f(x) = e^x - cos(x) - 2 using bisection method

To apply the bisection method to find the root of the function \( f(x) = e^x - \cos(x) - 2 \), we need to follow these steps:

1. **Identify an interval \([a, b]\)** where the function changes sign, i.e., \( f(a) \cdot f(b) < 0 \).
2. **Calculate the midpoint \( c = \frac{a + b}{2} \)**.
3. **Evaluate \( f(c) \)**.
4. **Determine the sub-interval**: If \( f(a) \cdot f(c) < 0 \), then set \( b = c \); otherwise, set \( a = c \).
5. **Repeat** steps 2-4 until the desired accuracy is achieved.

### Step 1: Find an Interval

Let's evaluate \( f(x) \) at some values to find an appropriate interval:

- \( f(0) = e^0 - \cos(0) - 2 = 1 - 1 - 2 = -2 \)
- \( f(1) = e^1 - \cos(1) - 2 \approx 2.718 - 0.540 - 2 \approx 0.178 \)

Since \( f(0) < 0 \) and \( f(1) > 0 \), we can choose the interval \([0, 1]\).

### Step 2: Bisection Process

Let's perform the bisection method for a few iterations:

1. **First Iteration**:
   - \( c = \frac{0 + 1}{2} = 0.5 \)
   - \( f(0.5) = e^{0.5} - \cos(0.5) - 2 \approx 1.649 - 0.877 - 2 \approx -1.228 \)
   - Since \( f(0) < 0 \) and \( f(0.5) < 0 \), update \( a = 0.5 \).

2. **Second Iteration**:
   - \( c = \frac{0.5 + 1}{2} = 0.75 \)
   - \( f(0.75) = e^{0.75} - \cos(0.75) - 2 \approx 2.117 - 0.731 - 2 \approx -0.614 \)
   - Update \( a = 0.75 \).

3. **Third Iteration**:
   - \( c = \frac{0.75 + 1}{2} = 0.875 \)
   - \( f(0.875) = e^{0.875} - \cos(0.875) - 2 \approx 2.398 - 0.639 - 2 \approx -0.241 \)
   - Update \( a = 0.875 \).

4. **Fourth Iteration**:
   - \( c = \frac{0.875 + 1}{2} = 0.9375 \)
   - \( f(0.9375) = e^{0.9375} - \cos(0.9375) - 2 \approx 2.551 - 0.586 - 2 \approx 0.034 \)
   - Since \( f(0.875) < 0 \) and \( f(0.9375) > 0 \), update \( b = 0.9375 \).

5. **Fifth Iteration**:
   - \( c = \frac{0.875 + 0.9375}{2} = 0.90625 \)
   - \( f(0.90625) \approx 2.460 - 0.616 - 2 \approx -0.156 \)
   - Update \( a = 0.90625 \).

Continue this process until the interval is sufficiently small (e.g., when \( |b - a| < \epsilon \) for a chosen \( \epsilon \)).

### Conclusion

You can continue this iteration to achieve your desired precision. The root lies within the interval and can be approximated as \( c \) converges.

Would you like further details on any specific part of this process, or do you have any questions? Here are some related questions you might consider:

1. How does the bisection method compare to other root-finding methods?
2. What is the significance of choosing the initial interval correctly?
3. Can the bisection method be applied to complex functions?
4. What are some limitations of the bisection method?
5. How would you implement the bisection method in a programming language? 

**Tip**: Always verify that \( f(a) \) and \( f(b) \) have opposite signs before starting the bisection process!

Explore the step-by-step process of finding the root of the function f(x) = e^x - cos(x) - 2 using the bisection method. This guide covers interval selection, iterative calculations, and the significance of choosing the right interval. Ideal for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation