重要的日子

7/1：♥️糖醋羊排CP成立！！♥️

Track

• sunclientlogin发票报销到账
• 开固态硬盘发票

Routine

爱心屋签到： aixinwu.sjtu.edu.cn/products/asw-store
每日二GRISSO💊

Starting from 5/12:

Non-routine

论文

Due

• 糖糖 党员会议 7/?
• lby会议室刷卡 7/24 下午
• 个人网站workshop 7/25 10：00
• 6003 project presentation 7/28
  • proposal
  • 代码
  • ppt
• 间隔棒机器人 7/31
  • arduino cloud 订阅
  • 订阅账单汇总
  • 说明书
  • 寄送
• expo志愿者 8/5
• 实验室网站调研部署 8/10(夏季学期结束前)
  • 登录界面
  • 用户点评
  • 多tab
  • 视频嵌入
• 给糖糖的画 8/13
• 青少年活动中心材料提交 9/1
• ICRA video clip 9/6
• ICRA 2026 9/15
• ICRA paper acceptance notification 1/31
• draumurvakna.github.io 一周年纪念日 2026/10/27
• 读书会吃饭 2026/3/23

No Due

• vmware horizon 报错 IT邮件
• 保温杯
• 釜山行
• 长沙玩
• 给糖糖补补
• “现在的贺卡好高级”
• claude api 报销，connected papers 报销
• 种下星露谷物语种子
• 软考
• ccf csp
• 给通哥画一张
• 穹彻智能实习
• Godot Learning [[Godot Learning Plan]]
• 琪琪水彩
• 流萤水彩
• 美罗城 云海肴 汽锅鸡
• 夏季expo 志愿者
• 《涉过愤怒的海》
• 《被讨厌的勇气》读书会p16
• 《被寄生的家庭》读书会p18
• 《蛙》读书会 p21
• 《断舍离》读书会 p20
• 潮汕砂锅粥
• 想买…
  • ipad（matebook e go）
  • 欧珑无极乌龙
  • bose大鲨2耳塞套
  • 综合效果器 ..?
  • 清音音箱
  • 220V户外电源

