fsolve multi variables help

I am trying to solve for angles in a system given 2 equations and 2 unknowns. First I defined ,y two functions in the form

function F=myfun(theta2, theta3, theta4)
F=[.2+.05.*cos(theta2)+.304.*cos(theta3)-.2.*cos(theta4);
    .1+.05.*sin(theta2)+.304.*sin(theta3)-.2.*sin(theta4)];
end

then I attempt to solve for the unknowns using fsolve

theta3(1)=170.69;
    theta4(1)=116.56;
    x0=0
for theta2=-90:5:270;
options = optimset('Display','iter');
x=fsolve(@myfun,X0,options);
end

what am I doing wrong? is this even possible?

Answer 1

[Star Strider]

I can’t figure out what you want to do.

The optimisation routines want the objective function argument to be a vector. The easiest way to do this with your function is to call it as:

@(b)myfun(b(1),b(2),b(3))

so:

x = fsolve(@(b)myfun(b(1),b(2),b(3)),X0,options);

with ‘X0’ now being a three-element vector.

I do not understand the loop. Because ‘X0’ never changes, the solution is the same for every iteration.

[Emilio Lopez]

I'm trying to solve a system of equations. that is, solve for theta3 and theta4 given theta2 i don't have initial theta3 and 4. my function F is defined in the format F(theta)=0.... so i'm trying to solve for the array of theta3 and theta 4 such that theta2=-90:5:270 thats why the for loop is needed, to change theta2 with every iteration but when i try to run my code, it tells me it needs inputs for theta3, theta 4 to execute

    F=[.2+.05.*cos(theta2)+.304.*cos(theta3)-.2.*cos(theta4);
    .1+.05.*sin(theta2)+.304.*sin(theta3)-.2.*sin(theta4)];

[Star Strider]

Now I understand.

Try this:

function F=myfun(theta2, theta3, theta4)
F=[.2+.05.*cosd(theta2)+.304.*cosd(theta3)-.2.*cosd(theta4);
    .1+.05.*sind(theta2)+.304.*sind(theta3)-.2.*sind(theta4)];
end

theta3 = 170.69;
theta4 = 116.56;
X0 = [theta3; theta4];
options = optimset('Display','iter');
theta2 = -90:5:270;
for k1 = 1:length(theta2)
    x(:,k1)=fsolve(@(b)myfun(theta2(k1),b(1),b(2)),X0,options);
end

figure(1)
plot(theta2, x)
grid
xlabel('\theta_2')
ylabel('\theta (°)')
legend('\theta_3', '\theta_4')

I tested that code. It works. (Note that I converted the trigonometric functions to their degree equivalents, since you define ‘theta2’ and the others in degrees.)

[Matt J]

This is virtually the same as my solution already presented below. The only difference is that I do not use a fixed initial guess theta3 = 170.69, theta4 = 116.56 for every theta2(k1). It is inadvisable to do so. Much faster convergence and much better resilience to local min can be obtained by doing a coarse initial sweep of F.

[Emilio Lopez]

you guys just saved my project grade, thanks! I could not figure it out.

Answer 2

[Matt J]

I suspect you might be after something like this. Note that I changed all of your sin() and cos() to sind() and cosd() since it looks like you are measuring angles in degrees.

    Theta2=(-90:5:270); N=numel(Theta2);

    [Theta3,Theta4]=ndgrid(1:2:360);

   for i=N:-1:1 %Get initial guesses through coarse sampling

      Error = sum(abs(  myfun(Theta2(i), Theta3(:).', Theta4(:).')  ));
      [minval,imin]=min(Error); 

       X0(i,:)=[Theta3(imin), Theta4(imin)];

   end

   options = optimset('Display','iter');

   for i=N:-1:1 %Refine the initial guesses

        thisFun=@(X) myfun(Theta2(i), X(1), X(2));
        x(i,:)=fsolve(thisFun,X0(i,:),options);

   end

function F=myfun(theta2, theta3, theta4)

    F=[.2+.05.*cosd(theta2)+.304.*cosd(theta3)-.2.*cosd(theta4);
      .1+.05.*sind(theta2)+.304.*sind(theta3)-.2.*sind(theta4)];