/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 Suppose that \(\alpha>1 .\) L... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 93

Suppose that \(\alpha>1 .\) Let \(x_{0}=1 .\) For \(n>1,\) let \(x_{n}=\) \(\left(\alpha+x_{n-1}\right) / 2 .\) Show that \(\left\\{x_{n}\right\\}\) is a bounded increasing sequence. To what number does \(\left\\{x_{n}\right\\}\) converge?

Short Answer

Expert verified
The sequence \( \{x_n\} \) is a bounded, increasing sequence that converges to \( \alpha \).

Step by step solution

01

Show Sequence is Increasing

To show the sequence \( \{x_n\} \) is increasing, we need to prove that \( x_{n} > x_{n-1} \) for all \( n > 1 \). Since \( x_n = \frac{\alpha + x_{n-1}}{2} \), we have \( x_n > x_{n-1} \) if \( \frac{\alpha + x_{n-1}}{2} > x_{n-1} \). This is equivalent to \( \alpha + x_{n-1} > 2x_{n-1} \), which simplifies to \( \alpha > x_{n-1} \). Initially, we have \( x_0 = 1 \), and given \( \alpha > 1 \), the sequence starts by satisfying this inequality and continues to do so, making it increasing.
02

Show Sequence is Bounded

To ensure the sequence is bounded above, we show \( x_n < \alpha \). Starting at \( x_0 = 1 < \alpha \), assume \( x_{n-1} < \alpha \). Then \( x_n = \frac{\alpha + x_{n-1}}{2} < \frac{\alpha + \alpha}{2} = \alpha \). By using mathematical induction, we prove \( x_n < \alpha \) for all \( n \), hence the sequence is bounded.
03

Determine the Limit of the Sequence

Since the sequence \( \{x_n\} \) is both increasing and bounded above, it converges by the Monotone Convergence Theorem. Let \( L \) be the limit. Then as \( n \rightarrow \infty \), we have \( x_n \rightarrow L \). Substituting into the recurrence relation, \( L = \frac{\alpha + L}{2} \). Solving for \( L \), we get \( 2L = \alpha + L \), which simplifies to \( L = \alpha \). Therefore, the sequence converges to \( \alpha \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Sequences
In sequences and series, a bounded sequence is a series of numbers that do not go beyond certain fixed values, making them easier to analyze. A sequence is considered bounded from above if there is a number that is greater than or equal to every number in the sequence. Similarly, it is bounded from below if there is a number smaller than or equal to every number in the sequence.
  • Upper Bound: A sequence \( \{x_n\} \) is said to be bounded above if there exists a number \( M \) such that \( x_n \leq M \) for all \( n \). For example, in the previous exercise, we see that the sequence is bounded above by \( \alpha \), since every term satisfies \( x_n < \alpha \).
  • Lower Bound: Similarly, the sequence \( \{x_n\} \) is bounded below if there exists a number \( m \) such that \( x_n \geq m \) for all \( n \). In our exercise, the sequence starts at \( x_0 = 1 \) and hence is also bounded below by 1, given \( \alpha > 1 \).
Recognizing whether a sequence is bounded is important. This knowledge helps in the process of determining if a sequence converges or has a limit. This leads us to the Monotone Convergence Theorem, which builds on these ideas.
Monotone Convergence Theorem
The Monotone Convergence Theorem is a powerful tool in calculus and real analysis. It provides criteria to determine the convergence of sequences. According to the theorem, any sequence that is both monotonic and bounded converges to a limit.
  • Monotonicity: A sequence is monotonic if it is either entirely non-increasing or non-decreasing. For example, in our exercise, the sequence \( \{x_n\} \) is increasing since each term is greater than the previous one, i.e., \( x_n > x_{n-1} \).
  • Convergence: Since the sequence \( \{x_n\} \) is shown to be both increasing and bounded above by \( \alpha \), the Monotone Convergence Theorem assures us that it must converge to some limit \( L \).
This theorem simplifies the task of finding limits, as once you establish monotonicity and boundedness, convergence is guaranteed without needing to evaluate infinitely many terms. This makes it invaluable for proofs and analysis of sequences alike.
Mathematical Induction
Mathematical induction is a fundamental proof technique in mathematics used to prove statements about natural numbers. It's particularly valuable when demonstrating properties of sequences.
  • Base Case: Begin by verifying that the property or statement holds true for the initial case. In our sequence example, \( x_0 = 1 \) and clearly \( 1 < \alpha \), starting the proof.
  • Inductive Step: Assume the statement is true for some arbitrary natural number \( n-1 \); this assumption is known as the "inductive hypothesis." Then, use this to prove that the statement must also be true for the number \( n \). In our case, assuming \( x_{n-1} < \alpha \), we show that \( x_n = \frac{\alpha + x_{n-1}}{2} < \alpha \), thus completing the step.
With these two steps, mathematical induction allows us to prove properties for infinitely many numbers in a sequence up to any desired bound. This method helps establish that sequences remain below a certain level, becoming an essential part of proving properties like boundedness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises \(47-56,\) use your intuition to decide whether the limit exists. Justify your answer by using the rigorous definition of limit. $$ \lim _{x \rightarrow-7}(|x|-3 x) $$

Graph \(f(x)=\left(x^{3}-3 x+1\right) /\left(x^{3}-3 x-1\right)\) in the viewing rectangle \([-3,3] \times[-15,15] .\) Locate the points, to four decimal places, at which \(f\) does not have a continuous extension.

An assertion is made about a function \(f\) that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function \(f\) that shows it is false. (Note: \(|f|\) is defined by \(|f|(x)=|f(x)| .)\) If \(f\) is continuous, then \(|f|\) is continuous.

Graph the given function \(f\) over an interval centered about the given point \(c,\) and determine if \(f\) has a continuous extension at \(c\). $$ f(x)=x /|x|, c=0 $$

A function \(f\) and a point \(c\) not in the domain of \(f\) are given. Analyze \(\lim _{x \rightarrow c} f(x)\) as follows. a. Evaluate \(f\left(c-1 / 10^{n}\right)\) and \(f\left(c+1 / 10^{n}\right)\) for \(n=2,3,4\). b. Formulate a guess for the value \(\ell=\lim _{x \rightarrow c} f(x)\). c. Find a value \(\delta\) such that \(f(x)\) is within 0.01 of \(\ell\) for every \(x\) that is within \(\delta\) of \(c\). d. Graph \(y=f(x)\) for \(x\) in \(\left[c-1 / 10^{4}, c\right) \cup\left(c, c+1 / 10^{4}\right]\) to verify visually that the limit of \(f\) at \(c\) exists. $$ f(x)=\frac{\cos (x)-1}{x^{2}}, c=0 $$
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.