vv

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@MapleEnthusiast The matrix has many symbolic parameters, so the pseudo inverse will be huge. But what do you want to do with it? For a system Ax=b, the pseudo inverse A' is generally used to compute the least squares solution A'b (as Carl said). Your system is homogeneous (b=0), so, A'b=0.

@MapleEnthusiast The More-Penrose inverse exists for any matrix (even non square or 0). But I don't see how could it be useful here.
As in my example, the matrix of your system is singular if the relations are considered. So, it does not make sense to solve it before replacing the dependent parameters. 

@MapleEnthusiast 

 

Take a simple similar situation.

You have the system

sys := {a*x + b*y + c*(a-b) = 0, x+y+c+d = 0};

{a*x+b*y+c*(a-b) = 0, x+y+c+d = 0}

(1)

and the parameters satisfy the relations

rels:=[a-b=0, c+d=0];

[a-b = 0, c+d = 0]

(2)

solve(sys,{x,y});

{x = -(a*c-2*b*c-b*d)/(a-b), y = -(a*d+b*c)/(a-b)}

(3)

simplify(%, rels);

Error, (in simplify/siderels:-Recurse) numeric exception: division by zero

 

sys1:=simplify(sys, rels); # now the system is homogeneous (exactly like yours)

{a*(x+y) = 0, x+y = 0}

(4)

solve(sys1, {x,y});

{x = -y, y = y}

(5)

Using a pseudo inverse cannot help here.

 

@nm Now I can't find this info either. Probably I read about it elsewhere.

@nm PDEtools:-Solve says clearly that the multiplicity is ignored.

@MapleEnthusiast In E is computed the regular inverse (and the result is simplified). The matrix C-A is invertible, so the pseudo inverse equals the regular inverse and using method=pseudo just complicates the computations.

@MapleEnthusiast Your system is a simple polynomial system and homogeneous linear  in the three unknowns. The rank being 2 (in general) it has a unique solution (modulo a multiplicative constant). Obviously It cannot be solved in another way.

So, your new questions make no sense.

P.S. You have very long names in your system. Not knowing anyway their meaning, it would have been much easier for us to see x1,x2,x3 (unknowns), a1,a2...,an (parameters). 

@zphaze  It's about maths, not other CAS-es (check again with Mathematica and Maxima!).
The equation (in z)   arctan(z) = a   has a real solution iff -Pi/2 < a < Pi/2. What is a in your case?

@RohanKarthik I would  have formulated the question this way:
Among all the terms A*x^a*y^b  satisfying a - b \in {0,-1}  of a polynomial f(x,y) ,
find the one for which a+b is minimal.

Or, solve:

min { a+b  |  x^a*y^b is a monomial in f(x,y) and a - b \in {0,-1} }.

@zphaze It is not possible to guess user's intentions so easily. f and f(x) may coexist in a legit expression:

plot(sin + sin(1), 0..2*Pi);

is equivalent to

plot(sin(x)+sin(1), x=0..2*Pi);

@ecterrab It is a bit strange to have 2 kinds of inert versions without being equivalent.
E.g. %int is transformed into Int by IntegrationTools but %gcd does not work under mod.
It should be chosen a wise and uniform decision about this. Of course, Gcd etc could be seen as "special" inert forms, and that's it!
 

 

@mthkvv 

restart;
p := 10^140:
a := modp1(ConvertIn((2*10^139+13)+x^1000-x^2,x),p):
save(p, a, "d:/tmp/pab.m");
restart;
read "d:/tmp/pab.m"; # ok
modp1(ConvertOut(a),p)[1] - (2*10^139+13); # 0

 

@acer Even after local gamma,  evalf(gamma(0)); ==> 0.57...   etc
due to `evalf/gamma`;

 

@janhardo You should be aware that Maple can help (in Calculus here) only at a formal level. To understand things you must study carefully a good textbook and be able to solve at least the simple exercises by hand. Not doing so, you are wasting your time.

@janhardo Is this a question, whether f(x) and f(y) are two functions?
You have only one f, in two variables. y is a variable, no y(x) here! 
Try to use proper terms and notations, otherwise everything becomes a mess.

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