Problem 41 - APEX Calculus

Question

Use the Bisection Method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval. $$ f(x)=e^{x}-2 \text { on }[0.65,0.7] $$

Step-by-Step Solution

Answer

The root is approximately 0.69.

Step 1: Initial Function Analysis

The function given is \( f(x) = e^x - 2 \). Our goal is to find the root of this function in the interval \([0.65, 0.7]\). Recall that a root of a function is where \( f(x) = 0 \).

Step 2: Evaluate Function at Endpoints

Compute \( f(0.65) = e^{0.65} - 2 \) and \( f(0.7) = e^{0.7} - 2 \). We have:\( f(0.65) \approx 1.915 - 2 = -0.085 \)\( f(0.7) \approx 2.0137 - 2 = 0.0137 \)Since \( f(0.65) \) and \( f(0.7) \) have opposite signs, a root exists between 0.65 and 0.7.

Step 3: First Bisection

Compute the midpoint of \([0.65, 0.7]\), which is:\( c = \frac{0.65 + 0.7}{2} = 0.675 \).Evaluate \( f(0.675) = e^{0.675} - 2 \).\( f(0.675) \approx 1.9641 - 2 = -0.0359 \).Since \( f(0.675) < 0 \) and \( f(0.7) > 0 \), the root is in \([0.675, 0.7]\).

Step 4: Second Bisection

Compute the midpoint of \([0.675, 0.7]\):\( c = \frac{0.675 + 0.7}{2} = 0.6875 \).Evaluate \( f(0.6875) = e^{0.6875} - 2 \).\( f(0.6875) \approx 1.9895 - 2 = -0.0105 \).Since \( f(0.6875) < 0 \) and \( f(0.7) > 0 \), the root is in \([0.6875, 0.7]\).

Step 5: Third Bisection

Compute the midpoint of \([0.6875, 0.7]\):\( c = \frac{0.6875 + 0.7}{2} = 0.69375 \).Evaluate \( f(0.69375) = e^{0.69375} - 2 \).\( f(0.69375) \approx 2.0021 - 2 = 0.0021 \).Since \( f(0.69375) > 0 \) and \( f(0.6875) < 0 \), the root is in \([0.6875, 0.69375]\).

Step 6: Refinement of Estimate

Continue the bisection process to further refine the estimate to within two decimal places. The interval narrows as:\([0.6875, 0.69375] \to [0.690625, 0.69375] \to [0.6921875, 0.69375] \to [0.69296875, 0.69375]\). Each evaluated midpoint brings us closer to the actual root.

Step 7: Final Approximation

After several iterations, you'll find that the value of \( x \) where \( f(x) \approx 0 \) within the precision of two decimal places is approximately 0.69. Further iterations if required will confirm the root to be precisely within the accuracy criteria of two decimal places.

Key Concepts

Root Approximation

Root approximation is a technique used in mathematics to find a value close to the root of a function. A root of a function is the value of the variable that makes the function equal to zero. In our example, we're using the Bisection Method to approximate the root of the function \( f(x) = e^x - 2 \).
This method helps identify a value for \( x \) such that \( f(x) \) is very close to zero. In many real-world problems, finding the exact root isn't always necessary or practical, so approximations work smoothly to provide a viable solution.

Numerical Methods

Numerical methods are mathematical tools designed to solve equations or problems that are difficult or impossible to solve analytically. They are particularly useful when dealing with complex functions or when high precision is required.
In the context of finding roots, numerical methods, like the Bisection Method, provide systematic approaches to dig deep into a function's behavior, offering step-by-step solutions. Unlike algebraic methods, which may require certain forms and simplifications, numerical methods can handle a variety of complex functions directly.

Exponential Function

The exponential function is one of the most fundamental functions in mathematics, denoted as \( f(x) = e^x \). It is characterized by the constant \( e \), approximately equal to 2.71828. This function grows rapidly, and its graph is characterized by a smooth, continuous curve that increases exponentially as \( x \) increases.
In the exercise at hand, \( f(x) = e^x - 2 \), the exponential function is modified slightly. The task is to find where this adjusted exponential function equals zero, indicating a root. Understanding exponential functions is crucial because of their prevalent usage in fields like finance, physics, and population dynamics.

Interval Halving

Interval halving is a critical concept in the Bisection Method. This technique involves repeatedly cutting an interval into halves to narrow down where the root of a function resides. Starting with an initial interval, the process evaluates the function's value at the midpoint.
Here's a simple breakdown of interval halving steps:

• Identify if a root might exist in your starting interval by checking endpoint signs.
• Calculate the midpoint to split the interval.
• Evaluate the function at this midpoint.
• Choose the sub-interval where the signs of the function values indicate a root.
• Repeat the process to refine the approximation.

This method is systematic and straightforward, making it easy for learners to grasp the iterative nature of finding approximations repeatedly.