Question

Suppose that \(R>0, x_{0}>0,\) and $$ x_{n+1}=\frac{1}{2}\left(\frac{R}{x_{n}}+x_{n}\right), \quad n \geq 0 $$ Prove: For \(n \geq 1, x_{n}>x_{n+1}>\sqrt{R}\) and $$ x_{n}-\sqrt{R} \leq \frac{1}{2^{n}} \frac{\left(x_{0}-\sqrt{R}\right)^{2}}{x_{0}} $$

Short answer

Using mathematical induction, we have shown that the properties: 1. \(x_n > x_{n+1} > \sqrt{R}\) for \(n \geq 1\). 2. \(x_n - \sqrt{R} \leq \frac{1}{2^n}\frac{(x_0-\sqrt{R})^2}{x_0}\) for \(n \geq 1\). hold for the sequence \(x_n\) defined by \(x_{n+1} = \frac{1}{2}(x_n + \frac{R}{x_n})\) for \(n \geq 1\), where \(x_0 > 0\), and \(R > 0\).

Step-by-step solution

Base case for Property 1 (\(n=1\))

We are given that \(x_1 = \frac{1}{2}(x_0 + (R/x_0))\). We want to show \(x_1 > \sqrt{R}\). We have \(x_1 - \sqrt{R} = \frac{1}{2}(x_0 + \frac{R}{x_0}) - \sqrt{R}\). Since \(x_0 > 0\), by multiplying both sides by \(2x_0\), we obtain \((x_1 - \sqrt{R})(2x_0) = x_0^2 + R - 2x_0\sqrt{R} = (x_0 - \sqrt{R})^2\), which is nonnegative. Therefore, \(x_1 > \sqrt{R}\).

Inductive step for Property 1

Assume \(x_n > x_{n+1} > \sqrt{R}\) for some \(n \geq 1\). We want to show that this is also true for \(n+1\). By the given formula: \(x_{n+2} = \frac{1}{2}(x_{n+1} + \frac{R}{x_{n+1}})\). Since \(x_n > x_{n+1}\), we have \(\frac{R}{x_n} > \frac{R}{x_{n+1}}\), so \(x_{n+2} = \frac{1}{2}(x_{n+1} + \frac{R}{x_{n+1}}) < \frac{1}{2}(x_n+\frac{R}{x_n}) = x_{n+1}\). On the other hand, we know \(x_{n+1} > \sqrt{R}\), so \(x_{n+1}^2 > R\), and \(\frac{R}{x_{n+1}} < x_{n+1}\). Then, \(x_{n+2} = \frac{1}{2}(x_{n+1} + \frac{R}{x_{n+1}}) > \frac{1}{2}(x_{n+1} + x_{n+1}) = \sqrt{R}\). Thus, \(x_{n+2} > x_{n+3} > \sqrt{R}\), completing the induction for Property 1.

Base case for Property 2 (\(n=1\))

We want to show that \(x_1 - \sqrt{R} \leq \frac{1}{2}\frac{(x_0-\sqrt{R})^2}{x_0}\). We know that \(x_1 = \frac{1}{2}(x_0 + (R/x_0))\). Then, \(x_1 - \sqrt{R}= \frac{1}{2}(x_0 + \frac{R}{x_0}) - \sqrt{R} = \frac{1}{2}\frac{(x_0-\sqrt{R})^2}{x_0}\). So the inequality holds for the base case when \(n=1\).

Inductive step for Property 2

Assume that \(x_n - \sqrt{R} \leq \frac{1}{2^n}\frac{(x_0-\sqrt{R})^2}{x_0}\) for some integer \(n\geq1\). We want to show that the inequality holds for \(n + 1\). First, note that \(x_{n+1} - \sqrt{R} = \frac{1}{2}(x_n + \frac{R}{x_n}) - \sqrt{R}\). By the inductive assumption, this is equal to \(\frac{1}{2}(x_n - \sqrt{R} + \frac{R}{x_n}) \leq \frac{1}{2}(\frac{1}{2^n}\frac{(x_0-\sqrt{R})^2}{x_0} + \frac{R}{x_n})\). Next, we show that \(\frac{R}{x_n} \leq \frac{1}{2^{n+1}}\frac{(x_0-\sqrt{R})^2}{x_0}\). From Property 1, we know \(x_n > \sqrt{R}\), so \(x_n - \sqrt{R} > 0\), and \(\frac{R}{x_n} \leq \frac{R}{\sqrt{R}} = \sqrt{R}\). Then, \(\frac{1}{2^n}\frac{(x_0-\sqrt{R})^2}{x_0} \geq \frac{1}{2^{n+1}}\frac{(x_0-\sqrt{R})^2}{x_0} \geq \sqrt{R} \geq \frac{R}{x_n}\). Hence, the inequality \(x_{n+1} - \sqrt{R} \leq \frac{1}{2^{n+1}}\frac{(x_0-\sqrt{R})^2}{x_0}\) holds, completing the induction for Property 2.

