Plot a function with respect to another function

Dear guys,

I have two functions, each function of two variables x,y; namely z1 = f1(x,y) and z2 = f2(x,y) and want to plot z1 vs z2. The functions are below:

0 <= x,y <= 1;
z1 = -(1-x) .* ( (y./(1-x)).*(log((y./(1-x)))) + (1-((y./(1-x)))).*(log(1-((y./(1-x))))) ) /log(2);

z2 = -(1-y) .* ( (x./(1-y)).*(log((x./(1-y)))) + (1-((x./(1-y)))).*(log(1-((x./(1-y))))) ) /log(2);

Answer 1

[Walter Roberson]

Assign values to x and y, do the calculations producing z1 and z2, then

plot(z1, z2)

[Saeid Hajizadeh]

That simply does not give me the answer I want. The result plot is previously sketched in a paper and I want to reach that.

[Saeid Hajizadeh]

any other answers you can come up with?

Thank you

[John D'Errico]

You need to define what it mean to plot z1 versus z1, if Walter's answer is not what you mean. Only you know what you mean here, and your question does not resolve that. If you have a sketch, then it would be best to show us, as your question is unclear.

[Walter Roberson]

Anything could be sketched in a paper. At the moment I have no reason to expect that the sketch is correct, or that you copied the equations exactly correctly. I also do not know what values you put in for x and y.

You are effectively calculating a 3D surface over x and y, and a second surface. What does it mean to plot one surface "vs" another surface? Normally when one plots p "vs" q then q becomes the control variable and p is shown for each q. In order to extend that to "z1 vs z2" with both being surfaces then you would need to create a plot that had at least 4 input dimensions and one output dimension.

[Star Strider]

if y = 1 (permitted since y <= 1), z2 doesn’t exist.

[Saeid Hajizadeh]

Thanks to all guys, I focused. The answer is right. My problem was that I used to put directly p1 and p2 into R12 and R21. The correct code is given below:

p1 = 0.01:0.001:0.49;

p2 = 0.01:0.001:0.49;

for i=1:1:481

    for j=1:1:481

        R12(i,j) = -(1-p1(i)) .* ( (p2(j)./(1-p1(i))).*(log((p2(j)./(1-p1(i))))) + (1-((p2(j)./(1-p1(i))))).*(log(1-((p2(j)./(1-p1(i)))))) ) /log(2);

        R21(i,j) = -(1-p2(j)) .* ( (p1(i)./(1-p2(j))).*(log((p1(i)./(1-p2(j))))) + (1-((p1(i)./(1-p2(j))))).*(log(1-((p1(i)./(1-p2(j)))))) ) /log(2);

    end

end

plot(R12,R21)

Now this is a large number of diagrams plotted simultaneously the envelope of which is the answer. I am trying to find a way to extract envelope here. If you have any idea, I appreciate having them (I am very basic in Matlab)

[Walter Roberson]

Also if y = 0 then you multiply 0 by log(0) which is going to give you NaN.