Recurrence Relations 1.4

1. Recurrence Relations 1.4

2. Find a formula connecting the variables in each table below : F = 3 × a + 1 F = 3a + 1 +3 +3 +3 +3 P = 5g – 2 +5 +5 +5 +5 E = 2x + 3

3. Find a formula for the nth term of each of the following sequences and hence find the 20th termof each sequence. • 3 , 7 , 11 , 15 , ........ • 14 , 26 , 38 , 50 , ....... • (c) 2 , 8 , 14 , 20 , ........ nthterm = 4n – 1 +4 +4 20thterm = 4 × 20 – 1 = 79 nthterm = 12n + 2 +12 +12 20thterm = 12 × 20 + 2 = 242 nthterm = 6n – 4 20thterm = 6 × 20 – 4 = 116

4. Recurrence Relations Notation We write the terms of a sequence as u1 , u2 , u3 , …….., un-1 , un , un+1 , ……... where u1 is the 1st term, u2 is the 2nd term etc…. and un is the nth term ( n being any whole number.)

5. Consider 5 9 13 17 ……. +4 +4 +4 A formula for the nth term, un in terms of n un = 4n + 1 u1 = 4×1 + 1, u2 = 4×2 + 1, u3 = 4×3 +1, ………… Or A REURRENCE RELATION calculating each term by using the previous term un+1 = un + 4 u1 = 5, u2 = 5 + 4, u3 = 9 + 4, u4 = 13 + 4, ………

6. Growth & Decay Adding on 21% gives us 121% or 1∙21 and this is called the GROWTH factor or multiplier Removing 15% leaves behind 85% or 0∙85 which is called the DECAY factor or multiplier Growth and decay factors allow us a quick method of tackling repeated % changes

7. Example 1 An oven contains 10 000 bacteria which are being killed off at a rate of 17% per hour by a particular disinfectant. (a) How many bacteria are left after 3 hours? (b) How many full hours are needed so that there are fewer than 4 000 bacteria? Let un be the number of bacteria remaining after n hours. Removing 17% leaves behind 83% so the DECAY factor is 0∙83 and un+1 = 0∙83 un

8. u0 = original value u0 = 10000 (a) u1 = 0∙83u0 = 0.83 × 10000 = 8300 u2 = 0∙83u1 = 0∙83 × 8300 = 6889 u3 = 0∙83u2 = 0∙83 × 6889 = 5718 So there are 5718 bacteria after 3 hours. u4 = 0∙83u3 = 0∙83 × 5718 = 4746 (b) u5 = 0∙83u4 = 0∙83 × 4746 = 3939 This is less than 4 000 so it takes 5 full hours to fall below 4000.

9. Example 2 The population of a town grows at a rate of 14% per annum. If P0 is the initial population and Pn is the population after n years. (a) Find a formula for Pn in terms of P0. (b) Find roughly how long it takes the population to treble. Adding on 14% gives us 114% so the GROWTH factor is 1∙14 and Pn+1 = 1∙14 Pn

10. P1 = 1∙14 P0 P2 = 1∙14 P1 = 1∙14 × 1∙14 P0 = (1∙14)2 P0 P3 = 1∙14 P2 = 1∙14 × (1∙14)2 P0 = (1∙14)3 P0 So in general we have Pn = (1∙14)n P0 If the population trebles then we need to have Pn > 3 P0 Dividing by P0 or (1∙14)n P0 > 3 P0 we get (1∙14)n> 3

11. We can use the calculator answer (ANS) button Insert 1∙14 by 1∙14 (=)ENTER 1∙14 × ANS Now each time you press the = button we multiply by another 1∙14 Count the number of times you press = to get a number greater than 3 From the above we can say it takes just over 8 years for the population to treble.

