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Chapter 9: Problem 59

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
Answer: The long-term amount of aspirin in the person's blood is 160mg.

Step by step solution

01

Determine the initial conditions and dosage information.

Given the initial dosage \(d_1 = 80mg\) and the half-life of aspirin as 24 hours, we can define the recurrence relation as \(d_n = d_{n-1} + 80\), where \(d_{n-1}\) is the amount of aspirin left in the person's blood from the previous day.
02

Calculate the amount of aspirin remaining after each day.

Since the half-life of aspirin is 24 hours, every 24 hours half of the drug in the blood is eliminated. Therefore, we can write the amount of aspirin remaining after n days as: \(d_n = \frac{1}{2}d_{n-1} + 80\). This is the recurrence relation for the amount of drug in the blood after the \(n\)-th dose. #b. Limit of the sequence#
03

Determine the limit of the sequence.

To find the limit of the sequence as \(n\) approaches infinity, we can first recognize that it is a geometric sequence with a common ratio of \(\frac{1}{2}\) and a constant term of 80. We can find the limit by using the formula for the sum of an infinite geometric series: L = \(\frac{initial\_term}{1 - common\_ratio} = \frac{80}{1 - \frac{1}{2}} = \frac{80}{\frac{1}{2}} = 160\). In the long run, there will be 160mg of aspirin in the person's blood. #c. Confirm the result#
04

Confirm the result of part (b) by finding the limit of \(\left\\{d_{n}\right\\}\) directly.

To confirm the result in part (b), we need to find the limit of the sequence \(\left\\{d_{n}\right\\}\) as \(n\) approaches infinity. We can rewrite the recurrence relation as \(d_n - \frac{1}{2}d_{n-1} = 80\), and as \(n\) approaches infinity, both \(d_n\) and \(d_{n-1}\) will approach the same limit, which we can denote as L: \(L - \frac{1}{2}L = 80\). Solving for L, we get: \(\frac{1}{2}L = 80\). \(L = 80 \times 2 = 160\). This confirms the result of part (b), that in the long run, there will indeed be 160mg of aspirin in the person's blood.

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Most popular questions from this chapter

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)

Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using. the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}},\) for \(n=0,1,2,3, \ldots .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n}\). b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\). c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}\) where \(p > 0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)
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