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Higher Unit 1

Outcome 4. Higher. Higher Unit 1. Recurrence Relations. Grow and Decay. Linear Recurrence Relation. Divergence / Convergence / Limits. Applications. Find a formula. Exam Type Questions. Outcome 4. Higher. Recurrence Relations. Sequences A 5 9 13 17 ……. B 3 6 12 24 …….

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Higher Unit 1

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

  2. Outcome 4 Higher Recurrence Relations Sequences A 5 9 13 17 ……. B 3 6 12 24 ……. C 2 3 5 8 13 …….. D 17 23 41 77 137 ……… E 2 3 5 7 11 ……… In the above sequences some have obvious patterns while others don’t however this does not mean that a pattern doesn’t exist.

  3. Recurrence Relations Outcome 4 Higher Notation Suppose we write the term 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.) The terms of a sequence can then be defined in two ways

  4. Recurrence Relations Outcome 4 Higher Either Using a formula for the nth term, un 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

  5. Recurrence Relations Outcome 4 Higher A 5 9 13 17 ……. Formula: un = 4n + 1 So u100 = 4 X 100 + 1 = 401 Recurrence Relation: un+1 = un + 4 with u1 = 5 u2 = u1 + 4 = 5 + 4 = 9 u3 = u2 + 4 = 9 + 4 = 13

  6. Recurrence Relations Outcome 4 Higher B 3 6 12 24 …… Formula: un = 3 X 2n-1 So u10 = 3 X 29 = 3 X 512 = 1536 Recurrence Relation: un+1 = 2un with u1 = 3. u2 = 2u1 = 2 X 3 = 6, u3 = 2u2 = 2 X 6 = 12, etc

  7. Recurrence Relations Outcome 4 Higher C 2 3 5 8 13 …….. No formula this time but we have a special type of recurrence relation called a FIBONACCISEQUENCE. Here u1 = 2 , u2 = 3 then we have u3 = u2 + u1 = 3 + 2 = 5 , u4 = u3 + u2 = 5 + 3 = 8 un+2 = un+1 + un In general ie apart from 1st two, each term is the sum of the two previous terms.

  8. Recurrence Relations Outcome 4 Higher D 17 23 41 77 137 ……… This sequence doesn’t have a recurrence relation but the terms can be found using the formula un = n3 - n + 17 Quite a tricky formula but it does work ... u1 = 13 - 1 + 17 = 17 u2 = 23 - 2 + 17 = 8 - 2 + 17 = 23 u10 = 103 - 10 + 17 = 1000 - 10 + 17 = 1007

  9. Recurrence Relations Outcome 4 Higher E 2 3 5 7 11 ……… 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.

  10. Outcome 4 Higher 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.

  11. Growth & Decay Outcome 4 Example 1 Higher An oven contains 10000 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 un represents 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

  12. Growth & Decay Outcome 4 Higher u0 = 10000 (a) u1 = 0.83u0 = 0.83 X 10000 = 8300 u2 = 0.83u1 = 0.83 X 8300 = 6889 u3 = 0.83u2 = 0.83 X 6889 = 5718 So there are 5718 bacteria after 3 hours. (b) u4 = 0.83u3 = 0.83 X 5718 = 4746 u5 = 0.83u4 = 0.83 X 4746 = 3939 This is less than 4000 so it takes 5 full hours to fall below 4000.

  13. Growth & Decay Outcome 4 Higher Example 2 The population of a town is growing 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

  14. Growth & Decay Outcome 4 Higher P1 = 1.14 P0 P2 = 1.14 P1 = 1.14 X 1.14 P0 = (1.14)2 P0 P3 = 1.14 P2 = 1.14 X (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 or (1.14)n P0 > 3 P0 Dividing by P0 we get (1.14)n> 3

  15. Growth & Decay Outcome 4 Higher We now use a bit of trial and error along with the ^ or xy 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.

  16. Outcome 4 Higher Linear Recurrence Relations Some recurrence relations take the form un+1 = kunwhere k is a real no. This leads to a formula for the nth term un = knu0where u0 is the starting value.

  17. Outcome 4 Higher Linear Recurrence Relations Many recurrence relations take the form un+1 = aun+ b where a & b are real nos. If we think about un+1 like y and un 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.