12. Examples to try • A population of bacteria grows at a rate of 20% per day. In an experiment the initial number of bacteria was 150. • a) Find a recurrence relation to model this experiment. • b) How many bacteria are there after 3 days? • c) How long will it take the population to double? 20% increase  120% left un+1 = 1∙2un uo = 150 150 (=)ENTER 1∙2 × ANS u1 = 180 u2= 216 u3= 259 u4= 311 Doubles after 4 days 259 after 3 days

13. A health report states that the level of harmful gasses in the atmosphere should be no more than 110 units. The current level is 150. • Environmental Health officials introduce a plan to reduce these gasses by 5% per annum. • How many years will it take for a safe level to be attained? 5% decrease  95% left un+1 = 0∙95un uo = 150 150 ENTER 0∙95 × ANS u1 = 143 u2= 135 u3= 129 u7= 104 Falls below 110 during 7th year

14. When an oil tanker runs aground it spills 25 000 tonnes of oil. The natural action of the waves will disperse the oil at a rate of 40% per week. • a) How long will it take to reduce the amount of oil to 1 000 tonnes? • b) After how many weeks will it be less than 100 tonnes. 40% decrease  60% left un+1 = 0∙6un uo = 25000 25000 ENTER 0∙6 × ANS u1 = 15000 u2= 9000 u3= 5400 u7= 700 Falls below 1000 during 7th week. Falls below 100 during 11th week. u11= 91

15. 4. Ethiopia’s population at the start of 2005 was 62 million. The rate of increase is 3% per annum. What is the population likely to be at the start of 2010? 3% increase  103% left un+1 = 1∙03un uo = 62 62 ENTER 1∙03 × ANS u1 = 63∙8 u2= 65∙8 u3= 67∙7 u4= 69∙8 u5= 71∙9 2006 2007 2008 2009 2010 71∙9 million at the start of 2010

16. An art dealer bought a painting for £2∙5 million at the beginning of 2003. She expects the value to increase by 6% per annum. • a) Find a recurrence relation for the value of the painting. • b) What is its expected value at the start of 2007? 6% increase  106% left un+1 = 1∙06un uo = 2∙5 2∙5ENTER 1∙03 × ANS u1 = 2∙575 u2= 2∙65 u3= 2∙73 u4= 2∙81 2004 2005 2006 2007 £2∙81 million at start of 2007

17. A virulent strain of ‘flu will affect 12% of the population per week if no preventative measures are taken. • a) What percentage remains healthy? • b) For a village of 5 000 people, how many will be healthy after 3 weeks. • c) How long will it take for half the population to be infected. 12% ill  88% healthy un+1 = 0∙88un uo = 5000 5000 ENTER 0∙88 × ANS u1 = 4400 u2= 3872 u3= 3407 After 3 weeks 3407 healthy u6= 2322 Half the population falls ill during 6th week

18. Linear Recurrence Relations A balloon contains 1 500ml of air and is being inflated by mouth. Each puff inflates the balloon by 15% but at the same time 100 ml of air escapes. (i) Find a linear recurrence relation to describe this. (ii) How much air is in the balloon after 5 puffs? (iii) If the volume reaches 3 litres then the balloon will burst. How many puffs will this take? (i) Suppose the starting volume is v0. Adding 15% gives us 115% or 1∙15 × previous amount But 100 ml then escapes

19. v0 = 1 500 Because 100 ml escapes = 1625 v1 = 1∙15v0 – 100 = 1768 v2 = 1∙15 × 1625 – 100 = 1934 v3 = 1∙15 × 1768 – 100 In general vn+1 = 1∙15vn – 100

20. In general vn+1 = 1∙15vn – 100 (ii) We can now use this formula as follows Using ANS function 1500 ENTER 1∙15 × ANS Each press of = gives the next term v0 = 1500 v5 = 2343 v1 = 1625 v2 = 1768 v8 = 3216 And so on Balloon bursts on 8th puff

21. A patient is given 160 ml of a drug. Every 6 hours 25% of the drug passes out of her bloodstream. To compensate she is given a further 20 ml dose every 6 hours. a) Write down a recurrence relation for the amount of drug in her bloodstream. b) How much drug will be in her bloodstream after 24 hours? 25% removed  75% left un+1 = 0∙75un + 20 uo = 160 160 ENTER 0∙75 × ANS + 20 u1 = 140 u2= 125 u3= 113∙75 u4= 105∙3 105 ml left in bloodstream

