1. Recurrence Relations Recurrence relations are a further method of modelling growth. They are used to predict the next value in a number pattern The relation tells you how to get from one value to the next. U n Notation for number patterns U 1 = value of 1 st term U 2 = value of 2 nd term etc U n = value of n th term U n–1 is the term before U n and U n+1 is the term after U n

2. In the number pattern 1, 4, 7, 10, 13, 16 U 1 = 1and U 4 = 10 To find the next number in the pattern add 3 to the term before. U 2 = U 1 + 3 = 1 + 3 = 4 U 3 = U 2 + 3 = 4 + 3 = 7 U 4 = U 3 + 3 = 7 + 3 = 10

3. To find the next term add 3 to the term before. nth term = (n–1) th term + 3 U n = U n–1 + 3 Next term = term before + 3 So if n = 4then U 4 = U 3 + 3 This could have been written as U n+1 = U n + 3 Next term = term before + 3 So if n = 4then U 5 = U 4 + 3

4. U n = 2U n–1 + 3U 1 = 3 Next term = 2  term before + 3 U 1 = 3 U 2 = 2U 1 + 3= 2  3 + 3 = 9 U 3 = 2U 2 + 3= 2  9 + 3 = 21 U 4 = 2U 3 + 3= 2  21 + 3 = 45

5. In questions involving investing money at a certain interest rate U 0 is usually used for the initial investment so that U 1 will give the value after 1 month or 1 year The pattern P n+1 = 1.05P n is in fact increasing the previous value by 5% so we can find out the value of £10 if it is invested at an annual interest rate of 5%. P n+1 = 1.05P n P 0 = 10 Next term = 1.05  term before P 0 is used in this case to show that P 1 is the value after 1 year P 0 = 10 P 1 = 1.05P 0 = 1.05  10 = 10.50 P 2 = 1.05P 1 = 1.05  10.5 = 11.025 P 3 = 1.05P 2 = 1.05  11.025 = 11.57625

6. Calculator 1)Type the starting value P 0 i.e. 10Enter 2)Type 1.05  AnsEnter 3) Keep pressing Enter to generate the pattern

7. Types of recurrence relations What do the following recurrence relations do? 1) U n = 1.10U n–1 Increases the previous U value by 10% each time 2) U n = 2U n–1 Doubles the previous U value each time 3) U n = 0.05U n–1 Finds 5% of the previous U value 4) U n = 1.05U n–1 –10 Increases the previous U value by 5% and subtracts 10 each time

8. More complex recurrence relations. 1) Paying off credit card bills. A car costs £3000 and the loan company charges 2% interest per month. You pay £300 off per month. How long does it take to repay? At the end of month 1 the loan has increased by 2%. Loan = 1.02  3000 = £3060 But £300 is paid off so £2760 is owed at the end of month 1 At the end of month 2 the loan has increased by 2%. Loan = 1.02  2760 = £2815.20 But £300 is paid off so £2515.20 is owed at the end of month 2

9. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

10. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Multiply by 1.02 to add 2% interest

11. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Subtract £300

12. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Multiply by 1.02 to add 2% interest

13. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Subtract £300

14. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Multiply by 1.02 to add 2% interest

15. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Subtract £300

16. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

17. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

18. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

19. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

20. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

21. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300

22. So to find out how much is owed the recurrence relation is U n = 1.02U n–1 – 300 Spreadsheet Link

23. Calculator 1)Type the starting value P 0 i.e. 10 Enter 2)Type 1.2  Ans – 300Enter 3) Keep pressing Enter to generate the pattern

24. The monthly interest rate can be easily changed to see the effect of different rates on the payment period. The graph on the below shows how the outstanding loan decreases.