
Find the Maclaurin
(a)
for:
series
(b) g[x)-sinx
f[x)-tx
Give your answersin the form ^
to establish
need
We
few derivatives
that
see
We
the first*
of
(a)
clearly
(0)
for all ke{0,1,2,...}
=
f\"(0)
=>
1
= 1
f'(0)
=>
f'\"{x) = e*
=
f(0)
=>
= e*
r(x)
= 1
series*
Use the Maclaurin
=>
f'(x) = ex
x = 0.
f<*>
= ex
f(x)
and
f(x)
them at
evaluate
akxk
1
= 1
f'\"(0)
So,
formula
\342\226\240pfrrt
.pvrrt
.p'Vn-\\
f(x)=M+m,+no)x2+no)x3+.
1!
3!
2!
0!
1
1
1
\342\200\224
+ \342\200\224X
+ \342\200\224Xz
+\342\200\224X\302\260
+\342\200\224X4
=
0!
= 1+
4!
yJ
'y'-x
Xzi
\342\226\240AiAX\"
+\342\200\224
+ \342\200\224
+ \342\200\224
+ \342\200\242\342\200\242\342\200\242
+ \342\200\224
+ \342\200\242
x
4!
3!
2!
_
3!
2!
1!
mI
%k
v
k=0 K!
to establish
need
We
few derivatives
that
see
there is
that
a
cycle
the
f(x)
of 4
continually
repeats
(b)
^(^
=
=> f(0)
006^
= -e\\r\\x
f\"(x)
f'\"(x)=
-C0SX
such
=>
f\"(0) = 0
=>
f'\"(0)
1!
0 1
+
= \342\200\224
+ -x
1!
0!
_
X5
+
~X~3T
that
this
is an
term,
general
alternating
will
We found in
=
f(5)(0)
0 .
3!
2!
\342\200\224
+
_
.
_
-1
1
0
\342\200\242
+ \342\200\224x4
+ \342\200\224x5
\342\200\224x5
2!
3!
x2
X7
X5
(2k
5!
4!
+ \"'
+ 1)!
them
Exercise 3C, questions1(b)(i)
and
(although
2(b)
that
we
both
did not
4
+ d.
= 0
= 1
1
these power series convergeforallx e R
relate the seriesto exand sinx).
S?i>*
f<4)(0)
~5!\"~7!\"
lH)k
series and
only have
generate
= -1
X'2k+1
we note*
so need (-1)^, and as we
odd powers and factorials,2/c+
1
So,
0!
the
=
f'(0)
w
f(x)=M+mx+no)x2+no)x3+..
To form
=0
=>
f(4)(x) = sinx =>
f(5)(x) = cosx =>
values
series formula*
Maclaurin
= sinx
f(x)
and
= f\302\253(0)
that^+4)(0)
Use
the first*
them at x = 0.
as f(4) (x) = f(x)
evaluate
We
of
Maclaurin
and
Taylor
series
As mentioned above,we cannotfind
\342\200\224
\\vl%.
y
for y
we can
However,
problemof/(0)
the
all
(and
ln(l
for
this avoids the
= 0) not existing.
x) as
+
at x
derivatives
series
Maclaurin
the
\342\200\224
orkedexample4.2
series
Maclaurin
the
Find
few derivatives
of
f(x) = ln(1
evaluate
them at
can
We
see
alternating signs
the Maclaurin
Use
x = 0.
=
f<4>(z)
+
-3l(1
f(5)(x) = 4!(1+
factorials
series formula
\342\200\242
= 0
f(0)
f'(0) = 1
f\"(0)= -1
r\"(x)=2i(\\+xr
of
a pattern
of
+ x)
f'(z) = (1+ z)\"1
=
f\"(x)
_(l + x)-2
and
f(x)
form ^akxk.
in the
answer
your
the first*
to establish
need
We
= ln(l + x), giving
for f(x)
*r
f(5)(0) =
x)-5
1
0
3!
3!
4!
-
sum
the
with
(-1 )k+
]
to
first
the
\342\200\242in
few
derivatives
at x
the series
how
Exercise3C,question
converges for
x < 1.
- u
wi\302\273
behave.
the
<
-1
Clearly,to have
ot
to be confident
We found in
of
any
at x,
function
converge
hope
of the
we can only
2(a)
to
that
that
this
power
series
function.
four non-zero termsofthe Maclaurin
cosx
(ii) sin3.x
(b) (i)
e*
(ii)
(c)
Vl
(a)
(i)
e~3x
1
(i)
+ x
(2-3x)
(d) (i)
-
5
series converging to the value
take
a value of x inside the
functions.
following
9
5!
4A
Find the first
Topic
4!
\342\200\224
x\302\260
interval of convergence
ofthe Maclaurin
series
of a function.
if
we
know
that
a
Maclaurin
series
However,
converges for
particular valuesof x, this does not mean that it converges to
the function it was derived
from!
We will see in the next section
how
we can
determine
when the Maclaurin seriesof a function
doesindeed
Exercise
x^+
,(-ifV
alternating series but*
starting at k = 1, we need
make
the first term positive
is an
Look for patterns
+ :
4
3
2
3!
2!
= \342\200\224
+ \342\200\224
x
xz+\342\200\224xD
1!
0\\
2!
this
41
So,
f(i)=^+no)x+no)x2+no),5.
0!
2!
Again
= 2\\
f\"\\0)
Option: Calculus
S>,
+ 0,.
tanx
(ii) secx
series
for
the
Findthe
f(x)
Showthat
f(x)
x
r<
f(x)
[6marks]
12
6
x4 in the Maclaurinexpansion
of
up to
series to find an
approximationin terms
In 2.
for
71
+
ln(cosx).
(a) Showthat
the
1L
=
-\\
21
(b)
Find the
(a)
Find
l +
[7 marks]
1
1X3
8
2!82
-x +
2
x2+
Find
=
for f(x)
series
Maclaurin
1X3X5
3!83
1
^4-x
is
,
+ Ix3x5x-x(2\302\253-l)
x3+-
expansion of
the Maclaurin
of convergence
interval
the
xn+-
n\\8n
radius of convergenceof this series.
your answer in the form ^jikxk.
(b)
for
=
Use this
of
2
term
the
Find
(b)
the term in od
is
x
(a)
up to
series
Maclaurin
the
ln(l+sinx)
Q
Maclaurin series for
[5 marks]
of the
terms
non-zero
three
first
= .x sin2.x.
/ (x) =
x
In(l
[8
+ x),
marks]
giving
of this powerseries.
[9
marks]
(a)
Find
(b)
Prove
S
the Maclaurin
this
that
power
expansion of
y/l
up to
+ x
series converges
the term in
for |jc|< 1.
[9
x3.
marks]
\\
y = f(x)
Explain
why neither
the function/(x):
,
x
(a)
-1 + -x
2
(b)
2
1-3jc-
+
of the followingcan beMaclaurin
series
of
x2
\342\200\224
+ \342\200\242\342\200\242\342\200\242
8
-..
\342\200\224+
4
4 Maclaurin
^3,
v*\\
+ 01.
and Taylor series
to the
Approximations
Maclaurin
series
you
value
for, say,
One
back
Look
the
at
expression for
^>-
truncation
error
^^ an alternatingseries^^
in
Section
3B
(Key
point 3.8).
calculator
the
of x are
it
power series.It cannotof course
this
many
enough
to give
of such
a sufficiently
to the issuefor
with
inputted,
series, so how many
answer?
accurate
in chapter
series
alternating
This is similar
3 of this option,
for the truncation error at n
but
the Maclaurin
series;
here, by truncating
series,
we are actually determining an nth degree
which
polynomial,
the function.
approximates
we
terms
of the
4.2
POINT
Maclaurin series:
\302\261r!mx,=m+mx+rmxl+...+
k\\
0!
1!
2!
n\\
k=0
to as
is referred
of
pn(x)
the
the nth degree
function
Maclaurin seriesfor
near
approximatethe function
can see how accurate the first
sin x are.
So,
the
for
have
c
xj^\\
+ d.
x~
degree
x (Worked
sin
to x
few
Maclaurin
\342\200\242
For
third
sin
x-x-
example
4.1(b)) to
(here we usex =
polynomials
0.5),
we
for
polynomial we
x:
give us sin 0.5~ 0.5,which
the
= 0
Maclaurin
This would
have
Option: Calculus
sin
first
Maclaurin polynomial,
f(x).
Using the
-
is
terms
an expression
found
where
The truncated
9
can
use
terms
infinitely
KEY
Topic
a
sin 0.5 or e3?
evaluate
simply
of^>-
calculator or computerfinds
way is to program the
series so that when values
possible
Maclaurin
the
how a
wondered
ever
Have
degree Maclaurin
xJ
is an
error
of just
polynomial we
over 4%.
would
This
~ 0.47916,
us sin0.5
give
which is only
a
error.
0.05%
\342\200\242
For
~ x
sinx
have
Maclaurin polynomial
fifth degree
the
:
I
5!
3!
This would
we
which
give us sin 0.5~ 0.4794270834,
is now
to 5DP.
accurate
becomesmore and more accurate
Clearly, our approximation
the degree of the polynomial;
with
the following
can quantify
the
truncation
error for any
Maclaurin series.
the higher
result, we
KEY
For a
the
4.3
POINT
Although
function f(x)
for
all derivatives
which
for the
expression
evaluated
error
Lagrange
at x = 0 exist:
Maclaurin
for
term
series
in
does not appear
k=o
the error
where
term
Rn
(x)
the
k\\
the more
is given
case br
by:
series,
R
J-f(n+l)
(x) =
\342\200\224xn+l
for
some
ce
]0,x[
given
+ l)!
(\302\253
This is
error
sometimes referred to
as the Lagrangeform
general
Taylor
examined
Section
\342\200\242in
(c)
booklet,
Formula
in
4D,
the
is
Formula
booklet.
of
the
term.
4 Maclaurin
r^
+\302\253.
and Taylor
series
The
result is beyond the scopeofthe
of this
proof
possible to understand the Lagrange
about the particular caseofa Oth order
it is
polynomial where
The Mean
cg]0,x[
by
the
there must
that
states
by thinking
Maclaurin
constant)
(i.e.
the error termis given
Theorem
Value
such
term
error
but
syllabus
first derivative.
exist some
that,
/(*)-/<0)
\342\200\224
x
0
=
=>f(x)
which is exactly
In
then
idea
This
of c, soinsteadwe
for
orked
the
an
in taking
error
we will
a function,
approximate
bound
to higher
generalises
the
estimating
nth
value
the particular
possible error
by taking c to be the worst
error)
to
polynomial
degree
know
not
= 0.
order derivatives.
the largest
find
will
point 4.3, with n
in Key
result
the
+ f'(c)x
f(0)
case
(an upper
possible.
4.3
example
(a) Find an expressionfor
term
error
the
in approximating
ex by its 2nd degreeMaclaurin
polynomial.
bound to 4DP on the errorwhen
an upper
Give
(b)
2nd
the
state
First
degree*
(a)
Maclaurin polynomial using the
The 2nd
degree
Maclaurin
e075.
to find
approximation
gives
polynomial
the
approximation
above
found
series
this
using
ex
\302\2531
+ x
x2
\342\200\224
+
2
Find the
error,
with
that*
of e* are
derivatives
all
noting
error term
ex
\302\2432(x)=f(5)(g)x5
ce]0,x[
3!
e\302\260xu
^5\\
at
Evaluate
x =
0.75
\342\200\242
(b)
Taking x
= 0.75 we
=
K2 (0.75)
As
the
an increasing function,
value of c will
possible
ec is
largest
the
give
largest
(the upper
For
bound) on the
.\\ Kz
(0.75)
<
c e ]0,0.75[
3!
^0.75/0753
:
= 0.1439
value
error
HINT
EXAM
a
possible
\342\200\242
have
series
alternating
particular
error
Lagrange
Topic 9
C
the appropriate
conditions),
the
result
for
the
truncation
error
using
(that
of x
value
term.
Calculus
Option:
XJ^\\
+ d.
satisfy
it
in
is easier
Key
point
to bound
3.8
the error
rather
than
at
the
^
^^\"-^
~~h
h
+
use Key point 4.3to bound
errors
from the
resulting
truncation of Maclaurin series,but we can also determine
for which valuesof x a Maclaurin
series
to
actually
converges
the function from which it was derived.
we
said
(Remember,
at the end of the last sectionthat a Maclaurin
series does not
We can
necessarily convergeto the function
for all x in the interval
of convergence.)
=
fM
let n
if we
^^-x^Rn(x)
kk=o
ft
/W(Q)
^
\\
if the
KEY
POINT
error term
*
Ki
k=0
only
Since
sense that
makes
\342\200\224>
<*>, it
derived
it was
which
from
tends to 0.
4.4
I
The
function
\342\226\240
as the
all x where
f(x) for
= 0
lim\\Rn(x)\\
orked
4.4
example
Prove that
the
We
is preciselythe same
series forf(x)
Maclaurin
(infinite)
series
Maclaurin
need to show
lim|/?n(x)|
for ex is
valid
that*
From
= 0
f{n)
for
xgK.
all
exam pie 4.1
Worked
(x)
= ex
= ex fora\\\\neU
f(^)(x)
+
Therefore,
f(\"+1)(c)x\"+1
K(4=
(n +
ce]0,x[
1)l
ecx\"+1
(n +
Simplify the
that
noting
modulus sign,*
ec>0,
(n +
.'.
ecx\"+1
lim
that
constant M e IR,
From
chapter
examination
question
=
we
e'\\x\\n+'
lim
fl\342\200\224>\302\260o
(n + 1)l
M\"+1
J^\\
<M
can
nJ
-
\"->-(n
+ 1)l
lir
foreomeM
of limits
the algebra
apply
+ 1)l
(\302\253
l)!>0
for some**
ec <M
lim
|^n(x)|=
and 1x^1= M\"1
Noting
1)l
1 Mixed
\342\200\242
Since
practice,
6, we know
| lm
\"->~(n
i i\"
\\x\\
\342\226\240\342\226\2400-
+ 1)!
that
lim|K\342\200\236(x)|<0
lim^
= 0
n\342\200\224>\302\260\302\260
ft?
4
r^
4 a.
...
Maclaurin
and
Tayl
continued.
EXAM HINT
And since,
by the
a useful
\342\226\240
=
0 is
lim
clearly,
Squeeze Theorem,
n\\
in
be
can
It
but
case
this
you should
how to
= 0
\\\\m\\Rn(x)\\
to know.
limit
quoted
> 0
\\fcn{x)\\
also know
Hence,
it.
prove
A similar
ex =
V
\342\200\224
for
e
all x
argument works for
cos x.
x or
sin
orked example 4.5
2k
Show
cos x: ^(\342\200\224l)
series for
the Maclaurin
that
*=0
We
(2^)'
to find
for4
f(n+1)(f),
= cosf, so we find the
first few derivatives
and
need
f(t)
is a
a
cycle of
4
f'\"(t)
repeats continually such
that
= smt
fW(t) = coet
that
values
= -coet
f\"(t)
There
pattern.
= coet
f (t) = -sint
f (t)
look for
x for
to cos
converges
7\342\200\224r-
= -sint
f(5)(t)
= f(k)(t)
f(k+4)(t)
\302\261sint
all derivatives
Since
either
that
or
\302\261sinf
\302\261cosf,
be
will
f(n+1)(f)
one
will
we
be*
f(\"+D(t)=
+zcoet
know
of these
We need
\\Rn(x)\\*
M*)l
= &+^(c)xn
(n
+ 1)l
|f(\"+1)(c)||x\"H
(n + 1)l
(n +
f(n+1)(f):
[\302\261cosf
we
that
know
Again
|f
<n+1)
(c)|
|x\"+1|
<
f\302\261sinf
Since
1)!
< 1
= 0
lim^\342\200\224\342\200\224
(n
Since
| 1m
+
1)!
|x|\"+1
:0\302\273
+
\302\253->\302\253(n
1)l
lim|^(x)|<0
And since,
by
clearly, \\Kn(x)\\ >
the Squeeze Theorem,
0
= 0
lim|K\342\200\236(*)|
Hence, cos
x=y
h
108
Topic 9
-
Option:Calculus
S>,
+ 0..
(-if
\342\200\224
for
(2\302\273:)i
allx
all xgR.
Worked examples4.4and
We see from
x for
always
of
the values
that
4.5
the functions ex
with
An example of a function
that
is not equal to its
x
cos
and
to the functions themselvescoincide
exactly
are equal
intervals
series of
Maclaurin
the
which
the
Maclaurin
of convergence of the respectivepowerseries.
This
will
be the case for the functionswe meet,but we still need to
showthis for
be able to
each
f(x) =
function.
particular
series
is
e
x*0
x
x =
0
0
Investigate the Maclaurin series
and
graph
of
this
function.
4B
Exercise
Find an upper boundon the
polynomials of given degree
functions for x - 0.5:
to approximate
(a) (i)
cosx,p4(x)
(ii) tanx,p3(x)
e*2,p2(x)
(ii)
(b)
(i)
(c)
(i) Jl + x~,p3(x)
when
error
the Maclaurin
using
the following
\\p2{x)
1
(ii)
->Pi(x)
(4-xf
Maclaurin
for e~x.
(a) Find the seconddegree
polynomial
(b) For what valuesof x > 0 will this polynomial
approximate
e~x to within 0.001? Give your answerto 4DP.
[6marks]
the
Using
error
when
/
(a) x
X5
.
estimate
to
1
.
sin x
5!
3!
x\302\260
(b)
x
3
< 1,
(a)
= ex
(b) Hence
an
find
in the form
[8 marks]
[6 marks]
Maclaurin
degree
to
approximation
a, b
where
\342\200\224,
the first
x.
(b) Show that
the
/
, giving your answer
e Z
three non-zero
an
upper
on the
bound
[10
marks]
series
series
-
y.2k+l
2k
for
place
terms of the Maclaurin
(_D^\342\200\224
k=\302\260
converges
polynomial forf(x).
b
arctan
for
6
<
the Lagrangeerrorterm,
error in this approximation.
Find
,\342\200\224
6
71 71
e
(c) Using
(a)
T 71 71
+ e~x
fourth
the
Find
r
tor xe\\
that
show
2! 3!
Let f(x)
for x
tanx
estimate
to
H
For \\y\\
an upper boundon the
using:
X3
x
term, find
error
Lagrange
\\x\\
< 1
and
+ \\
that it converges
to arctan x.
[11
marks]
4 Maclaurin
C
XJ^\\
+ d.
and Taylor
series
109
Maclaurin polynomial of ex
What degree
estimateof e to within
guarantee an
Show that
for sin
series
Maclaurin
the
2x convergesto the
[8 marks]
(a) Use the Lagrangeformofthe errorterm
error involved in approximating -ln(l
x+
-
x)
the
bound
at
x
- 0.5
by
+ \342\200\224.
3
/
By noting that
give an
J
to
x2 x3
\342\200\224
2
(b)
[6 marks]
allxeR.
for
function
0~6?
lxl
to
be taken
must
(a) Find
R3
N
(0.5)
(0-5)4
(0-5)5
-\342\200\224= ^\342\200\224\342\200\224+
+
improved bound
the
on the error.
of the
terms
three
first
\342\200\242\342\200\242\342\200\242
5
4
[9
marks]
Maclaurin Series for
(\\ +
xf.
terms of the expansionof (l + x) areneeded
to
a
value
of
1.182
accurate
to
within
10~6?
finding
(b) How many
guarantee
[10
Show that the
function
for
Maclaurin series for
Maclaurin
The
standard
following
met),
in the
appear
POINT
KEY
+ x)
(l
2
to the
converges
< 1.
|jc|
[10 marks]
series
of composite
Maclaurin
series (which
functions
we have
4.5
+ \342\200\242\342\200\242\342\200\242
x + \342\200\224
2!
\\n(\\ +
v
x)} =
+
x
sinx = x
cosx = ,l
2
3
x3
x5
1
3!
5!
x2
x4
1
4!
2!
arctanx
\342\200\224
x
1
5
3
can be
functions.
used to find Maclaurin seriesof morecomplicated
this
Sometimes
is straightforward.
orked example 4.6
We
just
series
Maclaurin
the
Using
need
to substitute
for cos
2X3
into
of cos(2.x3).
x, find the seriesexpansion
the#
series for cos
known
x
, ^
0
Option:
Calculus
+ G(.
6!
4!
2!
