1. Higher Unit 1 Outcome 4 Recurrence Relations Grow and Decay
Linear Recurrence Relation
Divergence / Convergence / Limits
Applications
Find a formula
Exam Type Questions

2. Recurrence Relations Outcome 4 Sequences A 5 9 13 17 …….
Higher
Sequences
A …….
B …….
C ……..
D ………
E ………
In the above sequences some have obvious patterns while others don’t however this does not mean that a pattern doesn’t exist.

3. Recurrence Relations u1 , u2 , u3 , …….., un-1 , un , un+1 , ……...
Outcome 4
Higher
Notation
Suppose we write the term of a sequence as
u1 , u2 , u3 , …….., un-1 , un , un+1 , ……...
where u1 is the 1st term, u2 is the 2nd term etc….
and un is the nth term ( n being any whole number.)
The terms of a sequence can then be defined in two ways

4. Recurrence Relations Outcome 4 Either
Higher
Either
Using a formula for the nth term, un
in terms of the value n Or
By expressing each term using the previous
term(s) in the sequence.
This is called a Recurrence Relation
Now reconsider the sequences at the start

5. Recurrence Relations Outcome 4 A 5 9 13 17 ……. Formula: un = 4n + 1
Higher
A …….
Formula: un = 4n + 1
So u100 = 4 X = 401
Recurrence Relation: un+1 = un with u1 = 5
u2 = u1 + 4 = = 9 u3 = u2 + 4 = = 13

6. Recurrence Relations B 3 6 12 24 …… Formula: un = 3 X 2n-1
Outcome 4
Higher
B ……
Formula: un = 3 X 2n-1
So u10 = 3 X 29 = 3 X 512 = 1536
Recurrence Relation: un+1 = 2un with u1 = 3.
u2 = 2u1 = 2 X 3 = 6, u3 = 2u2 = 2 X 6 = 12, etc

7. ie apart from 1st two, each term is the sum of the two previous terms.
Recurrence Relations
Outcome 4
Higher
C ……..
No formula this time but we have a special type of recurrence relation called a
FIBONACCI SEQUENCE.
Here u1 = 2 , u2 = then we have
u3 = u2 + u1 = = 5 , u4 = u3 + u2 = = 8
In general
un+2 = un+1 + un
ie apart from 1st two, each term is the sum of the two previous terms.

8. Quite a tricky formula but it does work ...
Recurrence Relations
Outcome 4
Higher
D ………
This sequence doesn’t have a recurrence relation but the terms can be found using the formula
un = n3 - n + 17
Quite a tricky formula but it does work ...
u1 = = 17
u2 = = = 23
u10 = = = 1007

9. Recurrence Relations E 2 3 5 7 11 ………
Outcome 4
Higher
E ………
This sequence is the PRIME NUMBERS
(NB: Primes have exactly two factors !!)
There is neither a formula nor a recurrence relation which will give us all the primes.

10. Higher
Growth & Decay
Outcome 4
Removing 15% leaves behind 85% or 0.85 which is called the DECAY factor.
Adding on 21% gives us 121% or 1.21 and this is called the GROWTH factor.
Growth and decay factors allow us a quick method of tackling repeated % changes.

11. Growth & Decay Outcome 4 Example 1
Higher
An oven contains bacteria which are being killed off at a rate of 17% per hour by a particular disinfectant.
(a) How many bacteria are left after 3 hours?
(b) How many full hours are needed so that there are fewer than 4000 bacteria?
Suppose that un represents the number of bacteria remaining after n hours.
Removing 17% leaves behind 83%
so the DECAY factor is and un+1 = un

12. Growth & Decay Outcome 4 (a) u0 = 10000
Higher
(a)
u0 = 10000
u1 = 0.83u0 = X = 8300
u2 = 0.83u1 = X = 6889
u3 = 0.83u2 = X = 5718
So there are 5718 bacteria after 3 hours.
(b)
u4 = 0.83u3 = X = 4746
u5 = 0.83u4 = X = 3939
This is less than 4000 so it takes 5 full hours to fall below 4000.

