/*! 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 94 Assume that each sequence conver... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume that each sequence converges and find its limit. \(a_{1}=0, \quad a_{n+1}=\sqrt{8+2 a_{n}}\)

Short Answer

Expert verified
The sequence converges to 4.

Step by step solution

01

Understand the Given Sequence Form

The sequence is given as \(a_1 = 0\) and \(a_{n+1} = \sqrt{8 + 2a_n}\). This means that each term in the sequence is derived from the previous term using this recursive formula.
02

Assume Sequence Convergence

We assume the sequence \((a_n)\) converges to a limit \(L\). Therefore, as \(n\) approaches infinity, both \(a_n\) and \(a_{n+1}\) approach \(L\). This gives us the equation \(L = \sqrt{8 + 2L}\).
03

Solve the Convergence Equation

Square both sides of the equation \(L = \sqrt{8 + 2L}\) to eliminate the square root, resulting in \(L^2 = 8 + 2L\).
04

Rearrange and Solve Quadratic Equation

Rearrange the quadratic equation as \(L^2 - 2L - 8 = 0\). Solve this using the quadratic formula \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), \(c = -8\).
05

Compute the Discriminant

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(-8) = 4 + 32 = 36\).
06

Determine the Roots

Now substitute into the quadratic formula: \(L = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}\). This gives two possible solutions: \(L = 4\) and \(L = -2\).
07

Confirm the Physical Plausibility

Since \(L = \sqrt{8 + 2L}\), \(L\) must be non-negative. Therefore, \(L = -2\) is not a valid solution. The sequence must converge to \(L = 4\).

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 91Ó°ÊÓ!

Key Concepts

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

Recursive sequence
A recursive sequence is one where each term is defined based on the preceding terms. This means there is a rule or formula that uses previous terms to generate the next term in the sequence. In the context of our exercise, the sequence starts with a known initial term, in this case, \( a_1 = 0 \), and follows a rule to get each subsequent term: \( a_{n+1} = \sqrt{8 + 2a_n} \). This recursive definition is crucial because it allows for the expression of terms without needing a closed-form solution — you just need to know the rule and the initial condition.

Recursive sequences can emerge in various mathematical and real-world applications, such as population models or computing Fibonacci numbers. Understanding how the sequence evolves can help determine its behavior, like whether it converges or diverges, and what its potential limits are.
Quadratic equation
A quadratic equation is a polynomial equation of degree two, generally expressed in the form \( ax^2 + bx + c = 0 \). The challenge often is to find the values of \( x \) that satisfy the equation, known as the roots of the equation. In the solution to this problem, rearranging the convergence condition \( L = \sqrt{8 + 2L} \) and eliminating the square root transforms it into the quadratic equation format as \( L^2 - 2L - 8 = 0 \).

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) provides a method to solve these equations by calculating the roots directly from the coefficients \( a \), \( b \), and \( c \). This formula is derived by completing the square method and is efficient for finding solutions to any quadratic equation.
Limit of a sequence
The limit of a sequence refers to the value that the terms of a sequence approach as the term number becomes very large (i.e., as \( n \) approaches infinity). If the terms get arbitrarily close to a number \( L \), we say that the sequence converges to \( L \). Determining the limit gives insights into the long-term behavior of the sequence.

In our exercise, the assumption that the sequence converges leads us to equate both \( a_n \) and \( a_{n+1} \) to the same limit \( L \). By solving the equation \( L = \sqrt{8 + 2L} \), we determine that \( L = 4 \). This confirms that as we go further in the sequence, the terms stabilize around the number 4.
Convergence of recursive sequences
The convergence of recursive sequences refers to whether the terms of a sequence defined by a recurrence relation approach a specific value (or limit) as the sequence progresses. Convergence is important because it allows us to understand the behavior of sequences in a broader context.

For a sequence to converge, it must become increasingly close to a fixed value known as the limit. In the original exercise, we assumed the sequence \(a_n\) converges to \( L \), which allowed us to simplify and solve for the limit using the recurrence relation \( L = \sqrt{8 + 2L} \). A crucial step in verifying convergence was recognizing that since \( L = \sqrt{8 + 2L} \), it implies that \( L \) must be non-negative, leading to the exclusion of \( L = -2 \) as a possible limit. This emphasizes ensuring solutions make logical sense given the sequence's definition and context.

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

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}.$$to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

A cubic approximation Use Taylor's formula with \(a=0\) and \(n=3\) to find the standard cubic approximation of \(f(x)=\) \(1 /(1-x)\) at \(x=0 .\) Give an upper bound for the magnitude of the error in the approximation when \(|x| \leq 0.1\)

Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=\frac{2^{n}-1}{3^{n}}\)

What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.

a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

See all solutions

Recommended explanations on Math Textbooks

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.