3
-
(2x3f
(2x3)4
(2x3)2
= 1-2x6+-x12-
Topic 9
already
Formula booklet:
zx =\\ +
These
marks]
\342\200\224
x16+-
45
Often this will involve
two
finding
and then combiningthem.
separate
Maclaurin
x and ex,
find
series
orkedexample4.7
We
for sin
series
Maclaurin
the
Using
the
series
of esinx
expansion
as
far as
the term in
x4.
the series for*
substituting
as far as the x4 term
going
start
by
sin x, only
We now use the
only going as
f
far
as
ex#
for
series
y-5 \\
1+
+-
+
X+
x4 and
then expand
^1 +
+ \342\200\224
+
x + \342\200\224
2
6
2
3
\"
+ \342\200\242
1+
4!
3!
2!
-
6
24
+
\342\200\242
3!
6
+
\342\200\242
^1 + x +
HINT
EXAM
It is
much quicker to
where possiblein
form
series
Maclaurin
in this
way so
known series.
combine
exam
the
also useresultson intervals
where
the expansion
to
the
function
we
know
these
will
be the intervals
(as
converges
of convergence of the functions we will meet)
to find the values
for which the expansion
ofcomposite
functions
are valid.
We can
orked
4.8
example
Find the Maclaurin
series
up
term in x3 for In
to the
y[\\ +
2x
and state the interval
2-3*
in
which
the expansionis valid.
the*
know
We
into
+
ln(Vl
=
2x)-ln(2-3x)
-ln(l + 2x)-ln(2-3x)
function
=
separate
functions
= -ln(l
find the*
series expansion
for
2 1+
-3x
+ 2x)-<Jln2+ ln
1+
+ 2x)-ln
-ln(l
this
in
form
Now
=
the
rearrange
original
so
+ x)
ln(l
+ 2x
2-3x
series expansion
for
Vl
In
each
-ln(l + 2x) =
(2x)-(2xf(2x)
= \342\200\224
2x-2x2
separately
2\\
+
\"
+
+
-3x
-+\342\200\242
\342\200\242\342\200\242
3
4 Maclaurin
C
XJ^\\
+ d.
and Taylor
series
'a
continued.
- x2 + -Ax5
= x
-
+ \342\200\242\342\200\242
-3x A3
-3x
-3x
-3x^
1+
In
-+
V1 +
2X
= x-
In
And*
2-3x
= ln,
m
-
+
for each
note that for
when
\342\200\224<
x
/
In
x2
59x3
2
3
24
\342\200\2241
< jc <
+
+
+
&
\342\200\242
&
\342\200\242
+
1,
-1<2x<1
\342\200\224
2
-3x
is valid
1+
-1 <
when
-3x
\342\226\240
\342\226\240<1
2
I
V
\342\200\242\342\200\242
1
<
2
smaller interval
2
i.e. when
2
< x<
\342\200\224
3
3
Vl +
Therefore,
Exercise
\342\200\224
2
5x
when
valid
is valid
1
i.e. when
to
both
is
+ x)
ln(1
then
we need the
be valid
Since
ln(1 + 2x)
of validity
and
In2 +
\342\200\224
+ \342\200\224
+
\\ZJ
function
-In2-
+
*
9x3 +
9x2
-3x
+ ..,
3
together
separately
+ \342\200\242\342\226\240
Ax5
We consider the
+ \342\200\242
3
everything
interval
-
4x3
.
+ \342\226\240
x2
put
finally
9x3
9x2
-3x
-
2x
is valid
In
2-3x
1
< x
when
<
1
\342\200\224
2
2
4C
In this exercise
in the
results
1. Find
(a)
the first
(i)
(b) (i)
(c)
(i)
four
non-zero
of the
terms
sin(3x4)
(ii)
cos
ln(2 + 3x)
(ii)
ln(l-2x)
e~T
(ii)
e*3
Maclaurin series for:
2. By combining Maclaurinseriesofdifferent
functions
series expansion as far as the term in x4 for:
(a) (i) ln(l + x)sin2x
(b)
(c)
(i)
(i)
1+ x
Findthe
(ii)
(ii)
the
sinx
\342\200\224
2x
(ii) ln(l-sinx)
sinx)
series
Maclaurin
find
ln(l-x)cos3x
\\
ln(l +
series
booklet.
Formula
given
standard Maclaurin
all the
assume
can
you
as far
as the
term in x4 for
e3x sin
2.x.
[4 marks]
Showthat
I
vl
+ x2 e
x
~1
- x
2
1
+ x2 \342\200\224x3+ \342\200\224x4.
3
112
Topic9
-
Option:
Calculus
1
r^
*a.
6
[5
marks]
Q
series for
Maclaurin
the
Find
(a)
your
form 2^akxk
in the
answer
ln(l + 4x2+ 4x), giving
(b) Statethe interval
of
of the
convergence
power series.
[6 marks]
0
x.
tan
for
non-zero terms of the Maclaurinseries
first two
the
Find
(a)
the Maclaurin seriesof etanx
the term in x4.
Hence find
(b)
including
up
to and
[6 marks]
cos x, find the series
to
the
termin x4.
expansion
x) up
Hence find the first two non-zero termsofthe expansion
(b)
of In (sec x) stating where the expansion is valid.
(c) Useyour result from (b) to find the first two non-zero
Q
(a)
series for
the Maclaurin
using
By
for In (cos
termsofthe series
(a) Findthe first
Q
of the
4 terms
=
f(x)
x.
tan
for
ln[(2 +
[8 marks]
Maclaurin series
for
xf(l-3x)]
(b) Findthe equationofthe tangentto f(x)
at
x =
0.
[10 marks]
(a) Findthe
^
series
Maclaurin
for
In
J
\\
stating the interval
power series.
terms of this seriesto estimate
the
first three
the
Use
X
of the
of convergence
(b)
J.
value
of x used.
2, stating the value
in your
(c) Provide an upper boundon the error
the
error
term.
approximation
using
Lagrange
of In
(d)
10.)
Refine
the
error
as a
on the
error by
the
considering
geometric series.
result
standard
the
Using
bound
upper
this not a valid
for ex, form
expansion?Doese^
a series for
have
e^.
a Maclaurin
Why is
series?
[13 marks]
series
Taylor
d\302\243|
We have seen that a Maclaurin
series is valid in an interval
=
centred
on x
0 and that close to this point, often just a few
terms
of the series are needed to give
a very
good approximation
= 0, even within
of the function. However,
further
away from x
the interval
terms
more
For
where
the
of the
series to
we found
example,
polynomial
for
sin
expansion
is valid,
you can needmany
get a reasonabledegreeofaccuracy.
above that the 5th degree Maclaurin
sin 0.5 correct to 5DPbut
x approximated
4 Maclaurin
r^
**.
and Taylor
series
'a
h
if
we
the same
to use
try
further
0, say at
from
away
polynomial to approximatea value
x \342\200\224
2, we
would
get
sin2^ 0.9092974
which is correctto only
worse
this
such as
for all x g R, we
higher higher
desireddegreeof
any reasonable
matter
away from
how
many
this
For
accuracy.
convergent
approximation
of the
terms
a Maclaurin
have
Maclaurin
order
is only
series
sin x that
overcome
can
and
whose
further
0 we go the
gets.
functions
For
The
IDP.
other
series valid
by simply taking
to get the
polynomials
such as ln(l + x)
functions,
difficulty
for -1 < x < 1,we cannot
make
at all outside this interval, no
polynomial we take.
To overcometheseproblems
we will try to convert the
Maclaurin seriesexpansion
to a power series centred on a
a rather
than x \342\200\224
0. In this way we will be able
general point x \342\200\224
to makea reasonable
of a function
approximation
polynomial
near to any given point.
So,
starting
with
g\\x)and
AJL
r-v
114
Topic 9
-
Calculus
Option:
S>,
+ 0,.
letting
g{x)
the
Maclaurin
g{0) +
-
\342\200\224
f(x
+ a)
series:
x+ -
x1+-
so that g^(x)
x5
-\\\342\200\224
= f{k)(x+ a)
we
have
f(x +
And
a)
=
+
f(a)
x +a
=
f(x)
+
f(a)
used
generalise
we have
results
the
of
in this
way
4.6
POINT
KEY
series
Maclaurin
for
^{x-ar+\302\243^(x-aY+.
Taylor series. All
as the
is known
This
\342\200\242
3!
x:
with
+
f^{x-a)
x2+-\342\200\224\342\200\224x-
2!
1!
finally replacing
+
\342\200\224\342\200\224x+J
I
approximations
Taylor
I
For a
function/(x)
at a exist:
|
/W = /(fl)
+
the
where
/
x
is the
+
- +
^^/(\(fl)")
error term
Rn
f{n+l)(c)
= -p
V^(x-a)w+1
form in which
evaluated
all derivatives
which
(x-a)/'(a)
Lagrange
Rr,(x)
This
for
+
is given
(x)
i?\342\200\236(x)
by
for some ce
this error termis given
i
r
\\a,x\\
in
the
booklet; of course to reduce this to the errorterm
series we need only let a = 0.
Formula
Maclaurin
of
a
orkedexample
4.9
(a)
the
Find
the 4th degree Taylor
(b) Using
error for
maximum
Start
by
series expansion
Taylor
the first*
finding
a
as an
polynomial
1
pattern.
(a)
series
for this
f{x) = \\nx
f(1)
f\"(x)
=>
=>
function, find
2\\x-3
f<4>(*)= -3lz^
f(5)(x) = 4!x-5
n > 1
= 0
=
f'(1)
1
=>
f'\"(1)
=>
fW(1)
=>
f<5)(1)
fW=M+m(,_1)+ni)(,_1)2+ni)(,_lf+...
11
0!
21
= 2!
=
a = 1
=
\342\200\224
+
-(x-1)
0!
=
31
(*-!)-*
1!
2!
(x-1)
+\342\200\224(x-1)
3!
(^_!)2J- + \302\261
(^_!)3'\342\200\224-\302\261
^.^ ^-
4!
+ ---+v
4 Maclaurin
-
^
-3!
= 4!
formula
with
*V
the
=> f\"(1)= -1
= -x-2
f'\"(x)=
have
the Taylor*
Apply
approximation
f'(x) = x-'
^oM-irVi)!
for
x-\\.
point
2 2
Here
we clearly
the
around
3\"
and
few derivatives
look for
xe
for f(x) = lnx
V>A
+ d.
\342\200\242\342\200\242\342\200\242
(x-1)
M)-1^.^
}
*v
and Taylor
+
\342\226\240
series
'a
continued.
Find
for
term1
the
error
this
truncation
look to
and
(b)
have:
\\Ne
,W=(,-0-fezl+fezl-fczl+K.M
place
an upper bound
on
Where,
this
M*) = \342\200\224^
1
oe
L,
^
2'2
_4\\c~5(x-f
5!
5c5
bound
produce
A*-if
lower1
Use the
for
c to
the upper
bound on
-1
AM5
-r\342\200\224
5c5
4D
Exercise
1. Find
the first
of the
of the
Taylor series for
each
following:
(a) (i)
(ii)
(b)
terms
non-zero
three
tanx about xsinx about
(i)
\\[x
(ii)
e*2
(c) (i)
44
71
x-
2
x = 1
about
x =
about
71
\342\200\224
4
cosecx about x-
71
\342\200\224
3
Use
to expressln(3+ x) asa seriesin
the term
Find a
of ex
in (x - 3)
(c)
Provide
an
upper
bound
3)
up
to and
[3
marks]
series expansion for
estimate the value
3 significant figures.
Hence
of (x -
in powers
\342\200\224
+ x
sin
of
sin
for the
up to
degree three.
)
V6
(b)
35\302\260
giving
your
answer
Topic 9
-
Option:Calculus
XJ^\\
+ d.
to
error in your estimate.
[6
116
as far
x,
[3 marks]
expansion
Taylor
including
(a)
\342\200\224
term in x3.
Findthe
Q
x
s expansion
Taylor
as the
about
cotx
(ii)
71
marks]
'a
Find an upperboundforthe errorin using
+
l-2(x-l)
1
1
to estimate \342\200\224
forxe
-4(x-l)
3(x-l)
4'4.
[5marks]
/(*) =
polynomial at a = 8 of
degree Taylor
*\302\243
<x< 9
for 7
that
Show
(b)
second
the
Find
(a)
|R2(x)|< 0.0004
that
Show
about any point x0
expansion
Taylor
marks]
has no Maclaurin expansionbut
= 4x
f(x)
[6
a
has
^ 0.
[5 marks]
Letf(x) = x*, forxgM+.
that
Show
(a)
1-lnx
/'(*)=
(b) Find the
term
that this
to cos
Use Taylor's
error
in
e up to the
value of this polynomial
and
confirm
sameas the maximumofj{x).
x for
expansion of cos x about
marks]
value
any
[8 marks]
that when h
derivative
the
[10
of x.
values
all
to show
theorem
approximating
=
/'(*)
is the
the Taylor
that
converges
x =
about
off(x)
expansion
Taylor
the maximum
Find
Prove
/
x2.
in
(c)
*
x
v
is small the
off(x)
\342\200\224
+ h) \342\200\224
f(%
fix
h)
\342\200\224
L
is approximately
h^
2h
at a
using
i
-h2f'\"(a).
[6 marks]
Applications
of
Representations
a number of contexts.
functions
We
look
here
series
can be
useful in
at two:
otherwise very difficult
to
not impossible to integrate as a standard
that is
a function
\342\200\242
integrating
(if
integrate
Taylor
using
At
function)
the
end
and
\342\200\242
limits
finding
of the
form
c
'0'
next
'
*
\342\200\224 \342\200\224
\302\260\302\260
have L'HopitalsRulebut sometimes
it is easierto dowith Taylor series, especially if the first part of
the questionhas already
asked
for a Taylor series!
For
We
of
the
will
look
a function
function;
we already
second
finding an expressionfor the integral
that cannot be integrated to give a standard
series method is just about the bestway
integral in this case.
the
representing
Series:
solutions
XJ^\\
+ d.
to
of
equations.
of
4 Maclaurin and
C
third
finding
approximations
the
we
of Taylor
application
differential
first at
the power
^>
the
chapter
will look at a
or
0
of
Taylor
series
i>
example 4.10
orked
Find I sin(x2)
series, giving your answerin the form ^akxk
a power
dx as
-
series
the
x we
for sin
series
Maclaurin
of
result
standard
the
Using
the#
e'mx=
can find
\342\200\236
all x<
for
(2fc+i)i
h
for sin(x2)
expansion
(-1)\"
x2k+'
c
-\342\200\224 \342\200\224-
>
sin(x2) = \302\243
(-1)V)2
k=t
(2/C+ 1)!
k
=
~-4k+2
(-1) x4
1
As
this
we can integrate
term
So,
eR
for allx
is valid
expansion
(-1)*x4t+2dx
term
by
i
allx
for
1)1
,=o (2*+
(2k+
-ff^X(-1/x_
SJ (2*+
dx
1)1
(-1)\"x4t+5
!(4k +
the
In
next
L'Hopitals
orked
worked
example we look at
Rule for a limit of the form
the
We
in
the
series for
Maclaurin
able
to be
hope
to
the
avoid
cos x, evaluate
x4
the
cancel
to
for cos
the
in
x*
coe x = 1
\342\200\242\342\200\242
coex-\\
situation
+ \342\200\242
+
6!
4!
2!
'0'
\342\200\224
x and
Maclaurin
that
the x4
has cancelled,
the
simplify
+:
in
the
+
1
r
4! 6!
b\\
limit
denominator1
we are free
r9
x*
fraction
to
x^O
So:
let
coex-\\
lim
-
+ :
1
\342\226\240
JL-JL
'4!~24
Topic 9
c
Option:Calculus
1-A
+ G(.
X2
\342\226\240
lim
*-><\\4!
118
+
6!
4!
2!
series
taking the
before
Now
to
X*
0
So, substitute
alternative
\342\200\242
\342\200\224
lim-
taking the
before
denominator
limit
an
4.11
example
COSX-1 +
Using
3)(2k + 1)!
6!
X*
h&\\
-
-1 + -
4E
Exercise
1.
Evaluate
sinx-x
, x ,.x
t.
(a) (1) hm
x^\302\260
x^\302\260
/
x
x
x^\302\260
x
x^O
Find Maclaurinseriesfor
rcos(x3)-l
the
sinx
+x
t. 3sinx-4xcosx
x^O
x
....
hm
(11)
x
(a)
p^ X
pX
lim
(ii)
xcosx-sinx
i.
hm
/.x
(l)
1-cosx
\342\200\224
t.
hm
x^O
(n)
X
p\302\261
\342\200\224
p*X
(c)
....
\342\200\224
X
lim-
(i)
(b)
series:
Maclaurin
using
in the
following
form ^jikX' k.
d,
J\342\200\224^\342\200\224
+ l)
'-
rln(x
(b) J \342\200\224^
x
dx
(a) Find the first
[8 marks]
the
about
(b)
x =
point
-.
find lim\342\226\240\"
X^3 ;
series,
Taylor seriesfor
3.
x2 -9
this
Using
of the
terms
non-zero
three
sin(x- 3)
[6 marks]
sin(x-3)
the Maclaurin seriesfor
(a) Find
/i
x
TT
(b)
Hence
(a)
Using
1
2
\342\200\224xz
(c)
Maclaurin
1
*
sin x-arctan
_
.
.
and cos x, show
of degree
polynomial
T
[7 marks]
for ln(l+x)
series
the term in x7.
4 for
x) is
ln(cos
4
x.
12
2
(b) Find
.
lim
Maclaurin
the
the
that
i.
i
evaluate
far as
x as
arctan
the Maclaurinpolynomial
ii.
degree
6 for
tan(x2).
tan(x2)
lim\342\200\224\342\200\224
-.
x^\302\260
evaluate
Hence,
of
[9 marks]
ln(cosx)
(a)
a power
series for
the Maclaurin
By using
series
.
i
of In
expansion
Hence find lim
1
(
\\l-2x3
,o
the term in xlz.
(b)
ln(l
+
^
J
x) or
otherwise, find
up to and including
\342\200\224
2x3
/
V1
\342\200\224
-.
the 4th degree
approximationto:
(a) By taking
[7 marks]
polynomial
[
for e *2, find
an
dx
e~*2
Jo
(b)
B(a)
(b)
Place bounds
on the error in
Find a powerseriesfor
How
error
many
of less
terms
Jo
of this
than 10~9?
this
dx
xv
[8 marks]
approximation.
in the
form
\\akxk.
^\"^
k=\\
series are requiredto ensure
an
[10 marks]
4 Maclaurin
and
Taylor
series
Summary
series
\342\200\242
A Maclaurin
is an
infinite polynomialwhichmatchesall
derivatives
the
of a
function
at zero.
\342\200\242
The
is given by:
for f(x)
Series
Maclaurin
f(x)
\342\200\242
The
nth
degree
Maclaurin
^
This is
\342\200\242
The
error
in using
f(x)
where
the
an
nth
=
can be
to enablelimits of the form -
0
,^
120
1
Topic
9
-
Option: Calculus
^
**\302\273*%*
for
\342\200\224xn+l
0 + 1)!
Maclaurin
Rn
and
(x)
(x-
used to
terms.
(c\\
f(n+l)
\302\261
error term
approximations
by:
n\\
some
series to
+
ce
allow expansionabout
+
oo
any
point,
a:
R\342\200\236(x)
by
f\302\260r
some
c e
J a,
provide a representationfor
\342\200\224
to be
by:
]0,x[
^^-f^(a)
is given
a)n+l
is given
polynomial
+ {x-a)f(a) + -
= f(a)
Lagrange
Maclaurin
degree
+ l)!
(\302\253
Taylor
of the function/(x)is given
series truncatedat n
K (*) -
\342\200\242
^x>+-
2!
generalise
approximations
Taylor
f\"(0)
k\\
Rn (x)
\342\200\242
+
+ f'(0)x
f(0)
polynomial
the full Maclaurin
term
=
found.
x\\ L
the
integral
of functions
and
Mixed
examination
Q
4
practice
/(x) = ln(l+ x).
find the first four non-zero terms of the Maclaurin
this expansion, find the exactvalue
of the alternating
first four
the
(a)
Find
(b)
Hence
(c)
Using
derivatives of
series:
the
Find
of (
x
of cscx in
expansion
harmonic
[7
marks]
[5
marks]
ascending powers
*) up to and includingthe term
7cV
(
in
\\
.
%
4j
I
(a) Find
^g
series
Taylor
forf(x).
