1. Unit 1.4 Recurrence Relations
Higher Mathematics
Unit 1.4
Recurrence Relations

2. 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.
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3. 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.
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4. 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 Þ term number i.e. first, second etc.
un Þ the value of the “current term”
un+1 Þ the value of the “next term”
un-1 Þ the value of the “previous term”
u0 Þ the value of the “starting term”
u1 Þ the value of the “first term” some times the “starting term”
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5. 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!!
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6. 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),

7. 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.
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8. 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 (u0 = 20)
They are known as Recurrence Relations as they form a recurring sequence based on the previous term.

9. 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 (u1 = 5)
What happens when we combine these ideas ?
(multiplying and adding) Þ un+1 = 2un (u1 = 5)
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10. 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.

11. 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.
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12. Generating recurrence relations on a Graphic Calculator.
Use the same recurrence relation as used in example 5.
Vn+1 = 0·6Vn (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.
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13. 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?
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14. Concept of a Limit. My Wall Day 4 Day 3 Day 1 Day 6 Day 5 Day 2
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15. 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
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.
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16. 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.
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17. 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
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18. a > 1 un=2·5un-1+ 20 as n gets larger u(n) moves to ∞ Continue  n10 2 45 3
132.5 4
351.25 5 6 7 8 9
as n gets larger u(n) moves to ∞
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19. a = 1 un=1un-1+ 20 as n gets larger u(n) moves to ∞10 2 30 3 50 4 70 5 90 6
110 7
130 8
150 9
170
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.
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20. 0 < a < 1
un=0.6un-1+ 20 n
u(n) 1 10 2 26 3
35.6 4
41.36 5
44.816 6 7 8 9 11
as n gets larger u(n) moves to This is a limit.
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21. a = 0
un=0un-1+ 20 n
u(n) 1 10 2 20 3 4 5 6 7 8 9 11
as n gets larger u(n) moves to This is a limit.
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22. −1 < a < 0
un=-0.4un-1+ 10 n
u(n) 1 2
9.2 3
6.32 4
7.472 5
7.0112 6 7 8 9 10 11
as n gets larger u(n) moves to …
This is known as an oscillating graph. It is a limit.
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23. when a = −1
un = −1un n
u(n) 1 9 2 11 3 4 5 6 7 8 10
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.
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24. And if a < -1?
un = −1.7un-1+20 n
u(n) 1 9 2
4.7 3
12.01 4
-0.417 5 6 7 8 10 11 12
Note that this graph heads off towards both + ∞ and – ∞ as successive values change sign.
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25. 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 !!

26. 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.
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27. 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 ) and when -1< a < 1, automatically gives a recurrence relation with a limit.
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28. 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.
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29. un = un+1 = un+2 = un+3 = ……… ® L !!!!!!!
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.)
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30. 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.

31. 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:
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32. 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.
Write a recurrence relation for the amount of chemical present at the end of each week.
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?
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33. Problems using Recurrence Relations
If 28% of the chemicals are removed then (100 – 28)% remain. This figure is used in the recurrence relation.
Chemical remaining is 72%, u0 = 15 tonnes
Relation un+1 = 0.72 un + 1
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 15 1
11.8 2
9.496 3 4 5 6 7 8 9

34. 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 = …
The authority would need to rethink their strategy as the amount of chemical pollutant will never fall to 3 tonnes.