@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!

Could you upload a worksheet please?

I am not sure yet (need to work with the worksheet), but I think that you need to use unapply on sigvm and then compose (@) the resulting function with a coordinate transformation function.

I'm having some trouble wrapping the 3D processor of my mind around your coordinate transformations: I think of the normal cylindrical coordinates as x = r*cos(theta), y= r*sin(theta), z=z. You seem to have some permutation of the x,y,z. (I am not saying that your transformations are wrong, just that they are difficult for me the visualize.)

It's `dsolve/numeric/DAE/sm_intersect`, not `dsolve/dae/numeric/sm_intesect`. The word order and capitalization matter. You can view the procedure via showstat(`dsolve/numeric/DAE/sm_intersect`). It's fairly short, but I don't know what it does. You may gain a little insight into the situation by issuing the command

trace(`dsolve/numeric/DAE/sm_intersect`);

before your dsolve command.

Maple does not have a general facility for regular functional equations (AFAIK), let alone a diffeo-functional equation (which may be the coinage of a new term) like this. Although I wouldn't be surprised if the answer in this particular case could be found after a few guess trials with different g(z).

As to the variable with respect to (wrt) which the derivative is taken: It is probably wrt z for two reasons:

1. The text you cite says that x is a natural number.
2. If it were wrt x, then there would be no need for z at all; it could just as well be G^x for some constant G.

The proper way to enter such an expression into Maple is with Eval (inert form) or eval (active form) rather than subs:

1/x!*Eval(Diff(g(z)^x, z$(x-1)), z=0) = n*x;

Are the x_n restricted to positive reals?

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
                        0      

Combining Axel's idea with my power series idea given in another Answer, we see that the power series expressed with lower limit n is also independent of n, and is the same as if the lower limit were 0. However, oddly enough, though the two expressions are equal, they are not identical as expressions.

S1:= sum(x^k/(6*k+1)!, k= 0..infinity);

S2:= sum(x^k/(6*k+1)!, k= n..infinity);

simplify(S1-S2);
                        0      

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

But it seems that the series starting at k = 0 is viewed as a power series evaluated at 1:

sum(x^k/(6*k+1)!, k= 0..infinity):
eval(%, x=1);

It is not clear what you want.

• Let A be the region enclosed by r = 1.
• Let B be the region enclosed by r = 2*cos(theta)

Then, proceeding left to right, there are three disjoint regions:

• Let R1 = A minus B
• Let R2 = A intersect B
• Let R3 = B minus A

What color do you want R1? What color do you want R2? What color do you want R3? What color do you want the boundary of A? What color do you want the boundary of B? Do you want to see the gridlines underneath the regions?

@litun Consider a simpler example: z(t) = t, G = 1 + t + t^2. Then G = z^0 + z^1 + z^2, so a_0(t) = a_1(t) = a_2(t) = 1. Or G = (1+t)*z^0 + t*z^1, so a_0(t) = 1 + t, a_1(t) = t, a_2(t) = 0. And there are numerous other possibilities. So the a_k(t) are not uniquely determined.

@tuGUTS Like I said, I don't know of any case where rsolve solves an equation with more than one independent variable; although I'm not absolutely sure that it will never solve them. Hopefully someone who knows more will comment on this.

This is just a seat-of-the-pants guess, but the answer to your equation may involve Stirling numbers.

Adri wrote:

...it is automatically filled with zeros on the empty places.

Zero is the default, but the fill value can be anything, including NULL.

b:= Matrix([a||(1..4)], fill= NULL);

The advantage of NULL is that it is automatically removed when converted to a list.

convert(b^%T, listlist);
  [[1, 5, 4, 2], [2, 3, 8, 2], [4, 4, 5], [5, 5, 4], [3, 7], [2, 8], [3], [4], [5]]