1. RECURRENCE RELATIONS

2. One way of defining a sequence of numbers is to use an iterative formula, or a recurrence relation. This is a formula which generates successive terms from the previous term or terms. Example 1: Find the first 6 terms of the sequence defined by: u 1 = 3 ; u n+1 = u n + 2. The recurrence relation tells us that: u 2 =u 1 + 2= 5 u 3 = u 2 + 2= 7 = 13 = 9 = 11 u 4 = u 5 = u 6 = u 3 + 2 u 5 + 2 u 4 + 2 Hence, the first 6 terms are:3, 5, 7, 9, 11, 13. We use the notation, u n to represent the “n th term”.

3. Example 2: The first two terms of a sequence are both 1. Given that u n+1 = u n + u n-1, find the first eight terms. The recurrence relation tells us that: u 3 = u 2 + u 1 = 1 + 1 =2 u 4 = u 3 + u 2 u 4 + u 3 u 5 + u 4 u 6 + u 5 u 7 + u 6 u 6 = u 5 = u 7 = u 8 = = 2 + 1 = = 3 + 2 = = 5 + 3 = = 8 + 5 = = 13 + 8 = 3 5 8 13 21 Hence, the first 8 terms are:1, 1, 2, 3, 5, 8, 13, 21. ( This is known as the Fibonacci sequence ).

4. Note that a recurrence relation can be written in many ways, so: The relation u n+1 = u n + 3 could be written as: u n = u n-1 + 3 It simply states that: “ Each term of this sequence is the previous term, plus 3”. Similarly: u n+1 = u n + u n-1 could be written as: u n+2 = u n+1 + u n or: u n = u n-1 + u n-2 Note also, great care must be taken with subscripts when handwriting these, to avoid confusing, for example u n+1 with u n +1.

5. Example 3: A sequence of numbers u 1, u 2, u 3, ….is defined by: u 1 = 1, u n+1 = ku n + 3; n ≥ 1, where k is a constant. a)Find u 2 and u 3 in terms of k. b) Given that u 3 = 13, find two possible values of k. a) From the recurrence relation: u 2 =ku 1 + 3= k + 3 u 3 =ku 2 + 3= k ( k + 3 ) + 3 = k 2 + 3k + 3 b) If u 3 = 13, then:13 = k 2 + 3k + 3 0 = k 2 + 3k – 10 0 = ( k + 5 )( k – 2 ) k = – 5 or 2.

6. Example 4: A sequence of positive numbers u 1, u 2, u 3, ….is defined by: u 1 = 1, u n+1 = + 2, n ≥ 1. a)Find u 2, u 3 and u 4. b) Given that the sequence converges to a limit, u, find the exact value of u. 1un1un a) From the recurrence relation: u 2 = 1u11u1 + 2 = 3 1u21u2 + 2 = 2 1313 1u31u3 + 2 = 2 3737 u 2 = 1 + 2u u = 1u1u + 2 u 2 – 2u – 1 = 0 = Since the terms are positive, u = u 3 =u 4 = b) If the sequence converges to u, then as n →, u n+1 = u n = u. 8

7. Summary of key points: This PowerPoint produced by R.Collins ; Updated Apr.2009 We use the notation, u n to represent the “n th term”. A recurrence relation can be written in many ways: Great care must be taken with subscripts when handwriting these, to avoid confusing, for example u n+1 with u n +1. e.g. u n+1 = 5u n + 2 could be written as: u n = 5u n-1 + 2