MST121 Chapter B1 Modelling With Sequences

MST121
Using Mathematics
Chapter B1
Modelling with sequences
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3.2
Contents
Introduction to Block B 4
Study guide 5
Introduction 6
1 Modelling car ownership 7
1.1 Constructing a model 8
1.2 Mathematical interlude: sums 11
1.3 Back to the model 14
2 Populations: exponential model 18
2.1 Purpose of general population models 18
2.2 The exponential model 18
2.3 Evaluating the exponential model 23
3 Populations: logistic model 28
3.1 Setting up the logistic model 28
3.2 Finding values for the parameters 32
3.3 Investigating the logistic model 35
4 Logistic recurrence sequences on the computer 39
5 Sequences and limits 42
5.1 Sequences from recurrence systems 42
5.2 Sequences from closed-form formulas 44
5.3 Sequences from sums 48
5.4 Some further sequences and series 50
Summary of Chapter B1 53
Learning outcomes 53
Solutions to Activities 55
Solutions to Exercises 58
Index 61
3
Introduction to Block B
Block B develops the theme of mathematical modelling, which was

introduced in Block A. As in Chapter A1, the first two chapters of
Block B are concerned with discrete models, that is, models in which the
independent variable takes separate or discrete values (as opposed to
values which range continuously over an interval of real numbers).
For example, Chapters B1 and B2 are both concerned with models for
populations, in which estimates for the population size are sought at
regular intervals (typically once a year).
Chapter B1 looks at two successive population models, of which the second
is a development of the first. This is in line with a general principle of
modelling, that one should start with a simple representation of the real
situation and then, if necessary, develop the model further. Chapter B1
permits you to revise to some extent what you learnt in Chapter A1 about
solving first-order linear recurrence systems. It also examines what is
meant by the convergence of a sequence to a limit, and introduces the
topic of series (sums of the terms of a sequence).
Chapter B2 develops the initial population model of Chapter B1 in a
different direction, by showing how to reflect the internal structure of a
population. For example, members of a human population can be
categorised as either juveniles or adults, and the variation in the sizes of
these two subpopulations is a matter of interest to planners. This is
modelled by using two dependent variables, one for each of the
subpopulation sizes, rather than the single one used in Chapter B1.
Each subpopulation size in a given year will depend significantly on both of
the subpopulation sizes in the previous year, leading to a more complex
type of recurrence system. Correspondingly, new mathematical entities are
required to describe the model and to obtain predictions from it. The pair
of subpopulation sizes is represented by a vector, and the array of
coefficients for the recurrence relation makes up a matrix. There is a
natural multiplication operation on matrices which is useful in this context
and also in many others. In terms of matrices, a closed-form solution of
the recurrence system can be found to describe the sequence of population
vectors.
Chapter B3 takes a step aside from the theme of discrete models, to
pursue further the topic of vectors. The algebraic approach to them
in Chapter B2 can be related to a geometric view, in which vectors are
seen as quantities which have both magnitude and direction. Thus wind
velocity is one example of a vector quantity. The ways in which vectors
can be combined make them a useful tool in many modelling situations.
Often this involves an application of trigonometry, which is developed and
practised further in the chapter. One important use for vectors is as a
model for the forces which cause objects to move or, when in balance, keep
objects from moving. The chapter and the block conclude with an
introduction to the subject of statics, which concerns the relationships
between forces which act on objects at rest.
4
Study guide
There are five sections in this chapter. They are

intended to be studied consecutively in five study
sessions. Each session requires two to three hours.
The pattern of study for each session might be as
follows.
Study session 1: Section 1.
Section 4 requires the use of the computer together
with Computer Book B.
Section 5 is independent of Section 4, so it can be
studied earlier if you wish. Subsection 5.4 will not
be assessed. Subsections 2.3 and 3.3 are just for
reading: they contain no activities.
The optional Video Band B(i)

Algebra workout – Sigma notation could be
viewed at any stage during your study of this
chapter.
5
Introduction
You saw in Chapter A1 that sequences can be used in a variety of

contexts. Here, we look in some depth at models based on sequences in
two particular situations. The first, discussed in Section 1, relates to
personal economics: ‘How long should you keep a car before selling it?’
The second, discussed in Sections 2–4, concerns the modelling of the
number of individuals in some animal population, which is relevant to
conservation, for example. Your work on these specific situations is
While these are important intended to develop skills that you will need when creating and using
skills, you are asked for the models of your own. These skills include choosing variables, setting up a
most part to appreciate their mathematical model, investigating the implications of the model, and
application rather than to interpreting the results of your calculations in terms of the original
master them at this stage. question that the model was intended to address.
In creating sequence models, it often proves natural to specify the sequence
through a recurrence system. For some sequences given by recurrence
systems, we can also describe the sequence by a closed form. Where such a
formula can be found, it may be possible to use it to answer the original
question. But even where such a formula cannot be found, the computer
enables sequences given by recurrence systems to be calculated. In the
model concerning car ownership, and in the first population model, we are
able to find closed forms. This enables us to work to a greater extent
algebraically (rather than numerically). In the second model for
populations, there is no known closed form. In that case, you are invited
(in Section 4) to use the computer to evaluate the sequences generated by
the recurrence system in the model, and are asked to take note of the
strategy used in carrying out such an investigation. You will see also that,
even in the absence of a closed form, it is sometimes possible to use
reasoning in order to cut down the scope of a numerical investigation.
In Sections 2–4 we pay particular attention to what happens to population
sequences in the long term. In Section 5 we step aside from specific
contexts, to look from a mathematical viewpoint at the question of how
sequences behave. In particular, we look at sequences whose values ‘settle
down’ close to a particular number in the long term. For example, the
values of the sequence given by
an = 2 + 10−n (n = 1, 2, 3, . . .)
are
2.1, 2.01, 2.001, 2.0001, . . . ,
and these are ‘settling down’ close to 2. Here, the number 2 is referred to
The concept of a limit is as the limit of the sequence.
important in explaining the
ideas of calculus, which you
will meet in Block C.
6
1 Modelling car owner ship
Being able to make use of mathematics to solve problems is an important

skill for a mathematician to have. This skill is not an easy one to acquire,
partly because there are no set procedures to follow. Even after you have
studied this section, you are not expected to be able to create a model for You will find that, in this
yourself for some new problem that is unrelated to any that you have seen course, your mathematical
before. The aim here is to focus on the processes involved when you are in modelling skills are assessed
such a situation. Only a little new mathematics is introduced in this in only minor ways. You
section, because the model constructed provides an application of the types should not, therefore, worry
about the modelling process
of sequences that you studied in Chapter A1. Apart from appreciating the
to the detriment of your study
process of modelling, therefore, you will be able to use this section to some
of the mathematics involved.
extent as revision of the work that you did in Chapter A1. You will also
see how to use a new notation to describe the sum of terms of a sequence.
In creating mathematical models, there are a number of non-mathematical
factors that need to be considered. For example, what features of the real
situation should you attempt to incorporate into the model? Exactly what
problem should you attempt to address through your model? Having
created the model, is it adequate for its intended purpose? In this section
you will see consideration of some of these wider issues in the context of a
model concerning car ownership.
In Chapter A1 you met a five-stage diagram of the modelling process (see Chapter A1, Section 7
Figure 1.1). This diagram is a reminder of the main processes involved in
modelling, and their logical order. However, it would be misleading to
suggest that modelling consists of five clear-cut stages, performed in a
fixed order. For example, when you start to create a model, you may find
that you have set an unrealistic target in defining its purpose. You may
then need to go back to the earlier stage, and refine your definition of
‘the purpose’.
Specify Create
purpose model
Do
mathematics
Evaluate Interpret
results
Figure 1.1 The modelling cycle
When you evaluate a model, you need to decide whether or not it is

adequate for its purpose. If it is not, then you need to decide why this is
the case. In doing so, you may examine the assumptions that you made in
creating your model. Does the process of evaluation suggest ways in which
you might modify these assumptions? If so, creation of a revised model,
and repetition of Stages 2–4 in the modelling cycle, may lead to a more
satisfactory outcome. In practice, several revisions may be needed. In the
current section, however, we shall forego any revision to the model first
presented.
7
CHAPTER B1 MODELLING WITH SEQUENCES
1.1 Constr ucting a model

Many car owners have wondered whether it is better to buy a car new or
used, and for how long to keep it. In this subsection, we construct an
initial model to investigate some aspects of this question. This is probably
not ‘the best model that can be thought up’, and the description given
here will, in the interests of clarity and cohesion, omit the false starts and
preliminary thoughts which occurred along the way. As the model is
developed, you may well think of alternative approaches!
Specify the purpose

There are many factors that people might consider important in deciding
which type of car to buy. Among these are passenger and luggage capacity,
fuel consumption, emission levels and engine size. To simplify matters, let
us suppose that the choice of car make and model has already been made.
Activity 1.1 Listing features
Do not spend more than a Assume that you have decided which type (make and model) of car you
minute or two on this activity. wish to buy, but not its age nor how long you will keep it. Note down
some features that you might take into account in deciding your strategy
for car purchase.
A solution is given on page 55.
Specify Before creating a mathematical model, the purpose for which it is intended
purpose must be clarified. It is sensible here to concentrate on features that lend
themselves to mathematical description. This suggests that we should
stick to aspects of the situation that can be measured; hence aspects such
as the colour and condition of the vehicle should be discounted for
modelling purposes.
Items which are measurable include the following:
age and mileage at which car is bought;
age and mileage at which car is sold;
cost of the car at purchase, and its resale value;
running costs, such as repairs and servicing, fuel, insurance and tax.
Although this list of features is restricted to measurable quantities, it is
still too complex to cope with as it stands. We need to pick out the item
that is considered to be of most importance, and then to identify what
other quantity or quantities it mainly depends upon. An item of
considerable interest to many is the cost of motoring, so it seems natural
here to concentrate upon that. Thus we aim to construct a model which
will show how to keep the cost of motoring as low as possible, in a sense yet
to be made precise.
How can this be achieved? If cost is to be the output, or dependent
variable, within the model, what inputs or independent variables does it
depend upon? If the values of the inputs can be altered as we please, then
it may be possible to select them in order to achieve some ‘lowest cost’ for
the motorist.
8
SECTION 1 MODELLING CAR OWNERSHIP
For a first model, it is appropriate to simplify as far as possible. Let us

therefore leave out of consideration the mileage of the car, and concentrate This is equivalent to assuming
instead on its age as the factor which determines its value at any time. that the car always registers a
Various ‘lowest cost’ questions can be posed, for example the following. ‘normal mileage’ for its age.
If the car is bought at three years old, at what age should it be sold?
If it is to be sold at ten years old, at what age should it be bought?
If it is to be kept for just two years, at what age should it be bought?
We shall fix on the following question.
Suppose that a car of given type is purchased when it is one year old.
At what age should it be sold in order to minimise the cost of owning
and running the car?
While this statement of purpose is more precise than ‘keep the cost of
motoring as low as possible’, we have not yet fully tied down what the
‘cost of owning and running the car’ involves. Each year, the car will
depreciate in value, which is to say that what you can expect to receive for
it at the end of the year will be less than what it was worth at the
beginning of that year. This decline in value is an annual ‘cost’ which may
vary from year to year. The running costs for the vehicle will also recur
annually but on a varying basis. The annual ownership and running cost is
the sum of these two items. We seek to minimise this cost as averaged over
the length of time between purchase and resale of the car.
This average cost is a measure which is more appropriate to the consumer
than the total cost of owning and running the vehicle. For example, a car
which is kept for 5 years and costs £10 000 overall works out at an average
of £2000 per year. Another car which is kept for 10 years and costs
£15 000 in all (more in total than the first car) averages only £1500 per
year, which is more economical than the first car.
We are now able to state the purpose of the model precisely, as follows.
Purpose of model
A car of given type is purchased when it is one year old. At what age
should it be sold in order to minimise the average annual cost of
owning and running the car?
Create the model Create

model
Now that the purpose of the model has been tied down, we define
appropriate variables to correspond to the major features identified above.
i (in years): age of the car (i = 0 when the car is new).
n (in years): length of time for which the car is owned (purchased at Note that i and n are
one year old, resold at age n + 1 years). integers.
an (in £): average annual cost of owning and running the car for n
years.
vi (in £): value of the car at age i years, where i = 0, 1, 2, . . . , n + 1 As you will see, v0 and c0 are
(what it costs to buy the car, or what its resale yields). relevant in the construction of
ci (in £): annual cost of servicing and repairs, from the beginning of
the model.
year i to the beginning of year i + 1, where i = 0, 1, 2, . . . , n.
9
In the above list, n is the independent variable (the input which can be
varied at will) and an is the dependent variable (the output, which is to be
minimised). The other variables, i, vi and ci , are needed as intermediate
quantities, permitting us eventually to express the output, an , in terms of
the input, n.
Note that ci includes only the running costs of servicing and repairs, and
not items such as fuel, insurance and tax. For a given type of car with
normal mileage, these latter items are assumed to be the same each year,
so they will not affect the purpose stated for our model.
At this point we need to make assumptions about how both the value of
the car and its servicing and repair costs vary with age. The depreciation
Sometimes a combination of in the value of any article with time is usually dealt with in one of two
these two approaches is ways:
adopted.
by assuming that the article has a definite fixed life, and that its value
declines to zero linearly over that life;
by assuming that the value of the article declines by a fixed proportion
each year.
Where used cars are concerned, the second approach seems to be closer to
actual valuations than the first. This observation does not extend to the
initial year of a car’s life, because of the substantial additional premium
which is paid to acquire a brand-new vehicle. However, since we are
considering a car bought at one year old, depreciation by a fixed
proportion will fit the bill here. Hence we have
vi+1 = rvi (i = 0, 1, . . . , n),
Data are introduced here only where r is a constant such that 0 < r < 1. Data gathered for one particular
for illustrative purposes. type of car, for illustrative purposes within this model, give r = 0.85 and
They would not normally v0 = 16 000. This value of r is equivalent to a depreciation rate of 15% per
appear this early in the annum. The value of v0 can be regarded as the initial cost of the car minus
modelling cycle. the ‘brand-new’ premium referred to above (or as what you might expect
to receive if you bought a new car and then immediately tried to resell it).
Hence, for this type of car, vi is given by
v0 = 16 000, vi+1 = 0.85vi (i = 0, 1, . . . , n). (1.1)
Reliable data on the costs of servicing and repairs are less easy to come by.
It is generally acknowledged that these annual costs of running a car
This is consistent with the increase with the age of the vehicle, and the simplest approach consistent
principle of simplicity within with this observation is to assume that these costs increase linearly with
mathematical modelling, age; that is,
which states that, in the
absence of any other guide as ci+1 = ci + d (i = 0, 1, . . . , n − 1),
to how to model a where d is a positive constant. Some (admittedly rough and ready) data
relationship between two
for the type of car being considered here lead to the values c0 = 600 and
variables, you should choose
d = 250. Hence ci is given by
the simplest form of function
which takes into account the c0 = 600, ci+1 = ci + 250 (i = 0, 1, . . . , n − 1). (1.2)
known facts.
Equations (1.1) and (1.2) will permit vi and ci to be expressed in terms of
car age, i years. How can we then relate the dependent variable of the
model, an , to n, vi and ci ?
The net loss involved in buying the car when one year old and selling it
n years later is v1 − vn+1 . Also, the total cost of running the car between
purchase and sale is
c1 + c2 + · · · + cn .
10
The total cost of owning and running the car for n years is therefore
v1 − vn+1 + c1 + c2 + · · · + cn ,
and the corresponding average annual cost over these years is
1
an = (v1 − vn+1 + c1 + c2 + · · · + cn ) (n = 1, 2, . . .). (1.3)
n
The mathematical model is encapsulated in equations (1.1)–(1.3), together
with the list of definitions of the variables which appear in these equations.
Before moving on to solve the equations, it is timely to list some of the
assumptions which have been made, explicitly or implicitly, in the course
of constructing the model.
The value of the car at a given age is the same for both purchase and
resale (there is no ‘dealer margin’ involved).
The value of the car diminishes by a fixed proportion each year.
The annual servicing and repair costs increase linearly with age.
The car is sold after a whole number of years.
Both car value and annual running costs depend only on the age of the
car (and not, for example, on its mileage).
Any effects of inflation are to be ignored, as are the annual costs of
fuel, insurance and tax.
In terms of the modelling cycle, it is time now to ‘do the mathematics’, in
order to solve equations (1.1)–(1.3). Before embarking on this task, there
are some mathematical preliminaries to consider.
1.2 Mathematical interlude: sums

