Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Many people take aspirin on a regular basis as a preventive measure for heart disease. Suppose a person takes \(80 \mathrm{mg}\) of aspirin every 24 hours. Assume also that aspirin has a half-life of 24 hours; that is, every 24 hours, half of the drug in the blood is eliminated. a. Find a recurrence relation for the sequence \(\left\\{d_{n}\right\\}\) that gives the amount of drug in the blood after the \(n\) th dose, where \(d_{1}=80\). b. Using a calculator, determine the limit of the sequence. In the long run, how much drug is in the person's blood? c. Confirm the result of part (b) by finding the limit of \(\left\\{d_{n}\right\\}\) directly.

Short Answer

Expert verified
The long-run amount of aspirin in the person's blood is 160 mg.

Step by step solution

01

Setting up the recurrence relation

To set up the recurrence relation for the amount of drug in the blood after each dose, we first consider the given information: 1. The person takes a dose of \(80 \mathrm{mg}\) of aspirin after every \(24\) hours. 2. The half-life of aspirin is \(24\) hours. We are given that \(d_1 = 80\). Now, let's consider \(d_2\). The person takes a new \(80 \mathrm{mg}\) dose after \(24\) hours, so the remaining drug in the blood after \(24\) hours is half of the previous dose. Therefore, after the second dose, the amount of drug in the person's blood will be \(0.5d_1 + 80\). In general, when the person takes the \(n\)th dose, the amount of drug in the person's blood would be half of what was left from the previous dose plus the new \(80 \mathrm{mg}\) dose. So we can write the recurrence relation as: \(d_{n} = 0.5d_{n-1} + 80\) for \(n \geq 2\). Now let's find the limit of this sequence.
02

Finding the limit using a calculator

To find the limit of this sequence, we can use a calculator to compute the first few terms of the sequence and see if it converges to a particular value. \(d_1 = 80\); \(d_2 = 0.5d_1 + 80 = 0.5(80) + 80 = 120\); \(d_3 = 0.5d_2 + 80 = 0.5(120) + 80 = 140\); \(d_4 = 0.5d_3 + 80 = 0.5(140) + 80 = 150\); \(d_5 = 0.5d_4 + 80 = 0.5(150) + 80 = 155\); ... After several iterations, we can see that the sequence seems to converge to \(160 \mathrm{mg}\). This means that in the long run, the person will have \(160 \mathrm{mg}\) of aspirin in their blood. Now, let's confirm this result by finding the limit directly.
03

Confirming the limit directly

To find the limit directly, we can set up the following equation $$ \lim_{n \to \infty} d_n = L $$ Since the limit exists, we know that $$ \lim_{n \to \infty} d_{n-1} = L $$ From the recurrence relation, we have $$ d_n = 0.5d_{n-1} + 80 $$ Taking the limit on both sides $$ \lim_{n \to \infty} d_n = 0.5\lim_{n \to \infty} d_{n-1} + 80 $$ Substituting the limits (\(L\)) we get $$ L = 0.5L + 80 $$ Solving for \(L\), $$ L - 0.5L = 80 \\ 0.5L = 80 \\ L = 160 $$ So, the limit, and thus the long-run amount of drug in the person's blood, is indeed \(160 \mathrm{mg}\), confirming our earlier result.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aspirin Dosage
When a person takes aspirin regularly, the body processes it at a certain rate. In this scenario, an individual takes an 80 mg aspirin dose every 24 hours. This creates a sequence where the amount of aspirin in the bloodstream changes over time. Essentially, how much aspirin remains in the blood after each dose depends on both the new dose and the amount remaining from previous doses.

Understanding this process is crucial for determining a safe and effective regimen. Too much aspirin in the body could be harmful, while too little might not provide the intended benefits. Thus, the determination of safe dosage intervals is based largely on the drug's half-life.

For calculations:
  • The amount of aspirin initially added is always 80 mg after each dose.
  • The remaining dose decreases due to the half-life effect.
Half-life
The half-life refers to the time it takes for the concentration of a substance in the body to reduce to half its initial amount. In the context of aspirin, the half-life is 24 hours. This means after 24 hours, only half of the aspirin remains in the bloodstream.

This characteristic plays a critical role in designing dosage schedules. It ensures that the remaining amount of aspirin does not accumulate to toxic levels. In our exercise, after each dose, the amount of aspirin decreases by half of what was present before the new dose is taken.
  • For instance, if one initially has 80 mg, the next day halves it to 40 mg, before adding the new dose.
  • This pattern persists, guiding the creation of the recurrence relation that models aspirin concentration over time.
Limit of a Sequence
In mathematics, the limit of a sequence is what the sequence tends to as the number of terms increases indefinitely. In our task, we're interested in the sequence of aspirin levels in the blood over days of continuous dosing, as represented by the recurrence relation: \[d_{n} = 0.5d_{n-1} + 80\]Finding the limit of this sequence helps us understand the long-term concentration of aspirin in the bloodstream. By calculating a few terms, we notice that the sequence approaches 160 mg, indicating that this is the limit.

Calculating the limit involves setting up and solving the equation: \[L = 0.5L + 80\]By solving, we find that the sequence converges to a limit (\(L\)) of 160 mg. This value suggests that, over time, the patient's bloodstream stabilizes at 160 mg of aspirin provided the regimen is maintained. This equilibrium represents a balance between the dose taken and the half-life effect.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100\)

Stirling's formula Complete the following steps to find the values of \(p>0\) for which the series \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges. a. Use the Ratio Test to show that \(\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}\) converges for \(p>2\). b. Use Stirling's formula, \(k !=\sqrt{2 \pi k} k^{k} e^{-k}\) for large \(k,\) to determine whether the series converges when \(p=2\). (Hint: \(1 \cdot 3 \cdot 5 \cdots(2 k-1)=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots(2 k-1) 2 k}{2 \cdot 4 \cdot 6 \cdots 2 k}\) (See the Guided Project Stirling's formula and \(n\) ? for more on this topic.)

Give an argument similar to that given in the text for the harmonic series to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5$$

Consider the sequence \(\left\\{x_{n}\right\\}\) defined for \(n=1,2,3, \ldots\) by $$x_{n}=\sum_{k=n+1}^{2 n} \frac{1}{k}=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2 n}.$$ a. Write out the terms \(x_{1}, x_{2}, x_{3}\). b. Show that \(\frac{1}{2} \leq x_{n}<1,\) for \(n=1,2,3, \ldots\). c. Show that \(x_{n}\) is the right Riemann sum for \(\int_{1}^{2} \frac{d x}{x}\) using \(n\) subintervals. d. Conclude that \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free