#2209 Lost solutions in to_poly_solve

None
closed
nobody
5
2013-01-01
2011-06-19
Aleksas
No

Lost solutions in to_poly_solve

(%i1) eq:sin(2*x)=cos(x);
(%o1) sin(2*x)=cos(x)
(%i2) eq1:trigexpand(eq);
(%o2) 2*cos(x)*sin(x)=cos(x)

Lost solutions:
(%i3) to_poly_solve([eq], [x]);
Loading maxima-grobner $Revision: 1.6 $ $Date: 2009/06/02 07:49:49 $
(%o3) %union([x=-(4*%pi*%z0+%pi)/2],[x=(4*%pi*%z1+%pi)/6])

Correct:
(%i4) to_poly_solve([eq1], [x]);
(%o4) %union([x=2*%pi*%z15-%pi/2],[x=2*%pi*%z17+%pi/2],
[x=2*%pi*%z19+%pi/6],[x=2*%pi*%z21+(5*%pi)/6])

Discussion

  • Aleksas

    Aleksas - 2011-06-19

    In to_poly_solve equation is not equivalent to
    lhs(equation)-rhs(equation)=0.

    For example:
    (%i1) eq:sin(2*x)=cos(x);
    (%o1) sin(2*x)=cos(x)
    (%i2) to_poly_solve([eq], [x]);
    Loading maxima-grobner $Revision: 1.6 $ $Date: 2009/06/02 07:49:49 $
    (%o2) %union([x=-(4*%pi*%z0+%pi)/2],[x=(4*%pi*%z1+%pi)/6])
    (%i3) to_poly_solve([lhs(eq)-rhs(eq)=0], [x]);
    (%o3) %union([x=2*%pi*%z15-%pi/2],[x=2*%pi*%z17+%pi/2],[x=2*%pi*%z19+%pi/6],[x=2*%pi*%z21+(5*%pi)/6])

     
  • Barton Willis

    Barton Willis - 2013-01-01
    • status: open --> closed
    • milestone: --> None
     
  • Barton Willis

    Barton Willis - 2013-01-01

    The solution sets are the same. To see this, solve sin(2x) = cos(x), and replace the arbitrary integer in the solution with 3k, 3k+1, and 3k+2.

     

Log in to post a comment.