111
2 3 4
,
Q
series
(b) Use
the degree5 Maclaurinpolynomial
e*2 sinx
of
the resultof part (a)to find
tx2sinx-x
lim
^^0
^V (a) Find the Maclaurin
series
x3
term in x3 for
to the
up
[6 marks]
2+x
(b) Use the Lagrangeerrorterm
that
show
the largest
error
occur
could
that
polynomial to approximatef(x) for 0 < x < 1is
using this
when
to
1
32*
[7 marks]
^S
(a)
the
Find
and the
terms
series for
Maclaurin
xex, stating the first
four
non-zero
general term.
rx
a Maclaurin
find
(b)
Hence
(c)
Hence show that
-
-\\
Q
Using Taylor's
1
3
2
^S (a) Find a Maclaurin
(b) Henceevaluate
expansion
e~*2
(a)
Vl-*2
to within
an error of 0.001.
+
[6marks]
\342\200\224
forallxeR
24
2
~
1+
3x4
x2
\342\200\224
2
5x6
+
+
8
16
+
35x8
128
+
lx3x5x-..(2A;-l)x2/:
^
\342\200\242\342\200\242\342\200\242
+
k\\2k
4 Maclaurin
1*
*p>\\
[8 marks]
that
Show
1
for e~*2.
theorem, show that
2
Q
[10 marks]
Jo
<cosx<l
1
correct
dt.
tef
6(4!)
expansion
dx
I
= l.
1\342\200\224
1
4(2!)
series
for
and Taylor
series
(b) Find the radius
(c) Hencefind
VM
(b)
x
=
(i)
(d)
*->o
estimate
What
(ii)
(a)
down
series
for
x to
arctan
the
= In
by f(x)
value
estimation.
error in your
for the
bound
defined
Write
for
made?
have you
upper
function/is
x7)
tt.
assumption
(iii) Findan
The
x3
series (up to the term in
Use your
Maclaurin
x-arctanx
t
t.
lim
evaluate
TT
(c) Hence
J
[12 marks]
j\"f(t) dt for some function/
find
the
By integrating an appropriate powerseries,
series of arctan x up to and includingthe termin x7.
arctan x
that
Show
(a)
,
stating the valuesof x for
for arcsinx
series
a power
power series.
of this
convergence
is convergent.
this
which
of
[14marks]
.
constant term in the Maclaurin
of the
f(x).
(b) Find the
series
Maclaurin
derivatives
three
first
up to
forf(x)
show that
and hence
off(x)
and includingthe x3
x2
x3
2
3
the
is:
term
+ \342\200\224
x + \342\200\224
(c)
Use this
series to find an approximatevalue
for
In 2.
(d) Use the Lagrangeform of the remainderto find
for the error in this approximation.
(e)
How
is this
good
upper bound
as an estimatefor
an
bound
upper
the
error?
actual
[17 marks]
IB
(\302\251
the series for
(a) Using
in
form
the
e*, write
the Maclaurin
down
2008)
Organization
expansion of e2x
Y^ifc
k=0
(b)
the Lagrange
that
Show
error term, Rn(x)
2n+1 e
Rn(x)<-\342\200\224-Xn+1ifO<*<-
0 + 1)!
(c)
that
show
Hence
using 6
e2x, gives
approximate
satisfies
i
2
terms of the expansionin
an error of lessthan
0.0001
(a)
to
in the
approximation
of
xe2x dx
Jo
(d)
How
less
many
than
Topic 9
-
Option:
Calculus
terms
10~9?
are required to
ensure the erroris
[12
marks]
'0>
In this
Differential
as equations
problems
equations
derivatives
these
\342\200\242
to solve
differential
involved
velocity
acceleration
(as
processes
many
you will
course
core
have looked
at
and
(as the rate of change of displacement)
the rate of change of velocity).We can model
in science by equations involving
the
rate of
cooling
of some variable,suchas populationgrowth
and
of bodies hotter than their surroundings.Newton's
known
Second
change
to
that
problems
Law actually
states that
of changeof momentum.Tofind the
these rates of change involves
solving
differential
In this chapter we first learn
to form differential
with emphasis on real-worldapplications,
and
methods for solving different
of differential
types
original
by separating
functions,
\342\200\242
to solve
a different
type of differential
well
equation
rate
using
by
\342\200\242
to solve
equations.
another
type
of differential
equations
cover
then
it by
equations.
through
multiplying
by
three
equation
a function
solutions
\342\200\242
to
approximate
to differential
a differential
solve
we
When
equation
derivatives
involving
this already
the form
the
for
case where
equations
To solve
this
integration:y
Because
of
\342\200\242
to
represent
equations
in
series
differential
graphically.
f(x).
dy
differential
-
equation
\\3x2 dx
equation
all that
\342\200\224
=
3x2.
dx
is needed
is
= x3 + c
there is
one
solution
to the differential equation. It could be
a family of solutions:
the
more
than
one
can be written
Taylor
dy
example,considerthe differential
As an
and
equation we work from an
to one without.
You have done
the equation
equations
using Euler'smethod
\342\200\224
=
dx
any
differential
up
Setting
a
substitution
from
variable
underlying
equations
the
find
the variables
the
to the
is equal
force
and
variables
involving
their
the
world
real
\342\200\242
to write
In
you
chapter
will learn:
of
constant
of integration
you find that
5
C
XJ^\\
+ d.
Differential
equations
123
The solution y
the
condition
differential
for
told
(0,2).
example
to
solution
equation.
and
the
general solutionto
We may also be given a boundary
the curve passes through a point,
allows us to narrow down the general
solution - in this casey = x3 + 2.
c is
\342\200\224
x3 +
that
This
particular
called the
Differential equations likethis areusedto solve
rates of change and you often
have
to
involving
differential equation from the informationgiven
There are several phrases you should
recognise:
KEY POINT
problems
a
form
in
a question.
5.1
dy
'rate of change' meansthe derivative
but
other
variables
can be used)
(sometimes
'proportional to x meansa constant
\342\200\224,
of x,
multiple
that
is kx (k is a constant)
'inverselyproportionalto x
'rate
orked
example
\342\200\224
means
x
means that the
of decrease'
is negative
derivative
5.1
The rate of hair lossofa manis proportional
to how much older than thirty he is. If he is born
20 000 hairs
and has 200 000 hairsat the age of 30, when will he be bald?
with
the
Define
variables,
units\342\200\242
including
Let x
minus
The
into an
information
the
Turn
sign
indicates
decreases
of hairs
equation
the
the equation by
boundary
age of
\342\200\242
dx
the
integration
conditions
to
find
\342\200\242
When
/.
kx2
x =
20000 = 0 +
20000 = c
x =
> 180000
30,H
-
C
Option: Calculus
XJ^\\
+ d.
+c
0 +
c
= 200000
30x30
+20000
= 450k
k=
.'.
+ 30kx
= 20000
0,H
^
When
2
200000 = -k\\ (302
9
in years
= -k(x-Z>0)
H =
\342\200\242
constants
Topic
man
x<30
when
Use
be the
hairs
when x>30 and
increases
Solve
of
number
number
the
that
the
H be
Let
400
H = -200x2
+12 OOOx
+ 20 000
continued,
Interpret the requirement
This
is a
by
using
*
baldness
for
\\Ne
need
find x
to
when
0 = -200x2
x=
quadratic equation. Solve it#
the formula or a calculator
Sinceage
must
1. Findthe general
2.
(c)
(i)
Find the
3sin2.x
dx
3^_2e2*=0
dx
dy
dx
differential
following
be
+ 20
000
(35F)
positive
61.6(3SF)
4cos
dx
v3y
(ii)
4e*\302\253-^
(ii)
corx\342\200\224=
= 0
dx
dy
o
a
= 3
equations:
4y = a
\342\200\224
(ii)
\342\200\224
=
2
cos2.x^-
the
of
solution
(i)
+12OOOx
5A
Exercise
(b)
0
6Wor-W2
z =
(a) (i)
H =
1
dx
particular solution of the following
differential
equations:
(a)
(i)
(ii)
\342\200\224
=
x2+4
dx
, y
= 2when.x
\342\200\224
=
\342\200\224;^^=,y
dx
= \342\200\224
whenx
2
V4-x2
(b) (ii) (.x2+l)\342\200\224=2.x,;/=
0 when*
dx
(c)
dy
(ii)
2x\342\200\224=
(i)
-e3x^-
(ii)
e2x
dx
1
dy
2
dx
= 1
x2+l,7
= 3,;/
dy
* \342\200\224
=
4,
y
(a)
(i) The value
of
an
= v2
= 1
when.x = l
= 0when;c = 0
= 0
1
when x = \342\200\224
2
dx
3.
=0
antique
vase, in $,
increases at a rate
proportional to the time,in years, since it has been
found. One year after it was found the value of the
vase was $500 and it was increasing at the rate of $30
per year. Find the value of the vase 10years after it
was found.
5
XJ^\\
+ d.
Differential
equations
'a
(ii) The distance,in
of a
m,
from the top of a
to the time, in s,
is at the top of the
falling object
towerincreases a rate proportional
it has been
the object
Initially
2 s the distance
tower.
from the top of the tower
is 20m.
the
distance
from the top of the tower
at
falling.
After
Find
3 s.
after
The rate of growth of a plant is inversely
(t +1), where t is the time,in weeks,
(i)
(b)
Initially the plant was
23
since
to
it was
planted.
its height
tall and
cm
proportional
increasingat the rateof5 cm/week.
after 10 weeks.
the
Find
was
of
height
the plant
of a
mass
The
(ii)
proportional
the baby had
baby increases at a rate
to its age in months.When
a mass of 4 kg, and
was 5 kg. Find the massofthe
6 months old.
(c)
of a
value
The
(i)
car decreases
at
baby
a rate
of the
age in years. The initial value
two years later was \302\2433000.
the value of the car which is t years
old,
old it
it is
when
to its
proportional
car was \302\2434500, and
Find
the
for
equation
old.
of a new gamesconsole
decreases
The price
month
one
2 months
when
its value
(ii)
inversely
at
a rate
proportional to the squarerootoftime,
since it was released. Initially the games
consolecost$350,and one month later it cost $300.
Find the equation
for the cost of the games console after
inversely
in
months,
t months.
of variables
Separation
The second type
of
differential
ableto solveis one that
can
^
Technically
this
separation
T
is not
mathematically
ax
the
valid. However,
-j-
resulting
correct answer. Is it
important
truth
or a
126
more
Topic 9
C
XJ^\\
To solve
\342\200\242
equations.
a differential
all the
equation by separation
x values on one sideand all
the other sideby
separate
integrate
+ d.
differential
as
5.2
POINT
full
Calculus
Option:
separable
\342\200\242
get the
useful method?
for you to
variables
get
= f(x)g(y)
solving such equations is called
shall refer to these differential
equations
and
\342\200\242
leads to the
integration
we
variables
KEY
of
need to be
separationof
for
method
The
which you
equation
in the form:
be written
\342\200\224
dx
both
as
multiplication
if it
sides.
were a
of variables:
the
or division
fraction
y values
on
orked
5.2
example
Showthat
the
differential equation
to the
solution
general
dy
-J--Xy-X
dx
written as y
can be
The
+ Aex2
-1
be written
can
equation
if y > 1.
dy
in#
form -f-
the
ax
= f(x)g(y)
y-1
divide
variables:
Separate
and
so use
variables
of
separation
by
:x(y-1)
dx
j
\342\200\242
by
Then
integrate
dy=\\xdx
j
dx
multiply
ln|y-l| = x +
\342\200\242
Rearrange
e'mce
I3ut
y-1>0
=
y-]
.
Since
ec is
relabel
a constant,
it
A*
as
differential
of
of the
=
example,we changedfromexecto Aex on
is a very common trickto convert
constant to a multiplicative constant.
the
to do
what
1
eo
ex
+ Aex2
we
equation
In the above
solution.
line. This
The next example
shows
conditions.
boundary
ex2
= Aex2
y
Notice that when solvingthis type
cannot simply add '+ c to the end
c
penultimate
an additive
from
when you
are given
orked example5.3
Solve
differential
the
givingyour
answer
cos2
equation
form y
in the
t~2y for
dx
0<
the
in
x < \342\200\224
given
2
-fdx
of
separation
form
so use
J>dy=f
J
variables
integration
XJ^\\
+ d.
4
-,
2
EXAM HINT
coe2 x
2c
\342\200\224
J
dx
coe2
\\e2ydy=\\eec2x
x
dx
is
another
just
\342\200\242
e2y
=tanx
the
To keep
constant.
algebra simplewe can
rewrite
it as
another
such as
c or C.
+ c
5
C
=
\342\200\224,
y
1
dx
e~2y
constant
Do the
x =
f(x).
A
= f(x)g(y)
when
that
\342\200\224
can be#
The equation
written
=
x\342\200\224
Differential
equations
'a
continued...
is a
This
to
use
place\342\200\242
conditions
boundary
to
c
find
to get y
as
71
\342\200\224
+
e[=2tan
=>c' =
c'
4
e-2
= 2tav\\x
.\\e2y
Rearrange
2tanx + c'
=^>e2y =
convenient
+
2y = \\n(2tanx
= f(x) \342\200\242
instructed
=> y
+ e-2)
1
\342\200\224
I n(2 ta
=
e-2
+ e-
nx
2)
HINT
EXAM
as sec2 x
Writing
recognisable integral. This
it into
turn
should
common
is a
a more
trick.
in
Notice that the last exampleaskedfor the equation
explicit form (y \342\200\224f(x)).If it had not done so it would have
been
to leave the answer in implicit form suchas
acceptable
+ e-2.
e2>/=2tanx
Exercise
5B
1.
Find
the
following differential
form y \342\200\224
f(x)
answer
in the
= 0whenx
=0
your
giving
equations,
of the
solution
particular
simplified
far as possible.
(i)
(a)
2.x2
dv
\342\200\224
=
dx
,y
3y
dy
(ii)
(b)
(i)
= 1
\342\200\224
=
4xy2,y
dx
^
= 2when;c = l
= iZ>;/
x
dx
dy
(ii)
when* = 0
\342\200\224
=
y
-3x2y,
- 3 when x -
0
dx
2.
the particular
You do
equations.
Find
t
(a)
v
/.x
(l)
....
(n)
128
Topic9
c
Calculus
Option:
xj^\\
+ d.
dy
-^-
=
solutions of the following
not need to give the equation
sinx
^
,
,;/ = 0 when.x =
dx
cosy
dy =
-^-
sec2.x
dx
sec2
y
^ t
,;/ = 0 whenx
71
\342\200\224
3
= 71
\342\200\224
3
differential
fory
explicitly.
as
(b) (i) 2(l + x)^- =
dx
(ii) (l +
-
2x^/1
dx
=0
= 0whenx
y2,y
\342\200\224
=
x2)
y2 ,y
= 0 when x = 0
2ex+2>/,;/= 0when.x= 0
-^ =
(i)
(c)
l +
dx
-J-=
(ii)
3.
the
Find
of the
solution
general
following differential
form y \342\200\224
simplified
f(x)
in the
answer
your
giving
= 2 when.x = 0
e*-^, 7
dx
equations,
as far
as
possible.
(a) (i)
(i)
(b)
=
2^ dx 3x2
(ii) -1^
x\342\200\224
(ii)
dx
-stcy
(c) (i)
=
+ 3)
x(y
(x-l)\302\243
Solve the
Q
differential
Give your
dN
=
that
Given
= xy
-
2y(1
k is
+ y
x) given that
answer in the form y -
where
\342\200\224kN,
+y2
{l-x2)^-
\342\200\224
=
dt
N =
-\\
dx
dx
x-Xy-\\.
2*
cscx\342\200\224
(ii)
equation
=
dx
y2
a positive
when
[6
f(x).
marks]
constant, show that
Ae~
[6 marks]
Find the generalsolution
dy 4 = y2, giving
x \342\200\224
that 2y
-
=
yy
\342\200\224
=
and
that
4 /\342\200\224\342\200\224
dx
\\l-x2
k is
where
+ \\/l-x2,
xyfk
equation
your answerin the form y
dx
Given that
differential
the
of
-
x =
\342\200\224
when
[6 marks]
f(x).
show
\342\200\224,
2
2
a constant to be found.
[7
marks]
differential
Homogeneous
equations
A
differential
homogeneous
'
dy
dx
For
t
example
dy
y2
\342\200\224^= ^\342\200\224
dx x2
and
is one
equation
Z
yx
x2
, \342\200\224^dy
dx
of the form:
=
+
y2
\342\200\224
are
,
both
,
,
homogeneous
2xy
because
dy
dx
f
dy =
_l(x2_
and ^-
^7
dx
V.xy
2
f_\\
xy
1
respectively.
xy
5
r^
**.
Differential
equations
129
'a
The first
can
we
already
second
a new
requires
separating variablesbut the
approach.
5.3
KEY POINT
differential
homogeneous
Any
solve by
to a variablesseparabledifferential
already) by makingthe
can be
equation
(if it
equation
variable
of
change
converted
is not
(or substitution):
y-vx
Here
v
variable
is a
we must
substitution
replacing
and not
a constant, so in
be sure to differentiate
making
the
this
when
product
\342\200\224:
dx
d ,
dy
v
-^--\342\200\224(vx)
dx
dx
-x\342\200\224 +
dx
(vxl)
V
;
dv
\342\200\224
X
hV
dx
We will use
this method to solvethe secondequation
given
above.
orked
Find
example
5.4
solution of
the general
Once we realisethat
It
homogeneous.
we use the
dx
we
variables we
x2 H~ \342\200\224
v2
dv
\342\200\224
=
2xy
can't
x, y
separate
consider whether
is
(see
above)
of variable
change
Change everything
into
> 0
the#
in tl le form
y2 =
Let y = vx
it is
and so
= vx
y
vand
x#
=
Ther\\s^dx
including
-f- and then
f(x).
simplify
\342\200\224(vx)
}
dxK
dv
ax
\342\200\224
+ v
x \342\200\224
dx
and eo,
dv
+ v =
X \342\200\224
x2+(vx)
7^^
dx
dv
X
\342\200\224^
2x[vx)
xz+vzxz
1 V \342\200\224
2vx2
dx
dv
+ v =
=> x \342\200\224
dx
130
Topic
9
-
C
Option: Calculus
XJ^\\
+ d.
\\
+ v2
2v
continued...
dv
=
=^>x\342\200\224
v
2v
dx
dv
X
\342\200\224/
\\-v2
\342\200\224
2v
dx
Separate the variables
and solve as normal
.'.
\342\200\242
dv=
\\
v2
\\ +
\\\342\200\224dx
jx
h-v2
=^ -ln|l-i/2|
= lnx
=>
= ln^-C
ln|l-^|
1
=^>
- v2 =
e
+ C
x
lnl
c
A
X
yz _
^-*
vwith
Replace
A
X
x2 -y2
=>
=^__~_y2
=
Ax
- x(x
- ^)
Exercise 5C
1.
Use
a substitution
general solution to
find the
\342\200\224
vxto
y
followinghomogeneousdifferential
your answers in implicit form.
(a)
=
(0
ax
0,) (i)
(c)
(i)
dx
\\x
...x
dy
-*-
(n)
\302\243.*\302\2612Z
x
dx
dy
=
x\342\200\224
4y
dx
(d) (i) ^
cbc
=^ +
e~*
2
xy-y
\342\200\224\342\200\224\342\200\224
x2
dx
\342\200\224
3x
x
\\xj
-
dy
(ii)
=
x\342\200\224
(n)
-r =
dx
+ y
2xy
+*
dx cos
x
leave
may
4y
ty_=(yX
(u)
y-\\-2
\302\243
You
equations.
the
V*,
(a)
Find
(b)
Find the
(a)
Using
,.rr
particular solution for
Find the
y
.
.,
equation
in the form y
(b)
dx
a substitution
differential
=
the equation\342\200\224
solution to
the general
= vx
dy
-*-
dx
= 3
whichy
\342\200\224.
x
when x
= e. [7marks]
find the generalsolution
x2
+ y2
\342\200\224
x2
+xy
. .
\342\200\224,
giving
your
to
the
answer
\342\200\224
f(x).
particular solutionwhich passesthrough
(ljl)-
the
point
[9 marks]
5
r^
**.
Differential
equations
(a)
(b)
the term
Define
Show that xy
'homogeneous
dy
\342\200\224
- x2
differential
equation.
+ y2 is a homogeneousdifferential
dx
equation.
