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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. Jack took a \(200-\mathrm{mg}\) dose of a strong painkiller at midnight. Every hour, \(5 \%\) of the drug is washed out of his bloodstream. Let \(d_{n}\) be the amount of drug in Jack's blood \(n\) hours after the drug was taken, where \(d_{0}=200 \mathrm{mg}.\)

Short Answer

Expert verified
Answer: The first five terms of the sequence are 200, 190, 180.5, 171.475, 162.90125. The explicit formula for the terms of the sequence is \(d_n = 200 \times 0.95^n\). The recurrence relation is given by \(d_{n+1} = 0.95d_n\) with initial condition \(d_0 = 200 \text{ mg}\). The estimated limit of this sequence is 0, as the amount of the drug decreases with time and converges to zero.

Step by step solution

01

Write out the first five terms of the sequence

To find the first five terms of the sequence, we need to compute the amount of drug in Jack's blood after each hour: \(d_0 = 200 \text{ mg}\) (initial dose) \(d_1 = d_0 - 5\% \times d_0 = 200 - 0.05 \times 200 = 190 \text{ mg}\) \(d_2 = d_1 - 5\% \times d_1 = 190 - 0.05 \times 190 = 180.5 \text{ mg}\) \(d_3 = d_2 - 5\% \times d_2 = 180.5 - 0.05 \times 180.5 = 171.475 \text{ mg}\) \(d_4 = d_3 - 5\% \times d_3 = 171.475 - 0.05 \times 171.475 = 162.90125 \text{ mg}\) So the first five terms of the sequence are \(200, 190, 180.5, 171.475, 162.90125\).
02

Find an explicit formula for the terms of the sequence

We can find an explicit formula for the terms of the sequence by noticing that the percentage reduction of the drug gets smaller due to the smaller amount of drug remaining in the bloodstream. We can express the decrease in the amount of drug as a multiplication factor \((1-0.05)^n\), where \(n\) is the number of hours passed. Thus, the explicit formula for the terms of the sequence can be given as: \(d_n = d_0 \times (1-0.05)^n = 200 \times 0.95^n\)
03

Find a recurrence relation that generates the sequence

A recurrence relation for this sequence would describe how to get the next term in the sequence from the current term. In our case, we are given that every hour, 5% of the drug is washed out of the bloodstream. So, for any term \(d_{n+1}\), 95% (100% - 5%) of \(d_n\) remains: \(d_{n+1} = 0.95d_n\) with initial condition \(d_0 = 200 \text{ mg}\).
04

Estimate the limit of the sequence or state that it does not exist

To estimate the limit of the sequence, we can either use a calculator (for example, computing the sequence terms for large values of \(n\)) or use a graphing utility (by plotting the sequence terms against the index \(n\)). As the amount of the drug decreases with time and never increases, the sequence is strictly decreasing, converging to a value that cannot be less than zero because the amount of the drug cannot be negative. So, the limit of the sequence exists and it converges to zero.

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Key Concepts

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

Explicit Formula
In mathematics, an explicit formula is a way of defining the terms of a sequence with a direct expression using its position number, typically denoted by a variable such as \( n \). Offering a clear and straightforward computation, explicit formulas are valuable as they allow instant access to any term without needing the previous terms.

Let's consider the situation of Jack and the painkiller, where the amount left in the bloodstream forms a sequence. Here, the explicit formula derived was \( d_n = 200 \times 0.95^n \). This means for any hour \( n \), you can directly calculate the remaining drug amount by simply using this formula.

The formula uses an exponential decay model, because each hour reduces the preceding amount by 5%, thus retaining 95% of the previous amount. It’s important because it provides an efficient way to compute any hour's drug concentration without manually calculating all prior terms.
Recurrence Relation
A recurrence relation is a way to define a sequence where each term is expressed in terms of one or more of the preceding terms. This is particularly useful when each term can be calculated from the previous one(s) with a simple computational rule.

In Jack's case, the recurrence relation is given as \( d_{n+1} = 0.95d_n \), with an initial condition \( d_0 = 200 \text{ mg} \). This basically says, to get the next term (or the drug amount in the next hour), take 95% of the current amount.

Unlike the explicit formula, you need to know the previous term(s) to use a recurrence relation. This can be less convenient for finding terms rapidly out of sequence order, but it mirrors processes in real-world applications, like this gradual drug elimination.
Limit of a Sequence
The limit of a sequence refers to the value that the terms approach as the sequence continues indefinitely. Determining this limit helps in understanding the asymptotic behavior of sequences.

In our scenario with the painkiller dose, the drug amount in Jack's bloodstream is continuously diminishing over time. By estimation or graphical plotting, it becomes apparent that this sequence tends toward zero, as the drug cannot persist in his blood indefinitely beyond the elimination rate.

This sequence is strictly decreasing, meaning each subsequent term is less than the one before, never dipping below zero since negative drug amounts are not feasible. Thus, the limit of Jack's sequence is zero—a clear indicator that the drug will completely clear after sufficient time.

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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) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd. } \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N,\) the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7,\) and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

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 \(\lim _{n \rightarrow \infty} x_{n}=\ln 2\)

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{p}}$$

For a positive real number \(p,\) how do you interpret \(p^{p^{p \cdot *}},\) where the tower of exponents continues indefinitely? As it stands, the expression is ambiguous. The tower could be built from the top or from the bottom; that is, it could be evaluated by the recurrence relations \(a_{n+1}=p^{a_{n}}\) (building from the bottom) or \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{0}=p\) in either case. The two recurrence relations have very different behaviors that depend on the value of \(p\). a. Use computations with various values of \(p > 0\) to find the values of \(p\) such that the sequence defined by (2) has a limit. Estimate the maximum value of \(p\) for which the sequence has a limit. b. Show that the sequence defined by (1) has a limit for certain values of \(p\). Make a table showing the approximate value of the tower for various values of \(p .\) Estimate the maximum value of \(p\) for which the sequence has a limit.

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