Transcript Higher Outcome 4

Higher Unit 1
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Higher
Outcome 4
Recurrence Relations
Grow and Decay
Linear Recurrence Relation
Divergence / Convergence / Limits
Applications
Find a formula
Exam Type Questions
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Recurrence Relations
Outcome 4
Higher
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Sequences
A
B
C
D
E
5
3
2
17
2
9
6
3
23
3
13
12
5
41
5
17
24
8
77
7
…….
…….
13
137
11
……..
………
………
In the above sequences some have obvious patterns
while others don’t however this does not mean that a
pattern doesn’t exist.
Recurrence Relations
Outcome 4
Higher
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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
Recurrence Relations
Outcome 4
Higher
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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
Recurrence Relations
Outcome 4
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A
5
Formula:
un = 4n + 1
So
9
13
17
…….
u100 = 4 X 100 + 1 = 401
Recurrence Relation:
u2 = u1 + 4 = 5 + 4 = 9
un+1 = un + 4 with u1 = 5
u3 = u2 + 4 = 9 + 4 = 13
Recurrence Relations
Outcome 4
Higher
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B
3
Formula:
So
6
12
24
……
un = 3 2n-1
X
u10 = 3 29 = 3 512 = 1536
X
Recurrence Relation:
u2 = 2u1 = 2 3 = 6,
X
X
un+1 = 2un
with u1 = 3.
u3 = 2u2 = 2 6 = 12, etc
X
Recurrence Relations
Outcome 4
Higher
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C
2
3
5
8
13
……..
No formula this time but we have a special type of
recurrence relation called a
FIBONACCI SEQUENCE.
Here
u1 = 2 , u2 = 3
then we have
u3 = u2 + u1 = 3 + 2 = 5 , u4 = u3 + u2 = 5 + 3 = 8
In general
un+2 = un+1 + un
ie apart from 1st two, each term is the sum of the
two previous terms.
Recurrence Relations
Outcome 4
Higher
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D
17
23
41
77
137 ………
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 = 13 - 1 + 17 = 17
u2 = 23 - 2 + 17 = 8 - 2 + 17 = 23
u10 = 103 - 10 + 17 = 1000 - 10 + 17 = 1007
Recurrence Relations
Outcome 4
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E
2
3
This sequence is the
5
7
11
………
PRIME NUMBERS
(NB: Primes have exactly two factors !!)
There is neither a formula nor a recurrence
relation which will give us all the primes.
Growth & Decay
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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.
Growth & Decay
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Higher
Example 1
Outcome 4
An oven contains 10000 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 0.83
and
un+1 = 0.83 un
Growth & Decay
Outcome 4
Higher
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(a)
u0 = 10000
u1 = 0.83u0 = 0.83 X 10000 = 8300
u2 = 0.83u1 = 0.83 X 8300 = 6889
u3 = 0.83u2 = 0.83 X 6889 = 5718
So there are 5718 bacteria after 3 hours.
(b)
u4 = 0.83u3 = 0.83 X 5718 = 4746
u5 = 0.83u4 = 0.83 X 4746 = 3939
This is less than 4000 so it takes 5 full hours to fall
below 4000.
Growth & Decay
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Higher
Example 2
Outcome 4
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 1.14
and
Pn+1 = 1.14 Pn
Growth & Decay
Outcome 4
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P1 = 1.14 P0
P2 = 1.14 P1 = 1.14 X 1.14 P0 = (1.14)2 P0
P3 = 1.14 P2 = 1.14 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
we get
(1.14)n P0 > 3 P0
(1.14)n > 3
Dividing by P0
Growth & Decay
Outcome 4
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We now use a bit of trial and error along with the
^ or xy buttons on the calculator.
If
If
If
If
n=5
n= 9
n= 7
n= 8
then
then
then
then
(1.14)5
(1.14)9
(1.14)7
(1.14)8
=
=
=
=
1.92…
3.25…
2.50…
2.85…
too small
too big
too small
too small but
closest to 3.
From the above we can say it takes just over 8 years
for the population to treble.
Linear Recurrence Relations
Outcome 4
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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.
Linear Recurrence Relations
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Outcome 4
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.
Linear Recurrence Relations
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Example
Outcome 4
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 1.15
X
previous amount,
Linear Recurrence Relations
Outcome 4
Higher
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however we also lose 100ml so we have…
V1 = 1.15V0 - 100
similarly
V2 = 1.15V1 - 100
and
V3 = 1.15V2 - 100
In general Vn+1 = 1.15Vn - 100
(ii)
We can now use this formula as follows
V0 = 1500
V1 = 1.15 X 1500 - 100
= 1625
Linear Recurrence Relations
Outcome 4
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V2
V3
V4
V5
=
=
=
=
1.15 X
1.15 X
1.15 X
1.15 X
1625 - 100 =
1769 - 100 =
1934 - 100 =
2124 - 100 =
1769
1934
2124
2343
So after 5
puffs the
balloon contains
2343ml of air.
(iii) continuing the above
V6 = 1.15 X 2343 - 100 = 2594
V7 = 1.15 X 2594 - 100= 2883
V8 = 1.15 X 2883 - 100 = 3216
The balloon bursts on the 8th puff.
BANG!!!
Linear Recurrence Relations
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Example
Outcome 4
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?
Linear Recurrence Relations
Outcome 4
Higher
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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 = 0.4A0 + 150
Similarly
A2 = 0.4A1 + 150
In general we get the recurrence relation
An+1 = 0.4An + 150
and this gives us the following sequence…...
Linear Recurrence Relations
Outcome 4
Higher
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A0 = 150
A1 = 0.4 X 150 + 150