有生之年系列

• 手工🎸
• 侍奉社

Archived

• 想游泳… 🔒Done! 2025-05-12 🕸️ Track > Routine > Non-routine > No Due
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• 计算机视觉pre 讲稿 5/28 🔒Done! 2025-05-28 🕸️ Track > Routine > Non-routine > Due
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• pixtral inference PR #1 5/28 🔒Done! 2025-05-28 🕸️ Track > Routine > Non-routine > Due
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• Seminar Series VI 5/30 🔒Done! 2025-05-30 🕸️ Track > Routine > Non-routine > Due
• 张vv 这就是中国 5/30 🔒Done! 2025-05-30 🕸️ Track > Routine > Non-routine > Due
• 6003 hw 1 6/1 🔒Done! 2025-06-02 🕸️ Track > Routine > Non-routine > Due
• 7606 quiz 1 6/4 🔒Done! 2025-06-04 🕸️ Track > Routine > Non-routine > Due
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• 羽毛球拉人 6/4 🔒Done! 2025-06-04 🕸️ Track > Routine > Non-routine > Due
• math hw2 6/4 🔒Done! 2025-06-04 🕸️ Track > Routine > Non-routine > Due
• update nyc 芝士蛋糕 6/4 🔒Done! 2025-06-04 🕸️ Track > Routine > Non-routine > Due
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• 显卡维修发票 🔒Done! 2025-06-05 🕸️ Track
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• 星露谷 笔记本安装6/7 🔒Done! 2025-06-07 🕸️ Track > Routine > Non-routine > Due
• 数据挖掘 hw3 6/8 🔒Done! 2025-06-07 🕸️ Track > Routine > Non-routine > Due
• 蛋糕烘培准备 6/8 🔒Done! 2025-06-09 🕸️ Track > Routine > Non-routine > Due
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• 重新打印数据挖掘考试纲要 6/11 🔒Done! 2025-06-11 🕸️ Track > Routine > Non-routine > Due
• 羽毛球 13:00 6/12 🔒Done! 2025-06-12 🕸️ Track > Routine > Non-routine > Due
• 数据挖掘考试 hand open-book 6/12 SY215 18:00 🔒Done! 2025-06-12 🕸️ Track > Routine > Non-routine > Due
• LLM task planning 论文发放给zxy 6/13 🔒Done! 2025-06-13 🕸️ Track > Routine > Non-routine > Due
• 芝士蛋糕原料收货6/13 🔒Done! 2025-06-14 🕸️ Track > Routine > Non-routine > Due
• 做蛋糕！！6/14 🔒Done! 2025-06-15 🕸️ Track > Routine > Non-routine > Due
• 青少年活动中心宣部长谈话 10:00 6/15 🔒Done! 2025-06-15 🕸️ Track > Routine > Non-routine > Due
• meet yezi 14:00 6/15 🔒Done! 2025-06-15 🕸️ Track > Routine > Non-routine > Due
• 请糕吃烤鱼(变成了东湖面馆。。。） 🔒Done! 2025-06-15 🕸️ Track > Routine > Non-routine > No Due
• 数学hw3 6/15 🔒Done! 2025-06-16 🕸️ Track > Routine > Non-routine > Due
• AR论文图片 6/15 🔒Done! 2025-06-17 🕸️ Track > Routine > Non-routine > Due
• 羽毛球 12:00 6/17 🔒Done! 2025-06-18 🕸️ Track > Routine > Non-routine > Due
• meet gjy 17:00 6/17 🔒Done! 2025-06-18 🕸️ Track > Routine > Non-routine > Due
• meet byy 6/20 10:30 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• 糖糖ppt 6/20 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• 实验室请吃饭 6/20 12:00 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• 🏓 6/21 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• 拿糖糖快递 6/21 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• hdzx xpy周日招生没法来,咨询钱老师 6/28 🔒Done! 2025-06-21 🕸️ Track > Routine > Non-routine > Due
• 数学hw4 6/25 🔒Done! 2025-06-23 🕸️ Track > Routine > Non-routine > Due
• 糖糖一轮笔试 6/24 14:30 🔒Done! 2025-06-25 🕸️ Track > Routine > Non-routine > Due
• 数学mid code generation 6/20 🔒Done! 2025-06-25 🕸️ Track > Routine > Non-routine > Due
• hdzx 学生面试试题 6/22 🔒Done! 2025-06-25 🕸️ Track > Routine > Non-routine > Due
• LDD 零件数统计 6/25 🔒Done! 2025-06-26 🕸️ Track > Routine > Non-routine > Due
• 数学mid 6/25 🔒Done! 2025-06-26 🕸️ Track > Routine > Non-routine > Due
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• Seminar 14:00 6/26 🔒Done! 2025-06-26 🕸️ Track > Routine > Non-routine > Due
• meet byy 16:30 6/26 🔒Done! 2025-06-26 🕸️ Track > Routine > Non-routine > Due
• Seminar Robotics Task Planning 6/26 🔒Done! 2025-06-26 🕸️ Track > Routine > Non-routine > Due
• 服务器安装软件 6/27 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• 面试试题打印 6/27 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• 面试座位表格 6/29 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• hdzx招生 6/29 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• 金力quiz2 ctpp 6/29 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• 金力quiz2 6/30 🔒Done! 2025-06-30 🕸️ Track > Routine > Non-routine > Due
• 羽毛球 12:00 7/7 🔒Done! 2025-07-07 🕸️ 重要的日子 > Track > Non-routine > Due
• 计算机视觉report 7/6 🔒Done! 2025-07-07 🕸️ 重要的日子 > Track > Non-routine > Due
• 开TPLINK发票 🔒Done! 2025-07-12 🕸️ 重要的日子 > Track
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• 若来rolife 🔒Done! 2025-07-13 🕸️ 重要的日子 > Track > Non-routine > No Due
• 双卡服务器修理 🔒Done! 2025-07-14 🕸️ 重要的日子 > Track > Non-routine > Due
  • 下单固态硬盘
  • 拷贝硬盘
  • 装机
• 数学 hw5 7/14 🔒Done! 2025-07-14 🕸️ 重要的日子 > Track > Non-routine > Due
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• ipad(940829) neovim installation 🔒Done! 2025-07-18 🕸️ 重要的日子 > Track > Non-routine > No Due♥️
• apply JI VPN 🔒Done! 2025-07-21 🕸️ 重要的日子 > Track > Non-routine > Due
• MATH6003J HW6 7/21 🔒Done! 2025-07-21 🕸️ 重要的日子 > Track > Non-routine > Due
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• 强化学习考试 7/23 🔒Done! 2025-07-22 🕸️ 重要的日子 > Track > Non-routine > Due
  • ctpp
• 羽毛球 晚7~8 7/22 🔒Done! 2025-07-23 🕸️ 重要的日子 > Track > Non-routine > Due
• 糖糖裤子退换♥️ 🔒Done! 2025-07-24 🕸️ 重要的日子 > Track > Non-routine > Due
  • 申请
  • 退
  • 收

