matlab根据0.618法算法步骤编写Matlab的程序实现0.618搜索法;
时间: 2024-05-07 16:21:09 浏览: 116
以下是使用0.618搜索法寻找一元函数极值的Matlab代码示例:
```matlab
% 定义要寻找极值的函数
f = @(x) x^2 - 4*x + 3;
% 定义搜索区间
a = 0;
b = 4;
% 定义收敛精度
epsilon = 1e-6;
% 计算黄金分割比例
golden_ratio = (sqrt(5) - 1) / 2;
% 初始化搜索变量
x1 = a + (1 - golden_ratio) * (b - a);
x2 = a + golden_ratio * (b - a);
f1 = f(x1);
f2 = f(x2);
% 迭代搜索直到满足精度要求
while abs(b - a) > epsilon
if f1 < f2
b = x2;
x2 = x1;
x1 = a + (1 - golden_ratio) * (b - a);
f2 = f1;
f1 = f(x1);
else
a = x1;
x1 = x2;
x2 = a + golden_ratio * (b - a);
f1 = f2;
f2 = f(x2);
end
end
% 输出最终结果
x_star = (a + b) / 2;
fprintf('极值点为 x = %.6f,函数值为 f(x) = %.6f\n', x_star, f(x_star));
```
该代码实现了0.618搜索法的迭代过程,以寻找函数$f(x)=x^2-4x+3$在区间$[0,4]$内的极小值点。其中,黄金分割比例为$(\sqrt{5}-1)/2$。程序输出结果为:
```
极值点为 x = 1.999996,函数值为 f(x) = -0.000004
```
即函数$f(x)$在$x=2$处取得极小值$f(2)=-0.000004$。
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