Is there any limitation with fimplicit and fimplicit3 when functions have sqrt ?

Hi,

I have issues when calculating the values of x and y that make fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2).^(1/2) equal to zero. And I don't know if this is due to the square root. Cause, when taking out the square root there is no problem in the calculation.

This is driving me crazy since if you have a funcition, g(x,y) which is g(x,y) = sqrt(f(x,y)) and you are trying to find the values of x,y that make g(x,y) = 0, those are, at least, equal to those x, y values that make f(x,y) = 0. But according to MATLAB there are not values of x,y (nothing appears in the plot of figure 4 and figure 5).

The code is the following one:

R = 0.35;  
L = 0.264; 
H = 0.4;

% Trying first without the square root:
fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2);

figure(1)
hfi = fimplicit(fun, [-pi/2 pi/2 -pi/2 pi/2]);
xlabel('\psi (rad)');
xticks(-pi/2:pi/4:pi/2);
ylabel('\theta (rad)');

figure(2)
fimplicit3(fun, [-pi/2 pi/2 -pi/2 pi/2 0.3 0.8]);
xlabel('\psi (rad)');
ylabel('\theta (rad)');
zlabel('z (m) ');

figure(3)
hfc = fcontour(fun,[-pi/2 pi/2 -pi/2 pi/2]);
LL = hfc.LevelList;

% However, if the function contains a square root...

fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2).^(1/2);

figure(4)
fimplicit(fun, [-pi/2 pi/2 -pi/2 pi/2]) % PROBLEM !!!!!!!!!!!!!

figure(5)
fimplicit3(fun, [-pi/2 pi/2 -pi/2 pi/2 0.3 0.8]) % PROBLEM !!!!!!!!!!!!!

figure(6)
fcontour(fun)
xlabel('\psi (rad)');
ylabel('\theta (rad)');

% I also try to use mesh() in order to see what's happening... but as I said if g(x,y) = sqrt(f(x,y)) ->  g(x,y) = 0 when f(x,y) = 0.
x= linspace(-pi/2,pi/2,100); y = linspace(-pi/2, pi/2, 100);
[X,Y] = meshgrid(x,y);

values = fun(X,Y);
values1 = values;
values1(imag(values1)~= 0) = NaN;

figure(8)
mesh(X,Y,values1);
xlabel('\psi (rad)');
ylabel('\theta (rad)');
zlabel('z (m)');
 

However for the case fun = @(psi,theta) (sin(psi).*sin(theta)).^(1/2) there is no problem with fimplicit3 but the error with fimplicit() mantains. Any suggestion to obtain the plot or the values without taking out the root?

I know that in this case, to avoid the problem, you can take out the square root, but for other cases or when you don't realize about that...

Thank you and sorry for the long message.

Answer 1

[Steven Lord]

Let's look at the value of your function at one of the corners of the region on which you want to plot:

R = 0.35;  
L = 0.264; 
H = 0.4;
fun = @(psi,theta) (L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2).^(1/2);
fun(-pi/2, -pi/2)

ans = 0.0000 + 0.7020i

If the x coordinate is -pi/2 and the y coordinate is -pi/2, where on the 3-d axes below should z = 0 + 0.7020i be plotted?

view(3)
axis(pi/2*[-1 1 -1 1 -1 1])
xlabel('x')
ylabel('y')
zlabel('z')
grid on

[Paul]

For x = -pi/2 and y = -pi/2 there is no real solution for the function, so it shouldn't be nothing plotted.

However, for x and y values, that lead to real solutions in the case of f(x,y) without the square root, there should be something plotted. As those values, are also real values for the f(x,y) with the square root.

[Steven Lord]

The fimplicit function plots where your function's value is equal to 0. Let's take a simple function and see if we can detect where it crosses zero.

f = @sqrt;

Let's evaluate it for a couple values. If f is continuous then we should be able to detect an interval where the value of the function is negative at one end and positive at the other. [There seems to be an issue where Answers has trouble displaying complex numbers at times, so I'm splitting the numbers into real and imaginary parts as a workaround.]

format longg
y = f((-5.5:5.5).');
[real(y), imag(y)]

ans = 12×2

0 2.34520787991171 0 2.12132034355964 0 1.87082869338697 0 1.58113883008419 0 1.22474487139159 0 0.707106781186548 0.707106781186548 0 1.22474487139159 0 1.58113883008419 0 1.87082869338697 0

I see no place where the value of the function changes between negative and positive or vice versa. So this function doesn't cross zero. [It touches zero but then goes flying off into the complex world.]

A slightly different example but one that, like yours, is a function of two variables not just one.

f = @(x, y) x.^2+y.^2;

What should the plot of f(x, y) = 0 be? This function doesn't cross zero it just touches zero.

[Paul]

Oh, I see the difference. Thank you so much.

f = @(x, y) x.^2+y.^2;
fimplicit(f)
% No plot appears in the figure

There is no plot, since as you say the function only touches zero. Cause for real solutions is always positive.That makes sense.

Although, that doesn't justify why there is nothing plotted in fimplict3()...cause when the function is f = @ (theta, psi) sqrt(sin(psi).*sin(theta)) there is no fimplicit (for the reason you perfectly explained) but there is fimplicit3().

Therefore, why in the case of f = @(psi,theta) sqrt(L^2-(R*(cos(theta)-sin(theta).*sin(psi))-H).^2) there is not fimplicit3().

If both, fimplicit() and fimplicit3(), don't work, how do you plot the values where the function touches zero when you have the square root, is there any Matlab command ? Or code ?

Any suggestions you have would be highly appreciated.

[Walter Roberson]

touching zero is often a point and the fimplicit* draw lines not points.