22. A job is advertised a starting salary of £24 000, with annual percentage increase of 7∙5% and an annual increment of £2 000. a) Find a recurrence relation for the total annual salary. b) Calculate the expected salary after 8 years. 7∙5% increase  107∙5% left un+1 = 1∙075un + 2000 uo = 24000 24000 ENTER 1∙075 × ANS + 2000 u2= 31885 And so on u1 = 27800 u8= £63 696∙21 Expected salary = £63 696∙21

23. The air pressure in a tractor tyre is 50 psi. Each week it loses 15% and 2 psi are pumped back in to compensate. The manufacturer states that it is dangerous to run the tyre at under 30 psi. a) Calculate the pressure after 3 weeks. b) Does the pressure ever drop below 30 psi? 15% loss  85% left un+1 = 0∙85un + 2 uo = 50 50 ENTER 0∙85 × ANS + 2 u2= 39∙8 u3 = 35∙9 u1 = 44∙5 Pressure after 3 weeks is 35∙9 psi If you keep pressing (=) answer is 13∙33∙ u5 = 29∙6

24. Linear Recurrence Relations These recurrence relations are called linear because they all fit a form that looks like : un+1 = aun + b Which is the same as y = mx + c un+1 = 1∙075un + 2000 un+1 = 0∙85un + 2 vn+1 = 1∙15vn – 100 un+1 = 0∙75un + 20

25. Linear Recurrence Relations Please note there are various ways to write the same recurrence relation. un = aun-1 + b un+1 = aun + b are the same since un+1 is the term after unand un is the term after un-1. Although we usually use u to represent the term in a sequence it is not necessary or essential. pn+1 = apn + b vn+1 = avn + b gn+1 = agn + b xn+1 = axn + b

26. Answers to Recurrenec Relations Worksheet

27. A farmer grows a variety of plum tree which ripens during the months of July and August. On the last day in July there was 2000 kg of ripe fruit ready to be picked. At the beginning of August the farmer hires some fruit pickers who manage to pick 75% of the ripe fruit each day. Also, each day 60 kg more of the plums become ripe. a) Find a recurrence for the weight of ripe plums left in the orchard. b) What is the estimated weight of ripe plums left in the orchard at the end of the day on the 7th August ? 75% picked  25% left W1 = 560kg End of 1st August Wn+1 = 0∙25Wn + 60 W2 = 200kg Enter 2000 0∙25 × ANS + 60 W7 = 80 kg

28. A large shoal of 300 fish are observed and it is noticed that every minute 30% of the fish leave the shoal and 25 join the shoal. a) Find a recurrence relation for the size of the shoal of fish. b) How many fish are in the shoal after 5 minutes and after 10 minutes? 30% leave  70% left N1 = 235 N2 = 190 Nn+1 = 0∙7 Nn + 25 Enter 300 N5 = 120 0∙7 × ANS + 25 N10 = 89

29. A pharmaceutical company is given permission to discharge a maximum of 60 kg of chemical waste into a section of river each day. The natural tidal nature of the river disperse 80% of the chemical each day. a) Find a recurrence relation for the amount of chemical in the river. b) How many kg of chemical waste are there in the river after 2 days and after 7 days? c) Due to environmental concerns the chemical waste in the section of river must not exceed 80 kg. Is it safe for the pharmaceutical company to continue discharging waste at this rate? Give a reason. Statement is essential, always worth a mark! 80% removed  20% left A1 = 72 kg An+1 = 0∙2An + 60 A2= 74∙4 kg Enter 60 0∙2 × ANS + 60 A7 = 75 kg In the long term level of waste levels off at 75 kg, so safe.

30. 120 rabbits are bred for sale. Each month the rabbit population increases by 15% and each month 30 rabbits are sold to customers. a) Write down a recurrence relation for the population of rabbits. b) Calculate how many rabbits there are after 6 months. c) What will happen to the rabbit population if it continues in this way? Statement is essential, always worth a mark! 15% increase  115% left R1 = 108 Rn+1 = 1∙15 Rn - 30 R6 = 15 Enter 120 1∙15 × ANS - 30 R10 = - 124 In the long term the rabbits will all be sold and the population disappears.

