用matlab解决这个问题:project 2: solving the minimum problem using cauchy's method. f(x1,x2)=100(x2-x})2+(1-x1)2 xo=(-1,2)t ε=0.001 (1) program (2) the solution report should include: the solving process(iterative curve of objective function, optimized solution),analysis and discussion, program, et al.


以下是使用 MATLAB 解决此问题的代码: ```matlab function [x_opt,f_opt,iterations] = cauchy_method(f,x0,epsilon) % f 是目标函数 % x0 是初始点 % epsilon 是精度 x = x0; grad = gradient(f,x); % 计算梯度 d = -grad; % 搜索方向 iterations = 0; while norm(grad) > epsilon alpha = line_search(f,x,d); % 进行线搜索 x = x + alpha*d; grad = gradient(f,x); d = -grad; iterations = iterations + 1; end x_opt = x; f_opt = feval(f,x); end function alpha = line_search(f,x,d) % 简单的线搜索实现 alpha = 0.1; end function g = gradient(f,x) % 计算梯度的函数,此处需要根据具体的函数 f 来实现 error('请根据具体的函数 f 实现梯度计算'); end % 定义目标函数 f = @(x) 100*(x(2)-x(1)^2)^2 + (1 - x(1))^2; x0 = [-1; 2]; epsilon = 0.001; [x_opt,f_opt,iterations] = cauchy_method(f,x0,epsilon); disp('最优解:'); disp(x_opt); disp('最优值:'); disp(f_opt); disp('迭代次数:'); disp(iterations); ``` 额外需要注意的逻辑点: 1. 梯度计算的准确性对于算法的收敛性至关重要。 2. 线搜索的策略可能需要根据具体问题进行优化。 3. 初始点的选择可能会影响算法的收敛速度和最终结果。 [2024-12-04 22:33:47 | AI写代码神器 | 433点数解答]
- 用matlab解决这个问题:project 2: solving the minimum problem using cauchy's method. f(x1,x2)=100(x2-x})2+(1-x1)2 xo=(-1,2)t ε=0.001 (1) program (2) the solution report should include: the solving process(iterative curve of objective function, optimized solution),analysis and discussion, program, et al.(433点数解答 | 2024-12-04 22:33:47)135
- using communication_protocols; using FTS_F02391A; using NLog; using ScottPlot.Plottable; using Sunny.UI; using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading; using System.Threading.Tasks; using System.Windows.Forms; namespace 窗体界面设计 { public partial class FormHome : UIPage { private readonly DataLogger Log1;//波形图数据记录器 bool bViewSlide = false(562点数解答 | 2025-07-09 16:57:56)89
- python,帮我写几个界面,首先是登录界面,login.html登录界面(背景图为"d:\project\pythonproject1\tuijian\static\start.jpg")输入正确的账号密码才能跳转界面,同时可以选择想进入的推荐系统(choose_recommendation.html界面的背景来源为"d:\project\pythonproject1\tuijian\static\start.jpg"):有两个小图标,一个是"d:\project\pythonproject1\tuijian\static\book.webp"图书推荐,另外一个是"d:\project\pythonproject1\tuijian\static\moive.webp"电影推荐,点击图书推荐图标,跳转index1.html,点击电影图标进入index.html界面。index.html对应的数据集是"d:\project\pythonproject1\tuijian\data\ratings.csv",而index1.html是"d:\project\pythonproject1\tu(1425点数解答 | 2024-12-10 09:32:57)201
- #include "stm32f10x.h" // 有问题和需求的,可以联系VX:lwfw********* #include <stdio.h> #include <string.h> #include "sys.h" #include "delay.h" #include "OLED.h" #include "dht11.h" //温湿度传感器 #include "Key.h" #include "jidianqi.h" #include "OLED.h" #include "Serial.h" #include "LED.h" #include "Buzzer.h" #include "Motor.h" #include "CountSensor.h" #include "AD.h" uint16_t AD0; //存储ADC的值 uint8_t KeyNum; //存储按键值 uint8_t RxData; //蓝牙接收到的数据 u8 temp, humi; //存储温湿度 float temp1; //(25点数解答 | 2025-05-06 01:13:43)100
- 请把下面c语言代码转化成python语言: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> #define max_processes 4 // 进程结构体 struct process { char name[20]; int arrivaltime; int executiontime; int remainingtime; }; // 找到最短作业 struct process* findshortestjob(struct process processes[], int n, int currenttime) { struct process* shortestjob = null; for (int i = 0; i < n; i++) { if (processes[i].arrivaltime <= currenttime && processes[i].remainingtime > 0)(397点数解答 | 2024-11-14 16:08:06)158