For the figure, the each step corresponds to a processed batch(16).

Resdcn + FPNx2 (↓:32,16,8,4)

18 layer

with warm-up and cosine schedule

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python train.py --log_name resdcn18_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch resdcn_18 \
                --lr 1e-5 \
                --batch_size 36 \
                --num_epochs 90 --num_workers 0 --log_interval 5

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python train.py --log_name resdcn50_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch resdcn_50 \
                --lr 1e-5 \
                --batch_size 24 \
                --num_epochs 90 --num_workers 0 --log_interval 5

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python train.py --log_name resdcn_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch resdcn_101 \
                --lr 5e-5 \
                --lr_step 40,70 \
                --batch_size 16 \
                --num_epochs 90 --num_workers 0 --log_interval 5

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python train.py --log_name resdcn_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch resdcn_101 \
                --lr 5e-5 \
                --lr_step 40,70 \
                --batch_size 16 \
                --num_epochs 90 --num_workers 0 --log_interval 5

HRNet (↓:4,4,4,4)

Use PSR dataset after remove bg samples (11729 samples)

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python train.py --log_name hrnet_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch hrnet \
                --lr 5e-5 \
                --lr_step 40,70 \
                --batch_size 16 \
                --num_epochs 90 --num_workers 0 --log_interval 5

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python train.py --log_name hg_all \
                --data_dir ~/cyl/Data/PSR_final \
                --arch hourglass_small \
                --lr 5e-5 \
                --lr_step 40,70 \
                --batch_size 16 \
                --num_epochs 90 --num_workers 0 --log_interval 5

Project Proposal: 2D Point Cloud Registration using the Iterative Closest Point (ICP) Algorithm

Course: VV570/MATH6003J: Introduction to Engineering Numerical Analysis 1

1. Introduction & Problem Statement

In robotic perception, sensors like LiDAR capture the environment as a “point cloud”—a set of data points in space. To enable tasks like mapping or localization, a robot must align multiple point clouds taken from different positions. This project will address this fundamental robotics problem by implementing the Iterative Closest Point (ICP) algorithm, a numerical optimization method used to find the optimal alignment between two point clouds2.

2. Objective

The primary objective of this project is to develop a working 2D implementation of the ICP algorithm. We will apply numerical analysis techniques to find the rigid transformation (rotation and translation) that minimizes the distance between two point clouds, demonstrating a practical application of concepts from the course to the field of robotics.

3. Methodology

Our approach will be to apply a computational tool to solve this problem3. We will use Python with the NumPy library for numerical computations and Matplotlib for visualization. The core steps of our implementation will be:

• Data Association: For each point in a source cloud, find the closest corresponding point in the target cloud.

