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www.mathsrevision.com Higher Unit 1 www.mathsrevision.com Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence / Convergence / Limits.

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Presentation on theme: "www.mathsrevision.com Higher Unit 1 www.mathsrevision.com Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence / Convergence / Limits."— Presentation transcript:

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2 Higher Unit 1 Recurrence Relations Grow and Decay Linear Recurrence Relation Divergence / Convergence / Limits Applications Find a formula Exam Type Questions Higher Outcome 4

3 Recurrence Relations Sequences A591317……. B361224……. C235813…….. D ……… E235711……… In the above sequences some have obvious patterns while others don’t however this does not mean that a pattern doesn’t exist. Higher Outcome 4

4 Notation Suppose we write the term of a sequence as u 1, u 2, u 3, …….., u n-1, u n, u n+1, ……... where u 1 is the 1 st term, u 2 is the 2 nd term etc…. and u n is the n th term ( n being any whole number.) The terms of a sequence can then be defined in two ways Recurrence Relations Higher Outcome 4

5 Either Using a formula for the nth term, u n in terms of the value n Or By expressing each term using the previous term(s) in the sequence. This is called a Recurrence Relation Now reconsider the sequences at the start Recurrence Relations Higher Outcome 4

6 Recurrence Relation: u n+1 = u n + 4 with u 1 = 5 Formula: u n = 4n + 1 A591317……. So u 100 = 4 X = 401 u 2 = u = = 9 u 3 = u = = 13 Recurrence Relations Higher Outcome 4

7 Recurrence Relation: u n+1 = 2u n with u 1 = 3. B361224…… Formula: u n = 3 X 2 n-1 So u 10 = 3 X 2 9 = 3 X 512 = 1536 u 2 = 2u 1 = 2 X 3 = 6, u 3 = 2u 2 = 2 X 6 = 12, etc Recurrence Relations Higher Outcome 4

8 C235813…….. No formula this time but we have a special type of recurrence relation called a FIBONACCI SEQUENCE. Here u 1 = 2, u 2 = 3 then we have u 3 = u 2 + u 1 = = 5, u 4 = u 3 + u 2 = = 8 In general u n+2 = u n+1 + u n ie apart from 1st two, each term is the sum of the two previous terms. Recurrence Relations Higher Outcome 4

9 D ……… This sequence doesn’t have a recurrence relation but the terms can be found using the formula u n = n 3 - n + 17 Quite a tricky formula but it does work... u 1 = = 17 u 2 = = = 23 u 10 = = = 1007 Recurrence Relations Higher Outcome 4

10 E235711……… This sequence is the PRIME NUMBERS (NB: Primes have exactly two factors !!) There is neither a formula nor a recurrence relation which will give us all the primes. Recurrence Relations Higher Outcome 4

11 Growth & Decay Removing 15% leaves behind 85% or 0.85 which is called the DECAY factor. Adding on 21% gives us 121% or 1.21 and this is called the GROWTH factor. Growth and decay factors allow us a quick method of tackling repeated % changes. Higher Outcome 4

12 Example 1 An oven contains 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 4000 bacteria? Suppose that u n represents the number of bacteria remaining after n hours. Removing 17% leaves behind 83% so the DECAY factor is 0.83 and u n+1 = 0.83 u n Growth & Decay Higher Outcome 4

13 (a) u 0 = u 1 = 0.83u 0 = 0.83 X = 8300 u 2 = 0.83u 1 = 0.83 X 8300 = 6889 u 3 = 0.83u 2 = 0.83 X 6889 = 5718 So there are 5718 bacteria after 3 hours. (b) u 4 = 0.83u 3 = 0.83 X 5718 = 4746 u 5 = 0.83u 4 = 0.83 X 4746 = 3939 This is less than 4000 so it takes 5 full hours to fall below Growth & Decay Higher Outcome 4

14 Example 2 The population of a town is growing at a rate of 14% per annum. If P 0 is the initial population and P n is the population after n years. (a) Find a formula for P n in terms of P 0. (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 P n+1 = 1.14 P n Growth & Decay Higher Outcome 4

15 Growth & Decay P 1 = 1.14 P 0 P 2 = 1.14 P 1 = 1.14 X 1.14 P 0 = (1.14) 2 P 0 P 3 = 1.14 P 2 = 1.14 X (1.14) 2 P 0 = (1.14) 3 P 0 So in general we have P n = (1.14) n P 0 If the population trebles then we need to have P n > 3 P 0 or (1.14) n P 0 > 3 P 0 Dividing by P 0 we get (1.14) n > 3 Higher Outcome 4

