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Worked solution

Find the limiting amount immediately before each injection

A patient receives 1010 units of a drug by injection every 2424 hours. The fraction remaining in the body after 2424 hours is e−1/5e^{-1/5}. If injections continue indefinitely, find the limiting number of units in the body just prior to an injection.

[Modified from Applied Calculus for Business, Economics, and the Social and Life Sciences (11th Edition)]

CJC · Sequences and Series · Example 37 (Self-Reading, Medicine)

Answer

10e1/5−1≈45.2\displaystyle\frac{10}{e^{1/5}-1}\approx45.2 units (3 s.f.).

Compare your reasoning

Separate the two moments in each cycle

nAnBn=qAn11010q210q+1010q2+10q310q2+10q+1010q3+10q2+10qn10qn−1+⋯+10q+1010qn+⋯+10q\begin{array}{c|l|l}n&A_n&B_n=qA_n\\\hline1&10&10q\\2&10q+10&10q^2+10q\\3&10q^2+10q+10&10q^3+10q^2+10q\\n&10q^{n-1}+\cdots+10q+10&10q^n+\cdots+10q\end{array}

Write q=e^(−1/5). Let A_n be the amount just after injection n and B_n the amount just before injection n+1. The source table places these two moments in separate columns.

Sum the pre-injection amounts

Bn=10q1−qn1−q⟹lim⁡n→∞Bn=10q1−qB_n=10q\frac{1-q^n}{1-q}\quad\Longrightarrow\quad\lim_{n\to\infty}B_n=\frac{10q}{1-q}

The first pre-injection term is 10q, not 10. Since 0<q<1, the geometric series converges.

Substitute the retention fraction

10e−1/51−e−1/5=10e1/5−1=45.166555…≈45.2 units\begin{aligned}&\frac{10e^{-1/5}}{1-e^{-1/5}}\\ & =\frac{10}{e^{1/5}-1}\\ & =45.166555\ldots\approx45.2\ \text{units}\end{aligned}

Multiply numerator and denominator by e^(1/5). This is the low point immediately before each injection, not a constant amount throughout the daily cycle.

How did this question go?

Common mistakes

The limiting amount just after injection is ten units larger, approximately 55.2 units. The question asks for the amount just before injection.