1. Higher Maths Revision Notes Recurrence Relations Get Started goodbye

2. Recurrence Relations know the meaning of the terms: sequence, nth term, limit as n tends to infinity use the notation u n for the nth term of a sequence define a recurrence relation of the form u n + 1 = mu n + 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 u n + 1 = mu n + c (m, c constants) in a mathematical model

3. A sequence is a list of terms. The terms can be identified using: 1 st, 2 nd, 3 rd, etc The general term is often referred to as the n th term. We are most interested in sequences where the n th term is a function of n. We already know how to find the formula for the n th 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 n th term is 7n … expect a 1 st term of 7. However, the 1 st term is 4 Since 4 = 7 – 3 then the actual n th term formula is: u n = 7n – 3 We can list the sequence if we have a formula for u n. e.g. u n = n 2 + 2n – 1 So u 1 = 1 2 + 2.1 – 1 = 2 u 2 = 2 2 + 2.2 – 1 = 7 u 3 = 3 2 + 2.3 – 1 = 14 u 4 = 4 2 + 2.4 – 1 = 23 We often use special terms for the terms of a sequence e.g. u n is often used for the n th term. This means the 1 st term is represented by u 1, The 2 nd by u 2 etc. Test Yourself?

4. When u n+1 is expressed as a function of u n then we have a recurrence relation. e.g. u n+1 = 3u n + 4 This relation will only pin down a particular sequence if we also know one term in the sequence, often u 1, but not always. e.g. Using the above, if u 1 = 2 then u 2 = 3.2 + 4 = 10; u 3 = 3.10 + 4 = 34 … giving the sequence 2, 10, 34, 106, … Whereas, if u 1 = 0 then u 2 = 3.0 + 4 = 4; u 3 = 3.4 + 4 = 16 … giving the sequence 0, 4, 16, 52, … Sometimes u 0 is used which is not strictly in the sequence. Test Yourself?

5. define a recurrence relation of the form u n + 1 = mu n + c (m, c constants) in a mathematical model define a recurrence relation of the form u n + 1 = mu n + c (m, c constants) in a mathematical model (i) Given the form and some terms (ii) Given a story to model choose

6. (i) Given the form and some terms Test Yourself? Example A recurrence relation is of the form u n+1 = au n + b. The 3 rd, 4 th, and 5 th terms are 9, 13 and 21 respectively. (a)Find the recurrence relation (b)List the first two terms. A recurrence relation is of the form u n+1 = au n + b. If you’re given enough information, you can form a system of two equations and solve it for a and b Using u 3 = 9 and u 4 = 13 and u n+1 = au n + b. 13 = 9a + b …  Using u 4 = 13 and u 5 = 21 21 = 13a + b …  Subtracting we  from  get 4a = 8  a = 2 Substituting in  gives 13 = 9.2 + b  b = 13 – 18 = –5

7. (ii) Given a story to model Test Yourself? Example An 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 u n to represent the number of vandalised sites on the n th Monday since the start of the campaign. Response u n+1 = 0·1u n + 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.

8. The condition for the limit of the sequence to exist. Test Yourself? The recurrence relation u n+1 = 0·5u n + 4 with u 1 = 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 u n 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 u n+1 = 2u n – 8 with u 1 = 9 produces 9, 10, 12, 16, 24, 40, 72, 136, … It diverges using any starting number … with the exception of u 1 = 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, u n+1 = au n + b, the sequence converges if –1 < a < 1 [i.e. if a is a proper fraction] What happens over time?

9. 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. u n+1 = au n + 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, u n+1 tends to u n. 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 u n itself may be a whole number to be sensible. This will affect the interpretation of the limit.