# Newton's Method: Step-by-Step Solution Given: - \( f(x) = x^3 - 4x - 9 \) - \( f'(x) = 3x^2 - 4 \) - Initial guess: \( x_0 = 2.5 \) - Perform 5 iterations. Newton's formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Let's compute each iteration step-by-step: --- ## **Step-by-Step Iterations** ### **Iteration 0** Initial guess: \( x_0 = 2.5 \) --- ### **Iteration 1** Calculate: - \( f(x_0) = (2.5)^3 - 4(2.5) - 9 = 15.625 - 10 - 9 = -3.375 \) - \( f'(x_0) = 3(2.5)^2 - 4 = 18.75 - 4 = 14.75 \) \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2.5 - \frac{-3.375}{14.75} \approx 2.5 + 0.2288 = 2.7288 \] --- ### **Iteration 2** - \( f(x_1) = (2.7288)^3 - 4(2.7288) - 9 \approx 20.324 - 10.915 - 9 = 0.409 \) - \( f'(x_1) = 3(2.7288)^2 - 4 \approx 3(7.449) - 4 = 22.347 - 4 = 18.347 \) \[ x_2 = 2.7288 - \frac{0.409}{18.347} \approx 2.7288 - 0.0223 = 2.7065 \] --- ### **Iteration 3** - \( f(x_2) = (2.7065)^3 - 4(2.7065) - 9 \approx 19.857 - 10.826 - 9 = 0.031 \) - \( f'(x_2) = 3(2.7065)^2 - 4 \approx 3(7.329) - 4 = 21.987 - 4 = 17.987 \) \[ x_3 = 2.7065 - \frac{0.031}{17.987} \approx 2.7065 - 0.0017 = 2.7048 \] --- ### **Iteration 4** - \( f(x_3) = (2.7048)^3 - 4(2.7048) - 9 \approx 19.823 - 10.819 - 9 = 0.004 \) - \( f'(x_3) = 3(2.7048)^2 - 4 \approx 3(7.321) - 4 = 21.963 - 4 = 17.963 \) \[ x_4 = 2.7048 - \frac{0.004}{17.963} \approx 2.7048 - 0.0002 = 2.7046 \] --- ### **Iteration 5** - \( f(x_4) = (2.7046)^3 - 4(2.7046) - 9 \approx 19.819 - 10.818 - 9 = 0.001 \) - \( f'(x_4) = 3(2.7046)^2 - 4 \approx 3(7.320) - 4 = 21.960 - 4 = 17.960 \) \[ x_5 = 2.7046 - \frac{0.001}{17.960} \approx 2.7046 - 0.00006 = 2.70454 \] --- ## **Summary Table** | Iteration | Approximate Root | |-----------|------------------| | 0 | 2.50000 | | 1 | 2.72881 | | 2 | 2.70651 | | 3 | 2.70478 | | 4 | 2.70456 | | 5 | 2.70454 | --- ## **Final Answer** **The approximate root after 5 iterations is:** \[ \boxed{2.70454} \] --- **If you need a MATLAB script for this, let me know!**

# Newton's Method: Comprehensive Solution ## **1. Introduction** Newton's Method, also known as the Newton-Raphson method, is an iterative numerical technique used to find approximate roots of real-valued functions. It utilizes the derivative of the function to improve the accuracy of the estimates with each iteration. The method is particularly effective for functions that are continuous and differentiable around the root. To apply Newton's Method, the following essential components are required: - The function \( f(x) \) whose root is sought. - The derivative \( f'(x) \) of the function. - An initial guess \( x_0 \) close to the actual root. ### **Explanation** The introduction provides foundational knowledge on Newton's Method, emphasizing its iterative nature and reliance on derivatives. This understanding is crucial for following the steps involved in the calculations, as it underpins the formula used in each iteration to refine the root estimate. --- ## **2. Relevant Formulas** ### **Newton's Iterative Formula** \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] - **Explanation**: This formula derives from the concept of tangent lines. The new approximation \( x_{n+1} \) is obtained by adjusting the current estimate \( x_n \) by the ratio of the function value to its derivative at that point. This adjustment aims to find where the tangent line intersects the x-axis, moving closer to the root. ### **Function and Derivative** - \( f(x) = x^3 - 4x - 9 \) - \( f'(x) = 3x^2 - 4 \) ### **Explanation** The function \( f(x) \) defines the problem, while \( f'(x) \) provides the necessary slope for the tangent line. Together, they are integral to applying Newton's Method successfully. --- ## **3. Detailed Step-by-Step Solution** ### **Step 1: Initialization** - Set the initial guess: \( x_0 = 2.5 \) ### **Explanation** Initialization is critical as it sets the starting point for the iterative process. The closer this initial guess is to the true root, the more efficient the convergence of the method. --- ### **Step 2: First Iteration** Calculate: - \( f(x_0) = (2.5)^3 - 4(2.5) - 9 = 15.625 - 10 - 9 = -3.375 \) - \( f'(x_0) = 3(2.5)^2 - 4 = 18.75 - 4 = 14.75 \) Now, apply Newton's formula: \[ x_1 = 2.5 - \frac{-3.375}{14.75} \approx 2.5 + 0.2288 = 2.7288 \] ### **Explanation** This step involves calculating both the function value and its derivative at the initial guess. The adjustments made using Newton's formula lead to a new approximation, which should be closer to the true root. --- ### **Step 3: Second Iteration** - \( f(x_1) = (2.7288)^3 - 4(2.7288) - 9 \approx 0.409 \) - \( f'(x_1) = 3(2.7288)^2 - 4 \approx 18.347 \) Now, apply the formula: \[ x_2 = 2.7288 - \frac{0.409}{18.347} \approx 2.7065 \] ### **Explanation** In this iteration, the function value at the new estimate \( x_1 \) is calculated. The method continues to refine the estimate, moving closer to the root with each iteration. --- ### **Step 4: Third Iteration** - \( f(x_2) = (2.7065)^3 - 4(2.7065) - 9 \approx 0.031 \) - \( f'(x_2) = 17.987 \) Apply the formula: \[ x_3 = 2.7065 - \frac{0.031}{17.987} \approx 2.7048 \] ### **Explanation** The reduction in the function value indicates that the approximation is approaching the root. Each iteration brings the estimate closer to convergence, as evidenced by the diminishing function values. --- ### **Step 5: Fourth Iteration** - \( f(x_3) = (2.7048)^3 - 4(2.7048) - 9 \approx 0.004 \) - \( f'(x_3) = 17.963 \) Use the formula: \[ x_4 = 2.7048 - \frac{0.004}{17.963} \approx 2.7046 \] ### **Explanation** Continuing to narrow down the approximation, the function value is now significantly smaller, indicating that the root is very close to the current estimate. --- ### **Step 6: Fifth Iteration** - \( f(x_4) = (2.7046)^3 - 4(2.7046) - 9 \approx 0.001 \) - \( f'(x_4) = 17.960 \) Apply the formula: \[ x_5 = 2.7046 - \frac{0.001}{17.960} \approx 2.70454 \] ### **Explanation** The final iteration shows that the method has converged to a very precise estimate of the root. The small function value confirms that the approximation is extremely close to the actual root. --- ## **Conclusion** The iterative process of Newton's Method has yielded an approximate root after five iterations: \[ \boxed{2.70454} \] This value represents the solution to the problem, demonstrating the effectiveness of Newton's Method in finding roots of equations. Each step has built upon the previous estimates, showcasing the iterative refinement characteristic of this numerical approach.