Transcript Document

Higher Mathematics
Unit 1.4
Recurrence Relations
The General Term of a
Sequence
A sequence is just a pattern of numbers defined by some given
rule.
This rule should allow you to determine any term of the
sequence quite clearly.
There are two ways of defining patterns or sequences of
numbers.
Continue

The General Term of a
Sequence
• The first is to give a general formula explicitly
which defines the nth term, un, of the sequence in
terms of n itself.
• The second is to give a formula which determines
the next value in the sequence from the previous
one. This is an implicit formula i.e. each value is
implied from the one before.
Continue

Definitions and Terminology
Often the most confusing aspect of this topic is the
terminology and notation used, in order to help you with
this here are some pointers!!
n

term number i.e. first, second etc.
un

the value of the “current term”
un+1 
the value of the “next term”
un-1 
the value of the “previous term”
u0

the value of the “starting term”
u1

the value of the “first term” some
times the “starting term”
Continue

Definitions and Terminology
Note :
Sometimes the starting value is referred to as u0 other
times as u1. You should be comfortable using either
notation.
Just remember that the 2nd term may therefore be u1 or u2
respectively!!
Continue

Explicit and Implicit
Formulae
There are two ways of defining patterns or sequences of
numbers:(a) Explicit Formulae (as in Worked Example 1)
(b) Implicit Formulae (known as Recurrence Relations),
Explicit Formulae
un is given (explicitly) in terms of n. This means it is easy
to find the actual value of any term e.g. u100, without
knowing the earlier ones.
This was shown in Worked Example 1.
Continue

Implicit Formulae
This means each term is determined (or implied) from
the previous term.
e.g. (a)
un+1 = 5un
(u0 = 2)
(b)
un
= un-1 – 3
(u0 = 100)
(c)
un+1 = 0·8un + 10
(u0 = 20)
They are known as Recurrence Relations as they form a
recurring sequence based on the previous term.
Linear Recurrence Relations
So far we have looked at the following types of recurrence
relations :(multiplying)
un+1 = 2un
(u1 = 5)
(adding)
un+1 = un + 3
(u1 = 5)
What happens when we combine these ideas ?
(multiplying and adding)
(u1 = 5)

un+1 = 2un + 3
Continue

Linear Recurrence Relations
This is called a Linear Recurrence Relation, and in
general looks like
un+1 = aun + b
(can you see why it`s called linear ?)
(y = mx + c ?)
Note :
It is much more difficult to find a general
(explicit) formula for un this time, though you
should not be asked to do this by the SQA.
Generating recurrence
relations on a Graphic
Calculator.
A graphic calculator provides an ideal platform to rapidly
calculate and display terms of a recurrence relation.
Here is how to do it.
Key the value of u0 and [ENTER] to get initial value.
Construct relation using ANS facility
[2nd][(-)]
Press [ENTER] to get next value.
Press [ENTER] to get next value.
Continue

Generating recurrence
relations on a Graphic
Calculator.
Use the same recurrence relation as used in example 5.
Vn+1 = 0·6Vn + 30
(V0 = 100)
Enter the
start value
i.e. 100
[ENTER]
Construct
relation
using ANS
facility
[2nd][(-)]
Press
[ENTER] to
get next
value
.
Press
[ENTER] to
get next
value
.
Continue

Concept of a Limit.
I work in an old shabby classroom and my Headteacher
agrees to let me paint a wall.
He says that I can only paint half of the unpainted wall each
day.
I agree!
How long will it take to completely paint the wall?
Continue

Concept of a Limit.
Day 4
Day 3
Day 5
Day 6
Day 1
Day 2
My Wall
Continue

Concept of a Limit.
So on day 1, I paint ½
On day 2, I paint ¼
On day 3, I paint ⅛ and so on.
So my total is given by
1 1 1 1
S      ...
2 4 8 16
S will never total 1 so I will never actually finish painting the
room, although it might appear that I do. There will always
be a little space if I do it properly. That is the idea of a limit
– the end is never reached, it only appears to be.
If you consider that you would probably just dab the little
space left at some point, that is the same as rounding your
answer.
Continue

Concept of a Limit.
Mathematically a limit is when each term of a sequence
approaches the same value.
Will every recurrence relation have a limit?
How can we tell?
We will consider some relations and look at the resulting
graphs to establish a pattern.
The existence of a limit should be obvious graphically.
Continue

Different values of a
We will look at different recurrence relations and compare
them graphically to consider whether a limit may exist or not.
We only need consider relations of the type: un+1 = aun + b
The values we need to consider are:
a>1
a=1
0<a<1
a=0
−1 < a < 0
a=−1
a<−1
Continue

a > 1
un=2·5un-1+ 20
u(n)=2.5u(n-1)+20
2500
n
u(n)
1
10
2
45
3
132.5
4
351.25
5
898.125
6
2265.31
7
5683.28
8
14228.2
9
35590.5
2000
u(n)
1500
1000
500
0
0
1
2
3
4
5
6
n
7
as n gets larger u(n) moves to ∞
Continue

a = 1
un=1un-1+ 20
u(n)=1u(n-1)+20
250
200
u(n)
150
100
50
0
0
2
4
6
n
8
10
12
n
u(n)
1
10
2
30
3
50
4
70
5
90
6
110
7
130
8
150
9
170
10
190
11
210
as n gets larger u(n) moves to ∞
Consider if the constant term = 0, i.e. don’t add 20.
When a = 1 the limit problem is ambiguous.
Continue

