Number to string variable

Hello everyone!

I have:

z1=12;

eq='(z-z1)^2';

Is it possible put value z1 into the eq variable (string)?

Thanks

Answer 1

[Star Strider]

It is if you have the Symbolic Math Toolbox:

syms z 
z1=12;
eq= (z-z1)^2;
eq = subs(eq)

produces:

eq =

(z - 12)^2

[Star Strider]

Nuno’s ‘Answer’ moved here...

I need the variable eq as string because i will use fsolve something i need something like:

x1=169;
x2=169;
y1=200;
y2=178;
z1=202;
z2=196;
x3=169;
x4=169;
y3=198;
y4=180;
z3=204;
z4=198;
equ1='(x-x1)^2+(y-y1)^2+(z-z1)^2-r^2'
equ2='(x-x2)^2+(y-y2)^2+(z-z2)^2-r^2'
equ3='(x-x3)^2+(y-y3)^2+(z-z3)^2-r^2'
equ4='(x-x4)^2+(y-y4)^2+(z-z3)^2-r^2'
sol=solve(equ1,equ2,equ3,equ4)

And i need put the value x1,x2,...,z4 into a variablees equ1,equ2,equ3 and equ4 (that are strings)

[Nuno]

but the variable eq is not a string

[Star Strider]

Actually, they’re not strings but symbolic expressions.

This is how I would do it:

syms x y z r

x1=169;
x2=169;
y1=200;
y2=178;
z1=202;
z2=196;
x3=169;
x4=169;
y3=198;
y4=180;
z3=204;
z4=198;
equ1=subs((x-x1)^2+(y-y1)^2+(z-z1)^2-r^2);
equ2=subs((x-x2)^2+(y-y2)^2+(z-z2)^2-r^2);
equ3=subs((x-x3)^2+(y-y3)^2+(z-z3)^2-r^2);
equ4=subs((x-x4)^2+(y-y4)^2+(z-z3)^2-r^2);
[x,y,z,r]=solve([equ1,equ2,equ3,equ4], [x,y,z,r])

The substitutions are correct, but the problem is that the solutions are empty symbolic variables. There seems to be no analytic solution to the system.

You might want to give it a go with fsolve:

x1=169;
x2=169;
y1=200;
y2=178;
z1=202;
z2=196;
x3=169;
x4=169;
y3=198;
y4=180;
z3=204;
z4=198;
equ = @(b) [(b(1)-x1).^2+(b(2)-y1).^2+(b(3)-z1).^2-b(4).^2;
            (b(1)-x2).^2+(b(2)-y2).^2+(b(3)-z2).^2-b(4).^2;
            (b(1)-x3).^2+(b(2)-y3).^2+(b(3)-z3).^2-b(4).^2;
            (b(1)-x4).^2+(b(2)-y4).^2+(b(3)-z4).^2-b(4).^2];

B = fsolve(equ, randi([80 120], 4, 1))

produces (with one set of initial estimates for [x,y,z,r]):

B =

     303.8357
     199.3227
     135.3456
     150.2771

It stops after 400 iterations because it is still searching for a solution, so experiment with the options structure and other fsolve variations if necessary to give the best parameter estimates and information about the solution.

EDIT —

If you want to print the equations out as strings, use this (or something like it):

x1=169;
y1=200;
z1=202;
equstr ='(x-%.0f)^2+(y-%.0f)^2+(z-%.0f)^2-r^2';         % Produces String For ‘equ’
fprintf(1, [equstr '\n'], x1, y1, z1)

that produces:

(x-169)^2+(y-200)^2+(z-202)^2-r^2