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