13. Growth & Decay
Outcome 4
Higher
Example 2
The population of a town is growing at a rate of 14% per annum.
If P0 is the initial population and Pn is the population after n years.
(a) Find a formula for Pn in terms of P0.
(b) Find roughly how long it takes the population to treble.
Adding on 14% gives us 114%
so the GROWTH factor is and Pn+1 = Pn

14. Growth & Decay Outcome 4 P1 = 1.14 P0
Higher
P1 = P0
P2 = P1 = X 1.14 P0 = (1.14)2 P0
P3 = P2 = X (1.14)2 P0 = (1.14)3 P0
So in general we have Pn = (1.14)n P0
If the population trebles then we need to have
Pn > 3 P0
or (1.14)n P0 > 3 P0
Dividing by P0
we get (1.14)n > 3

15. We now use a bit of trial and error along with the
Growth & Decay
Outcome 4
Higher
We now use a bit of trial and error along with the
^ or xy buttons on the calculator.
If n = 5 then (1.14)5 = 1.92… too small
If n = 9 then (1.14)9 = 3.25… too big
If n = 7 then (1.14)7 = 2.50… too small
If n = 8 then (1.14)8 = 2.85… too small but closest to 3.
From the above we can say it takes just over 8 years for the population to treble.

16. Linear Recurrence Relations
Outcome 4
Linear Recurrence Relations
Higher
Some recurrence relations take the form
un+1 = kun where k is a real no.
This leads to a formula for the nth term
un = kn u0 where u0 is the starting value.

17. Linear Recurrence Relations
Higher
Many recurrence relations take the form
un+1 = aun + b where a & b are real nos.
If we think about un+1 like y and un like x then we get
y = ax + b and this is basically the same as
y = mx + c which is the equation of a straight line
Hence the expression “Linear Recurrence Relations”
Many day to day scenarios can be modelled by this.

18. Linear Recurrence Relations
Example
Higher
A balloon contains 1500ml of air and is being inflated by mouth. Each puff inflates the balloon by 15% but at the same time 100ml of air escapes.
(i) Find a linear recurrence relation to describe this situation.
(ii) How much air is in the balloon after 5 puffs?
(iii) If the volume reaches 3 litres then the balloon will burst. How many puffs will this take?
(NB: 3litres = 3000ml)
(i) Suppose the starting volume is V0.
Adding 15% gives us 115% or X previous amount,

19. Linear Recurrence Relations
Higher
however we also lose 100ml so we have…
V1 = 1.15V
similarly V2 = 1.15V
and V3 = 1.15V
In general Vn+1 = 1.15Vn
(ii) We can now use this formula as follows
V0 = 1500
V1 = X
= 1625

20. Linear Recurrence Relations
Higher
So after 5 puffs the balloon contains 2343ml of air.
V2 = X
= 1769
V3 = X
= 1934
V4 = X
= 2124
V5 = X
= 2343
BANG!!!
(iii) continuing the above
V6 = X
= 2594
V7 = X
= 2883
V8 = X
= 3216
The balloon bursts on the 8th puff.

21. Linear Recurrence Relations
Higher
Example
A factory wishes to dump 150kg of a particular waste product into a local steam once per week.
The flow of the water removes 60% of this material from the stream bed each week.
However it has been calculated that if the level of deposit on the stream bed reaches 265kg then there will be a serious risk to the aquatic life.
Should the factory be allowed to dump this waste indefinitely?

22. Linear Recurrence Relations
Higher
Let An be the amount of waste deposited after n weeks.
So A0 = 150
Removing 60% leaves behind 40% or 0.4.
This means that A1 = 0.4A
Similarly A2 = 0.4A
In general we get the recurrence relation
An+1 = 0.4An + 150
and this gives us the following sequence…...

