@pawarabhijit7 You wrote:

My all values must be less than one because i have normalised all the terms

Does that mean that they are all less than one in absolute value? That's a stricter requirement than just being less than one.

And when you say that they are normalised, do you mean that the sum of the squares is one? If so, then that's an additional equation. Or do you mean the weaker condition that each has been divided by its own absolute value?

I didn't say that we needed three more equations; I said that we would need three more equations if we wanted to get a unique solution (or, indeed, any finite number of solutions). Do you have reason to suspect a finite number of solutions?

@pawarabhijit7 You wrote:

My all values must be less than one because i have normalised all the terms

Does that mean that they are all less than one in absolute value? That's a stricter requirement than just being less than one.

And when you say that they are normalised, do you mean that the sum of the squares is one? If so, then that's an additional equation. Or do you mean the weaker condition that each has been divided by its own absolute value?

I didn't say that we needed three more equations; I said that we would need three more equations if we wanted to get a unique solution (or, indeed, any finite number of solutions). Do you have reason to suspect a finite number of solutions?

Adri, The theorem that you cite is about systems of ODEs. The Asker is asking about systems of algebraic equations.

Adri, The theorem that you cite is about systems of ODEs. The Asker is asking about systems of algebraic equations.

@LEETZ That's the solution to the ODE. Note that it contains two arbitrary functions that are constant wrt x; thus it takes two (and only two) boundary conditions on x to solve for those functions.

@LEETZ That's the solution to the ODE. Note that it contains two arbitrary functions that are constant wrt x; thus it takes two (and only two) boundary conditions on x to solve for those functions.

@mehdi_mech I think that you are saying that you want to put in another variable, time, for which you had numeric values above. Is that right? Yes, that can be done, and we'd have to use solve (or one of it's equivalents) instead of fsolve. Yes, it will take substantially more time and memory. How much more? I don't know until we try it. It's possible that it would take so long that you will give up, or you may run out of memory. I just scanned the equations briefly. Are they all degree 3?

A good place to start is ?SolveTools,PolynomialSystem.

@mehdi_mech I think that you are saying that you want to put in another variable, time, for which you had numeric values above. Is that right? Yes, that can be done, and we'd have to use solve (or one of it's equivalents) instead of fsolve. Yes, it will take substantially more time and memory. How much more? I don't know until we try it. It's possible that it would take so long that you will give up, or you may run out of memory. I just scanned the equations briefly. Are they all degree 3?

A good place to start is ?SolveTools,PolynomialSystem.

The solution returned by solve for your set of 12 equations has 3 free variables; so, to get a unique solution you need 3 more equations.

You said that the solution returned by fsolve was not feasible, even though the solution is all real. So, obviously you have some additional constraints to feasibility in mind. Maybe these are inequalities. If you include these constraints in the set that you pass to solve, you might get better results. Or you can post your additional constraints here.

The solution returned by solve for your set of 12 equations has 3 free variables; so, to get a unique solution you need 3 more equations.

You said that the solution returned by fsolve was not feasible, even though the solution is all real. So, obviously you have some additional constraints to feasibility in mind. Maybe these are inequalities. If you include these constraints in the set that you pass to solve, you might get better results. Or you can post your additional constraints here.

@OffshoreEngineer

Here's an explanation of why my fix works: Your original sigvm had its parameters reversed compared to your other procedures xcart, ycart, zcart: The first parameter of sigvm was theta, and its second parameter was x. The code

unapply(sigvm(theta, x), [x, theta])

converts sigvm into a new procedure with the parameters in the other order.

Example:

F:= (x,y)-> exp(y)*sin(x);
(x, y) -> exp(y) sin(x)
F(ln(2), Pi/6);
                         /1   \           
                      exp|- Pi| sin(ln(2))
                         \6   /           
G:= unapply(F(x,y), [y,x]);
(y, x) -> exp(y) sin(x)
G(ln(2), Pi/6);
                               1

@OffshoreEngineer

Here's an explanation of why my fix works: Your original sigvm had its parameters reversed compared to your other procedures xcart, ycart, zcart: The first parameter of sigvm was theta, and its second parameter was x. The code

unapply(sigvm(theta, x), [x, theta])

converts sigvm into a new procedure with the parameters in the other order.

Example:

F:= (x,y)-> exp(y)*sin(x);
(x, y) -> exp(y) sin(x)
F(ln(2), Pi/6);
                         /1   \           
                      exp|- Pi| sin(ln(2))
                         \6   /           
G:= unapply(F(x,y), [y,x]);
(y, x) -> exp(y) sin(x)
G(ln(2), Pi/6);
                               1

@OffshoreEngineer Before you upload it, please test the Answer that I gave below.

@OffshoreEngineer 

What is the format of sigvm? My first guess is that it is a procedure that takes two arguments, x and theta, and returns a single value (on the HUE color scale). But it could return a list of three values (for the RGB or HSV color scales).

@tuGUTS 

There is no hope for solving this equation in Maple. I would guess that it far beyond the capability of any CAS.

It is not a differential equation (so dsolve doesn't apply), because it is evaluated at a particular point, z=0. It is not a regular equation (so solve doesn't apply), because you don't want to solve for a particular x. 

However, there are a few functions for which Maple can take an arbitrary-order derivative, although this facility is quite buggy. You can use this to test out a few simple cases for g(z).

T:= Eval(Diff(g(z)^x, z$(x-1)), z=0) = n*x*x!:
#I moved x! to other side of equation.

#Test g(z) = exp(z) (so g(z)^x = exp(z*x))
value(subs(g(z)^x= exp(z*x), T));
                   (x - 1)                   
                  x        = n x x!