0 < a < 1
n
u(n)
1
10
2
26
3
35.6
40
4
41.36
30
5
44.816
20
6
46.8896
7
48.1338
8
48.8803
9
49.3282
10
49.5969
11
49.7581
u(n)=0.6u(n-1)+20
60
50
u(n)
un=0.6un-1+ 20
10
0
0
2
4
6
n
8
10
12
as n gets larger u(n) moves to 50 . This is a limit.
Continue

a = 0
un=0un-1+ 20
u(n)=0u(n-1)+20
25
20
u(n)
15
10
5
n
u(n)
1
10
2
20
3
20
4
20
5
20
6
20
7
20
8
20
9
20
10
20
11
20
0
0
2
4
6
n
8
10
12
as n gets larger u(n) moves to 20 . This is a limit.
Continue

−1 < a < 0
n
u(n)
1
2
9
2
9.2
8
3
6.32
6
4
7.472
5
5
7.0112
3
6
7.19552
2
7
7.12179
8
7.15128
9
7.13949
10
7.14421
11
7.14232
u(n)=-0.4u(n-1)+10
10
7
u(n)
un=-0.4un-1+ 10
4
1
0
0
2
4
6
8
10
12
n
as n gets larger u(n) moves to 7.142857…
This is known as an oscillating graph. It is a limit.
Continue

when a = −1
un = −1un-1 + 20
n
u(n)
1
9
2
11
3
9
4
11
6
5
9
4
6
11
2
7
9
8
11
9
9
10
11
u(n)=-1u(n-1)+20
12
10
u(n)
8
0
0
2
4
6
n
8
10
12
This is an oscillating graph. It will not settle on a limit.
Consider if u0 = 10, though.
When a = −1 the limit problem is ambiguous.
Continue

And if a < -1?
n
u(n)
400
1
9
300
2
4.7
200
3
12.01
4
-0.417
5
20.7089
-200
6
-15.2051
-300
7
45.8487
8
-57.9428
9
118.503
10
-181.455
11
328.473
12
-538.404
u(n)=-1.7u(n-1)+20
100
0
u(n)
un = −1.7un-1+20
-100 0
2
4
6
8
10
12
14
-400
-500
-600
n
Note that this graph heads off towards both
+ ∞ and – ∞ as successive values change
sign.
Continue

Establishing if a recurrence
relation will have a Limit.
For every linear recurrence relation:-
un+ 1 = aun + b
un  L (Limit), as n   , if –1 < a < 1
and
un will NOT tend to a limit (other than 0) generally if a > 1
(or a < –1)
i.e A limit exists if a is a positive or negative fraction
Here`s how to find the limit very quickly and very
easily !!
Establishing if a recurrence
relation will have a Limit.
There are three reasons you could give an exam marker for
why you believe that the recurrence relation is producing
terms that appear to be approaching a limit.
Continue

Establishing if a recurrence
relation will have a Limit.
Reason 1
though the terms are getting smaller,
the actual amount they are decreasing by each time is
getting smaller also and is tending to zero.
Reason 2
the graph indicates that un is tending
to a limit (L) as n  .
Reason 3
the multiplication factor a = 0·7 (in
un+1 = 0·7un + 10) and when -1< a < 1, automatically
gives a recurrence relation with a limit.
Continue

Establishing if a recurrence
relation will have a Limit.
Important :In order to maximise the marks you get in an exam, you
must give one of the above reasons for why a limit exists
before actually proceeding to find the limit.
Continue

Finding the Limit (L).
Once you have established that a limit exists it is very easy
to find the limit (L).
Can you see that if the sequence tends to a limit, then
eventually after a while,
un = un+1 = un+2 = un+3 = ………
L
!!!!!!!
(i.e. all the terms (after a while) will effectively be the same
as each other (to a given number of decimal places) and will
have the value L.)
Continue

Finding the Limit (L).
To find L:
Simply replace both un and un+1 in the recurrence
relation by the letter L and solve the equation.
Problems using Recurrence
Relations
Problems involving recurrence relations often require the use
of common sense and a careful reading of the question!
A typical scenario requires the construction of an appropriate
relationship where the important quantity is the residue
rather than the quantity given.
This would typically be offered as a percentage. It means
that the residue is the value remaining after subtraction from
100%.
Let’s look at an example:
Continue

Problems using Recurrence
Relations
The local pond is badly polluted. It is thought to contain 15
tonnes of noxious chemicals which need to be removed.
The cleaning process reduces the chemicals by 28%
each week. The pond still suffers an additional 1 tonne of
chemicals each week.
a) Write a recurrence relation for the amount of chemical
present at the end of each week.
b) The local authority would like the content to fall below 3
tonnes. What advice should you give them about the
time needed for that to happen?
Continue

Problems using Recurrence
Relations
If 28% of the chemicals are removed then (100 – 28)%
remain. This figure is used in the recurrence relation.
a) Chemical remaining is 72%, u0 = 15 tonnes
Relation un+1 = 0.72 un + 1
b) Calculate a few terms to establish
the pattern….
The value of the linear coefficient (0.72)
is important. -1 < a < 1 shows that a limit
will exist for this situation. This is also
suggested by the table of values.
n
un
0
15
1
11.8
2
9.496
3
7.83712
4
6.642726
5
5.782763
6
5.163589
7
4.717784
8
4.396805
9
4.165699
Problems using Recurrence
Relations
To calculate the limit solve the problem
L = 0.72 L + 1
1L – 0.72L = 1
0.28L = 1
L = 1 ÷ 0.28
L = 3.571428571…
The authority would need to rethink their strategy as the
amount of chemical pollutant will never fall to 3 tonnes.