  18. Linear Recurrence Relations Example Outcome 4 Higher 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 V0. Adding 15% gives us 115% or 1.15 X previous amount,

  19. Linear Recurrence Relations Outcome 4 Higher however we also lose 100ml so we have… V1 = 1.15V0 - 100 similarly V2 = 1.15V1 - 100 and V3 = 1.15V2 - 100 In general Vn+1 = 1.15Vn - 100 (ii) We can now use this formula as follows V0 = 1500 V1 = 1.15 X 1500 - 100 = 1625

  20. BANG!!! Linear Recurrence Relations Outcome 4 Higher So after 5 puffs the balloon contains 2343ml of air. = 1769 V2 = 1.15 X 1625 - 100 = 1934 V3 = 1.15 X 1769 - 100 = 2124 V4 = 1.15 X 1934 - 100 V5 = 1.15 X 2124 - 100 = 2343 (iii) continuing the above V6 = 1.15 X 2343 - 100 = 2594 V7 = 1.15 X 2594 - 100 = 2883 V8 = 1.15 X 2883 - 100 = 3216 The balloon bursts on the 8th puff.

  21. Linear Recurrence Relations Outcome 4 Example Higher 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?

  22. Linear Recurrence Relations Outcome 4 Higher Let An be the amount of waste deposited after n weeks. So A0 = 150 Removing 60% leaves behind 40% or 0.4. This means that A1 = 0.4A0 + 150 Similarly A2 = 0.4A1 + 150 In general we get the recurrence relation An+1 = 0.4An + 150 and this gives us the following sequence…...

  23. Linear Recurrence Relations Outcome 4 Higher A0 = 150 A1 = 0.4 X 150 + 150 = 210 A2 = 0.4 X 210 + 150 = 234 A3 = 0.4 X 234 + 150 = 243.6 A10 = 0.4 X 249.974 + 150 = 249.990 When amount of waste reaches 250kg it stays at this. Check: If An = 250 then An+1 = 0.4 X 250 + 150 = 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.

  24. Outcome 4 Higher Divergence / Convergence/Limits As n   un   and we say that the sequence DIVERGES. 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

  25. Outcome 4 Higher Divergence / Convergence/Limits 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….. Check: if un = 8 un+1 = 0.5 X 8 + 4 = 8

  26. Outcome 4 Higher Divergence / Convergence/Limits As n   un  ± and we say that the sequence DIVERGES. un+1= -2un + 4 with u0 = 3 u0= 3 u1= -2 u2= 8 u3= -12 u10= 1708 u20= 1747628 u21= -3495252

  27. Outcome 4 Higher Divergence / Convergence/Limits 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 Check: if un = 22/3 un+1 =- 0.5 X22/3 + 4 = 22/3

  28. Divergence / Convergence/Limits Outcome 4 Higher Conclusions The linear recurrence relation un+1 = aun + 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 or a > 1 Then we say that the sequence diverges.

  29. Conclusion: if un+1 = aun + b converges to a limit then changing b changes the limit. Outcome 4 Higher un+1 = 0.5un + 4 with u0 = 3 Divergence / Convergence/Limits This is clearly heading to a limit of 20 Check: if un = 20 un+1 = 0.5 X 20 +10 = 20 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….

  30. Conclusion: if un+1 = aun + b converges to a limit then changing u0 does not affect the limit. Outcome 4 un+1 = 0.5un + 4 with u0 = 3 Higher Divergence / Convergence/Limits Again this is heading to a limit of 8 (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….

  31. Find the Limit Outcome 4 Higher Proof Suppose a limit exists for the recurrence relation un+1 = aun + b let the limit be L, then we have L = aL + b L = b L – aL = b (1 – a) Re arranging L(1 – a) = b

  32. Outcome 4 Example 1 Higher 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 We need to create a recurrence relation. First dose = u0 = 35 Burning off 70% leaves behind 30% or 0.3 After this another 35mg is taken so we have …..

  33. 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. Applications Outcome 4 Higher un+1 = 0.3un + 35 This sequence has a limit since 0 < 0.3 < 1 If we call the limit L then at this limit we have un+1 = un = L The equation un+1 = 0.3un + 35 now becomes L = 0.3L + 35 0.7L = 35 L = 35  0.7 = 350  7 = 50

  34. Outcome 4 Higher Example 2 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 (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?

  35. Applications Outcome 4 Higher (a) Problem 3.14% daily loss = ? Weekly loss. Losing 3.14% daily leaves behind 96.86% or 0.9686. Amount remaining after 1 week = (0.9686)7X A0 = 0.799854 X A0 = 0.80 X A0 or 80% of A0 This means that the car is losing 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

  36. Applications Outcome 4 Higher (b) Using An+1 = 0.8 An + 50 with A0 = 255 we get A0 = 255ml A1 = 254ml 1st week A2 = 253.2ml A3 = 252.6ml A4 = 252.0ml 4th week NB : even before adding the 50ml the level is above 200ml

  37. Applications Outcome 4 Higher (c) considering An+1 = 0.8 An + 50 Since 0 < 0.8 < 1 then a limit must exist and at this An+1 = An = L so An+1 = 0.8 An + 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.

