Solve contains wrong solution

edit

I beg to differ :

Maxima 5.28.0 http://maxima.sourceforge.net using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (a.k.a. GCL) 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) load(to_poly_solve);

Loading maxima-grobner $Revision: 1.6 $ $Date: 2009-06-02 07:49:49 $

(%o1) /usr/share/maxima/5.28.0/share/to_poly_solve/to_poly_solve.mac

(%i2) display2d:false;

(%o2) false

(%i3) f(x,t):=(%e^(xt))(x-1)^2;

(%o3) f(x,t):=%e^(xt)(x-1)^2

(%i4) map(factor,%solve([factor(diff(f(x,t),x))=0,factor(diff(diff(f(x,t),x),x))<0],x));

(%o4) %union([x = 1,-2 > 0],[x = (t-2)/t,2 > 0])

which is correct.

Note that I load()ed to poly solve, not to_poly_solver ; according to a comment in the former source, the latter is obsolete.

HTH,

                                          Emmanuel Charpentier

This is a bug (?) in to_poly_solve in Maxima.

(%i1) load(to_poly_solver);

Loading maxima-grobner $Revision: 1.6 $ $Date: 2009-06-02 07:49:49 $
(%o1) /Applications/MathApps/Sage-5.4.1-OSX-64bit-10.6.app/Contents/Resources/\
sage/local/share/maxima/5.26.0/share/contrib/to_poly_solver.mac
(%i3) display2d:false;

(%o3) false
(%i4) to_poly_solve([2*(x-1)*%e^(t*x)+t*(x-1)^2*%e^(t*x)=0,4*t*(x-1)*%e^(t*x)+t^2*(x-1)^2*%e^(t*x)+2*%e^(t*x)>0],[x]);

(%o4) %union([x = 1,2 > 0],[x = (t-2)/t,
                            t^2-2*(t-2)*t-4*t+4*(t-2)+(t-2)^2+2 > 0])

The current Maxima has

(%o4) %union([x = 1,2 > 0],[x = (t-2)/t,
                        t^2-2*(t-2)*t-4*t+4*(t-2)+(t-2)^2+2 > 0])

which is basically the same issue. I've reported this upstream here.

I would not agree with you that these solutions are 'wrong'. They are just presented in a very user unfriendly way. The 1st one, i.e.:

[x == 1, -2 > 0]

definitely is a solution - it just specifies a range of measure zero, since

sage: -2>0
False

so, it is not 'wrong' as you state - it is just 'totally useless' ;) The other solution is also perfectly ok since

sage: bool(-(t - 2)^2 + 2*(t - 2)*t - t^2 + 6)
True

so, the solution you want is

x == (t - 2)/t

which I think is ok ..... however I would definetely agree with you that

In[1]:= f[x_, t_] = Exp[x t] (x - 1)^2;
In[2]:= Solve[{D[f[x, t], x] == 0 , D[f[x, t], {x, 2}] < 0, {x, t}]
Out[2]= {{t -> ConditionalExpression[-(2/(-1 + x)), (t | x) \[Element] Reals]}}

from a well known competitor is far more user friendly