Re: SolveAlways documentation problem
- To: mathgroup at smc.vnet.net
- Subject: [mg78773] Re: [mg78745] SolveAlways documentation problem
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 9 Jul 2007 01:36:51 -0400 (EDT)
- References: <200707081013.GAA09510@smc.vnet.net>
On 8 Jul 2007, at 19:13, Chris Chiasson wrote:
> "SolveAlways[eqns,vars] is equivalent to Solve[!Eliminate[!
> eqns,vars]]."
>
> does not work with the first basic example given
>
> In[1]:= SolveAlways[a x+b==0,x]
> Out[1]= {{a->0,b->0}}
>
> In[2]:= Solve[!Eliminate[!(a x+b==0),x],x]
> Out[2]= {{}}
>
> --
> http://chris.chiasson.name/
>
The documentation is right:
Solve[! Eliminate[! (a x + b == 0), x]]
{{a -> 0, b -> 0}}
If you include x in Solve, than the answer will always be {{}} since
the equations that you get after Eliminate finishes its job do not
involve x (actually one could argue over whether the answer should be
{{}}, which means the equations are valid for all x, or {}, which
means there are no generic solutions, but this is not the issue here).
If you use LogicalExpand instead of Solve you will get:
LogicalExpand[ ! Eliminate[ ! a*x + b == 0, x]]
a == 0 && b == 0
which is the kind of answer that Reduce returns.
Andrzej Kozlowski
- References:
- SolveAlways documentation problem
- From: "Chris Chiasson" <chris@chiasson.name>
- SolveAlways documentation problem