(c)
Find the
particular solution for
which
x =
\342\200\224
4 when
y
giving your answerin the formy2=f(x).
(a)
(b)
the differential
Write
j
/
cbc
V-^y
Hence
find the
\\
(a)
dv
4jc \342\200\224
v ~\\~V
dx
2x+ y-l
(a)
find
your
answer
(b)
value of v
is the
(i)
The differential
x=
X-l>y = Y
x =
when
equation
+ 3
A-\342\200\224
=
Prove that if
/
dx
a differential
9
Although
toolsto
y
\342\200\224
not
solve
- c.
= 0, y
\342\200\224
3
giving
m
mar]csi
equation
l, show that
dy
\342\200\224
=
= 2e.
Show
i
=
h fiy)' dy 1.
r2
\342\200\224
1nx
]ny
dx
has
a particular
~
that
dv
f
, the
v
Option:
i-A
Calculus
1.
figures.
substitution
y-vx
marks]
produces
in variables
separable
differential
r$
have the
necessary
such as:
equation
the left hand side(LHS)
marks]
equations
we already
homogeneous,
a differential
form.
= ex
dx
Topic9
=
Jfclnv-v
-*- + 2xy
,2dy
x2
because
y
\\x
equation
Linear
\342\200\224
turns
[14
^
v
~h
\342\200\224
x
dx
value of k to three significant
find the
4jc
dv
\342\200\224
=
differential equation.
solution with x
solution with x = e,yy
Hence
y
whenx = e.
v-2
has
/(v)
If k
(ii)
form
the
in
solution of the differential
dv
dx
\342\200\224
[8 marks]
form /(x,y)
in the
A particular
=
x\342\200\224
x-
a homogeneous
into
the particular
Hence
dx
v
~h
\342\200\224
=
/^J.
\342\200\224
=
(b)
4jc
marks]
=
the substitution
that
Show
dv
[10
general solution of the equation
in thefbrm2ln|x|
^
equation
l,
is of
a convenient
form.
We notice
that
of x2 and -^- the derivative
derivative
is the
2x
dx
of y,
we have
which means
the differentiation
Therefore
can
we
has resulted from
the
product rule.
(x2y) using
equation equivalently as:
an
of
a product
write
the
that
expression
n
\342\200\224(x2y)-tx
dxK
Now, we
can integrate both sidesand
to get:
rearrange
x2y\342\200\224\\txdx
zx+C
X1
differential equationwherewe cannot
separate
the variables, the LHS will often
be more
However,
complicated.
this method does suggesta way to deal with these cases.
faced with a
When
Consider, in general, a first
that above:
differential
order
equation
to
similar
& +
p(x)y = Q(x)
where
and Q(x)
P(x)
linear differential
Note
that
functions of x. This
are just
equation.
is a function in
if there
as a
is known
dy
of\342\200\224,we
front
can
divide
this
form.
dx
through the
equation by that
We now wish to make the LHS
as above;to do this we multiply
function I(x):
/(*)\342\200\224
+
have
that
the form
the LHS in
proceed exactly
as
we
that
=>y
only
remaining
From
require.
=
The
- I(x)P(x)
1
=
f
\342\200\224\342\200\224]l(x)Q(x)dx
is to
question
decide on the function
We need:
r(x) = l(x)P(x)
Kx)
^\\?-^-dx=\\p(x)dx
J
I(x)
J
=> ln|/(x)| =
Jp(x) dx
=>I(x)
r^
can
l(x)Q(x)
this work.
This function
we
then
here we
above:
j-(l(x)y)
make
by a
l(x)Q(x)
such that V(x)
is chosen
if I(x)
product
the equation
through
=
to
it
of a
derivative
the
l(x)P(x)y
dx
Notice
to get
function
= ei
I(x) is known as the integratingfactor.
*c
...
I(x)
to
5.4
POINT
KEY
linear
a first order
Given
&
solve
and
equation:
= Q{x)
+ p(x)y
by the
through
multiply
differential
factor, l[x)
integrating
= eJp(x)<k,
differential equation.
the resulting
orked example 5.5
where
We
differential
the
Solve
to see
check
first
x =
\342\200\224
1 when
y
the 2
dx
whether
It
the LHS is
the*
(although
if
isn't
weren't there
it
would
we start by dividing through
to get the equation in the correct
for applying the integrating
the
Find
71
xe
sure not to miss the
coex dy
^
dy
e'mx
\342\200\236
=o
dx 2ye'mx
now
through
multiply
check that
-j\342\200\224
(ycos2x)
dx
'=
LHS
the
3
coex
coex
ax
by
form
\342\200\224
^>
(XX
\\x)
= JP(x)dx
= e]
_
~-2\\r\\eecx\\
_
Jn(secx)-
_
- sign on P(x)
by cos2x
is of the
-\"deecx
-(2tanx)y--
factor
cos2 x
pin
x
=coe2
We
71
=>-r-2y\342\200\224=\342\200\224
factor\342\200\242
integrating
l(x)
making
= 3 for
be).
Therefore
cosx
2y sinx
0.
a product.
of
derivative
dy
\342\200\224
-
cosx
equation
and*
5o,
coe2x dy
form
(2coe2xtanx)
,y
y
V
dx
dx
x
= \"Scoex
dx (2coexe'mx)y
dx (ycoe2x)
can
now
both sides*
integrate
y coe2
x =
J3 coe x dx
= Z)e'mx
And
finally
we
need
and rearrange
to
the
find
into
the
constant
form
y
C1
= f(x)
= 0,y
\\coe20
= '5emO
9
-
c
Option: Calculus
+ G(.
+c
=> c = \\
=^y
Topic
+ c
=1
Since x
:.
134
xeecx
cos2x-p--2cosxsinx
=> coe2
We
= 3coe2
x
ycoe2
= eec2
=
Z)e'\\r\\x
+ /\\
x(pe'mx +1)
=
\"5coex
5D
Exercise
1. Use
an integrating factor to find
the followinglineardifferential
-^- +
(i)
(a)
dx
dy
+
2y = ex
(ii) -^--4y
dx
-^- + ^-
(c) (1)
=\342\200\224
of
dy
-Jdx
which has y
Find the
=
2xy
dx
Find the
3
dx
(7?)
do
integrating
0
equation
differential
equation
+ y
the point
sinx
\342\200\224
cos2
x
and
(1,1).
that y
we ignore
factor?
= ^l
-2y-\\-\\
+
[10 marks]
\342\200\224
2 when
x
\342\200\224
0,
[12marks]
the constant of integration when finding
the
to
differential
the
equation
1
dy\342\200\224
\342\200\224*-
[12
x dx
S (a) Usethe
marks]
equation
terms of x.
\342\200\224+
2y
dx
[8 marks]
through
passes
\342\200\224
cos x
^
dy
the
of
Find the generalsolution
.x
= ex
[8
dx
Why
equation
ecosx
x
that
find y in
differential
[8 marks]
\342\200\224
\342\200\224
that
+ xy
Given
x3
particular solution of the lineardifferential
\342\200\224
x2
-
sin x
+ y
dx
+y
linear
\342\200\224
x-3
solution
dy
the
=
\342\200\224
\\.
Findthe general
\342\200\224
x
dx
general solutionto the differential
\342\200\224
-
x2
x
\342\200\224
e when
+ ^-
-^-
secx
(XzxLx)y=
dx
(n)
Find the particularsolution
of
to each
ex
dy
x2
x
dx
=
\342\200\224
\342\200\224
(ii)
ycotx-\\
dx
solution
general
equations:
\342\200\224
(b) (i)
the
z
substitution
\342\200\224
\342\200\224
to transform
marks]
the equation
dy +
-^xy = xy1
dx
into
(b)
Solve
(c)
Find
y
a linear
the
the
resulting
equation,
particular
solution
\342\200\224
\\ when
x and z.
writing z in terms of x.
to the original equationthat
differential equation in
x
\342\200\224
1.
[14
has
marks]
5
r
^
4a.
...
Differential
equations
I
135
J
the substitution z \342\200\224
or
y2
(a) Using
solve the
otherwise,
equation
x
dx
given
form y
\342\200\224 \342\200\224
-5.
4,
y
Give your
answer in the
\342\200\224
f(x).
to find the generalsolution
substitution
Use another
(b)
x
when
that
the
to
equation
dy
+ tan
cos y \342\200\224
x sin
dx
to
Approximations
Although we can now
a good
solve
y = sin x
[15
marks]
solutions
number
of first order
there
will still be some that
cannot
be
In these cases
of an exact algebraicexpression.
is possible to find an often
numerical
approximation
very
good
to the solution.
differential
solved
We
equations,
in terms
two
consider
will
based
\342\200\242
a method
curve
solution
a Taylor
forming
here:
on using the gradient of pointsaround
to move closer and closerto the solution
value of
particular
\342\200\242
such methods
it
the
at
a
x
to approximate
polynomial
the solution
curve.
Before
representing
a differential
Given
to such
solutions
methods, we will look at
differential equations graphically.
of these
first
the
developing
equation:
\302\243=/<*>>
we can find
the gradientat
putting thesevalues
we
Although
could
oix-a
pick
choose
integer-valued
For
example,
given the
point (a,b) by simply
- b into the equation.
any coordinates for this, it is easiestto
any
given
and y
points.
differential equation:
-J!-= X-y2+2
dx
we
can
find
the gradient at
^
dx
Topic
9
-
Option: Calculus
=
the point (2,3):
2-32+2
=
-5
'a
this
Continuing
up
a table
for a
process
range of coordinates,we can build
the gradient
showing
~
-1
0
1
0
-4
-3
0
y
2
3
1
2
0
3
4
3
-1
0
-1
2
-2
represent the gradientat
by drawing the tangent at that point:
here
from
graphically
we can
\\
\\
\\
\342\200\224
/
\\
\\
X
/
/
/
+
/
/
X
\\
each
point
-
i
X
KEY
points:
-2
1
-3
-1
And
various
at
i \\\\ -
\\
5.5
POINT
A plot of
of a
field
the tangents at all points (x,y) is calledthe
differential equation.
From the slopefield, we can
solution curves that correspond
To do so we just observe two
construct
then
to different
slope
approximate
initial conditions.
rules.
Solution curves:
1.
the
follow
direction
of the
tangents at each
point
2. donotcross.
5
c
xj^\\
+ d.
Differential
equations
f -
137
'a
The slight drawback hereis that, without
finding
enough
in some
detail, it can be difficult
tangents to form the slopefield
to get a good impression
of what the solutioncurves
should
look like, and of course, it is very
time
to generate
consuming
the tangents
of points.
number
a large
at
However, the processof generating
easier by
the
finding
the
between
relationship
can be
field
slope
all points with
made
the
samegradient.
5.6
POINT
KEY
on which all points have
known as an isocline.
A curve
the
In
dy
\342\200\224
=
dx
same
gradient
x
-
y2 +
is
c.
some constant
c for
dy
\342\200\224
=
above, with
example
2 the isoclineswill
be
by:
given
c
set
isoclines
find
To
the
\342\200\224
x \342\200\224
=> y2
y2+2
- 0
\342\200\242
c
\342\200\224
c.
isocline corresponding
on the
Therefore,
+ 2
\342\200\224x
(y2 = x + 2),the tangents
at
every
to:
will have
point
gradient 0
\342\200\242
c =
1 (y2
= x +1), the tangents
at
every
point
will have
gradient 1 and so on.
This
us to
allows
having
to
fill
Todothis,
we
each
isocline
gradient.
138
V
Topic
9
-
C
Option: Calculus
XJ^\\
+ d.
in
the
first
and
dense slope field quickly
of table we used above).
sketch a
sort
sketch
a few
draw arrows
(and
without
then go along
corresponding to the appropriate
isoclines, and
'a
this
Using
reconstruct
by
starting
moving
points
the family
construct
then
can
We
slope field and solutioncurves,
of the
idea
of solution curves:
we
can
relating to a particular initial condition
point (x0,y0) of the initial conditionand then
in the direction of the slopefield to other
step-by-step
the curve
at the
that lie
approximately on the solution
curve:
(xo,yo)
I Step
this
Obviously,
lengths
but
and
only
i
\\
\\
curve,
length
1|
-
\\
an approximation
gives
can be
the
=
accuracy
more
of them:
to the solution
improved by taking smallerstep
Oo,?/o)
IStep length =
0.51
5 Differential
S>,
+
0,.
equations
139
'a
This
follows.
as
summarised
KEY POINT
57
Given the differential
equation:
\342\200\224
=
at the
mo
=
\342\200\242
Move
-
y) with y
fix,
dx
\342\200\242
Start
method and canbe
as Euler's
is known
process
point (x0,y0)
and find the gradient
step h in the
a fixed
x-direction and calculatethe
from the
y-direction
k in the
distance
k =
field:
Using
x0
f(x09y0).
corresponding
gradient of the slope
\342\200\242
x =
when
y0
new
the
(xo,
point
m0h
k) repeatthe above.
(x0 +h,y0+
Vo)
-
\\
\\
I
/
Jr
= 0.25
length
Step
\\
\\
-
\\
the
Continuing
approximate
orked
the
above
generates
solution
curve.
a sequence
of points that
5.6
example
UsingEulers
a step
with
method
length
of 0.25, approximatethe solution,;/,
ofthe
equation
-J!-=X-y2+2
dx
when x
= 1,given
the
Find
that
at the
gradient
(0, 1) by
\342\200\224
1 when
y
substituting
differential
The x step length
point*
the
into
so this*
gives x1 immediately
corresponding
= mx
find the
y-distance and
add
140
Topic 9
C
Calculus
Option:
XJ^\\
+ d.
\342\200\224
0. Give
your
answer
to 3SE
x0=0,
y0=1
m0=0-\\2+2=\\
equation
is 0.25
Usingy
x
to
y0
Xi=0 + 0.25 = 0.25
yl=1
+
(1x0.25)
=1.25
differential
continued...
the
Find
at the
gradient
(0.25, 1.25) by
point#
so#
is 0.25
find x2 by adding
this
step
length
until
process
It
be
may
y1
the above
\342\200\242
reached.
to record the
information in a table to keep
track of the calculations,
but for
decimals
record
lengthy
only
few decimal
first
the
and use the
your calculator
in
X
y
m
0
1
1
0.25
1.25
0.6375
0.5
1.421 375
0.5-1.4213752+2
= 0.47327
0.75
places
1.421375 + (0.47327
= 1.54144
on
button
ANS
x 0.25)
+ (0.6875
= 1.421375
x = 1 is
helpful
= 03
the
add to
with
= 1.25
and
y-distance
continue
y2
+ 0.25
=0.25
x2
to x1
Using y = mxfind
Now
+ 2 = 0.6875
-1.252
equation
The x step length
corresponding
= 0.25
into
substituting
the differential
m,
subsequent
1.54144
1
+
calculations
(0.37395
x 0.25) 0.75-1.541442+2
= 0.37395
x 0.25)
= 1.63493
to 3SF*
Round
y(l)\302\2531.63
|n the
exam you
can start
a table
We
differential
to
knowledgeof
Given
equations,
from
series
Taylor
a differential
such as
th.s
second method of approximating
look at the
now
will
solutions
fitting.\"
using our
4 of this option.
this time
chapter
^^ngoteach
stage.
equation
x =
with
x0,y = y0
\302\243=/M
we can form a Taylor
pointx
=
y = y(Xo)
xQ given
the solution,
y, about
Taylor
<^[
series
looked at in
4D.
were
Section <^[
+
y^(X-Xof+.
y>(Xo)(X-Xo)+>^(X-Xof
this
Truncating
the
boundary condition:
+
where y(x) meansy
(usually
for
series
in the
3 or
evaluated
series to
at the
point x
\342\200\224
x .
a polynomial of somedesireddegree
an approximationto the solution
4) will produce
as
required.
The only difficulty
can be done
here is in finding
the derivatives,
by successivelydifferentiating
equation with respect to x:
This
differential
y\\y'\\y\"\\...
the
original
5
c
XJ^\\
+ 01.
Differential
equations
'a
d
fdy
=
y
dx
y\"=\342\200\224\\\342\200\224{f(x>yJ)
dx
vdx
as required, beforeevaluating
=
x0,y y0.
so on
=
and
atx
each
NB.To differentiate/^,;/) herewe will
technique of implicit differentiation.
The solution
of a differential
=
-\302\243
can
be
to use
the
5.8
POINT
KEY
need
derivative
x =
f(x>y)
by forming
approximated
x0,y =
a Taylor
y0
for y
polynomial
point x:
about the
y
equation:
=
+ y'(x0)(x-x0)
y(x0)
y\"M
+ ^(*-*o)'
2
,
3!
2!
whereyf\\yff\\
etc.
can
(*0)
+ ym
be found by
the original differential
equation
(x-x0f
+...
successively differentiating
respect to x.
with
orked example 5.7
a Taylor
Find
of degree
polynomial
3 to approximatethe solution
dy
\342\200\224
=
dx
this
use
and
a
Form
to find
polynomial
point x = l, the
To do
the#
initial
given
point.
will need
we
this
y2, x
= l,y = 0
an approximation of y(1.1)to 4DP.
about
polynomial
Taylor
x2 +
y(i)
=o
y'(l) = 12+02=1
y(l),y'(l),y\"(l)andy\"'(1)
We are
given
y(l)
we can
and
immediately evaluate y'(l)
x =
substituting
\\y
= 0
by
into the
differential equation
then
We
yw(l) by
need
first
to find
Use the
y\"(l) and##
the
differentiating
differential
rule
chain
equation
for
\342\200\224
dx
142
Topic 9
-
Option:Calculus
S>,
+ 0,.
of
(y2)
:2x
+
2yy'
the
differential
equation
continued...
Evaluate at x = l,y
Follow the same
=
chain
*
/.
*
procedure for
the product rule
Apply
=1
0,y'
+ (2x0xl)
(2xl)
= 2
the#
to yy
rule
=
=2
Similarly,
y'\"(l)
and
y\"(1)
=
+ yy\
+ 2(y'y'
2(l
+
(y')2
.-. y'\"(l) = 2(l +
+
yy'j
12+(0x2))
= 4
Now form the
\342\200\242
Taylor polynomial
Therefore the
Taylor
polynomial
of degree
+
y~y(i)+y'(i)(x-i)+^(x-i)2
=
0 +
1(x-1)
= (x-1)
+
^(x-
-i)s
+ -(x-1)2+-(x-1)3
+ -(x-1)5
(x-1)2
3
So
-1.1-1 +
y(1.1)
3 is
o
(1.1 -1)2
- 0.1107
+ -(1.1
3
-
if
Exercise
5E
1.
Use
value
approximate
differential
(a)
step length h - 0.1to find
of y when x = 0.4forthe following
s method with
Euler
an
equations:
(i)
(ii)
(b) (i)
^ = x2-y2,y =
dx
latx
= 0
dy
dx
= 2 at
x
\342\200\224
=
+ y), y
ln(x
-^-/
dx
=
dy
(c) (i) (x+ y)\342\200\224=
dx
(ii)
(x
+
=
y)\342\200\224
dx
3x2
= 0
= lat.x
sin(.x2),;/
(ii) ~^--y- 2ex,y=
dx
- 0
0
+ y2,y
e^02^,7
0
atx-
= 2atx
= 1
=0
atx = 0
5
*V
-
^
y*A
+ d.
Differential
equations
1
2. For the following
differential
isoclines
hence
and
and
(ii)
&-2X-,
\302\243-\302\243
ax
ax
dy
(b) (i)
\342\200\224
=
dx
that
=
(ii)
2x-y2
(ii)
y
dy
^
\342\200\224
=
xy +
dx
2x
^ = x2+ y-3
dx
polynomial of degree 3 about the pointx
the solution
of the following differential
a Taylor
Find
#y
dx
(c) (i) ^
3.
the slope field
these construct
by sketching
curves.
solution
(a) (i)
equation of the
find the
equations,
approximates
equations:
(a)
(i)
-f-
ax
= y2-x,
y0
(ii) (l + 2x)\342\200\224=
dx
x +
= l
at x0=0
\342\200\224
=
(b) (i)
(ii) sinx
\342\200\224
+ ycos.x
For the
differential
x0=0
\342\200\224
2
atx0=-7l
;/2, y0 = V2
=
dx
Q
at
2
y0=
cosx-siny-\\-x2,
dx
-
4y2, y0 =
atx0=
\342\200\224
4
equation
\302\243-(>+ix*-3)
(a)
Give the
(b)
Construct
equations of the isoclines.
the
slope
field.
