one way

PatrickT — Sat, 22 Dec 2012 00:49:01 Z

solve({eq1,eq2},{d,s});

{d = 2*a*c^2+2*a^2*b+2*a^2*c+4*a*c*b+2*b^2*a+2*b^2*c+2*b*c^2-a-b-c, s = 2*a*c^2+2*a^2*b+2*a^2*c+4*a*c*b+2*b^2*a+2*b^2*c+2*b*c^2}

eliminate

Preben Alsholm — Sat, 22 Dec 2012 01:57:15 Z

Question: Why didn't you expect solve to just express d in terms of s, a, b, and c? Should solve guess that you wanted to eliminate s?

restart;
eq1:=s=a+b+c+d;
eq2:=s=2*(a+b)*(a+c)*(b+c);
eliminate({eq1,eq2},s);
solve(%[2],d);
#Alternative:
solve(eq1,{d});
subs(eq2,%);
expand(%); #For comparison only
#But you could have eliminated e.g. c:
eliminate({eq1,eq2},c);
solve(%[2],d);

Explanation

Markiyan Hirnyk — Sat, 22 Dec 2012 02:06:50 Z

Analyzing the output of
restart;printlevel:=35:
eq1:=s=a+b+c+d;
eq2:=s=2*(a+b)*(a+c)*(b+c);
solve({eq1,eq2},d); ,
I draw the conclusion that Maple treats the system under consideration
as a system of two equations linear in d (pay attention to UncoupledLinear in theoutput).
Because eq2 has the form 0*d = s- 2*(a+b)*(a+c)*(b+c), Maple produces the false proposition
that the system has no solution.
Compare with
eq3 := 2*d = 4;
eq4 := 3*d = 6;
solve({eq3, eq4}, d);

                            {d = 2}
and with

eq5 := 4*d = s;
solve({eq3, eq5}, d);
NULL
PS. I answered the question in screen201.docx

Simple solution

Markiyan Hirnyk — Sat, 22 Dec 2012 02:46:28 Z

> restart; eq1 := s = a+b+c+d:

eq2 := s = (2*(a+b))*(a+c)*(b+c):

solve({eq1, eq2});

worth pondering

acer — Sun, 23 Dec 2012 08:04:44 Z

I believe that Preben's response is on the right track.

I'm not sure whether the solve help-pages make it clear enough (for systems of multivariate equations) just what is the definition of a solution according to the various possible calling sequences.

I suspect that a working definition of a "solution" to a call like,

    solve( eqs, vars );

is that evaluation at the returned result will satisfy all the equations. (It doesn't matter whether `eqs` are equations, or expressions which are taken to imply equations.)

Now, it may happen that there are other variables present in `eqs` which are not included in `vars`. The given calling sequence of the `solve` command appears to mean that any "solution" will only contain equations whose left hand sides are in `vars`. I didn't find this aspect described in any `solve` help-page, but then also I didn't find anything else like a tight set of definitions of a "solution" to a system of equations according to the various calling sequence possibilities. I will presume that this is an interpretation that agrees with `solve`'s design and behaviour for the remainder of this Comment.

So `solve({eq1,eq2},{d})` would return solutions containing only equations with `d` on the left hand side.

For the given example no single equation of the form `d=expr` can (by itself alone) satisfy both `eq1` and `eq2`. Or, more generally, it may be that no set consisting only of equations like var[i]=expr[i] for var[i] in `vars` would satisfy all equations in `eqs`.

In that case `solve` would return NULL, as no equation d=expr alone can satisfy both equations.

Consider the output from,

restart;

eq1:=s=a+b+c+d:
eq2:=s=2*(a+b)*(a+c)*(b+c):

eliminate({eq1,eq2},d);

      [                      /        2        2                    2
      [{d = s - a - b - c}, { -s + 2 a  b + 2 a  c + 4 a c b + 2 a c 
      [                      \                                       

                2        2          2\ ]
           + 2 b  a + 2 b  c + 2 b c  }]
                                     / ]

The first element of that output is of the form `d=expr`. Notice that `s` is not free in the (only) expression (implying an equation set to zero) in the set as the second entry of that output. This means that `solve({eq1,eq2},{d,s})` is a quite different request than is `solve({eq1,eq2},{d})`. Having the expression (in the set appearing as the second entry) in the above result from `eliminate` equal zero implies that `s=function_of_abc`. I see the second entry in the above result from `eliminate` as another way of encapsulating the notion that no single equation `d=expr` can alone satisfy both `eq1` and `eq2`, as there is a algebraic relationship amongst a,b,c, and s which does not involve d.

But if all this is correct interpretation then there is a useful bit of functionality missing from the `solve` command. It is the ability to force a non-trivial RHS in the equation `d=expr` in a result from calling `solve`. Let's suppose that you wish to avoid solutions containing `d=d`, if possible. If you simply invoke `solve(eqs)` and leave out the `vars` then there may be some choice of which variables appear in the solution trivially like `var[i]=var[i]` (as if they were parameters, say). One might want to specificy in a call to the `solve` command that `d=expr` appears with `expr` not identical to symbol `d`, and that any other variables could be inserted into the "solution" that would allow all original equations in `eqs` to be satisfied. If there were a choice as to what else to eliminate then `solve` might programmed choose. Eg. `s` might be chosen in the given example because doing so provides a simple explicit solution.

Why do I feel like this has been discussed before?

acer

A mountain of ungrounded words

Markiyan Hirnyk — Sun, 23 Dec 2012 10:39:49 Z

@acer In order to make such claims, the execution trace of the code under consideration by the trace or/and printlevel command is necessary. There is no such stuff in your comment.