• Numerical Optimization: Calculate the optimal rotation and translation that minimizes the mean squared error between the corresponding point pairs. This will be solved using Singular Value Decomposition (SVD).

• Iterative Refinement: Apply the calculated transformation to the source cloud and repeat the process until the alignment error converges below a set threshold.

We will start with the simplification of working in 2D and assume a reasonable initial alignment between the point clouds4.

4. Expected Outcomes & Justification of Results

By the end of this three-week project5, we will deliver:

• A Python implementation of the 2D ICP algorithm.

• Visualizations demonstrating the algorithm’s effectiveness by showing the point clouds before and after alignment.

• A plot of the alignment error over iterations, which will serve to justify our results by showing the convergence of the numerical method6.

Homework 5 Report

Zhichen Tang
124370910020

Exercise 1

In this first part, I looked at the logistic model for population growth. The main goal was to solve the equation $u’(t) = C u(t) (1 - u(t)/B)$ using a couple of different numerical methods. The parameters were set to $u_0=100$, $B=1000$, and $C=2/15$.

a)

I take the current value and add the slope at that point multiplied by the time step to get the next value.

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% Completed feuler.m
function [t, u, dt] = feuler( f, I, u0, N )
    dt = (I(2)-I(1))/N;
    t = linspace(I(1),I(2),N+1);
    u = zeros(1,N+1);
    u(1) = u0;
    for n = 1:N
        u(n+1) = u(n) + dt * f(t(n), u(n));
    end
end
````


### b)
First, I calculated a temporary "predicted" value (just like a standard Euler step). Then, I averaged the slope at the current point and the slope at the next point (using the predicted value) to get a more accurate step.

Matlab

% Completed heun.m
function [t, u, dt] = heun( f, I, u0, N )
dt = (I(2)-I(1))/N;
t = linspace(I(1),I(2),N+1);
u = zeros(1,N+1);
u(1) = u0;
for n = 1:N
u_predictor = u(n) + dt * f(t(n), u(n));
u(n+1) = u(n) + (dt/2) * ( f(t(n), u(n)) + f(t(n+1), u_predictor) );
end
end

<div class="post-content"><img src="/2025/07/14/[OBS]课程-Math6003 Hw5/ex1b.png" alt="" title=""></div>
### c) 

After implementing the methods, I ran them with a coarse time step (Deltat=5, corresponding to N=20) and plotted them against the exact solution.

It's pretty clear from the graph that the Heun method does a much better job. Its curve hugs the exact solution line much more tightly than the Forward Euler curve does. This makes sense because it's a higher-order method, so you'd expect it to be more accurate for the same step size.

<div class="post-content"><img src="/2025/07/14/[OBS]课程-Math6003 Hw5/ex1c.png" alt="" title=""></div>

### d)

Next, I repeated the comparison but with a much smaller time step (Deltat=0.05, or N=2000).


Absolutely. With the smaller time step, both methods improved dramatically. The plot shows that the lines for the Euler method, Heun's method, and the exact solution are basically right on top of each other. This really shows how decreasing the step size can significantly reduce the error in numerical solutions. However, the Heun method's absolute error is still lower than the Euler method.

<div class="post-content"><img src="/2025/07/14/[OBS]课程-Math6003 Hw5/ex1d.png" alt="" title=""></div>

---

## Exercise 2

### a)

To solve this, I first had to convert the single 2nd-order equation into a system of two 1st-order equations. I defined a new variable, V(t)=U′(t). This gives the first equation of the system.

