Chapter 2: Problem 4
Suppose \(x_{1}:=\frac{1}{2}\) and \(x_{n+1}:=x_{n}^{2}\). Show that \(\left\\{x_{n}\right\\}\) converges and find \(\lim x_{n}\). Hint: You cannot divide by zero!
Short Answer
Expert verified
The sequence \( \{x_n\} \) converges to 0.
Step by step solution
01
Understand the Recursive Relation
The sequence is recursively defined by \( x_1 = \frac{1}{2} \) and \( x_{n+1} = x_n^2 \). This means each term is the square of the previous term.
02
Calculate Initial Terms
To observe the behavior of the sequence, let's compute the first few terms: \( x_1 = \frac{1}{2} \), \( x_2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \), \( x_3 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \), and so on.
03
Hypothesize About Convergence
The terms are getting smaller with each iteration, trending towards a specific limit. We hypothesize that the sequence converges to 0.
04
Show Convergence to Zero
We need to prove that the sequence \( \{x_n\} \) converges to 0. As \( n \to \infty \), \( x_n \) will be given by \( x_{n+1} = (x_n)^2 \). If the sequence converges to L, then \( L = L^2 \). This equation has solutions \( L = 0 \) or \( L = 1 \). Since our sequence is clearly decreasing to 0, it cannot be approaching 1, thus \( L = 0 \).
05
Verify The Convergence with Inequality
Note that the sequence is decreasing because \( x_{n+1} = x_n^2 \leq x_n \) whenever \( x_n \leq 1 \). Also, each term is positive. Since the sequence is bounded below by 0 and is monotonic, it converges to 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sequence Convergence
Sequence convergence is a fundamental concept in real analysis. It refers to the behavior of a sequence as the number of terms increases indefinitely. Specifically, a sequence \( \{a_n\} \) is said to converge if there exists a limit \( L \) such that for any given small number (often called \( \epsilon \) in mathematics), there is a point in the sequence beyond which all terms are within \( \epsilon \) of \( L \).
This means that as \( n \to \infty \), the terms \( a_n \) get arbitrarily close to \( L \).
This means that as \( n \to \infty \), the terms \( a_n \) get arbitrarily close to \( L \).
- When the limit \( L \) exists, we say the sequence converges to \( L \).
- If no such \( L \) can be found, the sequence is divergent.
Recursive Sequences
A recursive sequence is one in which each term is defined as a function of its preceding terms. The sequence \( \{x_n\} \) given by the exercise is an excellent example, as it starts with an initial term \( x_1 = \frac{1}{2} \) and each subsequent term is formulated as \( x_{n+1} = x_n^2 \).
This kind of definition allows a sequence to be built systematically:
This kind of definition allows a sequence to be built systematically:
- Start with a known value or set of values.
- Apply a defined rule or operation repeatedly to generate new terms.
Monotonic Sequences
A sequence is called monotonic if its terms consistently increase or decrease. In the case of the exercise, \( \{x_n\} \) is a monotonically decreasing sequence because each term is a square of the previous one, and for terms less than 1, squaring reduces their size.
Monotonic sequences have some special properties that make them easier to analyze:
Monotonic sequences have some special properties that make them easier to analyze:
- If a sequence is both bounded and monotonic, it must converge.
- Bounded means there is a limit beyond which terms cannot go, either above or below.