Towards the end of the last subsection, the expression
c1 + c2 + · · · + cn
appeared, to denote the sum of the first n terms of a sequence
ci (i = 1, 2, . . .). In general, a sum of consecutive terms of a sequence of
numbers is called a series. Since series occur frequently in mathematics,
and it can become unwieldy to use the ‘+ · · · +’ device when several sums
are involved, an alternative and more compact notation exists to describe
them. This is referred to as ‘sum notation’ or as ‘sigma notation’, since it
makes use of the Greek upper-case letter Σ (sigma). Using this notation,
the sum above can be expressed as
n
� The right-hand side here is
c1 + c2 + · · · + cn = ci , read as ‘the sum from i equals
i=1
1 to n of ci ’.
in which�1 and n are called the limits of the sum. (Some books print the
sum as ni=1 ci when it appears in a line of text.)
Sometimes it is convenient to start a sum from a value of i other than 1, as
you will see in the following example.
11
Example 1.1 Using sigma notation
Express each of the following sums using sigma notation:

(a) the sum of the first n positive integers;
(b) the sum of the cubes of the integers between 3 and 9, inclusive;
Chapter A1, Subsection 4.2 (c) the result which you saw in Chapter A1 for the sum of a finite
geometric series, that is, � �
2 n 1 − rn+1
a + ar + ar + · · · + ar = a (r
= 1).
1−r
Solution
n
�
(a) The sum is 1 + 2 + · · · + n = i.
i=1
9
�
(b) The sum is 33 + 43 + · · · + 93 = i3 .
i=3
1
Note that ar = ar and that (c) Using sigma notation on the left-hand side, we have
ar0 = a × 1 = a. Also, the n
� � �
convention that 00 = 1 is 1 − rn+1

ari = a
1).
(r =
adopted so that the case i=0
1−r
r = 0 is covered.
Comment
In these sums i could be replaced by j, for example. The sum in part (a)
n
�
would then be j. Also, each of these sums could start at a different
j=1
value if the term being summed is adjusted accordingly. For example, the
6
�
sum in part (b) could be written as (i + 3)3 .
i=0
Here are some sums for you to manipulate.
Activity 1.2 Using sigma notation
(a) Use sigma notation to express the sum of the squares of the integers
between 5 and 13, inclusive.
(b) By first writing out the two sums involved, express
6
� 3
�
i
5 − 5i
i=1 i=1
as a single sum (with one sigma).

4
� 4
�
(c) Show that (600 + 250i) = 600 × 4 + 250 i.
i=1 i=1
Solutions are given on page 55.
The outcome of Activity 1.2(c) is a special instance of a result that is often

useful in simplifying the expression of sums. If xi (i = 1, 2, . . .) is a
sequence and a, b are constants, then we have
n
� n
�
(a + bxi ) = an + b xi . (1.4)
i=1 i=1
12
More generally, if the sum starts at m (where m ≤ n), then

n
� n
�
(a + bxi ) = a(n − m + 1) + b xi . (1.5)
i=m i=m
In Example 1.1(c) you were reminded of the formula for the sum of a finite
geometric series. There is also a formula for the sum of the finite
arithmetic series 1 + 2 + · · · + n, that is, for the sum of the first n positive
integers. This formula is derived by writing down the sum in the normal
order and then with this order reversed:
n
�
i = 1 + 2 + 3 + · · · + (n − 1) + n,
i=1
�n
i = n + (n − 1) + (n − 2) + · · · + 2 + 1.
i=1
On adding these two equations, with the addition of corresponding terms

on the right-hand side, we obtain
n
�
2 i = (n + 1) + (n + 1) + (n + 1) + · · · + (n + 1) + (n + 1)
i=1
� ��
n such terms in total
= n(n + 1),
so the required sum is
n
�
i = 12
n(n + 1). (1.6)
i=1
This result will be useful in the next subsection. Note the geometrical
interpretation of it, which is illustrated (for the case n = 5) in Figure 1.2.
The total shaded area is equal to the sum of the shaded column areas,
5
�
which is 1 + 2 + 3 + 4 + 5 = i, and also to half of the area of the whole
i=1
rectangle, which is 2
× 5 × 6. Figure 1.2 Shaded area
equals half of area of
rectangle
Activity 1.3 Summing positive integer s
Find the sum of the first 100 positive integers.

Intellectual precocity
It is said that the great German mathematician Carl Friedrich Gauss
(1777–1855), at the age of ten, was asked along with the rest of his
school class to carry out the sum which you did in Activity 1.3, having
been shown no way of doing it beyond the addition of successive
numbers. Their teacher intended it to be a lengthy task, but Gauss
came up with the correct answer almost immediately, by applying the
approach used above to derive equation (1.6).
‘Gauss had not been shown the trick for doing such problems rapidly.
It is very ordinary once it is known, but for a boy of ten to find it
instantaneously by himself is not so ordinary. This opened the door
through which Gauss passed on to immortality.’
(E. T. Bell, Men of Mathematics, Volume 1 (Pelican, 1953))
13
1.3 Back to the model

The mathematical model developed in Subsection 1.1 was described by
equations (1.1)–(1.3). On using sigma notation to rewrite the sum in
equation (1.3), we obtain the equations
v0 = 16 000, vi+1 = 0.85vi (i = 0, 1, . . . , n), (1.1)
c0 = 600, ci+1 = ci + 250 (i = 0, 1, . . . , n − 1), (1.2)
� n
�
1 �
an = v1 − vn+1 + ci (n = 1, 2, . . .). (1.7)
n i=1
Do Do the mathematics
mathematics
Equations (1.1) define a geometric sequence, and equations (1.2) define an
arithmetic sequence. Once a closed form has been found for each of these,
The closed forms for an the resulting expressions for vi and ci can be used on the right-hand side of
arithmetic sequence and for equation (1.7). With an then expressed in terms of n, we can seek to
a geometric sequence were minimise the value of an with respect to changes in n.
given in Chapter A1,
Subsections 2.2 and 3.2,
respectively. It is common Activity 1.4 Finding the closed-form solutions
practice to refer to such
formulas as ‘closed-form (a) Find the closed-form solution of the recurrence system (1.1).
solutions’ of the underlying
recurrence system.
(b) Find the closed-form solution of the recurrence system (1.2).
(c) Use equations (1.4) and (1.6) to simplify as far as possible the
n
�
expression for ci that follows from your answer to part (b).
i=1
Comment
Any finite arithmetic series can be summed in a manner similar to that
seen in the solution to part (c). Alternatively, such a sum can be found
directly by application of the approach used to derive equation (1.6).
From the result of Activity 1.4(a), we have

v1 − vn+1 = 16 000(0.85) − 16 000(0.85)n+1
= 16 000 × 0.85(1 − (0.85)n )
= 13 600(1 − (0.85)n ).
From the result of Activity 1.4(c), we have

n
�
ci = 25n(5n + 29).
i=1
On substituting these formulas into equation (1.7), we obtain

1� �
an = 13 600(1 − (0.85)n ) + 25n(5n + 29)
n
13 600
= (1 − (0.85)n ) + 25(5n + 29) (n = 1, 2, . . .).
n
With an now expressed in terms of n, we seek the smallest value of an .
This can be found by tabulating values of the sequence an , as shown
below. (The values are given correct to two decimal places.)
n 1 2 3 4 5 6 7 8 9 10
an 2890 2862 2849.3 2850.18 2863.12 2886.79 2920.02 2961.77 3011.11 3067.25
It appears that the smallest value of an is 2849.3, which occurs when n = 3.

14
Interpret the results Interpret

results
The prediction of the model is that, having acquired a one-year-old vehicle
of a certain type, the lowest average annual cost of owning and running it
thereafter is obtained by selling it after a further three years, for which the
average annual cost will be £2849.30 (not including fuel, insurance and
tax).
In order to interpret the results obtained from a mathematical model, it is
often helpful to draw a graph. Figure 1.3 shows graphically how an varies
with changes in n.
Figure 1.3 The sequence an
This graph indicates once more that the minimum value of an occurs for
n = 3, but it also shows that this minimum is barely below the value for
n = 4. This observation is substantiated by another look at the table. The
average annual cost for resale of the car after four years is less than £1
more than the minimum cost identified, and this difference is not enough
to decide whether to sell the car after three years or after four.
More generally, the variation of an around n = 3 is not huge. Perhaps you
would not object to selling the car after two years or after five, for the
average costs indicated. On the other hand, the increased costs involved in
selling after one year or beyond six years give pause for thought.
The model tells you that, on the basis of the various assumptions made
here, resale after anything between three and four years is the most
economical approach available, at an estimated average cost of about
£2850 per annum plus fuel, insurance and tax.
Evaluate the outcomes

Evaluate
On the face of it, the outcome of this modelling activity looks satisfactory.
We set out to find a ‘minimum cost of motoring’, in specified
circumstances, and have indeed found a feasible strategy for the car owner
which achieves such a minimum. However, this has been done within the
context of an idealised and simplified ‘world view’, geared particularly to
bringing mathematics to bear on the situation. The hope is that this
approach has revealed some essential truth about the situation being
studied, but this cannot yet be taken for granted. Before accepting the
conclusions from our model, it would be wise to check that the course of
action proposed (resale after three to four years) makes sense in reality.
15
We do not have sufficient information to make that check with reality here,
but it may well turn out (it is frequently the case) that the initial model is
judged as only partly successful once the comparison with reality has taken
place. If so, it may be necessary to try to improve the model, by taking
another trip around the five stages of the modelling cycle. An early step
would be to re-examine the assumptions which were made in constructing
the first model, in order to see whether any need amendment in the light
of experience gained.
For the model developed here, for example, the assumption of linearly
increasing annual costs for servicing and repairs may turn out to be too
simple. The data underlying this assumption are few, and more data on
annual running costs might suggest a suitable refinement.
Generalising the model

The problem posed in the ‘Purpose of model’ box on page 9 has been
solved, to the extent possible with the information available. Note,
however, that once a model has been constructed with such a specific
purpose in mind, it may come in useful for other related purposes. In this
case, we set up a model for one particular type of car, but the structure of
the model would be unchanged if the same approach were applied to a
different type of car.
The remainder of this section Thus, in place of equations (1.1), we have a formula for car value given by
will not be assessed.
vi+1 = rvi (i = 0, 1, . . . , n), (1.8)
In fact, it would have been where r is a constant, and in place of equations (1.2), we have a formula
better modelling practice to for servicing and repair costs given by
proceed with these more
general formulas in the first ci+1 = ci + d (i = 0, 1, . . . , n − 1), (1.9)
instance, instead of where d is a constant. Also, equation (1.7) continues to apply:
introducing illustrative data � n
�
values as was done in 1 �
an = v1 − vn+1 + ci (n = 1, 2, . . .). (1.7)
Subsection 1.1. n i=1
Activity 1.5 Doing the mathematics for the general case
This activity requires a (a) Find the closed-form solution of the recurrence relation (1.8).
generalisation of what you did
(b) Find the closed-form solution of the recurrence relation (1.9).
in Activity 1.4 and the main
text following that activity. (c) Use equations (1.4) and (1.6) to simplify as far as possible the
n
�
expression for ci which follows from your answer to part (b).
i=1
(d) Starting from equation (1.7) and your answers to parts (a) and (c),
express an as simply as possible in terms of the variable n and the
parameters v0 , r, c0 and d.
16
Comment
In the final formula obtained in part (d), parameter values can be
substituted for v0 , r, c0 and d to correspond to any given make and model
of car. The resulting specific version of the formula for an can then be used
to predict when such a car should be resold, after purchase at one year old,
to most economical effect.
Further generalisations of the model are possible, for example, to take into
account purchases at age other than one year. However, we shall not
pursue such extensions here.
Summar y of Section 1
This section has introduced:
the use of sigma notation to describe concisely sums of terms from a
sequence;
the word ‘series’ to describe a sum of consecutive terms of a sequence;
the formula 12 n(n + 1) for the sum of the first n positive integers;
the application of general modelling ideas to a specific problem
involving the economics of car ownership.
You will not be assessed directly on the modelling content of this section,
though the principles are important if you want to continue to further
studies in applied mathematics. The framework for the modelling cycle,
within the five key stages shown in Figure 1.1, was described concisely in
Chapter A1, Section 7. To some extent, at least, all of the aspects included
there were addressed in the course of the modelling done in this section.
Exercises for Section 1

Exercise 1.1
(a) The sum
1
2
+ ( 12 )2 + · · · + ( 12 )10
10
�
can be expressed as ( 12 )i , and also in the form
i=1
n
� n
�
1 1 i
( )
2 2
= 1
2
( 12 )i .
i=m i=m
What are the values of m and n?

