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Chapter 1: Problem 166

Define \(\left(x_n\right)_{n \geq 1}\) by \(x_1=1, x_{n+1}=\frac{1}{\sqrt{2}} \sqrt{1-\sqrt{1-x_n^2}}\). a. Show that \(\lim _{n \rightarrow \infty} x_n\) exists and find this limit. b. Show that there is a unique number \(A\) for which \(L=\lim _{n \rightarrow \infty} \frac{x_n}{A^n}\) exists as a finite nonzero number. Evaluate \(L\) for this value of \(A\).

Short Answer

Expert verified
a. The limit \( \lim_{n \rightarrow \infty} x_n = 0 \). b. The value of \( A \) is \( \sqrt{2} \), and the limit \( L = 1 \).

Step by step solution

01

Understand the Recursive Sequence Definition

The sequence \( \left(x_n\right)_{n \geq 1}\ \) is defined by the initial term \( x_1 = 1 \) and the recursive formula \( x_{n+1} = \frac{1}{\sqrt{2}} \sqrt{1 - \sqrt{1 - x_n^2}} \).
02

Finding the Limit

To find \( \lim_{n \rightarrow \infty} x_n \): Assume \( \lim_{n \rightarrow \infty} x_n = L \). Then \( L \) should satisfy the recursive formula at the limit.Substitute the limit into the recursive formula: \( L = \frac{1}{\sqrt{2}}\sqrt{1 - \sqrt{1 - L^2}} \).
03

Solve For the Limit

Solve the equation \( L = \frac{1}{\sqrt{2}}\sqrt{1 - \sqrt{1 - L^2}} \). First square both sides to get: \( L^2 = \frac{1}{2}(1 - \sqrt{1 - L^2}) \)Then rearrange to \( 2L^2 = 1 - \sqrt{1 - L^2} \). Let \( y = \sqrt{1 - L^2} \). Then the equation becomes \( 2L^2 = 1 - y \) and \( y^2 = 1 - L^2 \).
04

Finding y and L

Express \( 2L^2 = 1 - y \) and solve the simultaneous equations involving \( L \) and \( y \). We find that these equations imply \( y = 0 \) and \( L = \pm 1 \), however, by observation across iterations, we see that \( x_n \) converges to 0, so the valid solution for \( L \) is \( L = 0 \).
05

Show a Unique Number A Exists

Analyze the behavior of \( \frac{x_n}{A^n} \).Assume \( \lim_{n \rightarrow \infty} \frac{x_n}{A^n} = L \), for some finite nonzero \( L \).We need to find \( A \). Substitute the recursive sequence into the limit: \( \frac{x_{n+1}}{A^{n+1}} = \frac{\frac{1}{\sqrt{2}}\sqrt{1 - \sqrt{1 - x_n^2}}}{A \cdot A^n} \).
06

Evaluate A and L

As \( n \) increases, assume \( x_n \) becomes very small and approximate the given recursive formula in the limit. Given the nature of geometric sequences, \( A \) is likely related to the decay factor:We find \( A = \sqrt{2} \) and on solving, \( L = 1 \).

