Let's break down the solution step-by-step for each part of the question: --- ### (1.1) Algorithm for Halving the Interval (Bisection Method) **[10 marks]** #### **Step-by-Step Algorithm:** 1. **Input:** Function \( f(\theta) \), interval \([a, b]\), and tolerance \( \epsilon \). 2. **Check:** Ensure \( f(a) \) and \( f(b) \) have opposite signs (i.e., \( f(a) \cdot f(b) < 0 \)), indicating a root exists in \([a, b]\). 3. **Repeat up to max iterations or until the interval is within tolerance:** - Calculate midpoint: \( c = \frac{a + b}{2} \) - Evaluate \( f(c) \) - If \( |f(c)| < \epsilon \) or \( (b-a)/2 < \epsilon \): - **Stop:** \( c \) is the root (within tolerance) - Else: - If \( f(a) \cdot f(c) < 0 \): - Set \( b = c \) (root is in left half) - Else: - Set \( a = c \) (root is in right half) 4. **Output:** Approximate root \( c \) --- ### (1.2) Apply Bisection Method to \( f(\theta) = \theta^3 + 2\theta^2 - 3\theta - 1 \), interval \([1, 2]\) **[10 marks]** #### **Given:** - Initial interval: \([a, b] = [1.00000, 2.00000]\) - Tolerance: Up to five significant figures - Max iterations: 5 #### **Iterations Table:** Let’s denote \( f(\theta) = \theta^3 + 2\theta^2 - 3\theta - 1 \). | Iteration | \( a \) | \( b \) | \( c \) (midpoint) | \( f(a) \) | \( f(c) \) | Sign Change | New Interval | |-----------|------------|------------|--------------------|------------|------------|-------------|--------------| | 1 | 1.00000 | 2.00000 | 1.50000 | \(-1\) | \(2.125\) | \(-1 \cdot 2.125 < 0\)? No | [1.5, 2.0] | | 2 | 1.00000 | 1.50000 | 1.25000 | \(-1\) | \(0.328\) | \(-1 \cdot 0.328 < 0\)? No | [1.25, 1.5] | | 3 | 1.00000 | 1.25000 | 1.12500 | \(-1\) | \(-0.404\) | \(-1 \cdot -0.404 > 0\)? Yes | [1.0, 1.125] | | 4 | 1.12500 | 1.25000 | 1.18750 | \(-0.404\) | \(-0.053\) | \(-0.404 \cdot -0.053 > 0\)? Yes | [1.125, 1.1875] | | 5 | 1.18750 | 1.25000 | 1.21875 | \(-0.053\) | \(0.137\) | \(-0.053 \cdot 0.137 < 0\)? No | [1.1875, 1.21875] | **Note:** The actual function values should be calculated for each midpoint for accuracy (shown approx above). **Root after 5 iterations:** Approximate root \( \theta \approx 1.1875 \) (or midpoint of last interval). --- ### (1.3) Plot the Function \( f(\theta) \) and Indicate the Root **[5 marks]** You can plot using Python, MATLAB, or manually sketch. Here’s how you can do it in Python (matplotlib): ```python import numpy as np import matplotlib.pyplot as plt # Define the function def f(theta): return theta**3 + 2*theta**2 - 3*theta - 1 # Create theta values theta = np.linspace(0, 2.5, 400) y = f(theta) # Plot the function plt.figure(figsize=(8,5)) plt.plot(theta, y, label='$f(\\theta)$') plt.axhline(0, color='black', linewidth=0.7) plt.axvline(1.1875, color='red', linestyle='--', label='Approx. root') plt.xlabel('$\\theta$') plt.ylabel('$f(\\theta)$') plt.legend() plt.title('Plot of $f(\\theta)$ with Root Indicated') plt.grid(True) plt.show() ``` **On the plot:** Mark the vertical line at \( \theta \approx 1.1875 \) to indicate the root found. --- ### **Summary Table for Submission:** - **1.1:** Algorithm steps (as above) - **1.2:** Bisection method table, root after 5 iterations - **1.3:** Plot with root indication Let me know if you need the actual function values for each iteration or want the Python code output!