(c) Sketch the solutioncurves.
(d)
What
can
to
initial
For the
the
say about
you
the solution curvecorresponding
x =
condition
= \342\200\224M
Qyy
[9
marks]
differential equation
dy _ (x + y)
xy +
dx
with
y
\342\200\224
\\ when
method
Eulers
\342\200\224
use
x
0,
to find an approximatevalue
2
x=l.
when
ofy
with
correct to three decimalplaces.
Consider
the
differential
step
h =
length
0.2
answer
Give your
[9
marks]
equation
dx
boundary conditionx =
Use Eulers method with h =
with the
(a)
value
ofy
when
(b) Solvethe differential
(c)
(i)
Find the
part
lyy
0.1 to
Topic 9
1
^
-
Option:
*i.v%ft
Calculus
find an
approximate
x = 1.3.
equation.
percentage error in your
from
approximation
(a).
(ii) How can this error be decreased?
144
= 0.3.
[14
marks]
The function y
f'(x) =
x2y
with
f(x)
satisfies
/(o)
= 0.5.
\342\200\224
the differential
equation
Use Eulers methodwith step length h = 0.25 to find an
approximatevalue of/(l).
accurate?
(ii) How can your approximationbe mademore
Solve the differential equation and hencefind
the
actual
(a) (i)
(b)
value
of/(l).
(c) Sketch the graphof your
your approximationfrompart (a)
value of/(1).
0
For
the
(a)
(i)
differential
(l +
equation
to explain why
than the actual
use it
and
solution
is smaller
[17marks]
x)y' = y2
+ ex
Show that:
y
=:
y'(2y-\\) + zx
x
1+
\\2
x
l +
also that y = 1 at x = 0, find
polynomial about x - 0to approximate
(ii) Given
(b)
Hence find
at
solution.
an approximationto the solution
[12 marks]
equation
-^-
dx2
with boundary conditionsy
Find
Taylor
degree
the
x = 0.1.
Consider the differential
(a)
a 3rd
an expression
for
= 6xe~x
= 3
dy
\342\200\224
in
approximate
ofy
when
dy
\342\200\224
=
step
1 when
dx
terms
dx
(b) Use the Eulermethodwith
value
and
x =
0.
of x.
length
x\342\200\224\\.
0.2 to
find the
[12
marks]
5 Differential
r
^
+c
,,
equations
i
145
Summary
\342\200\242
'Rate
'Proportionalto x
a constant
means
but
(sometimes\342\200\224,
dx
of x, that
multiple
k
means
to x
proportional
Inversely
dy
the derivative
means
of change'
can be
variables
other
used).
is kx (k is a constant).
\342\200\224.
x
means that the
of decrease'
'Rate
\342\200\242
A differential
of the
equation
1
\342\200\242
A differential
differential
\342\200\224
vx turns
y
^
solved by separation
can be
f(x)g(y)
dy
=
the form \342\200\224
/
dx
(V\\
\342\200\224
is called
\\xj
differential
\342\200\242
A linear
\342\200\242
Linear
differential
of the
into
Drawing a slopefield
of variables:
a homogeneous
equation.
An isocline
the
all
differential equation.
the same
conditions.
initial
the
gradient.
equations at a given point states that
to differential
equation:
\342\200\224
=
withy
f(xyy)
dx
= y0 whenx
\342\200\224
x0
point [x0>y0) and find the gradient m0 = / [x0,y0).
a fixed step h in the x-directionand calculate
the corresponding
-
Start at the
-
Move
factor:
integrating
solution curvesto a differential
approximate
solutions
of a
field
slope
determined by
have
tangents
by the
through
l(x)Q(x).
curves are
solution
approximating
differential
=
construct
us to
enables
is a curve on which
Euler's method for
y)
variables separable
= Q(x).
dx
at all points (x,y) is calledthe
different
These
(x)
\342\200\224(/
dx
tangents
+
as \342\200\224
P(x)y
solved by multiplying
can be
equations
This turns the equation
plot
can be written
equation
equations into
differential
homogeneous
differentialequations.
given
\342\200\224
=
equation.
\342\200\242
A substitution
A
form
can be written in
which
equation
is negative.
derivative
distance
k in the
y-direction from the gradientofthe slopefield:
k) repeat the above.
Using the new point (x0 +h,y0+
k = m0h.
-
The solution of a differential
equation
\342\200\224
=
f(x,y)
can
be
y
where
by forming
approximated
y'\\y'\"->
=
y(x0)
etc.
+ y\\x0)(x-x0)
can be
w.r.t. x.
Topic
9
-
a Taylor
Option: Calculus
withx
= x0,y
polynomial
+
^(x-x0)2
found by successivelydifferentiating
= y0
for y
+
about the
point
xQ:
>^(x-x0f+...
the
original
differential
equation
Mixed
examination
on
Sketch
(a)
5
practice
graph
paper the
slope field for the differential
equation
dy
ax
and
points (x,y) wherex e {0,1,2,3,4}
y e {0,1,2,3,4}.
sketch
the curve that passes through the point
(b) On the slope field
(0,3).
Solve
the
differential
to
find
the
of
this
curve.
(c)
equation
equation
at the
Give
form y
in the
answer
your
\342\200\224
[14 marks]
f(x).
IB
(\302\251
(a) For
the
differential
equation
.
.
dy
^
+ ysmx + zy*
cosx^dx
and the
boundary condition y
this
Use
(b)
=
1 at
=0
^
0, find a 3rd degreeTaylor
close to x = 0.
the solution at
to approximate
polynomial
x=
the solution
to approximate
polynomial
2007)
Organization
x =
0.2. Give
your
[9
answer to 3 significantfigures.
differential
the
Solve
equation
=
x2 \342\200\224
3xy
dx
given that
A
y
that
curve
= 2. Give
at x
= 4
marks]
passes through
+ 2y2
in the
answer
your
the point (1,2)
is
form y
= f(x).
by the
defined
[9
marks]
differential
equation
\342\200\224
=
+ x2-y)
2x(l
dx
(a) (i) UseEulers methodto get
of 0.1.
steps
taking
an
value
approximate
ofy when
x = 1.3,
steps to four decimal placesin
Show intermediate
a
table.
(ii)
How can
a more accurateanswer
(b) Solvethe differential
form y = f(x).
Consider
the
giving
equation,
(a)
x
Use the
the value
answer
method?
in the
equation
-
Axy
-
e2
\342\200\224
0.
Euler method with
ofy
your
Euler s
using
marks]
differential
\342\200\224
4 when
obtained
[14
dy
dx
with y
be
when
x
step lengthh
\342\200\224
0.05
to find
an approximation
to
\342\200\224
0.2.
5 Differential
equations
'a
(b) Usethe integrating
hence
the
(c) How canthe
value
exact
find
to solve
method
factor
of y when
equation and
x=l.
values
the
between
difference
the differential
found
in (a)
and (b) be
decreased?
marks]
differential
the
Given
[17
equation
-*- = y
,2dy
xz
dx
condition y=latx=l:
with boundary
Show that
(a) (i)
+ y
\342\200\236
y'(l-x)
/
=\342\200\224\342\200\224^~.
a degree
Find
(ii)
+ xy
2 Taylor polynomial closeto x \342\200\224
1 to approximate
the
solution.
(iii) Hence find
(b)
=
y
(c)
f(x).
the true solutionat
5 decimal places.
the
Find
1.1.
equation exactly, giving your answerin the form
Hence find
(ii)
solution at x =
to the
approximation
the differential
Solve
(i)
an
error in your
percentage
x =
1.1, giving
your answer to
approximation, to
[17
2 significantfigures.
Given that
1 dy
y2 +1
2 dx
x2 +1
\342\200\224\342\200\224
=
marks]
and
that y =
0 when
x = 0, express
y
in terms of x.
Q
Solve
[9marks]
differential
the
equation
dy
XV
(l + x)\342\200\224\342\200\224
y -xt
x
V
Jdx
given
thaty
= 1 at
x
\342\200\224
0. Give
your
answer
in the
form y
\342\200\224
f(x).
[9 marks]
(a) Show that
dy
x^-dx
is a homogeneous differential
(b)
148
Topic
9
-
> 0,y
> 0, find
givingyour
answer
If x
Option: Calculus
XJ^\\
+ d.
the
+ :
lny-lnx
equation.
general
in the
x
y
solution
form yey,x =
to the
f(x).
differential equation,
[11 marks]
and
Summary
examination
mixed
practice
revisited
problem
Introductory
a wedding
Consider
cake with four layers:
fc^o
Each
has
layer
third a
of 1. The first
a thickness
radius of 1
and
the
layer
a radius
fourth
of the cakeand
the volume
needs
a radius
1
of 1, the seconda radiusof -,
of\342\200\224.
the
surface
area
(excluding
the bottom of the first
layer)
that
with icing.
covering
Now imagine there areinfinitely
to the cake. What can you say
many
layers
of the cake and the surfaceareathat needs
now?
icing
each layer of the cakeis a
We note that
the
2
1
4
3
Find
has
\\2
=
cylinder
so the
volumes
will be
about
the
volume
given by:
-71
y,=\342\200\236x|-jxi
2
V, =
V
7C
X |
-
I
Xl
=
Xl =
-TIXI-I
71
\342\200\224
\342\200\224
6 Summary
3T*a*
+*.
and
mixed
examination
practice
149
'a
,V = V1+V2+V3+V4
=
71
=
71
71
71
71
+ \342\200\224
+ \342\200\224
+ \342\200\224
22
42
32
.11
l
l
1
\342\200\224
\342\200\224
\342\200\224
+
+
+
42
32
22
this series
Continuing
which
we now
infinitely
layered
know to
many layers we get V
infinitely
be a convergentp-seriesand
^^\342\200\224,
k
k=\\
volume of this
that the
can conclude
we
so
-
is finite!
cake
If we did not know
could say:
with
cake
the
for
this
(or if
result
p-series
we did and wanted an upperbound
L1111111
.22 I1) U2
+ I(1)
2
I
TC|1
f
+ I(1)
4
82
72
volume)
we
4
+
J
+ J_+.
42
42
+-
= 71-11
+ -f- + -+-f2l2
4U
2)
=
62
52
J_+J_+J_+_L
+
<^1+|i+l
the
1
42
32
22
[
on
42)
- +
-- + ...
-+
4) 8U
4
+ .
U2
I+
...
+ I(1) + ..l
8
J
\\
1
-K
(sum
of a
geometric
sequence)
2;
v
= 271
way the
Either
volume is finite
(and
< 2tt).
In fact it
can be shown that
^i 1
^77
=
the surface areaof eachcylinder
If
we
deal
first
with
we get
together
the
the tops
circle
full
sameno matter
how
of each layer,
of radius
layers
many
which
Topic
9
-
c
Option: Calculus
xj^\\
+
01.
down
looking
1; that is
up of
the top and the sides.
from
above we
a total surface areaof 71
see that
x
7T
\342\200\224\342\200\242
6
when
l2 = 71. This
taken
will be
the
we have.
We can also establishthis in a moreformalway
the circular area of the top ofthe cylinder
minus
150
is made
=
6
k=lk
Now for
7lz
\342\200\224so
V
by
noting
the
that
circular
the top of each layer
is
area of the layer above.
made
up of
'a
T2 =
ri
=
-(irxay)
(*xay)
(7rxl^)-(7rx(i)2)
So,
Tx =
X l2 )
(tT
-
-1 7IX[
2\\
2\\
T2
=
icxil
*XI3
J
1
7TX
T3=\\kx
)
2
-
nx
Td=
A
-
/
2\\
-
7TX|
5
(
(l\\2^
.-.r = (7cxl2)-hex V
V
(all other
v5y
)
with
n layers,
terms between cancelout)
2\\
=
i-i
71
i
)
The
same
pattern
of cancellation
will occur
cake
the
for
leaving us with
just:
2\\
l-
r=7c
w
Then letting
n^oo
so we have
for the
layers
many
infinitely
+ l
cake,
T = limT\342\200\236
w
=
=
To
finish
we need
cylinder is given by
to add to
71
(l
(0)2)
2^
1
1-
= 7llim
+ 1
(algebra of limits)
71
this the surfaceareaofthe
sides
cylinders
(curved surface
area of
2nrh).
6
r^
of the
**.
Summary
and
mixed
examination
practice
151
Sx
= 271
X1 X1 =
S2
= 271
x
271
- x 1= \342\200\224
2
2
S3 = 271
1 ,
271
--xl
= \342\200\224
x 1 =\342\200\224
x
3
3
S4 =
271x -
x1
\342\200\224
=
4
4
/. 5
=^
+ 52
+ 53
271 271 271
= 271
+
+
+
this series
the
for
with
cake
+-
3
4
2
many layers we get,
infinitely
S
=
2ttYk=\\
know to
we now
which
areaof
sidesis
the
Therefore
eat but
of each
top
be a divergentp-seriesand
layer is finite (total
tt)
l
o d1 + -l + -l
271
^
Continuing
4
3
2
=
+ 54
the
so
total
K
we
conclude
surface
though the surface
including the areaofthe
that even
area when
infinite.
we have a cake with finite volume but infinite
could not find enough icingto cover...
is, a
cake that
algebraically; that is, if the sequence{an}converges
to
surface
area;
that
we
could
Summary
\342\200\242
Limits
as expected
behave
a limit
a and
the sequence [bn] to a limit b, then:
lim(
+
pan
=
limaw
lim(anbn)
= pa
+qlimbn
qbn)-pliman
+ qb forp,qe.
= ab
limbw
lima,,
a
- = \342\200\224
when
1limfr,
b
\342\200\242~~{bn)
b #
If the sequence{a\342\200\236}
diverges,
and liml
= 0
limU
a, n
\"-+*>{
\342\200\224
=
|
0
then
for some
constant c e ]
oo
J
usually necessaryto manipulatean into a form where the numerator and denominator
have
finite
limits
so that the algebra of limits can be applied.
This
can
often
be achieved by
dividing through by the highest power of n.
It is
both
If a sequence is sandwichedbetween
then
limit,
The
that
sequence
Theorem:
Squeeze
if we
have sequences
< \302\243l
< cM
aM
\"n \342\200\224
n
then
-L
\\imbn
Topic9
v
Option:
i-A
Calculus
other
both of which convergeto the same
sequences,
and must convergeto thelimittoo.
two
is squeezed
for
all n
e Z+
{<?\342\200\236},{bw}
and
limaM
and
{cn}
= limcM
that
such
= L
<
\302\260\302\260
If for
sequence
any
thelimitofthe
yn
If we find the limit
we write lim/(*).
If
so by
do
we
to exist at
f(xn) and
a sequence
that
x0
point.
Theorem and
lim/(*)
functions
Given
=
lim-
x^a g(x)
lim^-\342\200\224
x^a
g (x)
x\342\200\224>a
that
provided
= 0 or limf(x) =
the
Both
point x0 if
the value f(x0) must existfor
To show that
a function
The function
is said to be continuous
The
of a
derivative
Or equivalently
If not,
For a
-
then
A
is not
that
Theorem
Rolle's
Fora
function
X
at
x0
is differentiable
^
\302\260\302\260
x0
must
that
+c
(and
into
limit
continuous there.
x
for which
\342\200\224>
but
xQ
points of its domain.
by f'(x0)
f(x +h)-
)
f(x
\342\200\224-\342\200\224\342\200\224
= lim
exists.
i
at
that
a point
hence
not be
point.
x0:
lim f(x)
must already
exist)
x^x0
)
i
vertical.
that
point.
functions statesthat:
is continuous
,,
- oo '
f(x0)
at all
at a point, is continuousat
exist
'
form 0 x oo'or' <^
a sequence
find
is given
on an interval [a,b] and differentiable
a point
ce
]a,b[ such that
6
r
of a point a and
XQ
at
for differentiable
- f(b) then theremust
if f(a)
oo
-
is not differentiable
at
=
to be
\342\200\224-\342\200\224\342\200\224
if the
have a 'sharp point' at x0
to f(x)
f(x),
0
=
g (x)
of the
f(x)
f(x)
)
f(x
fix]-
^^^o
f(x) must be continuous
function
lim
function f(x) at the point x0
the function
the tangent
5
\342\200\224
:
and
x\342\200\224>a
limits
is continuous
if it
by f'(x0)= lim
must not
C
Q
x^x0
continous at a point xQ,
function f(x) to be differentiable
f(x)
the limit
exists.
the latter
at the
is continuous
and
limit
for
(x)
g
f(x)
\342\200\242
\342\200\224
neighbourhood
lim
x\342\200\224>a
be necessary to manipulate functions that have
Rule can then be applied.
quotients so that l'Hopitals
function
f(x)
coincide
as well.
'
<
functions
in the
differentiable
are
It may
A
xQ
from below,
that approaches
limits of functions of the form
that
= lim
lim
f(x)
x\342\200\224>a
then
\342\200\224>
and
lim
f(x)
X^Xq~
the algebraoflimitshold for
and g(x)
f(x)
g(x)^0, if either:
=
X^Xq+
Rule to find
use L'Hopital's
can
xn
If they
lim f(x) =
We
sequences
we write lim
above
from
two
find
a sequence
having
approaches
X^Xq
The Squeeze
by
y
same limit.
from aboveand thelimitfrombelowto existand
do then:
the limit
that
f(yn)
x
a point
to the
tend
not
do
function at point
of a
having
both
need
We
R is said to be
L e
\342\200\224>
L, then
thatf(xn)
of a function does not existat
which
for
\342\200\224>
x y but
have
\342\200\224>
x0 we
xn
function.
the limit
that
show
To
such that
{xn}
Summary
f'(c)
and
on
]a,b[,
= 0.
mixed
examination
practice
i
153
\342\200\242
The
is a generalisationof Rolles Theorem
that
Theorem
Value
Mean
For a
function f(x),
must
exist
that
\342\226\240f
11%\\
\342\200\242
The
c e ]a,b\\
a point
the area
under a curve).Fora
-f (
\\
.
f'(c)
b-a
links the conceptsof differentiation
of Calculus
Theorem
Fundamental
(to find
such that
and differentiateon ]a,b[,there
interval [a,b]
on an
is continuous
that:
states
function
continuous
f(x)
and
on the
integration
interval [a,b\\
\302\261rf(t)dt=f(x)
dxJa
and
for
function
any
such that g'(x)
g(x)
= f(x)
rb
=
J a f(x)dx
\342\200\242
The
g(b)-g(a).
integral
improper
/\342\200\242oo
dx =
f(x)
is convergent(exists)
\342\200\242
Convergence
known
-
a
(does
not
of improper integrals can be established
with
comparison
by
<
<
>
if
or
that
0
for
all
x
a
then:
is,
convergent
divergentintegrals;
f(x)
g(x)
f(x)
dx is convergentif
g(x)
dx
poo
J
Otherwise the integralis divergent
is finite.
limit
f(x) dx
or divergence
|\302\273oo
J
if the
*b
lim
1
poo
\342\200\242 \342\200\224
dx
only
converges
If j |/(x)|dx
if
is divergent
poo
Ja
for p
converges,
then
poo
I
g(x)
dx
other
is convergent.
f(x) dx is divergent.
> 1.
so does
I
f(x)
dx.
functions
integrals of increasing functions or decreasing
of
width
1.
many
rectangles
Improper
exist).
can
be
approximated
by the
sum of infinitely
-
For a decreasing
function
f(x)
for all
x > a, we have
For an
increasing function
upper
and
lower sum
such that:
k=a
k=a+\\
-
an
g(x) for all x > a, we
an
have
upper
and lower
sum such that:
fJg(k)<[g(x)dx<fjg(k)
k=a+\\
k=a
The
nth partial
sum, S , of a seriesis the sumofthe first
Sn =
An
series
infinite
to
converges
converges
+ un
u1+u2+-~
to S if the sequenceofits partialsums Sl,S2,S3,S4,...
S.
*k
k=\\
154
Topic 9
-
Option:Calculus
n terms:
\342\200\242
Test
Divergence
If limuk
^ 0
or if thelimitdoesnotexist,
\342\200\242
is divergent.
\\uk
Test
Comparison
of positive terms ^Jik
two series
Given
series
the
and
^bk
k=\\
-
to a limit S, ^Jik is alsoconvergent
is convergent
^bk
ak <
that
such
k=\\
for
bk
then if:
all k,
k=\\
to
T where
a limit
T <S
k=\\
-
so is ^bk
Jk-
is divergent,
^ak
k=\\
k=\\
\342\200\242
Limit
Test
Comparison
series of positive terms V%
Given two
and
V/\\,
k=\\
so does
converges
\342\200\242
\342\200\224
=
/ >
0, then
if one series
un
k=\\
and if one seriesdiverges
the other
lim
where
so
does
the other.
Test
Integral
/\342\200\242oo
a positive
Given
-
is
is convergent
then ^f(k)
convergent
x > 1,if J f(x) dx
function f(x),
decreasing
k=\\
-
then
divergent
^f(k)
is divergent
k=\\
\342\200\242
An
series
alternating
\342\200\242
Series
Alternating
alternately positive and negativeterms.
is one with
Test
an alternating series ^juk:
If for
k=\\
-
| uk+l |
then
\342\200\242
If
< | uk
the series
S =
^jUk is
I
for
all
k e
Z+ and lim | uk
|
= 0
is convergent.
the sum of an alternatingseriesthat
satisfies:
k=\\
I
then
than
uk+\\
|
< | uk |
taking the first n
the absolute value of the (n +
the error in
f\302\260r
aU
as an
terms
keZ+
lim | uk
and
approximation
|
= 0
to S (thetruncationerror)
is less
term:
l)th
\\S-Sn\\<\\an+l\\
\342\200\242
A series
^juk
is absolutely
convergent if the series^|
If a
series
^uk
w*
|
is convergent.
k=\\
k=\\
is convergentbut ^J
k=\\
uk
|
then
is divergent,
the series
is conditionally
k=i
convergent.
An absolutely
convergent series
is convergent.
6
Summary
and
mixed
examination
practice
Ratio Test
Given a series ^juk:
k=\\
-
< 1, then the seriesis absolutely
Uk+l
Iflim
hence
(and
convergent
convergent).
Uk
-
Uk+l
If lim
> 1, then the seriesis divergent.
Uk
= 1,
Iflim
k^oo
\342\200\242
A
Test is inconclusive.
infinite seriesof the form:
is an
series
power
the Ratio
then
^jik(x-b)k=ciQ+al(x-b)
+
+ a3(x-b)3
a2(x-b)2
H\342\200\224
k=0
It may
converge:
-
all
for
-
only for
\342\200\242
The
\\x
e R+.
such that a power seriesconverges
J?gR+
the radius
is called
> R
b\\
R
\\x-b\\<R, where R
R such that
number
largest
-
be
value
a single
x g
all
for
R
x e
- b <
\\
|x
for
of convergenceofthe powerseries.It may
R and
be
for
diverges
determined
by the
Ratio Test. If:
-
R
= oo
-
R
= 0
\342\200\242
The
then
then
converges for all xgI
the series
b.
converges only when x \342\200\224
the series
of a power seriesisthe setofall
of convergence
interval
converges. It always includesall pointssuch
of
that
\\x
-
< R
b\\
for
points
but may
which
the series
also include
end point(s)
any interval
contained
interval.
this
be differentiatedor integratedtermby
series can
A power
within its interval
of
convergence.
is an
series
\342\200\242
A Maclaurin
over
term
infinite polynomialwhich matchesallthe derivatives
of
a function
zero.
\342\200\242
The
for f(x)
series
Maclaurin
is given by:
no)
2!
^ik=Q
The
nth
degree
Maclaurin
of the functionf(x) is given
polynomial
,,,^.r(o)..\302\273.
+ /'(0)*
+'
\302\273
/W(Q)..\302\273-^.
4-^k=0
This
full Maclaurin
is the
\342\200\242
The
error
term
-**=/(0)
k\\
in using
an
nth
Rn(x)
=
l
K^-x\" ^n+i
(n
\342\200\242
generalise
approximations
Taylor
f(x)
Topic 9
-
Calculus
Option:
S>,
+ 0,.
= f(a)
n\\
n terms.
Maclaurin
degree
./(w)(Q)^
;
+
+\342\200\242\342\200\242\342\200\242
xn
2!
series truncated at
-xz
by:
is given
polynomial
for
Some
cg
+ 1)!
Maclaurin
series to
by:
]0,x[
allow expansionabout
+ (x-a)f'(a) + -+i-^^f(\"ka)
+
RnM
any
point,
a:
at
Lagrangeerror termRn
where the
Rn
&+l\\c)
=
(x)
is given
(x)
by
-
(x
c e
some
for
a)n+l
a, x\\
(n + \\)\\
\342\200\242
'0'
\342\200\224
form
of the
limits
enable
to
provide a representationfor
used to
can be
approximations
Taylor
Coo'
\342\200\224
to be
and
'Proportionalto x
a constant
means
but
(sometimes\342\200\224,
dx
of x, that
multiple
k
means
to x
proportional
Inversely
dy
the derivative
means
of change'
of functions
integral
and
found.
0
\342\200\242
'Rate
the
can be
variables
other
used).
is kx (k is a constant).
\342\200\224.
x
means that the
of decrease'
'Rate
\342\200\242
A differential
1
\342\200\242
A differential
\342\200\224
=
^
solved by separation
can be
f(x)g(y)
can be written in
which
equation
ov
form
of the
equation
is negative.
derivative
dy
=
the form \342\200\224
/
i
y i
\342\200\224
is
dx
of variables:
a homogeneous
called
\\xj
equation.
differential
\342\200\242
A substitution
differential equations into variablesseparable
turns homogeneous
y-vx
differential
equations.
differential
\342\200\242
A linear
\342\200\242
Linear
differential
can be written
equation
= Q(x).
dx
solved by multiplying
can be
equations
as \342\200\224
+
P[x)y
by the
through
factor:
integrating
l(x) = JP{x)dx
A
plot
of the
equation.
An isocline
y)
-
l(x)Q(x)-
construct
us to
enables
all
differential
the same
conditions.
initial
the
gradient.
equations at a given point states that
to differential
solutions
approximating
solution curvesto a differential
approximate
have
tangents
differential equation.
of a
field
slope
determined by
curves are
solution
is a curve on which
Euler's method for
the
dx
different
These
(x)
\342\200\224(l
at all points (x,y) is calledthe
tangents
Drawing a slopefield
given
into
the equation
turns
This
equation:
\342\200\224
=
withy
f(x,y)
dx
point (x0,y0)and find
= y0 whenx
= x0
-
Start
-
Movea fixed step h in the x-direction
and
calculate
the corresponding
=
k
from
the
ofthe
field:
m0h.
slope
y-direction
gradient
k) repeat the above.
Using the new point (x0 +h,y0+
-
at the
The solution of a differential
=
approximated
y
where
y'\\y\"\\
=
y(x0)
etc.
can
m0=
gradient
f (x0,y0).
distance
k in the
equation
-j7
can be
the
f{x>y)with
by forming a Taylor
+ y\\x0)(x-x0)
be found
x = x<\302\273y=
for y
polynomial
+
yo
about the
point
xQ:
+...
J-^(x-x0)
+^\342\200\224^(x-x0)
by successively differentiating
the
original
differential
mixed
examination
equation
w.r.t. x.
6 Summary
and
practice
'a
practice 6
examination
Mixed
differential
the
Solve
equation
-3x dy
+ ye~3x
dx
that
given
y
= e~x
Let f(x)
x =
\342\200\224
\\ when
x =
tan
0. Give
your answerin the form
the Maclaurin seriesfor
= 2fix)
W
J
3x2+ -x4
y
\342\200\224
f(x).
[13 marks]
x6
x
and
ex
cosx,
show that
\342\200\242
\342\200\242
+ \342\200\242
90
6
e
(b) Hence find
0,
2.x.
+ cos
(a) Assuming
71
for x e
cos x
2.x-2
+ cos
lim
[8
marks]
[6
marks]
[7
marks]
x\"
Find:
xP-a?
x^a xq \342\200\224
aq
(a) lim
-\\-x-'
tx
(b) lim-
xJ
x^O
By considering the gradientsof In
In n < n for n > 1.
(a) (i)
show
Hence
(ii)
that
vJ_
Zj
i
x and
x, show that
t. diverge ^es.
&=2
(b)
Use the
InJfc
Comparison Testto determinewhether
k=3
or diverges.
Consider
y
-2
the differential
whenx
dy
=
equation \342\200\224
dx ye~x
of y
2
with
the boundary
condition
= 0.
(a) Use Eulersmethodwith
value
converges
zli~T~
when
x
step
\342\200\224
0.3.
Give
h =
length
your
answer
0.1 to find an approximate
to three decimal
places.
(b)
d2y
d3y \342\200\224
to find the valuesof \342\200\224\342\200\224
and
dx2
dx3
Use differentiation
when
x - 0.
(i)
(ii) Find the Maclaurin
the term
in
Hence
(iii)
when
series
can
Topic 9
-
Option:Calculus
^3,
v*\\
+ 01.
up to
approximation for the value
improve
approximationsabove.
158
for y,
and including
x3.
find another
x = 0.3.
(c) Explain how you
expansion
the accuracy
ofy
of each of your
[20
marks]
3
(a)
the function
b such that
a and
Find
x<0
cosx+ a
/(*)=
\342\200\224
\342\200\224
0<x<
+1
xsin\\
x)
71
2
71
I
1
\342\200\224
x>
\342\200\224x2+b
I 2
71
is continuous
(b) Show that
everywhere,
at x
is differentiate
f(x)
\342\200\224
\342\200\224
but
at x
not
\342\200\224
0.
[9 marks]
71
^
Consider
.i
i.^r
i
differential
the
.
equation
.i
dy = ^(9x+y)(4x+y)
-^^-^
(a)
can be written
that this
Show
in
^- for x> 0.
\342\200\236
x2
dx
dy
[
\342\200\224
=
form
the
f\\
dx
(b)
Q
Hence
solve the
formy
- f{x).
(a)
Evaluate
(b)
Hence
the
equation,givingyour
y
\342\200\224
\\x
in the
answer
[13
\342\200\224dx.
integral
improper
show that the
'COS*
improper integralJ
dx exists
x
(is
[10 marks]
convergent).
Q
(a)
Find
(b)
Show
the
(a)
x5.
|x|< 2:
that, for
Find
up to the term in
for sinx
series
Maclaurin
sin.x-.xH
9
marks]
[9
3! 15
marks]
dx
f\302\260\302\260
oo
j
(b)
Consider
whether
determine
Hence
the differential
and
4-x2
.x =
\342\200\224
1 when
y
(a) Use Eulers methodswith
of y when x=l,
(b) (i)
By
first
equation.
(ii)
Give
0.
h \342\200\224
0.25
your
giving
finding
your
XJ^\\
+ d.
an approximate
value
decimal places.
factor, solve this differential
an integrating
answer
to find
to two
answer
in the
form y
\342\200\224
f(x).
Calculate,
correct to 2 decimal places,the value
when.x=
1.
6
C
[8 marks]
equation
dx
where | x | <2
or diverges.
converges
Zj~\342\200\224~ \342\200\224
1
k=2 , 'K In K
Summary
and
of y
mixed
examination
practice
(c)
of y
the graph
Sketch
why your
explain
value
for 0
\342\200\224
f(x)
approximate
<x
value
< 1. Useyour
sketch
is greater
than the
ofy
to
true
marks]
[24
ofy.
IB
(\302\251
2005)
Organization
00
Q
(a)
Show
(b)
Find
an
(Q
-\342\200\224
converges.
number of terms it is necessaryto taketo ensurethat
the minimum
sum is correct to
of this
approximation
integral J =
the improper
Consider
(-lf+1k
V
the series
that
1
7
A
1
Shmrtfaat
OI1UW
where
(b) Hence
A
find
(c) (i) Find the upperand
partial
(a)
part
for/.
~k2+?>k+2~
Hence evaluate
the upperand
\342\200\224
show that
(d) Hence
< In
12
ID
limx sin
(a) Find *->~
(b)
in
Determine
(a) Use
1
-x
2
(b)
1x3
2x5
Determine
divergent.
160
Topic
9
-
Option: Calculus
[15 marks]
< \342\200\224.
11
\342\200\224
-1
the radius
+
sums.
\342\200\224
\\\\\\)
^^/csinl
\342\200\224
I
,
x2+
+
ln(l
[6 marks]
or diverges.
converges
Taylors theorem to prove that
(a) Find
2
n +
2
lower
(b) By making an appropriatechoiceofx,
^3
few
\\xj
whether
^axa2 ...an<
first
1
1
1
A
(iii)
out the
and writing
that:
deduce
sums,
found.
sums
lower
(ii) By using the resultin
dx.
of I.
value
exact
the
mark]
x +2
to be
constants
B are
and
1,
1
x +
-\\-3x-\\-2
x2
[1
B
\342\200\224
LIldL
+2
+ 3x
x2
f0
(a)
1 decimalplace.
x)
< x.
hence
prove
that:
[9 marks]
n-
2-
n
of convergenceof the infinite
1x3x5
2x5x8
whether
,
x3+
the series
1x3x5x7
2x5x8x11
^sin|
k~1
series:
A
x4+...
y + kn
]
is convergent
or
[15marks]
ofthe powerseries:
(a) Determine the radiusof convergence
y(*-tf
ft
h
(b)
m
(a)
the set of all x points for which
interval
of convergence).
(the
the
Using
that
Show
for all
real numbers
1integrating the inequality
that
show
otherwise,
for all
or otherwise,
t
parts
Using
(d) or
(a) and
all positive
0 <
s <
integers
n:
|<1
4:
>i
[t, 2] or
t <
2:
>2-t
J
show that
nln
(e)
for
of part
(b) over the interval
real numbers t, such that 0 <
\\
Hence
[10marks]
s, suchthat
4-t
In
(d)
converges
4-5
S
(c) By
i
+ -
series
show that
Value Theorem,
Mean
f
1
nln 1+ (b)
the
Find
all positive
for
integers
n:
2n
1+i,>
2n+ l
n)
otherwise, showthat:
lim| 1+ -
=e
nj
[19
marks]
2002)
(\302\251IB Organization
^
(a) Find
the degree5 Maclaurin
f(x) =
(b)
(i)
Show
the function
ex2
that
f6) O)
(c)
for
polynomial
= (646*6+ 480x4+ 720x2+ 120)e^2
(ii)
Show
that/(6)(x)
(i)
Using
the answer
for all x > 0.
is increasing
to part
(a), find
an
estimate
for
2ex2dx
Jo
Give your answerto 5DP.
(ii)
Using
part
(b) and
the Lagrange
error term,placean
this estimate.
XJ^\\
+ 01.
bound
on
[18 marks]
6
c
upper
Summary
and
mixed
examination
practice
'a
lim
Answers
1
Chapter
(c) (i)
Exercise 1A
1.
(a)
(ii)
(i)
3
0
(c) (i)
(i)
3.
-2
(ii)
(ii) 4
(ii)
0
(b)
f(x)
lim
f(x)
does not
lim f(x)
does not exist
lim/(x)
= 0
lim f(x)
does not
lim /(x)
does not exist
(d) (i) lim/(x)
125
lim
2
1B
Exercise
=2
/(x)
(a)
0
(b) 4
(c)
0
(d) 0
=2
lim/(x)
= -7
lim/(jc)
= -8
4.
(a)
1
5.
(b)
0
does
/(x)
(e) (i) lim/(x)
x^x+
3. (c) 1
lim
f(x)
lim
f(x)
(ii) lim/(jc)
x^x+
(b)
not exist
=3
=3
x^xo
(b) 1
a =e
= 2
lim f(x)
lim
lnx<x
lim
(c) 3
(d)
exist
\342\200\224
(ii)
7.
exist
oof
5. 1
6. (a)
= -oo
lim f(x)
4. 0
1.
= does not exist
x^xo
27
(b)
lim
x^x+
\342\200\224
16
2.
= oo
x^xo
(b) (i)
(d)
f(x)
x^xo
/(x)
= 3
=2
= 2
x^xo
(i) e
(ii) That
limit
of function
lim
is
/(x)
= 2
function of limit.
1.
(a)
(i)
lim/(x):
2/
2. (a)
1C
Exercise
=1 /0*0
= 5
lim f(x) = 5
X^XQ
lim
/(x)
lim /(x)
(ii)
lim f(x)
X^XQ
lim
(b)
(i)
(ii)
/(x)
=5
=
1
~3
_1
~3
1
=
(b)
~3
(c)
lim/(x)
=
oo
lim /(*)
=
\342\200\224oo
/(x)
lim
f(x) =
x^x+
lim/(jc)
x^-3
(d)
not
exist
\342\200\224oo
Answers
^
\302\243
Sa
+ 0,.
2
4. a = -3
= 4
c=l
162
=-5
= -5,1
jc
3. (a)
b
= 6,
Yes
(e)
<does
lim
lim/(jc)
a\302\273i
lim/(jc) = -l
5. (b) No;
6.
(b)
4
7.
(a)
1
lim/(jc)
x- 2
1. (a)
(b) -1
(c) Doesnot
9
(b) 2
8.
IF
Exercise
exist
9. 0
10.
exist
not
Does
1D
Exercise
1.
(a)
(b) (i)
(c)
(ii) 1
(i) 2
(i)
(ii) 6
471
1
\302\253\302\273!
2.
(a)
-2
<
(d) 2
(e)
oo
3.
(a)
0
4.
(a) -1
5. (a)
(b)
6.
(b)
(b)
Secondset c )f
oo
(c)
differential
-4
4
9
8.
0
9.
-1
IE
Exercise
(ii) Yes
Yes
(a)
(i)
(b)
(i) Yes
(c) (i) No
(d) (i) No
(f)
2. (a)
Yes
(ii)
Yes
(ii) No
(ii) No
(ii) No
(i) 1
(ii)
(iii)
Yes
(f)
jc=
1
0
Does not exist
(iv)2
(vi)2
(b)
(ii)
(i) Yes
(e) (i)
3.
(c)
1
7.
i.
0
All x
(v)
2
e R exceptx = 0,1
L
X
k = -l
4. (a) a = -l,b
(b)
All
x g
=-2
(g)
R except x = -
6.
Consider
such as
C
a sequence
xfc=x0+-^\342\200\224
S>-
+ d.
4
y
x^O
'a
30
3.
5. (b) /(l)*/(4)
12. (c) 0.732
13.
Use
and
Theorem
Rolle's
induction.
(b)
1 root.
15.
xe]
All
UseRolles
except x = 0
(c) All xeR
3.(a)
examination
Mixed
1. (a) a = 2, b
2. 1
e R except jc
(c) All x
5.
(a) e
c =
5.b=16,
6. (b) sin
7.
-20
yx
\342\200\224
1.
=
f(x)
where
g(x) + kx,
1.
(a)
1G
=
(i)x
(b) (i) x = -
(b)
(i) -2sin(2*3)
t3+3
(ii)
x =
(ii)
x = 271,671
(ii)
* = 1
-l
= \302\261-t
3.
(ii) 2cos5x
(ii) -5^f2+3
(d) (i) f{b)-f(a)
(ii) t2+7
(ii) -3cos2(2a2)
(ii) g(y)
(ii) ~g(a)
(ii) 0
(ii) 0
(e)
(ii) -(h4
+ 5
e-*3
2. (a) (i) /(*)
2
(c) (i)*
e2
(i)
(d) (i)
Exercise
2
(a)
(c) (i) -r^t2+2t
question.
previous
e
(b)
That limit of function
is function
of limit
in
is function
g[x)
(b)
(i)
(c)
(i) 0
(a)
(i)
f(b)
(f(x))2
-
(b)
X
2. (a) (i) x = 2
(ii)
(b) (i) x = 0.881 (ii)
(c)
x =
jc
(ii)
jc
= 16
\\\302\243
4.
(a)
(i)
1 --
Answers
r^
(ii)
3*2ln(-j
5. (a) 2xsin(x4)-sin(x2)
(b)
164
4 a.
/'(*)
0
= 0.680
(b) No
(i) *:
1
2
Exercise 2A
11.
Let
(b) -
Chapter
0
-cos
(c) 2xsin \342\200\224
1
0
(c)
(c)
=0
10. (b) /'(0)
(b)
of limit
\302\261
except x-
xeR,
1 501
of function
11. (a) 1
9. (a) a = 3,b= -2
All
>
v
(b)
10. (c) 5
cos
x
2
(a)
9.
\342\200\224
\\xj
-
(c) That limit
is function
= 0,2
exceptx = -3, 1,3
4. All xgR,
(b)
(a)
1
except x = -l
1
4.