(b) What is the value of this sum?
Exercise 1.2
(a) Find the sum of the integers from 51 to 100, inclusive.
(b) Hence find the sum of the numbers 517, 527, 537, . . . , 1007.
(Hint : 517 = 7 + 10i, where i = 51.)
17
2 Populations: exponential model
2.1 Purpose of general population models

What will happen to the North Atlantic population of tuna if fishing
continues at its present levels? What strategies are effective in controlling
the spread of rabies in foxes? Will the blue whale escape extinction?
Can the world support 10 billion humans?
A mathematical model which aims to answer any of these questions about
animal populations would need to be based on detailed information about
the species concerned. However, to create a model to address any
particular question, one would draw on general models of the way an
animal population can change with time. The purpose of these models
might be stated thus.
Purpose of model
Specify Describe how an animal population may change with time, in a way
purpose that is applicable over a long period of time and for a variety of
populations.
We shall pay particular attention to the form of population variation that

is predicted for the long term. Will the population increase, perhaps more
and more rapidly? Or will it decrease, perhaps to extinction? Or will the
population stabilise, and if so at what level?
In Section 1 you saw a single model developed to answer a quite specific
question. In this and the next section, you will meet two successive models
set up in order to address the rather general purpose just stated. These
are ‘standard models’ used in the study of animal populations, and you are
not asked to dwell on the modelling processes which lead to their
construction, though you should be able to recognise that each of the
various modelling stages shown in Figure 1.1 is involved. The second
model, in Section 3, is a development of the first, in this section, and arises
from some perceived deficiencies of the first model.
2.2 The exponential model

For many animal populations, there is a pattern of variation within each
year. Population sizes are highest in the early autumn, after births during
spring and summer, and lowest at the end of winter, during which most
deaths occur. In studying long-term variations in population levels, the
main interest is in changes from one year to the next, rather than within a
For details of the distinction year. We therefore look here at discrete (rather than continuous) models of
between discrete and population change.
continuous models, see the
Summary of Block A at the
end of Chapter A3.
18
SECTION 2 POPULATIONS: EXPONENTIAL MODEL
For example, Figure 2.1 shows the size of a population of pheasants on

1 April and 1 October each year for 1937 to 1942. You can see that
numbers dropped during each winter, even though they increased from one
year to the next. We seek a model that predicts this year-on-year increase.
Figure 2.1 Population of ring-necked pheasants on Protection Island, USA, in

spring and autumn (1937–1942)
In setting up population models, a useful approach is to identify the

numbers of a species which leave or join the population each year. The
‘joiners’ include those who are born during the year, while the ‘leavers’
include those who die. Changes in population size may also be brought
about by geographical movements between separate populations of animals
of the same species (immigration and emigration). In this study, we
confine attention to populations where migration is not an important
factor. The models here assume that there is no migration, so we need to
focus on only the effects of births and deaths.
For human populations, a birth rate is usually given as a proportion of the
current population size (perhaps as a percentage, or as ‘births per
thousand’), and death rates are specified in a similar way. It seems natural
that for any animal population, the numbers of births and deaths will both
increase as the population size increases. As a first model of population
variation, it seems sensible to assume that each of the number of births
and the number of deaths is a fixed proportion of the current population
size. This model is illustrated in the following example.
Example 2.1 Modelling pheasants
Let Pn denote the population size of ring-necked pheasants on Protection

Island on 1 April, n years after 1937; thus P0 represents the population
size on 1 April 1937, which is 8. Assume that the number of births in each
subsequent year (from 1 April to 31 March) is 2.6Pn (that is, 260% of Pn ),
and the number of deaths in each subsequent year is 0.4Pn (40% of Pn ).
(a) Find a recurrence system that Pn must satisfy.
(b) State a closed form for Pn .
(c) What population sizes does this model predict for 1 April in each of
the years 1938, 1939, . . . , 1942 ?
(d) What form of population variation does this model predict in the long
term?
19
Solution
(a) During the (n + 1)th year, the ‘joiners’ will be the pheasants born
during the year, of which there are 2.6Pn , and the ‘leavers’ will be the
pheasants that die, of which there are 0.4Pn . The difference between
births and deaths gives the increase in the population size during the
year (since we are assuming that there is no migration), so
Pn+1 − Pn = 2.6Pn − 0.4Pn
= 2.2Pn .
Thus the sequence Pn is given by
In the population contexts P0 = 8, Pn+1 = 3.2Pn (n = 0, 1, 2, . . .).

considered from now on, the
subscript range will always be (b) This recurrence system describes a geometric sequence (as studied in
n = 0, 1, 2, . . . , as here, and it Chapter A1), with closed form
will usually be omitted in this Pn = 8(3.2)n (n = 0, 1, 2, . . .).
text.
(c) The predicted population sizes on 1 April are shown below.
(Each estimate is calculated using unrounded intermediate values,
then rounded to the nearest whole number.)
Year 1937 1938 1939 1940 1941 1942

Population size 8 26 82 262 839 2684
(d) This model predicts that the population size will increase, and will go
on increasing more and more rapidly (see Figure 2.2).
Figure 2.2 Graph of Pn = 8(3.2)n (n = 0, 1, . . . , 7)
Comment
This model predicts that the population size will increase very rapidly, and
this pheasant population did undergo rapid increase from 1937 to 1942
(as shown in Figure 2.1). However, the model predicts that this increase
will continue. For example, it predicts a population size on 1 April 1997 of
8(3.2)60 , which is about 1.6 × 1031 . Estimating the area of ground taken up
by a pheasant as 0.1 m2 , and using the fact that the island has area
1.6 × 106 m2 , this gives 1.6 × 1030 m2 of pheasants in an area of
1.6 × 106 m2 . This implies that the island will be covered about 1024 deep
in pheasants, which seems unlikely!
If the purpose of the model in this example is to predict the population
size in the long term, then it is certainly not reasonable. The population
cannot continue to grow indefinitely in the way predicted by the model.
20
We can generalise the model in Example 2.1 to other populations, by

keeping the assumptions that the birth and death rates are both constant,
but treating these values as parameters. We shall now investigate whether
such a generalised model predicts this same form of population growth in
the long term.
Suppose that P0 is the size of a particular population at the start of a
period of time for which the population is to be modelled, and that Pn
denotes this population size at n years after the starting time. Suppose
that the proportionate birth rate b and death rate c are both
(non-negative) constants, so that, during the year starting at time n, there
are bPn births and cPn deaths. Suppose, as before, that there is no
migration.
For the year starting at time n, the change in population size during the
year is Pn+1 − Pn , and this must equal ‘joiners’ minus ‘leavers’. So we have
Pn+1 − Pn = bPn − cPn ,
which is equivalent to
Pn+1 = (1 + b − c)Pn .
If 1 + b − c > 0 and P0 > 0,
then
This again describes a geometric sequence, for which the closed form is
Pn > 0 for n = 1, 2, 3, . . . .
Pn = (1 + b − c)n P0 .
If 1 + b − c = 0, then
Pn = 0 for n = 1, 2, 3, . . . .
The long-term behaviour of Pn will depend on the value of 1 + b − c.
(The possibility 1 + b − c < 0

cannot occur if P0 > 0
Activity 2.1 Predicted behaviour of exponential model because the number of
individuals who die during a
Describe the long-term behaviour of Pn = (1 + b − c)n P0 when year cannot exceed the
1 + b − c > 0. number who were either alive
at the start of the year or
A solution is given on page 56. born during the year.)
The three cases identified in Activity 2.1 are illustrated in Figure 2.3.
Figure 2.3 Plots of Pn = (1 + b − c)n P0 for (a) b > c, (b) b = c, (c) b < c
The parameters in this model are the starting population size, P0 , the
proportionate birth rate, b, and the proportionate death rate, c. Since the
last two of these occur in the combination b − c, it is often convenient to
replace this combination by a single symbol, for which we shall choose r.
21
This growth rate r gives the Thus r = b − c is the annual proportionate growth rate of the
net effect of births minus population. In terms of r, the recurrence relation for Pn is
deaths over each year, per
head of the population. Pn+1 = (1 + r)Pn ,
Although r is a growth rate, and the closed-form solution is
it can be negative.
Pn = (1 + r)n P0 .
The graphs in Figure 2.3 are The three cases b > c, b = c and b < c correspond respectively to r > 0,
those of Pn = (1 + r)n P0 in r = 0 and r < 0.
each of these cases.
The model just derived could be called a geometric model for population
variation, since it predicts that the population size rises or falls according
to a geometric sequence. However, it is more usual to describe it as the
exponential model, since the expression (1 + r)n P0 involves raising the base
number 1 + r to the exponent n.
Exponential model
The exponential model for population variation is based on the
assumption of a constant proportionate growth rate, r. The model is
described by either the recurrence relation
Pn+1 = (1 + r)Pn (n = 0, 1, 2, . . .), (2.1)
or its closed-form solution
Pn = (1 + r)n P0 (n = 0, 1, 2, . . .), (2.2)
where Pn is the population size at n years after some chosen starting
time.
Given data on population size over at least several years, it is possible to

select the parameters P0 and r in such a way that the corresponding
exponential model (given by equation (2.2)) ‘fits the data’ as closely as
possible. We shall not be going into the full details of how this is done, but
the following example and activity show what is involved in idealised cases.
Example 2.2 Modelling gannets
A colony of northern gannets at Cape St Mary, Newfoundland, Canada,

numbered 26 in the year 1889 and 160 in 1899.
(a) Assuming that this population can be modelled exactly over the
intervening period by an exponential model, find the value of the
annual proportionate growth rate, r, to two significant figures.
(b) Write down the corresponding formula for the population size Pn of
the colony at n years after the census date in 1889.
(c) Assuming that this model continues to hold after 1899, what
population size is predicted for the colony in the year 1914?
22
Solution
(a) Take n = 0 at the census date in 1889. Then n = 10 in 1899, so
P0 = 26 and P10 = 160. According to the exponential model
(equation (2.2)), we have Pn = (1 + r)n P0 , so, in particular,
P10 = (1 + r)10 P0 ; that is, 160 = (1 + r)10 × 26.
It remains to solve this equation for the proportionate growth rate, r.
On dividing through by 26 and then taking the 10th root, we obtain
� � �0.1
10 160 160
1+r = , so r = − 1 = 0.20 (to 2 s.f.).
26 26
Hence the proportionate growth rate is about 20% per year.
(b) The exponential model for the colony is therefore Pn = 26(1.20)n . As a check, the value of
26(1.20)10 should be close to
(c) The year 1914 corresponds to n = 25, for which the population size of
160, the given population size
the colony is predicted to be P25 = 26(1.20)25 2480.30. This must be
for 1899. In fact,
rounded to a whole number, to provide a meaningful prediction for a
population size. The prediction for the year 1914 is therefore 2480. 26(1.20)10 160.99.
(Note that even rounding to the nearest whole number may be We shall not pursue the
misleading. This prediction is based on a value of r which was rounded effects of such rounding. In
to two significant figures, so there is reason to question its accuracy. further discussion of this
If we use the exact value of r to calculate P25 , then the prediction model, we take r = 0.20.
becomes 2443 to the nearest integer.)
Activity 2.2 Modelling the US population (1790–1890)
The (human) US population was about 4 million in 1790, when the first
national census was taken, and 63 million in 1890.
intervening period by an exponential model, find to two significant
figures the value of the annual proportionate growth rate, r. (Take the
population to be measured in millions.)
(b) Write down the corresponding formula for the US population Pn
(in millions) at n years after the census date in 1790.
(c) Assuming that this model continues to hold for years after 1890, what
US population is predicted in the year 1950 ?
(d) In which year, according to the model, does the US population reach The approach needed to
500 million? answer this part is similar to
that for problems in
Chapter A3, Section 4: for
example, Activity 4.6(b) and
the main text before that
activity.
2.3 Evaluating the exponential model

To complete the modelling cycle for the general exponential model
developed in the previous subsection, we need to evaluate it. Is this model
satisfactory for the stated purpose, which was to describe how numbers in
a population may change with time, in a way that is applicable over a long
period of time and for a variety of populations? We consider in turn the
three cases identified in Activity 2.1 and illustrated in Figure 2.3.
23
If b < c (that is, r < 0), then the prediction is for a decreasing population
size. If a population has fewer births than deaths, then it can be expected
to decrease, and examples of real populations varying in this sort of way
can be given (see Figure 2.4). In this case, the mathematical solution
Pn = (1 + r)n P0 will eventually predict populations of size less than 1.
Such a prediction is not to be taken literally, of course. Values of Pn less
than 1 would suggest that the population has reached extinction. In
practice, a model implying such population decrease would be interpreted
as predicting extinction when the population size drops below some
minimum level from which there is no chance of recovery. There is nothing
obviously unrealistic about the general form of population change
predicted in this case.
Figure 2.4 Population (in breeding pairs) of red-backed shrikes in Great Britain
(1952–1989)
If b = c (that is, r = 0), then the prediction is for a constant population.

This may seem to be a peculiar ‘special case’, but in fact it is the closest to
reality in many instances. A fairly steady population is more typical than
rapid increase or decrease. An example is given in Figure 2.5(a).
Figure 2.5 Examples of fairly steady populations (in breeding pairs):

(a) pied flycatchers at Lemsjöholm, Germany (1956–1964),
(b) grey herons in England and Wales (1928–1986)
The graph in Figure 2.5(b) shows fairly steady population size over some
periods, interspersed with sudden drops and subsequent steady recoveries.
24
Here we must recall that the exponential model relies on the assumption of
constant proportionate birth and death rates. This assumption will fail to
hold where there are marked variations in weather or other conditions
which could affect the population. It is noticeable that the largest sudden
drop for the grey heron population of Figure 2.5(b) coincides with the
particularly severe UK winter of 1963.
If b > c (that is, r > 0), then the prediction is for a population that
increases. This increase continues indefinitely, and becomes more and more
rapid. The prediction in this case is not reasonable as regards long-term
behaviour. Any population which is subject to unlimited increase will
eventually reach an unsustainable size. What constitutes an
‘unsustainable’ level, and how long the population may take to reach that,
depends on the particular population being considered.
Many human populations have shown exponential increase over quite long
periods of time, as for example in the case of the US population considered
in Activity 2.2. Figure 2.6(a) shows the US population for the period
1790–1890, together with the exponential model whose growth rate was
obtained in Activity 2.2(a). The fit between the model and the data
appears good here. However, Figure 2.6(b) looks at the US population on
a different time scale, and includes further census data up to 1990. It is
evident that the previous exponential model is far from accurate in its
predictions after 1900. For example, the prediction from Activity 2.2(c) of
a population size 332 million in the year 1950 is more than twice as large
as the actual figure for that year, 151 million.
Figure 2.6 US population: (a) 1790–1890, (b) 1790–1990
The same inaccuracy applies for the gannet colony considered in

Example 2.2. The fit to data of the exponential model looks quite good at
relatively low population levels, but not thereafter (see Figure 2.7(a),
overleaf). The rate of increase slows down in later years (at higher
population levels). Such a slowing down of the rate of increase is also
apparent for the grey herons after 1963 (Figure 2.5(b)) and for the
populations shown in Figure 2.7(b) and (c). These three diagrams also
suggest that some steady population size may eventually be reached.
The explanation of these last examples is that rapid growth of a
population may occur when an animal species arrives in a new and
favourable habitat, but that such growth will not be sustained indefinitely.
25
Figure 2.7 Examples of increasing populations:

(a) population (in breeding pairs) of northern gannets at Cape St Mary, Newfoundland, Canada (1879–1939),
(b) estimates of the sea otter population off central California, USA (1914–1987),
(c) winter counts of elk in North Yellowstone National Park, USA (1968–1979)
Other types of population variation are also found with some species:
examples are shown in Figure 2.8. (For illustrative purposes, it is assumed
that the flour beetles of Figure 2.8(a) breed in clearly separated
generations.) Clearly, the exponential model has little application in such
cases.
Figure 2.8 Some other examples of population variation:

(a) numbers in successive generations of a laboratory population of a flour beetle,
A population density is the (b) population density of lemmings near Churchill, Canada (1929–1944)
number per unit area.
We conclude that the exponential model may be appropriate in certain
instances (declining populations, steady size, increases from a low level),
but that it has a serious flaw in predicting unlimited growth. It is this flaw
which we attempt to rectify by revising the model in the next section.
26
Origins of the exponential model

The exponential (or geometric) model for population variation is also
referred to as Malthusian, after Thomas Robert Malthus (1766–1834),
who was for many years Professor of Political Economy and Modern
History at the College of the East India Company in the UK. In 1798
he published An Essay on the Principle of Population, which expressed
his belief that the natural effect of human reproduction, if
unconstrained, was to produce geometric increases in population size.
He also thought that food production could increase only at an
arithmetic rate, which suggested a future in which starvation might be
the major factor in limiting populations. To avoid this bleak prospect,
he suggested that attention should no longer be paid to improving
living conditions for the poor, or to seeking advances in medical science
(so as not to lower death rates), while birth control and moral restraint
were to be encouraged. (However, Malthus himself had 11 children!)
The views of Malthus had considerable influence on social policy at the
time, and may well have been partly responsible for the undertaking of
the first UK census in 1801.
the exponential (or geometric) model for population variation,
Pn+1 = (1 + r)Pn , which is based on the assumption of a constant
proportionate growth rate r and has the closed-form solution
Pn = (1 + r)n P0 , where Pn is the population size at n years after some
chosen starting time;
a method for estimating, from data about the population, a value for
the parameter r in the exponential model;
evidence of the invalidity of the exponential model for large population
sizes, in that it predicts unlimited growth.

