Geometric Series: Calculating the drug level prior to a maximum dosage.

Question

A dose of $D$ milligrams of a drug is taken every 12 hours. Assume that the drug's half-life is such that every $12$ hours a fraction $r$, with $0<r<1$ of the drug remains in the blood. Let $d_1= D$ be the amount of the drug in the blood after first dose. It follows that the amount of the drug in the blood after the $n^{\mathrm{th}}$ dose is $$d_n= D\sum_{k=0}^{n-1}r^k.$$

At the steady state, $$d_\infty = \lim_{n\to\infty} d_n = \frac D{1-r}.$$

$d_\infty$ is the drug level just AFTER a dose, so it is the maximum drug level. Find the minimum drug level $d_{\min}$, just PRIOR to a steady state dose. Verify that $$d_\infty - d_{\min}=D.$$

I have no idea how to do this. Any ideas?

Answer 1

Hint: If in steady state you have $d_\infty$ just after a dose, after $12$ hours you have $rd_\infty=d_{min}$ because it is just before a dose.

Answer 2

Let $d(t)$ be the amount of the drug in the blood $t$ hours after a steady-state dose for $0\leqslant t<12$. Then $d(t)=d_\infty e^{-\lambda t}$ for some $\lambda>0$ (this is the definition of exponential decay). Then $$d_{\min} = \lim_{t\uparrow12}d(t)=\lim_{t\uparrow12}d_\infty e^{-\lambda t}=d_\infty e^{-12\lambda}.$$ From the given information about the half-life of the drug, we have $$d_\infty e^{-12\lambda}=rd_\infty, $$ and hence $$\lambda = \frac{\log\left(\frac1r\right)}{12}.$$ It follows that $$d_\min = d_\infty e^{-12\lambda} = d_\infty e^{\log r} = rd_\infty, $$ whence $$ \begin{align*} d_\infty - d_\min &= d_\infty - rd_{\infty}\\ &= d_\infty(1-r)\\ &= \left(\frac D{1-r}\right)(1-r)\\ &= D. \end{align*} $$

Answer 3

You defined $d_n$ to be the amount of drug in the blood just after the $n$th dose is taken. Where does the expression $$d_n = D\sum_{k = 0}^{n - 1} r^k$$ come from? The $m$th dose contributed $D$ milligrams of drug at the moment it was taken. The $n$th dose is taken $n - m$ time units later; by then, the contribution of the $m$th dose has decayed to $Dr^{n-m}$ milligrams. Since $d_n$ is the sum of the contributions from all previous doses, plus the current dose, $$d_n = \sum_{m=1}^n Dr^{n-m}.$$ Defining $k = n - m$, we get the expression you wrote.

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Let's say $d'_n$ is the amount of drug in the blood just before the $n$th dose is taken. Calculating $d'_n$ is just like calculating $d_n$, except we leave out the $n$th dose: $$d'_n = \sum_{m=1}^{n-1} Dr^{n-m}.$$ This time, since $m$ only goes up to $n-1$, we can define $k = (n - 1) - m$ and get $$d'_n = D\sum_{k=0}^{n-2} r^{k+1}.$$ Pulling out an $r$, $$d'_n = Dr\sum_{k=0}^{n-2} r^k$$ we see that $$d'_n = rd_{n-1}.$$

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Here's a graph of how the amount of drug in the blood changes over time. At the time of each does, $d_n$ is highlighted with a pink circle, and the $d'_n$ is highlighted with a purple square.

You defined $d_\infty$ to be the maximum amount of drug in the blood at steady state, and $d_\text{min}$ to be the minimum amount. Looking at the graph above, you should be able to see that $$d_\infty = \lim_{n \to \infty} d_n$$ and $$d_\text{min} = \lim_{n \to \infty} d'_n$$ We worked out earlier that $d'_n = rd_{n-1}$. From that, you can figure out that $$d_\text{min} = r d_\infty,$$ so $d_\infty - d_\text{min} = (1-r)d_\infty$. The rest is straightforward.