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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. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5$$

Short Answer

Expert verified
Based on the given recurrence relation and initial condition, the sequence appears to be converging to the limit L = 4.

Step by step solution

01

Calculate the first few terms of the sequence

Using the initial condition \(a_0 = 5\) and the recurrence relation formula \(a_{n+1} = \frac{1}{2} a_n + 2\), let's calculate the first few terms of the sequence. $a_0 = 5 \\ a_1 = \frac{1}{2}a_0 + 2 = \frac{1}{2}(5) + 2 = 4.5 \\ a_2 = \frac{1}{2}a_1 + 2 = \frac{1}{2}(4.5) + 2 = 4.25 \\ a_3 = \frac{1}{2}a_2 + 2 = \frac{1}{2}(4.25) + 2 = 4.125 \\ a_4 = \frac{1}{2}a_3 + 2 = \frac{1}{2}(4.125) + 2 = 4.0625$
02

Observe the pattern and make a conjecture

From the calculated terms of the sequence, it appears that the sequence is converging to a certain value, which is greater than 4. To find the limit, we can make the assumption that the sequence converges to a limit L, and then try to solve analytically for that limit.
03

Analytical solution for the limit

Assuming that the sequence converges to a limit L, we can write the equation: $$L = \frac{1}{2}L + 2$$ Now, let's solve this equation for L: $$L - \frac{1}{2}L = 2 \\ \frac{1}{2}L = 2 \\ L = 4$$
04

Conclusion

From our analytical solution, we can now conclude that the sequence converges to the limit L = 4. As the sequence gets larger, it gets very close to 4, and we can conjecture that the limit of the sequence is 4.

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Key Concepts

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

Convergence of Sequences
Understanding the behavior of sequences is a fundamental aspect of calculus. A sequence is simply a list of numbers in a certain order, often following a defined rule. When we talk about the convergence of a sequence, we are looking at whether the elements of the series are approaching a specific number, known as the limit, as the sequence progresses. For example, if we have a sequence where each number is getting closer and closer to the value 5, we might say that the sequence converges to 5.

One common type of sequence we encounter is defined by a recurrence relation, where each term is derived from the previous one(s) using a certain formula. To examine whether such a sequence converges, one method is to observe the pattern after computing several of its first terms, as we did in the exercise with the relation \( a_{n+1} = \frac{1}{2} a_n + 2 \) and initial condition \( a_0 = 5 \). If we notice that the terms are steadily getting closer to a single value, it suggests convergence.

In our example, as we compute the terms \( a_1, a_2, a_3, \ldots \), we observe that they are indeed getting closer to 4. This behavior hints that the sequence may have 4 as its limit. However, one must do a more rigorous analytical examination to confirm this initial conjecture formally.
Limit of a Sequence
The limit of a sequence is the value that the terms of the sequence approach as the sequence goes on indefinitely. In other words, as we progress through the sequence, the terms get arbitrarily close to this value, without necessarily ever reaching it. Mathematically, the limit of a sequence \( (a_n) \) as \( n \) goes to infinity is represented as \( \lim_{{n\to\infty}} a_n = L \), where \( L \) is the limit value.

In the exercise, we used an analytical approach to find the limit of the sequence defined by the recurrence relation. We made an assumption that the sequence converges to some limit \( L \), and then we plugged this value into the relation itself. Solving \( L = \frac{1}{2}L + 2 \) led us to discover that \( L = 4 \). This illustrates an important idea: if a sequence converges to a limit \( L \), plugging \( L \) into the recurrence formula should yield \( L \) itself, as the terms are getting infinitely close to \( L \).

Deducing the limit has powerful implications. It helps us understand long-term behavior without calculating every term, which would be impractical or even impossible for infinitely long sequences. It can also reveal insights into the system that the sequence might represent, like population growth, financial projections, or iterating algorithms in computer science.
Analytical Methods in Calculus
Calculus offers a wide array of analytical methods to solve different mathematical problems. Analytical methods involve logic and algebraic manipulation, as opposed to numerical methods which involve direct computation. In the context of sequences and their limits, analytical methods allow us to determine the behavior and the convergence point without having to compute every single term.

In our sequence example, after computing a few terms and conjecturing that the sequence might converge to 4, we employed analytical reasoning to solve the recurrence relation in the limit. This approach involves assuming that the sequence has a finite limit \( L \) and then solving for \( L \) algebraically. The solution was reached after rewriting the recurrence formula and isolating the variable \( L \), showing clear and logical steps leading to the conclusion that \( L = 4 \).

Using Recurrence Relations to Find Limits

Recurrence relations are particularly common in sequences, and analytical methods are often used to find their limits. By working out an equation that represents the recurrence and the limit as the same value, you can often solve for that limit. This method is not just about finding a number but understanding the underlying behavior of complex systems that can be encoded into sequences. It’s a powerful tool in a mathematician’s arsenal that transcends beyond mere computation to broader applications in modeling and predicting behaviors in the natural and social sciences.

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Most popular questions from this chapter

A fishery manager knows that her fish population naturally increases at a rate of \(1.5 \%\) per month, while 80 fish are harvested each month. Let \(F_{n}\) be the fish population after the \(n\) th month, where \(F_{0}=4000\) fish. a. Write out the first five terms of the sequence \(\left\\{F_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{F_{n}\right\\}\). c. Does the fish population decrease or increase in the long run? d. Determine whether the fish population decreases or increases in the long run if the initial population is 5500 fish. e. Determine the initial fish population \(F_{0}\) below which the population decreases.

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{1.001} \text { and } b_{n}=\ln n^{10}, n \geq 1$$

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n !}{n^{n}}\right\\}$$

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}_{n=2}^{\infty}.$$

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