Question

Suppose that \(s_{0}\) and \(A\) are positive numbers. Let $$ s_{n+1}=\frac{1}{2}\left(s_{n}+\frac{A}{s_{n}}\right), \quad n \geq 0 $$ (a) Show that \(s_{n+1} \geq \sqrt{A}\) if \(n \geq 0\). (b) Show that \(s_{n+1} \leq s_{n}\) if \(n \geq 1\). (c) Show that \(s=\lim _{n \rightarrow \infty} s_{n}\) exists. (d) Find \(s\).

Short answer

Question: Prove that for any positive real number A and a positive real number \(s_0\), the limit of the sequence \({s_n}\) defined by the recursive formula \(s_{n+1} = \frac{1}{2}(s_n + \frac{A}{s_n})\) is \(\sqrt{A}\). Answer: By proving that the sequence \({s_n}\) is decreasing for \(n \geq 1\) and bounded below by \(\sqrt{A}\), we showed that the limit exists. We then found this limit by solving the equation \(s = \frac{1}{2}(s + \frac{A}{s})\) and obtained that the limit is \(s = \sqrt{A}\).

Step-by-step solution

Proving \(s_{n+1} \geq \sqrt{A}\) by induction

Let's start with a base case: \(n=0\). \(s_1 = \frac{1}{2}(s_0 + \frac{A}{s_0})\) Since \(s_0 > 0\) and \(A > 0\), we have: \(s_1 > \frac{1}{2}(\sqrt{A} + \frac{A}{\sqrt{A}}) = \frac{1}{2}(2\sqrt{A}) = \sqrt{A}\) So, the inequality holds for \(n=0\). Now, let's assume that the inequality holds for some \(n=k\), i.e., \(s_k \geq \sqrt{A}\). We need to prove that it also holds for \(n=k+1\): \(s_{k+2} = \frac{1}{2}(s_{k+1} + \frac{A}{s_{k+1}})\) By the inductive hypothesis, we know that \(s_{k+1} \geq \sqrt{A}\), thus \(\frac{A}{s_{k+1}} \leq \sqrt{A}\), and: \(s_{k+2} \geq \frac{1}{2}(\sqrt{A} + \sqrt{A}) = \sqrt{A}\) Therefore, \(s_{n+1} \geq \sqrt{A}\) for all \(n \geq 0\) by induction.

Proving \(s_{n+1} \leq s_{n}\) for \(n \geq 1\)

Using the formula for the sequence: \(s_{n+1} = \frac{1}{2}(s_n + \frac{A}{s_n})\) Since we have already proved that \(s_n \geq \sqrt{A}\) for all \(n \geq 0\), we can state that: \(s_n \geq \sqrt{A} \Rightarrow \frac{A}{s_n} \leq \sqrt{A}\) Now we have: \(s_{n+1} = \frac{1}{2}(s_n + \frac{A}{s_n}) \leq \frac{1}{2}(s_n + s_n) = s_n\) So, for all \(n \geq 1\), the sequence is decreasing.

Proving the limit exists

We have found that the sequence \({s_n}\) is decreasing for \(n \geq 1\) and bounded below by \(\sqrt{A}\). A sequence is convergent if it is bounded and monotonic. So the limit of the sequence exists, i.e., \(s = \lim_{n\to\infty} s_n\) exists.

Finding the limit \(s\)

To find the limit \(s\), we will substitute \(s_{n+1}\) and \(s_n\) with \(s\) in the given equation, since the limit of the sequence equals its limiting value: \(s = \frac{1}{2}(s + \frac{A}{s})\) Now we solve this equation for \(s\): \(2s = s + \frac{A}{s} \Rightarrow s^2 - 2s + A = 0\) This is a quadratic equation with the discriminant \(\Delta = 2^2 - 4 \cdot 1 \cdot A = 4 - 4A\). Since \(A > 0\), the discriminant is negative. A square root of a negative number is imaginary, so \(s\) must be of the form \(s = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{2 \pm \sqrt{-4 + 4A}}{2}\). However, since we have proved that \(s \geq \sqrt{A}\), we must have: \(s = \frac{2 + \sqrt{-4 + 4A}}{2} = \sqrt{A}\) Thus, the limit of the sequence \(s_n\) is \(s = \sqrt{A}\).

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique commonly used in mathematics, particularly in proving properties of sequences and series. It operates on the principle that if a certain statement holds for the first term of a sequence and if the truth of that statement for one term implies its truth for the next term, then the statement is true for all terms of the sequence. In the context of the sequence given by the recurrence relation
\(s_{n+1} = \frac{1}{2}(s_n + \frac{A}{s_n})\),
mathematical induction is used to demonstrate properties such as the terms of the sequence being greater or equal to \(\sqrt{A}\).

Applying the Principle of Mathematical Induction

In the solution provided, the first part uses induction to show that each term in the sequence is at least as large as the square root of \(A\). Starting with the base case, where \(n=0\), the property is verified directly. Then, assuming the property holds for \(n=k\) (the inductive hypothesis), it's shown to hold for \(n=k+1\). This 'domino effect' assures us that the inequality \(s_{n+1} \geq \sqrt{A}\) is true for all natural numbers \(n\).

Real Analysis

Real analysis is a branch of mathematics that deals with the behavior and properties of real numbers, sequences, functions, and their limits. It provides the rigorous underpinnings of calculus and is fundamental for understanding convergence, continuity, and the structure of the real number line. The problem at hand, involving sequences and their convergence, is a classic study in real analysis.
Real analysis is concerned with several types of convergence for sequences, including pointwise, uniform, and absolute convergence. In this particular exercise, we are concerned with the concept of the limit of a sequence. Through the rules established in real analysis, sequences are analyzed to determine if they converge to a specific value as the sequence progresses to infinity.

Elements of Real Analysis in the Exercise

The exercise demonstrates a sequence's convergence using properties such as monotonicity and boundedness—key concepts in real analysis which ensure that a limit exists if a sequence is both bounded and monotonic.

Limits of Sequences

In mathematics, particularly in real analysis, the limit of a sequence is the value that the terms of the sequence approach as the index (often denoted by \(n\)) becomes very large. If such a limit exists, the sequence is said to be convergent; otherwise, it is divergent. The limit of a sequence is a fundamental concept that helps in the analysis of more complex mathematical structures like functions and series.

Convergence of the Sequence

For the sequence defined by \(s_{n+1} = \frac{1}{2}(s_n + \frac{A}{s_n})\), we are interested in proving its convergence and finding its limit as \(n\) approaches infinity. After establishing that the sequence is decreasing and bounded below by \(\sqrt{A}\), we invoke the Monotone Convergence Theorem from real analysis, which states that every bounded monotone sequence converges to a real limit. This fundamental theorem reinforces the conclusion that the limit \(\lim_{n \to \infty} s_n = \sqrt{A}\) exists.

Monotonic Sequences

A sequence is called monotonic if it is either entirely non-increasing (every term is less than or equal to the previous term) or non-decreasing (every term is greater than or equal to the previous term). Monotonic sequences have a very useful property in analysis: every bounded monotonic sequence is convergent. This property is especially significant in the context of the problem, where we are asked to show the convergence of a sequence.

Understanding Monotonicity in the Given Sequence

The solution steps demonstrate that the sequence \(\{s_n\}\) is non-increasing for \(n \geq 1\). By establishing that \(s_{n+1} \leq s_n\), we know that as \(n\) increases, the terms of the sequence do not increase. In combination with the proof that the sequence is bounded below by \(\sqrt{A}\), this information guarantees that the sequence converges, which is vital for real analysis and understanding the behavior of functions and their limits as they approach infinity.