AllxeR
practice
= l
All xeR,
(b)
(b)
Theorem
-y
27
2e-^2
1 --
4e-^
\342\200\224y
47
-\"\"0
Exercise 2B
1.
1
(a)
(b)
f\342\200\224i\342\200\224<r\342\200\224i\342\200\224^
J\302\260
+
x2+8*
+
(b)
4
17
f^k2+8k
(c) Diverges
<i\342\200\224L
+ 17
fc=0/c2+8/c
2.
(a)
3.
(b)
Diverges
Converges
(c) Converges
(d) Converges
(e)
(f)
Diverges
. , v
4- (c)
17
dx
<
>
J3
S^+e4
e*+e\"*
Converges
(b) p>\\
(a) p<0
4. 1
5.
5. 1
7.
4
(b)
8.
(b)
9.
2
(a)
X2-*<JV*dx<X2-*
<
(b);
v
1024
2*dx<
J'\302\260
512
In
examination
Mixed
1. (a) 2.
i-
3.
k2
w
Xe\"fc2<re\"*2ck<5/
(b)
> .
k=6
<
1
Jk + 2
3/3 +
ln(cos2
(a)
a =
(c)
0
l,b =
k=l
fc=2
-
(a)
< Jsr .
1
Jx +2
4.
+ 2
6.
7.
l1T<)2-r*x<l
^ K J10 Xv2
^
|;_\302\261<r_^<|:_\302\261
K
\\c2
fc=10
x
fc=ll
(b)
Xe\"fc2
<
Ji
<
fc
9. (b)
<
Xe\"fc2
= 21
(c)
Diverges
(b)
3
Converges
lr2
3
Chapter
U2+i
Exercise 3A
U2+i
y
d*
J\302\260\302\260e\"x2
fc=0
oo
(b)
dx
U2+u
6fc=i
s
/
\342\200\224
n,f(x) = X
<
8. (a)
(b)
1)
dx
fc=l
2. (a)
5
y\342\200\224
Jk
(0
1
(b) ln(cos23x+ l)
2C
Exercise
(b)
2
Integrate by parts
2
practice
Yes
12.
dx
f
k=2
1.
(a)
1.00,1.25,1.36,1.42
(b)
0.866,1.73,1.73,0.866
(c) 0.167,0.310,0.435,0.546
<
1^
(d) 0.368,0.503,0.553,0.571
k _\342\200\236k
+ 5),
2. (a) \342\200\224(3n
/lx \342\200\224
n(l9-n)
1
(b)
y
10
(/c + 2)3
S
(c)
1
divergent
^
-, divergent
/
V
1- (I' \\')
k )J
convergent
Answers
r^
**.
'a
(d)
divergent
6(1.2\"-l),
/
3.
Compare
to /
V
(a)
3(2fc-l)
^
\342\200\224
convergent
(ii)
Compare
to V1 J_
UJ
^k
(b)
(d)
\"3X^
(d) (i)
2X3*\"1
lnfe
\\*
to
Compare
(ii) Compare to
3x2fc
(e)
(i)
Test (lim =
Divergence
(ii) Divergence
3B
Exercise
(a)
Compare
(i)
(f)
^
to >
(i)
(lim =
Test
1)
<*>)
Test (lim = e)
Test (lim = e_1)
Divergence
(ii) Divergence
l
\342\200\224
^
(e)
1.
\342\200\224
(i)
\342\200\236\\
1-o
<
(C)
(c)
\342\200\224
kz
3.
(a) (i)
i
,
Converges
^
\342\200\224
(ii) Compare
1
(ii)
Diverges
V
(b) (i)
Compare
(b)
(ii) Compare
(c)
(i)
Compare
(i)
Converges
\342\200\224
|
\"1^k?
/\342\200\242
(ii) Converges
^k'
1
(ii) Compare
^
(c)
(i) Diverges
kJ
k3n
r
(ii) Converges
(d) (i) Compare
\\Kk2
k-\\
(ii) Compare
1
(d)
(i)
^
)
(-^
Converges
k2
\\K\302\260
J
-Ifi
f
\\_
(ii) Converges
(e)
(i)
k?
Compare
1 ^
<
(e)
(ii) Compare
(i) Diverges
yfk
VV/cy
(ii) Converges
(f) (i) Compare
^k'
4. (a) (i)
(b) (i)
(ii) Compare
2.
(a)
(i) Compare
5.
Diverges
(ii)
Converges
(ii) Diverges
Diverges
(c) (i) Converges
(d) (i) Converges
(ii)
Converges
(ii)
Converges
(e) (i)
(ii)
Converges
(a)
Diverges
(i) Converges,0.202
(ii) Converges,
(ii) Compare
^~Sk
k413
(b)
(i)
0.302
Diverges
(ii) Diverges
(b) (i)
DivergenceTest
lim
V
=
\342\200\224
2y
(c)
(i) Converges,7.71X
(d) (i)
(ii)
Divergence Test | lim
= \342\200\224
10~3
(ii) Diverges
Converges, 0.402
(ii) Converges,
(e) (i) Diverges
2.76
X 10~7
(ii) Converges,1.35x 10~22
166
*v
Answers
r^
*\302\253.
6.
(a)
(i) Absolutely
convergent
(ii) Absolutely
convergent
(b) (i)
(ii)
(i) Divergent
(b)
(c) (i)
(ii) Divergent
(c) (i)
convergent
Conditionally
(d) (i) Conditionally convergent
(ii) Conditionally convergent
(e) (i)
(d)
2. (a)
Diverges
(ii)
(c) ( )
Converges
(ii) Converges
(ii)
Converges
converges
absolutely for all xgM
R
= 5;
converges absolutely
= l;
converges
for all x e
(i) R =
2; convergesfor
(ii) R =
l; convergesfor
all x
for
e ]-5, 5 [
1[
]\342\200\2241,
all xg[3,7[
e [0, 2]
all x
Converges
(ii)
Converges
(e) (i)
(ii)
Diverges
Converges
(Limit
x e
for
absolutely
Converges
]\342\200\224l,l[
Diverges
(d) ( ) Diverges
Diverges
= oo;
\302\243
convergent
(a) ( )
(b) ( )
(a)
for all xel
convergent
Absolutely
(ii) Absolutely
8.
converges
(ii) R
convergent
(ii) Absolutely
= oo;
R
Diverges
= -1
Convergesconditionally
^ 1
Test: /
Comparison
at x
\342\200\224)
x = 1
at
(b) Converges absolutely
for
all
xel
(c) Convergesabsolutely
for
all
xel
k
(b)
9.
(Comparison Test:
Converges
(c)
Converges
(a)
x>e2
(Alternating
/
3. Convergesabsolutely
,TT7T)
4. ]-5,-3[
Series Test)
5. (a)
10.
(Comparison
Converges
X1
with
(a)
Converges
(b)
Diverges
Test:
(Comparison
^>
)
1
^/c3/2
Test)
(Ratio
= 3
\302\243
7. (a)
Converges
n
^3
+ fL
^l
3!
(b)
for
x e
= 2
(b) Converges absolutely
for
x e
(c) Converges absolutely
for
(b)
f^
= n+l
18.
(b)
(d)
]\342\200\224l,l[
\342\200\224
1
at x
examination
Mixed
=1
1.
23
(a)
(Comparison
Diverges
3
practice
^ 1
Test: /
\342\200\224)
k
Diverges
the
Express as
(b) Converges(Limit
of
sum
2.
fractions
two
19.
x e
1 1
2*2
k=n
X/W-X/W<r/Wdx
(c) 4X10-4
=
conditionally
Converges
f(k)<jj(x)dx<^f(k)
]-4,4[
x = -4,4
at x
Diverges
k
41
4!
xe[-7,l[
absolutely
Divergesat
15. Oil
(a)
+ fl+...
)
(c) Conditional
17.
eR
all xe]
for
6. (a)
(b) 1
14. (a)
?!
2!
Holds
11. (b) 4
13.
-5,-3
= i+JC+ ^L
\342\200\224
absolutely
Converges
12.
x =
,2
(b)
v '
^/'(JC)
v ;
(b)
all x
M
x g
All
at
Diverges
for
(a) Convergesfor
a > 1
(b) (i)
conditionally
3.
Diverges for a < 1
Converges
(ii) Converges
4.
(a)
c\302\253
(b)
2
Comparison
Test:
^
>
1
\342\200\224)
^k2
(b) Converges
85
(b)
<
72
absolutely
^V 1 < 263
\342\200\224
\342\200\224
\342\226\240
^m3
216
K2
6.
3C
Exercise
12
7.
1.
(a)
(i)
R =
(ii) R
0; only
= 0;
(c)
converges
only converges
at x
at
=0
x =
Diverges
(Test for Divergence)
-3
Answers
^
\302\243
S>,
+ 0,.
8. (b)
= 2
= B
A
4B
Exercise
2
<d,9.
(a)
(b)
1.
(Comparison Test: 3~2^
Divergent
convergent
Conditionally
Series
Y
(3\342\200\224
11.
Convergent
12.
(a)ln(ft + l)
13.
0<x<2
4.28 xlO\"2
(ii) 0.5625
2.44X10\"3
(c) (i)
(ii)
1.70
XlO\"4
(Alternating
2.
convergent (Integral
Test)
3.
+ \342\200\224(b)
(a) 1-jc
(a) 1.43XlO5
5. (a) 1+ ^
Diverges
0<x<0.1817
2
1.27
(b)
433
x4
x2
(b)
(ii)
0.187
(i)
(b)
)
Test)
(c) Absolutely
(a) (i) 1.25X10\"4
^4!
+
2!
\342\200\224
(b)
384
(c) 1.36 XlO\"4
14. (b) oo
(c)
6.
Diverges
+
x
(a)
\342\200\224
+ \342\200\242\342\200\242\342\200\242
5
3
7.
9. (a)
4
Chapter
17
32
L_J_
Factorise
4A
Exercise
(b)
the
from
4
i-_
1. (a) (i)
+
2
9
f\\\302\273
+ ...
24
243 7
x7
560
,
+\342\200\224x5
40
2
(b) (i) 1+ -+\342\200\224+
(ii) \\-3x +
8
48
+...
10.
9
2
2
\342\200\224
+ \342\200\224
+ ...
(c) (i) 1+ \342\200\224-
4
4
16
8
15
(n) l + -x2 +\342\200\224x4+
2.
4
2x2
5.
(b)
6.
(a)
(b)
30
\342\200\224
n(n
1)
+ \342\200\224-x2
4C
24
720
3x4\342\200\224xl2+\342\200\224
x20
(ii)
W
l-2x
(b)
4
3
45
3
9
2
8
560
x28+.
x3+...
9
+-x3
+..
8
xe+...
(ii)
^ + ^1
18
--x3
-2x-2x2
(c) (i) 1
+ ...
2
+ -x2
4
3
+
2
-4x4+...
+
.
48
8
+ \342\200\224
+ ...
(n) 1 + x3 + \342\200\224
2
6
972
= 4
\302\243
\\k
X
(b) ~1<X^1
\302\243(-l)Vt
k ~ 1
2. (a) (i)
2x2-x3\342\200\224xA+...
3
81
9
8. (a) Not equal to /(0).
at
(b) /(x) an increasingfunction
derivative
of series is negative
Answers
V>A
+ 01.
1
25
2
6
(ii) -x--x2+\342\200\224x3+2xA+.
(a) 1+ -jc\342\200\224x2 +\342\200\224x3+...
3
168
40
2
(b) (i) ln2 + -x--x2
15
2 12
243
(i)
k=2
7.
81
9
(a)
17
315
xA + \342\200\224
xe+...
3
4. (a)
l + nx
.
27
27 3
\342\200\224x2
+\342\200\224x3
+.
x5 +
(d) (i) x + -x3 +\342\200\224
2
1.
16
8
.... -13
+ -jc +
(n)
(a)
Exercise
-x2--x3+...
2
R3(0.5).
\342\200\224
+ ...
9
2
for
expression
720
81
,
3x--x3
(n)
_-_
x =
at
0
x =
but
first
(b) (i)
\\
+ -x2--x3+-x*+--
2
3
8
0
(n) x + 2x2
+ \342\200\224x3
+ \342\200\224x4
+ \342\200\242
6
3
(c)
X
(i) x
+
X
+
12
6
2
2. In3 + -x
X
x2+\342\200\224x3+...
+ -e3(x-3)2+-e3(x-3)3.
3. e3+e3(x-3)
(ii) -x
23
3. 2x + 6x2
5-
W
(b)
(a)
(b)
+
\342\200\242
-i<x<-
5.
6.
2
+ x +\342\200\224
+ \342\200\224
+ -x4+...
2
2
(b)
0.574
(c)
1.39 X10-6
i\302\260-
(a)
2 + \342\200\224
(jc-8)
+ ...
4E
Exercise
1.
\342\200\224
+ \342\200\224
+ ...
12
(a)
(i)
(ii) i
-i
2
(ii) 2
6
(b) (i) 2
(c) (i) -\\3
3
\342\200\224
x3+...
x2-
ln8--x-\342\200\224
8
2- (a)
8
(ii)
fc
X(-l)
3.
9.
(a)
- 0.687552; x
(c)
0.949
(d)
0.00625
3
=-
^
4.
(a) x
5
10. No Maclaurin seriesas derivatives
x = 0 do not exist.
1.
(a)
(b)
(i)
5.
at
i +
2fx--l
(b)
6. (a)
7. (a)
8. (a)
+...
+ 2fx--l
Mixed
a)
1.
2
(ii) e16+ 8e16
(x -
8
4)
+ 33e16
(x
-
A)2
(a)
'
*2
\302\243f
120V
+
3
... -1
x'
xJ
7
5
x2 +\342\200\224
+
...
(b)
6
-2
(c)
Sx9
2x3+2x6+
23
4x12+...
(b)
error
< 2.38
X
=
f'{x)
(\\ +
X2
2
+
Xy3;
X4
3
4
6
practice
x)-l;f\"{x)
X3
2
10
(b)
-1)(2* -1)!
examination
x
(b)
(-D*\"1
y
f^-K'-f)+l4-f)'+(c)
+
3
(b)
30
f'\"(x)= 2(l+
+...
(b)
\302\253>
\302\253)
xJ
S(2*c
\302\253\302\2534HKHJ+i+I(x-i)-I(x_i)2+...
**
.\\fc+l
2(-lf
3
4D
Exercise
X^/
(b) 6
;-\\<x<\\
2
,1 x
\"
(b)
6V
YIn
2
*6k
(a) (x-3)--(x-3)3+\342\200\224
(x-3)5+.
'
'
r2fc-l
(b)
J2
(2*)!
3
+ ln8
(b) y = \342\200\224x
00
(x-8)2+-
288
12
(c)
8
2 12
2
V3:aJ+.
12
8. (b) ee-\342\200\224(x-e)2+...
+ ...
(c) x + \342\200\224
8. (a)
xA
729
3
2
4
\342\200\242
+ ...
x + \342\200\224
\\
>x
1
.
1
11/4. (a) - + -V3:
2 2
Z\342\200\224jT\342\200\224x
k
k=\\
7. (a)
(b)
+
+ \342\200\224x3
+ 5x4
3
2
6.
12
6
2
81
18
3
\342\200\242
fi(x)
= -{\\ + xy2= -6(l
+ xY4
+ ...
ln2
Answers
r^
**.
4
5
Chapter
M*-!)+\302\245H)'-
5A
Exercise
+ ...
-{'\342\226\240
3.
a)
5x3
x+
41 x5
+
b)
+ \342\200\242
1.
120
6
(b) (i)
X
+
a)
X2
X3
8
16
+
4
x+
XA
X*
X*
\342\200\224
+ \342\200\224
+ \342\200\224
+ ...
1! 2!
+-
4!
Xn
4(2!)
3(1!)
(a)
= 12sin
3
(ii)
= 8e2+c
+
y
(i)
y =
ln(x2+l)-ln2
(ii)
y =
3
\342\200\224
x2+ \342\200\224
lnlxln\342\200\224
xn
= arcsin
\342\200\224
2)\\
(b)
K
id
(c) (i) y
5651
3.
7560
(a) (i)
11
b) R=l
*+
+
40
(ii)
45 m
(ii)
6.58
(ii)
350-50yft
(ii)
y--
kg
5B
Exercise
+ l)
k\\2k(2k
1.
1
(i)
(a)
y
= -x3/2
3
(b) (i) y = 2x*
I+T2\"
l-2x2
(ii) y = 3e
.3
+.
+
7
5
3
$1985
= -2e1\"2*+2
+
---;-1<a:<1
b) x
(ii) y
112
Ix3x5x---x(2fc-l);c2/c+1
+
o m-
= -2e\"3*+2
cm
4500-375t2
(i)
(c)
6
4
2
(i) 35.0
(b)
5x7
\342\200\224
H\342\200\224
+ ...;
n(n
4
x3
3x5
\342\200\224
+
+
- +2
(11)
- + ...
\342\226\2402k
k=o
y
(ii) y = tanx-x
y = arctan
-1<x<1
c)
-
(ii) y
(i)
(n-l)!
+ ...+
2
= -e2*+c
y
(c) (i) y = 3tanx +c
2.
b)
+ c
2
\342\200\242\342\200\242\342\200\242
2
b)
y =\342\200\224cos2x
X
I
.1
a)
(i)
(a)
2.
(i)
(a)
siny =
\342\200\224cosx
2
\342\226\240,v
,.,
d) (1)
10.
a)
304
4
(b)
m
9
1-x
3.
1
=
<C|
170
(ii)
arcsiny
v
=
+ x|
ln(l + x2)
6.
7.
1
10
Approximately
times
,2*
(d)
11
Answers
r*a*
(ii)
e^=e*+e2-l
(i)
y =
(i)
y = arcsin(ln|x|+c)
(ii)
y
(c) (i) y
4.
X77**
k=0 K'
(a)
(b)
(l-xf
(a)
(i)
2arctany = ln|l
2>
(1-x)
f'\"(x)
11-
tan y
(c) (i) e\"2'=-4e*+5
0
fix)
(e)
(ii)
{m)
b) /'(*)
(d)
\342\200\224
tan x \342\200\224
3
+*.
too large.
y
\302\261yjx3+c
(ii)
2 tan
= Aex(x-l)-3
(ii) y =
(In x2
+c)
fc = 3
Exercise 5C
1.
(a)
(i)
= -
x2 +c
= tan(c-cosx)
= e-(i-*)2
y =
r
-In
y-2x
y +x
= \\n\\x\\
+ c
l-x
c
\342\200\224
Sx
(ii) -In y
(b)
=
Ui
(i)
In
(ii)
\342\200\224
=
\\n\\x\\
+ t
-\\n\\x\\
4.
y =
5.
r =
+ c
6.
y =
8.
y:
ecosx(x+ c)
2
3
-\342\200\224+-
X1
(x + 2)cosx
X2
+c
ln|x|
y
(c) (i) --In
(ii) In
=
3-^1
= ln|x|+
L\\X_L
X\\
X
9.
ln|x| + c
X
+C
2(x2-l)
xz = -x
(a)
dx
z = l
(b)
+ ce2
c
(c) y=l
36
x3
e* =ln|x|
(i)
(d)
(ii) sin
10-
+c
{a)
'=Wt+t
+ c)
(b) siny = cosx(lnsecx
ln|x| + c
\342\200\224
=
\\x
2.
y =
(a)
(b) y 3.
xln| x |
y =
(a)
xln|x|+cx
4.
+
\\x
the form
5.
(a)
f
(b)
6.
is one
equation
(ii) 1.07
(c) (i)2.89
(ii)
(a)
(i)
of
Kx)
=^
x_y_
= arctan
|
2ln|x|
+ 4f(4x +
(b) (x-y
-^1-1^4
y +
+ ^-
\\
+ c
lf =16
7. (b) (ii) 3.79
5D
Exercise
1.
(a)
(i)
-ex+ce~2x
y=
1
-e*+ce4
(ii) y(b)
(i)
y
=
+ ccscx
\342\200\224cotx
x +c
....
(n) y =
cosx
. ...
(c)
In
+ _
(i) y = \342\200\224L_L
X
(n) y =
2.
1
y
c
1x1
X
c
\342\200\224+ 1
x2
x
y
1 ,
2
= -e^+-e2-x
2
3. y = x2ln|x-3| +
r^
cx2
**.
...
