/*! 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 53 Define the sequence \(a_{n}\) wi... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 53

Define the sequence \(a_{n}\) with \(a_{1}=\sqrt{2}\) and \(a_{n}=\sqrt{2+\sqrt{a_{n-1}}}\) for \(n \geq 2 .\) Show that \(\left\\{a_{n}\right\\}\) is increasing and bounded by 2 Evaluate the limit of the sequence by estimating the appropriate solution of \(x=\sqrt{2+\sqrt{x}}\)

Short Answer

Expert verified
The sequence \(a_{n}\) is increasing and bounded by 2. The limit of the sequence is 2.

Step by step solution

01

Prove the sequence is increasing

Assume \(a_n > a_{n-1}\), we want to prove \(a_{n+1} > a_n\). Substitute \(n+1\) and \(n\) into the sequence expression, we need to prove \(\sqrt{2+\sqrt{a_{n}}} > \sqrt{2+\sqrt{a_{n-1}}}\). After squaring both sides, it turns to \(2+\sqrt{a_{n}} > 2+\sqrt{a_{n-1}}\). It's obvious that \(\sqrt{a_{n}} > \sqrt{a_{n-1}}\) since square root function is an increasing function. Therefore, we prove that \(a_{n+1} > a_n\). So, the sequence is increasing.
02

Prove the sequence is bounded by 2

We will use mathematical induction. For \(a_1 = \sqrt{2} <2\), the base case is true. Now, assume \(a_n < 2\), we want to prove \(a_{n+1} < 2\). Substitute \(n+1\) into the sequence expression, we need to prove \(\sqrt{2+\sqrt{a_{n}}} < 2\). After subtract \(2\) from both sides, squaring both sides again, it turns to \(\sqrt{a_{n}} < 2\). Take square root both sides, we get \(a_{n} < 4\). Since our assumption is \(a_n).\ Therefore, we prove that \(a_{n+1} < 2\). So, the sequence is bounded by 2.
03

Find the limit of the sequence

To find the limit of the sequence, consider the expression given in the problem: \(x=\sqrt{2+\sqrt{x}}\). Square both sides of the equation to cancel out the root, we get \(x^2 = 2 + \sqrt{x}\). Solving the equation, rearrange \(x^2 - \sqrt{x} - 2 = 0\). The roots of the equation are \(x=1, 2\). Since \(a_n > 2 \). Evaluate the equation, we get \(x=2\). Hence, the limit of the sequence is 2.

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Key Concepts

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

Convergence of Sequences
Understanding the convergence of sequences is crucial for solving many mathematical problems. A sequence converges if it approaches a specific value as the sequence progresses to infinity. In this exercise, the sequence \( \{a_n\} \) displays convergence.

### Why Convergence MattersConvergence tells us about the long-term behavior of a sequence:
  • If a sequence converges, it reaches a steady-state.
  • This helps in predicting outcomes and understanding stability in various fields like economics and physics.
For the sequence given, it is shown to be bounded and increasing:
  • This ensures that it will not exceed a certain limit, making convergence possible.
      • Finally, solving the equation \(x = \sqrt{2 + \sqrt{x}}\) allows us to find that the sequence converges to 2.
Mathematical Induction
Mathematical induction is a powerful proof technique used to show that a statement holds true for all natural numbers. In this exercise, it proves that the sequence is bounded by 2.

### Understanding InductionInduction works in two main steps:
  • Base Case: Show the statement is true for the initial value (e.g., \(a_1 = \sqrt{2} < 2\)).
  • Inductive Step: Assume the statement is true for some arbitrary natural number \(n\), then prove it's true for \(n+1\).
In our problem, we use induction to prove \(a_{n+1} < 2\):
  • We assume \(a_n < 2\) and show \( \sqrt{2 + \sqrt{a_n}} < 2 \), confirming the sequence is bounded.
Recursive Sequences
Recursive sequences have each term defined in relation to preceding terms.

### Characteristics of Recursive Sequences
  • **Definition Dependency:** Each term is calculated using the previous term(s), like \(a_n = \sqrt{2 + \sqrt{a_{n-1}}}\).
  • **Complex Patterns:** They often require iterative methods to discover characteristics such as bounds or convergence.
In this exercise:
  • The recursive formula given helps us understand how \(a_n\) evolves over iterations.
  • Establishing a base case allows us to track the sequence, ensuring it's increasing and bounded.
Recursive sequences like this one help illustrate dynamic systems and are widely applicable in computer science and mathematics.

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Most popular questions from this chapter

Determine the radius and interval of convergence. $$\sum_{i=0}^{\infty} \frac{k}{2^{k}} x^{k}$$

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Prove that if \(\sum_{k=0}^{\infty} a_{k} x^{k}\) has radius of convergence \(r,\) with \(0 < r < \infty,\) then \(\sum_{k=0}^{\infty} a_{k} x^{2 k}\) has radius of convergence \(\sqrt{r}\)

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The energy density of electromagnetic radiation at wavelength \(\lambda\) from a black body at temperature \(T\) (degrees Kelvin) is given by Planck's law of black body radiation: \(f(\lambda)=\frac{8 \pi h c}{\lambda^{5}\left(e^{h c / 4 k T}-1\right)},\) where \(h\) is Planck's constant, \(c\) is the speed of light and \(k\) is Boltzmann's constant. To find the wavelength of peak emission, maximize \(f(\lambda)\) by minimizing \(g(\lambda)=\lambda^{5}\left(e^{h c / \lambda k T}-1\right) .\) Use a Taylor polynomial for \(e^{x}\) with \(n=7\) to expand the expression in parentheses and find the critical number of the resulting function. (Hint: Use \(\frac{h c}{k} \approx 0.014\).) Compare this to Wien's law: \(\lambda_{\max }=\frac{0.002898}{T} .\) Wien's law is accurate for small \(\lambda .\) Discuss the flaw in our use of Maclaurin series.
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