e^(x) matlab的正割法

Question

我正在尝试编写 f(x)=e^(-x)-sin(x) 的割线法以找到最小的正根。我的代码似乎出错了。

%// Initial values and tolerance
x(0) = 2;
x(1) = 10;
f = @(x) exp(-x)-sin(x);

error = 0.001;

%// Different iterations
for k=0:100

    x(k+1) = x(k) - (f(x(k)))*((x(k) - x(k-1))/(f(x(k)) - f(x(k-1))));

    if abs(x(k)-x(k-1)) < error
        return;
    end
end

Answer 1

为什么会出错？

正如您的问题的 cmets 中所述，vector(1D matrix) indices in matlab run from 1 to end，因此在您的情况下，您尝试访问代码中各个位置的无效索引：

%// Initial values and tolerance
x(0) = 2;                         %// <-- invalid index
x(1) = 10;
f = @(x) exp(-x)-sin(x);

error = 0.001;

%// Different iterations
for k=0:100
                     %// for k=0: x(k) and x(k-1) both invalid indices
                     %// for k=1: x(k-1) invalid index
    x(k+1) = x(k) - (f(x(k)))*((x(k) - x(k-1))/(f(x(k)) - f(x(k-1))));

    if abs(x(k)-x(k-1)) < error
        return;
    end
end

更正的代码

我们更正了上述无效索引案例（还将您的 for 循环更改为 while 循环）：

%// Function to (attempt) to find a root to
f = @(x) exp(-x)-sin(x);

%// Initial values and tolerance
x(1) = 2;
x(2) = 10;
error = 0.001;

%// Different iterations
k=2;
maxIt = 100;
while (abs(x(k)-x(k-1)) > error) && (k-1 <= maxIt)
    x(k+1) = x(k) - f(x(k))*(x(k) - x(k-1))/(f(x(k)) - f(x(k-1)));
    k = k + 1;
end

%// print result
disp(['x(end) = ' num2str(x(end)) ...
  ', f(x(end)) = ' num2str(f(x(end))) ...
  ', iterations: ' num2str(k-2)]);

产生结果

x(end) = 6.285, f(x(end)) = -4.9315e-10, iterations: 7