Problem 15 - A First Course in Numerical Methods

Question

For \(x>0\) consider the equation \(x+\ln x=0\) It is a reformulation of the equation of Example \(3.4\). (a) Show analytically that there is exactly one root, \(0

Step-by-Step Solution

Answer

Answer: To show that the equation \(x + \ln x = 0\) has exactly one root in \((0, \infty)\), we first define the function \(f(x) = x + \ln x\) then find its first derivative, \(f'(x) = 1 + \frac{1}{x}\). Since \(f'(x) > 0\) for all \(x > 0\), we know that the function is strictly increasing on \((0, \infty)\). Next, we analyze the limits of the function as \(x\) approaches 0 and infinity and see that \(\lim_{x \to 0^{+}} f(x) = -\infty\) and \(\lim_{x \to \infty} f(x) = \infty\), which means the function covers the entire range \((-\infty, \infty)\). Lastly, using the Intermediate Value Theorem, we can conclude that there is exactly one root \(x^*\) in the interval \((0, \infty)\) for the equation \(x + \ln x = 0\).

Step 1: Part (a): Showing that there is exactly one root

First, let's define \(f(x)=x+\ln{x}\). To check the monotonicity of this function, we need to look at its first derivative.

Step 1: Finding the first derivative

Derive \(f(x)\) with respect to \(x\): \[f'(x)=1+\frac{1}{x}\]

Step 2: Analyzing the first derivative

Observe that for \(x>0\), the first derivative is always positive \(f'(x)>0\). This implies that \(f(x)\) is a strictly increasing function on \((0, \infty)\).

Step 3: Finding the limits of the function as x approaches 0 and infinity

Now let's analyze the limits of the function: \[\lim_{x \to 0^{+}} f(x) = -\infty\] \[\lim_{x \to \infty} f(x) = \infty\]

Step 4: Applying the Intermediate Value Theorem

Using the limits, we see that the function covers the entire range \((-\infty, \infty)\). Since \(f(x)\) is strictly increasing on \((0, \infty)\), by the Intermediate Value Theorem, there is exactly one root \(x^*\) in this interval \((0, \infty)\) for the equation \(x+\ln x=0\).

Step 6: Part (b): Plotting the graph

Using a graphing tool like Desmos, plot the function \(f(x)=x+\ln x\) on the interval \([0.1, 1]\). Analyze the graph to see where the root lies.

Step 7: Part (c): Finding the root using numerical methods

Now we need to implement the bisection method, fixed point iteration, Newton's method, and the secant method, to find the root \(x^{*}\) of the equation \(x + \ln x = 0\). For each method, we will use \(|x_{k} - x_{k-1}| < 10^{-10}\) as the convergence criterion, print out the iterates, and analyze their convergence behavior. We will also provide the necessary justifications and explanations for each method. For the implementation of the methods, you can follow any programming language available to you. In this example, we will use MATLAB as requested, but you could easily adapt this to other programming environments.

Key Concepts

Bisection Method

The bisection method is a simple and reliable numerical technique used to find roots of continuous functions. It works by narrowing down the interval where the root lies until it converges to the solution. Start with an initial interval, say \([0.5, 0.6]\), where the function changes sign. This change of sign is crucial as it indicates at least one root exists in the interval.
Here's how the method operates:

• Check the midpoint of the interval.
• Determine which half-interval contains a root, by evaluating the function's sign at the midpoint.
• Repeat this process until the interval is sufficiently small, satisfying the given convergence criterion, \(|x_{k} - x_{k-1}| < 10^{-10}|\).

The bisection method is very robust because it always converges if the function is continuous and initially has opposite signs at the ends of the interval. However, it may not be the quickest method since it proceeds with a linear rate of convergence.

Fixed Point Iteration

Fixed point iteration is a numerical method that transforms an equation into a fixed-point form and iteratively solves for the root. This technique involves rewriting the problem, \(x + \ln x = 0\), to \(x = g(x)\). For effective application, \(g(x)\) should satisfy the conditions of the Fixed Point Theorem, ensuring convergence.
To ensure the root of the equation using fixed point iteration, the derived function \(g(x)\) must have a derivative whose absolute value is less than 1 over the interval of interest. Initializing with \(x_0 = 0.5\),

• Apply the function \(g(x)\) iteratively.
• Generate new approximations \(x_{n+1} = g(x_n)\).
• Check if the root meets the convergence criterion of \(|x_{k} - x_{k-1}| < 10^{-10}|\).

Fixed point iteration can be easy to implement and understand but may exhibit slow convergence, especially if the function does not meet the theorem's strict criteria.

Newton's Method

Newton's method, also known as the Newton-Raphson method, is a powerful numerical tool for finding successively better approximations to the roots of a real-valued function. It is highly regarded for its rapid (quadratic) convergence when an appropriate initial guess is chosen.
The method involves the use of the tangent of the function at a given point to estimate the root. Here's how you would apply Newton's method:

• Start with an initial guess, \(x_0 = 0.5\), close to where the root is expected.
• Derive the function \(f(x) = x + \ln x\) to find the tangent line of the curve.
• Update the approximation using the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) until the tolerance \(|x_{k} - x_{k-1}| < 10^{-10}|\).

Newton's method is efficient and preferred in many cases due to its speed of convergence. However, its efficacy heavily depends on the choice of the initial guess and requires derivative calculations, which can be complex for certain functions.

Secant Method

The secant method is similar to Newton's method but does not require the computation of derivatives, making it simpler when derivatives are difficult to find. Instead of tangents, secant lines are used to approximate the curve of the function.
Implementing the secant method involves:

• Starting with two initial guesses, \(x_0 = 0.5\) and \(x_1 = 0.6\).
• Using these points to form a secant line and calculate the new approximate root \(x_{n+1} = x_n - \frac{f(x_n)(x_n - x_{n-1})}{f(x_n) - f(x_{n-1})}\).
• Continue iterations until the convergence criterion, \(|x_{k} - x_{k-1}| < 10^{-10}|\), is met.

The secant method provides superlinear convergence and is a noteworthy choice when a derivative is unavailable or inconvenient to compute. Nevertheless, it generally converges slower than Newton's method.