1. Applications Example 1 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? *********************** 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 …..

2. u n+1 = 0.3u n + 35 This sequence has a limit since 0<0.3<1. If we call the limit L then at this limit we have u n+1 = u n = L The equation u n+1 = 0.3u n + 35 now becomes L = 0.3L + 35 0.7L = 35 L = 35  0.7= 350  7 = 50 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.

3. Example 2 (trickier!) 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. (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?

4. (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) 7 X A 0 = 0.799854 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

5. (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 (**) (**) even before adding the 50ml the level is above 200ml (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.