Gosse michel <michel.gosse at free.fr> wrote:
> (%i1) f(x):=a+k/(x-b);
> e1:f(-2)=-1;
> e2:f(1)=-1;
> e3:f(2)=5/2;
> solve([e1,e2,e3],[a,b,k]);
> (%o1) f(x):=a+k/(x-b)
> (%o2) k/(-b-2)+a=-1
> (%o3) k/(1-b)+a=-1
> (%o4) k/(2-b)+a=5/2
> (%o5) [[a=-1,b=2,k=0]]
> it seems it's a bug, because f(x)=-1 is clearly not a solution.
> The solution is good with f(-2)=1
>
> Maxima 5.20.1
"Richard Hennessy" <rich.hennessy at verizon.net> wrote:
> solve([e1,e2,e3],[a,b,k]); [[a = - 1, b = 2, k = 0]]
> [a,b,k]:map('rhs, first(%));
> [- 1, 2, 0]
> f(-1);
> - 1
>
> So f(x) = -1 is a solution.
>
> Rich
>
> f(x):=-1+0/(x-2);
I believe the problem Gosse sees here is that the solution f(x)=-1 is
apparently inconsistent with e3:f(2)=5/2.
Note that when [a=-1,b=2,k=0], f(2)=-1+0/0.
to_poly_solve() claims that there are no solutions:
(%i12) to_poly_solve([e1,e2,e3],[a,b,k]);
(%o12) %union()
Mark