Problem 97
Question
The sequence \(\left\\{a_{n}\right\\}\) is recursively defined. Find all fixed points of \(\left\\{a_{n}\right\\}\) $$ a_{n+1}=\frac{4}{a_{n}} $$
Step-by-Step Solution
Verified Answer
The fixed points of the sequence are 2 and -2.
1Step 1: Understand the Problem
To find the fixed points of the sequence \(\left\{a_{n}\right\}\), we need to determine the value(s) of \(a_n\) such that \(a_{n+1} = a_n\). This means we set the recursive formula equal to \(a_n\).
2Step 2: Set the Equation for Fixed Points
Start with the equation for a fixed point, \( a_{n+1} = a_n \). Substitute the recursive definition into this: \( \frac{4}{a_n} = a_n \).
3Step 3: Solve the Fixed Point Equation
Multiply both sides of the equation \( \frac{4}{a_n} = a_n \) by \(a_n\) to eliminate the fraction, resulting in the quadratic equation \( a_n^2 = 4 \).
4Step 4: Solve the Quadratic Equation
Solve the equation \( a_n^2 = 4 \) by taking the square root of both sides. This gives two possible solutions: \(a_n = 2\) and \(a_n = -2\).
5Step 5: Confirm the Solutions
Verify that both \(a_n = 2\) and \(a_n = -2\) are solutions by substituting them back into the original recursive equation.\ - If \(a_n = 2\), then \(a_{n+1} = \frac{4}{2} = 2\).\ - If \(a_n = -2\), then \(a_{n+1} = \frac{4}{-2} = -2\).\Both satisfy \(a_{n+1} = a_n\).
Key Concepts
Recursive SequenceFixed Point EquationQuadratic EquationSequence Analysis
Recursive Sequence
Recursive sequences involve finding each term based on the preceding terms using a specific rule or formula. In our given exercise, we have the sequence \(\{a_{n}\}\) defined by the recursive relation:
- \(a_{n+1} = \frac{4}{a_{n}}\)In this formula, each subsequent term \(a_{n+1}\) depends upon the current term \(a_{n}\). This means that once we know \(a_1\), we can find all subsequent terms. To engage with recursive sequences effectively, it's essential to clearly understand and apply this rule repeatedly for each index increment.
Recursive sequences can vary greatly depending on the defining formula, but they always rely on previous terms to determine subsequent terms, making them predictable once a starting point is established.
- \(a_{n+1} = \frac{4}{a_{n}}\)In this formula, each subsequent term \(a_{n+1}\) depends upon the current term \(a_{n}\). This means that once we know \(a_1\), we can find all subsequent terms. To engage with recursive sequences effectively, it's essential to clearly understand and apply this rule repeatedly for each index increment.
Recursive sequences can vary greatly depending on the defining formula, but they always rely on previous terms to determine subsequent terms, making them predictable once a starting point is established.
Fixed Point Equation
A fixed point in a sequence is a value that remains constant through recursive processes. This happens when the result of applying the recursive formula is the same as the input value. For the sequence \(\{a_{n}\}\), the fixed points are found by setting the recursive formula equal to the term itself:
- \(a_{n+1} = a_n\)Substitute the recursive definition into this fixed point formula:
- \(\frac{4}{a_n} = a_n\)A fixed point remains unchanged when the iterative function of the sequence is applied. This concept is crucial as it allows us to identify constants within dynamic processes, offering insight into the behavior and stability of different types of sequences.
- \(a_{n+1} = a_n\)Substitute the recursive definition into this fixed point formula:
- \(\frac{4}{a_n} = a_n\)A fixed point remains unchanged when the iterative function of the sequence is applied. This concept is crucial as it allows us to identify constants within dynamic processes, offering insight into the behavior and stability of different types of sequences.
Quadratic Equation
Solving the fixed point equation \(\frac{4}{a_n} = a_n\) requires rearranging it into a quadratic equation. Quadratic equations take the form \(ax^2 + bx + c = 0\). To eliminate the fraction, multiply both sides by \(a_n\), resulting in:
- \(a_n^2 = 4\)This is a simple quadratic where:- \(a = 1\)- \(b = 0\)- \(c = -4\)To solve the quadratic equation \(a_n^2 = 4\), take the square root of both sides. This yields two solutions:
- \(a_n = 2\)- \(a_n = -2\)Quadratic equations often provide multiple solutions, reflecting potential values within the context of the problem. Solving quadratics is a fundamental skill in mathematics, especially in sequence and calculus problems.
- \(a_n^2 = 4\)This is a simple quadratic where:- \(a = 1\)- \(b = 0\)- \(c = -4\)To solve the quadratic equation \(a_n^2 = 4\), take the square root of both sides. This yields two solutions:
- \(a_n = 2\)- \(a_n = -2\)Quadratic equations often provide multiple solutions, reflecting potential values within the context of the problem. Solving quadratics is a fundamental skill in mathematics, especially in sequence and calculus problems.
Sequence Analysis
Analyzing sequences, like the given recursive sequence, involves understanding the general behavior and characteristics of the sequence over time. For \(\{a_{n}\}\), the recursion \(a_{n+1} = \frac{4}{a_{n}}\) reveals fixed points which we solved to be 2 and -2.
Sequence analysis helps determine:
Sequence analysis helps determine:
- Convergence - whether terms approach a fixed value
- Divergence - whether terms move away from a fixed value
- Oscillation - whether terms fluctuate between values
Other exercises in this chapter
Problem 95
The sequence \(\left\\{a_{n}\right\\}\) is recursively defined. Find all fixed points of \(\left\\{a_{n}\right\\}\) $$ a_{n+1}=\frac{2}{5} a_{n}-\frac{9}{5} $$
View solution Problem 96
The sequence \(\left\\{a_{n}\right\\}\) is recursively defined. Find all fixed points of \(\left\\{a_{n}\right\\}\) $$ a_{n+1}=-\frac{1}{3} a_{n}+\frac{1}{4} $$
View solution Problem 98
The sequence \(\left\\{a_{n}\right\\}\) is recursively defined. Find all fixed points of \(\left\\{a_{n}\right\\}\) $$ a_{n+1}=\frac{7}{a_{n}} $$
View solution Problem 99
The sequence \(\left\\{a_{n}\right\\}\) is recursively defined. Find all fixed points of \(\left\\{a_{n}\right\\}\) $$ a_{n+1}=\frac{2}{a_{n}+2} $$
View solution