Exercise 2.1
According to UN estimates, the (human) population of the world was The world is one habitat for
1.65 billion in the year 1900 and 2.52 billion in 1950. which we can be sure that
there is no external migration!
intervening years by an exponential model, find to two significant Note that, in UN statistics,
figures the annual proportionate growth rate for the period 1900–1950. 1 billion = 1000 million.
(b) Assuming that this model continues to hold after 1950, what world
population is predicted for the year 2000 ?
Exercise 2.2
In fact, growth of the world population has been considerably faster since According to the UN, the
1950 than the model of Exercise 2.1 would suggest. Using the estimated world population reached
world population of 6.06 billion in the year 2000, repeat the task of 6 billion in October 1999.
Exercise 2.1(a) for the period 1950–2000.
27
3 Populations: logistic model
The general exponential model for population variation, developed in

Subsection 2.2, is summed up in the recurrence relation
Pn+1 = (1 + r)Pn (2.1)
and the corresponding closed-form solution
Pn = (1 + r)n P0 . (2.2)
This model was seen to work well enough for relatively small populations,
but to become invalid for large population sizes. We now seek to amend
the model in the case of a positive proportionate growth rate r, in order to
cope more realistically with the effects of continued population growth.
3.1 Setting up the logistic model

Equation (2.1) was based on the premise that both b (the proportionate
birth rate) and c (the proportionate death rate) were constant, regardless
of population size, so that the same was true of the proportionate growth
rate, r = b − c. Observation of various species suggests strongly that, all
other factors being equal, birth rates tend to decline or death rates to
increase (or both) as population size rises. These trends may be thought of
as the consequence of competition between individuals of the species
within a habitat of limited resources. They both point to a decline in the
proportionate growth rate as population size increases. We now seek to
build this feature into a revised model for population variation.
It may be timely at this point to recall exactly what ‘proportionate growth
Here ‘in year n’ refers to a rate’ means. If a population has size Pn in year n, and size Pn+1 in year
particular fixed census date n + 1, then the growth rate of that population over the year from n to
within the year concerned. n + 1 is Pn+1 − Pn . The proportionate growth rate is the ‘increase per
head’, which is the actual growth rate divided by the population size at
the start of the year:
Pn+1 − Pn
.
Pn
The exponential model assumes that the proportionate growth rate is a
constant, r, so that
As you can check, this Pn+1 − Pn
equation is just a = r.
Pn
rearrangement of
equation (2.1).
We now seek to alter this assumption, and put
Pn+1 − Pn
R is a real function. = R(Pn ), (3.1)
Pn
where R(P ) is a function of population size P . We noted two points above:
that the exponential model, based on R(P ) = r (constant), works well
when P is relatively small;

that the proportionate growth rate, R(P ), is seen to decrease, in
practical situations, as P increases.
28
SECTION 3 POPULATIONS: LOGISTIC MODEL
The simplest modelling step which takes account of these observations is to

assume that the proportionate growth rate, R(P ), is a decreasing
linear function of the form
R(P ) = mP + r,
where m < 0 (to meet the second point) and R(P ) r when P is
small (to incorporate the first one).
This function may be expressed as
� �
P
R(P ) = r 1 − , (3.2)
E
where r and E are positive parameters. (Recall that we are seeking to
amend the exponential model in the case r > 0.) The graph of this
function is given in Figure 3.1, which shows that r and E are the
intercepts of the line on the axes.
On substituting the expression for R(P ) from equation (3.2) into
equation (3.1), we obtain
� �
Pn+1 − Pn Pn
=r 1− ,
Pn E
from which we conclude that, according to our revised model, Pn satisfies
the recurrence relation Figure 3.1 Graph of
� � R(P ) = r(1 − P/E)
Pn
Pn+1 − Pn = rPn 1 − . (3.3)
E
This equation is called the logistic recurrence relation, and the
corresponding mathematical model is known as the logistic model for
population variation.
Logistic model
The logistic model for population variation is based on the
assumption of a proportionate growth rate R(P ) of the form
� �
P
R(P ) = r 1 − , (3.2)
E
where r and E are positive parameters.
The model is described by the recurrence relation
� �
Pn
Pn+1 − Pn = rPn 1 − (n = 0, 1, 2, . . .), (3.3)
E
where Pn is the population size at n years after some chosen starting
time.
We have now completed the second stage in the modelling cycle for the
logistic model. It is useful at this point to review how far we have come,
by looking in turn at each of the following questions.
What do the variables in equation (3.3) represent?
What assumptions are made in the model?
How can we interpret the model’s parameters in terms of the situation
being modelled?
29
Variables
The time variable, n years, can be chosen to be zero at any convenient
point in time. The population size at that time is P0 , and Pn then denotes
Sometimes Pn may be taken the population size n years later. In fact, this can be generalised to other
to denote the number of time intervals. We usually want to look at population sizes at yearly
breeding pairs, rather than intervals, because of the pattern of population variation within each year
individuals, or to denote that was mentioned at the start of Subsection 2.2, but there are occasions
population density. when a time interval other than one year is appropriate. This is
particularly likely when considering laboratory experiments, which are not
affected by nature’s annual cycle. For the flour beetles in Figure 2.8(a), it
would be natural to take Pn to be the population in the nth generation
from the start of the experiment. The steps involved in creating the model
do not depend on the choice of time interval, so this may be altered as
appropriate to suit the circumstances of the population being studied.
Assumptions
As in the exponential model, we have assumed that migration is not a
significant factor for the populations considered here. In moving from the
exponential to the logistic model, we assumed that the proportionate
growth rate is a decreasing linear function of population size. Since the
growth rate is births minus deaths, the linear decline in proportionate
growth rate could arise in several ways:
proportionate birth rate decreases linearly with population size, while
proportionate death rate remains constant;
proportionate death rate increases linearly with population size, while
proportionate birth rate remains constant;
proportionate birth rate decreases linearly with population size, and
proportionate death rate increases linearly with population size.
In each case, it is possible to derive the appropriate form of the logistic
recurrence relation from suitable information on the birth and death rates
of a population, as in the following example and activity.
Example 3.1 Modelling barnacle geese
In 1970 a winter refuge was created at Caerlaverock, Dumfries, Scotland,

for the population of barnacle geese which spends its winters there. The
size of this population grew steadily from about 3000 in 1970 to about
13 000 in the mid-1990s. Observations of the birth and death rates of the
barnacle geese suggest that the following information describes the
behaviour of the population during this period:
the annual proportionate death rate is constant at 0.11;
the annual proportionate birth rate decreases linearly with the
population size, P , according to the formula 0.31 − 1.5 × 10−5 P .
The population size on 1 January 1970 was 3200.
(a) Find a recurrence system for Pn , the population size of barnacle geese
n years after 1 January 1970.
(b) Show that the recurrence relation obtained is logistic, by identifying
the values of the parameters r and E from equation (3.3) which apply
in this case.
30
Solution
(a) The proportionate growth rate R(P ) at population size P is the
proportionate birth rate minus the proportionate death rate; that is,
R(P ) = (0.31 − 1.5 × 10−5 P ) − 0.11 = 0.20 − 1.5 × 10−5 P.
Hence the population growth for the year which starts at time n is
R(Pn )Pn = (0.20 − 1.5 × 10−5 Pn )Pn .
Since this is also the increase from Pn to Pn+1 , we have the recurrence
system
P0 = 3200, Pn+1 − Pn = (0.20 − 1.5 × 10−5 Pn )Pn .
(b) To show that this recurrence relation is logistic, the right-hand side
must be written in the form rPn (1 − Pn /E):
� �
−5 1.5 × 10−5
(0.20 − 1.5 × 10 Pn )Pn = 0.20Pn 1 − Pn
0.20
� �
= 0.20Pn 1 − 7.5 × 10−5 Pn .
This is indeed of the required form, with r = 0.20 and
1/E = 7.5 × 10−5 , that is, E = 13 300 (to 3 s.f.).
Activity 3.1 Revised model for the pheasants
A population of ring-necked pheasants was the subject of Figure 2.1 and

Example 2.1 (page 19), where an exponential model was assumed. The
following assumptions are now to be made in describing the behaviour of
this population:
the annual proportionate death rate is constant at 0.4; The assumption on the death
the annual proportionate birth rate decreases linearly with the rate is the same as in
Example 2.1, but the
population size, P , according to the formula 2.65 − 0.0015P .
assumption on the birth rate
The population size on 1 April 1937 was 8. has been altered.
(a) Find a recurrence system for Pn , the population size of ring-necked
pheasants n years after 1 April 1937.
(b) Show that the recurrence relation obtained is logistic, by identifying
the values of the parameters r and E from equation (3.3) which apply
in this case.
Parameter s
The parameter E may be interpreted from two different viewpoints. It
arose in Figure 3.1, as the intercept of the proportionate growth rate graph
(assumed linear) on the P -axis. Hence it is the population size at which the
proportionate growth rate is zero.
On the other hand, it can be seen directly from the logistic recurrence
relation (3.3) that if Pn attains the size E when n = m, then
� �
E
Pm+1 − E = rE 1 − = 0, so Pm+1 = E.
E
Hence Pm+2 = E, Pm+3 = E, . . . .
In other words, if Pn reaches size E, then it stays there.
31
These two views are, of course, consistent. Zero growth rate implies a
constant size of population, and vice versa. The parameter E is called the
equilibrium population level (or, in some contexts, the carrying
capacity).
More generally, the constant sequence
Pn = E (n = 0, 1, 2, . . .)
is a solution of the logistic recurrence relation. In fact, this sequence is the
only non-zero constant sequence which satisfies the logistic recurrence
relation. For if Pn = c (n = 0, 1, 2, . . .), where c is a constant, then from
equation (3.3) we have
� �
c
The constant sequence c − c = rc 1 − , so c = 0 or c = E.
E
Pn = 0 (n = 0, 1, 2, . . .)
We do not yet know whether other sequences generated by the logistic
represents a total absence of recurrence relation will settle down near some particular positive value in
the population. the long term, but if they do so, then that value must be E. The
population is then ‘in equilibrium’.
Thus, according to the logistic model, the equilibrium population level for
the barnacle goose population of Example 3.1 is 13 300, while the
equilibrium population level for the ring-necked pheasants in Activity 3.1
is 1500.
The meaning of the parameter r is apparent from the way in which the
logistic model was constructed: it is the proportionate growth rate of the
population at small population sizes. We can use the interpretation of E
given above to be more precise about what ‘small’ means here. If Pn is
small compared with the equilibrium population level, E, then Pn /E is
small compared with 1, so that the logistic recurrence relation (3.3) is
approximated closely by
Pn+1 = (1 + r)Pn ,
which is the exponential model once again. We have demonstrated that
this more primitive model is a limiting case of the logistic model. At low
population levels, population growth can be modelled satisfactorily by the
exponential model formula, Pn = (1 + r)n P0 , but this continues only while
the population size remains well below its equilibrium level, E.
3.2 Finding values for the parameter s

You have seen (in Example 3.1 and Activity 3.1) how to deduce the
appropriate form of logistic recurrence relation, by finding values for the
parameters r and E, in cases where linear expressions for the birth and
death rates (hence also for the growth rate) are known. Normally,
however, such expressions are not available in advance. This raises the
question of how values for these parameters can be estimated directly from
actual data about the population, in such a way that the corresponding
logistic model ‘fits the data’ as closely as possible. As with the exponential
model, we shall not go into full details about how this estimation is done,
but the following example and activity indicate how the parameters for the
logistic model may be estimated in some idealised cases.
32
Example 3.2 Revised model for gannets
For the population of gannets in Example 2.2 (page 22), it was shown that
the proportionate growth rate was r = 0.20 at relatively low population
levels. At higher population levels (later than 1919, see Figure 2.7(a)) this
growth rate is no longer appropriate.
The size of this population was 3200 in 1924, 3600 in 1929 and 4000 in
1934. The growth was therefore 400 in each of the 5-year periods before
and after 1929, when the population size was 3600. This suggests an
annual growth rate of about 80 for a population size of 3600.
Assuming that the behaviour of this population satisfies a logistic
recurrence relation, estimate the value of the equilibrium population
level, E.
Solution
The logistic recurrence relation is
� �
Pn
Pn+1 − Pn = rPn 1 − , (3.3)
E
where in this case r = 0.20 from consideration of the proportionate growth
rate at low population levels. Since the annual growth rate Pn+1 − Pn is
estimated to be 80 for the population size Pn = 3600, we have
� � � �
3600 3600
80 = 0.20 × 3600 1 − = 720 1 − .
E E
On solving this equation for E, we obtain
3600 80 8
=1− = , from which E = 4050.
E 720 9
This is the equilibrium level for the population.
Activity 3.2 Revised model for the US population
For the US population (1790–1890), you found in Activity 2.2 (page 23)
that the annual proportionate growth rate was r = 0.028. At higher
population levels (later than 1900, see Figure 2.6(b)) this is no longer
appropriate.
The size of the US population was 76 million in 1900, 92 million in 1910
and 106 million in 1920. Hence there was a growth of 30 million over a
20-year period with midpoint 1910, when the population was 92 million.
This suggests an annual growth rate of about 1.5 million for a population
size of 92 million.
Assuming that the behaviour of this population satisfies a logistic
recurrence relation, estimate the value of the equilibrium population
level, E. (As before, take the population size Pn to be measured in
millions.)
33
The approach used in the last example and activity can be applied only if
the value of r is found first from data on the population growth at low
population levels. When such data are not available, a strategy of the
following type may be adopted instead.
Example 3.3 Using growth data at two population levels
A population of deer is introduced to a nature reserve on an island, and its

growth is subsequently monitored. It is estimated that the annual
proportionate birth rate is 2.70 when P = 100 and 2.40 when P = 200,
while the annual proportionate death rate is 0.96 when P = 100 and 1.82
when P = 200. (In each case, P is the population size at the start of the
year.)
(a) Assuming that the behaviour of this population satisfies the logistic
model, estimate the values of the equilibrium population level, E, and
proportionate growth rate for low population sizes, r.
(b) If the population size is 150 at the start of a particular year, what size
of population is predicted for the start of the next year, according to
the model?
Solution
(a) The proportionate growth rate is
2.70 − 0.96 = 1.74 when P = 100,
2.40 − 1.82 = 0.58 when P = 200.
Also, the proportionate growth rate has the form r(1 − P/E) for a
population size P (equation (3.2)). Hence we have two simultaneous
equations involving r and E, namely,
� � � �
100 200
1.74 = r 1 − and 0.58 = r 1 − .
E E
There are various ways of solving these equations. For example, on
dividing through each equation by r and then rearranging, we obtain
1.74 100 0.58 200
+ = 1 and + = 1,
r E r E
You may find it helpful to which is a pair of simultaneous linear equations in the variables 1/r
express these equations in and 1/E. We can eliminate the 1/E term by subtracting the second
terms of x = 1/r and equation from twice the first equation:
y = 1/E. � �
3.48 200 0.58 200
+ − + = 2 − 1,
r E r E
which gives 2.9/r = 1, that is, r = 2.9.
On substituting this value for r into the first equation, we obtain
1.74 100 100 1.74 1.16
+ = 1; that is, =1− = = 0.4.
2.9 E E 2.9 2.9
Solving this equation gives E = 250.
34
(b) The logistic recurrence relation in this case is
� �
Pn
Pn+1 − Pn = 2.9Pn 1 − .
250
If Pn = 150 for some n, then we have
� �
150
Pn+1 = 150 + 2.9 × 150 1 − = 324.
250
(Note that, in this case, the population size has moved in one year
from well below the equilibrium population level, 250, to well above it.)
Activity 3.3 Modelling beetles
Experiments are conducted on a species of beetle which breeds with clearly

separated generations. In a series of experiments, different numbers of
beetles are introduced into identical laboratory cultures, and the numbers
in the next generation are counted. When the experiment is started with
100 beetles, the next generation numbers 220, whereas when the
experiment is started with 400 beetles, the next generation numbers 160.
(a) Write down the proportionate growth rate, (Pn+1 − Pn )/Pn , for n = 0, Recall that Pn is the
for each of P0 = 100 and P0 = 400. population size in the nth
generation from the start of
(b) Assuming that the behaviour of this population satisfies the logistic
the experiment.
model, estimate the values of the equilibrium population level, E, and
the proportionate growth rate for low population sizes, r.
(c) If the experiment is started with 200 beetles, how many will there be
in the next generation, according to the model?
3.3 Investigating the logistic model

