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Failure on symbolic solve

asked 2011-01-17 10:34:10 -0500

mouse gravatar image

updated 2011-01-17 14:01:54 -0500

kcrisman gravatar image

Trying to learn manipulation of symbolic expressions. However when I try this one I get a traceback. Is this a bug, limitation, or user error?

x,y,z = var('x y z')
assume(x>0)
assume(y>0)
assume(z>0)
eqn = (x == y**z)
solve(eqn, y)

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "_sage_input_68.py", line 10, in <module>
    exec compile(u'open("___code___.py","w").write("# -*- coding: utf-8 -*-\\n" + _support_.preparse_worksheet_cell(base64.b64decode("eCx5LHogPSB2YXIoJ3ggeSB6JykKYXNzdW1lKHg+MCkKYXNzdW1lKHk+MCkKYXNzdW1lKHo+MCkKZXFuID0gKHggPT0geSoqeikKc29sdmUoZXFuLCB5KQ=="),globals())+"\\n"); execfile(os.path.abspath("___code___.py"))
  File "", line 1, in <module>

  File "/tmp/tmpgB4hFo/___code___.py", line 8, in <module>
    exec compile(u'solve(eqn, y)
  File "", line 1, in <module>

  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/symbolic/relation.py", line 619, in solve
    ans = f.solve(*args,**kwds)
  File "expression.pyx", line 7518, in sage.symbolic.expression.Expression.solve (sage/symbolic/expression.cpp:27206)
  File "expression.pyx", line 7511, in sage.symbolic.expression.Expression.solve (sage/symbolic/expression.cpp:27084)
  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/interfaces/expect.py", line 1474, in __call__
    return self._obj.parent().function_call(self._name, [self._obj] + list(args), kwds)
  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/interfaces/expect.py", line 1373, in function_call
    return self.new(s)
  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/interfaces/expect.py", line 1154, in new
    return self(code)
  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/interfaces/expect.py", line 1090, in __call__
    return cls(self, x, name=name)
  File "/home/sage/sage/local/lib/python2.6/site-packages/sage/interfaces/expect.py", line 1517, in __init__
    raise TypeError, x
TypeError: Computation failed since Maxima requested additional constraints (try the command 'assume(>0)' before integral or limit evaluation, for example):
Is z an integer?
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answered 2011-01-17 14:21:31 -0500

kcrisman gravatar image

Here is a longish answer that will show you a few ways to get what you want... but won't exactly answer your question.

Apparently this is what Maxima is looking for (Sage uses Maxima for its assumptions and solving).

sage: x,y,z = var('x y z')
sage: assume(x>0)
sage: assume(y>0)
sage: assume(z>0)
sage: eqn = (x == y**z)
sage: eqn
x == y^z
sage: assume(z,'integer')
sage: solve(eqn,y)
[y == x^(1/z)]

I should point out that Maxima doesn't use its assumptions that much in its solving.

But Maxima then apparently doesn't care what the answer is, as long as x>0.

Maxima 5.22.1 http://maxima.sourceforge.net
using Lisp ECL 10.4.1
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) assume(y>0,z>0,x>0);
(%o1)                        [y > 0, z > 0, x > 0]
(%i2) eqn:x=y^z;
                                         z
(%o2)                               x = y
(%i3) solve(eqn,y);
Is z an integer?

n;
                                        1/z
(%o3)                             [y = x   ]

So this could be a bug - or feature - in Maxima. Of course, such functions are in general multivalued, so depending on your context this might not really be a good answer anyway. You might instead like

(%i5) load(to_poly_solver);

Loading maxima-grobner $Revision: 1.6 $ $Date: 2009/06/02 07:49:49 $
define: warning: redefining the built-in function prog1
define: warning: redefining the built-in function symbolcheck
define: warning: redefining the built-in function push
define: warning: redefining the built-in function pop
define: warning: redefining the built-in function tr_ev
(%o5) /Users/.../sage-4.6.2.alpha0/local/share/maxima\
/5.22.1/share/contrib/to_poly_solver.mac
(%i6) to_poly_solve(eqn,y);
                                          2 %i %pi %z5
                                          ------------
                                    1/z        z
(%o6)                  %union([y = x    %e            ])

where %z5 is an arbitrary integer. Normally this would come from solve(eqn,y,to_poly_solve=True) but the exception takes precedence in our current implementation.

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Hey Thanks. Other work is pounding on me today but I'll find some time to digest the answer. Appreciate the time you put into this. regards.

mouse gravatar imagemouse ( 2011-01-17 16:19:32 -0500 )edit

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Asked: 2011-01-17 10:34:10 -0500

Seen: 634 times

Last updated: Jan 17 '11