Problem 15 - Advanced Calculus

Question

Show that the sequence \(\left\\{x_{n} \mid\right.\) defined by the recurs?e formula: $$ x_{1}=1, \quad x_{n+1}=x_{n}+1 / x_{n} \quad \text { for } n>1 $$ obeys the inequality \(x_{n}>\sqrt{n}\) for all \(n \geq 2\).

Step-by-Step Solution

Answer

The statement is true. The recursive sequence \(x_{n}=x_{n-1}+1 / x_{n-1}\) with \(x_1=1\) does indeed satisfy the inequality \(x_{n}>\sqrt{n}\) for all \(n \geq 2\). The proof involves mathematical induction where both the base step and the induction step are successfully verified.

Step 1: Base Step

This is the establishing part of the proof. For \(n=2\), \(x_2 = x_1 + 1/x_1 = 1 + 1/1 = 2\). Clearly, \(x_2 > \sqrt{2}\), so the inequality holds for the base step.

Step 2: Induction Step

First, assume that the inequality \(x_{k} > \sqrt{k}\) holds for some positive integer \(k \geq 2\). Now, it needs to be proved that the inequality \(x_{k+1} > \sqrt{k+1}\) holds. Since \(x_{k+1} = x_k + 1/x_k\), replacing \(x_k\) with \(\sqrt{k}\) (since \(x_k > \sqrt{k}\)), we can see that \(x_{k+1} > \sqrt{k} + 1/\sqrt{k} = \sqrt{k} + \sqrt{1/k}\). After squaring both sides and simplifying, we get that \(x_{k+1}\) is larger than \(k+1\). Hence, the inequality \(x_{k+1} > \sqrt{k+1}\) holds and the statement is proven by mathematical induction.

Key Concepts

Recursive Sequences

A recursive sequence is a type of sequence where each term is defined based on previous terms. This feature allows you to build complex sequences from simple rules. The sequence in the original exercise demonstrates a common form of recursion. It starts with a known value, in this case, \( x_1 = 1 \). Each following term then is constructed using a specific formula based on its preceding term, as given by \( x_{n+1} = x_n + \frac{1}{x_n} \).

The key elements are:

• Initial term: The sequence begins with a defined starting value. Here, \( x_1 = 1 \) serves as a base.
• Recurrence relation: A formula that generates the subsequent terms depending on prior terms. The relation \( x_{n+1} = x_n + \frac{1}{x_n} \) is used here.

Understanding recursive sequences is crucial as they appear frequently in mathematics, providing a tool to define sequences incrementally based on previous results.

Inequalities

Inequalities are statements about the relative size or order of two objects. In mathematics, they are used to describe relations that are not precisely equal. In this exercise, the inequality \( x_n > \sqrt{n} \) suggests that each term in our sequence is greater than the square root of its index number.

Let's break down how inequalities function:

• Relation: The inequality establishes a relation that one expression is larger than the other.
• Simplification and Manipulation: Working with inequalities often involves simplifying expressions or manipulating the inequality to prove the desired relationship. This might include rearranging terms or applying operations equally to both sides of the inequality.

Understanding how to manipulate inequalities is fundamental in proving mathematical statements, such as showing that a certain property holds for all terms of a recursive sequence.

Mathematical Proof

Mathematical proofs are logical arguments used to establish the truth of a statement. They can be direct or, as in this exercise, use a technique called mathematical induction, which is particularly useful for statements about recursive sequences.

Here's how mathematical induction generally works:

• Base Case: First, you prove that the statement holds for the initial value of the sequence. This is the foundation of the induction process.
• Induction Step: Then, assuming the statement holds for some arbitrary value \( k \), you show it must also hold for \( k+1 \). This step is crucial as it establishes the domino effect of the proof, extending the truth of the statement to all subsequent terms.

Mastering mathematical induction involves understanding both these steps deeply, as they are combined to extend the proof to an entire range of natural numbers, one step at a time.