Higher Maths

Revision NotesRecurrence Relations Get Started

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Recurrence Relations

know the meaning of the terms: sequence, nth term, limit as n tends to infinity

use the notation un for the nth term of a

sequence

define a recurrence relation of the form un + 1 = mun + c (m, c constants) in a

mathematical model

know the condition for the limit of the sequence resulting from a recurrence

relation to exist

find (where possible) and interpret the limit of the sequence resulting from a recurrence relation in a mathematical model

interpret a recurrence relation of the form un + 1

= mun + c (m, c constants) in a

mathematical model

A sequence is a list of terms.

The terms can be identified using: 1st, 2nd, 3rd, etc

The general term is often referred to as the nth term.

We are most interested in sequences where the nth term is a function of n.

We already know how to find the formula for the nth term where the terms increase

by a constant amount.

e.g. 4, 11,18, 25, …

If we assume the terms continue to increase by 7 then we think, … the nth term is 7n … expect a 1st term of 7.

However, the 1st term is 4

Since 4 = 7 – 3 then the actual nth term formula is:

un = 7n – 3

We can list the sequence if we have a formula for un.

e.g.

un = n2 + 2n – 1

So u1 = 12 + 2.1 – 1 = 2

u2 = 22 + 2.2 – 1 = 7

u3 = 32 + 2.3 – 1 = 14

u4 = 42 + 2.4 – 1 = 23

We often use special terms for the terms of a sequence e.g.

un is often used for the nth term.

This means the 1st term is represented by u1,

The 2nd by u2 etc.

Test Yourself?

When un+1 is expressed as a function of un then we have a recurrence relation.e.g. un+1 = 3un + 4

This relation will only pin down a particular sequence if we also know one term in the sequence, often u1, but not always.

e.g. Using the above, if u1 = 2 then u2 = 3.2 + 4 = 10; u3 = 3.10 + 4 = 34 … giving the sequence 2, 10, 34, 106, …

Whereas , if u1 = 0 then u2 = 3.0 + 4 = 4; u3 = 3.4 + 4 = 16 … giving the sequence 0, 4, 16, 52, …

Sometimes u0 is used which is not strictly in the sequence.

Test Yourself?

define a recurrence relation of the form un + 1 = mun + c

(m, c constants) in a mathematical model

(i) Given the form and some terms

(ii) Given a story to model

choose

(i) Given the form and some terms

Test Yourself?

Example

A recurrence relation is of the form un+1 = aun + b.

The 3rd, 4th, and 5th terms are 9, 13 and 21 respectively.

• Find the recurrence relation

• List the first two terms.

A recurrence relation is of the form un+1 = aun + b.

If you’re given enough information, you can form a system of two equations and solve it for a and b

Using u3 = 9 and u4 = 13 and un+1 = aun + b.

13 = 9a + b …

Using u4 = 13 and u5 = 21

21 = 13a + b …

Subtracting we from get 4a = 8 a = 2

Substituting in gives 13 = 9.2 + b b = 13 – 18 = –5

(ii) Given a story to model

Test Yourself?

ExampleAn area initially has 5000 sites vandalised by graffiti artists.A campaign hopes to clean 90% of the sites during the working week.At the weekend the vandals deface another 100 sites.

Model the situation by a recurrence relation using un to represent the number of vandalised sites on the nth Monday since the start of the campaign.

Response

un+1 = 0·1un + 100

If a real-life situation is being modelled by a recurrence relationship take care.The model only gives a snap-shot of the actual function.

Many situations describe a two-stage process … the model only gives the values after both steps have been taken.

Don’t read anthing into apparent values at the end of the first steps.

The condition for the limit of the sequence to exist.

Test Yourself?

The recurrence relation un+1 = 0·5un + 4 with u1 = 264 produces the sequence 264, 136, 72, 40, 24, 16, 12, 10, 9, 8·5, 8·25, 8·125, … As n tends to infinity we see that un tends towards 8.In fact, using any starting number, this relation produces a sequence which will converge on 8.

On the other hand, the recurrence relation un+1 = 2un – 8 with u1 = 9 produces 9, 10, 12, 16, 24, 40, 72, 136, …It diverges using any starting number … with the exception of u1 = 8, where it ‘sticks’ at 8.

In both types, 8 is called a fixed point. • In the first type the sequence runs towards 8 … 8 is a limit.• In the second type the sequence runs away from 8 … 8 is not a limit.

How do you tell the types apart?In the relation, un+1 = aun + b, the sequence converges if –1 < a < 1 [i.e. if a is a proper fraction]

What happens over time?

Find and interpret the limit

Test Yourself?

Always state the grounds for a limit to exist.If you don’t, you may just be finding a fixed point … and every linear recurrence relation has a fixed point.

un+1 = aun + b has a limit since –1 < a < 1

How do we find the limit?If there exists a limit, L, then, as n tends to infinity, un+1 tends to un.

Solve the equation L = aL + b for L.

How do we interpret the limit?In any context, the recurrence relation which models it provides snapshots only of the situation. • Don’t make anything of ‘intermediate’ values (values deduced from the story mid-cycle), the model doesn’t promise any sense here.• Don’t try fractional values of n. n is a whole number. • In context un itself may be a whole number to be sensible.

This will affect the interpretation of the limit.