如何在matlab上用二分法求解x^3+2*x-1=0在[0,2]上的解,要完整的代码
时间: 2024-02-06 22:12:05 浏览: 17
完整代码如下:
```
f = @(x) x^3 + 2*x - 1;
a = 0;
b = 2;
tolerance = 1e-6;
while (b - a) > tolerance
c = (a + b) / 2;
if f(c) == 0
break;
elseif f(a) * f(c) < 0
b = c;
else
a = c;
end
end
x = (a + b) / 2;
```
其中,"f" 是方程的函数句柄,"a" 和 "b" 是区间的上下界,"tolerance" 是所需精度。在 while 循环中,不断将区间一分为二,直到满足所需的精度为止。最后,"x" 变量将包含方程的解。
相关问题
如何在matlab上用二分法求解x^3+2*x-1=0在[0,2]上的解
可以使用 MATLAB 中的 "fzero" 函数来使用二分法求解非线性方程的根。该函数需要一个函数句柄作为输入,表示要求解的方程,以及一个表示根的初始猜测值。以下是在 [0,2] 上求解方程 x^3+2*x-1=0 的代码:
```
f = @(x) x^3 + 2*x - 1;
x = fzero(f, [0, 2]);
```
其中,"f" 是方程的函数句柄,"[0, 2]" 是根的初始猜测值的区间。执行该代码后,变量 "x" 将包含方程的解。
matlab用二分法求解x^2+8*x+3=0,二分法进行编码
以下是使用 MATLAB 实现二分法求解 x^2 + 8*x + 3 = 0 的代码:
function [root, iter] = bisection(f, a, b, tol, maxiter)
% f: the function to be solved
% a, b: the interval to search for root
% tol: the tolerance for the root
% maxiter: the maximum number of iterations allowed
% check if the function has different signs at the endpoints
if f(a) * f(b) >= 0
error('The function has the same sign at the endpoints.');
end
% initialize the variables
iter = 0;
root = (a + b) / 2;
% iterate until the root is found or the maximum number of iterations is reached
while abs(f(root)) > tol && iter < maxiter
if f(a) * f(root) < 0
b = root;
else
a = root;
end
root = (a + b) / 2;
iter = iter + 1;
end
% check if the maximum number of iterations is reached
if iter == maxiter
warning('The maximum number of iterations is reached.');
end
end
% call the function with the given parameters
f = @(x) x^2 + 8*x + 3;
a = -10;
b = 10;
tol = 1e-6;
maxiter = 100;
[root, iter] = bisection(f, a, b, tol, maxiter);
% print the result
fprintf('The root is %f, found after %d iterations.\n', root, iter);