Problem 9 - Basic Analysis: Introduction to Real Analysis

Question

Let \(f(x):=x-\frac{x^{2}-2}{2 x}\) (you may recognize Newton's method for \(\sqrt{2}\) ). a) Prove \(f([1, \infty)) \subset[1, \infty) .\) b) Prove that \(f:[1, \infty) \rightarrow[1, \infty)\) is a contraction. c) Apply the fixed point theorem to find an \(x \geq 1\) such that \(f(x)=x,\) and show that \(x=\sqrt{2}\).

Step-by-Step Solution

Answer

The unique fixed point for \(f\) is \(x = \sqrt{2}\).

Step 1: Prove f([1, ∞)) ⊂ [1, ∞)

Let's start with proving that for any \(x \geq 1\), \(f(x) \geq 1\). First, express the function: \[ f(x) = x - \frac{x^2 - 2}{2x} = x - \frac{x}{2} + \frac{1}{x}. \] Thus, \[ f(x) = \frac{2x^2 - x^2 + 2}{2x^2} = \frac{2x - (x^2 - 2)/x}{2}. \] Now simplifying and rewriting, the expression becomes, \[ f(x) = \frac{x^2 + 2}{2x}. \]We need to show that \(\frac{x^2 + 2}{2x} \geq 1\) for \(x \geq 1\). Rearranging, we have: \[ \frac{x^2 + 2}{2x} \geq 1 \Rightarrow x^2 + 2 \geq 2x \Rightarrow (x-1)^2 \geq 0, \]which is always true. Thus, \(f(x) \geq 1\) for \(x \geq 1\), and this proves \(f([1, \infty)) \subset [1, \infty)\).

Step 2: Prove f is a contraction on [1, ∞)

To prove \(f\) is a contraction, we examine \(f'(x)\) and show its absolute value is less than 1 for all \(x \geq 1\). Determine \(f'(x)\):\[ f'(x) = 1 - \left(\frac{2x - x^2}{2x^2}\right)' = 1 - \frac{2 - x^2}{2x^3} + \frac{(x^2 - 2)x^2}{2x^4}. \] This simplifies to:\[ f'(x) = 1 - \frac{2}{x^2} + \frac{1}{x^2} = 1 - \frac{1}{x^2}. \]For \(x \geq 1\), \[ |f'(x)| = \left|1 - \frac{1}{x^2}\right| = 1 - \frac{1}{x^2} < 1. \] Thus, \(f\) is a contraction on \([1, \infty)\).

Step 3: Apply Fixed Point Theorem to find x such that f(x)=x

According to the Banach Fixed Point Theorem, a contraction mapping on a complete metric space has a unique fixed point. Since \(f\) is a contraction on \([1, \infty)\), it ensures the existence of a unique fixed point \(x\) such that \(f(x) = x\).Solving \(f(x) = x\), we have:\[ x = \frac{x^2 + 2}{2x} \, . \]Simplifying, we find:\[ x = \frac{x^2 + 2}{2x} \Rightarrow 2x^2 = x^2 + 2 \Rightarrow x^2 = 2 \Rightarrow x = \sqrt{2} \]. Thus, \(x = \sqrt{2}\) is the fixed point for \(f\).

Key Concepts

Contraction Mapping

A contraction mapping, within the realm of real analysis, is a function that brings points closer together. Specifically, it is a function \( f: X \to X \) on a given metric space \( (X, d) \) with a property that there exists a constant \( k \) where \( 0 \leq k < 1 \) such that for any two points \( x \) and \( y \) in \( X \), the distance between \( f(x) \) and \( f(y) \) is at most \( k \) times the distance between \( x \) and \( y \). This can be written as:

• \( d(f(x), f(y)) \leq k \cdot d(x, y) \)

This property ensures that if you repeatedly apply the function, the points will converge closer together.
In the context of the given exercise, the function \( f(x) = x - \frac{x^2 - 2}{2x} \) is shown to be a contraction on the interval \([1, \infty)\). This is by demonstrating that the derivative of \( f \), which affects how quickly the values change, is less than 1 in absolute value throughout the interval. Keep in mind:

• Contraction mappings are essential for ensuring convergence in iterative processes.
• They play a pivotal role in guaranteeing that fixed points can be found, making them a key tool in various numerical methods.

Banach Fixed Point Theorem

The Banach Fixed Point Theorem, also known as the Contraction Mapping Theorem, provides a foundational backing for finding fixed points in mathematical functions. It states that in a complete metric space, any contraction mapping will have one and only one fixed point. Here’s why this concept is invaluable:

• **Existence and Uniqueness**: The theorem not only promises the existence of a fixed point but also assures its uniqueness for any contraction mapping in a complete space.
• **Convergence**: Applying the contraction mapping iteratively will lead to a sequence that converges to the fixed point.

In the problem, the function \( f \) satisfies the criteria for a contraction on the set \([1, \infty)\). According to Banach's theorem, this guarantees a fixed point exists. The theorem simplifies the challenge of finding solutions by reducing it to checking contraction properties, thus bypassing complex algebraic manipulations. Given these principles, we take advantage of this theoretic guarantee to conclude that \( f(x) = x \) leads us to the fixed point \( x = \sqrt{2} \). This is useful because it connects theoretical results with practical problem-solving, particularly in analyzing and approximating roots of equations.

Newton's Method

Newton's Method, or Newton-Raphson method, is a powerful iterative process used to find approximations to the roots of real-valued functions. This method leverages the idea of linear approximation, using tangents to iteratively hone in on the root. Let's break it down:

• Given a function \( g \) whose root we are interested in, start with an initial guess \( x_0 \).
• The iterative formula is \( x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)} \).

For the specific function in this exercise, we see a manifestation of Newton's method for approximating \( \sqrt{2} \). The function \( f(x) = x - \frac{x^2 - 2}{2x} \) essentially simplifies to this method shown with each iterative step approaching closer to the actual root. Newton's method leverages derivatives, giving it a robust capability to swiftly converge to the root, especially if the initial guess is close to the true value.
It's important to note, however, that:

• The method can be sensitive to initial guesses.
• Convergence is not guaranteed; behavior can become erratic if the function's derivative vanishes or is undefined.

With an appropriate understanding, Newton's Method remains a dominant tool in numerical analysis, providing efficient and often easy-to-implement solutions to complex analytical problems.