16 Growth & Decay We now use a bit of trial and error along with the ^ or x y buttons on the calculator. If n = 5 then (1.14) 5 = 1.92… too small If n = 9 then (1.14) 9 = 3.25… too big If n = 7 then (1.14) 7 = 2.50… too small If n = 8 then (1.14) 8 = 2.85… too small but closest to 3. From the above we can say it takes just over 8 years for the population to treble. Higher Outcome 4

17 Linear Recurrence Relations Some recurrence relations take the form u n+1 = ku n where k is a real no. This leads to a formula for the n th term u n = k n u 0 where u 0 is the starting value. Higher Outcome 4

18 Linear Recurrence Relations Many recurrence relations take the form u n+1 = au n + b where a & b are real nos. If we think about u n+1 like y and u n like x then we get y = ax + b and this is basically the same as y = mx + c which is the equation of a straight line Hence the expression “Linear Recurrence Relations” Many day to day scenarios can be modelled by this. Higher Outcome 4

19 Example A balloon contains 1500ml of air and is being inflated by mouth. Each puff inflates the balloon by 15% but at the same time 100ml of air escapes. (i) Find a linear recurrence relation to describe this situation. (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? (NB: 3litres = 3000ml) (i) Suppose the starting volume is V 0. Adding 15% gives us 115% or 1.15 X previous amount, Linear Recurrence Relations Higher Outcome 4

20 however we also lose 100ml so we have… V 1 = 1.15V similarly V 2 = 1.15V and V 3 = 1.15V In general V n+1 = 1.15V n (ii) We can now use this formula as follows V 0 = 1500 V 1 = 1.15 X = 1625 Linear Recurrence Relations Higher Outcome 4

21 Linear Recurrence Relations V 2 = 1.15 X = 1769 V 3 = 1.15 X = 1934 V 4 = 1.15 X = 2124 V 5 = 1.15 X = 2343 So after 5 puffs the balloon contains 2343ml of air. (iii) continuing the above V 6 = 1.15 X = 2594 V 7 = 1.15 X = 2883 V 8 = 1.15 X = 3216 BANG!!! The balloon bursts on the 8th puff. Higher Outcome 4

22 Example A factory wishes to dump 150kg of a particular waste product into a local steam once per week. The flow of the water removes 60% of this material from the stream bed each week. However it has been calculated that if the level of deposit on the stream bed reaches 265kg then there will be a serious risk to the aquatic life. Should the factory be allowed to dump this waste indefinitely? Linear Recurrence Relations Higher Outcome 4

23 Let A n be the amount of waste deposited after n weeks. So A 0 = 150 Removing 60% leaves behind 40% or 0.4. This means that A 1 = 0.4A Similarly A 2 = 0.4A In general we get the recurrence relation A n+1 = 0.4A n and this gives us the following sequence…... Linear Recurrence Relations Higher Outcome 4

24 Linear Recurrence Relations A 0 = 150 A 1 = 0.4 X = 210 A 2 = 0.4 X = 234 A 3 = 0.4 X = A 10 = 0.4 X = When amount of waste reaches 250kg it stays at this. Check: If A n = 250 then A n+1 = 0.4 X = 250 This is below the danger level so factory could be allowed to continue dumping. We say that the sequence CONVERGES to a LIMIT of 250. Higher Outcome 4

25 Divergence / Convergence/Limits Consider the following linear recurrence relations (a) u n+1 = 2u n + 4 with u 0 = 3 u 0 = 3 u 1 = 10 u 2 = 24 u 3 = 52 u 10 = 7164 u 20 = As n   u n   and we say that the sequence DIVERGES. Higher Outcome 4

26 Divergence / Convergence/Limits (b) u n+1 = 0.5u n + 4 with u 0 = 3 u 0 = 3 u 1 = 5.5 u 2 = 6.75 u 3 = u 10 = U 20 = 7.999….. As n   u n  8 we say that the sequence CONVERGES to a limit of 8. Check: if u n = 8 u n+1 = 0.5 X = 8 Higher Outcome 4

27 Divergence / Convergence/Limits (c) (c) u n+1 = -2u n + 4 with u 0 = 3 u 0 = 3 u 1 = -2 u 2 = 8 u 3 = -12 u 10 = 1708 u 20 = u 21 = As n   u n  ±  and we say that the sequence DIVERGES. Higher Outcome 4

28 Divergence / Convergence/Limits (d) u n+1 = -0.5u n + 4 with u 0 = 3 u 0 = 3 u 1 = 2.5 u 2 = 2.75 u 3 = u 10 = u 20 = As n   u n  2 2 / 3 we say that the sequence CONVERGES to a limit of 2 2 / 3. Check: if u n = 2 2 / 3 u n+1 =- 0.5 X 2 2 / = 2 2 / 3 Higher Outcome 4