We have not, as yet, ‘done the maths’ to obtain closed-form solutions from
the logistic model, beyond finding the two constant sequences, Pn = E and
Pn = 0. In fact, unlike the situation with the exponential model, there are
only special cases for which closed-form solutions of the logistic recurrence
relation,
� �
Pn
Pn+1 − Pn = rPn 1 − , (3.3)
E
are known. Study of the population behaviour predicted by this model
therefore depends on either deductions from the form of the recurrence
relation itself or direct calculation of successive terms using the recurrence
relation. In this subsection we focus on what information can be obtained
by looking at the model from the graphical and algebraic standpoints. In
Section 4 we turn to direct calculation of terms, using the computer. Only
once these matters have been explored can we pronounce upon how well
the logistic model compares with reality.
35
Graphical approach
First, note that there are several ways in which the information given by
Essentially, this first way equation (3.3) can be graphed. One way is to plot proportionate growth
gives the graph in Figure 3.1, rate, R(Pn ) = (Pn+1 − Pn )/Pn , against Pn , which we illustrated when
but with P = 0 excluded. deriving the logistic recurrence relation. More useful, for current purposes,
is a plot of growth rate, Pn+1 − Pn , against Pn , which is shown in
Figure 3.2(a). The graph is that of the function given by the right-hand
side of equation (3.3), which is a quadratic function of Pn . It cuts the
horizontal axis at Pn = 0 and Pn = E.
Figure 3.2 Graphs from the logistic recurrence relation:

(a) plot of Pn+1 − Pn against Pn , (b) plot of Pn+1 against Pn
Note that population growth in a year means that Pn+1 − Pn > 0, while
population decline means that Pn+1 − Pn < 0. Hence there is population
growth where the graph of Figure 3.2(a) lies above the horizontal axis
(that is, where 0 < Pn < E) and population decline where the graph lies
below this axis (where Pn > E). The equilibrium population level, Pn = E,
appears as a point on the graph for which population growth is zero.
This maximum value is Also, the growth rate will be a maximum at the vertex of the parabola,
r× 1
− 1 1 where Pn = 12 E. The further that Pn is from 21 E, the smaller is the
2 E(1 2) = 4 rE,
corresponding growth rate.
as marked on Figure 3.2(a).
Another point to note is that altering the value of the parameter r has the
If the vertical axis were effect of scaling this parabola vertically, so that for a given value of Pn (and
labelled by y, then we would fixed E), the growth rate becomes larger in magnitude as r is increased.
call this a y-scaling.
The graph in Figure 3.2(b) is a plot of Pn+1 (next year’s population size)
against Pn (this year’s population size). This arises from rearranging
equation (3.3) as follows:
� �
Pn
Pn+1 = Pn + rPn 1 −
E
� � ��
Pn
= Pn 1 + r 1 − .
E
The right-hand side here is again a quadratic function of Pn , whose graph
passes through the point (E, E). The graph cuts the horizontal axis when
Pn+1 = 0, that is, when Pn = 0 and when 1 + r(1 − Pn /E) = 0. So the
intercepts on the horizontal axis are at Pn = 0 and Pn = E(1 + 1/r).
This latter value of Pn is of particular interest from the modelling point of
view since, according to the graph, if Pn = E(1 + 1/r) then Pn+1 = 0, and
if Pn > E(1 + 1/r) then Pn+1 < 0. Thus the population becomes extinct
within a year, according to the model, if its numbers ever rise as high as
E(1 + 1/r). As the parameter r is increased, this ‘ceiling on sustainability’
becomes closer to the equilibrium population level, E.
36
Algebraic approach
We have already employed some algebra above, while investigating what
useful information about the logistic model can be gleaned from graphs.
The following argument, however, depends upon algebraic manipulation
alone.
Starting once more from the logistic recurrence relation (3.3), we divide
both sides of the equation by E, to obtain
� �
Pn+1 Pn Pn Pn
− =r 1− .
E E E E
Now we put xn = Pn /E, so that also xn+1 = Pn+1 /E. This results in a
recurrence relation for the new variable xn , namely,
xn+1 − xn = rxn (1 − xn ), (3.4)
which looks simpler than the original, since the parameter E no longer
appears. Mathematically, it is a version of the logistic recurrence
relation (3.3) with E = 1. In modelling terms, the quantity xn represents
the population size Pn as a proportion of its equilibrium level, E.
The derivation of equation (3.4) shows that the parameter E is just a
scaling factor for any logistic recurrence sequence. So, in order to calculate
terms of a sequence Pn which satisfies the logistic recurrence relation (3.3),
with given values for the parameters r, E and for P0 , we could proceed as
follows:
calculate x0 = P0 /E;
calculate the required number of terms of the corresponding sequence
xn which satisfies the recurrence relation (3.4);
calculate Pn = xn E for each relevant value of n.
The graph of Pn against n is just a vertically scaled version of the graph of
xn against n, with scaling factor E.
It follows that, in order to understand the long-term behaviour of
sequences Pn which satisfy equation (3.3), for given r and E, we need to
understand only the long-term behaviour of sequences xn which satisfy
equation (3.4). This observation saves considerable labour where computer
investigation of equation (3.3) is concerned, since it tells us that there is no
need to experiment with different values of the parameter E.
Origins of the logistic model

The first person to have formulated the logistic model is reputed to be
the Belgian mathematician Pierre-François Verhulst (1804–49). He
argued against the previous view, based on the ideas of Malthus, that
populations tend to increase exponentially, by pointing out that there
were countervailing forces which acted to slow growth down.
Based on his model, formulated in 1846, he predicted that the Belgian
population would never exceed 9.4 million. Over 150 years later it is
just above 10 million, but when compared with predictions from the
exponential model, Verhulst’s estimate looks remarkably good.
37
the logistic model for population variation, with recurrence relation
� �
Pn
Pn+1 − Pn = rPn 1 − ,
E
which assumes that proportionate growth rate is a decreasing linear
function of population size;
methods for estimating, from data about a population, values for the
parameters r (proportionate growth rate at low population levels)
and E (equilibrium population level);
some graphical and algebraic analysis of the behaviour to be found
according to the logistic model.
Exercise 3.1
UN data suggest that the proportionate growth rate per decade of the
world population was 0.20 in 1960, when the population size was
3.02 billion, and 0.15 in 1990, when the population size was 5.27 billion.
(a) Assuming that the behaviour of the world population satisfies a
logistic model, estimate the values of the equilibrium population
level, E, and the parameter r.
(b) Using this model, and the world population estimate of 6.06 billion
for 2000, what population is forecast in the year 2010?
This is not a ‘standard Exercise 3.2

exercise’. It asks you to
analyse algebraically the
You saw at the end of Subsection 3.1 that where Pn /E is small compared
behaviour of the logistic with 1, the logistic model is approximated well by the exponential model.
model for population sizes This exercise concerns a similar approximation for the case where
close to the equilibrium 1 − Pn /E is small, that is, where Pn is close to its equilibrium level.
level E, and provides (a) Express the logistic recurrence relation,
predictions which can be � �
compared with numerical Pn
Pn+1 − Pn = rPn 1 − ,
results to be obtained in E
Section 4. However, if you are in terms of Qn , where Qn = E − Pn .
short of time then you may
prefer to omit this exercise. (b) Show that if Qn /E is small compared with 1, then Qn satisfies
approximately the geometric recurrence relation
The new variable Qn is (when
positive) the amount by Qn+1 = (1 − r)Qn .
which the population size Pn
(c) Use this result to describe the nature of logistic recurrence sequences
is below its equilibrium
value E. with terms close to Pn = E, in each of the following cases.
(i) 0 < r < 1 (ii) 1 < r < 2 (iii) r > 2
38
4 Logistic recur rence sequences on the computer
To study this section you will need access to your computer, together with
Computer Book B.
Bearing in mind the qualitative information gleaned in Subsection 3.3

about the logistic model, it is now time to calculate terms of logistic
recurrence sequences directly via the computer. The aim is to try and build
up an overview of such sequences, and to see how their behaviour depends
on the value of the parameter r. (As demonstrated in Subsection 3.3, there
is no need to consider separately the effect of varying the value of E.)
In Example 3.1 you saw that the growth in a certain barnacle goose
population can be modelled by a logistic recurrence relation with r = 0.20
and E = 13 300. Figure 4.1 shows the graph of a sequence generated by
this recurrence relation, but with P0 = 20 rather than the value of 3200 The corresponding picture for
quoted in Example 3.1. This graph, which is S-shaped overall, shows that P0 = 3200 is obtained by
the values of Pn initially increase more and more rapidly. As you saw in moving the vertical axis in
Subsection 3.1, this initial growth is similar to that of a geometric Figure 4.1 to the right, until
sequence, Pn = (1 + r)n P0 . The rate of growth reaches a maximum at the intercept of the curve on
it is 3200. The position of the
Pn = 12 E = 6650, as forecast earlier with reference to Figure 3.2(a), then
vertical axis then corresponds
tails off as Pn = E is approached. In the long term, values of Pn settle
to n = 0.
down close to E = 13 300 and become effectively constant.
Figure 4.1 Graph of logistic recurrence sequence with r = 0.20, E = 13 300,

P0 = 20
Are the features that are apparent here shared by sequences also generated
by the logistic recurrence relation but with other values of r? You can now
use your computer to examine this question.
��
39
Reviewing the results

Computer investigation of logistic recurrence sequences reveals an
These various forms of interesting variety of behaviour as the parameter r is increased. The
behaviour may also be seen in apparently unstructured behaviour shown by sequences when r = 2.75 and
sequences generated by other r = 3 is referred to as chaotic. The sequence behaviour of repeatedly
non-linear recurrence taking a number of different but repeating values, seen with r = 2.25 or
relations. r = 2.5, for example, is called cycling. For r = 2.25, the sequence Pn has
a 2-cycle, and for r = 2.5, Pn has a 4-cycle.
You may use this table for The table below provides a summary of information on the extent of
reference in future work various behaviour regimes for logistic recurrence sequences, as may be
(see Exercise 5.1), including inferred from computer investigation.
assignments.
Range of r Long-term behaviour of Pn
0<r≤1 Settles close to (converges to) E, with values
always just below E
1<r≤2 Settles close to E, with values alternating
between just above and just below E
2 < r ≤ 2.44 2-cycle, with one value above E and one value
below E
2.45 ≤ r ≤ 2.54 4-cycle, with two values above E and two
values below E
2.6 ≤ r ≤ 3 Chaotic variation between bounds (with some
exceptions)
Evaluating the logistic model

The remainder of this section In Subsection 3.1 we constructed a model of populations based on the
will not be assessed. logistic recurrence relation. We shall now complete the modelling cycle by
briefly evaluating this model, with reference to the examples of actual
population variation given in various figures of Section 2.
Many of the broad features of sequences generated by the logistic
recurrence relation do correspond to aspects of population examples seen
earlier.
For positive values of r less than 1, the sequence generated by the
logistic recurrence relation exhibits an S-shaped increase, provided
that the choice of P0 is small compared with E. Such a pattern is
shown in Figure 2.7(b). The examples in Figures 2.6 (US population),
2.7(a) and 2.7(c) also match the general form of part of the S-shape.
The graphs in Figures 2.6 and 2.7(a) have not yet approached close
to E, while Figure 2.7(c) starts with a value of P0 that is not small
relative to E, so the initial ‘geometric increase’ part of the graph is
missing. This is also the case with the graph for grey herons in
Figure 2.5(b), from 1963 onwards.
For r = 1.8, the sequence generated by the logistic recurrence relation
shows rapid increase, followed by oscillations, settling to an
equilibrium value. This is the same pattern as shown by the
population of beetles in Figure 2.8(a).
The graph for the lemming population in Figure 2.8(b) shows
systematic cycles, which might reflect the cyclic behaviour of
sequences generated by the logistic recurrence relation for values of r
in the range between 2 and 2.54.
40
SECTION 4 LOGISTIC RECURRENCE SEQUENCES ON THE COMPUTER
There are populations which show wide and unsystematic variation.

This could correspond to the chaotic behaviour of the sequences
generated for many values of r between 2.6 and 3.
As a final specific comparison, the graph for the pheasants in
Figure 2.1 (for either April or October data) shows rapid increase.
This could match the initial part of an S-shaped curve (for r < 1), or From the result of
resemble the rapid initial growth of a sequence generated for a larger Activity 3.1, the latter is
value of r. more likely.
Sequences generated by the logistic recurrence relation (with r < 3) can be

seen to correspond in their general features to several different patterns of
variation shown by actual animal populations. However, matching specific
numerical predictions based on the logistic model of populations to real
data shows the model to be less reliable. The comparison with reality
shows a qualitative rather than a quantitative match, in many cases. This
is hardly surprising, given the rather sweeping assumptions on which the
model is based. We have assumed that the environmental conditions are
unchanging, that there is no migration, and that the proportionate growth
rate (birth rate minus death rate) decreases linearly as the population size
increases. It is unlikely in any real case that all of these conditions will be
met closely. The assumption about the growth rate was based on
simplicity, rather than on any underlying biological reason, and will not
hold precisely in practice.
We have looked at only positive values for r. The logistic model can be
used, for P0 < E, with negative values of r, but it then has little to offer
which is not provided by the exponential model of decline.
One particular criticism of the growth rate assumption is that it predicts
extinction within a year for a population whose current size exceeds
E(1 + 1/r), and this is probably unrealistic. A population suffering from See Figure 3.2(b).
excessive overcrowding would be expected to survive, albeit at a much
lower level. Other models have been put forward to overcome this
drawback. One of these is the Ricker model, based on the recurrence William Ricker, born in 1908,
relation was until retirement Chief
Scientist to the Fisheries
Pn+1 = (1 + r)Pn exp(−aPn ), Board of Canada. He
where r has the same interpretation as in the logistic model, a is a positive developed this model in the
parameter and exp is the exponential function (introduced in Chapter A3, 1950s.
Subsection 3.2). The graphs of Pn+1 against Pn which can be drawn for
this model (the analogues of graphs such as that in Figure 3.2(b)) are
known as Ricker curves, and are much used to describe the behaviour of
fish populations and to determine sustainable fishing levels.
In this section you investigated sequences generated by the logistic
recurrence relation, Pn+1 − Pn = rPn (1 − Pn /E), via the direct evaluation
of terms by computer. The choice of starting value P0 does not usually
affect the long-term behaviour of the sequences. This behaviour alters
qualitatively as the parameter r is increased, as summarised in the table
on page 40. Some sequences have 2-cycles, 4-cycles or chaotic behaviour in
the long term.
41
5 Sequences and limits
In this section we address some issues which arose in the context of

investigating logistic recurrence sequences, though they have much wider
application. A sequence of numbers, un (n = 1, 2, 3, . . .), may or may not
‘settle down’ closer and closer to a single number in the long term (as n
becomes larger and larger). A sequence which does settle down in this way
is said to be convergent, and the number which it approaches more and
more closely is called the limit of the sequence. We are concerned here
with these ideas of convergence and limits for sequences.
Subsection 5.1 looks at what can be said in this regard about sequences
defined by recurrence systems, while Subsection 5.2 turns to sequences
described by closed-form formulas. Subsection 5.3 focuses on the topic of
sequences whose terms are the sums of terms of other sequences. These are
the series of which you saw examples in Subsection 1.2.
5.1 Sequences from recur rence systems