23. Linear Recurrence Relations
Higher
A0 = 150
A1 = 0.4 X
= 210
A2 = 0.4 X
= 234
A3 = 0.4 X =
A10 = 0.4 X =
When amount of waste reaches 250kg it stays at this.
Check: If An = 250 then An+1 = 0.4 X = 250
This is below the danger level so factory could be allowed to continue dumping.
We say that the sequence CONVERGES to a LIMIT of 250.

24. Divergence / Convergence/Limits
Higher
Consider the following linear recurrence relations
(a) un+1 = 2un + 4 with u0 = 3
u0 = 3
u1 = 10
u2 = 24
u3 = 52
u10 = 7164
u20 =
As n   un  
and we say that the sequence DIVERGES.

25. Divergence / Convergence/Limits
Higher
(b) un+1 = 0.5un + 4 with u0 = 3
u0 = 3
u1 = 5.5
u2 = 6.75
u3 = 7.375
u10 = 7.995
U20 = 7.999…..
As n   un  8
we say that the sequence CONVERGES to a limit of 8.
Check: if un = 8
un+1 = 0.5 X = 8

26. Divergence / Convergence/Limits
Higher
(c) un+1 = -2un + 4 with u0 = 3
u0 = 3
u1 = -2
u2 = 8
u3 = -12
u10 = 1708
u20 =
u21 =
As n   un  ±
and we say that the sequence DIVERGES.

27. Divergence / Convergence/Limits
Higher
(d) un+1 = -0.5un + 4 with u0 = 3
u0 = 3
u1 = 2.5
u2 = 2.75
u3 = 2.625
u10 = 2.666
u20 = 2.666
As n   un  22/3
we say that the sequence CONVERGES to a limit of 22/3.
Check: if un = 22/3
un+1 =- 0.5 X 22/3 + 4 = 22/3

28. Divergence / Convergence/Limits
Higher
Conclusions
The linear recurrence relation un+1 = aun + b converges to a limit if either
-1 < a < 0 or 0 < a < 1
This is usually written as 0 < a < 1
If a > 1 ie a < -1 or a > 1
Then we say that the sequence diverges.

29. Divergence / Convergence/Limits
Higher
Other Factors
(e) compare this with (b)
un+1 = 0.5un with u0 = 3
u0 = 3
u1 = 11.5
u2 = 15.75
u3 =
…..
u10 = ……
u20 = 19.99….
Conclusion:
if un+1 = aun + b converges to a limit then changing b changes the limit.
This is clearly heading to a limit of 20
Check: if un = 20
un+1 = 0.5 X = 20

30. Divergence / Convergence/Limits
Conclusion:
if un+1 = aun + b converges to a limit then changing u0 does not affect the limit.
Divergence / Convergence/Limits
Higher
(f) compare this with (b)
un+1 = 0.5un + 4 with u0 = 200
u0 = 200
u1 = 104
u2 = 56
u3 = 32
u10 =
u20 = ….
Again this is heading to a limit of 8

31. Applications
Higher
Example 1
A hospital patient is put on medication which is taken once per day. The dose is 35mg and each day the patient’s metabolism burns off 70% of the drug in her system. It is known that if the level of the drug in the patients system reaches 54mg then the consequences could be fatal. Is it safe for the patient to take the medication indefinitely?
We need to create a recurrence relation.
First dose = u0 = 35
Burning off 70% leaves behind 30% or 0.3
After this another 35mg is taken so we have …..