29 Divergence / Convergence/Limits Conclusions The linear recurrence relation u n+1 = au n + b converges to a limit if either -1 < a < 0 or 0 < a < 1 This is usually written as 0 < a < 1 If a > 1 ie a 1 Then we say that the sequence diverges. Higher Outcome 4

30 u n+1 = 0.5u n + 4 with u 0 = 3 Divergence / Convergence/Limits Other Factors (e) compare this with (b) u n+1 = 0.5u n + 10 with u 0 = 3 u 0 = 3 u 1 = 11.5 u 2 = u 3 = ….. u 10 = …… u 20 = 19.99…. This is clearly heading to a limit of 20 Check: if u n = 20 u n+1 = 0.5 X = 20 Higher Outcome 4 Conclusion: if u n+1 = au n + b converges to a limit then changing b changes the limit.

31 Divergence / Convergence/Limits (f) compare this with (b) u n+1 = 0.5u n + 4 with u 0 = 200 u 0 = 200 u 1 = 104 u 2 = 56 u 3 = 32 u 10 = u 20 = …. Again this is heading to a limit of 8 Higher Outcome 4 u n+1 = 0.5u n + 4 with u 0 = 3 Conclusion: if u n+1 = au n + b converges to a limit then changing u 0 does not affect the limit.

32 Find the Limit Proof Suppose a limit exists for the recurrence relation u n+1 = au n + b let the limit be L, then we have L = aL + b Re arranging Higher Outcome 4 L – aL = b L(1 – a) = b L = b (1 – a)

33 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? Applications Example 1 We need to create a recurrence relation. First dose = u 0 = 35 Burning off 70% leaves behind 30% or 0.3 After this another 35mg is taken so we have ….. Higher Outcome 4

34 Outcome 4 u n+1 = 0.3u n + 35 This sequence has a limit since 0 < 0.3 < 1 u n+1 = u n = L The equation u n+1 = 0.3u n + 35 now becomes L = 0.3L L = 35 L = 35  0.7= 350  7 = 50 Applications If we call the limit L then at this limit we have Higher Conclusion: the level of drug in the patients system will never exceed 50mg under these conditions. Since this is below the danger level it would be safe to continue indefinitely.

35 The brake fluid reservoir in a car is leaky. Each day it loses 3.14% of its contents. To compensate for this daily loss 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. Applications Example 2 (a) Find a recurrence relation to describe the above. (b) Determine the fluid levels after 1 week and 4 weeks. (c.) Is the process effective in the long run? Higher Outcome 4

36 Applications (a) Problem 3.14% daily loss = ? Weekly loss. Losing 3.14% daily leaves behind 96.86% or Amount remaining after 1 week = (0.9686) 7 X A 0 = X A 0 = 0.80 X A 0 or 80% of A 0 This means that the car is losing 20% of its brake fluid weekly So if A n is the fluid level after n weeks then we have A n+1 = 0.8 A n + 50 Higher Outcome 4

37 Applications (b) Using A n+1 = 0.8 A n + 50 with A 0 = 255 we get A 0 = 255ml A 1 = 254ml 1 st week A 2 = 253.2ml A 3 = 252.6ml A 4 = 252.0ml 4 th week NB : even before adding the 50ml the level is above 200ml Higher Outcome 4

38 Applications (c) considering A n+1 = 0.8 A n + 50 Since 0 < 0.8 < 1 then a limit must exist and at this A n+1 = A n = L so A n+1 = 0.8 A n + 50 ie L = 0.8L + 50 or 0.2L = 50 or L = 50  0.2= 500  2 = 250 In the long run the weekly level will be 250ml and won’t fall below 200ml so the driver should be OK with this routine. Higher Outcome 4

39 Given that u 6 = 48, u 7 = 44 and u 8 = 42 then find a & b. Finding a Formula Example A recurrence relation is defined by the formula u n+1 = au n + b u 8 = au 7 + b becomes44a + b = 42 u 7 = au 6 + b becomes48a + b = 44 Sim. equations Subtract up 4a = 2 so a = 0.5 Now put a = 0.5 into 44a + b = 42 to get 22 + b = 42 so b = 20 Higher Outcome 4

40 Example The nth term in a sequence is given by the formula u n = an + b Given that u 10 = 25 and u 12 = 31 then find a & b. Hence find u the 300 th term. Using u n = an + b u 10 = 10a + b becomes10a + b = 25 u 12 = 12a + b becomes12a + b = 31 Sim. equations subtract up2a = 6a = 3 Finding a Formula Higher Outcome 4

41 Now put a = 3 into 10a + b = 25 This gives us 30 + b = 25 So b = -5 The actual formula is u n = 3n - 5 So u 300 = 3 X = 895 Finding a Formula Higher Outcome 4

