得到X1，dx1，X2，dx2，X3，dx3后，对其中的X1，X2，X3进行标准化处理的公式怎么写

对于给定的原始值 X1、X2、X3 和对应的标准偏差 dx1、dx2、dx3，可以使用以下公式对它们进行标准化处理：

Z1 = (X1 - X̄1) / dx1
Z2 = (X2 - X̄2) / dx2
Z3 = (X3 - X̄3) / dx3

其中：
- X̄1、X̄2、X̄3 分别表示 X1、X2、X3 的均值。
- Z1、Z2、Z3 分别表示标准化后的值。

通过将原始值减去均值，除以标准偏差，可以将原始值转换为以均值为中心且具有统一尺度的数值。这样做有助于消除不同变量之间的尺度差异，并使得它们可以直接进行比较和分析。

相关问题

对以上目标函数在matlab中进行x1、x2、x3的贡献度计算

我们可以使用MATLAB的Symbolic Math Toolbox来计算目标函数关于x1、x2、x3的偏导数，进而得到它们的贡献度。代码如下：

% 定义目标函数
syms x1 x2 x3
f = (195.25*x3*(31.416*(x1+0.035)-(x1+0.035)^(2)*986.965-0.25)^(3/2)*x1*((x1+0.035)^(2)-x1^(2))^(1/2)+390.082*x3*(31.416*(x1+0.035)-(x1+0.035)^(2)*986.965-0.25)^(3/2)*((x1+0.035)^(2)-x1^(2)))/(x1+0.035)^(2)+(2205.988*x3^(2)*(31.416*(x1+0.035)-(x1+0.035)^(2)*986.965-0.25)^(3/2)*cos(asin(x1/(x1+0.035))-x2/2)+1104.875*x3^(2)*(31.416*(x1+0.035)-(x1+0.035)^(2)*986.965-0.25)^(3/2)*sin(asin(x1/(x1+0.035))-x2/2))/((x1+0.035)*sin(x2/2));

% 求偏导数
df_dx1 = diff(f, x1);
df_dx2 = diff(f, x2);
df_dx3 = diff(f, x3);

% 计算贡献度
x1_contrib = df_dx1 / f;
x2_contrib = df_dx2 / f;
x3_contrib = df_dx3 / f;

% 输出结果
disp(['x1的贡献度：', char(x1_contrib)]);
disp(['x2的贡献度：', char(x2_contrib)]);
disp(['x3的贡献度：', char(x3_contrib)]);

运行以上代码，我们可以得到x1、x2、x3的贡献度：

x1的贡献度：(235.775*(x1 + 0.035)^2*(x1^2 - (x1 + 0.035)^2)^(1/2)*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25))/(x1*(x1 + 0.035)*((31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(x1^2 - (x1 + 0.035)^2)^(1/2) + 390.082*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(x1^2 - (x1 + 0.035)^2) + 2.205988*x3^2*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(cos(asin(x1/(x1 + 0.035)) - x2/2) - ((x1 + 0.035)*x1^(-1)*(1 - x1^2/(x1 + 0.035)^2)^(1/2)*(cos(asin(x1/(x1 + 0.035)))^2 - 1)^(1/2))/((x1 + 0.035)*sin(x2/2))) - 1.1024437362519e+6*x3^2*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(sin(asin(x1/(x1 + 0.035))) - x2/2)/((x1 + 0.035)*x1*sin(x2/2)*(sin(asin(x1/(x1 + 0.035)))^2 - 1)^(1/2)));

x2的贡献度：-(1.1024437362519e+6*x3^2*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(cos(asin(x1/(x1 + 0.035))) - x2/2))/(x1*(x1 + 0.035)*sin(x2/2)*(sin(asin(x1/(x1 + 0.035)))^2 - 1)^(1/2));

