Chapter 4: Problem 39
Determine whether the sequence defined as follows has a limit. If it does, find the limit. \(a_{1}=3, a_{n}=\sqrt{2 a_{n-1}}, n=2,3, \ldots\)
Short Answer
Expert verified
Yes, the limit exists and it is 2.
Step by step solution
01
Understand the Sequence
The sequence is defined recursively with the first term given as \(a_1 = 3\) and each subsequent term \(a_n\) is defined as the square root of twice the previous term, \(a_{n} = \sqrt{2a_{n-1}}\). Our task is to determine if this sequence converges to a limit and, if so, find that limit.
02
Assume a Limit Exists
Assume the sequence \(a_n\) converges to a limit \(L\). Then, as \(n \to \infty\), both \(a_n\) and \(a_{n-1}\) approach \(L\) as well. Thus, we set the recursive formula equal to \(L\):\[ L = \sqrt{2L} \]
03
Solve the Limit Equation
Square both sides of the equation \(L = \sqrt{2L}\) to eliminate the square root:\[ L^2 = 2L \]Rearrange this into a standard quadratic equation:\[ L^2 - 2L = 0 \]Factor the quadratic:\[ L(L - 2) = 0 \]From this, we get two solutions: \(L = 0\) or \(L = 2\).
04
Verify the Feasibility of the Limits
Since \(a_1 = 3\) and every \(a_n > 0\), the sequence is always positive. Thus, \(L = 0\) is not feasible because the terms of the sequence are never zero. Therefore, the only possible limit is \(L = 2\).
05
Confirm Limit Stability
Check if the sequence actually approaches \(L = 2\). Starting from 3, calculate a few more terms:- \(a_1 = 3\)- \(a_2 = \sqrt{2 \times 3} = \sqrt{6} \approx 2.45\)- \(a_3 = \sqrt{2 \times \sqrt{6}} \approx 2.217\)- \(a_4 = \sqrt{2 \times 2.217} \approx 2.106\)The terms seem to decrease and get closer to 2, confirming \(L = 2\) is indeed the limit.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recursive Sequences
To fully grasp the concept of recursive sequences, think of them like a mathematical chain reaction. A recursive sequence is defined by a relation that uses previous terms to generate future ones. In the sequence from our exercise, each term is generated using the formula: \[ a_n = \sqrt{2a_{n-1}} \].
This means each new term is simply the square root of twice the previous term. It’s a little like a grocery list where each new item depends on the previous one. You start with an initial value, called the seed or first term (here it’s 3), and apply the rule over and over to create a series of terms.
Recursive sequences are key in many areas of mathematics and computer science, because they establish a step-by-step pattern. Understanding this pattern is crucial for predicting the behavior of the sequence and for finding limits.
This means each new term is simply the square root of twice the previous term. It’s a little like a grocery list where each new item depends on the previous one. You start with an initial value, called the seed or first term (here it’s 3), and apply the rule over and over to create a series of terms.
Recursive sequences are key in many areas of mathematics and computer science, because they establish a step-by-step pattern. Understanding this pattern is crucial for predicting the behavior of the sequence and for finding limits.
Limit of a Sequence
The limit of a sequence refers to the value that the terms of a sequence get closer to as the sequence progresses. In the exercise, we're tasked with determining whether the sequence has a limit, and if it does, what that limit is.
When analyzing sequences, imagining them as a race can be helpful. As the race progresses, each term gets a bit closer to a particular number, much like the runner approaches the finish line. In our example, through a series of calculations, we determined that the limit of the sequence \(a_n\) is 2.
Mathematically, if the difference between the terms of sequence \(a_n\) and a specific number \(L\) becomes arbitrarily small as \(n\) increases, \(L\) is the limit. This means that as you continue to apply the recursive formula, the values get closer and closer to \(L\). Understanding limits is fundamental in calculus and helps in analyzing the behavior of sequences and their convergence.
When analyzing sequences, imagining them as a race can be helpful. As the race progresses, each term gets a bit closer to a particular number, much like the runner approaches the finish line. In our example, through a series of calculations, we determined that the limit of the sequence \(a_n\) is 2.
Mathematically, if the difference between the terms of sequence \(a_n\) and a specific number \(L\) becomes arbitrarily small as \(n\) increases, \(L\) is the limit. This means that as you continue to apply the recursive formula, the values get closer and closer to \(L\). Understanding limits is fundamental in calculus and helps in analyzing the behavior of sequences and their convergence.
Quadratic Equations
Solving the limit of a recursive sequence often involves working with quadratic equations, as seen in this exercise. When we set the sequence's formula to approach a limit \(L\), the equation \(L = \sqrt{2L}\) appears.
Squaring both sides, we end up with a quadratic equation \(L^2 = 2L\). Rearranging gives us \(L^2 - 2L = 0\), reminiscent of a classic quadratic equation \(ax^2 + bx + c = 0\).
Using factoring, we find solutions to this quadratic: \(L(L-2) = 0\), which yields \(L = 0\) or \(L = 2\). In our context, however, only \(L = 2\) is possible as it's the only solution that matches the conditions of our sequence. Quadratic equations are a powerful tool for solving problems involving rates of change and growth, and they frequently appear in sequence analysis.
Squaring both sides, we end up with a quadratic equation \(L^2 = 2L\). Rearranging gives us \(L^2 - 2L = 0\), reminiscent of a classic quadratic equation \(ax^2 + bx + c = 0\).
Using factoring, we find solutions to this quadratic: \(L(L-2) = 0\), which yields \(L = 0\) or \(L = 2\). In our context, however, only \(L = 2\) is possible as it's the only solution that matches the conditions of our sequence. Quadratic equations are a powerful tool for solving problems involving rates of change and growth, and they frequently appear in sequence analysis.
Square Roots in Sequences
Square roots are not just operations for solving individual numbers; they can be vital parts of sequences. In our exercise, square roots play a fundamental role in forming each new term. The sequence given by \(a_n = \sqrt{2a_{n-1}}\) shows how a simple operation can build a complex pattern.
When using square roots in sequences, it's important to remember their properties. They only produce non-negative results and tend to decrease very large numbers by finding their 'balanced' midpoints. As you repeatedly apply a square root, the number becomes more stable, which leads to convergence.
Understanding square roots within sequences helps in predicting how sequences behave over time. It shows how operations can propagate through a sequence, driving it towards its limit or causing divergence.
When using square roots in sequences, it's important to remember their properties. They only produce non-negative results and tend to decrease very large numbers by finding their 'balanced' midpoints. As you repeatedly apply a square root, the number becomes more stable, which leads to convergence.
Understanding square roots within sequences helps in predicting how sequences behave over time. It shows how operations can propagate through a sequence, driving it towards its limit or causing divergence.
