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)