Suppose that Pn is a sequence generated by the logistic recurrence relation
� �
Pn
Pn+1 − Pn = rPn 1 − . (3.3)
E
In Section 4 you saw that, in certain cases, the sequence Pn settles down in
the long term to values that are effectively constant. When this happens,
we say that the sequence Pn is convergent or that it converges.
The value near which Pn settles in the long term is called the limit of the
sequence. For example, with P0 = 3200, E = 13 300 and r = 0.20, the
sequence Pn converges to the limit 13 300, as indicated by the graph in
Figure 5.1(a). This convergence is described concisely by the notation
If a sequence has a limit, then lim Pn = 13 300,
n→∞
that limit is unique.
or, equivalently, by
Pn → 13 300 as n → ∞,
a notation introduced in Chapter A1. (Some books print the limit as
A sequence which is not limn→∞ Pn when it appears in a line of text.) However, with the same
convergent is said to be values of P0 and E but with r = 2.9, Pn is not convergent, since it does not
divergent. settle near any particular value in the long term (see Figure 5.1(b)).
Figure 5.1 Graphs of logistic recurrence sequences with P0 = 3200 and E = 13 300 for (a) r = 0.20, (b) r = 2.9
42
SECTION 5 SEQUENCES AND LIMITS
You saw in Subsection 3.1 that if the logistic recurrence relation generates
a constant sequence, Pn = c say, then there are only two values that c can
take, namely 0 and E. If a sequence generated by the logistic recurrence
relation does converge, then for large enough n the terms of the sequence
are effectively equal. Thus in the long term, a convergent sequence is
effectively a constant sequence, so its limit must be one or other of the
values 0 or E. However, this says only that convergence to any other limit
is impossible. As the case above with r = 2.9 shows, there is no guarantee
of convergence even when potential limit values have been identified.
The above method of finding possible limit values applies to other types of
sequence generated by recurrence relations.
Example 5.1 Finding possible limit values for a sequence
Suppose that a sequence xn is generated by the recurrence relation Each non-zero value of x0
� � here will give a different
1 2
xn+1 = xn + (n = 0, 1, 2, . . .). (5.1) sequence.
2 xn
To what limit values might such a sequence xn converge?
Solution
Suppose that xn = c is a constant sequence generated by the given
recurrence relation. Then we have xn+1 = xn = c, and so
c = 12 (c + 2/c).
This is equivalent to
2c = c + 2/c; that is, c = 2/c.
√
So we obtain c2 = 2 and hence c = ± 2. These are the two possible limit
values for any sequence generated by the recurrence relation (5.1).
Activity 5.1 Finding possible limit values for a sequence
A population of ravens on an island has size 50 at the start of year 0.

Their proportionate birth and death rates are constant, being
respectively 0.3 and 0.4. There is no emigration from the island, but
8 ravens join the population from outside just before the start of each year.
(a) Show that the variation in this population can be modelled by the
recurrence system
P0 = 50, Pn+1 = 0.9Pn + 8 (n = 0, 1, 2, . . .), (5.2)
where Pn is the population size at the start of year n.
(b) Find any possible equilibrium level for this population. (This is
equivalent to seeking a value c for which the constant sequence Pn = c
is a solution to the recurrence relation.)
43
If a sequence is specified by a recurrence system, then the method of

Example 5.1 and Activity 5.1(b) may tell us the possible limit values to
which the sequence can converge. This is useful but limited information,
since there is no accompanying indication of whether or not convergence to
a possible limit actually occurs.
In general, sequences may be specified through a recurrence system or
through a closed form. Where available, a closed form can offer greater
opportunities for investigating whether or not a sequence converges, as you
will see in the next subsection.
In other cases, no closed form is available. This is true of any non-constant
sequence generated by the recurrence relation (5.1). Despite this, if x√0 is
not equal to zero, then any such√sequence will converge, to the limit 2 if
x0 is positive and to the limit − 2 if x0 is negative. In terms of the ‘lim’
notation introduced earlier, this may be expressed as
√ √
lim xn = − 2 if x0 < 0, lim xn = 2 if x0 > 0.
n→∞ n→∞
5.2 Sequences from closed-form formulas
Using the modulus
The modulus of a real In this subsection, the modulus of a real number is used to permit concise
number was defined in expression of certain types of inequality and to express the distance from
Chapter A3, Subsection 1.2. the origin to a point on the number line.
The distances from the origin to the points 5 and −4 on the number line
are 5 and 4, respectively. Since the modulus of a real number x is defined
by
�
x, if x ≥ 0,
|x| =
−x, if x < 0,
we have
|5| = 5 and |−4| = 4,
as shown in Figure 5.2.
Figure 5.2 Distances from the origin
More generally, the distance In general, the distance from the origin to a point a on the number line is
from any point p to another given by |a|.
point q is given by |p − q|.

Consider the inequalities −1 < a < 1 (that is, −1 < a and a < 1). These
say that the distance from a to the origin is less than 1. Thus
−1 < a < 1 and |a| < 1

are equivalent inequalities.
44
Manipulating closed forms

In Activity 5.1(b) you found a single equilibrium population value (80) for
an island colony of ravens whose population variation satisfies the
recurrence system (5.2). This leaves open the question of whether the size
of this population will actually converge to its equilibrium level.
Since the recurrence relation here is linear, it is possible to find a The closed-form formula for a
closed-form formula for Pn , which is linear recurrence system was
given in Chapter A1,
Pn = 80 − 30(0.9)n (n = 0, 1, 2, . . .). Section 4.
Given this formula, it is possible to deduce the long-term behaviour of the The long-term behaviour
population sequence Pn without any need for calculation. This is because of rn , for different values of r,
if n is large, then (0.9)n is small (that is, close to 0). With n sufficiently was the subject of
large, 30(0.9)n is also small, and we have 30(0.9)n → 0 as n → ∞. Chapter A1, Subsection 5.2.
According to the closed-form formula for Pn , therefore, the sequence In particular, if |r| < 1, then
converges, to the limit 80, which is the equilibrium level. Thus the raven rn → 0 as n → ∞.
population tends to this equilibrium level. A graph of this sequence is
Here we have r = 0.9.
shown in Figure 5.3.
Figure 5.3 Graph of Pn = 80 − 30(0.9)n

Notice how the graph of the sequence Pn settles near a horizontal straight
line as n becomes large. This behaviour corresponds to convergence of the
sequence, with the horizontal line being at the limit value.
A similar approach enables you to ascertain the long-term behaviour of
some other sequences defined by closed-form formulas. Not every such
sequence is convergent. For example, the sequences
an = n 2 and an = 2n Assume, here and in the
discussion below, that
both have values that increase without limit as the value of n increases. n = 1, 2, 3, . . . .
On the other hand, the values of an = −5n decrease as n increases, and
become arbitrarily large while remaining negative. In saying that the
terms of a sequence an ‘become arbitrarily large’, we mean that the terms As pointed out in
become arbitrarily far from 0. This description can be expressed in terms Chapter A1, Subsection 5.2,
of the distance |an |: ‘an becomes arbitrarily large’ means that |an | → ∞ such sequences are also called
as n → ∞. This is true of the sequence an = −5n, for which we also have unbounded.
an → −∞ as n → ∞. Thus none of the sequences
an = n 2 , an = 2n , an = −5n (5.3)
is convergent. The same applies for any other sequence defined by a closed
form where the terms, whether positive or negative, become arbitrarily
large in the long run.
45
There are other closed forms for sequences an where you can see that an
becomes smaller and smaller (arbitrarily small, that is, arbitrarily close
to 0) as n becomes large. Examples of this are
1 1 1
bn = 2
, bn = n and bn = − .
n 2 5n
These sequences are convergent, with limit 0. In each case, bn is of the
form 1/an , where terms of the sequence an (given respectively by
equations (5.3)) become arbitrarily large as n increases. This highlights a
general rule to use when considering the convergence of sequences.
Reciprocal Rule
If the terms of a sequence bn are of the form 1/an , where terms of the
sequence an become arbitrarily large as n increases, then the
sequence bn is convergent, and lim bn = 0.
n→∞
Another useful result concerns what happens to a sequence in the long run
if each of its terms is multiplied by the same number. For example, it has
already been remarked that the sequence an = 1/n2 converges to 0, since
its values eventually become arbitrarily small as n increases. Multiplying
all terms an by 50 makes them all larger by this factor, but this does not
prevent the resulting sequence, bn = 50/n2 , becoming arbitrarily small in
the long run.
This rule was used informally Constant Multiple Rule

in Chapter A1, If the terms of a sequence bn are of the form can , where the
Subsection 5.2.
sequence an is convergent with limit 0, and c is a constant, then the
sequence bn is also convergent, and lim bn = 0.
n→∞
These two rules may be used in combination. For example, the sequence
3 + n becomes arbitrarily large as n increases, so (by the Reciprocal Rule)
the sequence 1/(3 + n) converges to the limit 0. Thus (by the Constant
Multiple Rule) the sequence 100/(3 + n) also converges to the limit 0.
In more complicated formulas, you can look for quantities that become
small as n becomes large. For example, consider the sequence
5000
an = n. (5.4)
1 + 24(0.5)
Since 0.5 < 1, the quantity (0.5)n becomes arbitrarily small as n becomes
large. The same is true of 24(0.5)n (by the Constant Multiple Rule), so
1 + 24(0.5)n tends to 1. In the long run, the values of an become
arbitrarily close to 5000. In other words, the sequence defined by
equation (5.4) converges to the limit 5000.
In arguments such as this, it is useful to recall the known behaviour for
large n of the ‘basic sequences’ r n and np .
46
Long-term ‘basic sequence’ behaviour

The long-term behaviour of the sequence r n (n = 1, 2, 3, . . .) is as The statements here about rn
follows. follow from a table in
If |r| > 1, then |r n | → ∞ as n → ∞. (If r > 1 then r n → ∞,
Chapter A1, Subsection 5.2.
whereas if r < −1 then r n is unbounded and alternates in sign.)
If |r| < 1, then r n → 0 as n → ∞.
If r = 1, then r n = 1. If r = −1, then r n alternates between 1
and −1.
The long-term behaviour of the sequence np (n = 1, 2, 3, . . .) is as For example, if p = 3 then we
follows. have
If p > 0, then np → ∞ as n → ∞. n3 → ∞ as n → ∞,
If p < 0, then np → 0 as n → ∞. whereas if p = −2 then, since
p
If p = 0, then n = 1. n−2 = 1/n2 , we have
n−2 → 0 as n → ∞.
In the next example and activity, you are asked to decide whether or not
various sequences defined by closed-form formulas converge and, if they do,
to what limits. Reasoning of the type above should pay dividends here. If you find this difficult at
first, then there is always the
option of working out some
Example 5.2 Long-ter m behaviour from closed-form formulas numerical values of terms
(for n = 1, n = 10, n = 100,
For each of the sequences below, decide whether or not it converges and, if perhaps, using a calculator
it does, to what limit. In each case n = 1, 2, 3, . . . . where necessary) in order to
5 n2 try and spot what is going on.
(a) an = (0.6)n + 2(1.4)n (b) an = n (c) an =
17 − 3(2.1) 5 + 10n
Solution
(a) For large n, the sequence (0.6)n converges to 0 (it is rn with
r = 0.6 < 1), but the sequence (1.4)n (which is rn with r = 1.4 > 1) is
unbounded. Hence the sequence an does not converge.
(b) For large n, the sequence (2.1)n is unbounded, so the same is true of
17 − 3(2.1)n . Hence, by the Reciprocal Rule, 1/(17 − 3(2.1)n )
converges to 0, so by the Constant Multiple Rule (with c = 5), the
sequence an converges to the limit 0; that is, lim an = 0.
n→∞
(c) The given sequence can be written (dividing top and bottom of the
fraction by n) as
an = .
5/n + 10
Now 5/n converges to 0 for large n (1/n is of the form np with
p = −1), so 5/n + 10 converges to 10. Hence, for large n, the sequence
1
an will behave like 10 n, which is not convergent. We conclude that an
itself is not convergent either.
Comment
In part (c), since neither n nor n2 converges, it would appear that an is the
ratio of two sequences that do not converge. From this fact alone, we can
make no deductions about the convergence of an . To make progress, we
rewrite an in the form given, to produce a sequence in the denominator
that does converge.
47
Activity 5.2 Long-ter m behaviour from closed-form formulas
For each of the sequences below, decide whether or not it converges and, if
it does, to what limit. In each case n = 1, 2, 3, . . . .
1
(a) an =
2 + 3n
(b) an = 5 + n3
100
(c) an = n
4 + 20(0.6)
(d) an = 3 + (−1)n
60n
(e) an =
3 + 5n
Courses on analysis use more formal approaches to establish the long-term

behaviour of sequences. For example, the sequence Pn = 80 − 30(0.9)n
would be dealt with by identifying 80 − 30(0.9)n as the sum of two
sequences. This approach requires that 80 is thought of as the constant
sequence an = 80, which converges to 80, and an appeal to a sum rule as
well as the Constant Multiple Rule.
5.3 Sequences from sums

In Example 1.1(c), we recalled the formula for the sum of a finite
You first saw this in geometric series. Using sigma notation, this is
Chapter A1, Subsection 4.2. n
� � �
1 − rn+1
ari = a (r
= 1). (5.5)
i=0
1−r
Now the terms ari (i = 0, 1, 2, . . .) form a geometric sequence. From this
sequence, it is possible to construct a new sequence sn , each term of which
is a sum of consecutive terms from the sequence ar i . Thus we have
s0 = a, s1 = a + ar, s2 = a + ar + ar2 ,
and so on. The nth term of the new sequence is given by
n
�
sn = ari (n = 0, 1, 2, . . .),
i=0
which is the sum of a finite geometric series. Does the new sequence sn
have a limit? If it does, then lim sn exists and we write
n→∞
∞
�
lim sn = ari .
n→∞
i=0
This limit is the sum of an infinite geometric series,

a + ar + ar2 + ar3 + · · · .
48
Thanks to equation (5.5), it is possible to say almost immediately when

such a sum exists and, when it does, what its value is. If |r| > 1, then the
quantity rn+1 is unbounded for large n, so the same is true of sn . If r = 1,
then we have
sn = a
�
+ a + a��+ · · · + a� = (n + 1)a,
n+1 such terms
which is unbounded for large n. If r = −1, then the terms of sn take the
alternate values a and 0. Hence convergence can occur only if |r| < 1. In
this case, as recalled in Subsection 5.2, lim rn = 0. From equation (5.5), it
n→∞
follows that
�∞ � � The sequence rn+1 is the
1 − rn+1 a
ari = lim a = (|r| < 1). (5.6) sequence rn without the first
n→∞ 1−r 1−r
i=0 term r0 , so
lim rn+1 = lim rn .
n→∞ n→∞
Example 5.3 Sum of an infinite geometric series
What is the sum of the following infinite series?

