Higher Mathematics Unit 1.4
Linear Recurrence Relations
Contents
1.
1.
2.
3.
4.
5.
6.
7.
Dealing with Percentages.
Mathematical Models of Type u n1  au n  b .
Investigating u n1  au n  b .
Evaluating the Limit.
More Modelling.
Some General Points.
References/Resources.
Dealing with Percentages
At “C” level, questions can be done arithmetically, but at “H” level a more general approach is
required.
It is best to consider a specific problem such as:
“£1200 in a deposit account just now. Assuming 7% growth p.a., how much will there be in the
account after 1yr, 2yr, 3yr, 10yr, 50yr, n yr from now?”
The arithmetical approach of calculating 7% each time and adding it to the previous total is
hopelessly inefficient here, so we must adopt a more general approach:
7
 0.07
100
Hence, after 1yr,
7% 
Amount(£)  1200  7% of 1200
 11200  0.07 1200
 (1.07)1200
This shows that to increase £1200 by 7% we multiply by 1.07
Continuing the above argument, we get:
Amount after 2yr(£)  1.07  Amount after 1 yr
 (1.07)(1.07)1200
 (1.07) 2 1200
Continue in this way to obtain the general result
yr n
(1.07) n 1200
Do sufficient examples on this topic; able candidates will pick this up quickly (in fact, they will
know it from their S4 studies.), but weaker candidates will need more practice.
Decay
“10000 trees in a small wood. 3% die each year with no replacement. How many trees after 1yr,
2yr, 3yr, 10 yr, 100yr, n yr?”
The best analysis is
And, of course, 97%=0.97
3% die  97% survive.
So we have:
yr0
10000
yr1
(0.97)10000
yr2
(0.97) 2 10000
yr3
(0.97) 310000
And in general, after n years, the number of trees is given by (0.97) n 10000 .
Notes
2.
(i)
When using calculator to obtain a numerical answer, pupils should be encouraged to round
the final answer sensibly: e.g. in the “tree” example, above, with n =10, the computed
value of 7374.24….. would be rounded to 7374.
[An exception to this would be financial transactions involving banking, where answers
are usually given to the nearest pence.]
(ii)
It is instructive to make small changes to the percentage rate of growth/decay and see how
this affects the final answer. For example, in the case of the trees, above, changing the rate
of decay by just half a percentage point makes a significant difference to the final answer.
This demonstrates why it is sensible to round the final answer.
Mathematical Models Involving L.R.R.
Before convergence/divergence is discussed, pupils should have practice at forming equations.
A typical example is:
£1000 in the bank on January 1st . Interest is paid on Jan 2nd at rate of 4% per annum and £100 is
paid into account in December 31st each year.
How much money in account on Jan 1st ; 1yr, 2yr, 3yr, 4yr from now?
If u k is the amount of money in account on Jan 1st of yr k , find an expression for u k 1 in
terms of u k .
Insist as k etc being written as a subscript – otherwise U k 1  U k  1
Other topics which could be considered here are pollution, litter, population of vermin, etc, etc.
Examples may be found in the handouts, past papers and textbooks.
When making up questions, care must be taken to avoid ambiguity.
“Flow charts” can be useful: e.g. in the above example, we have:
(a)
(b)
Old amount
un
Increase by 4 %
1.04
Thus u n1  1.04 u n  100
Add£100
 100
New amount
3.
Investigating Convergence
A rigorous analysis of convergence requires knowledge of geometric series and therefore must be
deferred until S6.
At this stage, pupils can only “fiddle about”, and this “investigative” approach should be
controlled by the teacher, in order to avoid time wastage.
Make use of the “answer” button on the calculator. For example, with the L.R.R.
u n1  0.1u n  9 ; u 0  100 , we would do:
100
ENTER
.1
Ans
+
9
ENTER
ENTER
ENTER ……
The approach should be to consider first “small” values of a, with sequences such as
and
u n1  0.2 u n  30 ,etc, etc.
u n1  0.1u n  9
Then consider “big” a, with sequences such as
and
u n1  3 u n  2
u n1  1.9 u n  3
A few examples with values of a such as 0.7 or – 0.8 should then be considered.
From this the rule will be seen:
 1  a  1  convergence .
Significance of Starting Value
By taking a rapidly converging sequence such as u n1  0.1u n  9 , and different starting values,
such as 20, 100, 500000, -5634, etc, etc., it will be seen that the limit is independent of the starting
value.
4.
Calculating the Limit
From “number crunching”, the pupils should see that for a convergent sequence, “When  goes
into the calculator,  comes out.”
Thus   a  b , in the usual notation.
Answers may be set out as follows: u n1  0.1u n  18 . Here  exists since -1 < 0.1 < 1
For the limit, we have
  0.1   18
  0.1   18
b
0.9   18
or use ….  
1 a
18

0.9
  20
Notes
(a)
(b)
Give reason for the existence of  .
Give pupils practice at evaluating exact values of limits without a calculator.
Question
(a)
The terms of a sequence satisfy U k 1  k u n  5 . Find the value of k which
produces a sequence with a limit of 4.
(b)
5.
A Sequence satisfies the recurrence relation U n1  m u n  5 , U 0 =3
(i)
Express U 1 and U 2 in terms of m.
(ii)
Given that U 2 = 7, find the value of m which produces a sequence with
no limit
1.
4  k 45
2.
k
3.
U1  3m  5
4.
U 2  m(3m  5)  5
(m(3m  5)  5  7)
5.
3m 2  5m  2  0
6.
(3m  1)( m  2)  0
7.
m  2
1
4
4
(2)
(5)
5
1 k
More Modelling
Past papers are a fine source of examples.
The procedure is:
Read question carefully.
Set up model.
State, with reasons, whether sequence is convergent.
If appropriate, find limit algebraically.
When a question requires us to find the “long term situation”, we have to find the limit
algebraically.
Be careful to include questions such as the following:
(1998 P II, Q 8(a))
A gardener feeds her trees with “Bioforce, the wonder plant food”. It is known that in a week the
amount of plant food in the tree falls by about 25%.
The trees contain no Bioforce initially and the gardener applies 1g of Bioforce to each tree every
Saturday. Bioforce is only effective when there is continuously more than 2g of it in the tree.
Calculate how many weekly feeds will be necessary before the Bioforce becomes effective.
Here, we must calculate the amount of Bioforce in the tree just before and just after each feed.
The results may be recorded in a simple table and the conclusion should be stated precisely.
6.
Some General Points
(i)
(ii)
Exact values should be found algebraically.
Give pupils practice at simplifying fractions by multiplying numerator and denominator by
5
50

a suitable number.
e.g.
.
0 .7 7
Always give reason for existence of a limit, if one exists.
(iii)
If a question requires a sequence of iterates to be stated then a table is useful.
(iv)
Questions such as “ u n1  a u n  b . Find an expression for u3 in terms of u 0 ” have not
come up in recent times, but pupils should be given practice at them anyway.
And, of course, a question on this could easily come up in future!
Exam questions tend to be less wordy. Now include simultaneous equations.
(v)
e.g. U n1  AU n  b
U1  3
U2  6
U 3  10
7.
References/Resources
1.
2.
H.H.M. pages 69 – 84.
MIA pages 84 -94.
6  3a  b
10  6a  b