2. HIGHER MATHEMATICS Unit 1 - Outcome 4 Recurrence Relations

3. There are two ways of defining patterns of numbers, or sequences. 1. Using an EXPLICIT formula. This is when the formula is expressed in terms of a letter, and you can calculate any term in the sequence. e.g. u n = 3n + 5 u 1 = 3(1) + 5 = 8 u 2 = 3(2) + 5= 11 u 7 = 3(7) + 5= 26 u n = 4 n u1 u1 = 4 1 = 4 u4 u4 = 4 4 = 256 u6 u6 = 4 6 = 4096

4. 2. Using an IMPLICIT formula, which is usually called a RECURRENCE RELATION. This is when each term in the sequence is determined (or implied) from the previous term. e.g. u n+1 = u n + 4 u 0 = 2 STARTING VALUE u 1 = 2 + 4 u 2 = 6 + 4 = 6 = 10 u 3 = 10 + 4= 14 Sequence is : 2, 6, 10, 14, ……

5. u n+1 = 2u n - 3 u 0 = 5 u 1 = 2(5) - 3= 7 u 2 = 2(7) - 3= 11 u 3 = 2(11) - 3= 19 Sequence is : 5, 7, 11, 19, …… u n = 0.25u n-1 u 0 = 200 u 1 = 0.25(200)= 50 u 2 = 0.25(50)= 12.5 u 3 =0.25(12.5)= 4.125 Sequence is : 200, 50, 12.5, 4.125, ……

6. Finding An Explicit Formula Sometimes, it is possible to change a recurrence relation (or implicit formula) into an explicit formula. EXAMPLE 1 u n+1 = 3u n ; u 0 = 2 (multiplying) u 0 = 2 u 1 = 3u 0 = (3 x 2) u 2 = 3u 1 = 3 x (3 x 2) = 3 2 x 2 u 3 = 3u 2 = 3 x (3 2 x 2) = 3 3 x 2 u 9 = 3u 8 …………………………. = 3 9 x 2

7. u n = 3u n-1 …………………………. = 3 n x 2 u n = 2 x 3 n This is now an EXPLICIT formula, and we can calculate any term in the sequence by simple substitution. u 7 = 2 x 3 7 = 1458

8. EXAMPLE 2 (adding) u n+1 = u n + 3 ; u 1 = 5 u 1 = 5 u 2 = u 1 + 3= 5 + 3 u 3 = u 2 + 3= 5 + 3 + 3 = 5 + (2 x 3) u 4 = u 3 + 3= 5 +(2 x 3) + 3 = 5 + (3 x 3) u 5 = u 4 + 3= 5 +(3 x 3) + 3 = 5 + (4 x 3) u 6 = u 5 + 3……………………………… = 5 + (5 x 3) u 15 = ……………………………………. = 5 + (14 x 3) In general, u n = = 5 + ((n-1) x 3) u n = 5 + 3n - 3 u n = 3n + 2

9. Examples In Context 1.A car depreciates in value by 15% each year.Its value, C n, at the end of each year is given by the formula C n+1 = 0.85C n ; C 0 = 12000 (a)Find an explicit formula for C n in terms of n. (b) After how many years has it lost 70% of its original value? POINTS TO NOTE Value decreases by 15%, so 85% of value remains, hence 0.85 in formula. Original value of car is £12000, hence C o = 12000

10. (a) C 0 = 12000 C 1 = 0.85 x 12000 C 2 = 0.85 2 x 12000 C 3 = 0.85 3 x 12000 C n = 0.85 n x 12000 (b) 70% lost, so 30% of value remains 30% of £12000 = £3600 C 1 = 0.85 x 12000 = £10200 C 4 = 0.85 4 x 12000 = £6264.08 C 7 = 0.85 7 x 12000 = £3846.93 Car loses 70% of its value after 8 years. C 8 = 0.85 8 x 12000 = £3269.89

11. 2. In 1999, the population of Glasgow was 700000. The net rate of increase of the population, taking account of births, deaths and migration patterns, is 3%. Set up a recurrence relation to describe this situation, and calculate, to the nearest 100, the population of Glasgow at the end of 2004. POINTS TO NOTE Population increases by 3%, so population will be 103% of the population the previous year. Starting population, P 0, is 700000.

12. P n+1 = 1.03P n ; P 0 = 700000 (1999) 2000 P 1 = 1.03(700000) 2001 P 2 = 1.03 2 x 700000 2002 P 3 = 1.03 3 x 700000 2003 P 4 = 1.03 4 x 700000 2004 P 5 = 1.03 5 x 700000 = 811492 At the end of 2004, population of Glasgow is 811500.

13. Linear Recurrence Relations These are a combination of the two recurrence relations we have seen so far. u n+1 = 3u n + 4 ; u 1 = 5 u n+1 = 2u n - 1 ; u 0 = 2 u n+1 = mu n + c In general, It is very difficult to find an explicit formula for these, but this is something you will not be asked to do.

