Higher Unit 1 Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence /...

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Higher Unit 1Higher Unit 1

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Recurrence Relations

Grow and Decay

Linear Recurrence Relation

Divergence / Convergence / LimitsApplications

Find a formula

Exam Type Questions

Higher Outcome 4

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Recurrence Relations

Sequences

A 5 9 13 17 …….B 3 6 12 24 …….C 2 3 5 8 13 ……..D 17 23 41 77 137 ………E 2 3 5 7 11 ………

In the above sequences some have obvious patterns while others don’t however this does not mean that a pattern doesn’t exist.

Higher Outcome 4

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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 RelationsHigher Outcome 4

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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


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Recurrence Relation: un+1 = un + 4 with u1 = 5

Formula: un = 4n + 1

A 5 9 13 17 …….

So u100 = 4 X 100 + 1 = 401

u2 = u1 + 4 = 5 + 4 = 9 u3 = u2 + 4 = 9 + 4 = 13


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Recurrence Relation: un+1 = 2un with u1 = 3.

B 3 6 12 24 ……

Formula: un = 3 X 2n-1

So u10 = 3 X 29 = 3 X 512 = 1536

u2 = 2u1 = 2 X 3 = 6, u3 = 2u2 = 2 X 6 = 12, etc


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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.


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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


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E 2 3 5 7 11 ………

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.


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Growth & Decay

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.

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Example 1An 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

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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 = 4746u5 = 0.83u4 = 0.83 X 4746 = 3939This is less than 4000 so it takes 5 full hours to fall

below 4000.

Growth & Decay

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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 1.14 and Pn+1 = 1.14 Pn

Growth & Decay

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Growth & Decay

P1 = 1.14 P0P2 = 1.14 P1 = 1.14 X 1.14 P0 = (1.14)2 P0P3 = 1.14 P2 = 1.14 X (1.14)2 P0 = (1.14)3 P0So 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

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Growth & Decay

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 smallIf n = 9 then (1.14)9 = 3.25… too bigIf n = 7 then (1.14)7 = 2.50… too smallIf 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.

Higher Outcome 4

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Linear Recurrence Relations

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.

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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.

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Example

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,


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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

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V2 = 1.15 X 1625 - 100

= 1769V3 = 1.15 X 1769 -

100= 1934

V4 = 1.15 X 1934 - 100

= 2124V5 = 1.15 X 2124 -

100= 2343

So after 5 puffs the balloon contains

2343ml of air.

(iii) continuing the above

V6 = 1.15 X 2343 - 100

= 2594V7 = 1.15 X 2594 - 100

= 2883V8 = 1.15 X 2883 - 100

= 3216

BANG!!!

The balloon bursts on the 8th puff.

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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?


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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 + 150Similarly A2 = 0.4A1 + 150In general we get the recurrence relation

An+1 = 0.4An + 150

and this gives us the following sequence…...


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A0 = 150

A1 = 0.4 X 150 + 150

= 210A2 = 0.4 X 210 +

150= 234A3 = 0.4 X 234 +

150= 243.6A10 = 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 = 250This is below the danger level so factory could be allowed to continue dumping.

We say that the sequence CONVERGES to a LIMIT of 250.

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Divergence / Convergence/Limits

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 = 7340028

As n un

and we say that the sequence DIVERGES.

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(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 + 4 = 8

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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.

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(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

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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.

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4 with u0 = 3


Other Factors(e) compare this with (b) un+1 = 0.5un + 10 with u0 = 3

u0 = 3

u1 = 11.5

u2 = 15.75

u3 = 17.875

….. u10 = 19.98...

…… u20 = 19.99….

This is clearly heading to a limit of 20

Check: if un = 20

un+1 = 0.5 X 20 +10 = 20

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Conclusion:

if un+1 = aun + b converges to a limit

then changing b changes the limit.

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(f) compare this with (b) un+1 = 0.5un + 4 with u0 = 200

u0 = 200

u1 = 104

u2 = 56

u3 = 32

u10 = 8.1875

u20 = 8.0001….

Again this is heading to a limit of 8

Higher Outcome 4un+1 = 0.5un + 4 with u0 = 3

Conclusion:

if un+1 = aun + b converges to a limit

then changing u0 does not affect the limit.

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Find the Limit

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

Higher Outcome 4

L – aL = b

L(1 – a) = b

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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?

ApplicationsExample 1

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 …..

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Outcome 4

un+1 = 0.3un + 35

This sequence has a limit since 0 < 0.3 < 1

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

If we call the limit L then at this limit we have

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.

