Drug dose recurrence and the limit shortcut
A patient is given 8.0 mg of a drug at the same time each day. Immediately after the nth dose, the amount in the bloodstream is A_n mg. Between doses the body clears 65% of the drug, so the model is A_{n+1}=0.35A_n+8.0 with A_1=8.0. A student writes: 'For the long-term amount, set A_{n+1}=A_n=A. Then A=0.35A+8.0, so 0.65A=8.0 and A=12.307... mg. Therefore the amount never exceeds 12.31 mg.' A clinician asks whether the amount just after a dose ever exceeds 13 mg, and whether a safety threshold of 15 mg could be crossed. The student has not checked whether the sequence actually converges or whether it approaches the fixed point from below. The class must decide whether the shortcut is legitimate and what additional argument is needed.
1. Show that A_n is increasing and bounded above by 16, or otherwise justify that the limit exists. Then find the limit and decide whether the clinician's 13 mg question is answered by the value 12.307... alone.
Read the worked answer
Let L=160/13. From A_{n+1}=0.35A_n+8, subtract L: A_{n+1}-L=0.35(A_n-L), since L=0.35L+8. With A_1=8<L, it follows by induction that A_n<L for all n. Also A_{n+1}-A_n=8-0.65A_n>0 because A_n<L and 0.65L=8. Hence the sequence is increasing and bounded above by L, so it converges. Its limit must satisfy L=0.35L+8, giving L=160/13≈12.31 mg. The fixed-point shortcut alone does not justify 'never exceeds 12.31'; it only identifies a possible limit. The monotonicity argument from below is what rules out overshooting. Thus the amount just after a dose never exceeds 12.31 mg, so it never exceeds 13 mg.
2. Solve the recurrence explicitly and use it to find the smallest n for which A_n>12 mg. Comment on whether this changes the safety conclusion about the 15 mg threshold.
Read the worked answer
Solving the recurrence, A_n-L=0.35^{n-1}(A_1-L). Here A_1-L=8-160/13=-56/13, so A_n=160/13-(56/13)(0.35)^{n-1}. For A_n>12: 160-56(0.35)^{n-1}>156, so (0.35)^{n-1}<1/14. Taking logs gives n-1>ln(1/14)/ln(0.35)≈2.51, so n≥4. Check: A_2=10.8, A_3=11.78, A_4≈12.12, so the smallest n is 4. This does not change the safety conclusion: the limit 12.31 mg is below 15 mg, and the sequence approaches it from below, so the 15 mg threshold is never crossed.