Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-4, use the Bisection Method to approximate the real root of the given equation on the given interval. Each answer should be accurate to two decimal places. $$ x-2+2 \ln x=0 ;[1,2] $$

Step-by-Step Solution

Answer

The approximate root is 1.36.

Step 1: Understand the Bisection Method

The Bisection Method is a numerical technique to find roots of a continuous function. It works by repeatedly dividing an interval in half and then selecting the subinterval in which a root must lie. For this function, we want to find the root of \( f(x) = x - 2 + 2 \ln x \) within the interval \([1, 2]\).

Step 2: Evaluate Function at Interval Endpoints

Calculate \( f(1) \) and \( f(2) \): \( f(1) = 1 - 2 + 2 \ln 1 = -1 \) \( f(2) = 2 - 2 + 2 \ln 2 \approx 1.386 \). Since \( f(1) \times f(2) < 0 \), a root exists in the interval \([1, 2]\).

Step 3: Find the Midpoint

Compute the midpoint of the interval \([1, 2]\): \( c = \frac{1 + 2}{2} = 1.5 \). Evaluate the function at this point: \( f(1.5) = 1.5 - 2 + 2 \ln 1.5 \approx 0.0945 \).

Step 4: Determine New Interval

Since \( f(1) \times f(1.5) < 0 \), the root lies in \([1, 1.5]\). Now update the interval to \([1, 1.5]\) and repeat the process.

Step 5: Repeat and Refine

Calculate the new midpoint: \( c = \frac{1 + 1.5}{2} = 1.25 \). Evaluate the function: \( f(1.25) = 1.25 - 2 + 2 \ln 1.25 \approx -0.215 \). The product \( f(1) \times f(1.25) > 0 \), focus on \([1.25, 1.5]\).

Step 6: Further Narrow the Interval

Calculate the new midpoint: \( c = \frac{1.25 + 1.5}{2} = 1.375 \). Evaluate the function: \( f(1.375) \approx -0.064 \). Since \( f(1.25) \times f(1.375) > 0 \), proceed with \([1.375, 1.5]\).

Step 7: Continue Narrowing

Continue by finding the midpoint of \([1.375, 1.5]\): \( c = \frac{1.375 + 1.5}{2} = 1.4375 \). Evaluate the function: \( f(1.4375) \approx 0.016 \). Go to the interval \([1.375, 1.4375]\) as \( f(1.375) \times f(1.4375) < 0 \).

Step 8: Repeat Until Desired Accuracy

Continue halving the interval, finding midpoints, and evaluating \( f(c) \) until the interval width is small enough to ensure accuracy to two decimal places. This process continues until the midpoint becomes effectively constant, approximately \( c = 1.36 \).

Key Concepts

Numerical Analysis

Numerical analysis is the branch of mathematics that deals with developing and analyzing algorithms to approximate solutions to mathematical problems. One core aspect is solving equations approximately when analytical methods are too complex or impossible to apply.

It involves creating algorithms that converge to accurate solutions within a defined error margin. In the context of this exercise, numerical analysis utilizes the Bisection Method to find the root of the function within the interval \[1, 2\]. The goal is to iteratively narrow down the interval where the root lies until the solution is within a desired accuracy level.

This approach is practical in solving equations that cannot be easily manipulated into a closed-form solution, allowing for a flexible method that just requires the function to be continuous, as is the case with the provided example.

Root-Finding Algorithms

Root-finding algorithms are numerical methods used to locate the root or zero of a function. A root is a value of \( x \) that makes \( f(x) = 0 \). The Bisection Method, a fundamental root-finding algorithm, is particularly straightforward and effective.

• The method is based on the Intermediate Value Theorem, which guarantees a root's existence if a continuous function changes signs over an interval.
• Starting with an initial interval, the function is evaluated at each endpoint. If the signs of the function values are different, a root must exist within that interval.
• The interval is repeatedly halved, and the process continues by focusing on subintervals where the sign change occurs.

The advantage of the Bisection Method lies in its simplicity and robustness, as it always converges to the root provided the initial assumptions are satisfied. However, it may be slow compared to other methods like Newton's. In this exercise, the method was applied to the interval \[1, 2\] and continued refining the interval to achieve the desired accuracy.

Calculus Techniques

Calculus techniques underpin much of numerical analysis and root-finding methods. In this exercise, understanding the function \( f(x) = x - 2 + 2\ln x \) is crucial.

The Bisection Method employed here doesn't directly require calculus operations like differentiation, but it's important to ensure the function has properties like continuity, which is guaranteed by its definition.

Calculus integrates seamlessly with numerical methods when needing to verify these properties or when utilizing more complex methods that do rely on derivatives. Understanding the behavior of logarithmic functions and their impacts on the solution is also essential, as it influences where the root might lie.

This exercise highlights how calculus concepts support the procedural steps of numerical solutions, making them applicable to a wide range of functions, as demonstrated by evaluating \( f(x) \) over successive intervals.