14. Setting Up Linear Recurrence Relations EXAMPLE 1 1.Three thousand fish are introduced to a man made loch. Each day thereafter, 40% of the fish are caught by anglers. At the end of each day, 200 new fish are put into the loch. Set up a recurrence relation, u n+1, to describe this situation, and calculate how many fish there will be after 5 days. POINTS TO NOTE Starting population, u 0, is 3000. 40% lost, so 60% ( 0.6) remains

15. u n+1 =0.6u n + 200 u 0 = 3000 u 1 = 0.6(3000) + 200 = 2000 u 2 = 0.6(2000) + 200 = 1400 u 3 = 0.6(1400) + 200 = 1040 u 4 = 0.6(1040) + 200 = 824 u 5 = 0.6(824) + 200 = 694.4 After 5 days, population of fish is 694.

16. EXAMPLE 2 2.Chubby Bunny weighs 80kg. He is on a diet, and he thinks this will result in a weekly weight loss of 5% of his body weight. Sadly, he is addicted to Scotch pies, and this results in a weekly gain of 5kg. Set up a recurrence relation, w n+1, to describe his weight, and calculate Chubby Bunny’s weight four weeks into this diet. POINTS TO NOTE Starting weight, w 0, is 80kg. 5% lost, so 95% of weight remains.

17. w n+1 =0.95w n + 5w 0 = 80 w 1 = 0.95(80) + 5 = 81 w 2 = 0.95(81) + 5 = 81.95 w 3 = 0.95(81.95) + 5 = 82.85 w 4 = 0.95(82.85) + 5 = 83.70 After 4 weeks, Chubby Bunny's weight is 83.7kg(!)

18. LOOK at this recurrence relation u n+1 = 2u n + 3 ; u 0 = 5 u 0 = 5 u 1 = 13 u 2 = 29 u 3 = 61 u 4 = 125 u 5 = 253 u 6 = 509 + 8+ 16+ 32+ 64+ 128+ 256 As n gets bigger, the gap between successive terms increases. As n  So does u n

19. Now LOOK at this recurrence relation v n+1 = 0.6v n + 30 ; v 0 = 100 v 0 = 100 v 1 = 90 v 2 = 84 v 3 = 80.4 v 4 = 78.24 v 5 = 76.944 v 6 = 76.166 - 10- 6- 3.6- 2.16-1.296-0.778 As n gets bigger, the gap between successive terms decreases. As n  u n seems to be approaching a limiting value, or a LIMIT.

20. In general,for a recurrence relation of the form u n+1 = au n + b 1.It will approach a limit, L,provided that –1 < a < 1 2. It will NOT approach a limit, L,(except maybe zero) if a 1 Recurrence relations - Limits

21. Finding A Limit We have looked at this recurrence relation before. u n+1 = 0.6u n + 30 ; u 0 = 100 We decided that, as n , u n seemed to be approaching a limit. Finding this limit is easy peasy.

22. u n+1 = 0.6u n + 30 ; u 0 = 100 If a limit exists, then as n gets larger, the gap between successive terms gets smaller and smaller, until u n+1 and u n become so close that there is little difference between them. We can then use the same letter to denote them. The letter we use is L. L = 0.6L + 30 0.4L = 30 L = 75 as n , u n 75 Find the limit of: Since -1 < 0.6 < 1, a limit exists

23. Examples Requiring Calculation Of Limit The population of Edinburgh at the start of 1992 was 500000. Migration studies suggest that, each year, 2% of people born in Edinburgh leave to live somewhere else. During this same period, 7000 new people arrive to settle in the city. If this situation persists, what will eventually happen to the population of Edinburgh? EXAMPLE 1

24. P n+1 =0.98P n + 7000P 0 =500000 Since -1 < 0.98 < 1, a limit exists L = 0.98L + 7000 0.02L = 7000 L = 350000 Eventually, the population of Edinburgh will “settle” at 350000 (will not drop below 350000)

25. Chubby Bunny – who now weighs 90kg - has abandoned his old diet, and started the Atkin’s diet. He now loses one fifth of his body weight each month. Sadly, he is now hopelessly addicted to haggis suppers from his local chip shop, and this results in a monthly gain in weight of 15kg. If his weight drops below 76kg, he is going to buy himself an Aston Martin Vanquish. Will he get his car? EXAMPLE2

26. w n+1 =0.8w n + 15w 0 = 90 Since -1 < 0.8 < 1, a limit exists L = 0.8L + 15 0.2L = 15 L = 75 L < 76, so Chubby’s weight will eventually drop below target weight. He DOES get his car.

27. Recurrence Relations And Simultaneous Equations EXAMPLE 1 A recurrence relation is defined by the formula u n+1 = au n + b. Given that u 2 = 4, u 3 = 10 and u 4 = 28, find the values of a and b.

28. U n+1 = a U n + b U 2 = 4 U 3 = 10 U 4 = 28 U 3 = a U 2 + bU 4 = a U 3 + b 10 = a 4 + b 28 = a 10 + b 10 = 4a + b28 = 10a + b Now use simultaneous equations to find the values of a and b.

29. 4a + b = 10 10a + b = 28 6a = 18 a = 3 b = -2 U n+1 = 3 U n - 2