Chapter 10: Problem 93
In Exercises \(91-98\) , assume that each sequence converges and find its limit. $$ a_{1}=-4, \quad a_{n+1}=\sqrt{8+2 a_{n}} $$
Short Answer
Expert verified
The limit of the sequence is 4.
Step by step solution
01
Understand the Sequence Definition
The sequence is defined recursively by \( a_1 = -4 \) and \( a_{n+1} = \sqrt{8 + 2a_n} \). Each term depends on the previous one, and we are asked to find the limit as \( n \to \infty \).
02
Assume the Limit Exists
Assume that the sequence converges to a limit \( L \). That means for large \( n \), both \( a_n \) and \( a_{n+1} \) approach \( L \).
03
Set Up the Limit Equation
Since the limit of both \( a_n \) and \( a_{n+1} \) is \( L \), substitute \( L \) into the recursive formula: \( L = \sqrt{8 + 2L} \).
04
Solve the Limit Equation
Square both sides to eliminate the square root: \( L^2 = 8 + 2L \). Rearrange this equation to standard quadratic form: \( L^2 - 2L - 8 = 0 \).
05
Solve the Quadratic Equation
Use the quadratic formula \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -2 \), \( c = -8 \). Plug in the values: \( L = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \).
06
Choose the Appropriate Root
The solutions to the quadratic equation are \( L = 4 \) and \( L = -2 \). Since the sequence terms involve a square root, which only produces non-negative results, \( L = -2 \) cannot be correct if the sequence ever enters non-negative real numbers (which occurs since starting at \( -4 \) and iteratively applying the formula heads towards positive domain). Thus, \( L = 4 \).
07
Verify Convergence with Initial Terms
Substitute initial values. Starting from \( a_1 = -4 \), the next terms move towards positive values, as each iteration using \( a_n = \sqrt{8 + 2a_{n-1}} \) computes to an increasingly positive \( a_{n+1} \). This supports convergence to \( L = 4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recursive sequences
A recursive sequence is one where each term is defined in terms of previous terms. It relies on a starting point, often called the initial condition, and a repeated process to generate new terms.
For example, if we know \( a_1 = -4 \) and \( a_{n+1} = \sqrt{8 + 2a_n} \), we can calculate each subsequent term using the value of the previous one. This type of calculation is like walking step by step towards the sequence's behavior, one step at a time.
Recursive sequences are very useful because they can model processes or phenomena that have inherent repetition or dependency in their structure:
For example, if we know \( a_1 = -4 \) and \( a_{n+1} = \sqrt{8 + 2a_n} \), we can calculate each subsequent term using the value of the previous one. This type of calculation is like walking step by step towards the sequence's behavior, one step at a time.
Recursive sequences are very useful because they can model processes or phenomena that have inherent repetition or dependency in their structure:
- Each term is constructed by applying the sequence's rule repeatedly.
- Understanding the first few terms gives insight into the behavior of the entire sequence.
- It provides a foundation for more advanced concepts like convergence.
Convergence of sequences
The convergence of a sequence refers to the behavior of a sequence as its terms progress towards infinity. A sequence is said to converge if it approaches a specific value, called the limit, as more terms are added.
To determine if a sequence like the one given in the exercise converges, we use the property that if convergent, it will have the same value for both \( a_n \) and \( a_{n+1} \) when \( n \) is large. Let's assume it converges to \( L \). Then:
To determine if a sequence like the one given in the exercise converges, we use the property that if convergent, it will have the same value for both \( a_n \) and \( a_{n+1} \) when \( n \) is large. Let's assume it converges to \( L \). Then:
- \( a_n \to L \) and \( a_{n+1} \to L \), meaning as \( n \to \infty \), all terms move towards \( L \).
- Using this property, substitute \( L \) into the recursive formula to find the limit equation.
Quadratic equations
Quadratic equations appear frequently in various mathematical contexts, particularly when solving for limits in recursive sequences. A standard quadratic equation is expressed in the form \( ax^2 + bx + c = 0 \). Solving a quadratic equation often involves finding values for \( x \) that satisfy the equation.
In our case, after assuming the sequence's limit exists as \( L \), we derived a quadratic equation: \( L^2 - 2L - 8 = 0 \). To solve it, we used the quadratic formula:
In our case, after assuming the sequence's limit exists as \( L \), we derived a quadratic equation: \( L^2 - 2L - 8 = 0 \). To solve it, we used the quadratic formula:
- \( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- This formula calculates the possible solutions for \( L \).
- Understanding which solution fits the context, only non-negative roots were considered since our sequence eventually reached a positive number via square roots.