  38. Example Outcome 4 Higher A recurrence relation is defined by the formula Finding a Formula un+1 = aun+ b Given that u6 = 48 , u7 = 44 and u8 = 42 then find a & b . u8 = au7 + b becomes 44a + b = 42 Sim. equations u7 = au6 + b becomes 48a + b = 44 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

  39. Finding a Formula Outcome 4 Example Higher The nth term in a sequence is given by the formula un = an + b Given that u10 = 25 and u12 = 31 then find a & b. Hence find u300 - the 300th term. Using un = an + b u10 = 10a+ b becomes 10a + b = 25 Sim. equations u12 = 12a+ b becomes 12a + b = 31 subtract up 2a = 6 a = 3

  40. Finding a Formula Outcome 4 Higher Now put a = 3 into 10a + b = 25 This gives us 30 + b = 25 So b = -5 The actual formula is un = 3n - 5 So u300 = 3 X 300 - 5 = 895

  41. Outcome 4 Higher In an arithmetic series there is a constant difference between consecutive terms. Two Special Series eg 8, 14, 20, 26, …… here d = un+1 - un = 6 u1 = 8 = 8 + (0 X 6) u2 = 14 = 8 + (1 X 6) u3 = 20 = 8 + (2 X 6) u4 = 26 = 8 + (3 X 6) In general un = u1 + (n-1) X d So for the above u100 = u1 + 99d = 8 + (99 X 6) = 602

  42. Two Special Series Outcome 4 Higher In a geometric series there is a constant ratio between consecutive terms. eg 5, 10, 20, 40, …… here r = un+1 un = 2 u1 = 5 = 5 X 20 u2 = 10 = 5 X 21 u3 = 20 = 5 X 22 u4 = 40 = 5 X 23 In general un = u1X r(n-1) So for the above u100 = u1X r 99 = 3.17 X 1030 = 5 X 299

  43. www.maths4scotland.co.uk Higher Maths Strategies Sequences Click to start

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

  45. A recurrence relation is defined by where -1 < p < -1 and u0 = 12 a) If u1 = 15 and u2 = 16 find the values of p and q b) Find the limit of this recurrence relation as n  Hint Maths4Scotland Higher Put u1 into recurrence relation Put u2 into recurrence relation Solve simultaneously: (2) – (1) Hence -1 < p < 1, so a limit L exists State limit condition Limit = 16½ Use formula Previous Next Quit Quit

  46. Hint Maths4Scotland Higher • 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. • If he adopts the “20% pruning policy”, to what height will he expect the trees to grow in the long run. • 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. un = height at the start of year Construct a recurrence relation -1 < 0.8 < 1, so a limit L exists State limit condition Use formula Limit = 2.5 metres Use formula again Minimum prune = 25% m = 0.75 Previous Next Quit Quit

  47. Hint Maths4Scotland Higher 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 un and un+1 and represent the amounts that he owes at the starts of two successive months. Write down a recurrence relation involving un and un+1 b) Find the date and amount of the final payment. u0 = 2500 Construct a recurrence relation Calculate each term in the recurrence relation 1 Mar u0 = 2500.00 1 Apr u1 = 2237.50 1 May u2 = 1971.06 1 Jun u3 = 1700.62 1 Jul u4 = 1426.14 1 Aug u5 = 1147.53 1 Sept u6 = 864.74 1 Oct u7 = 577.71 1 Nov u8 = 286.38 1 Dec Final payment £290.68 Previous Next Quit Quit

  48. 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. Hint Maths4Scotland Higher Use formula for each sequence Sequence 1 Sequence 2 Cross multiply Equate the two limits Simplify Solve Limit = 25 Deduction Since limit exists a  1, so Previous Next Quit Quit

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

  50. 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 nth term exceeds 1000, and ii) the value of that term. Hint Maths4Scotland Higher First sequence has no limit since 3 is not between –1 and 1 Requirement for a limit 2nd sequence has a limit since –1 < 0.3 < 1 Sequence 2 u0 = 1 u1 = 2.6 u2 = 7.4 u3 = 21.8 u4 = 65 u5 = 194.6 u6 = 583.4 u7 = 1749.8 List terms of 1st sequence Smallest value of n is 8; value of 8th term = 1749.8 Previous Next Quit Quit

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