How can I speed up vpasolve?

I have the following equation that has no analytical inverse

,

where are constants. However, I would like to be able to find for a given z.

I am able to do this with the following code,

syms z thetaz
a = 1;
b = 2;
theta0 = 0;
eqn = (a*sin(theta0)-b*sin(thetaz))./(a*cos(theta0)+b*cos(thetaz)) == z;
v = 0.12; % given value of z
S = vpasolve(eqn,z == v);
vv = S.thetaz %associated value of thetaz

vv = 

but I have thousands of values of v so this takes a very long time.

I have tried using a matlabFunction to speed this up but I'm not sure how to input a z value to give a thetaz value back.

syms z thetaz 
a = 1;
b = 2;
theta0 = 0;
z = (a*sin(theta0)-b*sin(thetaz))./(a*cos(theta0)+b*cos(thetaz));
f1 = matlabFunction(z);
check = f1(0.12)

check = -0.0802

The above code thinks I would like to find z from thetaz, is there a way to find thetaz from z using a matlabFunction?

Alternatively is there a way to speed up vpasolve?

Many thanks.

Answer 1

[John D'Errico]

Seriously, it has no solution? Gosh, am I surprised. So what does this do?

syms z thetaz
a = 1;
b = 2;
theta0 = 0;
eqn = (a*sin(theta0)-b*sin(thetaz))./(a*cos(theta0)+b*cos(thetaz)) == z;

thsol = solve(eqn,thetaz,'returnconditions',true)

thsol = struct with fields:
        thetaz: [3×1 sym]
    parameters: [1×1 sym]
    conditions: [3×1 sym]

thsol.thetaz

ans = 

thsol.parameters

ans = 

k

thsol.conditions

ans = 

So there are infinitely many solutions. Each a multiple of 2*pi away from each other. We can arbitrarily choose the primary solutions, thus with k==0.

syms k
thsol = subs(thsol.thetaz,k,0)

thsol = 

Still not very interesting, but now if we do this:

thsol = matlabFunction(thsol);
thsol(0.12)

ans = 
   3.0818 - 0.0000i
  -0.1790 - 0.0000i
   3.1416 + 0.6931i

now we find three trivially easy to obtain solutions, two of them are real, one is complex.

But I guess no analytical solution exists, because you said so. :)

[Finn Allison]

Yes I should have thought of that! Thank you for your explanation and speedy reply.