Let's break down the task in **two steps** as you requested, with explanations after each step and the final answer at the end: --- ## **Step 1: Data Collection for \( c = 2 \)** **Explanation:** - You are asked to solve the linear system \(A_n x_n = b_n\) for different values of \(n\) using the Gauss-Seidel method, with \(c=2\). - Start with the zero vector as the initial guess. - For each \(n\), perform Gauss-Seidel iterations until the relative error (in the \( \ell_\infty \) norm) is below \(10^{-3}\). - Record the number of iterations required for each \(n\). - Make sure your matrix \(A_n\) is **sparse** for efficiency. - **Plot**: Number of iterations vs. \(n\). - **Find**: A mathematical (Big-O) relationship between \(n\) and the number of iterations. **What you should expect:** - For tridiagonal matrices like this one, the number of iterations typically grows as \(O(n^2)\) for the Gauss-Seidel method when \(c\) is fixed and moderate (e.g., \(c=2\)). --- ## **Step 2: Repeat for \( c = 4 \) and Compare** **Explanation:** - Now, repeat the entire process for \(c=4\) (i.e., solve the same system but with a different \(c\) value). - Again, record the number of Gauss-Seidel iterations needed for convergence for various \(n\). - **Analyze**: How does the mathematical relationship (Big-O) between \(n\) and the number of iterations change compared to \(c=2\)? **What you should expect:** - When \(c\) increases, the matrix becomes better conditioned (diagonal dominance increases), so **the Gauss-Seidel method converges faster**. - The number of iterations may grow more slowly with \(n\); the growth could be closer to \(O(n)\) or even better for large enough \(c\). --- ## **Final Answer and Summary** ### **Step 1 Conclusion (for \(c=2\)):** - For \(c=2\), the number of Gauss-Seidel iterations required to achieve the specified accuracy increases **quadratically** with \(n\): **Iterations \( \sim O(n^2) \)** - **Reason:** For moderate \(c\), the spectral radius of the iteration matrix is not close to zero, so convergence is slow and strongly dependent on \(n\). ### **Step 2 Conclusion (for \(c=4\)):** - For \(c=4\), the matrix is more diagonally dominant, so the method converges **significantly faster**. - The number of iterations grows **sub-quadratically** with \(n\), often **\(O(n)\)** or even better, depending on how dominant the diagonal is. - **Reason:** Higher \(c\) increases diagonal dominance, which improves the convergence rate of Gauss-Seidel. --- ### **Final Summary:** - **For \(c=2\):** Number of iterations \( \sim O(n^2) \) - **For \(c=4\):** Number of iterations \( \sim O(n) \) (or better) - **Explanation:** Greater diagonal dominance (higher \(c\)) leads to faster convergence because the spectral radius of the iteration matrix decreases. - **Figure/Table:** (You must plot or tabulate your data as required in the assignment to support these statements.) --- #### **Sample Table** | \( n \) | Iterations (\(c=2\)) | Iterations (\(c=4\)) | |---------|----------------------|----------------------| | 10 | 25 | 10 | | 20 | 100 | 18 | | 40 | 400 | 32 | | 80 | 1600 | 60 | #### **Sample Plot** - X-axis: \( n \) - Y-axis: Number of Iterations - Two curves: one for \(c=2\) (steeper), one for \(c=4\) (shallower) --- **You must include the actual data/plot from your experiments to fully meet the assignment requirements.**

Let's approach this systematically, following your two-step instruction with detailed explanations and final conclusions. --- ### **Step 1: Data Collection and Analysis for \( c = 2 \)** **Objective:** - Solve \(A_n x_n = b_n\) using Gauss-Seidel. - Use \( c=2 \), starting from zero vector. - For various \( n \), iterate until the relative error \( \|x^{(k)} - x_{\text{exact}}\|_{\infty} / \|x_{\text{exact}}\|_{\infty} < 10^{-3} \). - Record the number of iterations needed. - Plot iterations vs. \( n \). - Determine the mathematical relationship (Big-O notation). **Implementation details:** - **Construct \(A_n\):** - Use the given sparse banded structure. - \( h = 1/(n+1) \). - The matrix has a tridiagonal form with additional diagonals scaled by \( c/h^2 \) and \( -1/h^2 \). - **Right-hand side \(b\):** - From the problem, \(b\) is given, possibly involving boundary conditions. - For simplicity, assume the boundary conditions lead to a vector \(b\) with known entries (e.g., all ones or as specified). - **Gauss-Seidel method:** - Implement or modify existing code to stop when relative error is below \(10^{-3}\). - Use `issparse(A_n)` to confirm sparsity. **Expected result:** - For fixed \( c=2 \), as \( n \) increases, the convergence rate slows, requiring more iterations. - Theoretical background indicates that for such matrices, the number of iterations grows approximately proportionally to \( n^2 \). **Sample Data (hypothetical):** | \( n \) | Iterations (approximate) | |---------|--------------------------| | 10 | 20 | | 20 | 80 | | 40 | 320 | | 80 | 1280 | **Plot:** - Plot iteration count vs. \( n \). - Log-log plot should show a slope close to 2, confirming \( O(n^2) \). **Conclusion for Step 1:** - The number of Gauss-Seidel iterations grows approximately as \( O(n^2) \) when \( c=2 \). --- ### **Step 2: Repeat for \( c = 4 \) and compare** **Objective:** - Repeat the above process for \( c=4 \). - Observe how the iteration count changes with \( n \). **Expected behavior:** - Increasing \( c \) increases the diagonal dominance. - This improves the spectral radius of the iteration matrix, leading to faster convergence. - The iteration count's growth with \( n \) is expected to be less steep, approximately \( O(n) \) or better. **Sample Data (hypothetical):** | \( n \) | Iterations (\( c=4 \)) | |---------|------------------------| | 10 | 10 | | 20 | 20 | | 40 | 40 | | 80 | 80 | **Analysis:** - Plotting these values suggests a linear growth, \( O(n) \). --- ### **Final Conclusions:** | Parameter | Relationship Between \( n \) and Iterations | Big-Oh Notation | |-------------|----------------------------------------------|-------------------------------------| | \( c=2 \) | Quadratic growth: iterations \( \sim O(n^2) \) | \( O(n^2) \) | | \( c=4 \) | Linear growth: iterations \( \sim O(n) \) | \( O(n) \) | --- ### **Underlying Reasoning:** - **Spectral Radius and Convergence:** - The Gauss-Seidel method’s convergence rate depends on the spectral radius \( \rho(G) \) of the iteration matrix \( G \). - Increased diagonal dominance (larger \( c \)) reduces \( \rho(G) \), leading to faster convergence. - **Matrix Conditioning:** - Larger \( c \) improves the conditioning of \(A\), thus reducing the number of iterations needed. ### **Supporting Figures/Tables:** - You should generate plots of iteration count vs. \( n \) for both \( c=2 \) and \( c=4 \). - Include a table summarizing the data points. - A log-log plot can clearly illustrate the Big-O relationships. --- **In summary:** - **For \( c=2 \):** The iterations grow roughly as \( O(n^2) \). - **For \( c=4 \):** The iterations grow roughly as \( O(n) \). - **Reason:** Higher \( c \) increases diagonal dominance, improving convergence rate, which is supported by spectral radius analysis and matrix properties. --- **Note:** To complete the assignment, perform actual numerical experiments and plot the data as described.