1
2
+ ( 12 )2 + ( 12 )3 + · · ·
Solution
1
This is an infinite geometric series, with a = 2
and r = 21 , so by
equation (5.6), the sum is
∞
� 1
1 1 i 2
( ) = 1 = 1.
i=0
2 2
1− 2
Figure 5.4 shows a geometric interpretation of this result.
Figure 5.4 Illustration of an infinite geometric series with finite sum
Activity 5.3 Sum of an infinite geometric series

1− 1
2
+ ( 12 )2 − ( 12 )3 + · · ·
We derived equation (5.6) by taking the limit for large n of equation (5.5),
for |r| < 1. An alternative approach to summing an infinite geometric
series comes in handy when the series concerned is in the form of a
recurring decimal. This approach permits us to find the fraction which is
equivalent to any recurring decimal, as the next example illustrates.
49
Example 5.4 Fraction equivalent to recur ring decimal
Note that this decimal can be Find the fraction equivalent to 0.123 123 123 . . . .
considered as an infinite
geometric series, with Solution
a = 123 × 10−3 and r = 10−3 . Put s = 0.123 123 123 . . . . The repeating group, ‘123’, is 3 digits long, so
Recognition of these values, we multiply s by 103 , to obtain
followed by direct application
of equation (5.6), is not the 1000s = 123.123 123 123 . . . = 123 + s.
quickest way to find the Hence we have
equivalent fraction, though 123 41
you can check here that s= = .
equation (5.6) gives the same
999 333
result.
Activity 5.4 Fraction equivalent to recur ring decimal
Find the fraction equivalent to 0.454 545 . . . .

5.4 Some further sequences and series

This subsection will not be As you saw in Subsection 5.2, it is sometimes possible to tell from a
assessed. closed-form formula whether or not the corresponding sequence is
convergent. However, this is not always the case. For example, is the
sequence
� �
1 n
an = 1 + (n = 1, 2, 3, . . .) (5.7)
n
convergent? In the long term, there is an interplay between the quantity
1 + 1/n, approaching 1 more and more closely, and the fact that this
quantity (always greater than 1) is raised to the power n. The combined
effect is not easy to predict. In such cases, as with sequences given by
recurrence systems, calculation of enough terms of the sequence, perhaps
using a computer, may help to indicate whether or not the sequence
converges. Applying this approach to the sequence (5.7) gives the graph in
Figure 5.5, which does appear to indicate that the sequence converges.
This limit is established in Although the value of the limit cannot be read off accurately from the
courses on real analysis. The graph, it turns out that it is the base for natural logarithms,
natural logarithm function ln, e = 2.718 281 . . . .
with base e, was introduced in
Chapter A3, Subsection 4.3.
Figure 5.5 Approaching e
50
In Subsection 5.3 you saw how an infinite geometric series can be summed,
provided that the common ratio is less than 1 in magnitude. Some infinite
series which are not geometric can also be summed, and many fascinating
formulas arise in this way; for example,
∞
� (−1)i 1 1
=1− 3
+ 5
−
17 + · · ·
= 14 π,
i=0
2i + 1
∞
� (−1)i+1 1
=1− 2
+
13 − 1
4
+ · · ·
= ln 2.
i=1
i
∞
�
If ai (i = 1, 2, 3, . . .) is a sequence, then the sum ai can exist only if ai
i=1
becomes smaller and smaller in the long term, that is, only if lim ai = 0.
i→∞
However, this requirement of convergence to 0 for the sequence ai is not on
its own sufficient to guarantee the existence of a sum for the infinite series.
For example, the sequence 1, 12 ,
13 ,
14 , . . . converges to 0 but, as indicated
below, there is no sum for the infinite series 1 + 12 + 13 + 14 + · · · .
A non-existent sum
The following brief argument (by contradiction) shows that the infinite
series
1 1
1+ 2
+ 3
+
14 + · · ·
has no sum.
Suppose that the sum of the infinite series exists and is equal to s; that is,
1 1
s=1+ 2
+ 3
+
14 + · · · .
Multiply both sides by
12 , to obtain These manipulations of
1 1 1 infinite series can be shown to
2
s = 2
+ 4
+
16 + 1
8
+ ···. (5.8) be valid if the series have
The right-hand side is a subseries of the original. On subtracting it from sums.
the original, we have
1 1 1 1
2
s =1+ 3
+ 5
+ 7
+ ···. (5.9)
But now each term in series (5.9) is larger than its counterpart in
series (5.8): 1 > 21 ,
13 > 14 , and so on. Hence the sum of series (5.9) must be
greater than that of series (5.8) – but each is equal to 12 s! This
contradiction shows that the original series has no sum.
the concepts of convergence and limit for sequences, with the notation
lim an for the limit of a convergent sequence an ;
n→∞
for sequences defined by recurrence systems, the method of identifying
potential limit values by finding all constant sequences that satisfy the
recurrence relation;
for sequences an defined by closed-form formulas, methods of
reasoning based on how individual parts of the formula behave as n
becomes large;
the Reciprocal Rule and Constant Multiple Rule for sequences;
51
the behaviour for large n of the ‘basic sequences’ r n and np , as

summarised in the box on page 47;
a formula (a/(1 − r)) for the sum of any infinite geometric series,
∞
�
ari , with |r| < 1;
i=0
a method for finding the fraction equivalent to a given recurring
decimal.

Exercise 5.1
Consider the sequences defined by the logistic recurrence relation (3.3)
with the parameters stated below. Basing your answers on the
investigations in Section 4, decide in each case whether or not the sequence
converges and, if it does, to what limit. Also, for each case where the
sequence converges, state whether Pn ever exceeds E.
(a) r = 1.8, E = 200, P0 = 10
(b) r = 2.25, E = 1500, P0 = 8
(c) r = 0.4, E = 1000, P0 = 10
Exercise 5.2
The following sequences are given by closed-form formulas. In each case,
decide whether or not the sequence converges and, if it does, find the limit.
In each case, n = 1, 2, 3, . . . .
16n + 5
(a) an = 4 + (−0.5)n (b) an = n2 + 1/n (c) an =
2n − 1
Exercise 5.3
The following sequences are given by recurrence systems. In each case, use
the method of Example 5.1 and Activity 5.1(b) to find all possible limit
values for the sequence. Then use methods from Chapter A1 to find a
closed form for the sequence, and use this to determine the long-term
behaviour of the sequence. In each case, n = 0, 1, 2, . . . .
(a) x0 = 3, xn+1 = 2xn
(b) x0 = 3, xn+1 = 0.2xn
(c) x0 = 3, xn+1 = 0.3xn + 140
Exercise 5.4
1+ 1
9
+ ( 19 )2 + ( 19 )3 + · · ·
Exercise 5.5
Find the fraction equivalent to 0.513 513 513 . . . .
52
Summar y of Chapter B1
In this chapter you have seen sequences put to use in the development of
models for the economics of car ownership and for variations in animal
populations. In the context of the logistic model you saw that, while
numerical investigation has its uses, prior reasoning may profitably reduce
the scope of the task.
Logistic recurrence sequences exhibit an interesting variety of long-term
behaviour. The topic of what happens to sequences in the long term was
studied further from the mathematical point of view, with reference to
convergence and limits. These concepts may be applied both to sequences
and to series. The sum of either a finite or an infinite series may be
expressed concisely using sigma notation.
Lear ning outcomes

You have been working towards the following learning outcomes.
Terms to know and use

Sigma (sum) notation, series, proportionate growth rate, exponential
(geometric, Malthusian) model for populations, logistic model for
populations, logistic recurrence relation, equilibrium population level,
chaotic behaviour (of sequences), cycling, 2-cycle, 4-cycle, long-term
behaviour (of sequences), convergent sequence, converges, limit of a
sequence, arbitrarily large/small/close to, Reciprocal Rule and Constant
Multiple Rule for sequences, infinite (geometric) series.
Symbols and notation to know and use

n
� ∞
�
ai , lim an , ai .
n→∞
i=1 i=1
Modelling skills
Appreciate the roles of the various stages of the modelling cycle, as
summarised in Figure 1.1.
Interpret mathematical information obtained from a model in terms of
the original purpose or situation.
Evaluate a model in the light of what it predicts and how this
compares with the real situation being modelled.
Be aware of the possible need to revise a model on the basis of
evaluation of it.
53
Mathematical skills
Find the parameter values for the logistic model which correspond to
given information about the proportionate birth and death rates
(or growth rate) of a population.
Obtain information about a sequence generated by the logistic
recurrence relation, for given values of its parameters.
Recognise various possible forms of long-term behaviour for sequences,
in particular: convergence to a limit; cycling; chaotic behaviour; terms
which become arbitrarily large.
Apply and manipulate sigma notation, where appropriate.
n
�
Apply the formula i = 12 n(n + 1), where appropriate.
i=1
Apply, where appropriate, either of the formulas
n
� � �
i 1 − rn+1
ar = a 1),
(r =
i=0
1−r
∞
� a
ari = (|r| < 1).
i=0
1−r
Find the fraction which is equivalent to a given recurring decimal.
Mathcad skills
Calculate and graph terms of a logistic recurrence sequence, given a
starting value and values for the parameters.
Interpret the outcomes obtained from such calculations, in particular
as regards the long-term behaviour of the sequence.
Ideas to be aware of
The principle of simplicity (in mathematical modelling).
That the prior application of graphical or algebraic reasoning may
reduce the scope needed for a numerical investigation.
54
Solutions to Activities
Solution 1.1 (b) The recurrence system

You might, for example, consider: c0 = 600, ci+1 = ci +250 (i = 0, 1, . . . , n−1)
whether to buy new or used – if used, how old, has closed-form solution
and whether to buy from a dealer or privately;
ci = 600 + 250i (i = 0, 1, . . . , n).
the reliability of the type of car, both when new
and older, and its likely running costs; (c) From part (b), we have
n
� n
�
how the value of the car, to purchase or to resell,
ci = (600 + 250i)
varies with age; i=1 i=1
whether to borrow money to finance purchase. n
�
= 600n + 250 i (equation (1.4))
When looking at a particular vehicle, you would note i=1
features such as its mileage, colour, condition, etc.
= 600n + 250 × 12 n(n + 1) (equation (1.6))
2
Solution 1.2 = 125n + 725n
13 = 25n(5n + 29).
�
2 2 2 2
(a) The sum is 5 + 6 + · · · + 13 = i .
i=5
(b) Writing out the two sums, we have Solution 1.5

6
� 3
� We apply the results from Chapter A1.
5i − 5i
i=1 i=1 (a) The recurrence relation
2 3 4 5 6 2 3
= (5 + 5 + 5 + 5 + 5 + 5 ) − (5 + 5 + 5 ) vi+1 = rvi (i = 0, 1, . . . , n)
= 54 + 55 + 56 has closed-form solution
�6
= 5i . vi = v0 ri (i = 0, 1, . . . , n + 1).
i=4
(b) The recurrence relation
(c) The given sum is
ci+1 = ci + d (i = 0, 1, . . . , n − 1)
4
�
(600 + 250i) has closed-form solution
i=1
ci = c0 + di (i = 0, 1, . . . , n).
= (600 + 250 × 1) + (600 + 250 × 2)
+ (600 + 250 × 3) + (600 + 250 × 4) (c) From part (b), we have
n
� n
�
= 600 × 4 + 250(1 + 2 + 3 + 4)
4
ci = (c0 + di)
� i=1 i=1
= 600 × 4 + 250 i. n
i=1
�
= c0 n + d i (equation (1.4))
i=1
Solution 1.3 = c0 n + 12 dn(n + 1) (equation (1.6))
� �
From equation (1.6), the required sum is = n c0 + 12 d(n + 1) .
100
� (d) From the result of part (a), we have
1
i= 2 × 100 × 101 = 5050.
i=1 v1 − vn+1 = v0 r − v0 rn+1 = v0 r(1 − rn ).

From the result of part (c), we have
Solution 1.4 n
� � �
We apply the results from Chapter A1. ci = n c0 + 12 d(n + 1) .
i=1
(a) The recurrence system
Hence equation (1.7) provides the relationship
v0 = 16 000, vi+1 = 0.85vi (i = 0, 1, . . . , n) v0 r
an = (1 − rn ) + c0 + 12 d(n + 1) (n = 1, 2, . . .).
has closed-form solution n
vi = 16 000(0.85)i (i = 0, 1, . . . , n + 1).
55
Solution 2.1 This predicts that the 500 million population

level is passed between n = 174 and n = 175,
The long-term behaviour of
that is, between the anniversaries of the census
Pn = (1 + b − c)n P0 date in 1964 and 1965.
depends on the value of 1 + b − c, which is positive
for positive population sizes. Solution 3.1
If b > c, then we have 1 + b − c > 1, and Pn will (a) The proportionate growth rate R(P ) at
increase as n increases. This increase goes on forever, population size P is
becoming more and more rapid.
R(P ) = (2.65 − 0.0015P ) − 0.4
If b = c, then 1 + b − c = 1, and Pn stays constant = 2.25 − 0.0015P.
at P0 .
Hence the growth for the year which starts at
If b < c, then we have 0 < 1 + b − c < 1, and the time n is (2.25 − 0.0015Pn )Pn . Since this is also
value of Pn will decrease, getting closer and closer the increase from Pn to Pn+1 , we have the
to 0 as n increases. recurrence system
These three possible cases are illustrated in P0 = 8, Pn+1 − Pn = (2.25 − 0.0015Pn )Pn .
Figure 2.3, following the activity.
(b) The right-hand side of this recurrence relation
can be written as
Solution 2.2 � �
0.0015
2.25Pn 1 − Pn ,
(a) Take n = 0 at the census date in 1790. Then 2.25
n = 100 in 1890, so P0 = 4 and P100 = 63 (both
which is of the logistic form rPn (1 − Pn /E) with
in millions). According to the exponential model
r = 2.25 and E = 2.25/0.0015 = 1500.
(equation (2.2)), we have Pn = (1 + r)n P0 , so, in
particular,
P100 = (1 + r)100 P0 ;
Solution 3.2
The logistic recurrence relation is
that is,
� �
Pn
63 = (1 + r)100
× 4. Pn+1 − Pn = rPn 1 − ,
E
Solving this equation for r, we have
where in this case r = 0.028 from consideration of
� �0.01
63 the proportionate growth rate at low population
r= − 1 = 0.028 (to 2 s.f.), levels. Since the annual growth rate Pn+1 − Pn is
4
estimated to be 1.5 (million) for the population size
which is about 2.8% per year. Pn = 92 (million), we have
(In subsequent discussions of this model we take � � � �
92 92
r = 0.028.) 1.5 = 0.028 × 92 1 − = 2.576 1 − .
E E
(b) The exponential model for the US population in On solving this equation for E, we obtain
this period is therefore
92 1.5 1.076
Pn = 4(1.028)n . =1− = , so E 220.
E 2.576 2.576
(c) The year 1950 corresponds to n = 160, for which Hence the prediction, on the basis of the model and
the US population is predicted to be of data from 1790–1920, is for an equilibrium US
population level of 220 million.
P160 = 4(1.028)160 332 (million).
(d) To find when the population reaches 500 million,
Solution 3.3
we need to solve for n the equation
(a) The proportionate growth rate is
4(1.028)n = 500; that is, (1.028)n = 125.
220 − 100
Applying the natural logarithm function ln to = 1.2 when P0 = 100,
100
both sides and rearranging, we obtain

and
ln(125) 160 − 400
n= 174.8. = −0.6 when P0 = 400.

ln(1.028) 400
56
SOLUTIONS TO ACTIVITIES
(b) Since the proportionate growth rate has the Solution 5.2
form r(1 − P/E) for a population size P
(a) As n becomes large, 3n becomes arbitrarily
(equation (3.2)), we have
� � � � large, so this is also true of the sequence 2 + 3n.
100 400 By the Reciprocal Rule, therefore, the sequence
1.2 = r 1 − , −0.6 = r 1 − .
E E an = 1/(2 + 3n) converges, with limit 0; that is,
lim an = 0.
On dividing through each equation by r and n→∞
then rearranging, we obtain a pair of (b) As n becomes large, n3 and hence also 5 + n3
simultaneous linear equations in 1/r and 1/E: become arbitrarily large. The sequence
1.2 100 0.6 400 an = 5 + n3 does not converge.
+ = 1 and − + = 1.
r E r E
(c) The quantity (0.6)n becomes arbitrarily small as
Eliminating the 1/E term, by subtracting the
n becomes large. This is also true of 20(0.6)n , by
second equation from four times the first
the Constant Multiple Rule. Hence the sequence
equation, gives
4 + 20(0.6)n becomes close to 4 in the long term.
5.4 For large n, therefore, an = 100/(4 + 20(0.6)n )
= 3, that is, r = 1.8. becomes close to 100/4 = 25. In other words,
r
the sequence an converges, and lim an = 25.
On substituting this value for r into the first n→∞
equation, we obtain (d) The values given by this formula are (starting
1.2 100 with n = 1)
+ = 1;
1.8 E 2, 4, 2, 4, 2, . . . .
that is,
We obtain the value 2 whenever n is odd, and 4

100 1.2 1
whenever n is even. The values of an alternate
=1− = .
E 1.8 3 between 2 and 4, so the sequence does not

Solving this equation gives E = 300. converge.
(c) From part (b), the logistic recurrence relation is (e) The approach used in Example 5.2(c) is helpful
� � here. On dividing the top and bottom of the
Pn
Pn+1 − Pn = 1.8Pn 1 − . given expression for an by n, we have
300
60
If P0 = 200, then we have an = .
� � 3/n + 5
200
P1 = 200 + 1.8 × 200 1 − = 320.

