Look up ?coeftayl and related commands at the bottom of the help

int(1/(a-r*t),t); 
[discont(%, r), discont(%, a), discont(%, t)];
eval(%, r=b-c);
                              ln(a - r t)
                            - -----------
                                   r


                       [{0, a/t}, {r t}, {a/r}]


                                              a
                   [{0, a/t}, {(b - c) t}, {-----}]
                                            b - c

convert(sin(x), Sum, dummy=k);

                      infinity
                       -----       k  (2 k + 1)
                        \      (-1)  x
                         )     ----------------
                        /         (2 k + 1)!
                       -----
                       k = 0

In words this says that multiplication by 'a' is surjective, i.e. for all
y there is some x, such that y = a*x. Usually one would divide to get x.

But can that be done - what is for y = 1? Is there a name in Math for those
elements? Then use 'ifactor(256)' to get an idea for your task. 

But then you should kill that task by paper and pencil, not by coding.

sol := solve({5*A+3*B = 5, 8*A-9*B = 94}, {A, B});

                                  109      -430
                      sol := {A = ---, B = ----}
                                  23        69

# A,B are not assigned, just use them as variables
23*(A + B); 
                             23 A + 23 B

# now use the solution in case you want it
23*(A + B):
eval(%, sol);
                                -103/3

# then A,B still are variables
17*A + 19*B;
                             17 A + 19 B

Re-arranging as complex equation (real and imag part) You are searching for 

  S  := A[2]*exp(t[2]*w*I)      +  A[3]*exp(t[3]*w*I)      +1-A[2]-A[3]
  S1 := A[2]*t[2]*exp(t[2]*w*I) +  A[3]*t[3]*exp(t[3]*w*I)

This is a linear system in A[2], A[3] with a (unique) solution, which will
involve t[2], t[3] - but for any such pair of t.

From that you _may_ try to find those, for which the A are positive, but that
may work only if providing numerical values.


The other way round is to solve the above system for t[2], t[3], depending on
the A and that has many solution due to periodics, looking quite complicated.


But already viewing it as 2-dim linear in A it shows, that this may not give
what you have in mind (may be, only numerical values can help further on).

Instead of vector use Vector, you may wish to read the help towards that.

Your procG is just (a,b) -> max(a-b, 0), no?

Loops end with 'end do'. 

Now for your question: that variable i is local to the loop and one way would
be to write

  F:=Vector(2):
  for i from 1 to 2 do 
    F[i]:=unapply(i*'procG'(x,0), x); 
  end do;

That prevents the procG to be evaluated and should give what you want.


Perhaps I would write

  f:= j -> (x -> j*max(x,0));
  F:=Vector[row](2, f);

There are several issues with your code (you have no input arguments and if you
want to use such then do as given in the help, do not declare them as globals
and working with indexed variables needs knowledge about what Maple does then).

Just do (similar for the other task):

  10*(D(D(x)))(t) = x(t); 
  dsolve(%); 

                              1/2                  1/2
                            10    t              10    t
             x(t) = _C1 exp(-------) + _C2 exp(- -------)
                              10                   10

and then discuss the constants to satisfy your needs (both of the exp terms
achieve all values between 0 and infinity).

The equation does not have a solution modulo 23, thus can not be solved in Z.

restart;
f:=2017*x^2013+2015*y^2013-2033*z^2013-2011;

p:=23;
L:=Array(0..p-1,0..p-1,0..p-1, datatype=integer):
 
for i from 0 to p-1 do
for j from 0 to p-1 do
for k from 0 to p-1 do
  tmp:=subs(x=i,y=j,z=k, f);
  tmp:=tmp mod p;
  L[i,j,k]:=tmp;
  if tmp=0 then print(i,j,k); end if;
end do; end do; end do;
 
convert(L,list): convert(%, set);

               {1, 2, 4, 6, 8, 11, 13, 15, 18, 20, 22}

By brute force it shows that the value 0 is not being achieved.