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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.

ApplicationsExample 2

(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?

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Applications

(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 weeklySo if An is the fluid level after n weeks then we have

An+1 = 0.8 An + 50

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Applications

(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

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Applications

(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.

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Given that u6 = 48 , u7 = 44 and u8 = 42

then find a & b .

Finding a FormulaExampl

e A recurrence relation is defined by the formula

un+1 = aun + b

u8 = au7 + b becomes

44a + b = 42

u7 = au6 + b becomes

48a + b = 44

Sim.

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

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Example

The nth term in a sequence is given by the formulaun = 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

Sim. equations

subtract up 2a = 6 a = 3

Finding a Formula

Higher Outcome 4

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= 25 This gives us 30 + b = 25

So b = -5

The actual formula is un = 3n - 5

So u300 = 3 X 300 - 5 = 895

Finding a FormulaHigher Outcome 4

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eg 8, 14, 20, 26, ……here d = un+1 - un = 6

u1 = 8 = 8 + (0 X 6)

Two Special Series

In an arithmetic series there is a constant difference between consecutive terms.

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

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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

20u2 = 10 = 5 X

21u3 = 20 = 5 X

22u4 = 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

Two Special SeriesHigher Outcome 4

Higher Maths

Strategies

www.maths4scotland.co.uk

Click to start

Sequences

Maths4Scotland Higher

Sequences

The following questions are on

Non-calculator questions will be indicated

Click to continue

You will need a pencil, paper, ruler and rubber.


Hint

Previous NextQuitQuit

Put u1 into recurrence relation

Solve simultaneously:1

31 3p p

A recurrence relation is defined by 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

1n nu pu q

..... (1)15 12p q

Put u2 into recurrence relation ..... (2)16 15p q

(2) – (1) substitute into (1) 11q

Hence1

and 113

qp

State limit condition -1 < p < 1, so a limit L exists

Use formula1

cL

m

1

3

11

1L

Limit = 16½


HintPrevious NextQuitQuit

Construct a recurrence relation 1 0.8 0.5n nu u

State limit condition -1 < 0.8 < 1, so a limit L exists

Use formula1

cL

m

0.5

1 0.8L

Limit = 2.5 metres

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.

un = height at the start of year

Use formula again 1

cL

m

0.5

21 m

Minimum prune = 25%m = 0.75


Hint


Construct a recurrence relation 1 1.015 300n nu u

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

Calculate each term in the recurrence relation

1 Mar u0 = 2500.00

1 Apr u1 = 2237.50

1 May u2 = 1971.06

1 Jun u3 = 1700.62

1 Jul u4 = 1426.14

1 Aug u5 = 1147.53

1 Sept u6 = 864.74

1 Oct u7 = 577.71

1 Nov u8 = 286.38

1 Dec Final payment £290.68


Hint


Equate the two limits Cross multiply

Sequence 1

Since limit exists a 1, so

Use formula for each sequence1

cL

m

Limit = 25

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.

1 10n nu au 21 16n nv a v

10

1L

a

Sequence 2 2

16

1L

a

2

10 16

1 1a a

210 1 16(1 )a a

Simplify 210 16 6 0a a 25 8 3 0a a

Solve 5 3 1 0a a 3hence or

51a a

Deduction3

5a


Hint


Equate the two limits Cross multiply

Sequence 1

Use formula for each sequence1

cL

m

1 0.2

pL

Sequence 2

1 0.6

qL

0.8 0.4

p q 0.4 0.6p q

Rearrange0.6

0.4

qp

Two sequences are defined by the recurrence relations

If both sequences have the same limit, express p in terms of q.

1 0

1 0

0.2 , 1 and

0.6 , 1n n

n n

u u p u

v v q v

3

2

qp


Hint


Sequence 2

Requirement for a limit

4

1 0.3L

List terms of 1st sequence

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, andii) the value of that term.

1 0with3 0.4 1n nu u u 1 0with0.3 4 1n nv v v

First sequence has no limit since 3 is not between –1 and 1

2nd sequence has a limit since –1 < 0.3 < 1

4

0.7L 5

7

405

7Limit

u0 = 1

u1 = 2.6

u2 = 7.4

u3 = 21.8

u4 = 65

u5 = 194.6

u6 = 583.4

u7 = 1749.8

Smallest value of n is 8; value of 8th term = 1749.8


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Outcome 4

Higher Unit 1 Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence /...

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