不动点迭代

function xc = fpi( g, x0, tol )    x(1) = x0;    i = 1;    while 1        x(i + 1) = g(x(i));        if(abs(x(i+1) - x(i)) < tol)            break        end        i = i + 1;    end    xc = x(i+1);end

牛顿法：

function xk = funNewton(f, x0, max_steps, tol)    syms x    symbol_f = f(x);    dif_f = matlabFunction(diff(symbol_f));    clear x    x = x0;    for k = 1:max_steps       xk = x;       disp(['the ', num2str(k), ' time is ', num2str(x)])       %xk to save the last time value of x       x = x - f(x) / dif_f(x);       %newton solve        if(abs(xk - x) < tol)       %decide whether to break out            break;       end    endend

割线法：

function xc = CutLine( f, x0, x1, tol )    x(1) = x0;    x(2) = x1;    i = 2;    while 1        x(i + 1) = x(i) - (f(x(i)) * (x(i) - x(i - 1))) / (f(x(i)) - f(x(i - 1)));        if(abs(x(i + 1) - x(i)) < tol)            break;        end        i = i + 1;    end    xc = x(i + 1);end

Stewart平台运动学问题求解：

function out = Stewart( theta )    % set the parameter    x1 = 4;     x2 = 0;     y2 = 4;     L1 = 2;    L2 = sqrt(2);    L3 = sqrt(2);    gamma = pi / 2;    p1 = sqrt(5);    p2 = sqrt(5);    p3 = sqrt(5);    % calculate the answer    A2 = L3 * cos(theta) - x1;    B2 = L3 * sin(theta);    A3 = L2 * cos(theta + gamma) - x2;    B3 = L2 * sin(theta + gamma) - y2;        N1 = B3 * (p2 ^ 2 - p1 ^ 2 - A2 ^ 2 - B2 ^ 2) - B2 * (p3 ^ 2 - p1 ^ 2 - A3 ^ 2 - B3 ^ 2);    N2 =  -A3 * (p2 ^ 2 - p1 ^ 2 - A2 ^ 2 - B2 ^ 2) + A2 * (p3 ^ 2 - p1 ^ 2 - A3 ^ 2 - B3 ^ 2);    D = 2 * (A2 * B3 - B2 * A3);        out = N1 ^ 2 + N2 ^ 2 - p1 ^ 2 * D ^ 2;end

test our function at theta = - pi / 4 and theta = pi / 4

clear allclcformat shortdisp('f(- pi / 4) is ')out1 = Stewart(- pi / 4)disp('--------------')disp('f(pi / 4) is ')out2 = Stewart(pi / 4)