31. A patient is injected with 90 units of a drug in hospital. Every eight hours 30% of the drug passes out of the bloodstream. The patient is therefore given a further dose of 20 units of the drug at 8 hr intervals. • Find a recurrence relation for the amount of the drug in the bloodstream. • b) Calculate how many units of the drug you would expect there to be in the patient’s bloodstream after 48 hours. 30% loss  70% left D1 = 83 Dn+1 = 0∙7 Dn + 20 6 × 8hrs D6 = 69 units Enter 90 0∙7 × ANS + 20 Note that in the long term number of units of the drug will level off at 66∙7 units.

32. A family have taken out a £80000 mortgage to buy a cottage. The building society charge interest at 6% per annum. The family pay back £6000 each year. a) Find a recurrence relation for the amount of money owed to the building society. b) How much money do they owe after 5 years, after 10 years? c) During what year will the mortgage be paid off? 6% interest  106% left M1 = £78 800 M5 = £73 235 Mn+1 = 1∙06Mn – 6 000 M10 = £64 183 Enter 80 000 1∙06 × ANS – 6 000 M28 = - £2 233 Mortgage paid off during year 27

33. 3 = Divergence / Convergence/ Limits 2 × ANS + 4 As n   un   As n tends to infinity un tends to infinity Consider the following linear recurrence relations (a) un+1 = 2un + 4 with u0 = 3 u0 = 3 u1 = 10 u2 = 24 u3 = 52 u10 = 7164 u20 = 7340028 we say that the sequence DIVERGES.

34. 3 = 0∙5× ANS + 4 As n   un  8 we say that the sequence CONVERGES to a limit of 8. (b) un+1 = 0∙5un + 4 with u0 = 3 u0 = 3 u1 = 5∙5 u2 = 6∙75 u3 = 7∙375 u10 = 7∙995 u20 = 7∙999…..

35. 3 = -2 × ANS + 4 As n   un  ±  and we say that the sequence DIVERGES. (c) un+1 = -2un + 4 with u0 = 3 u0 = 3 u1 = -2 u2 = 8 u3 = -12 u10 = 1708 u20 = 1747628 u21 = -3495252

36. 3 = -0∙5× ANS + 4 As n   un  22/3 we say that the sequence CONVERGES to a limit of 22/3 (d) un+1 = -0∙5un + 4 with u0 = 3 u0 = 3 u1 = 2∙5 u2 = 2∙75 u3 = 2∙625 u10 = 2∙666 u20 = 2∙666

37. For un+1 = 0∙5un + 4 and un+1 = -0∙5un + 4 As n   un→ 8 or un  22/3 we say that the sequence CONVERGES to a limit For un+1 = 2un + 4 and un+1 = – 2un + 4 As n   un  ±  we say that the sequence DIVERGES

38. Conclusions The linear recurrence relation un+1 = aun + b converges to a limit only if – 1 < a < 1 Otherwise the sequence diverges.

39. When the linear recurrence relation un+1 = aun + b converges the answer remains the same no matter how many times you repeat the calculation. So un = un+1 = un+2 + un+3 …… If we let this limit be L, then un+1 = aun + b becomes L = aL + b Limit formula

40. Conclusion: if un+1 = aun + b converges to a limit then changing b changes the limit. Other Factors (e) compare this with (b) un+1 = 0∙5un + 10 with u0 = 3 u0 = 3 u1 = 11∙5 u2 = 15∙75 u3 = 17∙875 ….. u10 = 19∙98... …… u20 = 19∙99…. This is clearly heading to a limit of 20

41. Conclusion: if un+1 = aun + b converges to a limit then changing u0 does not affect the limit. (f) compare this with (b) un+1 = 0∙5un + 4 with u0 = 200 u0 = 200 u1 = 104 u2 = 56 u3 = 32 u10 = 8∙1875 u20 = 8∙0001…. Again this is heading to a limit of 8

42. A Formula for the Limit of a Converging Sequence un+1 = aun + b, converges if – 1 < a < 1 The limit, L, is given by the formula The limit depends on a and b but not on u0