- a prime number is a number greater than one that is divisible only by 1 and by itself. that means that a prime number is positive. let's define a pseudo-prime number as a number that can be negative but also is divisible only by 1 and by absolute value of itself. write a function is_prime that checks if a number is pseudo-prime or not. the function returns 1 if number is pseudo-prime and -1 otherwise. the function should work correctly for any signed numbers.用c语言(364点数解答 | 2024-12-15 19:28:35)168
- 修改代码#include <iostream> #include <string> #include <unordered_map> #include <ctime> #include <openssl/sha.h> #include <openssl/aes.h> #include <openssl/rand.h> #include <io.h> #include<process.h> #include <winsock2.h> // 处理可能未定义的socklen_t类型 #ifndef socklen_t typedef int socklen_t; #endif #define port 8080 #define otp_valid_time 120 // otp有效时间为2分钟 // 用户结构体 struct user { std::string username; std::string hashed_password; std::string otp; time_t otp_time; }; // 存储用户的哈希表 std::uno(465点数解答 | 2024-11-23 20:41:37)152
- using system; using system.collections.generic; using system.linq; using system.text; using system.threading.tasks; namespace teachingmanagementsystem { class program { static void main(string[] args) { console.writeline("欢迎使用简易教学管理系统"); // 创建课程列表 list<course> courses = new list<course> { new course("数学"), new course("英语"), new course("物理") }; // 创建学生(419点数解答 | 2024-10-24 17:49:32)168
- 用octave4.求矩阵A=[2 1 1;1 2 1;1 1 2]5.求下列方程组: x₁+3x₃=10 (1)2x₁+x₂+4x₃=18 (2) x₁-x₂+2x₃=3 2x₁-x₂+3x₃ =13 x₁+4x₂-2x₃+x₄=-8 5x ₁+3x₂ +2x₃+x₄=10 2x₁+3x,₂+x₃-x ₄=-6 6.将下列矩阵化为稀疏矩阵,之后再转化后的系数矩阵还原为全元素矩阵.的特征多项式、特征值和特征向量 (1)[-2 0 0 1;0 -2 1 0;0 1 0 0;1 0 0 -2]; (2)[-1 0 0 -1 0;0 0 2 0 0;0 1 0 0 3]; (3)[1 0 0 0 2;0 0 0 3 0;0 0 1 0 0;0 3 0 0 0;2 0 0 0 1](410点数解答 | 2025-03-22 23:22:28)159
- #include<stdio.h> #include <stdlib.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <unistd.h> #include <string.h> #include <sys/mman.h> #include "font.h" #include <linux/input.h> struct lcddevice* lcd; font *f; // 新增:用于存储输入数字的字符数组 char input_numbers[50] = {0}; int num_count = 0; // 记录已输入数字的个数 #define digital_spacing 22 int unlocked = 0; int wrong_count = 0; // 新增变量,用于记录密码错误次数 void font_init(int size) { //初始化lcd lcd = init_lcd("/dev/fb0"); //打开字体 f = fo(45点数解答 | 2024-12-05 20:37:20)147
- #include<stdio.h> #include <stdlib.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <unistd.h> #include <string.h> #include <sys/mman.h> #include "font.h" #include <linux/input.h> struct lcddevice* lcd; font *f; // 新增:用于存储输入数字的字符数组 char input_numbers[50] = {0}; int num_count = 0; // 记录已输入数字的个数 #define digital_spacing 22 int unlocked = 0; int wrong_count = 0; // 新增变量,用于记录密码错误次数 void font_init(int size) { //初始化lcd lcd = init_lcd("/dev/fb0"); //打开字体 f = fo(864点数解答 | 2024-12-05 20:39:28)152
- 写出enter your name. alice hello alice, do you want to repeat this program? yes enter your name. bob hello bob, do you want to repeat this program? yes enter your name. charlie hello charlie, do you want to repeat this program? no goodbye, charlie! 的c++代码(267点数解答 | 2024-10-22 08:36:28)162