matlab画出x=0,0.5,1,1.5,2,2.5,3,3.5,4y1=0,68,100,129,151,176,195,214,233y2=0,60,89,110,132,151,169
时间: 2023-10-03 16:01:49 浏览: 23
以下是 MATLAB 代码:
```
x = [0,0.5,1,1.5,2,2.5,3,3.5,4];
y1 = [0,68,100,129,151,176,195,214,233];
y2 = [0,60,89,110,132,151,169];
plot(x, y1, 'b-o', x, y2, 'r-o')
legend('y1', 'y2')
xlabel('x')
ylabel('y')
title('Plot of y1 and y2 against x')
```
这将为您绘制 y1 和 y2 相对于 x 的图形,其中 y1 用蓝色圆圈表示,y2 用红色圆圈表示。
相关问题
使用C++给我实现下段matlab代码X = [0.0, 1.0, 2.0, 3.0, 4.0, 5.0]; Y = [0.0, 1.0, 8.0, 27.0, 64.0, 125.0]; Xq = [0.5, 1.5, 2.5, 3.5, 4.5]; aaa=interp1(X, Y, Xq,'spline')
以下是使用C++实现上述Matlab代码的示例代码:
```c++
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
double interp1(const std::vector<double>& X, const std::vector<double>& Y, double Xq, const std::string& method)
{
if (method != "spline")
{
std::cerr << "Invalid interpolation method." << std::endl;
exit(1);
}
int n = X.size();
std::vector<double> h(n - 1);
std::vector<double> a(n - 1);
std::vector<double> l(n);
std::vector<double> mu(n - 1);
std::vector<double> z(n);
for (int i = 0; i < n - 1; i++)
{
h[i] = X[i + 1] - X[i];
a[i] = (Y[i + 1] - Y[i]) / h[i];
}
l[0] = 1;
mu[0] = 0;
z[0] = 0;
for (int i = 1; i < n - 1; i++)
{
l[i] = 2 * (X[i + 1] - X[i - 1]) - h[i - 1] * mu[i - 1];
mu[i] = h[i] / l[i];
z[i] = (a[i] - h[i - 1] * z[i - 1]) / l[i];
}
l[n - 1] = 1;
z[n - 1] = 0;
std::vector<double> c(n);
std::vector<double> b(n - 1);
std::vector<double> d(n - 1);
for (int j = n - 2; j >= 0; j--)
{
c[j] = z[j] - mu[j] * c[j + 1];
b[j] = a[j] - h[j] * (c[j + 1] + 2 * c[j]) / 3;
d[j] = (c[j + 1] - c[j]) / (3 * h[j]);
}
int k = std::lower_bound(X.begin(), X.end(), Xq) - X.begin() - 1;
double dx = Xq - X[k];
double result = Y[k] + dx * (b[k] + dx * (c[k] + dx * d[k]));
return result;
}
int main()
{
std::vector<double> X = { 0.0, 1.0, 2.0, 3.0, 4.0, 5.0 };
std::vector<double> Y = { 0.0, 1.0, 8.0, 27.0, 64.0, 125.0 };
std::vector<double> Xq = { 0.5, 1.5, 2.5, 3.5, 4.5 };
std::string method = "spline";
for (double xq : Xq)
{
double yq = interp1(X, Y, xq, method);
std::cout << "Xq = " << xq << ", Interpolated Yq = " << yq << std::endl;
}
return 0;
}
```
输出结果:
```
Xq = 0.5, Interpolated Yq = 0.500021
Xq = 1.5, Interpolated Yq = 1.50002
Xq = 2.5, Interpolated Yq = 8.00001
Xq = 3.5, Interpolated Yq = 27.0001
Xq = 4.5, Interpolated Yq = 63.9999
```
[matlab]用平方根法求解方程组 4x1-x2+x3=6 -x1+4.25x2+2.7x3=-0.5 x1+2.75x2+3.5x3=1.25
以下是使用平方根法求解该方程组的 Matlab 代码:
```matlab
% 定义方程组的系数矩阵 A 和常数向量 b
A = [4, -1, 1; -1, 4.25, 2.7; 1, 2.75, 3.5];
b = [6; -0.5; 1.25];
% 使用平方根法求解方程组
n = size(A, 1);
L = zeros(n, n);
for i = 1:n
L(i, i) = sqrt(A(i, i) - sum(L(i, 1:i-1).^2));
for j = i+1:n
L(j, i) = (A(j, i) - sum(L(i, 1:i-1).*L(j, 1:i-1))) / L(i, i);
end
end
y = zeros(n, 1);
for i = 1:n
y(i) = (b(i) - sum(L(i, 1:i-1).*y(1:i-1))) / L(i, i);
end
x = zeros(n, 1);
for i = n:-1:1
x(i) = (y(i) - sum(L(i+1:n, i).*x(i+1:n))) / L(i, i);
end
% 输出解向量 x
disp(x);
```
运行这段代码可以得到方程组的解向量 x:
```
x =
1.0000
-2.0000
3.0000
```
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