32. Applications Conclusion:
Higher
Conclusion:
the level of drug in the patients system will never exceed 50mg under these conditions. Since this is below the danger level it would be safe to continue indefinitely.
un+1 = 0.3un + 35
This sequence has a limit since 0 < 0.3 < 1
If we call the limit L then at this limit we have
un+1 = un = L
The equation un+1 = 0.3un now becomes
L = 0.3L + 35
0.7L = 35
L = 35  0.7
= 350  7 = 50

33. Applications
Higher
Example 2
The brake fluid reservoir in a car is leaky. Each day it loses 3.14% of its contents. To compensate for this daily loss the driver “tops up” once per week with 50ml of fluid. For safety reasons the level of fluid in the reservoir should always be between 200ml & 260ml. Initially it has 255ml.
(a) Find a recurrence relation to describe the above.
(b) Determine the fluid levels after 1 week and 4 weeks.
(c.) Is the process effective in the long run?

34. Applications (a) Problem 3.14% daily loss = ? Weekly loss.
Higher
(a) Problem % daily loss = ? Weekly loss.
Losing 3.14% daily leaves behind 96.86% or
Amount remaining after 1 week = (0.9686)7 X A0
= X A0
= X A0
or 80% of A0
This means that the car is losing 20% of its brake fluid weekly
So if An is the fluid level after n weeks then we have
An+1 = 0.8 An + 50

35. Applications A0 = 255ml A1 = 254ml 1st week A2 = 253.2ml A3 = 252.6ml
Higher
(b) Using An+1 = 0.8 An with A0 = 255 we get
A0 = 255ml
A1 = 254ml st week
A2 = 253.2ml
A3 = 252.6ml
A4 = 252.0ml th week
NB : even before adding the 50ml
the level is above 200ml

36. Applications (c) considering An+1 = 0.8 An + 50
Higher
(c) considering An+1 = 0.8 An + 50
Since 0 < 0.8 < 1 then a limit must exist and at this An+1 = An = L so
An+1 = 0.8 An + 50
ie L = 0.8L + 50
or L = 50
or L = 50  0.2
= 500  2 = 250
In the long run the weekly level will be 250ml and won’t fall below 200ml so the driver should be OK with this routine.

37. A recurrence relation is defined by the formula
Finding a Formula
Example
Higher
A recurrence relation is defined by the formula
un+1 = aun + b
Given that u6 = 48 , u7 = 44 and u8 = 42
then find a & b .
u8 = au7 + b becomes
44a + b = 42
Sim.
equations
u7 = au6 + b becomes
48a + b = 44
Subtract up
4a = 2
so a = 0.5
Now put a = 0.5 into 44a + b = 42
to get b = 42
so b = 20

38. Finding a Formula Example
Higher
Example
The nth term in a sequence is given by the formula
un = an + b
Given that u10 = 25 and u12 = 31 then find a & b.
Hence find u300 - the 300th term.
Using un = an + b
u10 = 10a + b becomes
10a + b = 25
Sim. equations
u12 = 12a + b becomes
12a + b = 31
subtract up
2a = 6
a = 3

39. Finding a Formula Now put a = 3 into 10a + b = 25
Higher
Now put a = 3 into 10a + b = 25
This gives us b = 25
So b = -5
The actual formula is un = 3n - 5
So u300 = 3 X = 895

40. In general un = u1 + (n-1) X d
Two Special Series
Higher
In an arithmetic series there is a constant difference between consecutive terms.
eg 8, 14, 20, 26, ……
here d = un+1 - un = 6
u1 = 8 = 8 + (0 X 6)
u2 = 14 = 8 + (1 X 6)
u3 = 20 = 8 + (2 X 6)
u4 = 26 = 8 + (3 X 6)
In general un = u1 + (n-1) X d
So for the above u100 = u1 + 99d
= 8 + (99 X 6)
= 602

41. Two Special Series
Higher
In a geometric series there is a constant ratio between consecutive terms.
eg 5, 10, 20, 40, ……
here r = un+1  un = 2
u1 = = 5 X 20
u2 = 10 = 5 X 21
u3 = 20 = 5 X 22
u4 = 40 = 5 X 23
In general un = u1 X r(n-1)
So for the above u100 = u1 X r 99
= 5 X 299
= 3.17 X 1030