2.32
(b) (i)1.57
+ 8)
y2=2x2(ln\\x\\
^dx
differential
(ii)
\342\200\224
=
dx
(c)
2.
ln|x|n\342\200\224
A homogeneous
(a)
1. (a) (i) 0.708
xtan(ln|x| + c))
(b) y = xtan
5E
Exercise
=
\342\200\224
x^O
x
1.45
'a
(c)
(d) Solution is constant or in
5.
(c)
1.629
6. (a)
(i)
equilibrium.
0.825
(b)
r =
(c)
(i)
x2
1
+ _ + e-o-3
_m^__
11.0%
(ii) Take smaller h.
7. (a) (i) 0.615
(ii)
use
smaller
step length
(b) y = -e3*3;/(1) = 0.698
(c)
8. (a)
The
tangent
(ii)
\\
is always below the
(b) 1.2165
+ -x2+-x3
+ 2x
2
2
9. (a) ^ = 4-3e\"*2
dx
1.
- -x2 + 3
y
3. (a) (i)
\\
1
2
2
3
3
4
2
3
A
\\
\\
X
\\
X
I
2
3
- + x + \342\200\224
xz + \342\200\224
x5
(i)
71
\342\200\224
+
r
~
/
+ n)-n(x
(n2 -2)(x
+ n) +
(3
47T2)(x
V2 +
(U)
+
4.
(a)
y =
+
JCV
\342\200\224-
/
/
/
/
/
/
-t- -/-
\\
0 on
7Tx3
4,
(b)
3. y =
Answers
**.
l-2x + \342\200\224
x
2
0.671
6-x2
y = x
y
\\
jc-3
r^
\342\200\224f\342\200\224*X
^-^
2. (a)
172
\342\200\224
\342\200\224
\\
7r)3
V2U-^]
lW2f
6
+
\\
gradient =
(b)yt
+
practice
I
(n)
2
4.58
V
+ c
+ -x2+-x3
+ x
X
,\302\273X
(b)
(a)
(b)
examination
Mixed
curve.
56
5
'a
4.
(i) 2.141
(a)
7. (b) y =
8. (a) 1
smaller steps
(ii) Take
(b) y = x2+e1~x2
9.
5.
4.46
(a)
(b) y = (x + 4)e2x2
;
(c) Take a smaller
6.
y
step
(a) (ii) l + 2(x-l)
+
y =
(ii)
1.204
-(x-if
(ii)
xe
Diverges
-
\342\200\224
+
\\2)
|
2
= L77
y(l)
(c) Gradient is decreasing;hencethe
produced
by Euler s method will
than the true curve.
x
69
(c) 0.026%
chords
track
higher
12. (b) 19
13.
2x
7. y:
(b)
(b) (i) y = yJA-x2\\arcsin[
x-l
(i)
5!
oo
(a) 1.83
11.
length
(iii) 1.205
(b)
+
3!
10. (a)
= 4.55
lnx + C
x\302\260x\302\260
\342\200\224
+ ...
x
(a)
-6x-
(i-*2)
A =
(a)
1,B =
-1
(W
h(\302\243
5e*
4(l +
9.
(b)
- e~
- 2xe~*
a)
(c)
x)
(iii)
ye^/x=Ax2
f\342\200\224-\342\200\224</<f\342\200\224-\342\200\224
\302\243lk2+3k +
sum
Upper
sum =
2.
(b) -3
3.
(a)
=
practice 6
examination
y
-(e3*+2)cosx
14.
\342\200\224
\342\200\224
1
(a)
(b) Diverges (DivergenceTest)
15.
no.
x.
Take
+ On
0,+On+...
-f1-ru2
^
<
4. (b)
Diverges
5.
2.674
(a)
(b) (i)
(ii)
16.
d2v
^4
dx2
7y
==
2,
d3y
-^-
dx3
= -2
~ 2 + 2x + x2
17. (a)
(b)
X3
-
a =
R
= l
xe[3,5[
3
(iii) 2.681
(a)
(a)
(b) Convergent
(c) For (a) take smaller step length;
more terms.
6.
+2
12
6
Chapter
1.
~0k2+3k
11
Lower
Mixed
=
2
0,b =
for
(b) take
19. (a)
(c)
\\
+ x2+
x4
\342\200\224
2
(i) 0.54479
(ii) 6.59X
-
10-
Answers
173
w
C
XJ^\\
+ d.
\\>
Glossary
Words
that
more
definitions of other terms are also defined in this glossary. The abstract
in terms of other,
defined terms can realistically
be explained
only
in bold in the
appear
nature of this
option
means
that some
concepts.
simple
Term
Definition
absolutely convergent
A
negativeterms which
still
converges
terms are made
all the
when
and
both positive
with
series
positive.
series
alternating
A
are alternately
terms
positive and negative.
Alternating Series Test
A
test
stating
. , , ,
is absolutely
h...
7
8
111
,
convergent because H 1
4
2
1
\\-.
is convergent.
whose
series
Example
, 111
1
1
111
1-- +
,
for convergence
of
that
an alternating
whose terms tend to
0 is
series,
series
^The
...
+
4
3
2
111
1
series
1
convergent.
is
h...
4
3
2
always
convergent.
boundarycondition
value
The
of the
unknown
at one point, which
find a particular
differential
Comparison
Test
A
test
function
us to
allows
solutionof
a
equation.
of series
for convergence
based on comparing it
to
If we know the
order to find its
Since
Jfc2
Comparison
Test for
A
test
to determine
whether an
improper integral converges
by
it
to
another
improper
comparing
t
-1
dx
2,77
test states
also
converges.
converges
because
1
cosx
< \342\200\224
and
integral.
1
^
and
\342\200\224\342\200\224
J
example
the comparison
>
7 1
that J \342\200\224dx
we know
converges.
convergent
conditionally
A convergent series which
diverges
when
all the terms are made
1
positive.
convergent
because
u
the
.,111
series H
u
but
1
2
3
i.
4
1-...
is conditionally
it is
1
2
3
1
convergent,
4
1-...
is
divergent.
A function
continuous
lim
f(x)
=
which satisfies
f(x0) for every
point x0.
The
function
shown
above is not
continuous because lim/(x)
/(1)
Glossary
c
XJ^\\
+ d.
= 3.
in
we need
= 0.
+ l<Jfc:
converges,
object
displacement
when
its position
1
that
improperintegrals
an
one boundary condition, for
another
series.
velocityof
= 2 > but
Term
Definition
converges (to a limit)
When
Example
sequence un
to a limit as n
close
arbitrarily
get
of a
terms
the
The
sequence
= \342\200\224
converges
un
to 0.
n
increases.
differential
An equation involving
an unknown
and its derivatives.
equation
function
Ay
\342\200\224
= 5
-3y
dx
differential
Test
Divergence
A
of series,
for convergence
if the terms
that
test
stating
do not
a
equation.
The series 1 + 2 + 3
+ ...
diverges.
of the series
then the series
to 0
tend
is an exampleof
diverges.
diverges
When
Euler's method
An
The sequence
does not
a sequence
converge.
approximate
sequenceof
of
Theorem
Calculus
to
diverges
infinity.
for solving
method
a differential equation where
the
solution
curve is approximated by a
Fundamental
= n
u
line
straight
segments, the
the line
shorter
The
more accurate
the
approximation.
segments.
The result relating differentiation
and integration: \342\200\224j/(t
)dt
=
\342\200\224
f(x).
dx}0
sin
= sin x
tat
a
general solution
A most
form of a function
general
a differential
satisfies
which
The series /
Definite
integrals
= Ae2x.
y
series
or both
is a
e~*dx
infinite.
diverges,
tend to
although its terms
where one
integrals
are
limits
2yis
harmonic
The
-.
k=i k
improper
=
of the equation
solution
general
dx
equation.
harmonic series
The
dy
-jL
0.
convergent definite
0
integral.
Integral
convergence of a series
A test for
Test
on
based
it to an
comparing
improper integral.
We can tell that
factor
A
because
converges
in solving
used
function
first
differential
order
a
equation:
I \342\200\224dx
interval of convergence
isocline
The
of values
interval
of x for which
a powerseriesconverges.
on which all
A curve
the same gradient.
l'Hdpital's
rule
The
that,
stating
both tend to
=
thenlim^
x^a
Lagrange
form
of the
An
error term
The series 1 + x
+ x + x2
of convergence
Isoclines are used to
field of a differential
sinx
_
= hm
cosx
*^\302\260
x
equation.
+ x3
+
... has
] -1,1[.
a slope
sketch
equation.
=cos0 = l
1
lim^.
x^a
When
for estimating
expression
or Taylor
the
integrate
g'(x)
g(x)
the error when
approximated
integrating
_
of functions
lim
x^O
if two functions /and g
oo
either
0 or
as x \342\200\224\302\273
a,
limits
about
result
factor
interval
points have
converges.
x
the equation by the
allows us to use the
Multiplying
product rule to
= JP(x)dx
l(x)
1
\342\200\224
n2
i
integrating
^
V
series
the
a function
by a
is
Maclaurin series
approximations.
sin(0.2)
is approximated
0.23
\342\200\224'-\342\200\224
by 0.2
sin(c)
the
A \342\200\236
\342\200\224^X0.24for
4!
error
is equal to
some ce
,
0,0.2.
Glossary
l^\\
+
0,.
Term
Definition
Limit ComparisonTest
A
Example
of series
for convergence
test
based on lookingat the limit of the
ratios of the terms of the two series.
As
17(^1)
^1>andi^
also
converges, then >
converges.
differential
linear
A
equation
differential
equation
dy
\342\200\224
both
and
terms.dx
lower sum
Ay
\342\200\224
=
appear as linear
y only
drawn under
of rectangles
sum
The
in which
the graph in order to approximate
the value of an improper integral.
Maclaurin series
A
a given
Theorem
Value
Mean
The
result stating
Maclaurin
A
terms of Maclaurinseriesup
polynomial
nth
The sum
sum
partial
a series.A series converges
the sequence of its partial
if
sums
solution
One possible function
which
y
-
5e2x
and
e]0,2[
degree Maclaurin
are
*
-\342\200\224,etc.
6
2
satisfies a differential
is x
sin x.
-_,S
S =1,S
and/(3) = 9
there
i
converges.
particular
= 4
f(2)
l
The partial sums of V
2^ JJ
of
n terms
first
since
for
polynomial
of the
5!
the 3rd
-is
3!
to
n.
power
...
1
X3
x
\342\200\224.
rk2
we can concludethat
for which/'fx) = 5.
by taking the
formed
polynomial
1
V
X5
X3
\342\200\224
x
For/(x) = x\\
function we can always
find
a
tangent parallel to a given chord.
nth degree
\342\200\224dx
is
W
3!
a
for
that
\302\260\302\260
7 1
sum for
The lower
sin x
function.
and differentiable
continuous
equation.
in order
constructed
series
power
to approximate
not a linear differential
y2 is
dx
y
=
\342\200\224
e2x are
two examples
equation.
of particular solutions of
the
equation
dx
power series
A series
A series
radiusof convergence
only
2 + -x-Ax3
of x.
powers
p-series
containing
natural
2
of the
The largest
form
2a 77
k=i kp
value of | x | for
Thep-series
a
The series1 +x +
Test
A
test
for convergence
basedon lookingat
the
of a series
ratios of its
The result
Theorem
stating
that
if a
continuous function takes equal
at two points,
then there
must
be a turning
point between
them. This is a special case of the
values
Mean Value
of variables
separation
for solving a differential
on integrating the
equation
with
to x and y
equation
respect
A method
separately.
c
XJ^\\
+ 01.
= x2,
concludethat
since/(l)
there
is
which/'Oc) = 0.
find
the
of a power
=/(-l) we can
x e] -1,1[ for
Theorem.
based
Glossary
Test can beusedto
of convergence
\302\245orf(x)
radius of
is 1.
The Ratio
radius
+ x3+...
1, so the
series.
terms.
Rolle's
x + x2
converges for | x \\ <
convergence
Ratio
> 1 and
diverges otherwise.
series converges.
power
whenp
converges
\342\200\242
which
+...
x2
dv
\342\200\224
= \342\200\224
dx
variables.
can
y3
be solved
by separation
of
Term
Definition
slope field
A
Example
various points
to represent
of tangents
at
in
order
plane
plot
in the
solutions of a first
can be
curves
Solution
following the direction
sketched by
of the
tangents.
differential
order
equation.
Squeeze
The result about limit of sequences
stating that, if two sequences
to the same limit,
and
converge
Theorem
sequence can
third
between
then
them,
<
Since\342\200\224
a
be 'squeezed'
the third
sequence convergesto
the
<
and
\342\200\224,
\342\200\224
and
n
n
n
n
both converge to
converges to 0.
0, it
\342\200\224
n
that
follows
n
same
limit.
A result showing how to
Taylor approximations
This
approximate a function
by a power
series and how to estimate the error
in such
truncation error
result
to Maclaurin
applies
approximation.
The differencebetweena partial
sum and the full sum of a series.
We
know
that
\342\200\224
= 2.
\\
l
\342\200\224 \342\200\224
= 0.875,
H
1-
4 8
error is 0.125.
2
The
the
of rectangles
drawn above
in order
to
approximate
graph
variables
differential
of an
Differential
separable
solved by the
equations
that
can
be
method of separation
ofvariables.
The
upper
sum for
\302\260\302\260
-1
is V\342\200\224.
[\342\200\224dx
\\
x2
2k2
integral.
improper
equations
the truncation
\302\260\302\260
-1
sum
the value
If we
^
the series by
approximate
upper sum
series
as a specialcase.
dv
x2
dx
y3
\342\200\224
= \342\200\224
can
be solved
by separation
of
variables.
Glossary
l^\\
+ 0..
Index
absolute
error term
79-81
convergence,
94, 155, 174
definition,
algebra
16-17
applied to functions,
to sequences, 5-7
Euler's
94, 155, 174
definition,
functions
Test, 76-79
Series
Alternating
definition, 94, 155,174
approximation
approximations
to Maclaurin
approximations
to solutions
of differential
equations,
general
141-43
harmonic
Test
Limit
Test, 72-73
infinite
50-55
definition,
29-32, 42
definition,
123
136-45
solutions,
approximations
127-28
condition,
124,
boundary
definition,
146, 157, 175
124,
linear,
132-36
mixed
exam practice,
separation
up,
setting
summary,
definition,
75, 84
integrals,
error term, 105-7
175
157,
175
153,
exercises, 23-24
comparison
test
for,
53, 62
Limit
65, 66,69
XJ^\\
of the
rule, 19-23
definition,
93, 155, 175
limits
lex
c
form
Lagrange
definition,
tests for, 69-79, 82-83
definition,
41, 152, 175
diverges,
Ind.
exercises,144
l'Hopitals
84-88
exercises,
138
definition, 146, 157,175
147-48
146
divergent series,
149-52
revisited,
isocline,
123-26
improper
divergent
introductory problem, 1-2
127
of variables, 126-29
DivergenceTest,69,
156, 175
95,
definition,
124
solution,
particular
89, 91
of convergence,
interval
129-32
homogenous,
133-34
146, 157, 175
exercises,135,143
to
solution,
69-88
93, 155, 175
factor,
integrating
exercises,33-34
general
96-98
tests for convergence/divergence,
Test, 73-75
Integral
175
definition,
functions,
equations,
practice,
93-95
summary,
converges (to a limit),
differential
66-69
power series, 88-93
69-84
differentiable
63-64
65
mixed exam
exercises,84-88
for,
series,
convergence,
convergent series
tests
practice,
summary, 62
integral,
improper
exam
mixed
24-27
exercises,27-29
convergent
174
53-54,
Theorem of Calculus,46-50
Fundamental
42, 174
definition,
62
175
154,
definition,
79-81
functions,
146
convergence and divergenceof, 50-56
94, 155, 174
definition,
continuous
ComparisonTest,
53-55
convergence,
129-31,
approximation of, 57-61
155, 175
Comparison
equations,
45-46,
integrals,
improper
for a series, 69-72
conditional
175
homogenous differential
exercises, 131-32
174
for improper integrals,
76
series,
definition,
141
124, 127-28,
condition,
93,
175
exercises,131-32,135-36
137-39
definitions,
127
124,
solution,
definition,
polynomial,
Comparison
154, 175
62,
140-41
isocline, 138
definition,
Theorem of Calculus,46-50
Fundamental
definition,
method,
boundary
of, 11-19
limit
exercises, 143-45
Taylor
29-34
differentiable,
136-37
slopefield,
24-29
continuous,
integrals, 57-61
series, 104-9
of improper
Euler's
146, 157, 175
exercises,143-45
76-79
series,
140-41
method,
definition,
applied
alternating
157, 175
105-7,
Lagrange,
Maclaurin series, 115,120,156
of limits
+ d.
Comparison
Test, 72-73
definition, 93, 155,176
of sequences
and functions, 3-4
continuous
differentiable
functions,
24-29
functions, 29-34
power, 88-91
11-19
limit
of a
function,
limit
of a
sequence, 4-8
p-series,76
Mean Value
Theorem,
mixed exam
practice, 43-44
Rolle's
Squeeze Theorem,8-11
8-11
157, 176
146,
133-34
141-43,
146
Taylor series, 113-16
58-60
117-18
of,
definition,
applications
exercises,61
for approximating
141-43
solutions to
exercises, 116-17,
119
Maclaurin
99-102
series,
110-12
functions,
definition,
Value
exam
infinite
Ratio
Test,
Fundamental Theoremof Calculus,
Value Theorem, 36-40
Theorem,
Squeeze
particular
58-60
definition,
exercises,61
154, 176
124
solution,
variables
176
definition,
177
sum, improper integrals,
62, 154, 177
upper
66-67, 78
93,
definition,
8-11
error
truncation
104-6
polynomial,
156, 176
120,
definition,
nth partial sum,
46-50
Rolle'sTheorem,34-36
definition,
Maclaurin
82-83
Mean
96-98
limits of sequencesand functions,
43-44
121-22
Maclaurin and Taylor
series,
mixed exam practice, 63-64
degree
73-75
theorems
equations, 147-48
series,
Test, 76-79
Limit Comparison Test, 72-73
158-60
practice,
differential
Test,
Integral
definition, 42, 154,176
exercises,39-40
mixed
equations,
Comparison Test,53-55,69-72
Test, 69, 75, 84
Divergence
36-39
Theorem,
Series
Alternating
120, 156, 176
112-13
exercises,102-3,109-10,
Mean
differential
tests
to, 104-9
approximations
of composite
nth
120, 156-57, 177
polynomial,
Taylor
improper integrals,
62, 154, 176
115
approximations,
definition,
factor,
integrating
lower sum,
152,
15-16
Taylor
exercises, 135-36
177
definitional,
for functions,
equations, 132-33
differential
definition,
equations, 137-39
157, 177
146,
definition,
41-42
summary,
field, differential
slope
34-36
Theorem,
Squeeze Theorem,
linear
36-40
differential
separable
definition,
146,
equations,
126, 130
157, 177
135
exercises, 125, 128,131-32,
polynomials
roots
finding
nth degree
of,
differential
Taylor,
Theorem
Rolle's
for,
35-36
Maclaurin, 104-6
141-43
equations,
power series, 88-91
156, 176
94-95,
definition,
exercises, 92-93
89, 91
89, 90, 91
of convergence,
interval
of convergence,
radius
p-series,76,
91,
150, 152
definition, 176
89, 91
of convergence,
radius
95, 176
definition,
Ratio Test, 82-83
and
of power
convergence
series, 89, 90
156, 176
94,
definition,
Rolle's Theorem,34-36
153, 176
42,
definition,
separation of variables,
definition,
sequence,limit
126-29
146, 157, 176
of, 4-8
series
alternating, 76-79
76-78
harmonic,
Maclaurin,
99-113
Index
r
^
+a.
...
\342\226\240
I.
#
Acknowledgements
and
authors
The
are grateful
publishers
for the permissionsgranted
to
reproduce
materials
the original or adaptedform.While
effort
has been made, it has not always
every
to
the
sourcesof
all
the
materials
u
sed,orto trace all copyright holders.
possible
identify
If any omissions arebrought
to our
notice, we will be happy to include the appropriate
in either
_^\"
^
been
on
acknowledgements
IB
exam
~
L
reprinting.
\302\251
International
questions
f
permissionto reproduce
International
Baccalaureate
Baccalaureate
Organization.
Organization
We gratefully
intellectual
*
acknowledge
property.
Cover image: Thinkstock
Diagrams
in
book
the
were
created
by Ben Woolley.
>
\342\20
180
Acknowledgements
w
C
o*
+ *.._*
i.
Os
*>