Re: Bug in Simplify?
- To: mathgroup at smc.vnet.net
- Subject: [mg33339] Re: Bug in Simplify?
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Sat, 16 Mar 2002 01:40:19 -0500 (EST)
- References: <a6pj7l$aoj$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Gianluca Gorni <gorni at dimi.uniud.it> wrote: > > As best I can tell, however, nothing that you said explains why > > we should expect Simplify[x/y, x==0 && y==0] to yield 0. This > > still puzzles me. > > It seems that Simplify will not look at denominators to see when > they vanish. For example > > Simplify[x/y == 1 && y == 0] > > will return the argument unevaluated. Also Solve has the same problem: > Solve[x/y == 1] will not warn you that x=y=0 is not a solution. > > A function that cares for vanishing denominators is Reduce: > > Reduce[x/y == 1 && y == 0] gives False > > Also, Reduce[x/y == 1] gives x == y && y != 0 Thanks for your response. > I don't know how relevant it is, but 2 years ago Adam Strzebonski, > in response to why Simplify[Sqrt[x^2], x == 1 + Sqrt[2]] did not > return x, wrote: > > > I am not convinced that Simplify should be > > optimized for assumptions of the form x==constant. I would > > expect that in most cases what one really wants is > > Simplify[expr/.x->constant] > > rather than > > Simplify[expr, x==constant]. However, just as we wish, In[1]:= Simplify[Sqrt[x^2], x \[GreaterEqual]0] Out[1]= x Hmm. That gives me a baroque idea. First, note that, as we wish, In[2]:= Simplify[x>=0 && x<=0 && y>=0 && y<=0] Out[2]= x==0 && y==0 Then, if you want Simplify to leave x/y as is (rather than giving 0), even though you have _in essence_ specified x==0 && y==0 in the assumptions, use In[3]:= Simplify[x/y, x>=0 && x<=0 && y>=0 && y<=0] Out[3]= x/y Best Regards, David Cantrell -- -------------------- http://NewsReader.Com/ -------------------- Usenet Newsgroup Service