higher maths revision notes recurrence relations get started goodbye
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Higher Maths
Revision NotesRecurrence Relations
Get Startedgoodbyegoodbye
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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
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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?
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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?
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define a recurrence relation of the form un + 1 = mun + c
(m, c constants) in a mathematical model
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(i) Given the form and some terms
(ii) Given a story to model(ii) Given a story to model
choose
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(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
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(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.
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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?
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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.
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