For actual searching in general this is not an efficient way (instead just 
count errors against 0 or break, instead of using Arrays).

Also use little Fermat, here giving x^2013 = x^11 mod 23.

Which gives 11*x^11 + y^11 + z^11 = 4 as variant & task (just for fun).

EQ; # Joe Riel named it p 
'subs(x = 0+2013*nx, y = 858+2013*ny, z = 1594+2013*nz, %)';
% mod 2013;

                 2013         2013         2013
           2017 x     + 2015 y     - 2033 z     - 2011 = 0

                                0 = 0

with(PDEtools);
eq:=f(x) = (x+a)*(b+c*exp(-x^2));
DPS:=dpolyform(eq, no_Fn);    # the differential polynomial form of it;

                                     2
          d         (-2 x a + 1 - 2 x ) f(x)
  DPS := [-- f(x) = ------------------------ + 2 x b (x + a)] &where
          dx                 x + a

        [f(x) + (-x - a) b <> 0]

ode:=op(1,DPS)[-1];
                                        2
             d         (-2 x a + 1 - 2 x ) f(x)
      ode := -- f(x) = ------------------------ + 2 x b (x + a)
             dx                 x + a

Have you tried the help for ?solve and for ?polar

The first 3 equations are linear (in 4 variables). The 4th is a polynomial of
degree 4 in T[go] and linear in T[gi]. While eq 5 and 6 are rationals and can
be made polynomial by multiplying with the denominator leading to polynomials
of degree 7 in T[ga] and T[gi].

Perhaps one could first solve the first 3 eqs, depending now on a parameter
in a linaer way (i.e. the solution is an affine line) and feed it to the others.

May be that Groebner bases are a way (converting coefficients to rationals
first is reasonable for that).

I played with it (since it is not even clear, why solutions should be just
isolated points). And if I understand it correctly, then the solution space
is a 1 dimensional curve (may be with components and may be using complex
solutions).

But it may depend on the 'application', which one would make sense.

http://www.mapleprimes.com/questions/132049-Transcendental-Equation#comment132068

For that I still get 1 solution, I think it is a cancellation problem 
(or may be, I do not get it):

Using convert/rational (see above) for your 10 Digits (you do, yes?) I get

tan(1/500000*(a-z)^(1/2))/tan(1/500000*(b*a-z)^(1/2))+
  4*z*(b*a-z)^(1/2)*(a-z)^(1/2)/(a-2*z)^2;
RootOf(%,z);
allvalues(%);
                       1/2
                (a - z)
            tan(----------)                 1/2        1/2
                  500000       4 z (b a - z)    (a - z)
           ----------------- + ---------------------------
                        1/2                     2
               (b a - z)               (a - 2 z)
           tan(------------)
                  500000

                                  a

where z = x^2, a = z0 = 2659236692000 and b  = 182314839/379890956

I want to avoid the pain of discussing, when do the imaginary solutions through
the square roots and the periodics for tan fit together for a real solution

I get only 1 solution (using Digits=10 to 'read' your floating point numbers)

 tan(2*(1.1575*10^12/(1-.5)^1.2-x^2)^(1/2)*10^(-6)*(1/2))/
   tan(2*(5.555*10^11/(1-.5)^1.2-x^2)^(1/2)*10^(-6)*(1/2)) =
 -4*x^2*(5.555*10^11/(1-.5)^1.2-x^2)^(1/2)*(1.1575*10^12/
   (1-.5)^1.2-x^2)^(1/2)/(1.1575*10^12/(1-.5)^1.2-2*x^2)^2;
 eq:=convert(%, rational);
 lhs(%) - rhs(%);
 EQ:=subs(x^2=z, %);
 RootOf(%, z);
 z0:=allvalues(%);
 sqrt(%);
 evalf(%);

That z0 := 2659236692000 is one of your constants in the given expression.

What are other solution you get?