Solve constrained equation with 2 variables - matlab

Question

EDIT:
I think what I need is to apply fmincon, but I don't know how. I want to do this:
Use fmincon to solve:
Minimize a, subject to the equality constraint
f(q,z) = 0
by varying the unknowns (q,z).
How do I use fmincon to implement this? Thanks!
Here is the entire function:
f(q,z) = 60*q^9*z^4 + 120*q^7*z^3 + 80*q^(17/2)*z^3 - 60*q^8*z^3 - 60*q^6*z^3 + 40*q^(15/2)*z^3 + 20*q^(11/2)*z^3 - 20*q^9*z^3 + 175*q^(13/2)*z^2 - 90*q^(15/2)*z^2 + 75*q^5*z^2 - 60*q^6*z^2 - 45*q^(11/2)*z^2 + 40*q^2*z^2 + 30*q^8*z^2 - 30*q^(3/2)*z^2 - 10*q^(17/2)*z^2 + 15*q^(9/2)*z^2 - 20*q^4*z^2 - 10*q^(7/2)*z^2 - 90*q^(11/2)*z + 86*q^(9/2)*z + 75*q^6*z - 45*q^(7/2)*z + 40*q^(3/2)*z - 30*q^7*z - 20*q^(13/2)*z - 24*q^(1/2)*z + 15*q^4*z + 15*q^3*z - 9*q^(5/2)*z - 9*q^5*z + 45*q^4 - 30*q^5 + 30*q^(1/2) - 29*q^3 + 21*q^(5/2) - 10*q^(9/2) - 9*q^2 - 24.
----- original post -------
I have a function of 2 variables q and z like
f(q,z) = 60*q^9*z^4 + 120*q^7*z^3 + 80*q^(17/2)*z^3 - 60*q^8*z^3 - 60*q^6*z^3 + ...
I need to find
f(q,z) = 0
with respect to z, but this is not possible analytically (resulting in RootOf). So instead of finding all points, I want to numerically find the smallest number z such that the value pair (q,z) fulfills f(q,z) = 0. Any ideas how to do this?

Answer 1

I'm afraid there is no minimum z, as f(q,z) = 0: q↓0 ⇒ z↓-∞.
Your original equation can be written as:
Writing all the terms following like powers of z as A, B, C, D and E, you basically have
This is a quartic equation in z, which can definitely be solved analytically. Granted, it's a bit awkward, but possible nonetheless.
Note that the fact that A = f(q), B = g(q), C = h(q), D = k(q) and E = m(q) does not matter, as your goal is to express z = F(q).
Here's one way to find the minimum z for which f(q,z)=0 holds:
function find_minimum_z
solution = fminsearch(#F, 3)
end
function R = F(q)
R = roots([
60.*q.^9
120.*q.^7 + 80.*q.^(17/2) - 60.*q.^8 - 60.*q.^6 + 40.*q.^(15/2) + 20.*q.^(11/2) - 20.*q.^9
175.*q.^(13/2) - 90.*q.^(15/2) + 75.*q.^5 - 60.*q.^6 - 45.*q.^(11/2) + 40.*q.^2 + 30.*q.^8 - 30.*q.^(3/2) - 10.*q.^(17/2) + 15.*q.^(9/2) - 20.*q.^4 - 10.*q.^(7/2)
-90.*q.^(11/2) + 86.*q.^(9/2) + 75.*q.^6 - 45.*q.^(7/2) + 40.*q.^(3/2) - 30.*q.^7 - 20.*q.^(13/2) - 24.*q.^(1/2) + 15.*q.^4 + 15.*q.^3 - 9.*q.^(5/2) - 9.*q.^5
45.*q.^4 - 30.*q.^5 + 30.*q.^(1/2) - 29.*q.^3 + 21.*q.^(5/2) - 10.*q.^(9/2) - 9.*q.^2 - 24
]);
R = min(R(imag(R)==0));
if isempty(R)
R = NaN; end
end
My trials indicate there is an asymptote at q=0 for which the minimum, real z that solves the quartic tends to -∞.
Indeed this is what you can see when you make a plot of F(q):

Answer 2

If all terms after the first term have a z³, then it's really quite simple:
f(q,z) = 60q⁹z⁴ + 120q⁷z³ + 80q¹⁷/²z³ - 60q⁸z³ - 60q⁶z³ + ...
= z⁴·(60q⁹) + z³·(120q⁷ + 80q¹⁷/² - 60q⁸ - 60q⁶ + ... )
f(q,z) = 0
⇒ z⁴ · (60q⁹) = -z³ · (120q⁷ + 80q¹⁷/² - 60q⁸ - 60q⁶ + ...)
⇒ -z · (60q⁹) = 120q⁷ + 80q¹⁷/² - 60q⁸ - 60q⁶ + ...
⇒ z = -(120q⁷ + 80q¹⁷/² - 60q⁸ - 60q⁶ + ...) / 60q⁹
= -(2q⁷ + 1⅓q¹⁷/² - q⁸ - q⁶ + ...) / q⁹