Then, I rearranged the original RLC equation to isolate U′′(t):

$$ U''(t) = -\frac{R}{L}U'(t) - \frac{1}{LC}U(t) + \frac{f}{LC} $$

By substituting V for U′ and V′ for U′′, I got the second equation. This allowed me to write the whole system in matrix form, X′=AX+b:

$$\begin{pmatrix} V' \\ U' \end{pmatrix} = \begin{pmatrix} -R/L & -1/(LC) \\ 1 & 0 \end{pmatrix} \begin{pmatrix} V \\ U \end{pmatrix} + \begin{pmatrix} f/(LC) \\ 0 \end{pmatrix} $$
### b)
Stability Analysis of Forward Euler The Forward Euler method is only stable if $|1 + \Delta t \lambda_i| \le 1$ for all eigenvalues ($\lambda_i$) of the matrix A. Using the given component values (L=0.01, C=10, R=0.1), the matrix A becomes: $$ A = \begin{pmatrix} -10 & -10 \\ 1 & 0 \end{pmatrix} $$ I calculated the eigenvalues to be $\lambda = -5 \pm \sqrt{15}$. To ensure stability, the time step $\Delta t$ has to be less than a critical value determined by the eigenvalue with the largest magnitude. The calculation showed that the method is stable only if: $$ \Delta t \le \frac{2}{5 + \sqrt{15}} \approx 0.2254 \text{ s} $$
### c) 
Simulating with Forward Euler I ran simulations for three different time steps. The results were a perfect illustration of the stability condition. 
<div class="post-content"><img src="/2025/07/14/[OBS]课程-Math6003 Hw5/ex2c.png" alt="" title=""></div>
As you can see: - With **N=43** ($\Delta t \approx 0.233$), the time step was too large, and the solution completely blew up. - With **N=46** ($\Delta t \approx 0.217$), the time step was just inside the stable range, and the solution correctly showed a damped wave. - With **N=500** ($\Delta t = 0.02$), the solution was also stable and much smoother. 
### d)
Simulating with Backward Euler Finally, I repeated the same simulations using the Backward Euler method. 
<div class="post-content"><img src="/2025/07/14/[OBS]课程-Math6003 Hw5/ex2d.png" alt="" title=""></div>
The difference is night and day. The Backward Euler method was **stable for all three time steps**, even the large one that made the Forward Euler method fail. This shows that implicit methods like Backward Euler are much more robust for systems like this, as they don't have the same strict stability requirements on the time step. While the accuracy still gets better with smaller steps, you can trust it to give a stable answer no matter what.

使用ssh作为命令行远程工具，启动远程的tensorboardx并且在本地的浏览器中打开。

远程运行：

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tensorboard --logdir <path> --port 6006

本地运行：

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ssh -N -f -L localhost:16006:localhost:6006 bohan@10.11.16.146

Ex 1

a

The result of hw4_1_a.m:

Vandermonde matrices are known to be ill-conditioned, with the condition number growing exponentially with $n$.

d

This tridiagonal matrix is well-conditioned. Its condition number grows polynomially with $n$ (like $O(n^2)$). The error $\epsilon_n$ should grow much more slowly than in the Vandermonde case, exhibiting polynomial rather than exponential growth. The residual $r_n$ will be a better, though not perfect, indicator of the true error compared to the previous case.

Comments on the obtained results:

• $\textbf{Error Growth:}$ The relative error $\epsilon_n$ for the tridiagonal system grows polynomially with $n$, which is significantly slower than the exponential growth observed for the Vandermonde matrix. This will be visible as a straight line on the loglog plot.
• $\textbf{Conditioning:}$ The dramatic difference in error behavior is due to the condition number. The Vandermonde matrix is severely ill-conditioned, whereas the tridiagonal matrix is well-conditioned.