43. Fish, like all animals need oxygen to survive. The fish in a certain tank use up 15% of the oxygen in the water each hour. However, due to the action of a pump, oxygen is added to the water at a rate of 1 part per metre3 each hour. The oxygen level in the tank should be between 5 and 7 parts per metre3 for the survival of the fish. Initially the concentration of oxygen in the tank is 6 ppm3 a) Write down a recurrence relation to describe the oxygen level . b) Say whether or not a limit exists, giving a reason. c) Determine, in the long term, whether the fish will survive. 15% used  85% left On+1 = 0∙85 On + 1 Limit exists since -1 < 0∙85 < 1 Fish survive since long term level is between 5 and 7 ppm3

44. An office worker has 70 folders on his desk ready to be filed. Each hour he manages to file 62% of the folders. However, another 20 are also added to his pile each hour. a) Letting Fn represent the number of folders on his desk after n hours, write down a recurrence relation to model this situation. b) How many folders are ready to be filed after 5 hours? c) In the long term how many folders should he expect on his desk ? 62% filed  38% left F1 = 46 F5 = 32 Fn+1 = 0∙38 Fn + 20 Enter 70 0∙38 × ANS + 20 Limit exists since -1 < 0∙38 < 1 In the long term the number of files levels off at 32

45. Trees are sprayed weekly with the pesticide, ’Killpest’, whose manufacturers claim it will destroy 65% of all pests. Between the weekly sprayings, it is estimated that 500 new pests invade the trees. A new pesticide, ’Pestkill’, comes onto the market. The manufacturers claim it will destroy 85% of existing pests but it is estimated that 650 new pests per week will invade the trees. Which pesticide will be more effective in the long term? 65% killed  35% left 85% killed  15% left Kn+1 = 0∙35 Kn + 500 Pn+1 = 0∙15 Pn + 650 In the long term Pestkill is marginally better

46. A Local Authority puts its Park cleaning out to tender. Kleenall claims it will remove 95% of all rubbish dropped each week. It is known that the public drop 25kg of rubbish each week. Pickit Ltd claim to remove 85% of all the litter dropped each week. They also say they will install litter bins which will reduce the amount of litter dropped each week to 20kg per week. If the two firms charge the same amount, which one should the council employ? 95% cleared  5% left 85% cleared  15% left Kn+1 = 0∙05 Kn + 25 Pn+1 = 0∙15 Pn + 20 In the long term Pickit Ltd is better

47. A chemical firm has applied to release 45 units of waste per week into the sea. It is estimated that the natural tidal action will disperse 60% of the waste each week. If a safe level is 80 units, is it safe to accept the application. 60% cleared  40% left Cn+1 = 0∙4 Cn + 45 In the long term level is constant at 75 units, so safe.

48. A hospital patient is put on medication which is taken once per day. The dose is 35mg and each day the patient’s metabolism burns off 70% of the drug in her system. It is known that if the level of the drug in the patients system reaches 54mg then the consequences could be fatal. Is it safe for the patient to take the medication indefinitely? Burning off 70% leaves behind 30% or 0∙3 un+1 = 0∙3un + 35  a = 0∙3, b = 35 In the long run level of drug in patients system levels out at 50 mg.

49. The brake fluid reservoir in a car is leaky. Each day it loses 3∙14% of its contents. The driver “tops up” once per week with 50ml of fluid. For safety reasons the level of fluid in the reservoir should always be between 200ml & 260ml. Initially it has 255ml. (a) Determine the fluid levels after 1 week and 4 weeks. (b) Is the process effective in the long run?  decay factor 0∙9686 3∙14% loss leaves 96∙86%

50. (a) Problem 3.14% daily loss = ? Weekly loss. Decay factor = 0∙9686  7 day factor = (0∙9686)7 Amount remaining after 1 week = (0.9686)7× A0 = 0∙799854 × A0 = 0∙80 × A0 or 80% of A0 The car is loses 20% of its brake fluid weekly So if An is the fluid level after n weeks then we have An+1 = 0∙8 An + 50