1. Higher Maths 1 4 Sequences UNIT OUTCOME SLIDE

2. Higher Maths 1 4 Sequences Number Sequences UNIT OUTCOME NOTE The consecutive numbers or terms in any sequence can be written as SLIDE u 0 , u 1 , u 2 , u 3 , u 4 , u 5 ... u n consecutive means ‘ next to each other’ u n +1 u n the n th number in a sequence the number after the n th number (the ‘next’ number) u n –1 the number before the n th number (the ‘previous’ number)

3. Higher Maths 1 4 Sequences Basic Recurrence Relations UNIT OUTCOME NOTE It is often possible to write the next number in a sequence as a function of the previous numbers. SLIDE u n = 2 u n –1 – 3 with u 0 = 4 4 , 5 , 7 , 11 , 19 , 35 , 67 … u n = a u n –1 + b or alternatively u n –1 × 2 – 3 u n The number sequence is as follows: is called a linear recurrence relation . u n +1 = a u n + b In number sequences, Example

4. Higher Maths 1 4 Sequences Sequences on a Calculator UNIT OUTCOME NOTE It is possible to use a scientific calculator to quickly generate consecutive numbers in a sequence. SLIDE Set up u 0 Example u n +1 = u n + 1 4 3 ( ) 2 – = Input the recurrence relation, using the answer button as u n with Step 1: Step 2: u 0 = -2 Repeatedly press equals to generate the number sequence. Step 3: 4 ÷ 3 × ANS + 1

5. Higher Maths 1 4 Sequences Convergence and Divergence. UNIT OUTCOME NOTE Some sequences never stop increasing, while others eventually settle at a particular number. SLIDE If the numbers in a sequence continue to get further and further apart, the sequence diverges . u n n u n n If a sequence tends towards a limit , it is described as convergent . u n n L L

6. Higher Maths 1 4 Sequences Limit of a Sequence UNIT OUTCOME NOTE For any linear recurrence relation SLIDE u n +1 = a u n + b -1 < a < 1 If it exists, the limit of a convergent sequence can be calculated from a simple formula. L = b 1 – a the sequence tends to a limit L if Limit of a Linear Recurrence Relation Convergence or divergence is not affected by u 0

7. Higher Maths 1 4 Sequences Solving Recurrence Relations UNIT OUTCOME NOTE It is possible to find missing values in a recurrence relation by using consecutive terms to form simultaneous equations. SLIDE Example A linear recurrence relation has terms u n +1 = a u n + b u 1 = 5 , u 2 = 9.5 and u 3 = 20.75 . Find the values of a and b . 9.5 = 5 a + b u 2 = a u 1 + b 20.75 = 9.5 a + b u 3 = a u 2 + b Solving equations and simultaneously gives u n +1 = a u n + b and u n +1 = 2.5 u n – 3 a = 2.5 and b = - 3 2 1 1 2

8. Higher Maths 1 4 Sequences Linked Recurrence Relations UNIT OUTCOME NOTE SLIDE Example Two libraries, A and B, have 100 books in total. a n + b n = 100 a n +1 = 0.85 a n + 0.45 b n b n = 100 – a n = 0.85 a n + 0.45 ( 100 – a n ) = 0.4 a n + 45 Every week, 85 % of the books borrowed from library A are returned there, while the other 15 % are returned to library B instead. Only 55 % of the books borrowed from library B are returned there, with the other 45 % returned to library A. Investigate what happens. L a = b 1 – a = 45 1 – 0.4 = 75 L b = 25

9. Higher Maths 1 4 Sequences Special Sequences UNIT OUTCOME NOTE There are some important sequences which it is useful to know. SLIDE u n +1 = a u n u n +1 = u n + b u n +2 = u n + u n +1 Sequence Recurrence Relation Example Arithmetic Geometric Fibonacci 2, 8, 14, 20, 26, 32 ... 3, 12, 48, 192, 768 ... with u 0 = 0 and u 1 = 1 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ... discovered 1202