• $\textbf{Residual as an Error Indicator:}$ For the well-conditioned tridiagonal matrix, the normalized residual $r_n$ and the relative error $\epsilon_n$ have much closer values. The residual is a far more reliable indicator of the true error in this case than it was for the Vandermonde system.

Ex 2

a

The task is to complete the files $jacobi.m$ and $gauss_seidel.m$. These files implement iterative methods based on a matrix splitting $A = P - (P-A)$, where $P$ is the preconditioner. The iterative update can be expressed as $x^{(k+1)} = x^{(k)} + z^{(k)}$, where $z^{(k)}$ is the solution to $Pz^{(k)} = r^{(k)}$, and $r^{(k)} = b - Ax^{(k)}$ is the residual.

For the $\textbf{Jacobi method}$, the preconditioner P is the diagonal of A. For the $\textbf{Gauss-Seidel method}$, P is the lower triangular part of A. The missing lines implement the calculation of the residual, the update step, and the storing of the residual norm at each iteration.

jacobi.m:

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function [x, iter, res]= jacobi(A,b,x0,nmax,tol)
% JACOBI iterative method
%   [X, Niter, Res] = JACOBI(A,B,X0,NMAX,TOL) attempts to solve the system of linear equations A*X=B
%   for X using the Jacobi method.

% Jacobi preconditioner
P=diag(diag(A)); %

% residual normalization
r0=norm(b);
if r0==0, r0=1; end

% first iteration
x=x0;
r=b-A*x; % The residual b - A*x
res(1)=norm(r); % The norm of the initial residual
iter=0;

while res(end)/r0 > tol && iter < nmax
    z=P\r; % Solve Pz=r
    x=x+z; % Update the solution
    r=b-A*x; % Recompute the residual
    iter=iter+1;
    res(iter+1)=norm(r); % Store the norm of the new residual
end

return

gauss_seidel.m:

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function [x, iter, res]= gauss_seidel(A,b,x0,nmax,tol)
% GAUSS-SEIDEL iterative method
%   [X, Niter, Res] = GAUSS_SEIDEL(A,B,X0,NMAX,TOL) attempts to solve the system of linear equations A*X=B
%   for X using the Gauss-Seidel method.

% Gauss-Seidel preconditioner
P=tril(A); %

% residual normalization
r0=norm(b);
if r0==0, r0=1; end

% first iteration
x=x0;
r=b-A*x; % The residual b - A*x
res(1)=norm(r); % The norm of the initial residual
iter=0;

while res(end)/r0 > tol && iter < nmax
    z=P\r; % Solve Pz=r using forward substitution
    x=x+z; % Update the solution
    r=b-A*x; % Recompute the residual
    iter=iter+1;
    res(iter+1)=norm(r); % Store the norm of the new residual
end

return

b

I solve the system for $n=4$ with the tridiagonal matrix $A$ having $2.5$ on the diagonal and $-1$ on the off-diagonals. The right-hand side is $b=(1.5,0.5,0.5,1.5)^{\top}$. The initial guess is $x^{(0)}=0$, tolerance is $10^{-10}$, and max iterations is $10^8$. I compare the number of iterations and computing time.

The matrix $A$ is strictly diagonally dominant since $|a_{ii}| = 2.5 > \sum_{j \neq i} |a_{ij}|$ for all rows. This property guarantees that both methods will converge.

For this well-conditioned matrix (as shown in part e), the number of iterations grows very slowly with $n$. The condition number remains small and constant. The relative error and the normalized residual are very close in value, indicating that the residual is a good estimator of the error.

d

The analysis from (c) is repeated for the matrix $A$ with 2 on the diagonal and -1 on the off-diagonals , and $b=(1,0,…,0,1)^{\top}$. The exact solution is again $x=(1,…,1)^{\top}$. This matrix is only weakly diagonally dominant, so I expect slower convergence.

The number of iterations will grow much more rapidly with $n$. This is because, as shown in part (e), the condition number of this matrix grows quadratically with $n$, making the problem progressively harder to solve.

