Problem 144 - Thomas Calculus in SI Units

Question

A decimal representation of \(e \quad\) Find \(e\) to as many decimal places as your calculator allows by solving the equation \(\ln x=1\) using Newton's method in Section 4.6.

Step-by-Step Solution

Answer

Using Newton's method, we find that \( e \approx 2.718281828 \).

Step 1: Understanding the problem

We are tasked to find the numerical value of the mathematical constant \( e \) using the equation \( \ln x = 1 \), which implies we seek \( x \) such that when the natural logarithm is applied to it, the result is 1. This \( x \) is actually the value of \( e \).

Step 1: Define the function

To apply Newton's method, we need to define the function \( f(x) \) related to the equation \( \ln x = 1 \). We use \( f(x) = \ln x - 1 \). The root of this function corresponds to the value of \( e \).

Step 2: Derive the function

Find the derivative of the function \( f(x) = \ln x - 1 \). The derivative is \( f'(x) = \frac{1}{x} \). Newton's method requires us to use both \( f(x) \) and \( f'(x) \).

Step 3: Apply Newton's Method

Newton's method updates the guess \( x_n \) using the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]. Start with an initial guess, say \( x_0 \), close to the expected value of \( e \), like 2. Use this to iterate and improve the approximation to \( e \).

Step 4: Perform Iterations

Start with \( x_0 = 2.0 \) and apply the Newton's method formula:\[ x_{n+1} = x_n - \frac{\ln x_n - 1}{1/x_n} \]. Calculate a few iterations: - \( x_1 = 2 - \frac{\ln 2 - 1}{1/2} = 2.5 \)- Continue iterating until the change between \( x_n \) and \( x_{n+1} \) is smaller than a very small number, indicating convergence.

Step 5: Convergence

After performing the iterations, observe the results:- Successive iterations should yield closer approximations of \( e \). Continue until desired precision is achieved based on calculator limitations.- Usually, the value will stabilize around 2.718281828 after a few iterations, matching the well-known value of \( e \).

Key Concepts

Natural Logarithm

The natural logarithm, represented by the symbol \( \ln(x) \), is a mathematical function that maps numbers to their corresponding logarithms with the base \( e \), where \( e \) is approximately 2.71828. Unlike logarithms with different bases, the natural logarithm is especially important in calculus because of its derivative properties and its role in exponential growth and decay.

Understanding \( \ln(x) \) is crucial when using Newton's method to find \( e \) because it directly involves solving the equation \( \ln x = 1 \). Here, \( x \) is the number we seek, and it turns out to be \( e \).

• \( \ln(e) = 1 \) because \( e^1 = e \).
• \( \ln(x) \) is undefined for \( x \leq 0 \). Therefore, \( x \) must be positive.
• The natural logarithm is closely linked to the derivative of exponential functions.

By understanding these key points, you can better comprehend how the natural logarithm is used to find values like \( e \) in various mathematical contexts.

Derivative

In calculus, a "derivative" is a measure of how a function changes as its input changes. The derivative of a function at any point gives the rate at which the function value changes with respect to changes in input.

For the function \( f(x) = \ln x - 1 \), the derivative \( f'(x) \) is crucial for applying Newton's method. The derivative of \( \ln x \) is \( \frac{1}{x} \), simplifying our function's derivative to \( f'(x) = \frac{1}{x} \).

This derivative is used in Newton's iterative formula:

• If \( f'(x) = 0 \), the method fails because you cannot divide by zero.
• The derivative being \( \frac{1}{x} \) reinforces the importance of choosing \( x \) values that are positive and not zero.

Understanding derivatives not only helps you apply Newton's method but also forms the foundational blocks of calculus, making them essential for understanding complex mathematical calculations.

Numerical Approximation

Numerical approximation involves finding an approximate solution to a problem that cannot be solved exactly. In the context of Newton's method and finding the value of \( e \), numerical approximation helps refine guesses iteratively until the solution is precise enough.

Newton's method is a powerful numerical approximation technique used to find roots of real-valued functions.

• Start with an initial guess \( x_0 \). For finding \( e \), a starting guess could be \( x_0 = 2 \).
• Update your guess using the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
• Repeat this process until the guesses converge to a stable value.
• The process is iterative and heavily relies on the initial guess being close to the actual value for rapid convergence.

This method showcases the power of numerical approximation in mathematics and engineering, particularly when exact solutions are not feasible due to complexity or limitations in analytical methods.