Use the method of successive approximation to solve dy/ dx = x+y+1, y(0) = 0. (Hint: You may want to construct the Picard’s sequence of functions y0(x), y1(x), . . . first.)
时间: 2023-06-03 21:04:32 浏览: 80
根据您的要求,我可以回答这个问题。首先,我们可以构造Picard序列求解该微分方程。令y0(x) = 0,则y1(x) = ∫(x,0) x t y0(t) dt = 0。接着,y2(x) = ∫(x, 0) x t y1(t) dt = (x^3)/2。同样地,我们可以继续构造后续的函数来逼近解。即使我们无法得到精确解,利用Picard序列的方法也可以得到足够精度的近似解。
相关问题
给出successive convex approximation的完整的CVX代码,目标函数为x^2+y^2-3xy
以下是使用successive convex approximation(SCA)方法来解决目标函数为x^2+y^2-3xy的完整CVX代码:
```matlab
cvx_begin
variable x
variable y
minimize (x^2 + y^2 - 3*x*y)
subject to
x + y == 1
cvx_end
for i = 1:10
cvx_begin quiet
variable x1
variable y1
minimize (x1^2 + y1^2 - 3*x*y - (grad([x;y])'*[x1;y1] + 1/2*norm([x1;y1]-[x;y],2)^2))
subject to
x1 + y1 == 1
cvx_end
x = x1;
y = y1;
end
% 输出结果
fprintf('最小值为%.4f\n', cvx_optval)
fprintf('x的值为%.4f\n', x)
fprintf('y的值为%.4f\n', y)
```
在这个代码中,我们首先定义了变量x和y,然后使用CVX来最小化目标函数x^2+y^2-3xy,同时满足约束条件x+y=1。然后,我们使用SCA方法来迭代地优化目标函数。在每个迭代步骤中,我们将先前的解作为初始点,然后使用CVX来求解一个局部凸近似问题,其中包括对原始目标函数的一阶泰勒展开式。在每次迭代之后,我们将当前解用作下一次迭代的初始点。最后,我们输出最小值以及x和y的值。
需要注意的是,SCA方法只能保证最终解是局部最小值,并且需要手动设置迭代次数。如果目标函数具有多个局部最小值,则SCA方法可能无法找到全局最小值。
write a C example use Newton-Raphson to get Fractrals for f(z)=z^3-1=0 and save the result into "Problem1.txt"
Sure, here's a C example that uses Newton-Raphson method to get fractals for f(z) = z^3 - 1 = 0:
```c
#include <stdio.h>
#include <math.h>
#define MAX_ITER 1000
#define EPSILON 1e-6
typedef struct {
double x;
double y;
} Complex;
void newton_raphson(Complex z, Complex *z_new) {
Complex numerator = {pow(z.x, 3) + 3 * z.x * pow(z.y, 2) - 1, 3 * pow(z.x, 2) * z.y - pow(z.y, 3)};
Complex denominator = {3 * pow(z.x, 2) + 3 * pow(z.y, 2), 6 * z.x * z.y};
z_new->x = z.x - numerator.x / denominator.x;
z_new->y = z.y - numerator.y / denominator.y;
}
int main() {
FILE *fp;
fp = fopen("Problem1.txt", "w");
if (fp == NULL) {
printf("Error opening file\n");
return 1;
}
for (double x = -2; x <= 2; x += 0.01) {
for (double y = -2; y <= 2; y += 0.01) {
Complex z = {x, y};
Complex z_new = {0, 0};
int iter = 0;
while (iter < MAX_ITER) {
newton_raphson(z, &z_new);
double distance = sqrt(pow(z_new.x - z.x, 2) + pow(z_new.y - z.y, 2));
if (distance < EPSILON) {
fprintf(fp, "%.2lf %.2lf\n", x, y);
break;
}
z = z_new;
iter++;
}
}
}
fclose(fp);
return 0;
}
```
This code performs a double loop over the region from -2 to 2 in both the x and y dimensions with a step size of 0.01. For each point (x, y), the code applies the Newton-Raphson method to find the root of the equation f(z) = z^3 - 1 starting from the initial point z = (x, y). The method is iterated until either the maximum number of iterations is reached (MAX_ITER) or the distance between successive iterations is below a small tolerance (EPSILON).
If the root is found, the point (x, y) is considered to be part of the fractal and is written to the file "Problem1.txt" in the format "x y".
Note that this code makes use of complex arithmetic and is fairly computationally intensive, so it may take a while to run depending on your machine.