matlab实现不动点迭代、牛顿法、割线法

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

牛顿法找根：
$$ f( x ) = ( 1 - \frac{3}{4x} ) ^ {\frac{1}{3} }$$

封装函数计算：
x_right = solve('(1 - 3 / (4 * x)) ^ (1 / 3)')

牛顿法实现：

function [y, dirv_y] = funNewton(x)
    y = (1 - 3 / (4 * x)) ^ (1 / 3);
    dirv_y = (1 - 3 / (4 * x)) ^ (- 2 / 3) / (4 * x ^ 2);
    % dirv_y is y's diff
end

clear all
clc
Error = 1e-6;
format long
x_right = solve('(1 - 3 / (4 * x)) ^ (1 / 3)')
%disp the right answer
x = 0.7;
for k = 1:50
   [y, dirv_y] = funNewton(x);
   %call the function to get the f(x) and it's diff
   xk = x;
   disp(['the ', num2str(k), ' time is ', num2str(x)])
   %xk to save the last time value of x
   x = x - y / dirv_y;
   %newton solve 
   if(abs(xk - x) < Error)
   %decide whether to break out
        break;
   end
end
xk
%output the value of x

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 all
clc

format short

disp('f(- pi / 4) is ')
out1 = Stewart(- pi / 4)

disp('--------------')

disp('f(pi / 4) is ')
out2 = Stewart(pi / 4)