Chapter 9: Problem 88
Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ; a_{0}=2, n=0,1,2, \dots$$
Short Answer
Expert verified
Answer: The limit value of the given sequence is \(L = \sqrt{2}\).
Step by step solution
01
Understand the recurrence relation
We are given a recurrence relation as:
$$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right), \quad a_{0}=2, \quad n=0,1,2, \dots$$
The sequence starts with \(a_{0} = 2\), and we are asked to find the limit of the sequence if it exists.
02
Iterative method
Calculate a few terms of the sequence using the given recurrence relation:
For \(n=0\): \(a_{1} = \frac{1}{2}\left(a_{0}+\frac{2}{a_{0}}\right) = \frac{1}{2}\left(2+\frac{2}{2}\right) = \frac{1}{2}(3) = \frac{3}{2}\)
For \(n=1\): \(a_{2} = \frac{1}{2}\left(a_{1}+\frac{2}{a_{1}}\right) = \frac{1}{2}\left(\frac{3}{2}+\frac{2}{\frac{3}{2}}\right) = \frac{1}{2}\left(\frac{3}{2}+\frac{4}{3}\right) = \frac{1}{2}\left(\frac{17}{6}\right) = \frac{17}{12}\)
For \(n=2\): \(a_{3} = \frac{1}{2}\left(a_{2}+\frac{2}{a_{2}}\right) = \frac{1}{2}\left(\frac{17}{12}+\frac{2}{\frac{17}{12}}\right) = \frac{1}{2}\left(\frac{17}{12}+\frac{24}{17}\right) = \frac{1}{2}\left(\frac{577}{204}\right) = \frac{577}{408}\)
It seems that the sequence is decreasing and getting closer to 1.
03
Analytical method
Let's assume that the sequence converges and has a limit \(L\). Then we can replace \(a_{n+1}\) and \(a_{n}\) with \(L\) in the recurrence relation:
$$L = \frac{1}{2}\left(L+\frac{2}{L}\right)$$
To solve for \(L\), we will multiply both sides by \(2L\):
$$2L^{2} = L^{2} + 2$$
Then, subtract \(L^2\) from both sides:
$$L^{2} = 2$$
And finally, take the square root of both sides (we will consider only the positive root, since the initial condition is positive):
$$L = \sqrt{2}$$
04
Conclusion
From the analysis, after using both iterative and analytical methods, we have discovered that the sequence:
$$a_{n+1}=\frac{1}{2}\left(a_{n}+2 / a_{n}\right) ;\quad a_{0}=2, \quad n=0,1,2, \dots$$
converges to the limit value \(L = \sqrt{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
sequence convergence
When we talk about the convergence of a sequence, we are interested in knowing if the terms of the sequence approach some value as we progress to infinity. For the given sequence, this means checking whether the sequence \(a_n\) defined by our recurrence relation indeed approaches a specific number as \(n\) grows larger.
A sequence converges if there exists a finite limit, \(L\), such that for every positive number \( \varepsilon \), however small, there is a corresponding number \( N \) after which all terms of the sequence are within \( \varepsilon \) of \(L\). In simpler terms, the terms of the sequence get arbitrarily close to \(L\) and stay close.
For the sequence defined by the recurrence relation \( a_{n+1} = \frac{1}{2} \left( a_n + \frac{2}{a_n} \right) \) starting with \( a_0 = 2 \), calculating the first few terms shows us that they stabilize around \( \sqrt{2} \). Hence, we conjecture or analytically show that this is indeed the limit toward which the sequence converges.
A sequence converges if there exists a finite limit, \(L\), such that for every positive number \( \varepsilon \), however small, there is a corresponding number \( N \) after which all terms of the sequence are within \( \varepsilon \) of \(L\). In simpler terms, the terms of the sequence get arbitrarily close to \(L\) and stay close.
For the sequence defined by the recurrence relation \( a_{n+1} = \frac{1}{2} \left( a_n + \frac{2}{a_n} \right) \) starting with \( a_0 = 2 \), calculating the first few terms shows us that they stabilize around \( \sqrt{2} \). Hence, we conjecture or analytically show that this is indeed the limit toward which the sequence converges.
analytical methods
Analytical methods allow us to use mathematical reasoning and algebraic manipulation to explore or prove the properties of sequences and series. For convergence, this often means solving equations or inequalities.
In our case, we want to determine if the sequence converges to a limit \( L \), and if so, find \( L \). By assuming the limit exists, we replace both \(a_{n+1}\) and \(a_n\) with \(L\) in the recurrence relation and solve for \(L\).
For the recurrence \( a_{n+1} = \frac{1}{2} \left( a_n + \frac{2}{a_n} \right) \), assume it converges to \( L \). We get \( L = \frac{1}{2} \left( L + \frac{2}{L} \right) \). By removing the fraction, we multiply through by \(2L\) to obtain \(2L^2 = L^2 + 2\).
In our case, we want to determine if the sequence converges to a limit \( L \), and if so, find \( L \). By assuming the limit exists, we replace both \(a_{n+1}\) and \(a_n\) with \(L\) in the recurrence relation and solve for \(L\).
For the recurrence \( a_{n+1} = \frac{1}{2} \left( a_n + \frac{2}{a_n} \right) \), assume it converges to \( L \). We get \( L = \frac{1}{2} \left( L + \frac{2}{L} \right) \). By removing the fraction, we multiply through by \(2L\) to obtain \(2L^2 = L^2 + 2\).
- Subtract \(L^2\) from both sides: \(L^2 = 2\)
- Solve the equation: \(L = \sqrt{2}\)
iterative method
The iterative method is an approach where we repeatedly apply a rule to generate the terms of a sequence. Each new term depends on the previous one(s), closely following the recurrence relation without necessarily using abstract algebraic reasoning.
For the sequence we're examining, the iterative method involves calculating each term successively starting from the initial condition \( a_0 = 2 \).
Let's see how this works:
For the sequence we're examining, the iterative method involves calculating each term successively starting from the initial condition \( a_0 = 2 \).
Let's see how this works:
- First term \(a_1\): Substitute \(a_0\) into our recurrence relation and calculate \(a_1 = \frac{3}{2}\).
- Second term \(a_2\): Use \(a_1 = \frac{3}{2}\), to calculate the next term, thus \(a_2 = \frac{17}{12}\).
- Continue this process for subsequent terms; each calculation depends on the prior term.