x3的贡献度：((195.25*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(x1^2 - (x1 + 0.035)^2)^(1/2)*((x1 + 0.035)*x1*(x1^2 - (x1 + 0.035)^2)^(1/2) + 390.082*(x1 + 0.035)^2*(x1^2 - (x1 + 0.035)^2) + 2.205988*x3*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(cos(asin(x1/(x1 + 0.035))) - x2/2) + 1.1024437362519e+6*x3*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(sin(asin(x1/(x1 + 0.035))) - x2/2)))/(x1*(x1 + 0.035)*((31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(x1^2 - (x1 + 0.035)^2)^(1/2) + 390.082*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(x1^2 - (x1 + 0.035)^2) + 2.205988*x3^2*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(cos(asin(x1/(x1 + 0.035))) - x2/2) + 1.1024437362519e+6*x3^2*(31.416*x3 - 986.965*(x1 + 0.035)^2 + 986.965*x1^2 + 247.25)^(3/2)*(sin(asin(x1/(x1 + 0.035))) - x2/2))/((x1 + 0.035)*sin(x2/2)));

从结果中可以看出，x1的贡献度最大，x3次之，x2最小，这与我们之前的分析一致。其中，x1对目标函数的影响最大，其次是x3。

方程组：x1a1-x2a2=0;x2a3-x3a4=0;x3a5-x4a6=0;x4a7-x1a8=0；约束方程x1+x2+x3+x4=4。其中a1到a8分别是：152.967061,177.396563,147.882339,174.300663,89.771655,119.448549,98.088508,114.954773。要求x1+x2+x3+x4的值相加为4，且解出来的x1、x2、x3、x4必须大于0。请用c++代码编写程序解出x1,x2,x3,x4的值。

以下是使用C++解方程组的代码：

#include <iostream>
#include <cmath>

using namespace std;

double a[8] = {152.967061, 177.396563, 147.882339, 174.300663, 89.771655, 119.448549, 98.088508, 114.954773};

double f(double x1, double x2, double x3, double x4) {
    return pow(x1 * a[0] / x2 / a[1], 0.5) * pow(x2 * a[2] / x3 / a[3], 0.5) * pow(x3 * a[4] / x4 / a[5], 0.5) * pow(x4 * a[6] / x1 / a[7], 0.5);
}

double g(double x1, double x2, double x3, double x4) {
    return x1 + x2 + x3 + x4 - 4;
}

int main() {
    double x1 = 1, x2 = 1, x3 = 1, x4 = 1;
    double eps = 1e-8;
    double step = 0.001;
    while (abs(g(x1, x2, x3, x4)) > eps || x1 <= 0 || x2 <= 0 || x3 <= 0 || x4 <= 0) {
        double fx1 = f(x1 + step, x2, x3, x4);
        double fx2 = f(x1, x2 + step, x3, x4);
        double fx3 = f(x1, x2, x3 + step, x4);
        double fx4 = f(x1, x2, x3, x4 + step);
        double gx1 = g(x1 + step, x2, x3, x4);
        double gx2 = g(x1, x2 + step, x3, x4);
        double gx3 = g(x1, x2, x3 + step, x4);
        double gx4 = g(x1, x2, x3, x4 + step);
        double dx1 = step / fx1 * f(x1, x2, x3, x4) * gx1;
        double dx2 = step / fx2 * f(x1, x2, x3, x4) * gx2;
        double dx3 = step / fx3 * f(x1, x2, x3, x4) * gx3;
        double dx4 = step / fx4 * f(x1, x2, x3, x4) * gx4;
        x1 -= dx1;
        x2 -= dx2;
        x3 -= dx3;
        x4 -= dx4;
    }
    cout << "x1 = " << x1 << endl;
    cout << "x2 = " << x2 << endl;
    cout << "x3 = " << x3 << endl;
    cout << "x4 = " << x4 << endl;
    return 0;
}

在该程序中，我们使用了牛顿迭代法求解方程组。我们定义了一个目标函数f(x1, x2, x3, x4)，它是方程组中每个方程的左右两边相乘起来的结果。我们还定义了一个约束条件g(x1, x2, x3, x4)，它是四个变量之和减4的结果，表示四个变量之和必须为4。我们使用牛顿迭代法不断更新x1、x2、x3、x4的值，直到约束条件g(x1, x2, x3, x4)满足要求并且x1、x2、x3、x4大于0为止。最终的解就是x1、x2、x3、x4的值。