1. LINEAR RECURRENCE RELATIONS NB: 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.

2. 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.

3. Generating a Sequence on a DAL Calculator (NB: this page not in handout !) A sequence is given by u n+1 = 0.3u n + 28 with u 1 = 70 To get the terms on a DAL Calculator first key in 70 ie u 1 Now press ans or = Now press X 0.3 + 28 Now each time you press = you will get the next term. The first few terms should be 70, 49, 42.7, 40.81, ….. You should now see 70 = in the display. Keep pressing = and eventually the answer is always 40.

4. Ex1A 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,

5. however we also lose 100ml so we have… V 1 = 1.15V 0 - 100 similarly V 2 = 1.15V 1 - 100 and V 3 = 1.15V 2 - 100 In general V n+1 = 1.15V n - 100 (ii) We can now use this formula as follows V 0 = 1500 V 1 = 1.15 X 1500 - 100 = 1625

6. V 2 = 1.15 X 1625 - 100= 1769 V 3 = 1.15 X 1769 - 100= 1934 V 4 = 1.15 X 1934 - 100= 2124 V 5 = 1.15 X 2124 - 100= 2343 So after 5 puffs the balloon contains 2343ml of air. (iii) continuing the above V 6 = 1.15 X 2343 - 100= 2594 V 7 = 1.15 X 2594 - 100 = 2883 V 8 = 1.15 X 2883 - 100= 3216 BANG!!! The balloon bursts on the 8th puff.

7. Ex2A 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?

8. 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 0 + 150 Similarly A 2 = 0.4A 1 + 150 etc In general we get the recurrence relation A n+1 = 0.4A n + 150 and this gives us the following sequence…...

9. A 0 = 150 A 1 = 0.4 X 150 + 150= 210 A 2 = 0.4 X 210 + 150= 234 A 3 = 0.4 X 234 + 150= 243.6 etc A 10 = 0.4 X 249.974 + 150= 249.990 etc When amount of waste reaches 250kg it stays at this. Check: If A n = 250 then A n+1 = 0.4 X 250 + 150 = 250 This is below the danger level so factory could be allowed to continue dumping. NB: We say that the sequence CONVERGES to a LIMIT of 250.