Now 3/n converges to 0 for large n, so 3/n + 5
300 converges to 5. Hence, as n increases, an
Hence if the experiment is started with 200
becomes arbitrarily close to 60/5 = 12. We
beetles, then there will be 320 in the next
conclude that an converges and that
generation, according to the model.
lim an = 12.
n→∞
Solution 5.1 Solution 5.3

(a) The population has 0.3Pn + 8 ‘joiners’ each year This is an infinite geometric series, with a = 1 and
and 0.4Pn ‘leavers’. Hence its annual growth r = − 12 . Hence, by equation (5.6), the sum is
rate is −0.1Pn + 8, which equals the change
∞
�
from Pn to Pn+1 ; that is, 1
(− 12 )i = = 23 .
Pn+1 − Pn = −0.1Pn + 8. i=0
1 − (− 21 )
With the addition of the starting value, P0 = 50,
this is equivalent to the recurrence system

Solution 5.4
P0 = 50, Pn+1 = 0.9Pn + 8 (n = 0, 1, 2, . . .), Put s = 0.454 545 . . . . The repeating group, ‘45’, is
as required. 2 digits long, so multiply s by 102 , to obtain
(b) An equilibrium level c for the population 100s = 45.454 545 . . . = 45 + s.
corresponds to a constant sequence Pn = c Hence we have s = 45
= 5
99 11 .
which satisfies the recurrence relation. If there is
such a constant sequence, then we have
c = 0.9c + 8. Hence 0.1c = 8, so c = 80. The
equilibrium level for the population is 80.
57
Solutions to Exercises

(a) The sum may be written as (a) Take n = 0 at the census date in 1900. Then
10 n = 50 in 1950, so P0 = 1.65 and P50 = 2.52
�
1
+ ( 12 )2 + · · · + ( 12 )10 = ( 12 )i (both in billions). According to the exponential
2
i=1 model (equation (2.2)), we have
9
� Pn = (1 + r)n P0 , so, in particular,
1
� 1 �i
= 2 2 , P50 = (1 + r)50 P0 ;
i=0
so m = 0 and n = 9. that is,
(b) By the result of Example 1.1(c), with a = 1, 2.52 = (1 + r)50 × 1.65.

r = 12 and n = 9, the value of this sum is On solving this equation for r, we obtain
9 � �0.02
� 1 − ( 12 )10 2.52
1
( 12 )i = 1
× r= − 1 = 0.0085 (to 2 s.f.).
2
i=0
2
1 − 12 1.65
= 1 − ( 12 )10 This is just under 1% per year.
0.999 02. (b) With this value for r, the world population in
2000 (when n = 100) is predicted to be
Solution 1.2 P100 = 1.65(1.0085)100 3.85 (billion).
(a) The sum of the integers from 51 to 100,
inclusive, is
Solution 2.2
100
� 100
� 50
�
i= i− i. Take n = 0 this time in 1950, so that n = 50 in 2000,
i=51 i=1 i=1 P0 = 2.52 and P50 = 6.06 (both in billions). Then, as
before, we have P50 = (1 + r)50 P0 , which on this
Now, by equation (1.6), we have
occasion gives
100
�
i= 1
× 100 × 101 = 5050, 6.06 = (1 + r)50 × 2.52.
2
i=1 On solving this equation for r, we obtain
50
� � �0.02
1 6.06
i= 2 × 50 × 51 = 1275. r= − 1 = 0.018 (to 2 s.f.).
i=1 2.52
Thus the given sum is equal to This growth rate of almost 2% (for 1950–2000) is
5050 − 1275 = 3775. about twice as high as that found in the solution to
Exercise 2.1(a) for the period 1900–1950.
(b) The numbers given form a finite arithmetic
sequence whose terms can be expressed as
Solution 3.1
7 + 10i (i = 51, 52, . . . , 100).
(a) The proportionate growth rate has the form
Using equation (1.5), we obtain the sum of these
r(1 − P/E) for a population size P
terms:
(equation (3.2)). Hence (with P measured in
100
� 100
� billions) we have
(7 + 10i) = 7 × 50 + 10 i � � � �
3.02 5.27
i=51 i=51
0.20 = r 1 − , 0.15 = r 1 − ,
= 350 + 10 × 3775 (from (a))
E E
= 38 100.
which lead to
0.20 3.02 0.15 5.27
+ =1 and + = 1.
r E r E
58
SOLUTIONS TO EXERCISES
Eliminating the 1/r term, by subtracting 3 times (i) If 0 < r < 1, then 0 < 1 − r < 1, so (1 − r)n
the first equation from 4 times the second, gives is positive and decreases towards 0 as n
21.08 − 9.06 increases. Since Qn is the amount by which Pn
= 1; that is, E = 12.02. is below E, this predicts that the values of Pn
E
are always less than E, but approach E more
On substituting this value for E into the first
and more closely as n increases.
equation, we obtain
(ii) If 1 < r < 2, then −1 < 1 − r < 0, so
0.20 3.02
+ = 1; (1 − r)n is alternately positive and negative,
r 12.02
decreasing in magnitude towards 0 as n

that is,
increases. Hence Pn oscillates either side of E,
0.20 3.02 9 but again tends towards E.
=1− = .
r 12.02 12.02 (iii) If r > 2, then 1 − r < −1, so (1 − r)n

Solving this equation gives r 0.267.
oscillates in sign and increases in magnitude as n
(The estimated equilibrium population of the increases. Correspondingly, Qn moves further
world is therefore about 12 billion, which is away from 0, and Pn moves further from E.
roughly twice the level in 2000.)
(b) The logistic recurrence relation in this case is Solution 5.1
� � As you saw in Section 4, the convergence or
Pn
Pn+1 − Pn = 0.267Pn 1 − , otherwise of a logistic recurrence sequence depends
12.0
only on the value of the parameter r and, if it does
where n is measured in decades and Pn in converge, the limit is E. Solutions can be obtained
billions. If Pn = 6.06 (in 2000), then we have by reference to the table on page 40.
� �
6.06 (a) With r = 1.8, we have 1 < r < 2. The sequence
Pn+1 = 6.06 + 0.267 × 6.06 1 − 6.86.
12.0 converges, to the limit E = 200, with values
Thus, according to the model, the world eventually alternating above and below E.
population in 2010 is predicted to be 6.86 billion. (b) With r = 2.25 > 2, the sequence is not
convergent. (The long-term behaviour is, in fact,
Solution 3.2 that of a 2-cycle.)
(c) With r = 0.4 < 1, the sequence converges, to the
(a) If Qn = E − Pn , then we have Pn = E − Qn
limit E = 1000, with values always below E.
(and Pn+1 = E − Qn+1 ). Therefore the logistic
recurrence relation gives
(E − Qn+1 ) − (E − Qn )
Solution 5.2
� �
E − Qn (a) As n becomes large, (−0.5)n becomes arbitrarily
= r(E − Qn ) 1 − . close to 0. Hence an = 4 + (−0.5)n tends to 4.
E
In other words, the sequence an converges to the
Hence limit 4.
Qn − Qn+1 = r(E − Qn )(Qn /E); (b) As n becomes large, 1/n becomes arbitrarily
that is, close to 0 (by the Reciprocal Rule), but n2
becomes arbitrarily large. Hence an = n2 + 1/n
Qn+1 − Qn = −rQn (1 − Qn /E). becomes arbitrarily large, so the sequence an is
(b) If Qn /E is small compared with 1, then the not convergent.
above recurrence relation can be approximated (c) The approach used in Example 5.2(c) and
by Qn+1 − Qn = −rQn , or Activity 5.2(e) can be applied here. On dividing
Qn+1 = (1 − r)Qn , the top and bottom of the given expression
for an by n, we have
as required.
16 + 5/n
(c) The recurrence relation in part (b) defines a an = .
2 − 1/n
geometric sequence with closed form
Now 5/n and −1/n both converge to 0 for
Qn = (1 − r)n Q0 , large n, so 16 + 5/n converges to 16 and 2 − 1/n
where we assume that Q0 > 0 (that is, P0 < E). converges to 2. Thus the sequence an converges
to the limit 16/2 = 8.
59

(a) Suppose that a constant sequence xn = c This is an infinite geometric series, with a = 1 and
satisfies the given recurrence relation, r = 19 . Hence, by equation (5.6), the sum is
xn+1 = 2xn . This gives c = 2c; that is, c = 0. ∞
� 1
Thus the only possible limit value is 0. ( 19 )i = = 98 .
i=0
1 − 19
The recurrence system defines a geometric
sequence. From Chapter A1, Section 3, the
closed-form solution is
Solution 5.5
xn = 2n x0 = 3 × 2n .
Put s = 0.513 513 513 . . .. The repeating group, ‘513’,
As n increases, the values of xn grow arbitrarily is 3 digits long, so multiply s by 103 , to obtain
large, so this sequence is not convergent.
1000s = 513.513 513 513 . . . = 513 + s.
(b) Proceeding as in part (a), we obtain the
equation c = 0.2c for possible values of c in a Hence we have
constant sequence xn = c. Again, the only 513 57 19
s= = = .
possible limit value is 0. 999 111 37
This is a geometric sequence once more, with
closed form
xn = (0.2)n x0 = 3(0.2)n .
In this case, the values of xn tend to 0 as n
increases, so the sequence xn converges to the
limit 0.
(c) As before, we seek values of c for which the
constant sequence xn = c satisfies the given
recurrence relation, xn+1 = 0.3xn + 140. This
gives c = 0.3c + 140; that is, 0.7c = 140. Thus
c = 200, so the only possible limit value is 200.
This is a linear recurrence system, so we apply
the appropriate formula from Chapter A1,
Section 4. This says that the closed-form
solution for the linear recurrence system
x0 = a, xn+1 = rxn + d (r = 1)
is given by
� �
d d
xn = a + rn − .
r−1 r−1
In the current case, we have a = x0 = 3, r = 0.3
and d = 140. Hence the closed-form solution is
� �
140 140
xn = 3 + (0.3)n −
0.3 − 1 0.3 − 1
= −197(0.3)n + 200.
Now (0.3)n becomes arbitrarily small as n
becomes large so, by the Constant Multiple
Rule, the first quantity in the closed form also
becomes arbitrarily small. Thus
−197(0.3)n + 200 tends to 200 as n increases.
In other words, the sequence xn converges to the
limit 200.
Note that in each of parts (b) and (c), the limit
found via the closed form does agree with the
possible limit value established beforehand by
seeking a constant sequence.
60
Index
basic sequences 46
carrying capacity 32
chaotic behaviour 40
constant multiple rule 46
convergent sequence 42
cycling behaviour 40
divergent sequence 42
equilibrium population level 32
exponential model 22
4-cycle 40
geometric model 22
infinite geometric series 48
limit of a sequence 42
limits of a sum 11
logistic model 29
logistic recurrence sequence
long-term behaviour 40
Malthusian model 27
modelling cycle 7
parameter estimation 22, 32
population density 26
principle of simplicity 10
proportionate growth rate 22, 28
reciprocal rule 46
Ricker model 41
series 11
sigma notation 11
sum notation 11
2-cycle 40
61

MST121 Chapter B1 Modelling With Sequences

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MST121

Modelling with sequences

About this course

The Open University, Walton Hall, Milton Keynes, MK7 6AA.

All rights reserved. No part of this publication may be reproduced, stored in a

1 Modelling car ownership 7

1.1 Constructing a model 8

1.2 Mathematical interlude: sums 11

1.3 Back to the model 14

2 Populations: exponential model 18

2.1 Purpose of general population models 18

2.2 The exponential model 18

2.3 Evaluating the exponential model 23

3 Populations: logistic model 28

3.1 Setting up the logistic model 28

3.2 Finding values for the parameters 32

3.3 Investigating the logistic model 35

4 Logistic recurrence sequences on the computer 39

5 Sequences and limits 42

5.1 Sequences from recurrence systems 42

5.2 Sequences from closed-form formulas 44

5.3 Sequences from sums 48

5.4 Some further sequences and series 50

Block B develops the theme of mathematical modelling, which was

There are ﬁve sections in this chapter. They are

The optional Video Band B(i)

You saw in Chapter A1 that sequences can be used in a variety of

Being able to make use of mathematics to solve problems is an important

Figure 1.1 The modelling cycle

When you evaluate a model, you need to decide whether or not it is

1.1 Constr ucting a model

Specify the purpose

Activity 1.1 Listing features

purpose must be clariﬁed. It is sensible here to concentrate on features that lend

themselves to mathematical description. This suggests that we should

as the colour and condition of the vehicle should be discounted for

Items which are measurable include the following:

age and mileage at which car is bought;

age and mileage at which car is sold;

cost of the car at purchase, and its resale value;

Although this list of features is restricted to measurable quantities, it is

that is considered to be of most importance, and then to identify what

other quantity or quantities it mainly depends upon. An item of

considerable interest to many is the cost of motoring, so it seems natural

here to concentrate upon that. Thus we aim to construct a model which

How can this be achieved? If cost is to be the output, or dependent

variable, within the model, what inputs or independent variables does it

For a ﬁrst model, it is appropriate to simplify as far as possible. Let us

Create the model Create

1.2 Mathematical interlude: sums

Example 1.1 Using sigma notation

Express each of the following sums using sigma notation:

convention that 00 = 1 is 1 − rn+1

Here are some sums for you to manipulate.

Activity 1.2 Using sigma notation

as a single sum (with one sigma).

The outcome of Activity 1.2(c) is a special instance of a result that is often

More generally, if the sum starts at m (where m ≤ n), then

On adding these two equations, with the addition of corresponding terms

Find the sum of the ﬁrst 100 positive integers.