e

I use the given formula for the eigenvalues of a symmetric tridiagonal matrix $M$ with $a$ on the diagonal and $b$ on the off-diagonals:
$$ \lambda_{i}(M)=a+2b\cos\left(\frac{\pi i}{n+1}\right), \quad i=1,…,n $$
The condition number is $\kappa_2(A) = |\lambda_{max}|/|\lambda_{min}|$.

Case 1: Matrix from part (c)
Here, $a=2.5$ and $b=-1$. The eigenvalues are $\lambda_i = 2.5 - 2\cos(\frac{\pi i}{n+1})$.

• As $n\to\infty$, $\lambda_{max} \to 2.5 - 2(-1) = 4.5$.
• As $n\to\infty$, $\lambda_{min} \to 2.5 - 2(1) = 0.5$.

The condition number $\kappa(A)$ approaches a constant, $\kappa(A) \to 4.5 / 0.5 = 9$. This is consistent with the numerical results from part (c), explaining the fast and stable convergence.

Case 2: Matrix from part (d)
Here, $a=2$ and $b=-1$. The eigenvalues are $\lambda_i = 2 - 2\cos(\frac{\pi i}{n+1}) = 4\sin^2(\frac{\pi i}{2(n+1)})$.

• As $n\to\infty$, $\lambda_{max} \to 4\sin^2(\pi/2) = 4$.
• For large $n$, $\lambda_{min} = 4\sin^2(\frac{\pi}{2(n+1)}) \approx 4(\frac{\pi}{2(n+1)})^2 = \frac{\pi^2}{(n+1)^2}$.
  The condition number $\kappa(A) \approx \frac{4}{\pi^2/(n+1)^2} = O(n^2)$. This quadratic growth in the condition number is consistent with the numerical results from part (d), explaining the significant increase in iterations with $n$.

Exercise 1

Problem Analysis

The objective is to find the coefficients $(\alpha, \beta, \gamma)$ for the finite difference formula:
$$
D_{h}f(\overline{x})=\frac{\alpha f(\overline{x})+\beta f(\overline{x}-h)+\gamma f(\overline{x}-2h)}{h} \quad
$$
The analysis begins by substituting the Taylor series expansions for $f(\overline{x}-h)$ and $f(\overline{x}-2h)$ around the point $\overline{x}$ into the formula.

The expansions are:
$$
f(\overline{x}-h) = f(\overline{x}) - hf’(\overline{x}) + \frac{h^2}{2}f’’(\overline{x}) - \frac{h^3}{6}f’’’(\overline{x}) + O(h^4)
$$
$$
f(\overline{x}-2h) = f(\overline{x}) - 2hf’(\overline{x}) + 2h^2f’’(\overline{x}) - \frac{4h^3}{3}f’’’(\overline{x}) + O(h^4)
$$
Substituting these into the formula and grouping terms by derivatives of $f(\overline{x})$ results in:
$$
D_{h}f(\overline{x}) = \frac{\alpha + \beta + \gamma}{h} f(\overline{x}) + (-\beta - 2\gamma) f’(\overline{x}) + \left( \frac{\beta}{2} + 2\gamma \right)h f’’(\overline{x}) + O(h^2)
$$
To approximate $f’(\overline{x})$, the coefficient of $f’(\overline{x})$ must be $1$, and the coefficient of $f(\overline{x})$ must be $0$. This yields two necessary equations:

1. $\alpha + \beta + \gamma = 0$
2. $-\beta - 2\gamma = 1$

a) Solution for First-Order Accuracy

For the formula to be of order 1, the two fundamental equations must be satisfied.

From equation (2), $\beta$ can be expressed in terms of $\gamma$:
$$
\beta = -1 - 2\gamma
$$
Substituting this into equation (1):
$$
\alpha + (-1 - 2\gamma) + \gamma = 0 \implies \alpha = 1 + \gamma
$$
Therefore, the family of coefficients $(\alpha, \beta, \gamma)$ that provides at least first-order accuracy is given by:
$$
(\alpha, \beta, \gamma) = (1+\gamma, -1-2\gamma, \gamma), \quad \forall \gamma \in \mathbb{R}
$$

b) Solution for Second-Order Accuracy

To achieve second-order accuracy, the truncation error must be $O(h^2)$. This requires the coefficient of the $h$ term in the error expansion to also be zero.

$$\frac{\beta}{2} + 2\gamma = 0$$

This creates a system of three linear equations to be solved for the unique values of $(\alpha, \beta, \gamma)$:

1. $\alpha + \beta + \gamma = 0$
2. $-\beta - 2\gamma = 1$
3. $\frac{\beta}{2} + 2\gamma = 0$

