matlab二分法代码

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侨光辉0a1

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摘要您好，亲

我是生活导师小强，非常高兴能够为您解答。以下是 MATLAB 中二分法的示例代码：

function [root,iter] = bisection(f,a,b,tol,maxiter)

% f: the function to be solved

% a, b: the initial interval [a,b]

% tol: the tolerance

% maxiter: the maximum number of iterations allowed

% Initialization

iter = 0;

fa = feval(f,a);

% Check if the initial interval is valid

if sign(feval(f,b)) == sign(fa)

error('f(a) and f(b) must have opposite signs')

end

% Loop until convergence or the maximum number of iterations is reached

while (b-a)/2 > tol && iter < maxiter

% Compute the midpoint and evaluate the function at the midpoint

c = (a + b)/2;

fc = feval(f,c);

% Update the interval [a,b]

if fc == 0

% The root has been found

root = c;

return

elseif sign(fc) == sign(fa)

a = c;

fa = fc;

else

b = c;

end

% Update the iteration counter

iter = iter + 1;

end

% Output the root and the number of iterations

root = (a + b)/2;

end

咨询记录 · 回答于2024-01-08

matlab二分法代码

您好，亲 我是生活导师小强，非常高兴能够为您解答。以下是 MATLAB 中二分法的示例代码： function [root,iter] = bisection(f,a,b,tol,maxiter) % f: the function to be solved % a, b: the initial interval [a,b] % tol: the tolerance % maxiter: the maximum number of iterations allowed % Initialization iter = 0; fa = feval(f,a); % Check if the initial interval is valid if sign(feval(f,b)) == sign(fa) error('f(a) and f(b) must have opposite signs') end % Loop until convergence or the maximum number of iterations is reached while (b-a)/2 > tol && iter < maxiter % Compute the midpoint and evaluate the function at the midpoint c = (a + b)/2; fc = feval(f,c); % Update the interval [a,b] if fc == 0 % The root has been found root = c; return elseif sign(fc) == sign(fa) a = c; fa = fc; else b = c; end % Update the iteration counter iter = iter + 1; end % Output the root and the number of iterations root = (a + b)/2; end

这是一个函数，需要传递四个参数： - 待解的函数 f - 初始区间 [a, b] - 容忍误差 tol - 最大迭代次数 maxiter 函数将返回求解的根以及迭代次数。使用该函数的示例代码如下： f = @(x) x^3 - 2*x - 5; a = 2; b = 3; tol = 1e-6; maxiter = 100; [root,iter] = bisection(f,a,b,tol,maxiter); fprintf('The root is %.6f, found in %d iterations\n',root,iter); 在这个例子中，我们使用二分法求解函数 f(x) = x^3 - 2x - 5 在区间 [2,3] 中的根，容忍误差为 1e-6，最大迭代次数为 100。函数返回的结果是根的近似值和求解过程中迭代的次数。

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