= 210
A2 = 0.4 X 210 + 150
= 234
A3 = 0.4 X 234 + 150
= 243.6
A10 = 0.4 X 249.974 + 150 = 249.990
When amount of waste reaches 250kg it stays at this.
Check: If An = 250 then
An+1 = 0.4 X 250 + 150 = 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.
Divergence / Convergence/Limits
Outcome 4
Higher
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Consider the following linear recurrence relations
(a)
un+1 = 2un + 4 with u0 = 3
u0 = 3
As n  
un  
u1 = 10
u2 = 24
and we say that the
sequence DIVERGES.
u3 = 52
u10 = 7164
u20 = 7340028
Divergence / Convergence/Limits
Outcome 4
Higher
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(b)
un+1 = 0.5un + 4 with u0 = 3
u0 = 3
u1 = 5.5
As n  
un  8
u2 = 6.75
we say that the sequence
u3 = 7.375
CONVERGES to a limit of 8.
u10 = 7.995
U20 = 7.999…..
Check: if un = 8
un+1 = 0.5 X 8 + 4 = 8
Divergence / Convergence/Limits
Outcome 4
Higher
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(c)
un+1 = -2un + 4 with u0 = 3
u0 = 3
u1 = -2
u2 = 8
u3 = -12
u10 = 1708
u20 = 1747628
u21 = -3495252
As
n
un  ±
and we say that the
sequence DIVERGES.
Divergence / Convergence/Limits
Outcome 4
Higher
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(d)
un+1 = -0.5un + 4 with u0 = 3
u0 = 3
As n  
un  22/3
u1 = 2.5
we say that the sequence
u2 = 2.75
CONVERGES to a limit of 22/3.
u3 = 2.625
u10 = 2.666
u20 = 2.666
Check: if un = 22/3
un+1 =- 0.5 X 22/3 + 4 = 22/3
Divergence / Convergence/Limits
Higher
Outcome 4
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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.
Conclusion:
Divergence / Convergence/Limits
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if un+1 = aun + b converges
Outcome
to a limit 4then changing b
changes the limit.
Other Factors
un+1 = 0.5un + 4
with u0 = 3
(e)
compare this with (b)
un+1 = 0.5un + 10 with u0 = 3
u0 = 3
This is clearly heading
u1 = 11.5
to a limit of 20
u2 = 15.75
Check: if un = 20
u3 = 17.875
…..
un+1 = 0.5 X 20 +10 = 20
u10 = 19.98...
……
u20 = 19.99….
Conclusion:
Divergence / Convergence/Limits
Higher
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(f)
if un+1 = aun + b converges
Outcome
to a limit4 then uchanging
n+1 = 0.5un + 4
0 does
withthe
u0 = 3
compare thisuwith
(b)not affect
limit.
un+1 = 0.5un + 4 with u0 = 200
u0 = 200
u1 = 104
Again this is heading
u2 = 56
to a limit of 8
u3 = 32
u10 = 8.1875
u20 = 8.0001….
Find the Limit
Outcome 4
Higher
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Proof
Suppose a limit exists for the recurrence relation
un+1 = aun + b
let the limit be L, then we have
L = aL + b
Re arranging
L – aL = b
L(1 – a) = b
L=
b
(1 – a)
Applications
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Higher
Outcome 4
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 …..
Applications
Conclusion:
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Higher
the levelOutcome
of drug in
4 the patients
system will never exceed 50mg
under
Since
un+1 = these
0.3un +conditions.
35
this is below the danger level it
This sequence would
has a limit
since
0 < 0.3 < 1
be safe
to continue
indefinitely.
If we call the limit L then
at this limit we have
un+1 = un = L
The equation
un+1 = 0.3un + 35 now becomes
L = 0.3L + 35
0.7L = 35
L = 35  0.7= 350  7 = 50
Applications
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Higher
Example 2
Outcome 4
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?
Applications
Higher
Outcome 4
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(a) Problem 3.14% daily loss = ? Weekly loss.
Losing 3.14% daily leaves behind 96.86% or 0.9686.
Amount remaining after 1 week = (0.9686)7 X A0
= 0.799854 X A0
= 0.80 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
Applications
Outcome 4
Higher
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(b) Using
An+1 = 0.8 An + 50
with
A0 = 255 we get
A0 = 255ml
A1 = 254ml
1st week
A2 = 253.2ml
A3 = 252.6ml
A4 = 252.0ml
4th week
NB : even before adding the 50ml
the level is above 200ml
Applications
Outcome 4
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(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 0.2L = 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.
Finding a Formula
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Example
Outcome 4
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.
u7 = au6 + b becomes
48a + b = 44
equations
Subtract up
4a = 2 so a = 0.5
Now put a = 0.5 into 44a + b = 42 to get 22 + b = 42
so b = 20
Finding a Formula
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Outcome 4
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
u12 = 12a + b becomes
12a + b = 31
subtract up
2a = 6
Sim.
equations
a = 3
Finding a Formula
Outcome 4
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Higher
Now put a = 3 into 10a + b = 25
This gives us
30 + b = 25
So
The actual formula is
So
b = -5
un = 3n - 5
u300 = 3 X 300 - 5 = 895
Two Special Series
Outcome 4
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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
Two Special Series
Outcome 4
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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 = 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
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Sequences
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Higher
A recurrence relation is defined by un1  pun  q where -1 < p < -1 and u0 = 12
a) If u1 = 15 and u2 = 16 find the values of p and q
b) Find the limit of this recurrence relation as n  
Put u1 into recurrence relation
15  12 p  q
..... (1)
Put u2 into recurrence relation
16  15 p  q
..... (2)
Solve simultaneously:
Hence
p
Previous
1  3p