Key concepts

Real Analysis

Real Analysis is a branch of mathematics that deals with real numbers and real-valued functions. It serves as the foundation for understanding various mathematical concepts, including sequences, series, and continuity. An essential aspect of Real Analysis is examining the behavior of sequences and their limits. The sequences are an ordered list of numbers that follow a particular rule for generating the subsequent terms. This concept is vital in proving statements regarding convergence, boundedness, or rate of change.

In the context of our given problem, Real Analysis emerges when analyzing the behavior of the sequence defined by the recursive formula:

• \(x_{n+1} = \frac{1}{2}\left(\frac{R}{x_{n}} + x_{n}\right)\)

This formula informs us how each new term in the sequence depends on the previous one. By applying Real Analysis, we investigate the long-term behavior—such as convergence to the limit—of this sequence as we confirm that each term is properly placed between specific bounds. Moreover, mathematical tools from Real Analysis allow us to rigorously show that certain properties hold for the entire sequence regardless of the starting point \(x_0\).
Understanding these underlying principles makes Real Analysis essential for comprehending how sequences such as iterative sequences behave within specified conditions, like ensuring every term stays above \(\sqrt{R}\).

Iterative Sequences

Iterative Sequences refer to sequences defined by a formula that expresses each term as a function of the preceding term(s). These sequences build upon their initial term, evolving step-by-step as a new term appears according to a specific iterative rule. Because of their recursive nature, iterative processes are widely used in mathematical studies and computational algorithms.

In this exercise, the sequence \((x_n)\) is iteratively defined by the formula:

• \(x_{n+1} = \frac{1}{2}\left(\frac{R}{x_{n}} + x_{n}\right)\)

This iterative rule serves to progressively "correct" the term \(x_n\), nudging it closer to \(\sqrt{R}\). Such sequences are often studied to analyze convergence properties, and here, the goal is to show that the terms will approach precisely \(\sqrt{R}\) from above, as well as diminish the difference with \(\sqrt{R}\) following a specific rate described by the given inequality.
This iterative technique is influential in numerical methods, like algorithms for calculating square roots, due to its simplicity and effectiveness. By following the iterative sequence, we ensure that each step brings the sequence closer to its theoretical limit. Iterative sequences are practical demonstrations of mathematical ideas put into dynamic processes.

Inductive Hypothesis

The Inductive Hypothesis serves as a pivotal tool for proofs in mathematics, primarily featuring in Mathematical Induction. It is a statement assumed to be true for a particular term in the sequence to help prove its correctness for the next term. This process is key in demonstrating the universality of a mathematical property—especially within recursive or iterative sequences.

In our problem, the inductive process involves two parts:

• Base Case - We confirm that the desired property holds for the initial value, often \(n=1\) in sequences.
• Inductive Step - Assume the property is true for \(x_n\), then prove it must hold for \(x_{n+1}\).

Here, the inductive hypothesis helps show two crucial properties: that each term remains bounded both above and below, by employing the inequalities \(x_n > x_{n+1} > \sqrt{R}\), and that the terms approach \(\sqrt{R}\) at a specific rate of approximation. This induction ensures that what holds true for initial steps holds true for the entire sequence.
Mathematically, the Inductive Hypothesis transforms a seemingly infinite chain of reasoning into a two-step process manageable and applicable to entire sequences. Thus, it simplifies complex mathematical ideas into understandable and applicable proofs, ultimately reinforcing the credibility of the iterative methods used.