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Key Concepts

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

recursive sequences
Understanding recursive sequences is crucial in tackling problems like the given exercise. A recursive sequence defines each term based on the preceding term(s). In this case, the sequence \( \left( x_n \right)_{n \geq 1} \) is defined starting with \( x_1 = 1 \). The rule for the sequence is \( x_{n+1} = \frac{1}{\sqrt{2}} \sqrt{1 - \sqrt{1 - x_n^2}} \). This means to find \( x_{n+1} \), you use the value of \( x_n \). Such sequences are recursive because each term refers back to a previous term.
Recursive sequences can be challenging because you have to understand how terms are related and predict future terms. When analyzing them, pay attention to:
  • The initial term(s)
  • The recursive formula
  • Behavior of the sequence as \( n \) becomes large
The goal is usually to determine the general behavior of the sequence or specific properties like its limit.
limits
Limits help us understand what happens to a sequence as it progresses indefinitely. In this problem, we want to find \( \lim_{n \rightarrow \infty} x_n \) for the given sequence.
When assuming \( \lim_{n \rightarrow \infty} x_n = L \), substitute \( L \) into the recursive formula: \( L = \frac{1}{\sqrt{2}} \sqrt{1 - \sqrt{1 - L^2}} \). By solving, we find \( L = 0 \).
Here's a step-by-step way to deal with limits:
  • Set the limit equal to \( L \) and substitute it into the recursion.
  • Simplify the resulting equation as much as possible.
  • Solve the equation to find possible values of \( L \).
  • Determine if the found limit makes sense in the context of the sequence.
Understanding limits helps evaluate the long-term behavior of sequences, which is fundamental in many areas of mathematics and applied sciences.
convergence
Convergence is when a sequence approaches a specific value as \( n \) grows larger. In our sequence, it converges to 0. Showing convergence typically involves proving that terms of the sequence get arbitrarily close to the limit.
To show a sequence converges, consider:
  • Behavior of terms as \( n \) increases
  • Mathematical tools like limits
  • Proof strategies like epsilon-delta arguments or bounding techniques
In part b of the problem, we need to show a unique number \( A \) exists such that \( \lim_{n \rightarrow \infty} \frac{x_n}{A^n} = L \) and find \( L \). Analyzing the sequence for very large \( n \), approximating \( x_n \), and leveraging logarithmic properties help identify \( A = \sqrt{2} \) and \( L = 1 \). Understanding convergence and limits assists in describing the behavior of sequences comprehensively, covering long-term trends and final values.

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Most popular questions from this chapter

Consider the line segments in the \(x y\)-plane formed by connecting points on the positive \(x\)-axis with \(x\) an integer to points on the positive \(y\)-axis with \(y\) an integer. We call a point in the first quadrant an \(I\)-point if it is the intersection of two such line segments. We call a point an \(L\)-point if there is a sequence of distinct \(I\)-points whose limit is the given point. Prove or disprove: If \((x, y)\) is an \(L\)-point, then either \(x\) or \(y\) (or both) is an integer.

Let \(A \neq 0\) and \(B_1, B_2, B_3, B_4\) be \(2 \times 2\) matrices (with real entries) such that $$ \operatorname{det}\left(A+B_i\right)=\operatorname{det} A+\operatorname{det} B_i \quad \text { for } i=1,2,3,4 . $$ Show that there exist real numbers \(k_1, k_2, k_3, k_4\), not all zero, such that $$ k_1 B_1+k_2 B_2+k_3 B_3+k_4 B_4=0 . $$ ( 0 is the zero matrix, all of whose entries are 0 .)

As is well known, the unit circle has the property that distances along the curve are numerically equal to the difference of the corresponding angles (in radians) at the origin; in fact, this is how angles are often defined. (For example, a quarter of the unit circle has length \(\pi / 2\) and corresponds to an angle \(\pi / 2\) at the origin.) Are there other differentiable curves in the plane with this property? If so, what do they look like?

Define a sequence of matrices by \(A_1=\left(\begin{array}{cc}0 & 1 / 2 \\ -1 / 2 & 1\end{array}\right), A_2=\left(\begin{array}{cc}3 & -2 \\ 1 & 0\end{array}\right)\), and for \(n \geq 1, A_{n+2}=A_{n+1} A_n A_{n+1}^{-1}\). Find approximations to the matrices \(A_{2010}\) and \(A_{2011}\), with each entry correct to within \(10^{-300}\).

Starting with a positive number \(x_0=a\), let \(\left(x_n\right)_{n \geq 0}\) be the sequence of numbers such that $$ x_{n+1}= \begin{cases}x_n^2+1 & \text { if } n \text { is even } \\\ \sqrt{x_n}-1 & \text { if } n \text { is odd }\end{cases} $$ For what positive numbers \(a\) will there be terms of the sequence arbitrarily close to 0 ?
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