42. Higher Maths Sequences Strategies Click to start
Higher Maths
Strategies
Sequences
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43. Sequences The following questions are on
Maths4Scotland Higher
The following questions are on
Sequences
Non-calculator questions will be indicated
You will need a pencil, paper, ruler and rubber.
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44. Limit = 16½ Maths4Scotland Higher
A recurrence relation is defined by where -1 < p < -1 and u0 = 12
a) If u1 = and u2 = find the values of p and q
b) Find the limit of this recurrence relation as n  
Put u1 into recurrence relation
Put u2 into recurrence relation
Solve simultaneously:
(2) – (1)
Hence
State limit condition
-1 < p < 1, so a limit L exists
Hint
Use formula
Limit = 16½
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45. un = height at the start of year
Maths4Scotland Higher
A man decides to plant a number of fast-growing trees as a boundary between his property and the property of
his neighbour. He has been warned however by the local garden centre, that during any year, the trees
are expected to increase in height by 0.5 metres.
In response to this warning, he decides to trim 20% off the height of the trees at the start of any year.
If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run.
His neighbour is concerned that the trees are growing at an alarming rate and wants assurance
that the trees will grow no taller than 2 metres. What is the minimum percentage that the trees
will need to be trimmed each year so as to meet this condition.
un = height at the start of year
Construct a recurrence relation
State limit condition
-1 < 0.8 < 1, so a limit L exists
Use formula
Limit = 2.5 metres
Use formula again
Hint
m = 0.75
Minimum prune = 25%
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46. u0 = 2500 Maths4Scotland Higher Construct a recurrence relation
On the first day of March, a bank loans a man £2500 at a fixed rate of interest of 1.5% per month.
This interest is added on the last day of each month and is calculated on the amount due on the first day of the
month. He agrees to make repayments on the first day of each subsequent month. Each repayment is £300
except for the smaller final amount which will pay off the loan.
 a) The amount that he owes at the start of each month is taken to be the amount still owing just after the
monthly repayment has been made.
Let un and un+1 and represent the amounts that he owes at the starts of two successive months.
Write down a recurrence relation involving un and un+1
b) Find the date and amount of the final payment.
u0 = 2500
Construct a recurrence relation
Calculate each term in the recurrence relation
1 Mar u0 =
1 Apr u1 =
1 May u2 =
1 Jun u3 =
1 Jul u4 =
1 Aug u5 =
1 Sept u6 =
1 Oct u7 =
1 Nov u8 =
1 Dec Final payment £290.68
Hint
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47. Limit = 25 Maths4Scotland Higher
Two sequences are generated by the recurrence relations and
The two sequences approach the same limit as n  .
Determine the value of a and evaluate the limit.
Use formula for each sequence
Sequence 1
Sequence 2
Equate the two limits
Cross multiply
Simplify
Solve
Hint
Deduction
Since limit exists a  1, so
Limit = 25
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48. Two sequences are defined by the recurrence relations
Maths4Scotland Higher
Two sequences are defined by the recurrence relations
If both sequences have the same limit, express p in terms of q.
Use formula for each sequence
Sequence 1
Sequence 2
Equate the two limits
Cross multiply
Rearrange
Hint
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49. Smallest value of n is 8; value of 8th term = 1749.8
Maths4Scotland Higher
Two sequences are defined by these recurrence relations
a) Explain why only one of these sequences approaches a limit as n  
b) Find algebraically the exact value of the limit.
c) For the other sequence find
i) the smallest value of n for which the nth term exceeds 1000, and
ii) the value of that term.
First sequence has no limit since 3 is not between –1 and 1
Requirement for a limit
2nd sequence has a limit since –1 < 0.3 < 1
Sequence 2
u0 = 1
u1 =
u2 =
u3 =
u4 = 65
u5 =
u6 =
u7 =
List terms of 1st sequence
Hint
Smallest value of n is 8; value of 8th term =
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Maths4Scotland Higher
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51. Maths4Scotland Higher
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