From equation (3), we find that $\beta = -4\gamma$. Substituting this into equation (2):
$$
-(-4\gamma) - 2\gamma = 1 \implies 2\gamma = 1 \implies \gamma = \frac{1}{2}
$$
With the value for $\gamma$, $\beta$ can be found:
$$
\beta = -4 \left( \frac{1}{2} \right) = -2
$$
Finally, substituting $\beta$ and $\gamma$ into equation (1) yields $\alpha$:
$$
\alpha + (-2) + \frac{1}{2} = 0 \implies \alpha = \frac{3}{2}
$$
The unique values which give a formula of order 2 are:
$$
\left(\alpha, \beta, \gamma\right) = \left(\frac{3}{2}, -2, \frac{1}{2}\right)
$$

Exercise 2

a) Midpoint Formula Calculation

The first task is to compute the integral of the function $f(x)=e^{-x}sin(x)$ on the interval $[a,b]=[0,2]$. The computation uses the composite midpoint formula with $M=10$ sub-intervals.

The resulting numerical approximation of the integral is 0.468592. The exact integral is 0.466630. So the error is 0.001962.

b) Convergence of the Midpoint Formula

This section analyzes the convergence of the composite midpoint formula for the same function and interval. The number of sub-intervals is varied as $M=10^{1},10^{2},10^{3},…10^{5}$.

The absolute error, $|I(f)-Q_{h}^{pm}(f)|$, is computed for each corresponding step size, $h=(b-a)/M$. This error is the difference between the numerical result and the exact value of the integral, $I=(1-e^{-2}(sin(2)+cos(2)))/2$.

A graph of the error versus the step size $h$ is then created using a logarithmic scale for both axes. In such a plot, the slope of the resulting line corresponds to the order of convergence.

The plot displays a straight line with a negative slope. A linear fit to the logarithmic data reveals a slope of approximately 2. This visually confirms that the method exhibits 2nd-order convergence, which aligns with the theoretical error bounds for the composite midpoint rule.

c) Convergence of Trapezoidal and Simpson’s Formulas

The analysis is repeated using the composite trapezoidal formula and the composite Simpson’s formula. The errors for both methods, $|I(f)-Q_{h}^{trap}(f)|$ and $|I(f)-Q_{h}^{simp}(f)|$, are plotted on the same graph in logarithmic scale.

The resulting graph contains two distinct straight lines, one for each method.

• Trapezoidal Rule: The error line for the trapezoidal rule has a slope of approximately 2. This confirms its theoretical 2nd-order convergence.
• Simpson’s Rule: The error line for Simpson’s rule is significantly steeper, with a slope of approximately 4. This demonstrates its theoretical 4th-order convergence.

A comparison of the results shows that for any given step size $h$, the error from Simpson’s rule is substantially smaller than the error from the trapezoidal rule, highlighting the superior accuracy of the higher-order method for smooth functions.

d) Analysis for a Non-Smooth Function

Finally, the convergence analysis is repeated for all three methods on a new function, $f(x)=\sqrt{|x|^{3}}$, over the interval $[a,b]=[-2,2]$. The exact value for this integral is $I=\frac{16}{5}\sqrt{2}$.

Comments on the Obtained Results

The function $f(x)=\sqrt{|x|^{3}}$ is not sufficiently smooth at the point $x=0$ within the integration interval. While the function and its first derivative are continuous, its second derivative, $f’’(x) \propto |x|^{-1/2}$, is unbounded at $x=0$.

The theoretical error estimates that predict 2nd and 4th-order convergence for these methods are based on the assumption that the function’s higher-order derivatives are continuous and bounded. Since this condition is violated, a degradation in the observed convergence rates is expected.

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