42 eg8, 14, 20, 26, ……here d = u n+1 - u n = 6 u 1 = 8 = 8 + (0 X 6) Two Special Series In an arithmetic series there is a constant difference between consecutive terms. u 2 = 14 = 8 + (1 X 6) u 3 = 20 = 8 + (2 X 6) u 4 = 26 = 8 + (3 X 6) In general u n = u 1 + (n-1) X d So for the above u 100 = u d = 8 + (99 X 6)= 602 Higher Outcome 4

43 In a geometric series there is a constant ratio between consecutive terms. eg5, 10, 20, 40, ……here r = u n+1  u n = 2 u 1 = 5 = 5 X 2 0 u 2 = 10 = 5 X 2 1 u 3 = 20 = 5 X 2 2 u 4 = 40 = 5 X 2 3 In general u n = u 1 X r (n-1) So for the above u 100 = u 1 X r 99 = 5 X 2 99 = 3.17 X Two Special Series Higher Outcome 4

44 Higher Maths Strategies Click to start Sequences

45 Maths4Scotland Higher Sequences The following questions are on Non-calculator questions will be indicated Click to continue You will need a pencil, paper, ruler and rubber.

46 Maths4Scotland Higher Hint PreviousNext Quit Put u 1 into recurrence relation Solve simultaneously: A recurrence relation is defined by where -1 < p < -1 and u 0 = 12 a)If u 1 = 15 and u 2 = 16 find the values of p and q b)Find the limit of this recurrence relation as n   Put u 2 into recurrence relation (2) – (1) Hence State limit condition -1 < p < 1, so a limit L exists Use formula Limit = 16½

47 Maths4Scotland Higher Hint PreviousNext Quit Construct a recurrence relation State limit condition -1 < 0.8 < 1, so a limit L exists Use formula Limit = 2.5 metres A man decides to plant a number of fast-growing trees as a boundary between his property and the property of his neighbour. He has been warned however by the local garden centre, that during any year, the trees are expected to increase in height by 0.5 metres. In response to this warning, he decides to trim 20% off the height of the trees at the start of any year. (a)If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run. (b)His neighbour is concerned that the trees are growing at an alarming rate and wants assurance that the trees will grow no taller than 2 metres. What is the minimum percentage that the trees will need to be trimmed each year so as to meet this condition. u n = height at the start of year Use formula again Minimum prune = 25% m = 0.75

48 Maths4Scotland Higher Hint PreviousNext Quit Construct a recurrence relation On the first day of March, a bank loans a man £2500 at a fixed rate of interest of 1.5% per month. This interest is added on the last day of each month and is calculated on the amount due on the first day of the month. He agrees to make repayments on the first day of each subsequent month. Each repayment is £300 except for the smaller final amount which will pay off the loan. a) The amount that he owes at the start of each month is taken to be the amount still owing just after the monthly repayment has been made. Let u n and u n+1 and represent the amounts that he owes at the starts of two successive months. Write down a recurrence relation involving u n and u n+1 b) Find the date and amount of the final payment. u 0 = 2500 Calculate each term in the recurrence relation 1 Mar u 0 = Apr u 1 = May u 2 = Jun u 3 = Jul u 4 = Aug u 5 = Sept u 6 = Oct u 7 = Nov u 8 = Dec Final payment £290.68

49 Maths4Scotland Higher Hint PreviousNext Quit Equate the two limits Cross multiply Sequence 1 Since limit exists a  1, so Use formula for each sequence Limit = 25 Two sequences are generated by the recurrence relations and The two sequences approach the same limit as n  . Determine the value of a and evaluate the limit. Sequence 2 Simplify Solve Deduction

50 Maths4Scotland Higher Hint PreviousNext Quit Equate the two limits Cross multiply Sequence 1 Use formula for each sequence Sequence 2 Rearrange Two sequences are defined by the recurrence relations If both sequences have the same limit, express p in terms of q.

51 Maths4Scotland Higher Hint PreviousNext Quit Sequence 2 Requirement for a limit List terms of 1 st sequence Two sequences are defined by these recurrence relations a)Explain why only one of these sequences approaches a limit as n   b)Find algebraically the exact value of the limit. c)For the other sequence find i)the smallest value of n for which the n th term exceeds 1000, and ii)the value of that term. First sequence has no limit since 3 is not between –1 and 1 2 nd sequence has a limit since –1 < 0.3 < 1 u 0 = 1 u 1 = 2.6 u 2 = 7.4 u 3 = 21.8 u 4 = 65 u 5 = u 6 = u 7 = Smallest value of n is 8; value of 8 th term =

52 Maths4Scotland Higher Previous Quit You have completed all 6 questions in this presentation Back to start

53 Are you on Target ! Update you log book Make sure you complete and correct ALL of the Recurrence RelationsRecurrence Relations questions in the past paper booklet. Outcome 4


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