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
end
end

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)
04-30 08:26