Plotting help

Hi, I'm pretty new to Matlab, and I tried to execute this script:

eq1 = '((((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))+((((x-(i*y))-((3.46163-(i*(0)))))/((x-(i*(y)))+(3.46163-(i*(0)))))*(exp(-2*i*((pi*2)/1796)*(x-(i*(y)))*z))))/(1+(((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))*(((x-(i*(y)))-(3.46163-(i*(0))))/((x-(i*(y)))+(3.46163-(i*(0)))))*(exp(-2*i*((pi*2)/1796)*(x-(i*(y)))*z)))))-.30441';

eq2 = '((((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))+((((x-(i*y))-((3.46153-(i*(0)))))/((x-(i*(y)))+(3.46153-(i*(0)))))*(exp(-2*i*((pi*2)/1798)*(x-(i*(y)))*z))))/(1+(((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))*(((x-(i*(y)))-(3.46153-(i*(0))))/((x-(i*(y)))+(3.46153-(i*(0)))))*(exp(-2*i*((pi*2)/1798)*(x-(i*(y)))*z)))))-.30440';

eq3 = '((((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))+((((x-(i*y))-((3.46143-(i*(0)))))/((x-(i*(y)))+(3.46143-(i*(0)))))*(exp(-2*i*((pi*2)/1800)*(x-(i*(y)))*z))))/(1+(((1.5-(x-(i*(y))))/(1.5+(x-(i*(y)))))*(((x-(i*(y)))-(3.46143-(i*(0))))/((x-(i*(y)))+(3.46143-(i*(0)))))*(exp(-2*i*((pi*2)/1800)*(x-(i*(y)))*z)))))-.30439';

ezplot(eq1), hold on;

ezplot(eq2), hold on;

ezplot(eq3), hold off;

but it didn't work (some of you want to sternly tell me off right now, I know). Here is the error message:

??? Error using ==> char
Cell elements must be character arrays.

Error in ==> ezplot at 158
      fmsg = char(f);

Error in ==> nonlintest2 at 4
ezplot(eq1), hold on;

Error in ==> run at 74
    evalin('caller',[script ';']);

So, as a new user, I don't know how to use this error report very well, and looking online has gotten me some approximate info, but nothing that truly explains my situation to me. Can anyone help me with this please? I'd love to learn how to make this work.

[the cyclist]

Why did you enclose eq1, etc., in single quotes?

[Fangjun Jiang]

@cyclist, to use ezplot()

Answer 1

[Walter Roberson]

It is a bug in ezplot in constructing the error message to tell you that ezplot cannot be used to plot equations in more than 2 variables.

[Alex]

Any solutions for that bug? What would you recommend that I do in this situation?

[Walter Roberson]

If you really wanted to, you could edit ezplot() so that you change line 158 to

if isa(f{1},'inline'); fmsg = char(f{1}); else fmsg = char(f); end

Of course all that will do is enable the program to reach the correct error() message; you would still need to adjust your program to try/catch to account for the possibility of that error.

[Alex]

So, as a new user, what does any of that mean?

[Walter Roberson]

It means that you should not use ezplot() to plot a function with three or more variables. Your equations all obviously involve x and y, and each of them has a single occurrence of z (which you can see near the beginning of the third text line of each, above.)

Perhaps you should double-check that you have not made an error in transcribing the equations: perhaps that z is intended to be x or y.

[Alex]

Well, I developed these equations myself, so I know that it's supposed to be "z". Z does not necessarily represent a direction either, it's just a placeholder value. I'm trying to solve simultaneous non-linear equations, and I wanted to use a graphical analysis to check answers, but neither of these things are working for me. Any suggestions?

[Walter Roberson]

You are expecting to be able to create a graphical representation of something that involves a symbolic constant? And over a complex field? You do realize that the graphs are non-linear in z, so it isn't even as simple as drawing a graph and adding a "multiplied by z" notation to it?

Perhaps it would make sense to substitute a specific value in for z before attempting the plot? You could do that with

eq1a = strrep(eq1,'z','1.234987');

and plotting that ?

[Walter Roberson]

Those equations are difficult to solve, even individually by substituting values in to free variables.

[Alex]

Yeah, that's the difficulty that I'm having with this. I don't know a lot about the language of Matlab, nor it's graphing capabilities, hence the post. If I had the optimization toolbox I could try using fsolve, but I don't, which leaves me in a pickle.