Let's answer (c) and (d) step by step as requested: --- ### **Step 1: (c) Write a computer program for the bisection method** Here's a **Python code** to implement the bisection method for \( f(x) = \frac{\cos(x)}{3} - x - \frac{1}{2} \) on the interval \([-1, 1]\), with stopping criterion \( |x_k - x_{k-1}| < 10^{-7} \). ```python import math def f(x): return math.cos(x)/3 - x - 0.5 a = -1 b = 1 xk_old = a k = 0 print(f"{'k':<5}{'x_k':<20}") while True: xk = (a + b) / 2 print(f"{k:<5}{xk:<20.8f}") if k > 0 and abs(xk - xk_old) < 1e-7: break if f(a) * f(xk) < 0: b = xk else: a = xk xk_old = xk k += 1 ``` **Explanation:** - At each iteration, the midpoint \( x_k \) is computed and printed. - The interval is updated based on the sign of \( f(x) \). - The loop stops when \( |x_k - x_{k-1}| < 10^{-7} \). - Each printed \( x_k \) has at least 8 decimal digits. --- ### **Step 2: (d) Error Estimation and Iteration Count** The theoretical error bound for the bisection method after \( k \) iterations is: \[ |x_k - x^*| \leq \frac{b_0 - a_0}{2^k} \] To achieve \( |x_k - x^*| < 10^{-7} \) with \( b_0-a_0 = 2 \): \[ \frac{2}{2^k} < 10^{-7} \implies 2^k > 2 \times 10^7 \implies k > \log_2(2 \times 10^7) \] Calculate: \[ \log_2(2 \times 10^7) = \log_2(2) + \log_2(10^7) = 1 + 7 \log_2(10) \approx 1 + 7 \times 3.32193 = 1 + 23.2535 = 24.2535 \] So, **at least 25 iterations** are needed. **Explanation:** - The formula gives the minimum number of iterations to guarantee the error is less than \( 10^{-7} \). - The number of iterations observed in the program in (c) should match or exceed this estimate. --- ### **Final Answers** **(c) The Python code above implements the bisection method and prints each iteration number and midpoint \( x_k \) with 8 decimal digits.** **(d) At least 25 iterations are needed for the error to be less than \( 10^{-7} \). This matches the result you will observe in the output of the program in (c), confirming the consistency between the estimation and the numerical implementation.**