Higher Mathematics

Unit 1.4

Recurrence Relations

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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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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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 term number i.e. first, second etc.un the value of the “current term”

un+1 the value of the “next term”

un-1 the value of the “previous term”

u0 the value of the “starting term”

u1 the value of the “first term” some

times the “starting term”

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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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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),

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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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 + 10 (u0 = 20)

They are known as Recurrence Relations as they form a recurring sequence based on the previous term.

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 + 3 (u1 = 5)

What happens when we combine these ideas ? (multiplying and adding) un+1 = 2un + 3

(u1 = 5)

 

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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.

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.

Press [ENTER] to get next value. Continue

Generating recurrence relations on a Graphic

Calculator.Use the same recurrence relation as used in example 5.Vn+1 = 0·6Vn + 30 (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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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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Concept of a Limit.

My Wall

Day 1

Day 2

Day 3Day 4

Day 5Day 6

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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.

1 1 1 1...

2 4 8 16S

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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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Different values of aWe 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 > 1a = 1

0 < a < 1a = 0

−1 < a < 0a = − 1a < − 1 Continue

a > 1 u(n)=2.5u(n-1)+20

0

500

1000

1500

2000

2500

0 1 2 3 4 5 6 7

n

u(n

)

as n gets larger u(n) moves to ∞

n u(n)

1 10

2 45

3 132.5

4 351.25

5 898.125

6 2265.31

7 5683.28

8 14228.2

9 35590.5

un=2·5un-1+ 20

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a = 1u(n)=1u(n-1)+20

0

50

100

150

200

250

0 2 4 6 8 10 12

n

u(n

)

n u(n)

1 10

2 30

3 50

4 70

5 90

6 110

7 130

8 150

9 170

10 190

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.

un=1un-1+ 20

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0 < a < 1u(n)=0.6u(n-1)+20

0

10

20

30

40

50

60

0 2 4 6 8 10 12

n

u(n

)

n u(n)

1 10

2 26

3 35.6

4 41.36

5 44.816

6 46.8896

7 48.1338

8 48.8803

9 49.3282

10 49.5969

11 49.7581

as n gets larger u(n) moves to 50 . This is a limit.

un=0.6un-1+ 20

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a = 0u(n)=0u(n-1)+20

0

5

10

15

20

25

0 2 4 6 8 10 12

n

u(n

)

n u(n)

1 10

2 20

3 20

4 20

5 20

6 20

7 20

8 20

9 20

10 20

11 20

as n gets larger u(n) moves to 20 . This is a limit.

un=0un-1+ 20

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−1 < a < 0u(n)=-0.4u(n-1)+10

0

1

2

3

4

5

6

7

8

9

10

0 2 4 6 8 10 12

n

u(n

)

This is known as an oscillating graph. It is a limit.

n u(n)

1 2

2 9.2

3 6.32

4 7.472

5 7.0112

6 7.19552

7 7.12179

8 7.15128

9 7.13949

10 7.14421

11 7.14232as n gets larger u(n) moves to 7.142857…

un=-0.4un-1+ 10

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when a = −1u(n)=-1u(n-1)+20

0

2

4

6

8

10

12

0 2 4 6 8 10 12

n

u(n

)

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.

n u(n)

1 9

2 11

3 9

4 11

5 9

6 11

7 9

8 11

9 9

10 11

un = −1un-1 + 20

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And if a < -1?u(n)=-1.7u(n-1)+20

-600

-500

-400

-300

-200

-100

0

100

200

300

400

0 2 4 6 8 10 12 14

n

u(n

)

Note that this graph heads off towards both + ∞ and – ∞ as successive values change sign.

n u(n)

1 9

2 4.7

3 12.01

4 -0.417

5 20.7089

6 -15.2051

7 45.8487

8 -57.9428

9 118.503

10 -181.455

11 328.473

12 -538.404

un = −1.7un-1+20

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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 !!

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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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 + 10) and when -1< a < 1, automatically

gives a recurrence relation with a limit.

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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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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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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.

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:Continue

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.

a) Write a recurrence relation for the amount of chemical present at the end of each week.

b) 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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Problems using Recurrence Relations

If 28% of the chemicals are removed then (100 – 28)% remain. This figure is used in the recurrence relation.

a) Chemical remaining is 72%, u0 = 15 tonnesRelation un+1 = 0.72 un + 1

b) Calculate a few terms to establish the pattern….

The value of the linear coefficient (0.72) is important. -1 < a < 1 shows that a limitwill exist for this situation. This is alsosuggested by the table of values.

n un

0 15

1 11.8

2 9.496

3 7.83712

4 6.642726

5 5.782763

6 5.163589

7 4.717784

8 4.396805

9 4.165699

Problems using Recurrence Relations

To calculate the limit solve the problem

L = 0.72 L + 1 1L – 0.72L = 1

0.28L = 1L = 1 ÷ 0.28L = 3.571428571…

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