p
1
3
substitute into (1)
q  11
1
and q  11
3
State limit condition
Use formula
(2) – (1)
c
L
1 m
-1 < p < 1, so a limit L exists
L 
Quit
11
1
Limit = 16½
1
3
Hint
Quit
Next
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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.
(a)
If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run.
(b)
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.
Construct a recurrence relation
un1  0.8un  0.5
-1 < 0.8 < 1, so a limit L exists
State limit condition
Use formula
L
c
1 m
L 
0.5
1  0.8
Use formula
again
L
c
1 m
2 
0.5
1 m
Previous
un = height at the start of year
Quit
Limit = 2.5 metres
m = 0.75 Minimum prune = 25%
Hint
Quit
Next
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Higher
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.
Construct a recurrence relation
un1  1.015un  300
u0 = 2500
Calculate each term in the recurrence relation
1 Mar
u0 = 2500.00
1 Aug
u5 = 1147.53
1 Apr
u1 = 2237.50
1 Sept
u6 = 864.74
1 May
u2 = 1971.06
1 Oct
u7 = 577.71
1 Jun
u3 = 1700.62
1 Nov
u8 = 286.38
1 Jul
Previous
u4 = 1426.14
1Quit
Dec
Quit
Final
payment £290.68
Hint
Next
Maths4Scotland
Higher
Two sequences are generated by the recurrence relations un1  aun  10 and vn1  a2vn  16
The two sequences approach the same limit as n  .
Determine the value of a and evaluate the limit.
Use formula for each sequence
L
Sequence 1
10
1 a
Equate the two limits
Simplify
Solve
L
10
16

1 a
1  a2
5a  3 a 1  0
Previous
c
1 m
Sequence 2
10a 2  16a  6  0
Deduction
L
16
1  a2
Cross multiply

5a2  8a  3  0
hence
a 1
Since limit exists a  1, so a 
Quit
3
5
or
a
10 1  a 2   16(1  a)
3
5
Limit = 25
Quit
Hint
Next
Maths4Scotland
Higher
Two sequences are defined by the recurrence relations
un1  0.2un  p, u0  1 and
vn1  0.6vn  q, v0  1
If both sequences have the same limit, express p in terms of q.
L
Use formula for each sequence
Sequence 1
L
p
1  0.2

p
L
Sequence 2
p
q

0.8
0.4
Equate the two limits
Rearrange
c
1 m
0.6q
0.4

q
1  0.6
Cross multiply
p
0.4 p  0.6q
3q
2
Hint
Previous
Quit
Quit
Next
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Higher
Two sequences are defined by these recurrence relations
un1  3un  0.4
with
u0  1
vn1  0.3vn  4
with
v0  1
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
L
Sequence 2
List terms of
1st sequence
4
1  0.3
L
4
0.7
Limit 
40
5
 5
7
7
u0 = 1
u3 = 21.8
u6 = 583.4
u1 = 2.6
u4 = 65
u7 = 1749.8
u2 = 7.4
u5 = 194.6
Smallest value of n is 8; value of 8th term = 1749.8
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